Differential Equations For Dummies

Differential Equations For Dummies
Differential
Equations
FOR
DUMmIES
‰
by Steven Holzner, PhD
Differential
Equations
FOR
DUMmIES
‰
Differential
Equations
FOR
DUMmIES
‰
by Steven Holzner, PhD
Differential Equations For Dummies®
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Library of Congress Control Number: 2008925781
ISBN: 978-0-470-17814-0
Manufactured in the United States of America
10 9 8 7 6 5 4 3 2 1
About the Author
Steven Holzner is an award-winning author of science, math, and technical
books. He got his training in differential equations at MIT and at Cornell
University, where he got his PhD. He has been on the faculty at both MIT and
Cornell University, and has written such bestsellers as Physics For Dummies
and Physics Workbook For Dummies.
Dedication
To Nancy, always and forever.
Author’s Acknowledgments
The book you hold in your hands is the work of many people. I’d especially
like to thank Tracy Boggier, Georgette Beatty, Jessica Smith, technical
reviewer Jamie Song, PhD, and the folks in Composition Services who put the
book together so beautifully.
Publisher’s Acknowledgments
We’re proud of this book; please send us your comments through our Dummies online registration
form located at www.dummies.com/register/.
Some of the people who helped bring this book to market include the following:
Acquisitions, Editorial, and
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Project Editor: Georgette Beatty
Composition Services
Project Coordinator: Erin Smith
Acquisitions Editor: Tracy Boggier
Layout and Graphics: Carrie A. Cesavice,
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Copy Editor: Jessica Smith
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Erin Calligan Mooney
Indexer: Broccoli Information Management
Technical Editor: Jamie Song, PhD
Editorial Manager: Michelle Hacker
Editorial Assistants: Joe Niesen, Leeann Harney
Cartoons: Rich Tennant
(www.the5thwave.com)
Publishing and Editorial for Consumer Dummies
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Publishing for Technology Dummies
Andy Cummings, Vice President and Publisher, Dummies Technology/General User
Composition Services
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Debbie Stailey, Director of Composition Services
Contents at a Glance
Introduction .................................................................1
Part I: Focusing on First Order Differential Equations......5
Chapter 1: Welcome to the World of Differential Equations .........................................7
Chapter 2: Looking at Linear First Order Differential Equations................................23
Chapter 3: Sorting Out Separable First Order Differential Equations........................41
Chapter 4: Exploring Exact First Order Differential Equations
and Euler’s Method........................................................................................................63
Part II: Surveying Second and Higher Order
Differential Equations .................................................89
Chapter 5: Examining Second Order Linear Homogeneous
Differential Equations....................................................................................................91
Chapter 6: Studying Second Order Linear Nonhomogeneous
Differential Equations..................................................................................................123
Chapter 7: Handling Higher Order Linear Homogeneous Differential
Equations ......................................................................................................................151
Chapter 8: Taking On Higher Order Linear Nonhomogeneous
Differential Equations..................................................................................................173
Part III: The Power Stuff: Advanced Techniques ..........189
Chapter 9: Getting Serious with Power Series and Ordinary Points........................191
Chapter 10: Powering through Singular Points ..........................................................213
Chapter 11: Working with Laplace Transforms ..........................................................239
Chapter 12: Tackling Systems of First Order Linear Differential Equations ...........265
Chapter 13: Discovering Three Fail-Proof Numerical Methods ................................293
Part IV: The Part of Tens ...........................................315
Chapter 14: Ten Super-Helpful Online Differential Equation Tutorials....................317
Chapter 15: Ten Really Cool Online Differential Equation Solving Tools ................321
Index .......................................................................325
Table of Contents
Introduction..................................................................1
About This Book...............................................................................................1
Conventions Used in This Book .....................................................................1
What You’re Not to Read.................................................................................2
Foolish Assumptions .......................................................................................2
How This Book Is Organized...........................................................................2
Part I: Focusing on First Order Differential Equations.......................3
Part II: Surveying Second and Higher Order
Differential Equations.........................................................................3
Part III: The Power Stuff: Advanced Techniques ................................3
Part IV: The Part of Tens........................................................................3
Icons Used in This Book..................................................................................4
Where to Go from Here....................................................................................4
Part I: Focusing on First Order Differential Equations ......5
Chapter 1: Welcome to the World of Differential Equations . . . . . . . . .7
The Essence of Differential Equations...........................................................8
Derivatives: The Foundation of Differential Equations .............................11
Derivatives that are constants............................................................11
Derivatives that are powers................................................................12
Derivatives involving trigonometry ...................................................12
Derivatives involving multiple functions ..........................................12
Seeing the Big Picture with Direction Fields...............................................13
Plotting a direction field ......................................................................13
Connecting slopes into an integral curve .........................................14
Recognizing the equilibrium value.....................................................16
Classifying Differential Equations ................................................................17
Classifying equations by order ...........................................................17
Classifying ordinary versus partial equations..................................17
Classifying linear versus nonlinear equations..................................18
Solving First Order Differential Equations ..................................................19
Tackling Second Order and Higher Order Differential Equations ............20
Having Fun with Advanced Techniques ......................................................21
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Differential Equations For Dummies
Chapter 2: Looking at Linear First Order Differential Equations . . . . .23
First Things First: The Basics of Solving Linear First Order
Differential Equations ................................................................................24
Applying initial conditions from the start.........................................24
Stepping up to solving differential
equations involving functions.........................................................25
Adding a couple of constants to the mix...........................................26
Solving Linear First Order Differential Equations
with Integrating Factors ............................................................................26
Solving for an integrating factor.........................................................27
Using an integrating factor to solve a differential equation ...........28
Moving on up: Using integrating factors in differential
equations with functions .................................................................29
Trying a special shortcut ....................................................................30
Solving an advanced example.............................................................32
Determining Whether a Solution for a Linear First Order
Equation Exists ...........................................................................................35
Spelling out the existence and uniqueness theorem
for linear differential equations ......................................................35
Finding the general solution ...............................................................36
Checking out some existence and uniqueness examples ...............37
Figuring Out Whether a Solution for a Nonlinear
Differential Equation Exists.......................................................................38
The existence and uniqueness theorem for
nonlinear differential equations......................................................39
A couple of nonlinear existence and uniqueness examples ...........39
Chapter 3: Sorting Out Separable First Order
Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41
Beginning with the Basics of Separable Differential Equations ...............42
Starting easy: Linear separable equations ........................................43
Introducing implicit solutions ............................................................43
Finding explicit solutions from implicit solutions ...........................45
Tough to crack: When you can’t find an explicit solution ..............48
A neat trick: Turning nonlinear separable equations into
linear separable equations ..............................................................49
Trying Out Some Real World Separable Equations....................................52
Getting in control with a sample flow problem ................................52
Striking it rich with a sample monetary problem ............................55
Break It Up! Using Partial Fractions in Separable Equations....................59
Table of Contents
Chapter 4: Exploring Exact First Order Differential
Equations and Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .63
Exploring the Basics of Exact Differential Equations ................................63
Defining exact differential equations .................................................64
Working out a typical exact differential equation ............................65
Determining Whether a Differential Equation Is Exact..............................66
Checking out a useful theorem ...........................................................66
Applying the theorem ..........................................................................67
Conquering Nonexact Differential Equations
with Integrating Factors ............................................................................70
Finding an integrating factor...............................................................71
Using an integrating factor to get an exact equation.......................73
The finishing touch: Solving the exact equation ..............................74
Getting Numerical with Euler’s Method ......................................................75
Understanding the method .................................................................76
Checking the method’s accuracy on a computer.............................77
Delving into Difference Equations................................................................83
Some handy terminology ....................................................................84
Iterative solutions ................................................................................84
Equilibrium solutions ..........................................................................85
Part II: Surveying Second and Higher Order
Differential Equations..................................................89
Chapter 5: Examining Second Order Linear
Homogeneous Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .91
The Basics of Second Order Differential Equations...................................91
Linear equations...................................................................................92
Homogeneous equations.....................................................................93
Second Order Linear Homogeneous Equations
with Constant Coefficients ........................................................................94
Elementary solutions ...........................................................................94
Initial conditions...................................................................................95
Checking Out Characteristic Equations ......................................................96
Real and distinct roots.........................................................................97
Complex roots.....................................................................................100
Identical real roots .............................................................................106
Getting a Second Solution by Reduction of Order ...................................109
Seeing how reduction of order works..............................................110
Trying out an example .......................................................................111
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Differential Equations For Dummies
Putting Everything Together with Some Handy Theorems ....................114
Superposition......................................................................................114
Linear independence .........................................................................115
The Wronskian ....................................................................................117
Chapter 6: Studying Second Order Linear Nonhomogeneous
Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .123
The General Solution of Second Order Linear
Nonhomogeneous Equations ..................................................................124
Understanding an important theorem.............................................124
Putting the theorem to work.............................................................125
Finding Particular Solutions with the Method of
Undetermined Coefficients......................................................................127
When g(x) is in the form of erx ..........................................................127
When g(x) is a polynomial of order n ..............................................128
When g(x) is a combination of sines and cosines ..........................131
When g(x) is a product of two different forms ...............................133
Breaking Down Equations with the Variation of Parameters Method ....135
Nailing down the basics of the method ...........................................136
Solving a typical example..................................................................137
Applying the method to any linear equation ..................................138
What a pair! The variation of parameters method
meets the Wronskian......................................................................142
Bouncing Around with Springs ’n’ Things ................................................143
A mass without friction .....................................................................144
A mass with drag force ......................................................................148
Chapter 7: Handling Higher Order Linear Homogeneous
Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .151
The Write Stuff: The Notation of Higher Order
Differential Equations ..............................................................................152
Introducing the Basics of Higher Order Linear
Homogeneous Equations.........................................................................153
The format, solutions, and initial conditions .................................153
A couple of cool theorems ................................................................155
Tackling Different Types of Higher Order Linear
Homogeneous Equations.........................................................................156
Real and distinct roots.......................................................................156
Real and imaginary roots ..................................................................161
Complex roots.....................................................................................164
Duplicate roots ...................................................................................166
Table of Contents
Chapter 8: Taking On Higher Order Linear Nonhomogeneous
Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .173
Mastering the Method of Undetermined Coefficients
for Higher Order Equations.....................................................................174
When g(x) is in the form erx ...............................................................176
When g(x) is a polynomial of order n ..............................................179
When g(x) is a combination of sines and cosines ..........................182
Solving Higher Order Equations with Variation of Parameters..............185
The basics of the method..................................................................185
Working through an example............................................................186
Part III: The Power Stuff: Advanced Techniques...........189
Chapter 9: Getting Serious with Power Series
and Ordinary Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .191
Perusing the Basics of Power Series..........................................................191
Determining Whether a Power Series Converges
with the Ratio Test ...................................................................................192
The fundamentals of the ratio test...................................................192
Plugging in some numbers ................................................................193
Shifting the Series Index..............................................................................195
Taking a Look at the Taylor Series .............................................................195
Solving Second Order Differential Equations with Power Series ...........196
When you already know the solution ..............................................198
When you don’t know the solution beforehand .............................204
A famous problem: Airy’s equation..................................................207
Chapter 10: Powering through Singular Points . . . . . . . . . . . . . . . . . .213
Pointing Out the Basics of Singular Points ...............................................213
Finding singular points ......................................................................214
The behavior of singular points .......................................................214
Regular versus irregular singular points.........................................215
Exploring Exciting Euler Equations ...........................................................219
Real and distinct roots.......................................................................220
Real and equal roots ..........................................................................222
Complex roots.....................................................................................223
Putting it all together with a theorem..............................................224
Figuring Series Solutions Near Regular Singular Points..........................225
Identifying the general solution........................................................225
The basics of solving equations near singular points ...................227
A numerical example of solving an equation
near singular points........................................................................230
Taking a closer look at indicial equations.......................................235
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Differential Equations For Dummies
Chapter 11: Working with Laplace Transforms . . . . . . . . . . . . . . . . . .239
Breaking Down a Typical Laplace Transform...........................................239
Deciding Whether a Laplace Transform Converges ................................240
Calculating Basic Laplace Transforms ......................................................241
The transform of 1..............................................................................242
The transform of eat ............................................................................242
The transform of sin at ......................................................................242
Consulting a handy table for some relief ........................................244
Solving Differential Equations with Laplace Transforms ........................245
A few theorems to send you on your way.......................................246
Solving a second order homogeneous equation ............................247
Solving a second order nonhomogeneous equation .....................251
Solving a higher order equation .......................................................255
Factoring Laplace Transforms and Convolution Integrals .....................258
Factoring a Laplace transform into fractions .................................258
Checking out convolution integrals .................................................259
Surveying Step Functions............................................................................261
Defining the step function .................................................................261
Figuring the Laplace transform of the step function .....................262
Chapter 12: Tackling Systems of First Order Linear
Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .265
Introducing the Basics of Matrices ............................................................266
Setting up a matrix .............................................................................266
Working through the algebra ............................................................267
Examining matrices............................................................................268
Mastering Matrix Operations......................................................................269
Equality................................................................................................269
Addition ...............................................................................................270
Subtraction..........................................................................................270
Multiplication of a matrix and a number.........................................270
Multiplication of two matrices..........................................................270
Multiplication of a matrix and a vector ...........................................271
Identity.................................................................................................272
The inverse of a matrix......................................................................272
Having Fun with Eigenvectors ’n’ Things..................................................278
Linear independence .........................................................................278
Eigenvalues and eigenvectors ..........................................................281
Solving Systems of First-Order Linear Homogeneous
Differential Equations ..............................................................................283
Understanding the basics..................................................................284
Making your way through an example ............................................285
Solving Systems of First Order Linear Nonhomogeneous Equations .....288
Assuming the correct form of the particular solution...................289
Crunching the numbers.....................................................................290
Winding up your work .......................................................................292
Table of Contents
Chapter 13: Discovering Three Fail-Proof Numerical Methods . . . . .293
Number Crunching with Euler’s Method ..................................................294
The fundamentals of the method .....................................................294
Using code to see the method in action ..........................................295
Moving On Up with the Improved Euler’s Method ..................................299
Understanding the improvements ...................................................300
Coming up with new code .................................................................300
Plugging a steep slope into the new code .......................................304
Adding Even More Precision with the Runge-Kutta Method ..................308
The method’s recurrence relation....................................................308
Working with the method in code ....................................................309
Part IV: The Part of Tens ............................................315
Chapter 14: Ten Super-Helpful Online Differential
Equation Tutorials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .317
AnalyzeMath.com’s Introduction to Differential Equations ...................317
Harvey Mudd College Mathematics Online Tutorial ...............................318
John Appleby’s Introduction to Differential Equations...........................318
Kardi Teknomo’s Page .................................................................................318
Martin J. Osborne’s Differential Equation Tutorial..................................318
Midnight Tutor’s Video Tutorial.................................................................319
The Ohio State University Physics Department’s
Introduction to Differential Equations...................................................319
Paul’s Online Math Notes ............................................................................319
S.O.S. Math ....................................................................................................319
University of Surrey Tutorial ......................................................................320
Chapter 15: Ten Really Cool Online Differential
Equation Solving Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .321
AnalyzeMath.com’s Runge-Kutta Method Applet ....................................321
Coolmath.com’s Graphing Calculator .......................................................321
Direction Field Plotter .................................................................................322
An Equation Solver from QuickMath Automatic Math Solutions...........322
First Order Differential Equation Solver....................................................322
GCalc Online Graphing Calculator .............................................................322
JavaView Ode Solver....................................................................................323
Math @ CowPi’s System Solver...................................................................323
A Matrix Inverter from QuickMath Automatic Math Solutions ..............323
Visual Differential Equation Solving Applet ..............................................323
Index........................................................................325
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Differential Equations For Dummies
Introduction
F
or too many people who study differential equations, their only exposure
to this amazingly rich and rewarding field of mathematics is through a
textbook that lands with an 800-page whump on their desk. And what follows
is a weary struggle as the reader tries to scale the impenetrable fortress of
the massive tome.
Has no one ever thought to write a book on differential equations from the
reader’s point of view? Yes indeed — that’s where this book comes in.
About This Book
Differential Equations For Dummies is all about differential equations from
your point of view. I’ve watched many people struggle with differential equations the standard way, and most of them share one common feeling:
Confusion as to what they did to deserve such torture.
This book is different; rather than being written from the professor’s point
of view, it has been written from the reader’s point of view. This book was
designed to be crammed full of the good stuff, and only the good stuff. No
extra filler has been added; and that means the issues aren’t clouded. In this
book, you discover ways that professors and instructors make solving problems simple.
You can leaf through this book as you like. In other words, it isn’t important
that you read it from beginning to end. Like other For Dummies books, this one
has been designed to let you skip around as much as possible — this is your
book, and now differential equations are your oyster.
Conventions Used in This Book
Some books have a dozen confusing conventions that you need to know
before you can even start reading. Not this one. Here are the few simple conventions that I include to help you navigate this book:
2
Differential Equations For Dummies
Italics indicate definitions and emphasize certain words. As is customary
in the math world, I also use italics to highlight variables.
Boldfaced text highlights important theorems, matrices (arrays of numbers), keywords in bulleted lists, and actions to take in numbered steps.
Monofont points out Web addresses.
When this book was printed, some Web addresses may have needed to break
across two lines of text. If that happens, rest assured that I haven’t put in any
extra characters (such as hyphens) to indicate the break. So when using one
of these Web addresses, type in exactly what you see in this book, pretending
as though the line break doesn’t exist.
What You’re Not to Read
Throughout this book, I share bits of information that may be interesting to
you but not crucial to your understanding of an aspect of differential equations. You’ll see this information either placed in a sidebar (a shaded gray
box) or marked with a Technical Stuff icon. I won’t be offended if you skip
any of this text — really!
Foolish Assumptions
This book assumes that you have no experience solving differential equations.
Maybe you’re a college student freshly enrolled in a class on differential equations, and you need a little extra help wrapping your brain around them. Or
perhaps you’re a student studying physics, chemistry, biology, economics,
or engineering, and you quickly need to get a handle on differential equations
to better understand your subject area.
Any study of differential equations takes as its starting point a knowledge of
calculus. So I wrote this book with the assumption in mind that you know
how to take basic derivatives and how to integrate. If you’re totally at sea
with these tasks, pick up a copy of Calculus For Dummies by Mark Ryan
(Wiley) before you pick up this book.
How This Book Is Organized
The world of differential equations is, well, big. And to handle it, I break that
world down into different parts. Here are the various parts you see in this book.
Introduction
Part I: Focusing on First Order
Differential Equations
I start this book with first order differential equations — that is, differential
equations that involve derivatives to the first power. You see how to work
with linear first order differential equations (linear means that the derivatives
aren’t squared, cubed, or anything like that). You also discover how to work
with separable first order differential equations, which can be separated so
that only terms in y appear on one side, and only terms in x (and constants)
appear on the other. And, finally, in this part, you figure out how to handle
exact differential equations. With this type of equation you try to find a function whose partial derivatives correspond to the terms in a differential equation (which makes solving the equation much easier).
Part II: Surveying Second and Higher
Order Differential Equations
In this part, I take things to a whole new level as I show you how to deal with
second order and higher order differential equations. I divide equations into
two main types: linear homogeneous equations and linear nonhomogeneous
equations. You also find out that a whole new array of dazzling techniques
can be used here, such as the method of undetermined coefficients and the
method of variation of parameters.
Part III: The Power Stuff: Advanced
Techniques
Some differential equations are tougher than others, and in this part, I bring
out the big guns. You see heavy-duty techniques like Laplace transforms and
series solutions, and you start working with systems of differential equations.
You also figure out how to use numerical methods to solve differential equations. These are methods of last resort, but they rarely fail.
Part IV: The Part of Tens
You see the Part of Tens in all For Dummies books. This part is made up of
fast-paced lists of ten items each; in this book, you find ten online differential
equation tutorials and ten top online tools for solving differential equations.
3
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Differential Equations For Dummies
Icons Used in This Book
You can find several icons in the margins of this book, and here’s what they
mean:
This icon marks something to remember, such as a law of differential equations or a particularly juicy equation.
The text next to this icon is technical, insider stuff. You don’t have to read it
if you don’t want to, but if you want to become a differential equations pro
(and who doesn’t?), take a look.
This icon alerts you to helpful hints in solving differential equations. If you’re
looking for shortcuts, search for this icon.
When you see this icon, watch out! It indicates something particularly tough
to keep an eye out for.
Where to Go from Here
You’re ready to jump into Chapter 1. However, you don’t have to start there if
you don’t want to; you can jump in anywhere you like — this book was written to allow you to do just that. But if you want to get the full story on differential equations from the beginning, jump into Chapter 1 first — that’s where
all the action starts.
Part I
Focusing on First
Order Differential
Equations
I
In this part . . .
n this part, I welcome you to the world of differential
equations and start you off easy with linear first order
differential equations. With first order equations, you
have first order derivatives that are raised to the first
power, not squared or raised to any higher power. I also
show you how to work with separable first order differential equations, which are those equations that can be separated so that terms in y appear on one side and terms in
x (and constants) appear on the other. Finally, I introduce
exact differential equations and Euler’s method.
Chapter 1
Welcome to the World of
Differential Equations
In This Chapter
Breaking into the basics of differential equations
Getting the scoop on derivatives
Checking out direction fields
Putting differential equations into different categories
Distinguishing among different orders of differential equations
Surveying some advanced methods
I
t’s a tense moment in the physics lab. The international team of highpowered physicists has attached a weight to a spring, and the weight is
bouncing up and down.
“What’s happening?” the physicists cry. “We have to understand this in terms
of math! We need a formula to describe the motion of the weight!”
You, the renowned Differential Equations Expert, enter the conversation
calmly. “No problem,” you say. “I can derive a formula for you that will
describe the motion you’re seeing. But it’s going to cost you.”
The physicists look worried. “How much?” they ask, checking their grants
and funding sources. You tell them.
“Okay, anything,” they cry. “Just give us a formula.”
You take out your clipboard and start writing.
“What’s that?” one of the physicists asks, pointing at your calculations.
8
Part I: Focusing on First Order Differential Equations
“That,” you say, “is a differential equation. Now all I have to do is to solve it,
and you’ll have your formula.” The physicists watch intently as you do your
math at lightning speed.
“I’ve got it,” you announce. “Your formula is y = 10 sin (5t), where y is the
weight’s vertical position, and t is time, measured in seconds.”
“Wow,” the physicists cry, “all that just from solving a differential equation?”
“Yep,” you say, “now pay up.”
Well, you’re probably not a renowned differential equations expert — not yet,
at least! But with the help of this book, you very well may become one. In this
chapter, I give you the basics to get started with differential equations, such
as derivatives, direction fields, and equation classifications.
The Essence of Differential Equations
In essence, differential equations involve derivatives, which specify how a
quantity changes; by solving the differential equation, you get a formula for
the quantity itself that doesn’t involve derivatives.
Because derivatives are essential to differential equations, I take the time in
the next section to get you up to speed on them. (If you’re already an expert
on derivatives, feel free to skip the next section.) In this section, however,
I take a look at a qualitative example, just to get things started in an easily
digestible way.
Say that you’re a long-time shopper at your local grocery store, and you’ve
noticed prices have been increasing with time. Here’s the table you’ve been
writing down, tracking the price of a jar of peanut butter:
Month
Price
1
$2.40
2
$2.50
3
$2.60
4
$2.70
5
$2.80
6
$2.90
Chapter 1: Welcome to the World of Differential Equations
Looks like prices have been going up steadily, as you can see in the graph
of the prices in Figure 1-1. With that large of a price hike, what’s the price of
peanut butter going to be a year from now?
2.90
2.80
Price
2.70
2.60
2.50
Figure 1-1:
The price of
peanut
butter by
month.
2.40
1
2
3
4
5
6
Time
You know that the slope of a line is ∆y/∆x (that is, the change in y divided by
the change in x). Here, you use the symbols ∆p for the change in price and ∆t
for the change in time. So the slope of the line in Figure 1-1 is ∆p/∆t.
Because the price of peanut butter is going up 10 cents every month, you
know that the slope of the line in Figure 1-1 is:
∆p
= 10¢/month
∆t
The slope of a line is a constant, indicating its rate of change. The derivative
of a quantity also gives its rate of change at any one point, so you can think of
the derivative as the slope at a particular point. Because the rate of change of
a line is constant, you can write:
dp ∆p
=
= 10¢/month
∆t
dt
In this case, dp/dt is the derivative of the price of peanut butter with respect
to time. (When you see the d symbol, you know it’s a derivative.)
And so you get this differential equation:
dp
= 10¢/month
dt
9
10
Part I: Focusing on First Order Differential Equations
The previous equation is a differential equation because it’s an equation that
involves a derivative, in this case, dp/dt. It’s a pretty simple differential equation, and you can solve for price as a function of time like this:
p = 10t + c
In this equation, p is price (measured in cents), t is time (measured in months),
and c is an arbitrary constant that you use to match the initial conditions of
the problem. (You need a constant, c, because when you take the derivative of
10t + c, you just get 10, so you can’t tell whether there’s a constant that should
be added to 10t — matching the initial conditions will tell you.)
The missing link is the value of c, so just plug in the numbers you have for
price and time to solve for it. For example, the cost of peanut butter in month 1
is $2.40, so you can solve for c by plugging in 1 for t and $2.40 for p (240 cents),
giving you:
240 = 10 + c
By solving this equation, you calculate that c = 230, so the solution to your
differential equation is:
p = 10t + 230
And that’s your solution — that’s the price of peanut butter by month. You
started with a differential equation, which gave the rate of change in the price
of peanut butter, and then you solved that differential equation to get the
price as a function of time, p = 10t + 230.
Want to see the solution to your differential equation in action? Go for it! Find
out what the price of peanut butter is going to be in month 12. Now that you
have your equation, it’s easy enough to figure out:
p = 10t + 230
10(12) + 230 = 350
As you can see, in month 12, peanut butter is going to cost a steep $3.50,
which you were able to figure out because you knew the rate at which the
price was increasing. This is how any typical differential equation may work:
You have a differential equation for the rate at which some quantity changes
(in this case, price), and then you solve the differential equation to get
another equation, which in this case related price to time.
Note that when you substitute the solution (p = 10t + 230) into the differential
equation, dp/dt indeed gives you 10 cents per month, as it should.
Chapter 1: Welcome to the World of Differential Equations
Derivatives: The Foundation of
Differential Equations
As I mention in the previous section, a derivative simply specifies the rate at
which a quantity changes. In math terms, the derivative of a function f(x),
which is depicted as df(x)/dx, or more commonly in this book, as f'(x), indicates how f(x) is changing at any value of x. The function f(x) has to be continuous at a particular point for the derivative to exist at that point.
Take a closer look at this concept. The amount f(x) changes in a small distance
along the x axis ∆x is:
f(x + ∆x) – f(x)
The rate at which f(x) changes over the change ∆x is:
f ^ x + ∆x h - f ^ x h
∆x
So far so good. Now to get the derivative dy/dx, where y = f(x), you must let
∆x get very small, approaching zero. You can do that with a limiting expression, which you can evaluate as ∆x goes to zero. In this case, the limiting
expression is:
f ^ x + ∆x h - f ^ x h
dy
= lim
∆x
dx ∆x " 0
In other words, the derivative of f(x) is the amount f(x) changes in ∆x, divided
by ∆x, as ∆x goes to zero.
I take a look at some common derivatives in the following sections; you’ll see
these derivatives throughout this book.
Derivatives that are constants
The first type of derivative you’ll encounter is when f(x) equals a constant, c.
If f(x) = c, then f(x + ∆x) = c also, and f(x + ∆x) – f(x) = 0 (because all these
amounts are actually the same), so df(x)/dx = 0. Therefore:
f ^xh = c
df ^ x h
=0
dx
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12
Part I: Focusing on First Order Differential Equations
How about when f(x) = cx, where c is a constant? In this case, f(x) = cx, and
f(x + ∆x) = cx + c ∆x.
So f(x + ∆x) – f(x) = c ∆x and (f(x + ∆x) – f(x))/∆x = c. Therefore:
df ^ x h
=c
dx
f ^ x h = cx
Derivatives that are powers
Another type of derivative that pops up is one that includes raising x to the
power n. Derivatives with powers work like this:
f ^xh = x n
df ^ x h
= n x n -1
dx
Raising e to a certain power is always popular when working with differential
equations (e is the natural logarithm base, e = 2.7128 . . ., and a is a constant):
f ^ x h = e ax
df ^ x h
= a e ax
dx
And there’s also the inverse of ea, which is the natural log, which works like
this:
f ^ x h = ln ^ x h
df ^ x h 1
=x
dx
Derivatives involving trigonometry
Now for some trigonometry, starting with the derivative of sin(x):
f ^ x h = sin^ x h
df ^ x h
= cos^ x h
dx
And here’s the derivative of cos(x):
f ^ x h = cos^ x h
df ^ x h
= -sin^ x h
dx
Derivatives involving multiple functions
The derivative of the sum (or difference) of two functions is equal to the sum
(or difference) of the derivatives of the functions (that’s easy enough to
remember!):
f ^ x h = a^ x h ! b ^ x h
df ^ x h d a ^ x h d b ^ x h
=
!
dx
dx
dx
Chapter 1: Welcome to the World of Differential Equations
The derivative of the product of two functions is equal to the first function
times the derivative of the second, plus the second function times the derivative of the first. For example:
f ^ x h = a^ x h b ^ x h
df ^ x h
d b^ x h
d a^ x h
= a^ x h
+ b^ x h
dx
dx
dx
How about the derivative of the quotient of two functions? That derivative is
equal to the function in the denominator times the derivative of the function
in the numerator, minus the function in the numerator times the derivative
of the function in the denominator, all divided by the square of the function
in the denominator:
d a^ x h
d b^ x h
a^ x h
df ^ x h b ^ x h dx - a ^ x h dx
f ^xh =
=
2
dx
b^ x h
b^ x h
Seeing the Big Picture
with Direction Fields
It’s all too easy to get caught in the math details of a differential equation,
thereby losing any idea of the bigger picture. One useful tool for getting an
overview of differential equations is a direction field, which I discuss in more
detail in Chapter 2. Direction fields are great for getting a handle on differential equations of the following form:
dy
= f _ x, y i
dx
The previous equation gives the slope of the equation y = f(x) at any point x. A
direction field can help you visualize such an equation without actually having
to solve for the solution. That field is a two-dimensional graph consisting of
many, sometimes hundreds, of short line segments, showing the slope — that
is, the value of the derivative — at multiple points. In the following sections,
I walk you through the process of plotting and understanding direction fields.
Plotting a direction field
Here’s an example to give you an idea of what a direction field looks like.
A body falling through air experiences this force:
F = mg – γ v
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14
Part I: Focusing on First Order Differential Equations
In this equation, F is the net force on the object, m is the object’s mass, g is
the acceleration due to gravity (g = 9.8 meters/sec2 near the Earth’s surface),
γ is the drag coefficient (which adds the effect of air friction and is measured
in newtons sec/meter), and v is the speed of the object as it plummets
through the air.
If you’re familiar with physics, consider Newton’s second law. It says that
F = ma, where F is the net force acting on an object, m is its mass, and a is its
acceleration. But the object’s acceleration is also dv/dt, the derivative of the
object’s speed with respect to time (that is, the rate of change of the object’s
speed). Putting all this together gives you:
F = ma = m dv = mg - c v
dt
Now you’re back in differential equation territory, with this differential equation for speed as a function of time:
dv = g - c v
m
dt
Now you can get specific by plugging in some numbers. The acceleration due
to gravity, g, is 9.8 meters/sec2 near the Earth’s surface, and let’s say that the
drag coefficient is 1.0 newtons sec/meter and the object has a mass of 4.0 kilograms. Here’s what you’d get:
dv = 9.8 - v
dt
4
To get a handle on this equation without attempting to solve it, you can plot
it as a direction field. To do so you create a two-dimensional plot and add
dozens of short line segments that give the slope at those locations (you can
do this by hand or with software). The direction field for this equation
appears in Figure 1-2. As you can see in the figure, there are dozens of short
lines in the graph, each of which give the slope of the solution at that point.
The vertical axis is v, and the horizontal axis is t.
Because the slope of the solution function at any one point doesn’t depend
on t, the slopes along any horizontal line are the same.
Connecting slopes into an integral curve
You can get a visual handle on what’s happening with the solutions to a differential equation by looking at its direction field. How? All those slanted line
segments give you the solutions of the differential equations — all you have
to do is draw lines connecting the slopes. One such solution appears in
Figure 1-3. A solution like the one in the figure is called an integral curve of
the differential equation.
Chapter 1: Welcome to the World of Differential Equations
50
45
v
40
35
30
25
Figure 1-2:
A direction
field. 20
1
2
3
4
5
6
7
t
8
9
10
1
2
3
4
5
6
7
t
8
9
10
50
45
v
40
35
30
Figure 1-3: 25
A solution in
a direction
field. 20
15
16
Part I: Focusing on First Order Differential Equations
Recognizing the equilibrium value
As you can see from Figure 1-3, there are many solutions to the equation that
you’re trying to solve. As it happens, the actual solution to that differential
equation is:
v = 39.2 + ce–t/4
In the previous solution, c is an arbitrary constant that can take any value.
That means there are an infinite number of solutions to the differential
equation.
But you don’t have to know that solution to determine what the solutions
behave like. You can tell just by looking at the direction field that all solutions
tend toward a particular value, called the equilibrium value. For instance, you
can see from the direction field graph in Figure 1-3 that the equilibrium value
is 39.2. You also can see that equilibrium value in Figure 1-4.
50
45
v
40
35
30
Figure 1-4:
An
equilibrium 25
value in a
direction
field. 20
1
2
3
4
5
6
7
t
8
9
10
Chapter 1: Welcome to the World of Differential Equations
Classifying Differential Equations
Tons of differential equations exist in Math and Science Land, and the way
you tackle them differs by type. As a result, there are several classifications
that you can put differential equations into. I explain them in the following
sections.
Classifying equations by order
The most common classification of differential equations is based on order.
The order of a differential equation simply is the order of its highest derivative. For example, check out the following, which is a first order differential
equation:
dy
= 5x
dx
Here’s an example of a second order differential equation:
d 2 y dy
+
= 19x + 4
dx
dx 2
And so on, up to order n:
9
dn y
d n -1 y
d2 y
dy
- 16
+ 12
- 19x + 4 = 0
n - 1 + . . . + 14
dx n
dx
dx
dx 2
As you might imagine, first order differential equations are usually the most
easily managed, followed by second order equations, and so on. I discuss
first order, second order, and higher order differential equations in a bit more
detail later in this chapter.
Classifying ordinary versus
partial equations
You can also classify differential equations as ordinary or partial. This classification depends on whether you have only ordinary derivatives involved or
only partial derivatives.
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18
Part I: Focusing on First Order Differential Equations
An ordinary (non-partial) derivative is a full derivative, such as dQ/dt, where
you take the derivative of all terms in Q with respect to t. Here’s an example
of an ordinary differential equation, relating the charge Q(t) in a circuit to the
electromotive force E(t) (that is, the voltage source connected to the circuit):
L
dQ 1
d2Q
+R
+ Q = E ^t h
dt
C
dt 2
Here, Q is the charge, L is the inductance of the circuit, C is the capacitance
of the circuit, and E(t) is the electromotive force (voltage) applied to the circuit. This is an ordinary differential equation because only ordinary derivatives appear.
On the other hand, partial derivatives are taken with respect to only one variable, although the function depends on two or more. Here’s an example of a
partial differential equation (note the squiggly d’s):
α2
2 2 u _ x, t i 2u _ x, t i
=
2t
2x 2
In this heat conduction equation, α is a physical constant of the system that
you’re trying to track the heat flow of, and u(x, t) is the actual heat.
Note that u(x, t) depends on both x and t and that both derivatives are partial
derivatives — that is, the derivatives are taken with respect to one or the
other of x or t, but not both.
In this book, I focus on ordinary differential equations, because partial differential equations are usually the subject of more advanced texts. Never fear
though: I promise to get you your fair share of partial differential equations.
Classifying linear versus
nonlinear equations
Another way that you can classify differential equations is as linear or nonlinear. You call a differential equation linear if it exclusively involves linear
terms (that is, terms to the power 1) of y, y', y", and beyond to y(n). For example, this equation is a linear differential equation:
L
dQ 1
d2Q
+R
+ Q = E ^t h
dt
C
dt 2
Chapter 1: Welcome to the World of Differential Equations
Note that this kind of differential equation usually will be written this way
throughout this book. And this form makes the linear nature of this equation
clear:
LQ" + R Q l+ 1 Q = E ^ t h
C
On the other hand, nonlinear differential equations involve nonlinear terms in
any of y, y', y", up to y(n). The following equation, which describes the angle of
a pendulum, is a nonlinear differential equation that involves the term sin θ
(not just θ):
d 2 θ + g sin θ = 0
L
dt 2
Handling nonlinear differential equations is generally more difficult than handling linear equations. After all, it’s often tough enough to solve linear differential equations without messing things up by adding higher powers and
other nonlinear terms. For that reason, you’ll often see scientists cheat when
it comes to nonlinear equations. Usually they make an approximation that
reduces the nonlinear equation to a linear one.
For example, when it comes to pendulums, you can say that for small angles,
sin θ ≈ θ. This means that the following equation is the standard form of the
pendulum equation that you’ll find in physics textbooks:
d2θ + g θ= 0
L
dt 2
As you can see, this equation is a linear differential equation, and as such,
it’s much more manageable. Yes, it’s a cheat to use only small angles so that
sin θ ≈ θ, but unless you cheat like that, you’ll sometimes be reduced to using
numerical calculations on a computer to solve nonlinear differential equations; obviously these calculations work, but it’s much less satisfying than
cracking the equation yourself (if you’re a math geek like me).
Solving First Order Differential Equations
Chapters 2, 3, and 4 take a look at differential equations of the form f'(x) =
f(x, y); these equations are known as first order differential equations
because the derivative involved is of first order (for more on these types
of equations, see the earlier section “Classifying equations by order.”
19
20
Part I: Focusing on First Order Differential Equations
First order differential equations are great because they’re usually the most
solvable. I show you all kinds of ways to handle first order differential equations in Chapters 2, 3, and 4. The following are some examples of what you
can look forward to:
As you know, first order differential equations look like this: f'(x) = f(x, y).
In the upcoming chapters, I show you how to deal with the case where
f(x, y) is linear in x — for example, f'(x) = 5x — and then nonlinear in x,
as in f'(x) = 5x2.
You find out how to work with separable equations, where you can
factor out all the terms having to do with y on one side of the equation
and all the terms having to do with x on the other.
I also help you solve first order differential equations in cool ways, such
as by finding integrating factors to make more difficult problems simple.
Direction fields, which I discuss earlier in this chapter, work only for equations of the type f'(x) = f(x, y) — that is, where only the first derivative is
involved — because the first derivative of f(x) gives you the slope of f(x) at
any point (and, of course, connecting the slope line segments is what direction fields are all about).
Tackling Second Order and Higher Order
Differential Equations
As noted in the earlier section “Classifying equations by order,” second order
differential equations involve only the second derivative, d 2y/dx 2, also known
as y". In many physics situations, second order differential equations are
where the action is.
For example, you can handle physics situations such as masses on springs or
the electrical oscillations of inductor-capacitor circuits with a differential
equation like this:
y" – ay = 0
In Part II, I show you how to tackle second order differential equations with a
large arsenal of tools, such as the Wronskian matrix determinant, which will tell
you if there are solutions to a second (or higher) order differential equation.
Other tools I introduce you to include the method of undetermined coefficients
and the method of variation of parameters.
Chapter 1: Welcome to the World of Differential Equations
After first and second order differential equations, it’s natural to want to keep
the fun going, and that means you’ll be dealing with higher order differential
equations, which I also cover in Part II. With these high-end equations, you
find terms like d ny/dxn, where n > 2.
The derivative d ny/dxn is also written as y(n). Using the standard syntax, derivatives are written as y', y", y''', yiv, yv, and so on. In general, the nth derivative
of y is written as y(n).
Higher order differential equations can be tough; many of them don’t have
solutions at all. But don’t worry, because to help you solve them I bring to
bear the wisdom of more than 300 years of mathematicians.
Having Fun with Advanced Techniques
You discover dozens of tools in Part III of this book; all of these tools have been
developed and proved powerful over the years. Laplace Transforms, Euler’s
method, integrating factors, numerical methods — they’re all in this book.
These tools are what this book is all about — applying the knowledge of hundreds of years of solving differential equations. As you may know, differential
equations can be broken down by type, and there’s always a set of tools developed that allows you to work with whatever type of equation you come up
with. In this book, you’ll find a great many powerful tools that are just waiting
to solve all of your differential equations — from the simplest to the seemingly
impossible!
21
22
Part I: Focusing on First Order Differential Equations
Chapter 2
Looking at Linear First Order
Differential Equations
In This Chapter
Beginning with the basics of solving linear first order differential equations
Using integrating factors
Determining whether solutions exist for linear and nonlinear equations
A
s you find out in Chapter 1, a first order differential equation simply has
a derivative of the first order. Here’s what a typical first order differential equation looks like, where f(t, y) is a function of the variables t and y (of
course, you can use any variables here, such as x and y or u and v, not just t
and y):
dy
= f _ t, y i
dt
In this chapter, you work with linear first order differential equations — that
is, differential equations where the highest power of y is 1 (you can find out
the difference between linear and nonlinear equations in Chapter 1). For
example:
dy
=5
dt
dt = y + 1
dt
dt = 3y + 1
dt
I provide some general information on nonlinear differential equations at the
end of the chapter for comparison.
24
Part I: Focusing on First Order Differential Equations
First Things First: The Basics of Solving
Linear First Order Differential Equations
In the following sections, I take a look at how to handle linear first order differential equations in general. Get ready to find out about initial conditions,
solving equations that involve functions, and constants.
Applying initial conditions from the start
When you’re given a differential equation of the form dy/dt = f(t, y), your goal
is to find a function, y(t), that solves it. You may start by integrating the equation to come up with a solution that includes a constant, and then you apply
an initial condition to customize the solution. Applying the initial condition
allows you to select one solution among the infinite number that result from
the integration. Sounds cool, doesn’t it?
Take a look at this simple linear first order differential equation:
dy
=a
dt
As you can see, a is just a regular old number, meaning that this is a simple
example to start with and to introduce the idea of initial conditions. How can
you solve it? First of all, you may have noticed that another way of writing
this equation is:
dy = a dt
This equation looks promising. Why? Well, because now you can integrate
like this:
y
t
# dy = # a dt
y0
x0
Performing the integration gives you the following equation:
y – y0 = at – at0
You can combine y0 – at0 into a new constant, c, by adding y0 to the right side
of the equation, which gives you:
y = at + c
Chapter 2: Looking at Linear First Order Differential Equations
That was simple enough, right? And guess what? You’re done! The solution to
this differential equation is y = at + c.
So, for example, if a = 3 in the differential equation, here’s the equation you
would have:
dy
=3
dt
The solution for this equation is y = 3t + c.
Note that c, the result of integrating, can be any value, which leads to an infinite set of solutions: y = 3t + 5, y = 3t + 6, y = 3t + 589,303,202. How do you track
down the value of c that works for you? Well, it all depends on your initial conditions; for example, you may specify that the value of y at t = 0 be 15. Setting
this initial condition allows you to state the whole problem — differential
equation and initial condition — as follows:
dy
=3
dt
y(0) = 15
Substituting the initial condition, y(0) = 15, into the solution y = 3t + c gives
you the following equation:
y(t) = 3t + 15
Stepping up to solving differential
equations involving functions
Of course, dy/dt = 3 (the example from the previous section) isn’t the most
exciting differential equation. However, it does show you how to solve a differential equation using integration and how to apply an initial condition. The
next step is to solve linear differential equations that involve functions of t
rather than just a simple number.
This type of differential equation still contains only dy/dt and terms of t,
making it easy to integrate. Here’s the basic form:
dy
= g ^t h
dt
where g(t) is some function of t.
Here’s an example of this type of differential equation:
dy
= t 3 - 3t 2 + t
dt
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26
Part I: Focusing on First Order Differential Equations
Well, heck, that’s easy too; you simply rearrange to get this:
dy = t3 dt – 3t2 dt + t dt
Then you can integrate to get this equation:
4
2
y = t - t 3+ t + c
4
2
Adding a couple of constants to the mix
The next step up from equations such as dy/dx = a or dy/dt = g(t) are equations of the following form, which involve y, dy/dt, and the constants a and b:
dy
= ay - b
dt
How do you handle this equation and find a solution? Using some handy algebra, you can rewrite the equation like this:
dy/dt
=a
y - _ b/a i
Integrating both sides gives you the following equation:
ln | y – (b/a) | = at + c
where c is an arbitrary constant. Now get y out of the natural logarithm,
which gives you:
y = (b/a) + deat
where d = ec. And that’s it! You’re done. Good job!
Solving Linear First Order Differential
Equations with Integrating Factors
Sometimes integrating linear first order differential equations isn’t as easy as
it is in the examples earlier in this chapter. But it turns out that you can often
convert general equations into something that’s easy to integrate if you find
an integrating factor, which is a function, µ(t). The idea here is to multiply the
differential equation by an integrating factor so that the resulting equation
can easily be integrated and solved.
In the following sections, I provide tips and tricks for solving for an integrating
factor and plugging it back into different types of linear first order equations.
Chapter 2: Looking at Linear First Order Differential Equations
Solving for an integrating factor
In general, first order differential equations don’t lend themselves to easy
integration, which is where integrating factors come in. How does the method
of integrating factors work? To understand, say, for example, that you have
this linear differential equation:
dy
+ 2y = 4
dt
First, you multiply the previous equation by µ(t), which is a stand-in for the
undetermined integrating factor, giving you:
µ^t h
dy
+ 2 µ^t h y = 4 µ^t h
dt
Now you have to choose µ(t) so that you can recognize the left side of this
equation as the derivative of some expression. This way it can easily be
integrated.
Here’s the key: The left side of the previous equation looks very much like
differentiating the product µ(t)y. So try to choose µ(t) so that the left side of
the equation is indeed the derivative of µ(t)y. Doing so makes the integration
easy.
The derivative of µ(t)y by t is:
d 8 µ^t h y B
dy d µ ^ t h y
= µ^t h +
dt
dt
dt
Comparing the previous two equations term by term gives you:
d µ^t h
= 2 µ^t h
dt
Hey, not bad. Now you’re making progress! This is a differential equation you
can solve. Rearranging the equation so that all occurrences of µ(t) are on the
same side gives you:
d µ ^ t h /dt
=2
µ^t h
Now the equation can be rearranged to look like this:
d µ^t h
= 2 dt
µ^t h
Fine work. Integration gives you:
ln |µ(t)| = 2t + b
where b is an arbitrary constant of integration.
27
28
Part I: Focusing on First Order Differential Equations
Now it’s time for some exponentiating. Exponentiating both sides of the
equation gives you:
µ(t) = ce2t
where c is an arbitrary constant.
So that’s it — you’ve solved for the integrating factor! It’s µ(t) = ce2t.
Using an integrating factor to solve
a differential equation
After you solve for an integrating factor, you can plug that factor into the
original linear differential equation as multiplied by µ(t). For instance, take
your original equation from the previous section:
µ^t h
dy
+ 2 µ^t h y = 4 µ^t h
dt
and plug in the integrating factor to get this equation:
ce 2t
dy
+ 2 ce 2t y = 4 ce 2t
dt
Note that c drops out of this equation when you divide by c, so you get the
following equation (because you’re just looking for an arbitrary integrating
factor, you could also set c = 1):
e 2t
dy
+ 2 e 2t y = 4 e 2t
dt
When you use an integrating factor, you attempt to find a function µ(t) that,
when multiplied on both sides of a differential equation, makes the left side
into the derivative of a product. Figuring out the product allows you to solve
the differential equation.
In the previous example, you can now recognize the left side as the derivative
of e2t y. (If you can’t recognize the left side as a derivative of some product, in
general, it’s time to go on to other methods of solving the differential equation).
In other words, the differential equation has been conquered, because now
you have it in this form:
d _ e 2t y i
= 4e 2t
dt
You can integrate both sides of the equation to get this:
e2ty = 2e2t + c
Chapter 2: Looking at Linear First Order Differential Equations
And, finally, you can solve for y with your handy algebra skills:
y = 2 + ce–2t
You’ve got yourself a solution. Beautiful.
The use of an integrating factor isn’t always going to help you; sometimes,
when you use an integrating factor in a linear differential equation, the left
side isn’t going to be recognizable as the derivative of a product of functions.
In that case, where integrating factors don’t seem to help, you have to turn to
other methods. One of those methods is to determine whether the differential equation is separable, which I discuss in Chapter 3.
Moving on up: Using integrating factors in
differential equations with functions
Now you’re going to take integrating factors to a new level. Check out this
linear equation, where g(t) is a function of t:
dy
+ ay = g ^ t h
dt
This one’s a little more tricky. However, using the same integrating factor
from the previous two sections, eat (remember that the c dropped out), works
here as well. After you multiply both sides by eat, you get this equation:
e at
dy
+ a e at y = e at g ^ t h
dt
Now you can recast this equation in the following form:
d _ e at y i
= e at g ^ t h
dt
To integrate the function g, I use s as the variable of integration. Integration
gives you this equation:
e at y =
#e
as
g ^ s h ds + c
You can solve for y here, which gives you the following equation:
y = e - at
#e
as
g ^ s h ds + ce - at
And that’s it! You’ve got your answer!
Of course, solving this equation depends on whether you can calculate the
integral in the previous equation. If you can do it, you’ve solved the differential
equation. Otherwise, you may have to leave the solution in the integral form.
29
30
Part I: Focusing on First Order Differential Equations
Trying a special shortcut
In this section, I give you a shortcut for solving some particular differential
equations. Ready? Here’s the tip: In general, the integrating factor for an
equation in this form:
dy
+ ay = g ^ t h
dt
is this:
µ ^ t h = exp
# a dt
In this equation, exp(x) means ex.
As an example, try solving the following differential equation with the shortcut:
dy 1
+ y=4+t
dt
2
Assume that the initial condition is
y = 8, when t = 0
This equation is an example of the general equation solved in the previous
section. In this case, g(t) = 4 + t, and a = 1⁄2.
Using a, you find that the integrating factor is et/2, so multiply both sides of
equation by that factor:
e t/2
dy e t/2
+
y = 4e t/2 + te t/2
dt
2
Now you can combine the two terms on the left to give you this equation:
d _ e t/2 y i
= 4e t/2 + te t/2
dt
All you have to do now is integrate this result. The term on the left and the
first term on the right are no problem. The last term on the right is another
story.
You can use integration by parts to integrate this term. Integration by parts
works like this:
b
# f ^ x h g l ^ x h dx = f ^ b h g ^ b h
a
b
- f ^ah g ^ah -
# f l^ x h g ^ x h dx
a
Chapter 2: Looking at Linear First Order Differential Equations
Applying integration by parts to the last term on the right, and integrating
the others, gives you:
et/2 y = 8et/2 + 2 t et/2 – 4et/2 + c
where c is an arbitrary constant, set by the initial conditions. Dividing by eat
gives you this equation:
y = 4 + 2t + ce–at
By applying the initial condition, y(0) = 8, you get
y(0) = 8
8=4+c
Or c = 4. So the general solution of the differential equation is:
y = 4 + 2t + 4e–t/2
In Chapter 1, I explain that direction fields are great tools for visualizing differential equations. You can see a direction field for the previously noted
general solution in Figure 2-1.
25
20
y
15
10
Figure 2-1:
The
direction
field of the
general
solution.
5
0
0
1
2
3
4
5
6
x
7
8
9
10
31
32
Part I: Focusing on First Order Differential Equations
Connecting the slanting lines in a direction field gives you a graph of the solution. You can see a graph of this solution in Figure 2-2.
25
20
y
15
10
Figure 2-2:
The graph of
the general
solution.
5
0
1
0
2
3
4
5
6
x
7
8
9
10
Solving an advanced example
I think you’re ready for another, somewhat more advanced, example. Try
solving this differential equation to show that you can have different integrating factors:
t
dy
+ 2y = 4t 2
dt
where y(1) = 4.
To solve, first you have to find an integrating factor for the equation. To get it
into the form:
dy
+ ay = g ^ t h
dt
you have to divide both sides by t, which gives you this equation:
dy 2
+ t y = 4t
dt
Chapter 2: Looking at Linear First Order Differential Equations
To find the integrating factor, use the shortcut equation from the previous
section, like this:
µ ^ t h = exp
# a dt = exp # 2t dt
Performing the integral gives you this equation:
µ ^ t h = exp
# 2t dt = e
2 ln t
= t2
So the integrating factor here is t2, which is a new one. Multiplying both sides
of the equation by the integrating factor, µ(t) = t2, gives you:
t2
dy
+ 2ty = 4t 3
dt
Because the left side is a readily apparent derivative, you can also write it in
this form:
d _ yt 2 i
= 4t 3
dt
Now simply integrate both sides to get:
yt2 = t4 + c
Finally you get:
y = t 2 + c2
t
where c is an arbitrary constant of integration.
Now you can plug in the initial condition y(1) = 4, which allows you to see
that c = 3. And that helps you come to this solution:
y = t 2 + 32
t
And there you have it. You can see a direction field for the many general solutions to this differential equation in Figure 2-3.
33
34
Part I: Focusing on First Order Differential Equations
5
4
3
y
2
1
0
–1
Figure 2-3:
The
direction
field of a
more
advanced
solution.
–2
–3
–4
–5
–2
–1
0
1
x
2
You can see this function graphed in Figure 2-4.
8
7
y
6
5
4
Figure 2-4:
The graph of 3
a more
advanced
solution. 2
–2
–1
0
x
1
2
Chapter 2: Looking at Linear First Order Differential Equations
Determining Whether a Solution for a
Linear First Order Equation Exists
I show you how to deal with different kinds of linear first order differential
equations earlier in this chapter, but the fact remains that not all linear differential equations actually do have a solution.
Luckily, a theorem exists that tells you when a given linear differential equation with an initial condition has a solution. That theorem is called the existence and uniqueness theorem.
This theorem is worth knowing. After all, if a differential equation doesn’t have
a solution, what use is it to search for a solution? In other words, this theorem
represents another way to tackle linear first order differential equations.
Spelling out the existence and uniqueness
theorem for linear differential equations
In this section, I explain what the existence and uniqueness theorem for linear
differential equations says. Before I continue, however, note that a continuous
function is a function for which small changes in the input result in small
changes in the output (for example, f(x) = 1/x is not continuous at x = 0).
Without further ado, here’s the existence and uniqueness theorem:
If there is an interval I that contains the point to, and if the functions p(x)
and g(x) are continuous on that interval, and if you have this differential
equation:
dy
+ p^ x h y = g ^ x h
dx
then there exists a unique function, y(x), that is the solution to that differential equation for each x in interval I that also satisfies this initial condition:
y(to) = yo
where yo is an arbitrary initial value.
In other words, this theorem says that a solution exists and that the solution
is unique.
35
36
Part I: Focusing on First Order Differential Equations
Finding the general solution
Thinking about the theorem in the previous section begs the question: What
is the general solution to the following linear differential equation?
dy
+ p^ x h y = g ^ x h
dx
Note that this differential equation has a function p(x) and g(x), which provides a more complex situation. So you can’t use the simple form I explain in
the earlier section “Adding a couple of constants to the mix,” where a and b
are constants like this:
dy
= ay - b
dx
The solution here is:
y = (b/a) + ceat
Now you face a more complex situation, with functions p(x) and g(x). A general solution to the general equation does exist, and here it is:
y=
# µ ^ s h g ^ s h ds + c
µ^t h
where the integrating factor is the following:
µ ^ t h = exp
# p ^ t h dt
The integrals in these equations may not be possible to perform, of course.
But together, the equations represent the general solution.
Note that for linear differential equations, the solution, if there is one, is completely specified, up to a constant of integration, as in the solution you get in
the earlier section “Solving an advanced example”:
y = t 2 + c2
t
where c is a constant of integration.
You can’t necessarily say the same thing about nonlinear differential
equations — they may have solutions of completely different forms, not just
differing in the value of a constant. Because the solution to a linear differential equation has one form, differing only by the value of a constant, those
solutions are referred to as general solutions. This term isn’t used when discussing nonlinear differential equations, which may have multiple solutions
of completely different forms. I discuss nonlinear first order differential equations later in this chapter.
Chapter 2: Looking at Linear First Order Differential Equations
Checking out some existence
and uniqueness examples
In this section, I include a few examples to help you understand the existence
and uniqueness theorem for linear differential equations.
Example 1
Apply the existence and uniqueness theorem to the following equation to
show that there exists a unique solution:
dy
= 4x
dx _ y - 5 i
Just kidding! This equation isn’t linear because the term (y – 5) is in the denominator of the right side. And, of course, because the equation isn’t linear, the
existence and uniqueness theorem doesn’t apply. Did you catch that?
Example 2
Try this differential equation (which I promise is linear!). Does a unique solution exist?
dy
+ 2y = 4x 2
dx
where y(1) = 2.
The equation is already in the correct form:
dy
+ p^ x h y = g ^ x h
dx
where p(x) = 2 and g(x) = 4x2.
Note that p(x) and g(x) are continuous everywhere, so there’s a general solution that’s valid on the interval, – ∞ < x < ∞.
In particular, the initial condition is y(1) = 2, which is definitely inside the
interval that p(x) and g(x) are continuous (everything is inside that interval).
So, yes, there exists a solution to the initial value problem.
Example 3
Now take a look at this equation, which is similar to the example in the previous section, and determine whether a unique solution exists:
x
dy
+ 2y = 4x 2
dx
where y(1) = 2.
37
38
Part I: Focusing on First Order Differential Equations
The next step is to put the equation into this form:
dy
+ p^ x h y = g ^ x h
dx
Here’s what the equation should look like:
dy 2y
+ x = 4x
dx
In other words:
2
p^ x h = x
and
g(x) = 4x
Note that p(x) and g(x) aren’t continuous everywhere. In particular, p(x) is
discontinuous at x = 0, which makes the interval in which p(x) and g(x) are
continuous on the interval 0 > x and 0 < x.
Because the initial condition here is y(1) = 2, the point of interest is x = 1,
which is inside the interval where p(x) and g(x) are continuous. Therefore, by
the existence and uniqueness theorem, the initial value problem indeed has a
unique solution. Cool, huh?
Figuring Out Whether a Solution for a
Nonlinear Differential Equation Exists
In the previous sections of this chapter, I cover linear first order differential
equations in detail. But you may be wondering: Is there such a thing as a nonlinear differential equation? You bet there is! A nonlinear differential equation
simply includes nonlinear terms in y, y', y", and so on. Nonlinear equations
are pretty tough, so I don’t delve into them a lot in this book. But I do want to
discuss one important theorem related to solving these equations.
You see, the existence and uniqueness theorem (which you use for linear
equations, and which I cover earlier in this chapter) is analogous to another
theorem that’s used for nonlinear equations. I explain this theorem and show
some examples in the following sections.
Chapter 2: Looking at Linear First Order Differential Equations
The existence and uniqueness theorem for
nonlinear differential equations
Here’s the existence and uniqueness of solutions for nonlinear equations:
Say that you have a rectangle R that contains the point (to, yo) and that the
functions f and df/dy are continuous in that rectangle. Then, in an interval
to – h < t < to + h contained in R, there’s a unique solution to the initial
value problem:
dy
= f _ t, y i , y _ t 0 i = y 0
dt
Note that this theorem discusses the continuity of both f and df/dy instead of
the continuity of both p(x) and g(x). Like the first theorem in this chapter,
this theorem guarantees the existence of a unique solution if its conditions
are met.
Here’s another note: If the differential equation in question actually is linear,
the theorem reduces to the first theorem in this chapter. In that case, f(t, y) =
–p(t)y + g(t) and df/dy = –p(t). So demanding that f and df/dy be continuous is
the same as saying that p(t) and g(t) be continuous.
Here’s a side note that many differential equations books won’t tell you: The
first theorem in this chapter guarantees a unique solution, but it’s actually a
little tighter than it needs to be in order to guarantee just a solution (which
isn’t necessarily unique). In fact, you can show that there’s a solution — but
not that it’s unique — to the nonlinear differential equation merely by proving
that f is continuous.
A couple of nonlinear existence and
uniqueness examples
In the following sections, I provide two examples that put the nonlinear existence and uniqueness theorem into action.
Example 1
Determine what the two theorems in this chapter have to say about the following differential equation as far as its solutions go:
dy 5x 2 + 9x + 6
=
dx
2 _ y - 4i
where y(0) = –1.
39
40
Part I: Focusing on First Order Differential Equations
Well, as you can see, this is a nonlinear equation in y. So the first theorem,
which deals only with linear differential equations, has nothing to say about it.
That means you need the nonlinear theorem. Note that for this theorem:
f _ x, y i = 5x + 9x + 6
2 _ y - 4i
2
and
df = - 5x 2 + 9x + 6
2
dy
2 _ y - 4i
These two functions, f and df/dy, are continuous, except at y = 4.
So you can draw a rectangle around the initial condition point, (0, –1) in which
both f and df/dy are continuous. And the existence and uniqueness theorem
for nonlinear equations guarantees that this differential equation has a solution in that rectangle.
Example 2
Now determine what the existence and uniqueness theorems say about this
differential equation:
dy
= y 1/5
dx
where y(1) = 0.
Clearly, this equation isn’t linear, so the first theorem is no good. Instead you
have to try the second theorem. Here, f is:
f(x, y) = y1/5
and df/dy is:
- 4/5
df = y
dy
5
Now you know that f(x, y) is continuous at the initial condition point given by:
y(1) = 0
But df/dy isn’t continuous at this point. The upshot is that neither the first
theorem nor the second theorem have anything to say about this initial value
problem. On the other hand, a solution to this differential equation is still
guaranteed because f(x, y) is continuous. However, it doesn’t guarantee the
uniqueness of that solution.
Chapter 3
Sorting Out Separable First Order
Differential Equations
In This Chapter
Figuring out the fundamentals of separable differential equations
Applying separable differential equations to real life
Advancing with partial fractions
S
ome rocket scientists call you, the Consulting Differential Equation
Expert, into their headquarters.
“We’ve got a problem,” they explain. “Our rockets are wobbling because we
can’t solve their differential equation. All the rockets we launch wobble and
then crash!”
They show you to a blackboard with the following differential equation:
2
dy
= x
dx 2 - y 2
“It’s not linear,” the scientists cry. “There’s a y2 in there!”
“I can see that,” you say. “Fortunately, it is separable.”
“Separable? What does that mean?” they ask.
“Separable means that you can recast the equation like this, where x is on
one side and y is on the other,” you say while showing them the following
equation on your clipboard:
(2 – y2) dy = x2 dx
“You can integrate the equation with respect to y on one side, and x on the
other,” you say.
“We never thought of that. That was too easy.”
42
Part I: Focusing on First Order Differential Equations
That’s what this chapter covers: separable first order differential equations.
(First order equations, as I note in Chapter 1, have derivatives that go up only
to the first order.) I explain the basics of separable equations here, such as
determining the difference between linear and nonlinear separable equations
and figuring out different types of solutions, such as implicit and explicit. I also
introduce you to a fancy method for solving separable equations involving
partial fractions. Finally, I show you a couple of real world applications for
separable equations. When you’re an expert at these equations, you too can
solve problems for rocket scientists.
Beginning with the Basics of Separable
Differential Equations
Separable differential equations, unlike general linear equations in Chapter 2,
let you separate variables so only variables of one kind appear on one side,
and only variables of another kind appear on the other. Say, for example, that
you have a differential equation of the following form, in which M and N are
functions:
M _ x, y i + N _ x, y i
dy
=0
dx
And furthermore, imagine that you could reduce this equation to the following form, where the function M depends only on x and the function N depends
only on y:
dy
M ^xh + N _ yi x = 0
This equation is a separable equation; in other words, you can separate the
parts so that only x appears on one side, and only y appears on the other.
You write the previous equation like this:
M(x) dx + N(y) dy = 0
Or in other words:
M(x) dx = –N(y) dy
If you can separate a differential equation, all that’s left to do at that point is
to integrate each side (assuming that’s possible). Note that the general form
of a separable differential equation looks like this:
M ^xh + N _ yi
dy
=0
dx
Chapter 3: Sorting Out Separable First Order Differential Equations
However, nothing here says that N(y) has to be linear in y. For example, consider this separable differential equation that isn’t linear:
x + y2
dy
=0
dx
And if you’re still not convinced, check out this one, which is also separable
but not linear:
x 9 + _1 - y 3 i
dy
=0
dx
In the following sections, I ease you into linear separable equations before
tackling nonlinear separable equations. I also show you a trick for turning
nonlinear equations into linear equations. (It’s so cool that it’ll impress all
your friends!)
Starting easy: Linear separable equations
To get yourself started with linear separable equations, say that you have
this differential equation:
dy
- x 2= 0
dx
This equation qualifies as linear. This also is an easily separated differential
equation. All you have to do is put it into this form:
dy = x2 dx
And now you should be able to see the idea behind solving separable differential equations immediately. You just have to integrate, which gives you this
equation:
3
y= x +c
3
where c is an arbitrary constant. There’s your solution! How easy was that?
Introducing implicit solutions
Not all separable equation solutions are going to be as easy as the one in the
previous section. Sometimes finding a solution in the y = f(x) format isn’t terribly easy to get. Mathematicians refer to a solution that isn’t in the form
y = f(x) as an implicit solution. Coming up with such a solution is often the
best you can do, because solving a separable differential equation involves
43
44
Part I: Focusing on First Order Differential Equations
integrating both sides of the equation, and there’s no guarantee that the integration will come out cleanly. (The form y = f(x) is known as an explicit solution; I show you how to find an explicit solution from an implicit solution in
the next section.)
Try this differential equation to see what I mean:
2
dy
= x
dx 2 - y 2
How about it? One of the first things that should occur to you is that this isn’t
a linear differential equation, so the techniques in the first part of this chapter won’t help. However, you’ll probably notice that you can write this equation as:
(2 – y2) dy = x2 dx
As you can see, this is a separable differential equation because you can put y
on one side and x on the other. You can also write the differential equation
like this:
- x 2 + _2 - y 2i
dy
=0
dx
You can cast this particular equation in terms of a derivative of x, and then
you integrate with respect to x to solve it. After integration, you wind up with
the following:
-x 2=
d _- x 3 / 3 i
dx
Note that:
3
dy d _ 2y - y / 3 i
_ 2 - y 2 i dx =
dx
because of the chain rule, which says that:
df = df dy
dx dy dx
So now you can write the original equation like this:
- x 2 + _2 - y 2i
3
dy
y3
= d e - x + 2y - o = 0
dx dx 3
3
If the derivative of the term on the right is 0, it must be a constant this way:
e
3
- x 3 + 2y - y = c
o
3
3
Chapter 3: Sorting Out Separable First Order Differential Equations
Finally, multiplying by 3 gives you the following implicit solution to your original separable equation:
–x3 + 6y – y3 = c
To see how the solutions look graphically, check out the direction field for this
differential equation in Figure 3-1. (I introduce direction fields in Chapter 1.)
4
3
2
y
1
0
–1
Figure 3-1:
The –2
direction
field of a
nonlinear –3
separable
equation. –4
–4
–3
–2
–1
0
1
x
2
3
4
Finding explicit solutions from
implicit solutions
The implicit solution in the previous section, with terms in y and y3, isn’t terribly easy to cram into the y = f(x) format. In this section, you discover that
you can find an explicit solution to a separable equation by using a quadratic
equation, which is the general solution to polynomials of order two.
Try another, somewhat more tractable problem. Solve this differential
equation:
dy 9x 2 + 6x + 4
=
dx
2 _ y - 1i
where y(0) = –1.
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46
Part I: Focusing on First Order Differential Equations
If this differential equation were of the following form:
dy
= 9x 2 + 6x + 4
dx
there would be no problem. After all, you would just integrate. But you’ve
probably noticed that pesky 2(y – 1) term in the denominator on the right
side. Fortunately, you may also realize that this is a separable differential
equation because you can put y on one side and x on the other. Simply write
the equation like this:
2(y – 1) dy = (9x2 + 6x + 4) dx
Now you integrate to get this equation:
y2 – 2y = 3x3 + 3x2 +4x + c
Using the initial condition, y(0) = –1, substitute x = 0 and y = –1 to get the
following:
1+2=c
Now you can see that c = 3 and that the implicit solution to the separable
equation is:
y2 – 2y = 3x3 + 3x2 +4x + 3
If you want to find the explicit solution to this and similar separable equations, simply solve for y with the quadratic equation because the highest
power of y is 2. Solving for y using the quadratic formula gives you:
y = 1 ! 3x 3 + 3x 2 + 4x + 4
You have two solutions here: one where the addition sign is used and one
where the subtraction sign is used. To match the initial condition that y(0) =
–1, however, only one solution will work. Which one? The one using the subtraction sign:
y = 1 - 3x 3 + 3x 2 + 4x + 4
In this case, the solution with the subtraction is valid as long as the expression under the square root is positive — in other words, as long as x > –1.
You can see the direction field for the general solutions to this differential
equation in Figure 3-2.
Chapter 3: Sorting Out Separable First Order Differential Equations
2
1
0
y
–1
–2
Figure 3-2:
The
direction –3
field of a
separable –4
equation
with initial
conditions. –5
–2
–1
0
1
x
2
3
As I note in Chapter 1, connecting the slanting lines in a direction field gives
you a graph of the solution. You can see a graph of this particular function in
Figure 3-3.
2
1
0
y
–1
–2
Figure 3-3:
A graph of
the solution –3
of a
separable –4
equation
with initial
conditions. –5
–2
–1
0
1
x
2
3
47
48
Part I: Focusing on First Order Differential Equations
Tough to crack: When you can’t find an
explicit solution
Most of the time, you can find an explicit solution from an implicit solution.
But every once in a while, getting an explicit solution is pretty tough to do.
Here’s an example:
dy
y sin x
=
dx _1 + 2y 2 i
where y(0) = 1.
As you get down to work (bringing to bear all your differential equation skills!),
the first thing that may strike you is that this equation isn’t linear. But, you’ll
also likely note that it’s separable. So simply separate the equation into y on
the left and x on the right, which gives you this equation:
_1 + 2 y 2 i dy
= sin x dx
y
This equation subsequently becomes
dy
y + 2y dy = sin x dx
Now you can integrate to get this:
ln|y| + y2 = –cos x + c
Next, take a look at the initial condition: y(0) = 1. Plugging that condition into
your solution gives you this equation:
0 + 1 = –1 + c
or
c=2
So your solution to the initial separable equation is:
ln|y| + y2 = –cos x + 2
This is an implicit solution, not an explicit solution, which would be in terms
of y = f(x). In fact, as you can see from the form of this implicit solution, getting an explicit solution would be no easy task.
Chapter 3: Sorting Out Separable First Order Differential Equations
However, never fear the implicit solution! You still can use numerical or graphical methods to deal with such solutions. For instance, take a look at the direction field for this differential equation, which indicates what the integral curves
look like, in Figure 3-4.
3
2
y
1
Figure 3-4: 0
The
direction
field of a –1
separable
equation
with a hard- –2
to-find
explicit
solution. –3
–3
–4
–2
–1
0
1
x
2
3
4
A neat trick: Turning nonlinear separable
equations into linear separable equations
In this section, I introduce you to a neat trick that helps with some differential equations. With it, you can make a linear equation out of a seemingly nonlinear one. All you have to do to use this trick is to substitute the following
equation, in which v is a variable:
y = xv
In some cases, the result is a separable equation.
As an example, try solving this differential equation:
dy 2y 4 + x 4
=
dx
xy 3
49
50
Part I: Focusing on First Order Differential Equations
At first glance, this equation doesn’t look separable. In fact, even if you break
it out into two fractions, it still doesn’t look separable:
3
dy 2y
= x + x3
dx
y
What do you do now? Keep reading to find out.
Knowing when to substitute
You can use the trick of setting y = xv when you have a differential equation
that’s of the following form:
dy
= f _ x, y i
dx
when f(x, y) = f(tx, ty), where t is a constant.
You can see that substitution is possible, because substituting tx and ty into
this differential equation gives you the following result:
dy 2t 4 y 4 + t 4 x 4
=
dx
txt 3 y 3
which breaks down to:
dy 2y 4 + x 4
=
dx
xy 3
Substituting y = xv into this differential equation gives you:
2 ^ xv h + x 4
v + x dv =
3
dx
x ^ xvh
4
This equation now can be simplified to look like this:
4
x dv = v +3 1
dx
v
You now have a separable equation!
Separating and integrating
Continuing with the example from the previous section, you can now separate the terms, which gives you:
dx = v 3 dv
x
v4+ 1
Chapter 3: Sorting Out Separable First Order Differential Equations
After you integrate both sides, you get the following equation:
ln^ x h =
ln_ v 4 + 1i
+c
4
where c is a constant of integration. Bearing in mind that, where k is a constant:
ln(x) + ln(k) = ln(kx)
and that:
n ln(x) = ln(x n )
you get:
v4 + 1 = (kx)4
where:
c = –ln(k)
Where does all this get you? You’re ready to substitute with the following:
y
v= x
This substitution gives you:
4
4
y
d x n + 1 = ^ kx h
So:
y4 + x4 = mx8
where m = k4. And solving for y gives you the following:
y = (mx8 – x4)1/4
And there’s your solution. Nice work!
51
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Part I: Focusing on First Order Differential Equations
Trying Out Some Real World
Separable Equations
In the following sections, I take a look at some real world examples featuring
separable equations.
Getting in control with a sample
flow problem
To understand the relevance of differential equations in the real world, here’s
a sample problem to ponder: Say that you have a 10-liter pitcher of water, and
that you’re mixing juice concentrate into the pitcher at the same time that
you’re pouring juice out. If the concentrate going into the pitcher has 1⁄4 kg of
sugar per liter, the rate at which the concentrate is going into the pitcher,
which I’ll call rin, is 1⁄100 liter per second, and the juice in the pitcher starts off
with 4 kg of sugar, find the amount of sugar in the juice, Q, as a function of
time, t.
Because this problem involves a rate — dQ/dt, which is the change in the
amount of sugar in the pitcher — it’s a differential equation, not just a simple
algebraic equation. I walk you through the steps of solving the equation in
the following sections.
Determining the basic numbers
When you start trying to work out this problem, remember that the change in
the amount of sugar in the pitcher, dQ/dt, has to be the rate of sugar flow in
minus the rate of sugar flow out, or something like this:
dQ
= _ rate of sugar flow ini - _ rate of sugar flow outi
dt
Now you ask: What’s the rate of sugar flow in? That’s easy; it’s just the concentration of sugar in the juice concentrate multiplied by the rate at which
the juice concentrate is flowing into the pitcher, which I’ll call rin. So, your
equation looks something like this:
r
_ rate of sugar flow ini = 4in kg/sec
Now what about the flow of sugar out? The rate of sugar flow out is related to
the rate at which juice leaves the pitcher. So if you assume that the amount of
juice in the pitcher is constant, then rin = rout = r. That, in turn, means the rate
Chapter 3: Sorting Out Separable First Order Differential Equations
of sugar flow out is the rate at which the juice leaves the pitcher, r, multiplied
by the concentration of sugar in the juice, which is Q divided by the capacity
of the pitcher (10 liters), or Q/10. Here’s what your equation would look like:
Q
_ rate of sugar flow outi = 10 r kg/sec
So that means:
dQ
Qr
= _ rate of sugar flow ini - _ rate of sugar flow outi = r dt
4 10
where the initial condition is:
Q0 = 4 kg
Solving the equation
The equation at the end of the previous section is separable, and separating
the variables, each on their own side, gives you this equation:
dQ Qr r
+
=
dt
10 4
Now that, you might say, is a linear differential in Q. And you’d be right. So you
know that the equation is both linear and separable.
You can handle this differential equation using the methods in Chapter 2. For
instance, to solve, you find an integrating factor, multiply both sides by the
integrating factor, and then see if you can figure out what product the left
side is the derivative of and integrate it. Whew! It sounds rough, but note that
the equation is of the following form:
dy
+ ay = b
dt
The solution to this kind of differential equation is already found in Chapter 2;
you use an integrating factor of eat. The solution to this kind of equation is:
y = (b/a) + ce–at
So you can see that the solution to the juice flow problem is:
Q(t) = 2.5 + ce–rt/10
Because r = 1⁄100 liter per second, the equation becomes:
Q(t) = 2.5 + ce–t/1000
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54
Part I: Focusing on First Order Differential Equations
And because the initial condition is:
Q0 = 4 kg
you know that:
Q = 2.5 + 1.5 e–t/1000
Note the solution as t → ∞ is 2.5 kg of sugar, and that’s what you’d expect.
Why? Because the concentrate has 1⁄4 kg of sugar per liter, and 10 liters of
water are in the pitcher. So 10/4 = 2.5 kg.
The direction field for different values of Q0 appears in Figure 3-5. Notice that
all the solutions tend toward the final Q of 2.5 kg of sugar, as you’d expect.
20
15
Q
10
Figure 3-5:
The
direction
field of a
flow
problem
solution.
5
0
0
300
600
900
1200
1500
1800 t 2100
You can see a graph of this solution in Figure 3-6.
2400
2700
3000
Chapter 3: Sorting Out Separable First Order Differential Equations
5
4
Q
3
2
Figure 3-6: 1
The graph of
the solution
of a flow
problem. 0
0
300
600
900
1200
1500
1800 t 2100
2400
2700
3000
Striking it rich with a sample
monetary problem
You may not have realized that differential equations can be used to solve
money problems. Well they can! And here’s a problem to prove it: Say that
you’re deciding whether to deposit your money in the bank. You can calculate how your money grows, dQ/dt, given the interest rate of the bank and the
amount of money, Q, that you have in the bank. As you can see, this is a job
for differential equations.
Figuring out the general solution
Suppose your bank compounds interest continuously. The rate at which your
savings, Q, grows, is:
dQ
= rQ
dt
where r is the interest rate that your bank pays.
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Part I: Focusing on First Order Differential Equations
This equation says that the rate at which your money grows is equal to the
interest rate multiplied by the current amount of money you have. That’s an
equation for the rate at which your money grows, not the actual amount of
money.
Say that you have Q0 money at t = 0:
Q(0) = Q0
How much money would you have at a certain time in the future? That’s easy
enough to figure out. Separate the variables, each on their own side, like this:
dQ
= r dt
Q
Then integrate:
ln|Q| = rt
Finally, exponentiate both sides, which gives you the following equation:
Q = cert
To match the initial condition:
Q(0) = Q0
the solution becomes:
Q = Q0ert
So, in other words, your money would grow exponentially. Not bad.
Compounding interest at set intervals
Now I want you to examine the result from the previous section a little, deriving it another way so that it makes more sense. If your bank compounded
interest once a year, not continuously, after t years, you’d have this much
money:
Q = Q0(1 + r)t
That’s because if your interest was 5 percent, after the first year, you would
have 1.05Q0; at the end of the second year, 1.052Q0, and so on.
Chapter 3: Sorting Out Separable First Order Differential Equations
What if your bank compounded interest twice a year? Would you have this
much at the end of t years:
Q = Q0(1 + r)2t
No, you wouldn’t. Why? Because that would pay you r percent interest twice
a year. For example, if r = 8 percent, the previous equation would pay you 8
percent of your total savings twice a year. Instead, banks divide the interest
rate they pay you by the number of times they compound per year, like this:
Q = Q 0 c1 + r m
2
2t
In other words, if the bank compounds twice a year, and the annual interest
rate is 8 percent, six months into the year it pays you 4 percent, and at the
end of the year it pays another 4 percent.
In general, if your bank compounds interest m times a year, after t years,
you’d have:
r m
Q = Q 0 c1 + m
mt
If you take the limit as m → ∞ — that is, as your bank starts to compound
continuously — you get this equation:
r m
Q = mlim
Q 0 c1 + m
"3
mt
But that’s just the expansion for ert. So, as the bank compounds continuously,
you get:
mt
r m = Q e rt
Q = mlim
Q 0 c1 + m
0
"3
And this result confirms the answer you got from solving the differential
equation in the previous section.
So if you had $25 invested, and you left it alone at 6 percent for 60 years,
you’d have:
Q = Q0ert = 25e0.06(60)
or:
Q = Q0ert = 25e0.06(60) = $914.96
Hmm, not such a magnificent fortune.
57
58
Part I: Focusing on First Order Differential Equations
Adding a set amount of money
How about if you add a set amount every year to the equation in the previous
section? That would be better, wouldn’t it? Say that you add $5,000 a year. In
that case, remember that the set amount would change the differential equation for your savings, which was this:
dQ
= rQ
dt
The equation would change to this, where k is the amount you contribute
regularly:
dQ
= rQ + k
dt
If you deposit regularly, k > 0; if you withdraw regularly, k < 0. Ideally, you
should add or subtract k from your account continuously over the year to
make your solution exact, but here you can just assume that you add or subtract k once a year.
Putting this new equation into standard separable form gives you this:
dQ
- rQ = k
dt
This equation is of the following form:
dy
+ ay = b
dt
The solution to this kind of equation is:
y = (b/a) + ce–at
In this case, that solution means:
Q = cert – k/r
What’s going on here? It looks like you have the solution for leaving money in
the bank without adding anything minus the amount you’ve added. Can that
be right? The answer is in c, the constant of integration. Here, the initial condition is:
Q(0) = Q0
which means that:
Q(0) = cer0 – k/r = c – k/r = Q0
Chapter 3: Sorting Out Separable First Order Differential Equations
or to simplify things:
c = Q0 + k/r
So your solution turns out to be:
Q = cert – k/r = (Q0 + k/r)ert – k/r
Working this out gives you:
Q = _ Q 0 + k/r i e rt - k/r = Q 0 e rt + kr _ e rt - 1i
That looks a little better! Now the first term is the amount that you’d earn if
you just left Q0 in the account, and the second term is the amount resulting
from depositing or withdrawing k dollars regularly.
For example, say you started off with $25, but then you added $5,000 every
year for 60 years. At the end of 60 years at 6 percent, you’d have:
5, 000 0.06 (60)
- 1i
Q = Q 0 e rt + kr _ e rt - 1i = 25e 0.06 (60) +
_e
0.06
After calculating this out, you’d get:
Q = 25e 0.06 (60) +
5, 000 0.06 (60)
- 1i = $914.96 + $2, 966, 519 = $2, 967, 434
_e
0.06
Quite a tidy sum.
Break It Up! Using Partial Fractions
in Separable Equations
When a term in a separable differential equation looks a little difficult to integrate, you can use the method of partial fractions to separate it. This method
is used to reduce the degree of the denominator of a rational expression.
For example, using the method of partial fractions, you can express:
6
x 2 + 2x - 8
as the following equation:
6
= 1 - 1
x 2 + 2x - 8 x - 2 x + 4
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Part I: Focusing on First Order Differential Equations
Note that the power of the denominator has been reduced by one. You’ll
often see the method of partial fractions used when solving differential equations involving fractions, because using this method makes integrating the
resulting two terms a lot easier.
Here’s an example:
dy
2xy
=
dx x 2 - y 2
You’re probably screaming at your book right now. That’s not separable, you
say. However, perhaps you’ve also noticed that this equation is of the following form:
dy
= f _ x, y i
dx
And f(x, y) = f(tx, ty), where t is a constant. So you now have this equation:
dy
2txty
2xy
=
=
dx t 2 x 2 - t 2 y 2 x 2 - y 2
Yep, you guessed it: This equation calls for the old trick of substituting y = vx.
(See the earlier section “A neat trick: Turning nonlinear separable equations
into linear separable equations” for details.) Substituting y = vx into this differential equation gives you the following equation:
2x ^ xv h
v + x dv = 2
dx x - ^ xvh 2
or to simplify matters:
v + x dv = 2v 2
dx 1 - v
By subtracting the term on the right and adding to the left, you get the
following:
v _ v 2 + 1i
x dv +
=0
dx
_ v 2 - 1i
Flipping the fractions gives you:
dx + v 2 - 1 dv = 0
x
v _ v 2 + 1i
Chapter 3: Sorting Out Separable First Order Differential Equations
Time to use the method of partial fractions. In this case, you get the following
equation:
dx + v 2 - 1 dv = dx + d 2v - 1 n dv = 0
x
x
v2+ 1 v
v _ v 2 + 1i
or to simplify:
dx + 2v dv - dv = 0
x
v
v2+ 1
Ah, much better. Now you can integrate! Integrating all these terms gives you:
ln|x| + ln|v2 + 1| – ln|v| = c
or:
ln|x| + ln|v2 + 1| = ln|v| + c
Exponentiating both sides gives you this equation:
x(v2 + 1) = kv
where k = ec.
Substituting v = y/x gives you the following:
x _ y 2 + x 2 i ky
= x
x2
The previous equation can then be simplified to the implicit solution of the
following:
y2 + x2 = ky
or to simplify even further:
y2 – ky + x2 = 0
Finally, you can solve using the quadratic equation to get the following
explicit solution:
y=
k ! k 2 - 4x 2
2
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Part I: Focusing on First Order Differential Equations
Chapter 4
Exploring Exact
First Order Differential
Equations and Euler’s Method
In This Chapter
Understanding the fundamentals of exact differential equations
Figuring out whether differential equations are exact
Turning nonexact equations into exact equations with integrating factors
Crunching numbers with Euler’s method
Digging into difference equations
N
ot all first order differential equations are linear (see Chapter 2) or
separable (see Chapter 3). So sometimes you have to use other tools to
solve first order differential equations. One of those tools is knowing how
to solve exact differential equations. That’s what I introduce you to in this
chapter. And for those really intractable (a.k.a. difficult) differential equations, you also get an introduction to working with mathematical methods. In
particular, you find out how to use Euler’s method to approximate solutions
to just about any differential equation (assuming it has a solution!). And you
find out that Euler’s method is a type of difference equation (don’t worry;
I explain everything).
Exploring the Basics of Exact
Differential Equations
One of the most powerful methods for working with differential equations is
seeing whether they’re exact, and if they are, you can tackle them. I explain
the basics in the following sections.
64
Part I: Focusing on First Order Differential Equations
Defining exact differential equations
To solve an exact differential equation, you have to find a function whose
partial derivatives correspond to the terms in the differential equation. For
example, assume that you have a differential equation of this form:
M _ x, y i + N _ x, y i
dy
=0
dx
where M and N are functions.
Now suppose that you can find a function f(x, y) such that the following equations are true:
2f _ x, y i
= M _ x, y i
2x
2f _ x, y i
= N _ x, y i
2y
Note that the previous two equations are partial derivatives with respect
to x, so I include the symbol ∂. (Check out Chapter 1 for more about partial
derivatives.)
The differential equation that you’re trying to solve becomes:
2f _ x, y i 2f _ x, y i dy
+
=0
2x
2y
dx
which is equal to (note: ordinary derivatives now):
df _ x, y i
=0
dx
So the solution, after integration, is:
f(x, y) = c
The previous solution is at least the implicit solution; you may have to move
things around a bit to get the actual explicit solution. (As I explain in Chapter 3,
implicit solutions aren’t in the form y = f(x), but explicit solutions are.)
Here’s the point: If you can find a function f such that a differential equation
can be reduced to the following form:
df _ x, y i
=0
dx
then the differential equation is said to be exact.
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
Working out a typical exact
differential equation
Exact differential equations can be tough to solve, so in this section, I provide
a typical example so you can get the hang of it. Check out this differential
equation:
4x + 2y 2 + 4xy
dy
=0
dx
This equation isn’t linear, and it also isn’t separable. However, you may suspect that the equation is exact. So, if that’s the case, how do you solve it?
You might note that this function has some interesting properties:
f(x, y) = 2x2 + 2xy2
In particular, note that ∂f/∂x (note: this is the partial derivative with respect
to x) is:
2f = 4x + 2y 2
2x
Those are the first two terms of the equation you’re looking to solve. Also,
note that
2f = 4xy
2y
which looks a lot like the third term in the original equation. So you could
write the equation like this:
2f + 2f dy = 0
2x 2y dx
You’re making progress! Note that because y is a function of f, you can use the
chain rule (see Chapter 3) and switch to ordinary derivatives, meaning that
you can write the original equation like this:
df = 0
dx
where
f(x, y) = 2x2 + 2xy2
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Part I: Focusing on First Order Differential Equations
The ordinary derivative I just gave you is easy to integrate, so you get the following equation:
2x2 + 2xy2 = c
where c is a constant of integration. That’s the implicit solution. Now you can
divide by 2, absorbing that 2 into c, which gives you:
x2 + xy2 = c
So that’s your implicit solution. But it’s easy to solve for y, as you can see in
this equation:
2
y = c -x x
Determining Whether a Differential
Equation Is Exact
As you can see from the previous section, knowing when a differential equation is exact is helpful. But just when is it exact? Keep reading to find out.
Checking out a useful theorem
How can you tell, in a systematic way, whether a differential equation is
exact? You’re in luck: It turns out that there’s a handy theorem to determine
whether a differential equation is exact, and here it is:
If the functions M, N, oM/oy, and oN/ox are continuous in a rectangle R,
then this differential equation:
M _ x, y i + N _ x, y i
dy
=0
dx
is an exact differential equation in the rectangle R if and only if:
oM _ x, y i oN _ x, y i
=
oy
ox
at every point in R.
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
In other words, if you have the differential equation
M _ x, y i + N _ x, y i
dy
=0
dx
there exists a function f(x, y) in a rectangle R such that:
2f _ x, y i
= M _ x, y i
2x
and
2f _ x, y i
= N _ x, y i
2y
if and only if:
2M _ x, y i 2N _ x, y i
=
2y
2x
in the rectangle R.
So how do you solve exact differential equations? You have to solve the partial differential equation:
2f _ x, y i
= M _ x, y i
2x
and
2f _ x, y i
= N _ x, y i
2y
And if you can find it, the implicit solution is f(x, y) = c.
Applying the theorem
So how about an example that shows how to apply the theorem in the previous section? Take a look at this differential equation:
2xy + _1 + x 2 i
dy
=0
dx
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Part I: Focusing on First Order Differential Equations
In other words:
M(x, y) = 2xy
and
N(x, y) = (1 + x2)
Note that
2M _ x, y i 2N _ x, y i
=
= 2x
2y
2x
So now you know that the differential equation is exact. And now you need to
find M(x, y) and N(x, y) such that:
2f _ x, y i
= M _ x, y i
2x
and
2f _ x, y i
= N _ x, y i
2y
When you find f(x, y), the solution to the original equation is
f(x, y) = c
Start by finding f(x, y). Because
M(x, y) = 2xy
that means
2f _ x, y i
= 2xy
2x
Integrating both sides of the previous equation gives you:
f(x, y) = x2y + g(y)
where g(y) is a function that depends only on y, not on x. So what’s g(y)? You
know that:
2f _ x, y i
= N _ x, y i
2y
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
and
N(x, y) = (1 + x2)
or
2f _ x, y i
= _1 + x 2 i
2y
Because you know that
f(x, y) = x2y + g(y)
you get this equation:
2f _ x, y i
2g
y = 1 + x2
= x 2+
2y
2y _ i
By canceling out the x2, you get
2g
y =1
2y _ i
Fortunately, this equation is easy enough to integrate. Integrating it gives
you this:
g(y) = y + d
where d is a constant of integration. And because
f(x, y) = x2y + g(y)
you get:
f(x, y) = x2y + y + d
Whew. You now know that the solution to f(x, y) = x2y + g(y) is:
f(x, y) = c
So
x2y + y = c
where the constant of integration d has been absorbed into the constant of
integration c.
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Part I: Focusing on First Order Differential Equations
As you recall from the earlier section “Defining exact differential equations,”
finding f(x, y) gives you the implicit solution to the exact differential solution
you’re trying to solve. Happily, x2y + y = c is easy to solve for y in terms of x,
giving you:
y=
c
_1 + x 2 i
And that, my friends, is the explicit solution to your original equation. Cool!
Conquering Nonexact Differential
Equations with Integrating Factors
The earlier sections in this chapter cover exact differential equations. But
what about those equations that don’t look exact but can be converted into
exact equations? Yes, you read that right! Sometimes, differential equations
are intractable and not exact. For example, that’s the case with this differential equation:
y-x
dy
=0
dx
You start off by checking this differential equation for exactness, meaning
that you have to cast it in this form:
M _ x, y i + N _ x, y i
dy
=0
dx
So, in the case of the example:
M(x, y) = y
and
N(x, y) = –x
The differential equation is exact if:
2M _ x, y i 2N _ x, y i
=
2y
2x
But these two aren’t equal; rather:
2M _ x, y i
=1
2y
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
and
2N _ x, y i
= -1
2x
As you can see, the original equation isn’t exact. However, the two partial
derivatives, ∂M/∂x and ∂N/∂y, differ only by a minus sign. Isn’t there some
way of making this differential equation exact?
Yup, you’re right. There is a way! You just have to use an integrating factor.
As I discuss in Chapter 2, integrating factors are used to multiply differential
equations to make them easier to solve. In the following sections, I explain
how to use an integrating factor to magically turn a nonexact equation into
an exact one.
Finding an integrating factor
How can you find an integrating factor that makes differential equations
exact? Just follow the steps in the following sections.
Multiplying by the factor you want to find
Using the example from earlier, here I show you how to find an integrating
factor that makes differential equations exact. Say you have this differential
equation:
M _ x, y i + N _ x, y i
dy
=0
dx
Multiplying this equation by the integrating factor µ(x, y) (which you want to
find) gives you this:
µ _ x, y i M _ x, y i + µ _ x, y i N _ x, y i
dy
=0
dx
This equation is exact if:
2 a µ _ x, y i M _ x, y i k
2y
2 a µ _ x, y i N _ x, y i k
=
2x
which means that µ(x, y) must satisfy this differential equation:
V
R
2µ _ x, y i
2µ _ x, y i S a 2M _ x, y i k 2N _ x, y i W
W µ _ x, y i = 0
M _ x, y i
- N _ x, y i
+S
2y
2x
2y
2x
WW
SS
X
T
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72
Part I: Focusing on First Order Differential Equations
Well, yipes. That doesn’t seem to have bought you much simplicity. In fact,
this equation looks more complex than before. What you have to do now is to
assume that µ(x, y) is a function of x only — that is, µ(x, y) = µ(x) — which
also means that:
2µ ^ x h
=0
2y
This turns the lengthy and complex equation into the following:
N
J
2µ ^ x h K 2M _ x, y i 2N _ x, y i O
+
µ x =0
O ^ h
K
2x
2y
2x
P
L
Or in other words:
- N _ x, y i
N _ x, y i
N
J
2µ ^ x h K 2M _ x, y i 2N _ x, y i O
=
µ x
O ^ h
K
2x
2y
2x
P
L
Dividing both sides by N(x, y) to simplify means that:
J 2M _ x, y i 2N _ x, y i N
2µ ^ x h
1
Oµ ^ x h
K
=
O
2x
2y
2x
N _ x, y i K
P
L
Well, that looks somewhat better. You can see how to finish the example in
the next section.
Completing the process
Remember the differential equation that you’re trying to solve? If not, here it is:
y-x
dy
=0
dx
In this case,
M(x, y) = y
and
N(x, y) = –x
Plug these into your simplified equation from the previous section, which
becomes:
2µ ^ x h -1
= x `1 - ^-1hj µ ^ x h
2x
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
or:
2µ ^ x h - 2
= x µ^ x h
2x
Now this equation can be rearranged to get this one:
2µ ^ x h
= -2 2xx
µ^ x h
Integrating both sides gives you:
ln|>(x)| = –2 ln|x|
And finally, exponentiating both sides results in this equation:
! µ ^ x h = 12
x
For this differential equation:
2M _ x, y i
=1
2y
and
2N _ x, y i
= -1
2x
Because these equations differ by a sign, you may suspect that you need the
negative version of µ(x), meaning that:
µ ^ x h = - 12
x
So that, at last, is your integrating factor. Whew!
Using an integrating factor
to get an exact equation
To carry on from the example in the previous section: Multiplying your original
equation by your integrating factor gives you this new differential equation:
-y
dy
+1
x dx = 0
x2
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74
Part I: Focusing on First Order Differential Equations
Is this exact? To find out, consider that
M _ x, y i =
-y
x2
and
N _ x, y i = 1
x
The differential equation is exact if:
2M _ x, y i 2N _ x, y i
=
2y
2x
You can now see that these two terms are equal, because:
2M _ x, y i -1
= 2
2y
x
and
2N _ x, y i -1
= 2
2x
x
So the integrating factor, –1/x2, did the trick, and the equation is now exact.
The finishing touch: Solving the
exact equation
The last step is to solve the exact equation as I explain earlier in this chapter.
In other words, you have to solve it by finding a function f(x, y) = c such that:
2f _ x, y i
= M _ x, y i
2x
and
2f _ x, y i
= N _ x, y i
2y
As noted in the previous section,
M _ x, y i = -
y
x2
which means that
2f _ x, y i
y
=- 2
2x
x
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
Integrating both sides of the equation gives you:
y
f _ x, y i = x + g _ y i
And because, using the info from the previous section:
2f _ x, y i
= N _ x, y i = 1
x
2y
you can see that
g(y) = d
where d is a constant of integration. So f(x, y) must be:
y
f _ x, y i = x + d
As you recall from the earlier section “Defining exact differential equations,”
the implicit solution of an exact differential equation is:
f(x, y) = c
So:
y
x =c
where the constant d has been absorbed into the constant c. Now you know
that the explicit solution to the exact equation is:
y = cx
Cool. The solution turned out to be simple.
Getting Numerical with Euler’s Method
Face it: Sometimes, you just can’t find a solution to some differential equations. Or, at best, you may only be able to find an implicit one. So what do
you do? In that case, it could be time to turn to numerical methods — that is,
using a computer.
I know, using a computer feels a lot like a cop-out, but sometimes you have
no other choice. Chapters 1, 2, and 3 all contain examples where I used a
computer to calculate direction fields and plot some actual solutions. Now, in
the following sections, I introduce one simple numerical method of solving
differential equations. This method is called Euler’s method.
75
76
Part I: Focusing on First Order Differential Equations
So who was this Euler guy?
Leonhard Paul Euler (1707–1783) was a Swiss
mathematician who lived, for the most part, in
Russia and Germany. He was an important
genius who contributed to physics, calculus,
and graph theory. He’s also responsible for
much of the mathematical terminology that
people take for granted, such as the notation
used for mathematical functions.
Understanding the method
Euler’s method basically says “Hey, you may not have the actual curve that
represents the solution to your differential equation, but you have the slope
of that curve everywhere (because the slope — the rate of change of the
curve — is the derivative).”
To better understand what I mean, say that you have the following general
differential equation:
dy
= f _ x, y i
dx
And suppose that you have a point, (x0, y0), that’s on the solution curve.
Because of your differential equation, you know that the slope of the solution
curve at that point is f(x0, y0).
Now imagine that you want to find the solution at a point, (x, y), a short distance away. Here’s how you might find y:
y = y0 + ∆y
Anything with the symbol ∆ signifies change. So, ∆y means a change in y.
Because the slope, m, is defined as ∆y/∆x, this equals (for small ∆x):
y = y0 + m ∆x
And because ∆x = x – x0, you have:
y = y0 + m (x – x0)
Because the slope, m, is equal to the derivative at (x0, y0), and because of
your original equation, m = f(x0, y0), you get:
y = y0 + f(x0, y0) (x – x0)
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
So if you keep (x – x0) small, this approximation should be close. All it’s doing
is using the slope to extrapolate (x0, y0) to a new point nearby, (x, y). This
process is illustrated in Figure 4-1.
ƒ(x0, y0)
y
y0
Figure 4-1:
Euler’s
method at
work.
x0
x
If you call ∆x by the term h, as it’s often referred to when using Euler’s
method, you get this:
y1 = y0 + f(x0, y0) h
And in general, you can find any point along the solution curve like this,
when using Euler’s method:
yn+1 = yn + f(xn, yn) h
Checking the method’s accuracy
on a computer
In this section, I put Euler’s method to work, finding a solution to the following differential equation:
dy
=x
dx
where
y(0) = 0
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78
Part I: Focusing on First Order Differential Equations
In fact, you already know the solution by simply integrating:
2
y= x
2
Because you know the solution, you can now check the accuracy of Euler’s
method. And you’re in luck, because here I show you how to develop a short
program that uses Euler’s method to solve differential equations. I developed
this program using the Java programming language (which you can get for
free at java.sun.com; just click the Java SE link under the Downloads tab).
You don’t have to know Java to use this book — the programming in this section is only a demonstration. But because numerical methods of solving differential equations require a computer, I use a programming language to get
things working. You can skip this part if you aren’t interested.
Defining initial conditions and functions
The Java program starts by defining the initial condition point — which is a
solution point for the differential equation, by definition — (x0, y0). Then it
defines the step size, h, and the number of steps you want to take, n. So, for
example, if you have the initial point at (0, 0), and you want the solution at
(10, y), your step size is 0.1, and you need n = 100 steps:
double
double
double
double
x0 = 0.0;
y0 = 0.0;
h = 0.1;
n = 100;
At this point, the program also defines a Java function, f(x, y), that returns the
value of the derivative at any point (x, y):
public double f(double x, double y)
{
return x;
}
Because you know the exact solution, you’ll also set up a Java function to
return the exact solution at any point so you can compare it to the Euler’s
method value:
public double exact(double x, double y)
{
return x * x / 2;
}
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
The actual work is done in a Java function named “calculate.” In this function,
each step using Euler’s method is calculated and displayed, like this:
public void calculate()
{
double x = x0;
double y = y0;
double k;
System.out.println(“x\t\tEuler\t\tExact”);
for (int i = 1; i < n; i++){
k = f(x, y);
y = y + h * k;
x = x + h;
System.out.println(round(x) + “\t\t” + round(y) +
“\t\t” + round(exact(x, 0)));
}
}
The program also has a function named “round,” which rounds values to two
decimal places for the printout:
public double round(double val)
{
double divider = 100;
val = val * divider;
double temp = Math.round(val);
return (double)temp / divider;
}
Examining the entire code
The following is the whole code, e.java, which is a short program for calculating differential equation solutions using Euler’s method. The parts you have to
change when you want to solve your own differential equations are in bold:
public class e
{
double
double
double
double
x0 = 0.0;
y0 = 0.0;
h = 0.1;
n = 100;
public e()
{
79
80
Part I: Focusing on First Order Differential Equations
}
public double f(double x, double y)
{
return x;
}
public double exact(double x, double y)
{
return x * x / 2;
}
public static void main(String [] argv)
{
e de = new e();
de.calculate();
}
public void calculate()
{
double x = x0;
double y = y0;
double k;
System.out.println(“x\t\tEuler\t\tExact”);
for (int i = 1; i < n; i++){
k = f(x, y);
y = y + h * k;
x = x + h;
System.out.println(round(x) + “\t\t” + round(y) +
“\t\t” + round(exact(x, 0)));
}
}
public double round(double val)
{
double divider = 100;
val = val * divider;
double temp = Math.round(val);
return (double)temp / divider;
}
}
An example at work
In this section, I put the program to work. As it stands, e.java is set up with
the following differential equation:
dy
=x
dx
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
where
y(0) = 0
The program starts at (x0, y0) = (0, 0), which you know is a point on the solution curve, and then it calculates 100 steps, each of ∆x = 0.1.
You start by using the Java compiler, javac.exe, to compile the code:
C:\>javac e.java
Then, you run the compiled code, e.class, using java.exe like this: java e. The
program then prints out the current x value, the Euler approximation of the
solution at that x value, and the exact solution, as you see here:
C:\>java e
x
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
Euler
0.0
0.01
0.03
0.06
0.1
0.15
0.21
0.28
0.36
0.45
0.55
0.66
0.78
0.91
1.05
1.2
1.36
1.53
1.71
1.9
2.1
2.31
2.53
2.76
3.0
3.25
3.51
3.78
4.06
4.35
4.65
4.96
5.28
Exact
0.01
0.02
0.05
0.08
0.13
0.18
0.24
0.32
0.4
0.5
0.6
0.72
0.85
0.98
1.13
1.28
1.45
1.62
1.81
2.0
2.21
2.42
2.65
2.88
3.13
3.38
3.65
3.92
4.21
4.5
4.81
5.12
5.45
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82
Part I: Focusing on First Order Differential Equations
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
5.61
5.95
6.3
6.66
7.03
7.41
7.8
8.2
8.61
9.03
9.46
9.9
10.35
10.81
11.28
11.76
12.25
12.75
13.26
13.78
14.31
14.85
15.4
15.96
16.53
17.11
17.7
18.3
18.91
19.53
20.16
20.8
21.45
22.11
22.78
23.46
24.15
24.85
25.56
26.28
27.01
27.75
28.5
29.26
30.03
30.81
31.6
32.4
33.21
34.03
34.86
5.78
6.13
6.48
6.85
7.22
7.61
8.0
8.41
8.82
9.25
9.68
10.13
10.58
11.04
11.52
12.0
12.5
13.0
13.52
14.04
14.58
15.12
15.68
16.24
16.82
17.4
18.0
18.6
19.22
19.84
20.48
21.12
21.78
22.44
23.12
23.8
24.5
25.2
25.92
26.64
27.38
28.12
28.88
29.64
30.42
31.2
32.0
32.8
33.62
34.44
35.28
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
8.5
8.6
8.7
8.8
8.9
9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
35.7
36.55
37.41
38.28
39.16
40.05
40.95
41.86
42.78
43.71
44.65
45.6
46.56
47.53
48.51
36.12
36.98
37.84
38.72
39.6
40.5
41.4
42.32
43.24
44.18
45.12
46.08
47.04
48.02
49.0
So there you have it — Euler’s method looks pretty accurate for your differential equation’s solution. At x = 9.9, it’s off only by 1% (49.00 – 48.51). Not bad.
Tip: You can get even better accuracy by using smaller step sizes.
You have to be careful when it comes to step size. If you have a really steep
slope, you have to use a truly tiny step size to get the most accurate results.
Delving into Difference Equations
Euler’s method, which is discussed in the previous section, brings up an
interesting topic: difference equations. That’s correct, you read right —
“difference” equations. I know what you’re thinking: An explanation is in
order! Allow me to start at the beginning. Derivatives are built to work like
this, where ∆x → 0 and y = f(x):
f ^ x + ∆x h - f ^ x h
dy
= lim
∆x
dx ∆x " 0
However, sometimes, you don’t want ∆x to go to zero. In other words, sometimes it just makes sense to keep the step size, ∆x, discrete and nonzero. For
example, you can set up a differential equation to calculate the interest you
pay on a loan, but because ∆x → 0, the differential equation calculates the
interest as if it were compounded continuously. But what if the interest is
actually compounded monthly or annually? In that case, you should use ∆x =
1 month or 1 year, not ∆x → 0.
When ∆x doesn’t go to zero, you have a difference equation, not a differential
equation. I take some time to look at difference equations in the following
sections.
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84
Part I: Focusing on First Order Differential Equations
Some handy terminology
Euler’s method is a good example of a difference equation, because every y
value depends on the previous y value:
yn+1 = yn + f(xn, yn) h
Here’s how it looks written as a difference equation, where the value yn+1 is
some function of yn:
yn+1 = f(n, yn)
Note: This is the kind of equation you’d use if interest was compounded
annually in a savings account instead of continuously. In that case, the future
value of the savings account, yn+1, depends on the current value, yn.
Just as with differential equations, you can apply some terminology to difference equations. In particular, the previous equation is a first order difference
equation. Why? Because it depends on yn, not on earlier terms (yn-1, yn-2, and
so on). The equation is linear if f(n, yn) is linear in yn. If f(n, yn) isn’t linear in yn,
the equation is nonlinear.
You can also have initial conditions for difference equations, such as the first
value being set to a constant:
y0 = c
Iterative solutions
Because a difference equation is defined using successive terms:
yn+1 = f(n, yn)
the solution is all the terms y0, y1, y2 . . . yn+1.
To make solving these kinds of equations a little more manageable, assume
that the function f in the original equation only depends on yn, not on n itself,
so you get:
yn+1 = f(yn)
So:
y1 = f(y0)
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
and:
y2 = f(y1)
which means that:
y2 = f(y1) = f(f(y0))
This equation is sometimes written as:
y2 = f(y1) = f(f(y0)) = f 2(y0)
which is often called the second iterate of the difference equation’s solution,
written as f 2(y0).
Here’s the third iterate:
y3 = f(y2) = f(f(y1)) = f(f(f(yo)))
The third iterate can also be written using the f n( ) notation, like this:
y3 = f(y2) = f(f(y1)) = f(f(f(y0))) = f 3(y0)
As long as yn+1 = f(yn), you can write the nth iterate as:
yn = f n(y0)
Equilibrium solutions
It’s often important to know what happens as n → ∞. Does the series converge?
Does it diverge? The answer tells you whether you have a viable solution.
Sometimes, all the yn have the same value. In this case, the solution is said to
be an equilibrium solution. The series converges to that equilibrium solution.
So, every term is the same:
yn = f n(y0)
and:
yn = f(yn)
I introduce some examples in the following sections.
85
86
Part I: Focusing on First Order Differential Equations
Working without a constant
To better understand equilibrium solutions, say, for example, that you have
savings in a bank that pays an interest rate, i, annually. The amount of money
you have in year, n + 1, is written like so:
yn+1 = (1 + i) yn
The solution of this difference equation is easy, because i is constant. Each
year’s savings is simply (1 + i ) multiplied by the previous year’s savings. So
your equation looks like this:
yn = (1 + i)n y0
This means that the limiting value as n → ∞ depends on i.
For instance, if i is less than 0, you get:
lim y n = 0
n"3
If i is equal to 0, you get:
lim y n = y 0
n"3
Otherwise, you get:
lim y n = nonexistent
n"3
That is to say, the equilibrium solution is i = 0.
Working with a constant
Just for kicks, change the circumstances from the previous section. Say that
you increase your savings each year by adding a constant value, c. Your equation would now look like this:
yn+1 = (1 + i) yn + c
The solution to this difference equation is:
yn = (1 + i)ny0 + (1 + (1 + i) + (1 + i)2 + (1 + i)n-1)c
If i isn’t equal to 0, you can write the equation as:
y n = ^1 + i h y 0 +
n
^1 + i h - 1
c
i
n
Chapter 4: Exploring Exact First Order Differential Equations and Euler’s Method
For example, if you started with y0 = $1,000 and i = 5%, and you add $1,000 a
year for 10 years, you’d end up with an equation that looks like this:
y n = ^1 + i h y 0 +
n
^1 + i h - 1
c = 1.62 _1, 000i + 1.62 - 1 _1, 000i
i
.05
n
or:
y n = 1.62 _1, 000i + 1.62 - 1 _1, 000i = 1, 620 + 12, 400 = $14, 020
.05
Not a bad return!
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88
Part I: Focusing on First Order Differential Equations
Part II
Surveying Second
and Higher Order
Differential
Equations
I
In this part . . .
t’s time to up the ante with second order — and
higher — differential equations. A second order differential equation is one that involves a second derivative;
higher order equations involve three or more.
In this part, you discover that there are dazzling new techniques to bring to bear with second and higher order
equations, such as the popular method of undetermined
coefficients and variation of parameters. Get set to exercise your brain!
Chapter 5
Examining Second Order
Linear Homogeneous
Differential Equations
In This Chapter
Focusing on the fundamentals of second order linear differential equations
Getting a grip on constant coefficients
Exploring characteristic equations
Taking a crack at reduction of order
Understanding the Wronskian and other cool theorems
I
n Part I, I tell you what you need to know about a variety of first order differential equations. Now you’re striking out into uncharted territory —
second order differential equations. These kinds of equations are based heavily in physics — in wave motion, electromagnetic circuits, heat conduction,
and so on. They’re also fun and interesting. In this chapter, I walk you through
the fundamentals of second order linear homogeneous differential equations,
with a few useful tips and theorems thrown in along the way.
The Basics of Second Order
Differential Equations
To better understand second order differential equations, first take a look at
the following equation:
d2y
= f _ x, y i
dx 2
92
Part II: Surveying Second and Higher Order Differential Equations
As you can see, this equation has a second derivative, which makes it a
second order differential equation. But you know what? That’s not quite good
enough. The f(x, y) is okay for first order differential equations, but it isn’t
good enough for second order ones because the function f may also depend
on dy/dx. So the following is the general form of a second order differential
equation:
d2y
dy
= f d x, y,
n
dx
dx 2
In the following sections, I introduce several important types of second order
differential equations, including linear equations and homogeneous equations.
In this chapter, I focus on the second order linear homogeneous differential
equation (try saying that ten times fast!); if you can solve the homogeneous
form of a differential equation, you can always solve the same differential
equation as a nonhomogeneous differential equation (or at least give the solution in integral form). In other words, finding the solution to the homogeneous
differential equation is the fundamental part of solving second order differential equations. (I show you how to solve second order linear nonhomogeneous
differential equations in Chapter 6. As I note there, solving nonhomogeneous
equations usually involves solving the corresponding homogeneous equation
as well.)
Linear equations
I restrict this chapter to second order linear differential equations that have
the following form:
y" + p(x)y' + q(x)y = g(x)
where:
y'' =
d2y
dx 2
yl=
dy
dx
and:
However, a typical second order linear equation is sometimes written as the
following (who says that mathematicians don’t like to mix things up a bit?):
P(x)y" + Q(x)y' + R(x)y = G(x)
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
where P(x), Q(x), R(x), and G(x) are functions. The transition between the
original equation and the alternate is easy. Check it out:
p^ x h =
Q^ x h
P ^xh
q^ xh =
R^ x h
P ^xh
g^xh =
G^ xh
P ^xh
In this chapter, I show you only how to solve the typical second order linear
equation for regions where p(x), q(x), and g(x) are continuous functions, meaning that their values don’t make discontinuous jumps. I also usually provide
initial conditions as well, such as:
y(x0) = y0
But that condition isn’t enough to specify the solution of a second order
linear differential equation. You also need to specify the value of y'(x0). That
value is usually something like this:
y'(x0) = y'0
Second order differential equations that aren’t in the form of y" + p(x)y' +
q(x)y = g(x) are called nonlinear.
Homogeneous equations
The equation y" + p(x)y' + q(x)y = g(x) is referred to as homogeneous if g(x) =
0; that is, the equation is of the following form:
y" + p(x)y' + q(x)y = 0
Alternately, using the P(x), Q(x), R(x), G(x) terminology from the previous
section, a homogeneous equation can be written as:
P(x) y" + Q(x)y' + R(x)y = 0
If second order linear differential equations can’t be put into either of these
forms, the equation is said to be nonhomogeneous. To find out how to solve
linear nonhomogeneous differential equations, check out Chapter 6.
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Second Order Linear Homogeneous
Equations with Constant Coefficients
You may think that second order linear homogeneous differential equations
are intimidating. But they really aren’t, if you know some fundamentals.
The best place to start solving second order differential equations is with
equations where P(x), Q(x), and R(x) are constants, a, b, and c. So, for example, you get this equation when you include the constants:
ay" + by' + cy = 0
Okay, so you’ve narrowed things down. Now you’re talking about second
order linear homogeneous differential equations with constant coefficients.
Despite the hairy name, these equations are pretty easy to solve. In fact, you
can always solve an equation of this type using some elementary solutions
and initial conditions. Take a look at the examples in the following sections.
Elementary solutions
To get you up to speed, I begin with this typical equation:
y" – y = 0
Yes, this qualifies as a second order linear homogeneous differential equation.
Why? Just figure that a = 1, b = 0, and c = –1. There you have it.
To solve this differential equation, you need a solution y = f(x) whose second
derivative is the same as f(x) itself, because subtracting the f(x) from f"(x)
gives you 0.
You can probably think of one such solution: y = ex. Substituting y = ex into the
equation gives you this:
ex – ex = 0
As you can see, y = ex is indeed a solution.
In fact, y = c1ex (where c1 is a constant) is also a solution, because y" still
equals c1ex, which means that substituting y = c1ex into the original equation
gives you:
c1ex – c1ex = 0
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
So y = c1ex is also a solution. In fact, that solution is more general than just
y = ex, because y = c1ex represents an infinite number of solutions, depending
on the value of c1.
You can go further still by noting that y = e–x is also a solution, because:
y" – y = e–x – e–x = 0
Again, note that if y = e–x is a solution, then y = c2e–x (where c2 is a constant) is
also a solution because:
y" – y = c2e–x – c2e–x = 0
And here’s the fun part: If y = c1ex and y = c2e–x are both solutions, then the
sum of these two must also be a solution:
y = c1ex + c2e–x
In other words, if y = f1(x) and y = f2(x) are solutions to a second order linear
homogeneous differential equation, then:
y = f1(x) + f2(x)
is also a solution.
Initial conditions
In this section, I give you a look at some initial conditions to fit with the
solutions from the previous section. For instance, say that you have these
conditions:
y(0) = 9
and
y'(0) = –1
To meet these initial conditions, you can use the form of the solution y = c1ex +
c2e–x, which means that y' = c1ex – c2e–x. Using the initial conditions, you get:
y(0) = c1ex + c2e–x = c1 + c2 = 9
y'(0) = c1ex – c2e–x = c1 – c2 = –1
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So you get these two equations:
c1 + c2 = 9
c1 – c2 = –1
These are two equations in two unknowns. To solve them, write c1 + c2 = 9 in
this form:
c2 = 9 – c1
Now substitute this expression for c2 into c1 – c2 = –1, which gives you this
equation:
c1 – 9 + c1 = –1
or:
2c1 = 8
So
c1 = 4
Substituting the preceding value of c1 into c1 + c2 = 9 gives you:
c1 + c2 = 4 + c2 = 9
or:
c2 = 5
These values of c1 and c2 give you the following solution:
y = 4ex + 5e–x
That was simple enough, and you can generalize it as the solution to any linear
homogeneous second order differential equation with constant coefficients.
Checking Out Characteristic Equations
Here’s the general second order linear homogeneous equation, where a, b,
and c are constants:
ay" + by' + cy = 0
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
The solution of this equation is of the form y = erx, where r isn’t yet determined.
Plugging y = erx into the equation gives you:
ar2erx + brerx + cerx = 0
Dividing by erx (which is always nonzero) gives you:
ar2 + br + c = 0
This equation is called the characteristic equation for ay" + by' + cy = 0, and it’s
a quadratic equation. If the roots of the characteristic equation are r1 and r2,
the general solution of ay" + by' + cy = 0 is:
y = c1 er x + c 2 er
1
2
x
Earlier in this chapter, I treat this equation as one solution of ay" + by' + cy =
0. But the truth is that it’s the unique solution, as is indicated by a theorem
later in this chapter (see the later section “Putting Everything Together with
Some Handy Theorems”). Because ar 2 + br + c = 0 is a quadratic equation, the
following are three possibilities for r1 and r2:
r1 and r2 are real and distinct
r1 and r2 are complex numbers (complex conjugates of each other)
r1 = r2, where r1 and r2 are real
You can take a look at these cases in the following sections.
Real and distinct roots
When it comes to characteristic equations, one possibility is that r1 and r 2 are
real, and not equal to each other. I explain the basics of understanding this
concept and provide a clarifying example in the following sections.
The basics
When solving a general problem, where:
y(x0) = y0
and:
y'(x0) = y'0
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you get the following:
y 0 = c1 er
1
x0
+ c2 er
2
x0
and
yl 0 = c1 r 1 e r
1
x0
+ c2 r1 er
2
x0
Then you can solve for c1 and c2 in these equations, which gives you:
c1 =
yl 0 - y 0 r 2 - r
r1 - r 2 e
1
x0
c2 =
y 0 r 1 - yl 0 - r
r1 - r 2 e
2
x0
and:
An example
How about an example to bring the concept of real and distinct roots into
focus? Try this second order linear homogeneous differential equation:
y" + 5y' + 6y = 0
with the initial condition that:
y(0) = 16
and:
y'(0) = –38
To solve, make the assumption that the solution is of the form y = cerx.
Substituting that equation into y" + 5y' + 6y = 0 gives you:
cr2erx + 5crerx + 6cerx = 0
Next you divide by cerx to get:
r 2 + 5r + 6 = 0
which is the characteristic equation for y" + 5y' + 6y = 0. You can also write
this equation as:
(r + 2) (r + 3) = 0
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
So the roots of the characteristic equation are:
r1 = –2
and
r2 = –3
which means that the solution to y" + 5y' + 6y = 0 is:
y = c1e–2x + c2e–3x
where c1 and c2 are determined by the initial conditions:
y(0) = 16
and:
y'(0) = –38
Substituting the initial conditions into your solution gives you these equations:
y(0) = c1 + c2 = 16
and:
y'(0) = –2c1 – 3c2 = –38
From the first equation, you can see that c2 = 16 – c1, and substituting that
into the second equation gives you:
–2c1 – 3c2 = –2c1 – 48 + 3c1 = c1 – 48 = –38
or:
c1 = 10
Substituting this value of c1 into y(0) = c1 + c2 = 16 gives you:
c1 + c2 = 10 + c2 = 16
So that means:
c2 = 6
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Now you’ve found c1 and c2, which means that the general solution to y" + 5y' +
6y = 0 is:
y = 10e–2x + 6e–3x
You can see this solution graphed in Figure 5-1.
9
8
7
6
y
5
4
3
Figure 5-1:
2
The solution
to an 1
equation
with real 0
and distinct
roots. –1
0
1
2
x
3
Complex roots
Besides real and distinct roots (see the previous section), another possibility,
when it comes to characteristic equations, is having complex roots. In this
case, the quadratic formula yields two complex numbers. I explain the basics
of this concept and walk you through an example in the following sections.
The basics
Check out the following quadratic equation:
r=
- b ! b 2 - 4ac
2a
The discriminant, b2 – 4ac, is negative, which means that you’re taking the
square root of a negative number, so you get complex numbers. In particular,
the imaginary part of the two roots varies by their sign (+ or –), as shown
by the example equation.
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
In other words, the roots are of the following forms:
r1 = λ + iµ
and
r2 = λ – iµ
where λ and µ are both real numbers, and i is the square root of –1.
As you can see, the characteristic equation has complex roots. What does
that mean for the solution to the differential equation? As you find out earlier
in this chapter, the solutions to the differential equation are:
y1 = er
1
x
and
y 2 = er
2
x
So:
y1 = e(λ + iµ)x
and:
y2 = e(λ – iµ)x
Now you have to deal with complex numbers as exponents of e. You may be
familiar with the following handy equations:
eiax = cos ax + i sin ax
and:
e–iax = cos ax – i sin ax
You’re making progress! You have now removed i from the exponent. Putting
these two equations to work gives you these forms for the solutions, y1 and y2:
y1 = e(λ + iµ)x = eλx(cos µx + i sin µx)
and:
y2 = e(λ – iµ)x = eλx(cos µx – i sin µx)
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However, you still have that pesky factor of i; and you want to get rid of it.
Why? The general differential equation that y1 and y2 are solutions of only has
real coefficients:
ay" + by' + cy = 0
Here’s the key: You could get rid of those factors of i if you could just divide
them out. After all, i is just a constant. For instance, if you had the solution in
a form like this:
yn = i eλxcos µx
you could replace i with an arbitrary constant of integration, c, which would
look like this:
yn = c eλxcos µx
Now try adding and subtracting y1 and y2. As you recall from the earlier section “Elementary solutions,” if two functions are a solution to a linear differential equation, the sum and difference of those functions are also solutions. So
you can introduce the new solutions m(x) and n(x), the sum and difference of
y1 and y2:
m(x) = y1(x) + y2(x)
and:
n(x) = y1(x) – y2(x)
First, calculating m(x) gives you the following equation:
m(x) = y1(x) + y2(x) = eλx(cos µx + i sin µx) + eλx(cos µx – i sin µx)
which you can convert to:
m(x) = y1(x) + y2(x) = 2 eλxcos µx
Your solution now looks fine — there’s no pesky i lurking around. Now how
about calculating n(x)? Here’s what that looks like:
n(x) = y1(x) – y2(x) = eλx(cos µx + i sin µx) – eλx(cos µx – i sin µx)
which works out to:
n(x) = y1(x) – y2(x) = 2i eλx sin µx
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
Your solution is looking even better, because now m(x) and n(x) have the following forms:
m(x) = c1 eλxcos µx
and:
n(x) = c2 eλx sin µx
Yes, it’s true that n(x) as written is multiplied by i: 2i eλx sin µx. But that’s the
beauty of the whole thing: 2i is just a constant, so it can be replaced by c2.
And that substitution gets rid of the pesky i.
So, finally, you can write the solution as:
y(x) = m(x) + n(x)
or:
y(x) = c1 eλx cos µx + c2 eλx sin µx
where you get λ and µ from the roots of the differential equation’s characteristic equation:
ar 2 + br + c = 0
where the roots are:
r1 = λ + iµ
and
r2 = λ – iµ
An example
If you feel like you’ve grasped the basics of complex roots, take a look at an
example. Try solving this differential equation:
2y" + 2y' + y = 0
where:
y(0) = 1
and
y'(0) = 1
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You probably already have a good idea what to do here. You simply need to
find the characteristic equation and then the roots of that equation. Here’s
the characteristic equation:
2r 2 + 2r + 1 = 0
You can find the roots of this characteristic equation by using the quadratic
equation from the previous section. Your work should give you these roots:
-2 ! 4 - 8
4
This equation works out to be:
r = –1⁄2 ± (1⁄2)i
So the two roots are:
r1 = –1⁄2 + (1⁄2)i
and:
r2 = –1⁄2 – (1⁄2)i
Because the two roots are of the form:
r1 = λ + iµ
and
r2 = λ – iµ
then:
λ = –1⁄2
and:
µ = 1⁄2
So the solution is:
y(x) = c1 e–x/2 cos x⁄2 + c2 e–x/2 sin x⁄2
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
To find c1 and c2, you simply have to apply the initial conditions (substituting
in for y(0) and y'(0)), which means that:
y(0) = c1 = 1
and:
y'(0) = 1⁄2c + 1⁄2c = 1
1
2
Substituting c1 = 1 gives you:
y'(0) = 1⁄2 + 1⁄2c = 1
2
So c2 = 1. That makes the general solution to 2y" + 2y' + y = 0:
y(x) = e–x/2 cos x⁄2 + e–x/2 sin x⁄2
And that’s it — no imaginary terms involved. Cool. You can see a graph of
this solution in Figure 5-2.
1
y
0
Figure 5-2:
The solution
to an
equation
with
complex
roots. –1
–1
0
1
2
3
4
5
6
x
7
8
9
10
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Part II: Surveying Second and Higher Order Differential Equations
Identical real roots
In the previous sections, I cover the cases where the characteristic equation
has real and distinct roots as well as complex roots. All that’s left is the case
where the characteristic equation has two real roots that are identical to each
other. You get the fundamentals and an example in the following sections.
The basics
When the roots of a characteristic equation are identical, the discriminant of
the quadratic equation, b2 – 4ac, equals zero, which means that:
r1 = –b/2a
and
r2 = –b/2a
However, now you have a problem. Why? Because second order differential
equations are supposed to have two expressions in their solution. Because
r1 = r2, you get this:
y1 = c1e–bx/2a
y2 = c2e–bx/2a
These solutions differ only by a constant, so they aren’t really different at all;
you could write either y1 + y2 or y1 – y2 as:
y = ce–bx/2a
So, as you can see, you really only have one solution in this case. How can you
find another? Here’s the traditional way of solving this issue: So far, you’ve
been finding solutions to second order linear differential equations by assuming the solution is of the following form:
y(x) = cerx
However, consider this: What if you replaced the constant with a function of x,
m(x), instead? Doing so would let you handle more general second order linear
differential equations. By doing this, you get:
y(x) = m(x)erx
In this case, r = –b/2a, so you get this form of the solution:
y(x) = m(x)e–bx/2a
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
Differentiating the equation gives you:
y l^ x h = ml^ x h e - bx/2a - b m^ x h e - bx/2a
2a
You also need y"(x) to substitute into the differential equation. Doing so
gives you:
2
y'' ^ x h = m'' ^ x h e - bx/2a - b ml^ x h e - bx/2a + b 2 m^ x h e - bx/2a
2a
4a
Substituting y(x), y'(x), and y"(x) into the following differential equation:
ay" + by' + cy = 0
gives you (after rearranging the terms):
2
2
a m'' ^ x h + ^ b - b h ml^ x h + c b - b + c m m^ x h = 0
4a 2a
So b – b = 0 in the second term, along with combining the fractions in the
third term, gives you:
2
a m'' ^ x h + c c - b m m^ x h = 0
4a
Here’s the trick: Note that c – b2/4a is the discriminant of the characteristic
equation, and in this case (the case of real identical roots), the discriminant
equals zero. So you have:
a m"(x) = 0
or simply, dividing by a:
m"(x) = 0
Wow, that looks pretty easy after all. After integrating, you get this final form
for m(x):
m(x) = d1x + d2
where d1 and d2 are constants.
Because the second solution you’ve been looking for is y2(x) = m(x)erx =
d1xerx + d2erx, here’s the general solution of the differential equation in the
case where the characteristic equation’s roots are real and identical:
y(x) = c1xe–bx/2a + c2e–bx/2a
where c1 and c2 are constants.
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This trick, it turns out, can often be used to find a second solution to a
second order linear homogeneous differential equation if you already know
one solution. In fact, trying a solution of the form y(x) = m(x)f(x) is so useful
that it has been given a name: the method of reduction of order. I discuss this
method in more detail later in this chapter.
An example
Want to see an example so that the identical real root concept can really take
hold? Take a look at this differential equation:
y" + 2y' + y = 0
where:
y(0) = 1
and:
y'(0) = 1
The characteristic equation is:
r 2 + 2r + 1 = 0
You can factor this equation into:
(r + 1) (r + 1) = 0
So the roots of the characteristic equation are identical, –1 and –1.
Now the solution is of the following form:
y(x) = c1xe–bx/2a + c2e–bx/2a
To figure out c1 and c2, use the initial conditions. Substituting into the equation gives you:
y(0) = c2 = 1
Differentiating the y(x) equation gives you y'(x); you also can substitute c2 = 1
to get:
y l^ x h = c 1 e - bx/2a -
bc 1 - bx/2a b - bx/2a
xe
e
2a
2a
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
From the initial conditions, y'(0) equals:
y'(0) = c1 – 1 = 1
So c1 = 2, giving you the following general solution:
y(x) = 2xe–x + e–x
And that’s that. You can find a graph of this solution in Figure 5-3.
1
y
Figure 5-3:
The solution
to an
equation
with
identical
real roots. 0
0
1
2
3
4
x
5
6
Getting a Second Solution
by Reduction of Order
If you know one solution to a second order differential equation, you can get
a first order differential equation for the second solution — or rather, for the
derivative of the second solution, which you can then integrate to get the
actual second solution. That’s where the term reduction of order comes from.
The method is very cool, as you find out in the following sections.
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Seeing how reduction of order works
What is the reduction of order method? Basically it says that if you already
know a solution, y1(x), to a differential equation of this form:
y" + p(x)y' + q(x) = 0
you can substitute the following expression into the equation to search for a
second solution:
y(x) = m(x)y1(x)
Note that the original equation isn’t restricted to differential equations with
constant coefficients. However, it is homogeneous (in other words, it equals
zero).
Differentiating y(x) = m(x)y1(x) with respect to x gives you:
y'(x) = m'(x)y1(x) + m(x)y1'(x)
You also need the second derivative of y(x) = m(x)y1(x), which looks like this:
y"(x) = m"(x)y1(x) + m'(x)y1'(x) + m'(x)y1'(x) + m(x)y1"(x)
The middle two terms are the same, so the equation becomes:
y"(x) = m"(x)y1(x) + 2m'(x)y1'(x) + m(x)y1"(x)
Putting y'(x) and y"(x) into y" + p(x)y' + q(x) = 0 gives you the following
equation:
y1m" + (2y1' + py1)m' + (y1" + py1' + qy1) = 0
I bet you recognize the expression in the second set of parentheses here. And
guess what? That’s the point! That expression is y1" + py1' + qy1, which is just
y1 substituted into the original differential equation. But y1 is a solution of
that homogeneous differential equation, so the expression in parentheses
equals 0!
So you end up with this equation:
y1m" + (2y1' + py1)m' = 0
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
It doesn’t look pretty, but examine what you have here: This is actually a first
order differential equation in m'(x). If you substitute n(x) = m'(x) into the
equation, you get this:
y1n'(x) + (2y1' + py1)n(x) = 0
This final equation is a first order differential equation in n(x), and it can
often be solved as a first order linear differential equation or as a separable
differential equation (see Chapters 2 and 3 for more about these types of
equations).
Trying out an example
In this section, you can see reduction of order at work in an example. Take a
look at this second order differential equation:
2x 2y" + xy' – y = 0
How do you tackle this one? I walk you through the process in the following
sections.
The first solution
In the equation 2x 2y" + xy' – y = 0, note how each successive derivative y' and
y" is multiplied by a new power of x. This suggests that differentiating a solution of this differential equation generates a new power of x in the denominator (and that the x and x 2 cancel out). So try a first solution of the form:
y1 = 1/x n
Substituting this solution into the equation gives you:
2x 2n(n+1)/x n+2 – xn/x n+1 – 1/x n = 0
By cancelling 1/x n, you get:
2n(n+1) – n – 1 = 0
or:
2n2 – n – 1 = 0
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Part II: Surveying Second and Higher Order Differential Equations
The quadratic equation gives you these roots:
1!3
4
Taking the root r = 1, you see that:
y1 = 1/x
is a solution of 2x 2y" + xy' – y = 0. (You can take a look at the other root, r = –1⁄2,
in the later section “A little shortcut.”)
The second solution
In the previous section, you find the first solution, y1 = 1/x. Now you use the
reduction of order method to find a second solution. This means that you’re
now looking for a solution of the following form:
y2 = m(x)/x
The first derivative of this solution is:
y2' = m'(x)/x – m(x)/x 2
The second derivative is:
y2" = m"(x)/x – 2m'(x)/x2 + 2m(x)/x 3
Substituting y' and y" into 2x 2y" + xy' – y = 0 gives you (after the algebra settles):
2xm" – m' = 0
To make this equation a little easier to work with, set m'(x) = n(x), which
gives you this:
2x dn = n
dx
Separating the variables results in the following equation:
dn = dx
n
2x
or:
ln(n) = ln(x)/2
Exponentiating both sides of the equation gives you:
n(x) = kx1/2
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
where k is a constant. Because n(x) = m'(x), you know that:
m(x) = cx3/2 + d
where c and d are constants.
This solves for m(x). Now because, by y2 = m(x)/x:
y2 = m(x) y1 = m(x)/x
you know that:
y2 = c1x1/2 + c2/x
Note that the second term here can be absorbed together with y1 (which is
1/x). So the complete solution is (I’ve absorbed the constants together as
needed):
y = y1 + y2 = c1x1/2 + c2/x
A little shortcut
You may have noticed that in the previous example, you didn’t actually need
to use reduction of order. I’ll show you what I mean. The differential equation
you were solving looked like this:
2x 2y" + xy' – y = 0
You guessed that the solution was of the following form:
y = 1/x n
Substituting this into the differential equation gave you:
2n2 – n – 1 = 0
which has these roots:
1!3
4
So the roots are 1 and –1⁄2, which means that you did in fact have two solutions already:
y1 = c1/x
and:
y2 = c2x1/2
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Part II: Surveying Second and Higher Order Differential Equations
Putting Everything Together with
Some Handy Theorems
A couple of formal theorems, which I explain in the following sections, lock
down exactly how to find the general solutions to second order linear homogeneous differential equations and formalize the information from earlier
sections.
Superposition
I’ll start off with the theorem on superposition of solutions, which formally
says this:
If you have the second order linear homogeneous differential equation:
y " + p(x)y' + q(x) = 0
and have two solutions, y1(x) and y2(x), then any linear combinations of
these solutions:
y = c1 y1 + c2 y2
where c1 and c2 are constants, is also a solution of the differential equation.
In plain English, this theorem simply says that if you have two solutions to a
second order linear differential equation, a linear combination of those two
solutions is also a solution. You actually use this theorem throughout the first
part of this chapter as you solve second order linear homogeneous differential equations for two solutions, y1 and y2, and give the general solution as a
linear combination of the two.
Here’s another example; say that you have the following second order linear
differential equation:
y" – y = 0
You can guess that the solution is of the form y = enx and that substituting it in
gives you:
n2enx – enx = 0
or:
n2 – 1 = 0
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
Therefore:
n = ±1
So the two solutions you’ve worked out are:
y1 = c1ex
and
y2 = c2e–x
According to this theorem, the following linear superposition of these solutions is also a solution:
y = c1ex + c2e–x
You can verify your solution by substituting it into the original equation:
y" – y = c1ex + c2e–x – (c1ex + c2e–x ) = 0
Linear independence
Earlier in this chapter, you work problems to find the solutions y1 and y2 to
second order linear homogeneous differential equations, but it turns out that
there’s a restriction on them — they have to be linearly independent. What,
exactly, does that mean?
If you have two functions, f(x) and g(x), they’re linearly dependent in an interval I if for all x in I you can find two constants, c1 and c2, such that this equation is true:
c1f(x) + c2 g(x) = 0
In other words, f(x) just differs from g(x) by a constant.
If the functions f and g aren’t linearly dependent in the interval I, they’re linearly independent in the interval I. In other words, if it’s impossible to find
two constants such that the previous equation is true for f(x) and g(x) in the
interval I, then f(x) and g(x) are linearly independent in the interval I.
Practically speaking, if one function isn’t just a multiple of another function,
the two are linearly independent.
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Here’s the formal theorem:
If you have the second order linear homogeneous differential equation:
y" + p(x)y' + q(x) = 0
and have two linearly independent solutions, y1(x) and y2(x), then any
linear combinations of these solutions:
y = c1 y1 + c2 y2
where c1 and c2 are constants, is the general solution of the differential
equation. Every solution of the differential equation can be expressed in
the form of y = c1 y1 + c2 y2.
In other words, you can’t just use any two solutions, y1 and y2, to a second
order linear homogeneous differential equation. Instead, those two solutions
have to be linearly independent so that a linear combination of them is a general solution of the differential equation.
Clear as mud, right? Well, here’s an example to wrap your brain around. Say
that you have the following differential equation:
y" + 6y' + 8y = 0
What’s the general solution of this equation? Well, guessing that the solutions
are of the form y = erx, you get the following equation by substituting the solution into the differential equation:
r 2erx + 6rerx + 8erx = 0
The characteristic equation is:
r 2 + 6r + 8 = 0
which can be factored into:
(r + 4) (r + 2)
So the roots of the characteristic equation are –4 and –2, and your solutions
are of the form:
y1 = c1e–4x
and
y2 = c2e–2x
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
These solutions are linearly independent — they don’t just differ by a multiplicative constant — so by the theorem of linear independence, the general
solution to the differential equation has the following form:
y = c1e–4x + c2e–2x
The Wronskian
You may have heard of the Wronskian, especially if you aspire to be a
Differential Equations Wizard (and doesn’t everyone?). But if you haven’t
heard of it, you’re probably wondering what it is. Briefly put, the Wronskian
is the determinant of an array of coefficients that lets you determine the
linear independence of solutions to a differential equation.
Okay, that was a mouthful. What does it mean in layman’s terms? I explain
what you need to know and provide some examples in the following sections.
Arrays and determinants
Say that you have a second order linear homogeneous differential equation
with the following general solution:
y = c1y1 + c2 y2
And assume that you have these initial conditions
y(x0) = y0
and
y'(xo) = y'o
The initial conditions mean that the constants c1 and c2 have to satisfy these
equations:
yo = c1y1(x0) + c2 y2(x0)
and:
y'0 = c1y'1(x0) + c2 y'2(x0)
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Solving this system of two equations for c1 and c2 gives you:
c1 =
y 0 y l 2 _ x 0 i - y l 0 y 2_ x 0 i
y 1_ x 0 i y l 2 _ x 0 i - y l 1_ x 0 i y 2 _ x 0 i
c2 =
- y 0 y l 1 _ x 0 i + y l 0 y 1_ x 0 i
y 1_ x 0 i y l 2 _ x 0 i - y l 1_ x 0 i y 2 _ x 0 i
and:
At some point in time, someone got the bright idea that this system looks like
the division of two array determinants. As a refresher on determinants, say
you have this array:
a b
c d
In the example, the determinant is:
a b
= ad - cb
c d
Using this notation, you can write the expressions for c1 and c2 in determinant
form:
c1 =
y0
yl 0
y 2_ x 0 i
yl 2_ x 0 i
y 1_ x 0 i y 2 _ x 0 i
y l 1_ x 0 i y l 2 _ x 0 i
and:
c2 =
y 1_ x 0 i y 0
y l 1_ x 0 i y l 0
y 1_ x 0 i y 2 _ x 0 i
y l 1_ x 0 i y l 2 _ x 0 i
In order for c1 and c2 to make sense in these expressions, the denominator
must be nonzero, so:
W=
y 1_ x 0 i y 2 _ x 0 i
= y 1_ x 0 i y l 2 _ x 0 i - y l 1_ x 0 i y 2 _ x 0 i ! 0
y l 1_ x 0 i y l 2 _ x 0 i
Here’s the Wronskian (W) of the solutions y1 and y2:
W = y1(x0)y'2(x0) – y1'(x0)y2(x0)
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
Who was Wronski?
Jósef Maria Hoëné-Wronski (1778–1853) was a
Polish philosopher. He studied and published in
many fields, such as philosophy, math, and
physics, and he also was an inventor and a
lawyer.
Despite deep flashes of insight into math and
some other fields, Wronski spent much of his life
pursuing ultimately unsuccessful topics, such as
perpetual motion machines. Unfortunately, during
his lifetime, his work was largely ridiculed and
ignored. However, since his death, people have
come to unearth significant nuggets of thought
from his writings.
You may also see the Wronskian referred to as the Wronskian determinant. It’s
named for a Polish mathematician, Jósef Maria Hoëné-Wronski (and you
thought mathematicians had bland names). Check out the nearby sidebar
“Who was Wronski?” for more info.
The formal theorem
The Wronskian is a measure of linear independence. Consider this, for example: If you have two solutions, y1 and y2, and their Wronskian is nonzero, then
those solutions are linearly independent. And that, my friend, means that
every solution to the differential equation is a linear combination of those
two solutions.
And that brings me to a formal theorem:
If you have the second order linear homogeneous differential equation:
y" + p(x)y' + q(x) = 0
and have two solutions, y1(x) and y2(x), where their Wronskian is nonzero
at a point x0, then any linear combinations of these solutions:
y = c1 y1 + c2 y2
where c1 and c2 are constants, is the general solution of the differential
equation. Every solution of the differential equation can be expressed in
the form of y = c1 y1 + c2 y2.
Here’s the gist of this formal theorem: If the Wronskian of your two solutions
is nonzero at a point x0, then a linear combination of those solutions contains
every solution of the differential equation. In other words, if you have two
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solutions, y1 and y2, and their Wronskian is nonzero, then this is the general
solution of the corresponding differential equation:
y = c1y1 + c2 y2
Note: The y1 and y2 form is what’s called a fundamental solution to the differential equation.
Example 1
How about some examples to get a hang of the Wronskian? Earlier in this
chapter, you tackled this differential equation:
y" + 2y' + y = 0
The characteristic equation is:
r 2 + 2r + 1 = 0
and you can factor that equation this way:
(r + 1) (r + 1) = 0
which means that the roots of the differential equation are identical, –1 and
–1. So the solution is of the following form:
y(x) = c1xe–x + c2e–x
Now take a look at the Wronskian, which looks like this:
y 1_ x 0 i y 2 _ x 0 i
W=
= y 1_ x 0 i y l 2 _ x 0 i - y l 1_ x 0 i y 2 _ x 0 i
y l 1_ x 0 i y l 2 _ x 0 i
Because:
y1 = c1xe–x
and
y2 = c2e–x
the Wronskian equals:
W = y1(x0)y'2(x0) – y1'(x0)y2(x0) = –c1xe–x c2e–x – (c1e–x – c1xe–x)c2e–x
or:
W = –c1c2e–2x
Chapter 5: Examining Second Order Linear Homogeneous Differential Equations
And because the Wronskian is always nonzero, you don’t even have to substitute x0 in; y1 and y2 form a fundamental solution of the differential equation, as
long as c1 and c2 are both nonzero.
Example 2
Here’s another example. Say that you have two solutions, y1 and y2, to a
second order linear homogeneous differential equation such that:
y1 = er
1
x
and
y 2 = er
2
x
Your task now is to show that y1 and y2 form a fundamental solution to the differential equation if r1 ≠ r2.
Time to check the Wronskian:
W=
y 1_ x 0 i y 2 _ x 0 i
= y 1_ x 0 i y l 2 _ x 0 i - y l 1_ x 0 i y 2 _ x 0 i
y l 1_ x 0 i y l 2 _ x 0 i
Because y1 = er1x and y2 = er2x, the Wronskian equals:
W = y 1_ x 0 i y l 2 _ x 0 i - y l 1 _ x 0 i y 2 _ x 0 i = e r x r 2 e r x - r 1 e r x e r
1
2
1
2
x
or:
W = _ r 2 - r 1i e (r
1
+ r2 ) x
Don’t forget: You don’t have to substitute x0 into the Wronksian, because it’s
clearly nonzero everywhere as long as r1 ≠ r2. So y1 and y2 form a fundamental
solution set of the differential equation.
Because:
y1 = er
1
x
and
y 2 = er
2
x
you can see that what the Wronskian is trying to say is that y1 and y2 are linearly independent. In other words, you can’t multiply y1 by a constant so that
it equals y2 for all x in an interval, such as –2 < x < 2, as long as r1 ≠ r2.
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Example 3
Here’s one final Wronskian example. Earlier in this chapter, you solve this differential equation:
2x2y" + xy' – y = 0
By guessing that the solution was of the following form:
y1 = 1/x n
you came up with this characteristic equation:
2n2 – n – 1 = 0
and found this solution:
y = c1x 1/2 + c2/x
Is this the general solution? You can find out by bringing the Wronskian to the
rescue:
W = y 1_ x 0 i y l 2 _ x 0 i - y l 1_ x 0 i y 2 _ x 0 i = - c 1 x 1/2 c 2 /x 2 -
c 1 - 1/2
x c 2 /x
2
or:
W = - c 1 c 2 x - 3/2 -
c 1 c 2 - 3/2
x
2
So:
W = - 3 c 1 c 2 x - 3/2
2
If c1 and c2 ≠ 0, then because W ≠ 0 for x > 0, you can conclude that:
y = c1x 1/2 + c2/x
is indeed the general solution.
Chapter 6
Studying Second Order
Linear Nonhomogeneous
Differential Equations
In This Chapter
Focusing on the facts of linear nonhomogeneous second order equations
Mapping out the method of undetermined coefficients
Perusing variation of parameters
Applying nonhomogeneous equations to physics
C
hapter 5 is about second order linear homogeneous differential equations of this form:
y" + p(x)y' + q(x)y = 0
This equation is homogeneous because it equals zero (and there’s no term
that only relies on x). In this chapter, however, you tackle differential equations of the nonhomogeneous form:
y" + p(x)y' + q(x)y = g(x)
where g(x) ≠ 0.
Exciting, isn’t it? To start, I provide you with a theorem that gives you the
power you need to solve these kinds of equations. After that, I describe some
great techniques for working with these equations. I even throw in a few
physics examples to show you how these equations apply to real life.
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The General Solution of Second Order
Linear Nonhomogeneous Equations
Luckily for mathematicians everywhere, a quick and easy theorem giving the
general solution of a second order linear nonhomogeneous differential equation exists. And this theorem is the one that you’ll refer to over and over
again in this chapter. In the following sections, I explain the theorem and how
you can put it to work.
Understanding an important theorem
Without further ado, here’s the theorem of the general solution of a second
order linear nonhomogeneous differential equation:
The general solution of the nonhomogeneous differential equation:
y" + p(x)y' + q(x)y = g(x)
is:
y = c1 y1(x) + c2 y2(x) + yp(x)
where c1 y1(x) + c2 y2(x) is the general solution of the corresponding homogeneous differential equation:
y" + p(x)y' + q(x)y = 0
(for example, y1 and y2 are a fundamental set of solutions to the homogeneous equation) and where yp(x) is a specific solution to the nonhomogeneous equation.
This very important theorem basically says that to find the general solution
to a nonhomogeneous differential equation, you need the sum of the general
solution of the corresponding homogeneous differential equation added to a
particular solution of the nonhomogeneous differential equation. A particular
solution is any solution of the nonhomogeneous differential equation. Quite a
mouthful isn’t it? Not to worry; I explain how to use this theorem in the next
section.
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
Putting the theorem to work
You need to follow these steps to find the general solution of a second order
linear nonhomogeneous equation:
1. Find the corresponding homogeneous differential equation by setting
g(x) to 0.
2. Find the general solution, y = c1 y1(x) + c2 y2(x), of the corresponding
homogeneous differential equation.
This solution is referred to as yh.
3. Find a single solution to the nonhomogeneous equation.
This solution is sometimes referred to as the particular (or specific)
solution, yp.
4. The general solution of the nonhomogeneous differential equation is
the sum of yh + yp.
To see these steps in action, take a look at this nonhomogeneous differential
equation:
y" – y' – 2y = e3x
To solve this equation, start by getting the homogenous version of it, like this:
y" – y' – 2y = 0
You can assume that the solution is of the form y = ert. So, when you substitute that into the differential equation, you get the following characteristic
equation (see Chapter 5 for more details):
r2 – r – 2 = 0
You can then factor the characteristic equation this way:
(r + 1)(r – 2) = 0
Now you can see that the roots, r1 and r2, of the characteristic equation are –1
and 2, giving you:
y1 = e–x
and:
y2 = e2x
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So, the general solution to the homogeneous differential equation is given by:
y = c1e–x + c2e2x
You’re halfway there! Now you need a particular solution to the original nonhomogeneous differential equation:
y" – y' – 2y = e3x
Note that g(x) has the form e3x here, so you can assume that the particular
solution has this form:
yp(x) = Ae3x
In this case, A is an arbitrary coefficient.
Substituting this form into the nonhomogeneous equation gives you:
9Ae3x – 3Ae3x – 2Ae3x = e3x
or:
9A – 3A – 2A = 1
So
4A = 1
and
A = 1⁄4
The particular solution is:
3x
y p^ x h = e
4
Because the general solution to the nonhomogeneous equation is equal to
the sum of the general solution of the corresponding homogeneous differential equation and a particular solution of the nonhomogeneous differential
equation, you get this equation as the general solution:
y = yh + yp
or:
3x
y = c 1 e - x + c 2 e 2x + e
4
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
Finding Particular Solutions with the
Method of Undetermined Coefficients
Are there any techniques that allow you to find particular solutions to nonhomogeneous differential equations? Yes, there are! I’ll start with the method
of undetermined coefficients, which is actually a method you use earlier in
this chapter (you just didn’t know the fancy name there).
You start by finding a particular solution to the nonhomogeneous differential
equation:
y" + p(x)y' + q(x)y = g(x)
The method of undetermined coefficients notes that when you find a candidate
solution, y, and plug it into the left-hand side of the equation, you end up with
g(x). Because g(x) is only a function of x, you can often guess the form of
yp(x), up to arbitrary coefficients, and then solve for those coefficients by
plugging yp(x) into the differential equation.
This method works because you’re grappling only with g(x) here, and the
form of g(x) can often tell you what a particular solution looks like. For
instance, if g(x) is in the form of
erx, try a particular solution of the form Aerx, where A is a constant.
Because derivatives of erx reproduce erx, you have a good chance of finding a particular solution this way.
a polynomial of order n, try a polynomial of order n. For example, if
g(x) = x2 + 1, try a polynomial of the form Ax2 + B.
a combination of sines and cosines, sin αx + cos βx, try a combination
of sines and cosines with undetermined coefficients, A sin αx + B cos βx.
Then plug into the differential equation and solve for A and B.
In the following sections, you take a look at some examples that put to work
the method of undetermined coefficients, and then you solve actual differential
equations using the theorem I explain earlier in this chapter.
When g(x) is in the form of erx
You see the following equation in its entirety in the earlier section “Putting
the theorem to work,” but here’s how to figure out the particular solution:
y" – y' – 2y = e3x
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From the form of the term on the right, g(x), you can guess that a particular
solution, yp(x), has the form:
yp(x) = Ae3x
Substituting this into the differential equation gives you:
9Ae3x – 3Ae3x – 2Ae3x = e3x
Dropping e3x out of each term gives you:
9A – 3A – 2A = 1
So:
4A = 1
and
A = 1⁄4
The particular solution for the differential equation is:
3x
y p^ x h = e
4
When g(x) is a polynomial of order n
The example in this section points out how to deal with a particular solution
in the form of a polynomial. Your mission, should you choose to accept it, is
to find the general solution of the following nonhomogeneous equation:
y" = 9x2 + 2x – 1
where:
y(0) = 1
and
y'(0) = 3
The general solution to the homogeneous equation
The homogeneous equation is simply:
y" = 0
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
And you can integrate to get this equation:
y' = c1
Integrating again gives you the general solution to the homogeneous differential equation, yh:
yh = c1x + c2
The particular and general solutions to the nonhomogeneous equation
To find the general solution to the nonhomogeneous equation, you need a
particular solution, yp. The g(x) term in the original equation is 9x2 + 2x – 1,
so you can assume that the particular solution has a similar form:
yp = Ax 2 + Bx + C
where A, B, and C are constant coefficients that you have to determine. But
here’s the issue: The supposed form of yp has terms in common with yh, the
general solution of the homogenous equation:
yh = c1x + c2
yp = Ax 2 + Bx + C
Both of these have an x term and a constant term. When yh and yp have terms
in common — differing only by a multiplicative constant — that isn’t good,
because those are really part of the same solution. When you add yh and yp
together, you get this:
y = yh + yp = Ax 2 + (c1 + B)x + (c2 + C)
which can be rewritten:
y = yh + yp = Ax 2 + cx + d
where c and d are constants. As you can see, the terms Bx + C in yp really
don’t add anything.
The way to handle this issue is to multiply yp by successive powers of x until
you don’t have any terms of the same power as in yh. For example, multiply yp
by x to get:
yp = Ax 3 + Bx 2 + Cx
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This equation still isn’t good enough, however. After all, the Cx term overlaps
with the c1x term in yh. So, you have to multiply by x again to get:
yp = Ax4 + Bx 3 + Cx 2
This solution has no terms in common with the homogeneous general solution, yh, so you’re in business.
Substituting yp = Ax4 + Bx 3 + Cx 2 into y" = 9x2 + 2x – 1 gives you:
12Ax 2 + 6Bx + 2C = 9x 2 + 2x – 1
Comparing coefficients of like terms gives you:
12A = 9
6B = 2
2C = –1
which means that:
A = 3⁄4
B = 1⁄3
C = –1⁄2
So the particular solution is:
y p= 3 x4+ 1 x3- 1 x2
4
3
2
Therefore, the general solution is:
y = y h + y p = c1 x + c 2 + 3 x 4 + 1 x 3 - 1 x 2
4
3
2
Or, after rearranging to make things look pretty, you get:
y = y h + y p = 3 x 4 + 1 x 3 - 1 x 2 + c1 x + c 2
4
3
2
where you can get c1 and c2 by using the initial conditions. Substituting y(0) =
1 gives:
y(0) = 1 = c2
So c2 = 1. Taking the derivative of your general solution gives you:
y' = 3x3 + x2 – x + c1
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
And substituting the initial condition y'(0) = 3 gives you:
y'(0) = 3 = c1
Here’s the general solution with all the numbers filled in:
y = 3 x 4 + 1 x 3 - 1 x 2 + 3x + 1
4
3
2
When g(x) is a combination
of sines and cosines
Combinations of sines and cosines can be a bit tricky, but with the help of
the following sections, you can solve equations featuring these combinations
in a snap.
Example 1
Find a particular solution to the following non-homogeneous equation:
y" – y' – 2y = sin 2x
This example looks as though a particular solution may be of the form yp =
Asin2x + Bcos2x. There’s one way to find out if you’re correct; you can substitute this solution into your equation to get:
(–6A + 2B) sin 2x + (–6B – 2A) cos 2x = sin 2x
You can now equate the coefficients of sin 2x and cos 2x to get these
equations:
–6A + 2B = 1
and
–6B – 2A = 0
Multiplying –6B – 2A = 0 by –3 and adding the result to 6A + 2B = 1 gives you
20B = 1. So B = 1⁄20. Substituting that number into –6B – 2A = 0 gives you the following equation:
- 6 - 2A = 0
20
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So A = –3⁄20. Therefore, a particular solution of the original nonhomogeneous
equation is:
y p = - 3 sin 2x + cos 2x
20
20
Example 2
Here’s another example using sines and cosines. Try finding a particular solution of the following equation:
y" + 16y = 4 cos 4x
Assume that the solution looks like this:
yp = A cos 4x + B sin 4x
So far, so good. Now, however, when you plug this solution into the differential equation, you get the following result:
(16A – 16A) cos 4x + (16B – 16B) sin 4x = 4 cos 4x
Hang on a minute here! The coefficients of cos 4x and sin 4x on the left are
both zero! So, there are no solutions that are a combination of a sine and a
cosine that solve the differential equation. You can see why if you take a look
at the corresponding homogeneous differential equation:
y" + 16y = 0
The general solution to this homogeneous equation is:
y = c1 cos 4x + c2 sin 4x
This general solution has the same form as your attempted particular solution:
A cos 4x + B sin 4x. But because that’s the homogeneous solution, it isn’t going
to be a particular solution to the nonhomogeneous equation. Instead, you have
to find another version of yp — one that will result in sin 4x and cos 4x terms
when differentiated. The simplest form (besides yp = A cos 4x + B sin 4x) that
can do that is:
yp = Ax cos 4x + Bx sin 4x
Give it a try. Here’s what y'p looks like:
y'p = A cos 4x – 4Ax sin 4x + B sin 4x + 4Bx cos 4x
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
And y"p looks like this:
y"p = –4A sin 4x – 4A sin 4x – 16Ax cos 4x + 4B cos 4x + 4B cos 4x – 16Bx
sin 4x
Substituting y"p into the original nonhomogeneous differential equation gives
you this result:
–4A sin 4x – 4A sin 4x – 16Ax cos 4x + 4B cos 4x + 4B cos 4x – 16Bx sin 4x +
16Ax cos 4x + 16Bx sin 4x = 4 cos 4x
Wow! Thankfully, this collapses to:
–8A sin 4x + 8B cos 4x = 4 cos 4x
So, A = 0 and B = 1⁄2, giving you the following particular solution:
y p = x sin 4x
2
And that’s that. The general solution to this differential equation is therefore:
y = c 1 cos 4x + c 2 sin 4x + x sin 4x
2
When g(x) is a product
of two different forms
Here’s a neat trick: If your g(x) term is a product of erx with sin x and cos x or a
polynomial, you should attempt a particular solution that’s a similar product,
using coefficients whose values have yet to be determined.
Ready to see this handy trick at work? Take a look at this differential equation:
4y" + y = 5ex cos 2x
Assume that a particular solution is the product of ex and cos 2x and sin 2x:
yp = Aex cos 2x + Bex sin 2x
Here’s the derivative y'p:
y'p = Aex cos 2x – 2Aex sin 2x + Bex sin 2x + 2Bex cos 2x
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And here’s the second derivative y"p:
y"p = Aex cos 2x – 2Aex sin 2x – 2Aex sin 2x – 4Aex cos 2x + Bex sin 2x + 2Bex
cos 2x + 2Bex cos 2x – 4Bex sin 2x
or:
y"p = –3A ex cos 2x – 4Aex sin 2x – 3Bex sin 2x + 4Bex cos 2x = ex cos 2x (–3A +
4B) + ex sin 2x (–4A – 3B)
Substituting this into your original differential equation gives you the following:
4(–3Aex cos 2x – 4Aex sin 2x – 3Bex sin 2x + 4Bex cos 2x) + (Aex cos 2x + Bex
sin 2x) = 5ex cos 2x
or:
–12Aex cos 2x – 16Aex sin 2x – 12Bex sin 2x + 16Bex cos 2x + Aex cos 2x + Bex
sin 2x = 5ex cos 2x
Collecting terms gives you this equation:
(–11A+ 16B)ex cos 2x + (–16A – 11B)ex sin 2x = 5ex cos 2x
So, matching the coefficients of cos 2x and sin 2x gives you these equations:
–11A + 16B = 5
and
–16A – 11B = 0
From this equation, A = –11B/16, and by substituting it into the previous
equation you get this:
–121B + 256B = 80
So, 135B = 80, or B = 16⁄27. Figuring out A yields:
–16A – 176⁄27 = 0
And A = –11⁄27. Therefore the particular solution, yp, is:
y p = -11 e x cos 2x + 16 e x sin 2x
27
27
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
Breaking Down Equations with the
Variation of Parameters Method
What if g(x) isn’t in one of the forms that I discuss in the previous sections?
What if you can’t get the method of undetermined coefficients to work? In
either case, you can always turn to the variation of parameters method.
So what’s this variation of parameters technique? It’s a clever one! Say that
you have the following differential equation:
y" + p(x)y' + q(x)y = g(x)
Now assume that you know the solution to the corresponding homogeneous
equation:
y" + p(x)y' + q(x)y = 0
The general homogenous solution is:
yh = c1 y1 + c2 y2
Here’s the first trick: You now replace the constants c1 and c2 with functions
u1(x) and u2(x) to find a particular solution. This is what the equation looks like:
yp = u1(x)y1 + u2(x)y2
Then you try to find the functions u1(x) and u2(x) such that this equation is a
particular solution of the nonhomogeneous differential equation (not the
homogeneous differential equation).
Here’s the second trick: Substituting this equation into your original nonhomogeneous equation is going to give you one equation in two unknowns,
u1(x) and u2(x), as well as their first two derivatives. You need a second constraint on the values of u1(x) and u2(x) as well. And because you’re looking
only for a single particular solution, you can choose that constraint on u1(x)
and u2(x) to make the math easier.
In the following sections, I explain the basics of the variation of parameters
method and take you step by step through some interesting examples. I also
discuss the relationship of this method to one called the Wronskian (see
Chapter 5 for details).
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Nailing down the basics of the method
First up: You need y' and y" to substitute into your original nonhomogeneous
equation. Start with y', the first derivative of yp = u1(x)y1 + u2(x)y2:
y' = u'1(x)y1 + u1(x)y'1 + u'2(x)y2 + u2(x)y'2
Here’s where the second trick comes in (the one designed to make the math
easier). Because you get to choose the second constraint on u1(x) and u2(x),
choose the constraint so that:
u'1(x)y1 + u'2(x)y2 = 0
The second trick makes the math easier because now the first derivative
becomes:
y' = u1(x)y'1 + u2(x)y'2
Quite a bit simpler, right? Now for y":
y" = u'1(x)y'1 + u1(x)y"1 + u'2(x)y'2 + u2(x)y"2
Now substitute y, y', and y" into the nonhomogeneous equation. The resulting
equation is going to look messy, but don’t fret because something good
happens:
u1(x)[y1" + p(x)y1' + q(x)y1] + u2(x)[y2" + p(x)y2' + q(x)y2] + u'1(x)y1' +
u'2(x)y2' = g(x)
Note that the first two terms are zero (that’s the good thing that happens),
because y1 and y2 are solutions of the homogeneous equation. This means
that you’re left with:
u'1(x)y1' + u'2(x)y2' = g(x)
You now have two equations in u'1(x) and u'2(x):
u'1(x)y1(x) + u'2(x)y2(x) = 0
u'1(x)y1'(x) + u'2(x)y2'(x) = g(x)
Using these equations, you can solve for u'1(x) and u'2(x). Then you can integrate them, and you’ll have a particular solution to the original nonhomogeneous differential equation, because where y1 and y2 are linearly independent
solutions of the homogeneous equation:
yp = u1(x)y1(x) + u2(x)y2(x)
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
Solving a typical example
To help solidify the basics in your mind, put them to work with the following
differential equation:
y" + 4y = sin2 2x
The homogenous equation is:
y" + 4y = 0
The general solution to the homogeneous equation is the following (I’ll skip the
details here; you can find out how to complete this step earlier in this chapter):
yh = c1 cos 2x + c2 sin 2x
So:
y1 = cos 2x
and
y2 = sin 2x
This means that you’ll be searching for a particular solution of the following
form:
yp = u1(x) cos 2x + u2(x) sin 2x
where you need to determine u1(x) and u2(x). The method of variation of
parameters tells you that:
u'1(x)y1(x) + u'2(x)y2(x) = 0
u'1(x)y1'(x) + u'2(x)y2'(x) = g(x)
Substituting in for y1 and y2 gives you these equations:
u'1(x) cos 2x + u'2(x) sin 2x = 0
–2u'1(x) sin 2x + 2u'2(x) cos 2x = sin2 2x
Solving these equations for u'1(x) and u'2(x) gives you:
3
ul 1^ x h = - sin 2x
2
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Part II: Surveying Second and Higher Order Differential Equations
and:
2
ul 2 ^ x h = sin 2x cos 2x
2
You can integrate to find u1(x) and u2(x):
3
u 1^ x h = cos 2x - cos 2x
4
12
and:
3
u 2 ^ x h = sin 2x
12
Why aren’t you using any constants of integration here? Because you need
only one particular solution, so you can choose the constants of integration
to equal zero, which means that the particular solution is (after the algebra
and trig dust settles):
2
2
y p ^ x h = cos 2x + sin 2x
6
12
So, the general solution of your original nonhomogeneous equation is:
y = yh + yp
which means it looks like this:
2
2
y = c 1 cos 2x + c 2 sin 2x + cos 2x + sin 2x
6
12
Beautiful.
Applying the method to
any linear equation
As you can see from the previous sections, the method of variation of parameters can be useful. It’s broader in application than the method of undetermined coefficients that I discuss earlier in this chapter. Why? The method of
undetermined coefficients works only for a few forms of g(x).
On the other hand, you can use the method of variation of parameters for all
linear differential equations (linear in y). For second order differential equations like the ones in this chapter, you get a system of two equations to solve;
for a system of three differential equations, you get three equations to solve;
and so on. The problem comes in the integration of u'1(x), u'2(x), and so on,
because the integration may not be possible.
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
Take a look at an example that brings this issue to light. Here’s a whopper of
a differential equation:
d 2 y 2 dy 2y 1
- x
+
= ln^ x h
dx x 2 x
dx 2
Why is this equation such a whopper? Well, for starters, it isn’t separable
(see Chapter 3 for an explanation of separable equations). And to top it off,
you can’t use the method of undetermined coefficients. But it’s linear, so you
can use the method of variation of parameters, as you find out in the following sections.
The general solution of the homogeneous equation
First, take a look at the homogeneous equation:
d 2 y 2 dy 2y
- x
+
=0
dx x 2
dx 2
You may notice that this equation looks a little more manageable than the
original one. After looking at the form of this differential equation and noting
that each successive term has another power of x in the denominator, you
would likely decide to try a solution of the form y = x n. And by substituting
into the differential equation, you get:
n(n–1)x n–2 – 2nx n–2 + 2x n–2 = 0
After dividing by x n–2 and doing a little algebra, you get:
n2 – n – 2n + 2 = 0
So:
n2 – 3n + 2 = 0
You can solve this with the quadratic equation to get:
n= 3!1
2
So n1 = 1 and n2 = 2, which means that you have two linearly independent
solutions of the homogeneous differential equation:
y1 = x
and
y2 = x 2
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You get the following as the general solution of the homogeneous differential
equation:
yh = c1x + c2 x 2
The particular and general solutions of the nonhomogeneous equation
Now you have to find the particular solution, yp, of the original nonhomogeneous differential equation. Why? Because its general solution is the sum of
the general solution of the homogeneous differential equation and the particular solution of the nonhomogeneous differential equation:
y = yh + yp
Say you have a linear differential equation of the following form:
y" + p(x)y' + q(x)y = g(x)
In this case, it’s a doozy of an equation, and:
2
p^ x h = x
q ^ x h = 22
x
g^xh = 1
x ln^ x h
Okay, so you can’t use the method of undetermined coefficients. Never fear;
this is where the method of variation of parameters comes in. Accordingly,
you plan a solution of the following form:
yp = u1(x)x + u2(x)x 2
The method of variation of parameters gives you:
u'1(x)y1(x) + u'2(x)y2(x) = 0
u'1(x)y1'(x) + u'2(x)y2'(x) = g(x)
So:
u'1(x)x + u'2(x)x 2 = 0
and:
ul 1^ x h + ul 2 ^ x h 2x = 1
x ln^ x h
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
From:
u'1(x)x + u'2(x)x2 = 0
you get:
u'1(x) = –u'2(x)x
And substituting that into the second equation gives you:
- ul 2 ^ x h x + ul 2 ^ x h 2x = 1
x ln^ x h
So with a little combining, you get:
ul 2 ^ x h = 12 ln^ x h
x
Substituting that equation into u'1(x)x + u'2(x)x 2 = 0 gives you the following:
ul 1^ x h = - 1
x ln^ x h
Now you have u'1(x) and u'2(x), and you have to integrate them. (Tip: You can
find the integrals in most standard calculus books.) Here are the answers (as
in the previous section, neglecting any constants of integration, because you
need only one particular solution):
u 1^ x h = - 1 ln 2 ^ x h
2
and:
1
u 2^ x h = - 1
x ln^ x h - x
Substituting that into the following equation:
yp = u1(x)x + u2(x)x 2
gives you:
y p = - x ln 2 ^ x h - x ln^ x h - x
2
Alright. You’re almost there! Because:
y = yh + yp
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Part II: Surveying Second and Higher Order Differential Equations
you know that:
y = c 1 x + c 2 x 2 - x ln 2 ^ x h - x ln^ x h - x
2
In fact, you can absorb the final –x term into c1x, giving you the general
solution:
y = c 1 x + c 2 x 2 - x ln 2 ^ x h - x ln^ x h
2
What a pair! The variation of parameters
method meets the Wronskian
As noted in the previous sections, the method of variation of parameters
allows you to tackle linear differential equations, such as this second order
differential equation:
y" + p(x)y' + q(x)y = g(x)
The method of variation of parameters relies on the solution to the homogeneous equation:
y" + p(x)y' + q(x)y = 0
The solution to the homogeneous equation is:
yh = c1 y1(x) + c2 y2(x)
The method of variation of parameters says that you then try to find a particular solution of the following form:
yp = u1(x)y1(x) + u2(x)y2(x)
Substituting yp into the differential equation gives you these two equations:
u'1(x)y1(x) + u'2(x)y2(x) = 0
u'1(x)y1'(x) + u'2(x)y2'(x) = g(x)
You can formally solve these equations for u'1(x) and u'2(x) as follows:
ul 1^ x h =
- y 2^ x h g ^ x h
y 1^ x h y l 2 ^ x h - y l 1^ x h y 2 ^ x h
ul 2^ x h =
y 1^ x h g ^ x h
y 1^ x h y l 2 ^ x h - y l 1^ x h y 2 ^ x h
and:
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
In fact, it turns out that the denominator is the Wronskian (introduced in
Chapter 5), W, for y1, y2, and x, W(y1, y2)(x):
W = y1(x)y'2(x) – y1'(x)y2(x)
So, you can write the equations for u'1(x) and u'2(x) like this instead:
ul 1^ x h =
- y 2^ x h g ^ x h
W _ y 1 , y 2 i^ x h
ul 2^ x h =
y 1^ x h g ^ x h
W _ y 1 , y 2 i^ x h
and:
Note that dividing by the Wronskian is okay because y1 and y2 are a set of linearly independent solutions, so their Wronskian is nonzero. This means that
you can solve for u1(x) (at least theoretically) like this:
u 1^ x h = -
y 2^ x h g ^ x h
dx + c 1
1 , y 2 i^ x h
# W _y
And you can solve for u2(x) like this:
u 2^ x h =
y 1^ x h g ^ x h
dx + c 2
1 , y 2 i^ x h
# W _y
Of course, there’s no guarantee that you can perform the integrals in these
equations. But if you can, you can get u1(x) and u2(x), such that a particular
solution to the differential equation is:
yp = u1(x)y1(x) + u2(x)y2(x)
The general solution is:
y = c1 y1(x) + c2 y2(x) + u1(x)y1(x) + u2(x)y2(x)
Bouncing Around with Springs ’n’ Things
Second order differential equations play a big part in elementary physics.
They’re used in describing the motion of springs and pendulums, electromagnetic waves, heat conduction, electric circuits that contain capacitors and
inductors, and so on. I provide a couple of examples of second order differential equations in the following sections.
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A mass without friction
Here I show you the differential equation describing the motion of a mass on
the end of a spring. Say, for example, that you have the situation shown in
Figure 6-1, where a mass is moving around (without friction) on the end of a
spring. In the following sections, I show you how to solve this physics problem with the help of a nonhomogeneous equation.
A
Figure 6-1:
A spring
with a mass
moving
without
friction.
B
F
F
C
Turning the physics into a differential equation
The force that the spring in Figure 6-1 exerts on the mass is proportional to
the amount that the spring is stretched, and the constant of proportionality
is called the spring constant, k. Thus the force that the spring exerts on the
mass is:
F = –ky
where k is the spring constant (which you have to measure for every spring
you want to use) and y is the distance away from the equilibrium position
(where the spring is unstretched).
The minus sign is included to indicate that the force is a restorative force,
meaning that it always pulls toward the equilibrium position. So the force is
always in the opposite of the direction you’re pulling. If x, the distance from
the equilibrium position, is positive, the force exerted by the spring is negative, pulling back toward the equilibrium position.
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
The force exerted by the spring is proportional to the length by which you’re
pulling the spring. As you may know from elementary physics, the force on
an object is equal to its mass multiplied by its acceleration:
F = ma
where m is the object’s mass, and a is its acceleration.
Here’s where the differential equation comes in, because acceleration is the
second order derivative of distance with respect to time (symbolized by t):
a=
d2y
dt 2
You can write the equation for force like this:
F=m
d2y
dt 2
In keeping with the notation I use throughout this book, I’ll write the previous
equation like this, where the second derivative of distance with respect to
time is given by y":
F = my"
Note that in physics, differentiating by time is often given by putting a dot
above the variable being differentiated, and differentiating twice with respect
to time is given by placing two dots above the variable being differentiated,
like this: ë.
The mass is accelerated by the spring, so F = ma is equal to F = –ky. So you
have this:
my" = –ky
You’re back in differential equation territory again, so now you’ll take over
from the physicists. Here’s what this equation looks like in a form you’re
more used to:
my" + ky = 0
This equation is okay, but it’s a homogeneous differential equation. And this
is, after all, a chapter on nonhomogeneous differential equations. To solve
this dilemma, you can add a periodic force, acting on the mass, say F0cosω0t.
Adding this force makes this a nonhomogeneous differential equation:
my" + ky = F0 cos ω0t
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You’re driving the mass with this new force, F0 cos ω0t. This is a periodic force,
with period (that is, the time it takes to complete a cycle):
2π
T= ω
0
What’s going to happen now that the mass is subject to the spring force and
the new driving force? You can find out by solving my" + ky = F0 cos ω0t.
The general solution to the homogeneous equation
To solve my" + ky = F0 cos ω0t, you need to take a look at the corresponding
homogeneous equation:
my" + ky = 0
Put this equation into standard form, like so:
ky
y'' + m = 0
In other words, the equation looks like this:
ky
y'' = - m
Now you need something that changes sign upon being differentiated twice,
which means that you need to use sines and cosines. So assume that:
y1 = cos ωx
and
y2 = sin ωx
Plugging y1 and y2 into the y" equation, you get:
k
ω2= m
or:
ω=
k
m
So the general homogeneous solution is:
yh = c1 cos ωx + c2 sin ωx
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
The particular and general solutions to the nonhomogeneous equation
Now you have to find a particular solution to the nonhomogeneous equation:
my" + ky = F0 cos ω0t
which you can write as:
ky F
y'' + m = m0 cos ω 0 t
Using the method of undetermined coefficients that I describe earlier in this
chapter, you can guess that the particular solution is of the following form:
yp = A cos ω0t
where A is yet to be determined.
You can dispense with the B sin ω0t term because the differential equation
involves only y" and y, and g(x) is a cosine. So the B in B sin ω0t would have to
be zero.
Plugging your attempted solution into the nonhomogeneous equation yields:
2
k A cos ω t = F 0 cos ω t
- A ω 0 cos ω 0 t + m
0
0
m
or, since k/m = ω2:
2
F
- A ω 0 cos ω 0 t + A ω 2 cos ω 0 t = m0 cos ω 0 t
Dividing by cos ω0t to simplify gives you:
2
F
- A ω 0 + A ω 2 = m0
Or:
2
F
A a ω 2 - ω 0 k = m0
So:
A=
F0
2
m aω 2 - w 0 k
yp=
F 0 cos ω 0 t
2
m aω 2 - ω 0 k
And:
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Part II: Surveying Second and Higher Order Differential Equations
The general solution to the forced spring differential equation is:
y = c 1 cos ωx + c 2 sin ωx +
F 0 cos ω 0 t
2
m aω 2 - ω 0 k
You can see the graph of a representative solution, with c1 = c2 = 1, ω = 1, ω0 = 1⁄2,
and F0/m(ω2 – ω02) = 1, in Figure 6-2.
3
2
y
1
0
Figure 6-2: –1
Graphing
the math
behind a –2
mass
without
friction. –3
0
5
10
15
20
25
x
30
35
40
A mass with drag force
In this section is another example, but this time you include a drag force on
the mass. (You don’t have to read this example if you don’t want to. If you do,
be forewarned that the algebra gets a little involved.)
A drag force acting on a mass is referred to as damping. For example, the mass
may be moving through water or heavy fluid, and the damping force tends to
slow it down. The damping force is usually proportional to the speed of the
mass (the faster it goes, the more damping force there is), and the constant of
proportionality is γ, the damping coefficient. So here’s what the differential
equation of motion looks like with damping:
my" + γy' + ky = 0
Chapter 6: Studying Second Order Linear Nonhomogeneous Differential Equations
You now have a term in y', the speed of the mass on a spring. So how do you
solve this one? Well, you can try a solution of the form y = ert (which includes
sines and cosines if r is complex). Plugging this solution for y into the previous equation gives you:
mr 2ert + γrert + kert = 0
Canceling out ert gives you this characteristic equation:
mr 2 + γr + k = 0
The roots look like this:
- γ ! _ γ 2 - 4mk i
r1, r 2 =
2m
1/2
Look at the case where the discriminant, γ 2 – 4mk, is less than zero. In that
case:
r1, r 2 =
γ ! i _ 4mk - γ 2 i
2m
1/2
So substituting r1 and r2 into ert, the solution can be written as:
y = e–γt/2m(A cos ωt + B sin ωt)
where:
ω=
_ 4km - γ 2 i
1/2
2m
The solution to the example is an interesting result. It indicates that the
motion of the mass is sinusoidal (like a sine wave), but it’s also multiplied by
an exponential that decays with time. So in this case, the mass oscillates with
a diminishing amplitude, tending toward zero motion. The usual way that the
solution is written is to let A = C cos δ and B = C sin δ, where δ is called the
phase angle. Now you can write the solution this way:
y = Ce–γt/2m cos (ωt – δ)
Writing the solution this way gives you the result in an even neater form. As
you can see, in this case, the solution has a decaying sinusoidal form — the
cos (ωt – δ) is multiplied by e–γt/2m, which goes to zero in time.
This reaction is just what you’d expect from a damped mass on a spring — it
would start oscillating, as you’d expect, and then slowly its motion would die
away. Think of a mass on a spring immersed in oil, for example.
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You can see a representative graph showing the mass’s motion in time in
Figure 6-3.
3
2
y
1
0
–1
Figure 6-3:
Graphing
the math –2
behind a
mass with
drag force. –3
0
5
10
15
20
25
x
30
35
40
Chapter 7
Handling Higher Order
Linear Homogeneous
Differential Equations
In This Chapter
Discovering the notation of higher order differential equations
Introducing the fundamentals of higher order linear homogeneous equations
Working with real and distinct roots
Checking out complex roots
Dealing with duplicate roots
S
ome technicians from the local nuclear power plant storm into your
richly appointed office and say, “We need you to solve a differential equation, and quick.”
“Sure,” you say, “just let me finish lunch.”
“No,” they say, “we need the solution now.” You notice that they keep looking
over their shoulders nervously.
“And if you don’t get your solution now?” you ask, annoyed.
“Boom,” they say.
You put down your sandwich; this might be a rush job after all. “Okay,” you
say with a sigh, “let me see the differential equation.”
They put a piece of paper in front of you, and then you take a look:
y''' – 6y" + 11y' – 6y = 0
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Part II: Surveying Second and Higher Order Differential Equations
“It’s a third order differential equation,” they say.
“I can see that,” you tell them.
“We only know how to solve up to second order,” they wail.
“Many first and second order methods are applicable to higher order differential equations,” you say, getting out your clipboard.
“Can you speed things up?” they ask, fidgeting.
“No problem,” you say. “The solution looks like this:”
y = c1ex + c2e2x + c3e3x
“Wow,” they say. “That was quick work.” Then the techs start hurrying out
the door without paying you so much as a penny. Good thing you love working differential equations.
This chapter introduces higher order (also called nth order) differential equations, and by higher order, I mean any order higher than two. With the help of
this chapter, you can solve higher order differential equations in a snap, even
if they aren’t a matter of life and death! I focus on tips and tricks for solving
different kinds of higher order linear homogeneous equations (in other
words, those that equal zero).
The Write Stuff: The Notation of Higher
Order Differential Equations
Before I get into the nitty-gritty of higher order differential equations in the
following sections, you need a primer on their notation. Why? Well, they’re
slightly different from first and second order equations. For instance, here’s
an example of a higher order differential equation:
d4y
-y=0
dx 4
As you can see, this involves the fourth derivative of y with respect to x, so
it’s a fourth order differential equation. If you’ve read the first few chapters of
this book, you may expect that you can also write the equation like this:
y'''' – y = 0
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
Surprise! Instead, this equation is usually written as:
y(4) – y = 0
Note the terminology: Instead of writing y'''', you write y(4). Derivatives up
to third order, y''', are commonly written using primes, but when it comes to
fourth order and up, go with the y(4) notation. After all, what would you
rather see:
y(9) – y = 0
or:
y''''''''' – y = 0
In some books, you may see higher order derivatives given with Roman
numerals, like this:
y ix – y = 0
But because Roman numerals are on their way out in common usage today,
I stick with the y(9) version here.
Introducing the Basics of Higher Order
Linear Homogeneous Equations
Are you ready? It’s time to dig into higher order linear homogeneous differential equations. I start from the beginning in the following sections, with information on their format, solutions, and initial conditions. I also provide some
handy theorems to help you on your way.
The format, solutions, and initial
conditions
A general higher order linear differential equation looks like this:
dny
d n -1 y
d n-2 y
dy
+ ... + p n - 1^ x h
+ p n^ x h y = g ^ x h
n + p 1^ x h
n - 1 + p 2^ x h
dx
dx
dx
dx n - 2
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This differential equation is linear in all derivatives of y with respect to x (it
involves only terms to the power 1), and it’s nonhomogeneous because it
equals the function g(x).
The homogeneous version of this differential equation looks like this:
dny
d n -1 y
d n-2 y
dy
+ ... + p n - 1^ x h
+ p n^ x h y = 0
n + p 1^ x h
n - 1 + p 2^ x h
dx
dx
dx
dx n - 2
where g(x) = 0.
Because homogeneous equations equal zero, working with them is a nice way
to ease into the world of higher order equations. You take a look at the higher
order homogeneous differential equation in this chapter, and then you delve
in to the nonhomogeneous version in Chapter 8.
Say, for example, that you have some solutions of the homogeneous
equation — y1, y2, and so on. If y1, y2, and so on are all solutions, then a linear
combination of them is also a solution. For instance:
y = c1 y1 + c2 y2 + . . . cn-1 yn–1 + cn yn
In this solution, c stands for various constants. (See the later section “The
general solution of a higher order linear homogeneous equation” for more
information.)
In addition, higher order differential equations can have initial conditions,
just as first order and second order differential equations can. For a higher
order differential equation, you can have this many initial conditions:
y(x0) = y0, y'(x0) = y'0 . . . y(n–1)(x0) = y(n–1)0
Solving differential equations of higher order where n = 3 or more is a lot like
solving differential equations of first or second order, except that you need
more integrations and have to solve larger systems of simultaneous equations to meet the initial conditions. To satisfy all these initial conditions, you
end up with a series of simultaneous equations, one for y(x0) = y0, one for
y'(x0) = y'0, and so on:
c1 y1 + c2 y2 + . . . cn–1 yn-1 + cn yn = y0
c1 y'1 + c2 y'2 + . . . cn–1 y'n–1 + cn y'n = y'0
c1 y"1 + c2 y'2 + . . . cn–1 y"n–1 + cn y"n = y"0
c1 y (n–1)1 + c2 y (n–1)2 + . . . cn–1 y (n–1)n-1 + cn y (n–1)n = y (n–1)0
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
A couple of cool theorems
In the following sections, I give you a couple of helpful theorems about higher
order linear homogeneous differential equations. Put them to good use!
The general solution of a higher order linear homogeneous equation
You may wonder whether the solution you’ve found to a homogeneous equation, y = c1 y1 + c2 y2 + . . . cn–1 yn–1 + cn yn, is a general solution. In other words, you
want to figure out whether it encompasses every solution of the differential
equation. And that brings me to a theorem about the general solution of a
higher order linear homogeneous differential equation.
If you have n solutions, y1, y2, . . . yn, of a general linear homogeneous differential equation of order n:
dny
d n -1 y
d n-2 y
dy
+ . . . + p n - 1^ x h
+ p n^ x h y = 0
n + p 1^ x h
n - 1 + p 2^ x h
dx
dx
dx
dx n - 2
then a linear combination of those solutions:
y = c1 y1 + c2 y2 + . . . cn–1 yn–1 + cn yn
encompasses all solutions if y1, y2, . . . yn are linearly independent.
What does it mean to be linearly independent? Well, the functions f1, f2, f3 . . . fn
are linearly dependent if there exists a set of constants c1, c2, c3 ... cn (not all of
which are zero) and an interval I such that:
c1f1 + c2f2 + . . . cn–1fn–1 + cnfn = 0
for every x in I. The f1, f2, f3 . . . fn are linearly independent if they aren’t linearly
dependent. It’s as simple as that.
If you have n linearly independent solutions for a linear homogeneous differential equation of order n, you have a fundamental set of solutions for the differential equation. The general solution to the linear homogeneous differential
equation is a linear combination of the functions in the fundamental set of
solutions.
In other words, it’s the same story as I discuss in Chapter 5 for second order
linear homogeneous differential equations — only here it’s generalized for
higher orders.
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Solutions as related to the Wronskian
You can cast the theorem in the previous section in terms of the Wronskian,
which is the determinant of this matrix for a higher order differential equation (see Chapter 5 for full details about the Wronskian):
y1
y l1
W _ y 1 , y 2 ,... y n i =
ym2
y2
yl2
ym2
y (n - 1) 1 y (n - 1) 2
y3
yl 3
ym3
y (n - 1) 3
...
...
...
...
yn
yl n
ym n
y (n - 1) n
Here’s the theorem from the previous section written in terms of the
Wronskian:
If you have n solutions, y1, y2, . . . yn, of a general linear homogeneous differential equation of order n:
dny
d n - 1 + p ^ x h d n - 2 + . . . + p ^ x h dy + p ^ x h y = 0
n + p 1^ x h
2
n -1
n
dx
dx
dx n - 1
dx n - 2
and if their Wronskian, W(y1, y2 . . . yn)(x) ≠ 0 in an interval I for at least
one point x0 in that interval, then all solutions of the homogeneous equation are encompassed by linear combinations of those solutions.
Tackling Different Types of Higher Order
Linear Homogeneous Equations
In the following sections, you can check out several different cases of higher
order linear homogeneous equations: those with real and distinct roots, real
and imaginary roots, complex roots, and duplicate roots.
Real and distinct roots
The following sections walk you through one third order equation and one
fourth order equation, both of which have real and distinct roots.
A third order equation
Start by taking a look at the third order differential equation you solved at the
beginning of this chapter:
y''' – 6y" + 11y' – 6y = 0
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
Assume these initial conditions:
y(0) = 9
y'(0) = 20
y"(0) = 50
This differential equation has constant coefficients (see Chapter 5 for more
information), so you can start by assuming a solution of the following form:
y = e rt
Plugging this solution into your original equation gives you:
r 3ert – 6r 2ert + 11rert – 6ert = 0
Canceling out the ert yields:
r 3 – 6r 2 + 11r – 6 = 0
The latter is the characteristic equation for the original homogeneous equation (check out Chapter 5 for more on characteristic equations). How do you
solve this equation for the roots?
Here’s when you start seeing one of the difficulties of higher order differential
equations — they’re like first and second order differential equations, only
more so. That is, a second order differential equation gives you a second order
characteristic equation, which you can solve with the quadratic equation.
But what if you’re faced with a differential equation of order 5? There is no
“quintic” equation — you’re on your own when it comes to finding the roots.
In the case of the characteristic equation in this example, you can (luckily)
factor it into this:
(r – 1) (r – 2) (r – 3) = 0
So the roots are:
r1 = 1
r2 = 2
r3 = 3
The roots are real and distinct, and the linearly independent solutions are:
y1 = ex
y2 = e2x
y3 = e3x
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So the general solution to the original homogeneous equation is:
y = c1ex + c2e2x + c3e3x
Now turn to the initial conditions. In addition to the form for y, you also need
y' and y" to meet the initial conditions:
y' = c1ex + 2c2e2x + 3c3e3x
and
y" = c1ex + 4c2e2x + 9c3e3x
From the initial conditions, y(0) = 9, y' (0) = 20, and y"(0) = 50, here are your
three simultaneous equations in c1, c2, and c3:
y(0) = c1 + c2 + c3 = 9
y'(0) = c1 + 2c2 + 3c3 = 20
y"(0) = c1 + 4c2 + 9c3 = 50
Once again, here’s another place where you see the difference between first
and second order differential equations and those of higher order. With first
order differential equations that have initial conditions, solving for c1 is trivial.
With second order differential equations that have initial conditions, you end
up with a 2 x 2 system of equations — and solving that is easy. However, starting with third order differential equations that have initial conditions, the n x n
system of simultaneous equations can be a little more challenging.
Solving a 3 x 3 system of simultaneous equations for c1, c2, and c3 involves
some tedious algebra. You can calculate this out if you want to invest the
time, or you can just use a computer. If you want to solve the system online,
there are various services to do so. One such Web site is math.cowpi.com/
systemsolver, which is a handy tool. You can simply select the type of
system you’re dealing with (such as 3 x 3, 4 x 4, and 5 x 5), plug in the right
numbers, and voila! You have an answer.
The solutions of your equations, simply stated, are:
c1 = 2
c2 = 3
c3 = 4
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
So the solution of the original homogeneous equation with the initial conditions applied is:
y = 2ex + 3e2x + 4e3x
Cool. Good work!
A fourth order equation
Now you’re ready to try a differential equation of fourth order. Take a look at
this beauty:
y(4) + 10y''' + 35y" + 50y + 24 = 0
Yep, that’s a whopper. And it has initial conditions, too:
y(0) = 10
y'(0) = –20
y"(0) = 50
y'''(0) = –146
The differential equation has constant coefficients, so you can try a solution
of the following form:
y = erx
Substituting this solution into your original homogeneous equation gives you:
r4erx + 10r 3erx + 35r 2erx + 50rerx + 24erx = 0
And dividing by erx gives you this characteristic equation:
r 4 + 10r 3 + 35r 2 + 50r + 24 = 0
Okay, now you’re in a pickle. What are the roots of this equation? Well, by just
looking at it, you can figure out that it can be factored this way:
(r + 1) (r + 2) (r + 3) (r + 4) = 0
Just kidding — you can’t really tell that just by looking at the equation; I only
knew because I’m the one who made up the problem. So, if you have a characteristic equation like this one that’s tough to factor, you can turn to online
equation solvers. One of my favorites is www.hostsrv.com/webmab/app1/
MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=
solve&s3=basic (yes, I know — that’s a heck of an URL). A quick tip: When
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you enter variables raised to any power, be sure to add a caret (for instance,
you enter r 2 as r^2). After you enter your equation, click the Solve button,
and if your equation is factorable, you’ll get the roots.
The roots of your characteristic equation are:
r1 = –1
r2 = –2
r3 = –3
r4 = –4
In other words, you have these three linearly independent solutions:
y1 = e–x
y2 = e–2x
y3 = e–3x
y4 = e–4x
So the general solution to the original homogeneous equation is:
y = c1e–x + c2e–2x + c3e–3x + c4e–4x
To meet the initial conditions, you need y', y", and y''':
y' = –c1e–x –2c2e–2x – 3c3e–3x – 4c4e–4x
y" = c1e–x + 4c2e–2x + 9c3e–3x + 16c4e–4x
y''' = –c1e–x – 8c2e–2x – 27c3e–3x – 64c4e–4x
Substituting the initial conditions gives you:
y(0) = c1 + c2 + c3 + c4 = 10
y'(0) = –c1 – 2c2 – 3c3 – 4c4 = –20
y"(0) = c1 + 4c2 + 9c3 + 16c4 = 50
y'''(0) = –c1 – 8c2 – 27c3 – 64c4 = –146
Well, this is another fine mess — you have a 4 x 4 system of simultaneous
equations in c1, c2, c3, and c4. You can invest the time to solve it algebraically,
or you can use a program to solve this system, such as math.cowpi.com/
systemsolver (which I mention in the previous section).
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
You can see that:
c1 = 4
c2 = 3
c3 = 2
c4 = 1
which makes the general solution, including initial conditions, to the original
homogeneous equation look like this:
y = 4e–x + 3e–2x + 2e–3x + e–4x
Real and imaginary roots
Linear homogeneous equations with constant coefficients often have both
real and imaginary roots. Check out this equation, for instance:
y(4) – y = 0
Here are the initial conditions:
y(0) = 3
y'(0) = 1
y"(0) = –1
y'''(0) = –3
Because this is a linear homogeneous differential equation with constant
coefficients, you decide to try a solution of the following form:
y = ert
Plugging the solution into the original equation gives you:
r4ert – ert = 0
After canceling out ert, you get this equation:
r4 – 1 = 0
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As you may have noticed, this is a more manageable characteristic equation
than the one in the previous section. You can easily factor this characteristic
equation into:
(r 2 – 1) (r 2 + 1) = 0
So the roots of this characteristic equation are:
r1 = 1
r2 = –1
r3 = i
r4 = –i
It looks as though you have real and imaginary roots for r3 and r4. You can
handle complex roots with the following relations:
e(α + iβ)x = eαx(cos βx + i sin βx)
and:
e(α – iβ)x = eαx(cos βx – i sin βx)
Here, α = 0 for r3 and r4, and you get:
eiβx = cos βx + i sin βx
and:
e–iβx = cos βx – i sin βx
So y3 and y4 can be expressed as a linear combination of sines and cosines.
Thus you have these solutions:
y1 = ex
y2 = e–x
y3 = cos x
y4 = sin x
The general solution to the original homogeneous equation is:
y = c1ex + c2e–x + c3 cos x + c4 sin x
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
What happened to the i multiplying sin βx? As I discuss in Chapter 5, the i can
be absorbed into the constants c3 and c4, because i is, after all, just a constant.
Now you have to handle the initial conditions. To do that, you also need y',
y", and y''':
y' = c1ex – c2e–x – c3sin x + c4cosx
y" = c1ex + c2e–x – c3cos x – c4sin x
y''' = c1ex – c2e–x + c3sin x – c4cos x
Substituting the initial conditions into the equation gives you:
y(0) = c1 + c2 + c3 = 3
y'(0) = c1 – c2 + c4 = 1
y"(0) = c1 + c2 – c3 = –1
y'''(0) = c1 – c2 – c4 = –3
And there you have it again — a 4 x 4 system of simultaneous equations.
You can solve this system by doing the algebra, or you can make it easy
on yourself and check out an online simultaneous equation solver, such as
math.cowpi.com/systemsolver. Go ahead; I’ll wait for you to come up
with the answers . . . .
Ready? So you have:
c1 = 0
c2 = 1
c3 = 2
c4 = 2
which means that the general solution with initial conditions is:
y = e–x + 2 cos x + 2 sin x
It’s lucky that c1 = 0. Why? Otherwise the solution would grow exponentially.
You can see a graph of the solution in Figure 7-1, where the exponential term
dies away in time.
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4
3
y
2
1
0
Figure 7-1:
The solution
to an
equation
with real
and
imaginary
roots.
–1
–2
–3
–4
0
1
2
3
4
5
6
7
8
9
10 11 12 13 x 14 15 16 17 18 19 20
Complex roots
The previous section handles the case where you have both real and imaginary
roots. What about the case where you have complex roots? For example, take a
look at this fine equation:
y(4) + 4y = 0
Because this is a fourth order linear homogeneous differential equation with
constant coefficients, you can try a solution of the following form:
y = erx
Plugging the solution into the original equation gives you:
r4erx + 4erx = 0
Next, dividing by erx gives you this characteristic equation:
r4 + 4 = 0
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
So now you need the fourth root of –4, which isn’t something you’re likely to
find on your calculator. Once again, these equations come to the rescue:
e(α + iβ)x = eαx(cos βx + i sin βx)
and:
e(α - iβ)x = eαx(cos βx – i sin βx)
You can think of –4 as –4 + 0i, so:
–4 = 4 cos π + 4i sin π = 4eiπ
Note that this relation is determined only up to multiples of 2π, so this is
actually:
–4 = 4 cos (π + 2nπ) + 4i sin (π + 2nπ) = 4ei(π + 2nπ)
where n is an integer.
You can therefore find the fourth root of –4 this way:
(–4)1/4 = 41/4 ei(π/4 + nπ/2)
Expanding the exponent gives you:
(–4)1/4 = 41/4 (cos (π/4 + nπ/2) + i sin (π/4 + nπ/2))
And using different values of n gives you the fourth roots of –4:
r1 = 1 + i
r2 = –1 + i
r3 = –1 – i
r4 = 1 – i
Great! You’ve made progress. Now you know that the fundamental set of solutions is e(1 + i)x, e(–1 + i)x, e(–1 – i)x, and e(1 – i)x. By using these equations:
e(α + iβ)x = eαx(cos βx + i sin βx)
and
e(α – iβ)x = eαx(cos βx – i sin βx)
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the fundamental set of solutions can be given this way in the general solution:
y = c1ex cos x + c2ex sin x + c3e–x cos x + c4e–x sin x
And that’s the general solution of your original homogeneous equation. Cool.
Duplicate roots
Duplicate roots are just what they sound like: one or more roots that are
repeated as you figure out the general solution of a homogeneous equation.
I describe several types of duplicate roots in the following sections.
A fourth order equation with identical real roots
Take a look at this differential equation, which is a fourth order homogeneous
differential equation with constant coefficients:
y(4) + 4y''' + 6y" + 4y' + y = 0
You can try a solution of the following form:
y = erx
Plugging the solution into the original homogeneous equation gives you this
characteristic equation:
r4 + 4r 3 + 6r 2 + 4r + 1 = 0
Holy mackerel, you might think. What are the roots of that thing? You can
factor this equation using an online equation factoring program, such as
hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?
site=quickmath&s1=equations&s2=solve&s3=basic.
If you’re determined to rely on your own brain to factor such an equation,
here’s a trick that sometimes helps: Convert the equation from base r to base
10. That is, r 2 + 2r + 1 becomes 100 + 20 + 1 = 121, which can be easily factored
into 11 x 11. Converting 11 from base 10 back to base r gives you (r + 1) (r + 1),
so the roots are –1 and –1. You can use this quick trick to astound your friends.
No matter which method you use, you’ll sooner or later figure out that you
can factor r4 + 4r 3 + 6r 2 + 4r + 1 = 0 this way:
(r + 1) (r + 1) (r + 1) (r + 1)
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
So the roots of the characteristic equation are –1, –1, –1, –1 — all repeated
real roots. Does that mean the solutions are:
y1 = e–x
y2 = e–x
y3 = e–x
y4 = e–x
Hardly. You need linearly independent solutions to get a fundamental set of
solutions from which to build your general solutions. And because these are
all e–x, they’re obviously not linearly independent. (I talk about linear independence in more detail in the earlier section “The general solution of a
higher order linear homogeneous equation.”)
What can you do here? Well, clearly you can’t have this as the general solution:
y = c1e–x + c2e–x +c3e–x +c4e–x
Why? Because that’s really equivalent to:
y = ce–x
where c = c1 + c2 +c3 + c4
So what can you do? Easy: You add powers of x until you have as many linearly
independent solutions as you need. You can convert y = c1e–x + c2e–x + c3e–x + c4e–x
into a true general solution by introducing factors of x, x2, and x3 like this:
y = c1e–x + c2xe–x +c3x2e–x +c4x3e–x
A fifth order equation with identical real roots
Take a look at this whopper, which is a fifth order homogeneous differential
equation:
y(5) – y(4) – 2y''' + 2y" + y' – y = 0
Because it’s homogeneous and has constant coefficients, you can try a solution in the form y = erx, giving you:
r 5erx – r 4erx – 2r 3erx + 2r 2erx + rerx – erx = 0
After dividing by erx, you get this equation:
r 5 – r 4 – 2r 3 + 2r 2 + r – 1= 0
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Your response may sound something like this: “I’m supposed to factor that?
Are you crazy?” Never fear; you can turn for help to an online equation
solver, such as hostsrv.com/webmab/app1/MSP/quickmath/02/
pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic.
The roots are –1, –1, 1, 1, 1, so r 5 – r 4 – 2r 3 + 2r 2 + r – 1 = 0 can be factored
into this:
(r – 1) (r – 1) (r – 1) (r + 1) (r + 1) = 0
or:
(r – 1)3 (r + 1)2 = 0
As you can see, this equation looks a lot more manageable than r 5 – r 4 – 2r 3 +
2r 2 + r – 1 = 0. Because –1 is a root of the characteristic equation, this is a
solution:
y1 = e–x
In fact, –1 is a double root, which also makes this a solution:
y2 = xe–x
Because you can also have 1 as a root, you have this solution:
y3 = ex
Finally, you may also notice that 1 is a triple root, so in that case you also have:
y4 = xex
and
y5 = x 2ex
So the general solution to the original homogeneous equation is:
y = c1e–x + c2xe–x + c3ex + c4xex + c5 x 2ex
Notice that once again, the general solution is a linear combination of n linearly independent solutions, where n is the order of the differential equation.
And note that also, you multiply solutions by x, x2, and so on depending on
their multiplicity as roots of the characteristic equation. Here, one root had
multiplicity 2 and the other multiplicity 3.
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
Identical imaginary roots
Are you curious about the case where the roots of a characteristic equation
are duplicate and imaginary? For example, take a look at this differential
equation with constant coefficients:
y(4) + 8y" + 16y = 0
You can try a solution of the following form:
y = erx
Substituting this solution into your original homogeneous equation gives you:
r 4erx + 8r 2erx +16erx = 0
So your characteristic equation is:
r 4 + 8r 2 + 16 = 0
You can factor this into:
(r 2 + 4) (r 2 + 4) = 0
So the roots are 2i, 2i, –2i, and –2i. As you can see, you have duplicate imaginary roots here.
You can use these equations as I described earlier in this chapter:
e(α + iβ)x = eαx(cos βx + i sin βx)
and:
e(α – iβ)x = eαx(cos βx – i sin βx)
Here are the first two solutions, y1 and y2:
y1 = cos 2x
and
y2 = sin 2x
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What about y3 and y4? You can use the same technique as in the previous section: Multiply by progressively higher powers of x until you get all the linearly
independent solutions you need. In this case, all you need is one power of x
to give you:
y3 = x cos 2x
and
y4 = x sin 2x
So the general solution to the homogeneous equation is:
y = c1 cos 2x + c2 sin 2x + c3x cos 2x + c4x sin 2x
Identical complex roots
Here’s another example; it’s a fourth order equation with constant coefficients:
y(4) – 8y''' + 32y" – 64y' + 64y = 0
This equation is no problem. You just assume a solution of the form y = erx
and plug it into the original equation to get:
r 4erx – 8r 3erx + 32r 2erx – 64rerx + 64erx = 0
So your characteristic equation is:
r 4 – 8r 3 + 32r 2 – 64r + 64 = 0
Solving this characteristic equation gives you these roots:
r1 = 2 + 2i
r2 = 2 + 2i
r3 = 2 – 2i
r4 = 2 – 2i
So 2 + 2i is a root of multiplicity 2, and so is 2 – 2i. That leads to the following
general solution:
y = b1e(2 + 2i)x + b2 xe(2 + 2i)x + b3e(2 – 2i)x + b4 xe(2 – 2i)x
Chapter 7: Handling Higher Order Linear Homogeneous Differential Equations
where b1, b2, b3, and b4 are constants. You can rewrite this equation as the
following:
y = e2x(b1e2ix + b2xe2ix) + e2x(b3e–2ix + b4xe–2ix)
or:
y = e2x(b1e2ix + b3e–2ix) + e2xx(b2e2ix + b4e–2ix)
Once again, you can turn to these relations:
eiβx = cos βx + i sin βx
and:
e–iβx = cos βx – i sin βx
Using these equations gives you:
y = e2x(b1(cos 2x + i sin 2x) + b3 (cos 2x - i sin 2x)) + e2xx(b2(cos 2x + i sin 2x) +
b4(cos 2x - i sin 2x))
Combining terms and consolidating constants gives you this form for the
general solution:
y = e2x(c1 cos 2x + c2 sin 2x) + e2xx(c3 cos 2x + c4 sin 2x)
And there you go, that’s the general solution where the characteristic equation has the roots 2 + 2i, 2 + 2i, 2 – 2i, and 2 – 2i.
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Chapter 8
Taking On Higher Order
Linear Nonhomogeneous
Differential Equations
In This Chapter
Breaking down higher order equations with the method of undetermined coefficients
Using the variation of parameters to solve higher order equations
T
he door to your office opens. A group of rocket scientists enters, looking
embarrassed. “We need you to solve a differential equation,” they say.
“I thought you were math specialists,” you retort.
“We thought so too. But this one has us stumped. It specifies the shape of the
rocket. Without it, we can’t take off.” They slide a sheet of paper onto your
desk. “Nobody has to know that we came to you for help, right?” They look
over their shoulders nervously.
“Right,” you say, taking a look at the sheet of paper, on which you see this
differential equation:
y''' + 6y" + 11y' + 6y = 336e5x
The equation has the following initial conditions:
y(0) = 9
y'(0) = –7
y"(0) = 47
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Part II: Surveying Second and Higher Order Differential Equations
“It’s a nonhomogeneous third order differential equation,” the rocket scientists say. “You’re stumped, right? We knew it!”
“Not so fast!” you say. “The solution is:”
y = 5e–x + 2e–2x + e–3x + e5x
The rocket scientists are stunned. “How did you do that?” they ask.
“Easy,” you say. “I read Chapter 8 of Differential Equations For Dummies. I
highly recommend it. Here’s my bill for solving your equation.”
The rocket scientists take the bill, and, reading it, raise their eyebrows.
“That’s astronomical,” they say.
“Well, you are rocket scientists, aren’t you?” you ask.
This chapter shows you how to handle nonhomogeneous linear differential
equations of order n, where n = 3, 4, 5, and so on. They look like this:
dny
d n -1 y
d n-2 y
dy
+ p 1^ x h
+ . . . + p n - 1^ x h
+ p n^ x h y = g ^ x h
n - 1 + p 2^ x h
dx n
dx
dx
dx n - 2
In a higher order (or nth order) linear nonhomogeneous equation, the p variables are various functions. And in this case, unlike in Chapter 7, where I discuss higher order homogeneous equations, g(x) ≠ 0. When you’re done with
this chapter, you’ll be able to handle nonhomogeneous equations with ease.
Mastering the Method of Undetermined
Coefficients for Higher Order Equations
I first discussed the method of undetermined coefficients in Chapter 6 for
second order differential equations like this one:
y" + p(x)y' + q(x)y = g(x)
In this chapter, however, I generalize that method to help you get a handle on
differential equations of arbitrarily high order, not just two.
Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations
The method of undetermined coefficients is all about finding a particular solution to a nonhomogeneous equation, yp. This method says that when you find
a candidate solution, yp, and plug it into the left-hand side of the equation, you
end up with g(x). Because g(x) is only a function of x, it’s often possible to
guess the form of yp(x), up to arbitrary coefficients, and then solve for those
coefficients by plugging yp(x) into the differential equation.
The form of g(x) can often tell you what a particular solution looks like (just
as it can when you’re dealing with second order equations). In particular, if
g(x) is in the form of:
erx, try a particular solution of the form Aerx, where A is a constant.
Because derivatives of erx reproduce erx, you have a good chance of finding a particular solution this way.
A polynomial of order n, try a polynomial of order n.
A combination of sines and cosines, sinαx + cosβx, try a combinations
of sines and cosines with undetermined coefficients, Asinαx + Bcosβx.
Then you can plug into the differential equation and solve for A and B.
I explain these different forms in the following sections, but before I get to
them, here’s a summary of the method of undetermined coefficients for a
higher order differential equation:
1. Find the general solution, yh = c1 y1 + c2 y2 + . . . + cn yn of the associated
homogeneous differential equation.
2. If g(x) is of the form erx, a polynomial, a combination of sines and
cosines, or a product of any of these, assume that the particular solution is of the same form, using coefficients whose values have yet to
be determined.
3. If g(x) is the sum of terms, g1(x), g2(x), g3(x), and so on (as they are in a
polynomial), break the problem into various subproblems like this:
dny
d n -1 y
d n-2 y
dy
+
+ . . . + p n - 1^ x h
+ p n ^ x h y = g 1^ x h
p
x
^
h
n + p 1^ x h
2
dx
dx
dx n - 1
dx n - 2
dny
d n -1 y
d n-2 y
dy
+
+ . . . + p n - 1^ x h
+ p n^ x h y = g 2^ x h
p
x
^
h
n + p 1^ x h
2
dx
dx
dx n - 1
dx n - 2
dny
d n -1 y
d n-2 y
dy
+
+ . . . + p n - 1^ x h
+ p n^ x h y = g 3^ x h
p
x
^
h
n + p 1^ x h
2
dx
dx
dx n - 1
dx n - 2
The particular solution of the nonhomogeneous equation is the sum of
the solutions of those subproblems.
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Part II: Surveying Second and Higher Order Differential Equations
4. Substitute yp into the differential equation, and solve for the undetermined coefficients.
5. Find the general solution of the nonhomogeneous differential equation, which is the sum of yh and yp:
y = yh + y p
6. Use the initial conditions to solve for c1, c2 . . . cn.
When g(x) is in the form erx
Take a look at the nonhomogeneous differential equation you solve so brilliantly for the rocket scientists at the beginning of this chapter:
y''' + 6y" + 11y' + 6y = 336e5x
Here are the equation’s initial conditions:
y(0) = 9
y'(0) = –7
y"(0) = 47
As with second order differential equations, the general solution to this nonhomgeneous differential equation is the sum of the solution to the corresponding homogeneous differential equation, in which g(x) equals 0:
y''' + 6y" + 11y' + 6y = 0
and a particular solution to the full nonhomogeneous differential equation. In
other words:
y = yh + yp
where yh is the general solution to the homogeneous differential equation,
and yp is a particular solution to the nonhomogeneous differential equation.
The general solution to the homogeneous equation
The first order of business is to solve the homogeneous differential equation
y''' + 6y" + 11y' + 6y = 0. This is a third order linear homogeneous differential
equation with constant coefficients (like in Chapter 7), so you decide to try a
solution of the following form:
y = erx
Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations
Plugging this solution into the homogeneous equation gives you:
r 3erx + 6r 2erx + 11rerx + 6erx = 0
And dividing by erx (to make things a bit simpler) gives you the following
characteristic equation (see Chapter 5 for more about this type of equation):
r 3 + 6r 2 + 11r + 6 = 0
If you’re feeling extra sharp today and can do the algebra yourself, or if you
have an online equation solver at your beck and call like the one at www.
hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=
quickmath&s1=equations&s2=solve&s3=basic, you can see that the roots
of this equation are:
r1 = –1
r2 = –2
r3 = –3
You can factor r 3 + 6r 2 + 11r + 6 = 0 this way:
(r + 1) (r + 2) (r + 3) = 0
So the general solution to the homogeneous differential equation y''' + 6y" +
11y' + 6y = 0 is:
yh = c1e–x + c2e–2x + c3e–3x
The particular solution to the nonhomogeneous equation
After you find the general solution to the homogeneous equation, you have to
find a particular solution, yp, to the nonhomogeneous equation. In this case,
g(x) has the following form:
g(x) = 336e5x
The method of undetermined coefficients says that you should try to match
the form of g(x) so that differentiating yp gives you the same form, up to the
value of multiplicative constants. Because differentiating simple exponents
gives you the same exponent back with possibly a different coefficient, the
method of undetermined coefficients says that in this case you should try yp
of this form:
yp = Ae5x
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Part II: Surveying Second and Higher Order Differential Equations
Substituting this solution into the equation gives you:
125Ae5x + 150Ae5x + 55Ae5x + 6Ae5x = 336e5x
or:
125A + 150A + 55A + 6A = 336
Do the math and you get:
336A = 336
So:
A=1
which means that the particular solution, yp, of the nonhomogeneous equation is given by this:
yp = e5x
And because:
y = yh + yp
the general solution to the nonhomogeneous equation is:
y = c1e–x + c2e–2x + c3e–3x + e5x
Applying initial conditions
To handle the initial conditions of the original nonhomogeneous equation,
you need to find y, y' and y":
y = c1e–x + c2e–2x + c3e–3x + e5x
y' = –c1e–x – 2c2e–2x – 3c3e–3x + 5e5x
y" = c1e–x + 4c2e–2x + 9c3e–3x + 25e5x
Plugging in x = 0 gives you the following:
y(0) = c1 + c2 + c3 + 1 = 9
y'(0) = –c1 – 2c2 – 3c3 + 5 = –7
y"(0) = c1 + 4c2 + 9c3 + 25 = 47
Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations
Time to do a little simplifying:
y(0) = c1 + c2 + c3 = 8
y'(0) = –c1 – 2c2 – 3c3 = –12
y"(0) = c1 + 4c2 + 9c3 = 22
This is a 3 x 3 system of simultaneous equations. You can work it out step
by step, or you can use a handy equation system solver like the one at
math.cowpi.com/systemsolver. Either way you work it, you find that c1,
c2, and c3 are:
c1 = 5
c2 = 2
c3 = 1
So the general solution to your original nonhomogeneous equation with the
given initial conditions is:
y = 5e–x + 2e–2x + e–3x + e5x
Just as you told the rocket scientists at the beginning of the chapter. Nice work!
When g(x) is a polynomial of order n
Take a look at this differential equation stumper that I put together for you:
y''' – y' = x + 60e–4x + 9 sin x
It’s the triple play: A third degree differential equation that’s equal to a polynomial, an exponential, and a trig function — all at the same time. Seriously,
though, solving this equation isn’t as difficult as you may think; just check
out the following sections.
The general solution to the homogeneous equation
Start by looking for the general solution to the corresponding homogeneous
differential equation:
y''' – y' = 0
This equation looks like a linear homogeneous differential equation with constant coefficients, so you can plug in the handy y = erx:
r 3erx – rerx = 0
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Part II: Surveying Second and Higher Order Differential Equations
So the characteristic equation, after dropping erx, is:
r3 – r = 0
One root is clearly r = 0, and dividing by r gives you:
r2 – 1 = 0
The roots of this equation are 1 and –1, and so the solution to the homogeneous differential equation is:
yh = c1 + c2ex – c3e–x
The particular solution to the nonhomogeneous equation
Okay, so far so good. But now you have to find a particular solution, yp. In this
case, g(x) is:
g(x) = x + 60e–4x + 9 sin x
Clearly, you have a polynomial on your hands. What to do? The way to
handle this situation is to break g(x) into three parts:
g1(x) = x
g2(x) = 60e–4x
g3(x) = 9 sin x
Now you have three differential equations to solve:
y''' – y' = x
y''' – y' = 60e–4x
y''' – y' = sin x
And the corresponding particular solutions are yp1, yp2, and yp3, respectively,
which means that the general solution of the nonhomogeneous equation is:
y = yh + yp1 + yp2 + yp3
Alright, start by looking for yp1, with g1(x) = x. To do so you have to solve
y''' – y' = x.
Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations
Because g1(x) = x here, you may think of using yp1 = Ax, but unfortunately, x is
already a solution of the homogeneous differential equation. So yp1 = Ax is
out. Instead, you can try:
yp1 = Ax 2
Plugging that solution into y''' – y' = x gives you:
–2Ax = x
So:
A = –1⁄2
And therefore:
2
y p1 = Ax 2 = - x
2
Okay, now for yp2, which is the solution to y''' – y' = 60e–4x. Because g2(x) = e–4x
here, you can try a solution of the following form:
yp2 = Ae–4x
Plugging yp2 into y''' – y' = 60e–4x gives you:
–64Ae–4x + 4Ae–4x = 60e–4x
or:
–60Ae–4x = 60e–4x
So:
A = –1
and
yp2 = –e–4x
Great! You’re making progress. The last thing you have to do is find yp3, the
third particular solution. That means solving y''' – y' = sin x, for which you
may be tempted to try a solution like this:
yp3 = Acos x + Bsin x
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Part II: Surveying Second and Higher Order Differential Equations
However, take a look at the equation you’re trying to solve here:
y''' – y' = sin x
The first and third derivatives of Acos x yield terms in sin x, which is what
you’re looking for here, but the first and third derivatives of Bsin x yield
terms in cos x, which means that B = 0. So try a solution like yp3 = Acos x.
Plugging yp3 into y''' – y' = sin x gives you:
Asin x + Asin x = sin x
After dropping sin x you get:
A+A=1
So:
A = 1⁄2
which makes yp3 equal to:
y p3 = cos x
2
You now have all the particular solutions to the nonhomogeneous equation
(and the general solution to the homogeneous equation from the previous
section), so you can finally put everything together! Because:
y = yh + yp1 + yp2 + yp3
you get:
2
y = c 1 + c 2 e x - c 3 e - x - x - e - 4x + cos x
2
2
When g(x) is a combination
of sines and cosines
Here’s another higher order problem to work through; this time, the solution
is a combination of sines and cosines, and the problem is of the fourth order:
y(4) + 2y" + y = 8 sin x – 16 cos x
Easy, right? All you have to do is find the general solution to the homogeneous equation, followed by the particular and general solutions to the nonhomogeneous equations. Read on for details.
Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations
The general solution to the homogeneous equation
The homogeneous differential equation, in which g(x) equals 0, is:
y(4) + 2y" + y = 0
This is a linear homogeneous differential equation with constant coefficients.
This means you can try a solution of the form y = erx. Plugging this solution
into the homogeneous equation gives you:
r 4erx + 2r 2erx + erx = 0
Or, for simplicity’s sake:
r4 + 2r 2 + 1 = 0
You can factor this into:
(r 2 + 1) (r 2 + 1) = 0
So the roots of the characteristic equation are i, i, –i, and –i. Two solutions to
the homogeneous differential equation are:
y1 = eix
and
y2 = e–ix
And because i and –i are repeated roots, you also have:
y3 = x eix
and
y4 = x e–ix
So:
yh = b1eix + b2e-ix + b3 x eix + b4 x e–ix
where b1 – b4 are constants. And because:
eiβx = cos βx + i sin βx
and:
e–iβx = cos βx – i sin βx
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Part II: Surveying Second and Higher Order Differential Equations
you can express the general solution to the homogeneous equation as:
yh = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x
The particular solution to the nonhomogeneous equation
Now it’s time to find a particular solution, yp, of your original nonhomogeneous
equation. In this case,
g(x) = 8 sin x – 16 cos x
You might consider trying a particular solution of the following form:
yp = A sin x + B cos x
But that particular solution isn’t linearly independent from yh, which has sin x
and cos x terms in it. (Linear independence is important to find a complete
set of solutions; see Chapter 5 for more information.) So you might consider
a solution of the form yp = A x sin x + B x cos x. But once again, this solution
isn’t linearly independent with respect to yh, which has terms in x sin x + x
cos x already. So you’re left with the following for yp:
yp = A x 2 sin x + B x 2 cos x
Substituting this solution into your nonhomogeneous equation, and collecting terms gives you:
–8 A sin x – 8 B cos x = 8 sin x – 16 cos x
So, comparing terms, you can see that:
A = –1
and:
B=2
Now you know that the particular solution is:
yp = –x 2 sin x + 2 x 2 cos x
The general solution to the nonhomogeneous equation is:
y = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x – x 2 sin x + 2 x 2 cos x
Wow, a pretty lengthy solution. Impressive!
Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations
A handy trick for finding particular
solutions in sine and cosine form
The method of undetermined coefficients (which
is discussed earlier in this chapter) is based on
the fact that when g(x) is of a certain form, you
can often guess the form of the particular solution up to arbitrary coefficients. In fact, you can
play a neat trick this way. If, for example, you
have a sixth order homogeneous differential
equation (yowza!) and know that x 2e–x sin x is a
solution, can you determine the other solutions?
That is, given that:
y1 = x 2 e–x sin x
can you find y 2 – y 6 so that all solutions are linearly independent? The short answer is this:
Yes, you can. Because x 2 e–x sin x is a solution,
so are these two:
y2 = e–x sin x
y3 = x e–x sin x
Also, because x 2 e–x sin x is a solution, so is x 2 e–x
cos x, because complex roots of the characteristic equation come in conjugate pairs. This
means that you can find the other three solutions:
y4 = e–x cos x
y5 = x e–x cos x
y6 = x2 e–x cos x
So the general solution to the homogeneous differential equation that has x 2 e–x sin x as a solution is:
y = x 2 e–x sin x + e–x sin x + x e–x sin x + e–x cos
x + x e–x cos x + x 2 e–x cos x
And you can determine all that starting with just
one solution, y1 = x 2 e–x sin x.
Solving Higher Order Equations with
Variation of Parameters
The method of undetermined coefficients, which I discuss earlier in this
chapter, is good only for certain forms of g(x). For more general differential
equations of a higher order, you can try the method of variation of parameters.
This method was first introduced in Chapter 6 for differential equations of
the second order, but here I generalize it for equations of order n.
The basics of the method
To get a good grasp on this method, imagine that you have this general linear
nonhomonegeous differential equation of order n:
dny
d n -1 y
d n-2 y
dy
+ p 1^ x h
+ . . . + p n - 1^ x h
+ p n^ x h y = g ^ x h
n - 1 + p 2^ x h
dx n
dx
dx
dx n - 2
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Part II: Surveying Second and Higher Order Differential Equations
The corresponding homogeneous differential equation is:
dny
d n -1 y
d n-2 y
dy
+ . . . + p n - 1^ x h
+ p n^ x h y = 0
n + p 1^ x h
n - 1 + p 2^ x h
dx
dx
dx
dx n - 2
If the general solution to the homogeneous differential equation is:
yh = c1y1 + c2 y2 + c3 y3 + . . . + cn yn
then the variation of parameters says to look for a particular solution of the
following form:
yp = u1(x)y1 + u2(x)y2 + u3(x)y3 + . . . + un(x)yn
where u1(x), u2(x), and so on are functions.
The method of variation of parameters for differential equations of order n
says that to find yp, you can solve this system of simultaneous equations for
u'1(x), u'2(x), and so on like this:
u'1y1 + u'2 y2 + . . . u'n yn = 0
u'1y'1 + u'2 y'2 + . . . u'n y'n = 0
u'1y(n–2)1 + u'2 y(n–2)2 + . . . u'n y(n–2)n = 0
u'1y(n–1)1 + u'2 y(n–1)2 + . . . u'n y(n–1)n = g(x)
Then you integrate u'1(x), u'2(x), and so on to find u1(x) and u2(x), which in
turn gives you yp.
Working through an example
Here’s an example that can help you get your feet wet with the method of variation of parameters. Try solving this differential equation using the method:
y(4) = 6x
To do so, you first need the solution to the homogeneous differential equation:
y(4) = 0
You can solve this equation by integration to get:
y1 = 1
y2 = x
y3 = x 2
y4 = x 3
Chapter 8: Taking On Higher Order Linear Nonhomogeneous Differential Equations
The homogeneous equation’s general solution is:
yh = c1 + c2x + c3x 2 + c4x 3
Now you insert u1(x), u2(x), and so on for the constants to find the particular
solution of the nonhomogeneous solution:
yp = u1(x) + u2(x)x + u3(x)x 2 + u4(x)x 3
Here’s where the method of variation of parameters kicks in, giving you these
simultaneous equations in u'1(x), u'2(x), u'3(x), and u'4(x) by integration:
u'1 + u'2x + u'3x 2 + u'4x 3 = 0
u'2 + u'32x + u'43x 2 = 0
u'32 + u'46x = 0
u'46 = 6x
True, this is a set of four simultaneous equations in four unknowns, but it
isn’t so difficult to solve. Why? Because the equations get progressively simpler. Starting at the bottom, for example, you can see that after you cancel
the 6 on each side, you get this:
u'4(x) = x
You can then substitute that result into the previous equation:
u'32 + u'46x = 0
to get:
u'32 + 6x 2 = 0
So:
u'3(x) = –3x 2
You can find the others the same way. Here are u'1(x), u'2(x), u'3(x), and u'4(x):
u'1 = –x 4
u'2 = 3x 3
u'3 = –3x 2
u'4 = x
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Part II: Surveying Second and Higher Order Differential Equations
Integrating these gives you:
5
u1 = - x
5
4
u 2 = 3x
4
u3 = –x 3
2
u4 = x
2
Because:
yp = u1(x)y1 + u2(x)y2 + u3(x)y3 + . . . + un(x)yn
the particular solution, yp, equals:
5
5
5
y p = - x + 3x - x 5 + x
4
2
5
Or, to even out the denominators:
5
5
5
5
y p = - 4x + 15x - 20x + 10x
20
20
20
20
So:
5
5
5
5
y p = - 4x + 15x - 20x + 10x
20
20
20
20
which gives you the long-awaited result:
5
yp= x
20
Now you know that the general solution of the nonhomogeneous equation is:
5
y h = c1 + c 2 x + c 3 x 2 + c 4 x 3 + x
20
And there you have it — you arrived at the general solution using the method
of variation of parameters. (Note: You can get the same result by simply integrating the differential equation four times.)
Part III
The Power Stuff:
Advanced
Techniques
T
In this part . . .
his part is where I help you pull out the power tools.
Here, you use series solutions, Laplace transforms,
and systems of differential equations. In addition, you
figure out how to use numerical methods to solve differential equations — this is the last-chance method, but it
rarely fails.
Chapter 9
Getting Serious with Power
Series and Ordinary Points
In This Chapter
Checking out the basics of power series
Trying the ratio test
Shifting the index value of a series
Surveying the Taylor Series
Putting your knowledge to use by solving second order equations
I
n Parts I and II of this book, I describe a variety of useful methods for solving first order, second order, and higher order differential equations. But
sometimes, those methods, as cool as they are, just won’t work. You have to
solve some differential equations (such as those involving what differential
equations experts call ordinary points) with a power series — that is, a summation of an infinite number of terms. I know this work sounds intimidating,
but believe it or not, it’s sometimes the easiest way to go. I show you what
you need to know in this chapter.
Perusing the Basics of Power Series
Power series are (often infinite) sums of terms. Here’s an example of a
common power series:
3
y=
!a
n=0
n
xn
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Part III: The Power Stuff: Advanced Techniques
In this series, an and x n are constants. The infinity sign on top of the sigma
indicates that n goes from 0 to infinity, and the sigma is the notation for a
summation. This series is shorthand for the following infinite expansion,
where the coefficients (a0, a1, a2, and so on) are constants:
y = a0 + a1x + a2x2 + a3x3 + . . .
The trouble with an infinite expansion, of course, is that it might diverge. That
is, it might become infinite as you add more and more terms. Power series
that become infinite aren’t of much help to anyone, so in this chapter, you
work only with series that stay finite — those that converge to a particular
value. A power series is said to converge for a particular x if this limit is finite:
m
lim
!a
n"3 n=0
n
xn
If this limit is infinite, the series doesn’t converge. In fact, your series might
also converge absolutely. A series is said to converge absolutely if the summation of the absolute values of its terms converges (note the use of absolute
value notation):
3
y=
!a
n
xn
n=0
If a series converges absolutely, it also converges (of course).
Determining Whether a Power Series
Converges with the Ratio Test
So how do you know whether a series converges? That’s an easy question:
You use the ratio test. I discuss the basics of this test and provide a few
numerical examples in the following sections.
The fundamentals of the ratio test
The ratio test compares successive terms of a series to see whether the series
will converge. If the ratio of the (n + 1)th term to the nth term is less than 1 for
a fixed value of x, the series is said to converge for that x. The series diverges
if the ratio is greater than 1. For example, suppose you had this series:
!a _x - x
3
y=
n
n=0
0
i
n
Chapter 9: Getting Serious with Power Series and Ordinary Points
The ratio of the (n + 1)th term to the nth term is:
a n + 1_ x - x 0 i
a n_ x - x 0 i
n +1
n
To find out whether the series converges or diverges, take a look at the limit
of the ratio:
a n + 1_ x - x 0 i
lim
n"3
a n_ x - x 0 i
n +1
n
This limit can also be written this way:
a n + 1_ x - x 0 i
lim
n"3
a n_ x - x 0 i
n +1
= x - x 0 lim
n
an +1
an
= x - x0 L
So the series is said to converge absolutely for a particular x if |x – x0| < 1/L.
The series diverges if |x – x0| > 1/L. And if |x – x0| = 1/L, the ratio test is
inconclusive and can’t be used.
In the large world of mathematics, there’s a number that’s either positive or
zero, called the radius of convergence, ρ, such that the series converges
absolutely if |x – x0| < ρ and diverges if |x – x0| > ρ. The region in which
|x – x0| < ρ in which the series converges is called the interval of convergence.
Plugging in some numbers
The ratio test makes a lot more sense with numbers plugged into it. So in the
following sections, I walk you through several examples. That way you can
see for yourself just how useful this test is.
Example 1
Here’s an easy example to begin with. Where, if anywhere, does the following
series converge absolutely?
! ^ -1h ^ x - 3h
3
n
n
n=0
The first step is to look at the limit of the ratio of the (n + 1)th term to the
nth term:
^ -1h
lim
n"3
^ x - 3h
n +1
^ -1h ^ x - 3 h
n
n +1
n
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This ratio works out to:
^ -1h
lim
n"3
^ x - 3h
n +1
^ -1h ^ x - 3 h
n
n +1
n
= x-3
As you can see, the ratio is |x – 3|, and that ratio must be less than one. So
the range in which the series converges absolutely is |x – 3| < 1, or 2 < x < 4.
And the series diverges if x < 2 or x > 4.
Example 2
Ready for another example? Take a look at this series:
! ^ x 4+ 1h
3
n
n
n=0
Determine the radius of convergence and the interval of convergence for this
series. To do so, first apply the ratio test, which gives you this limit:
x+1
4 n ^ x + 1h
=
lim
n
n"3
4
4 n + 1 ^ x + 1h
n +1
As you can see, this series converges absolutely for |x + 1| / 4 < 1 or |x + 1|
< 4. So, in this case, the radius of convergence is 4, and the series converges for
–5 < x < 3, which is the interval of convergence. That wasn’t so difficult, was it?
Example 3
Here’s a final example showing how to use the ratio test. Take a look at this
series:
! ^ 2xn+ 1h
3
n
2
n=0
As always, you use the ratio test, like so:
^ n + 1h
lim
n"3
n2
2
^ 2x + 1h
n +1
^ 2x + 1h
<1
n
which equals:
^ n + 1h
2x + 1 lim
n"3
n2
2
<1
Chapter 9: Getting Serious with Power Series and Ordinary Points
The limit evaluates to 1 as n → ∞, so you get this:
|2x + 1|< 1
or:
|x + 1⁄2|< 1⁄2
The radius of convergence is 1⁄2, and the interval of convergence is –1 < x < 0.
Shifting the Series Index
A method that can come in handy when working with differential equations is
called shifting the series index. For example, say that you have this series,
which starts at an index value of 3:
!a _x - x
3
n
0
i
n
n=3
If you want this series to start at n = 0 instead of n = 3, you can simply shift
the index by 3, like this:
3
!a
n+3
_ x - x 0i
n+3
n=0
Here’s another example. Say that you have this series:
! ^ n + 2 h_ x - x
3
0
i
n-2
n=2
And say that you want to shift this so only powers of n were involved, not
n – 2. You can make this shift by replacing the dummy variable n with n + 2,
giving you:
! ^ n + 4 h_ x - x
3
0
i
n
n=0
Taking a Look at the Taylor Series
You can express continuous functions (those functions that don’t take discontinuous jumps) as a series: the Taylor series. The Taylor series says that a
function can be expressed as an expansion around a point, x0, like this:
f ^xh =
3
!
n=0
f (n) _ x 0 i
_ x - x 0i
n!
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Part III: The Power Stuff: Advanced Techniques
In this series, n! is n factorial, or n · (n – 1) · (n – 2) . . . 3 · 2 · 1. If a function
has a Taylor series expansion at x = x0 with a nonzero radius of convergence
(see the earlier section “The fundamentals of the ratio test” for more about
this radius), the function is said to be analytic at x = x0.
A few types of Taylor series are especially important. Recognize this particular series?
3
n
! xn!
= ex
n=0
It’s your old friend ex!
Here’s sin x:
! ^^-21nh +x1h !
3
n
2n + 1
= sin x
n=0
And don’t forget cos x:
2n
! ^ -^12hnhx! = cos x
n=0
3
n
Solving Second Order Differential
Equations with Power Series
This section is all about tackling second order differential equations (which
I introduce in Chapters 5 and 6) with power series. Say, for example, that you
have a linear homogeneous second order differential equation like this:
P ^xh
d2y
dy
+ Q^ x h
+ R^ x h y = 0
dx
dx 2
Throughout the examples in this chapter, assume that P(x), Q(x), and R(x)
are all polynomials with no common factors. That’s the easiest type of problem to solve with power series. However, this method is also applicable when
P(x), Q(x), and R(x) are general analytic functions, such as sin x or cos x.
When working with power series, you divide the problems that fit the form
of the previous equation into two types — those where you don’t end up
dividing by zero and those where you do. In this chapter, I focus on ordinary
points. Ordinary points are points x0 where P(x0) isn’t equal to zero:
P(xo) ≠ 0
Chapter 9: Getting Serious with Power Series and Ordinary Points
Because P(x) is continuous, it follows that there’s an interval around x0 in
which P(x) isn’t 0. Because P(x) is a nonzero polynomial, you can divide by it
to get the following (but remember that P(x), Q(x), and R(x) have no common
factors):
d2y
dy
+ p^ x h
+ q^ xh y = 0
dx
dx 2
where:
p^ x h =
Q^ x h
P ^xh
q^ xh =
R^ x h
P ^xh
and:
This equation is the type that you’re going to look at in the following sections, and you’re going to use ordinary points where nothing goofy happens
(like functions suddenly going to infinity). In addition, you can impose initial
conditions on this type of equation, such as:
y(0) = c1
and
y'(0) = c2
Points where functions go to infinity are called singular points (you can take a
look at those in Chapter 10). At singular points, P(x0) = 0, and at least one of
Q(x0) and R(x0), isn’t 0. So at least one of p(x) or q(x) becomes unbounded as
x → x0. In this chapter, the functions are much better behaved.
So how do you solve second order differential equations using power series?
The most basic way is to simply substitute a power series like this one:
!a _x - x
3
n
0
i
n
n=3
into a differential equation. Then you can try to solve for the coefficients
(which will be constants in this chapter). Because p(x) and q(x) are restricted
to be polynomials in this chapter, working with power series will be even
easier as you try to figure out the solution.
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Part III: The Power Stuff: Advanced Techniques
When you already know the solution
Take a look at this second order differential equation:
d2y
+y=0
dx 2
In your eagle-eyed solving mode, you no doubt recognize this equation to be
a favorite among second order differential equations. You probably realize
that the solution is:
y = c1y1(x) + c2 y2(x)
where:
y1(x) = sin(x)
and
y2(x) = cos(x)
So you’ve solved the problem, but hold off crowing just yet, because you’re
going to tackle this problem using a power series. This problem is usually the
first one you tackle, because you already know the solution, and you already
know what sin(x) and cos(x) look like in terms of power series (see the earlier section “Taking a Look at the Taylor Series”). Here’s sin(x):
! ^^-21nh +x1h !
n
3
2n + 1
= sin x
n=0
And here’s cos x:
! ^ -^12hnhx!
n
3
2n
= cos x
n=0
You’re ready to start solving your original differential equation using a power
series of the following form:
!a _x - x
3
n
n=0
0
i
n
Chapter 9: Getting Serious with Power Series and Ordinary Points
Centering the power series on a selected point
When solving a differential equation using a power series, you first have to
select a point on which to center the power series (x0 in the previous series).
In the case of your original example equation, do what you’ll usually do
(unless there’s a compelling reason otherwise): Choose x0 to be 0. So your
series will be an expansion of the solution around the ordinary point 0:
3
!a
n
xn
n=0
Take a deep breath and then substitute this series into the original differential equation you’re trying to solve. To do so, you need to find the second
derivative of the series. Keep reading to find out what to do.
Finding the derivatives of a series
How do you find the second derivative of a series? As you may have guessed,
you work term by term. So you start off with a solution y of the following form:
3
y=
!a
n
xn
n=0
To find y", start by finding y'. Here’s what the terms of the series look like:
y = a0 + a1x + a2 x 2 + a3x 3 + . . .
So after differentiating term by term you get
y' = a1 + 2a2 x + 3a 3 x 2 + . . .
The general nth term here is:
nan x n–1
So y' equals:
3
y l= ! na n x n - 1
n =1
Note that this series starts at n = 1, not n = 0 as the series for y does, because
you took the derivative of the series.
You can find y" by differentiating y' = a1 + 2a2 x + 3a 3 x 2 + . . . . You get this result
for y":
y" = 2a2 + 6a3x + . . .
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The general term here is:
n(n – 1)an x n–2
So you can give y" as the following:
y'' = ! n^ n - 1h a n x n - 2
3
n=2
Substituting power series into the differential equation
The original differential equation looks like this:
d2y
+y=0
dx 2
So you can substitute the power series for y and y" to get this result:
! n^ n - 1h a
3
3
n
x n-2 +
n=2
!a
n
xn= 0
n=0
which is your differential equation in series form.
Ensuring the same index value
To compare the series in the differential equation, make sure they start at the
same index value, n = 0. You can shift the first series here by replacing n with
n + 2 to get this result (I explain how to shift a series index in detail earlier in
this chapter):
! ^ n + 2h^ n + 1h a
3
3
n+2
xn+
n=0
!a
n
xn= 0
n=0
When you do some simplifying, you get:
! 8^ n + 2h^ n + 1h a
3
n+2
x n + an x nB = 0
n=0
You can factor out xn as well:
! 8^ n + 2h^ n + 1h a
3
n+2
+ anB x n = 0
n=0
Believe it or not, you’re making progress here. Seriously!
Using the recurrence relation to find even coefficients
Because the series shown in the previous section equals 0, and because it
must work for all x, each term must equal 0. In other words, you get this:
(n + 2)(n + 1)an + 2 + an = 0
Chapter 9: Getting Serious with Power Series and Ordinary Points
which is called a recurrence relation. When you have a differential equation’s
recurrence relation, you practically have the solution in your pocket. Alright,
not really, but the recurrence relation goes a long way.
In this case, the recurrence relation says that, for n = 0:
(2)(1)a2 + a0 = 0
So:
a2 =
-a0
^ 2h^1h
and for n = 2:
(4)(3)a4 + a2 = 0
So:
a4 =
-a2
^ 4 h^ 3 h
Substituting a2 from the first equation into the second equation gives you this
result:
a4 =
a0
^ 4 h^ 3 h^ 2h^1h
But (4)(3)(2)(1) = 4!, so you get:
a4 =
a0
4!
Similarly, for a6:
(6)(5)a6 + a4 = 0
or:
a6 =
-a4
^ 6 h^ 5 h
Substituting for a4 in terms of a0 gives you:
a6 =
-a0
^ 6 h^ 5 h^ 4!h
But (6)(5)(4!) = 6!, so you have:
a6 =
-a0
6!
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Part III: The Power Stuff: Advanced Techniques
To summarize, you have the following:
a2 =
-a0
2!
a4 =
a0
4!
a6 =
-a0
6!
So now you can relate the even coefficients in general. If n = 2m (in which m
stands for a positive integer or zero):
^ -1h a 0
^ 2mh !
m
a n = a 2m =
m = 0, 1, 2, 3 . . .
Because you set a0 based on the initial conditions for a given problem, you
can now find the even coefficients of the solution.
Using the recurrence relation to find odd coefficients
Now you can move on to the odd coefficients. Turn back to the recurrence
relation in the previous section for the solution:
(n + 2)(n + 1)an+2 + an = 0
You can see that for n = 1 you get:
(3)(2)a3 + a1 = 0
So:
a3 =
- a1
^ 3 h^ 2h
Taking a clue from the even coefficients in the previous section, you can see
that (3)(2) = 3!, so you get this:
a3 =
- a1
3!
Now, for n = 3 in the recurrence relation, you get:
(5)(4)a5 + a3 = 0
or:
a5 =
-a3
^ 5 h^ 4 h
Chapter 9: Getting Serious with Power Series and Ordinary Points
Substituting a3 in terms of a1 into this equation gives you:
a5 =
a1
^ 5 h^ 4 h^ 3!h
a5 =
a1
5!
or:
Substituting n = 5 into the recurrence relation gives you:
(7)(6)a7 + a5 = 0
or:
a7 =
-a5
^7 h^ 6 h
And substituting a5 into this equation gives you:
a7 =
- a1
^7 h^ 6 h^ 5!h
which means that:
a7 =
- a1
7!
To sum up, the odd coefficients you have so far are:
a3 =
- a1
3!
a5 =
a1
5!
a7 =
- a1
7!
So now you can relate the odd coefficients of the solution this way in general,
if n = 2m + 1:
^ -1h a 1
^ 2m + 1h !
m
a n = a 2m + 1 =
m = 0, 1, 2, 3 . . .
Putting together the solution
Using the two equations for even and odd coefficients from the previous sections, you get the following general solution to your differential equation in
series terms:
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Part III: The Power Stuff: Advanced Techniques
3
y = a0
!
m=0
m
^ -1h x 2m + 1
^ 2m + 1h !
m
3
+ a1
^ -1h x 2m
^ 2mh !
!
m=0
That’s the solution to the differential equation in series terms. In this case,
the two series are recognizable as cos(x) and sin(x). Here’s sin(x):
3
!
n=0
^ -1h x 2n + 1
= sin x
^ 2n + 1h !
n
And here’s cos x:
3
!
n=0
^ -1h x 2n
= cos x
^ 2nh !
n
So you can rewrite the solution as:
y = a0 cos(x) + a1 sin(x)
The terms a0 and a1 are arbitrary constants (just like c0 and c1), and they’re
set by matching the initial conditions.
When you don’t know the
solution beforehand
How do you use a power series to solve a differential equation for which you
don’t already know the solution? For instance, what if you have an equation
like this:
d2y
dy
-x
+ 2y = 0
dx
dx 2
Give it a try, using a power series like this:
3
y=
!a
n
xn
n=0
The following sections will guide you through the process.
Differentiating and substituting power series
into the differential equation
Differentiating the power series from the previous section gives you the following equation:
3
y l= ! na n x n - 1
n =1
Chapter 9: Getting Serious with Power Series and Ordinary Points
And differentiating this gives you:
y'' = ! n^ n - 1h a n x n - 2
3
n=2
Now for the fun part: substitution! Substituting these equations into the original differential equation gives you this result:
! n^ n - 1h a
3
n
x n-2
n=2
3
- x ! na n x n - 1
n =1
3
+2 !an x n = 0
n=0
Using the recurrence relation to find coefficients
Now it’s time to equate powers of x on the left side of the equation to 0 on the
right side. Combining the coefficients for powers of x gives you this equation:
[2a2 + 2a0] + [6a3 + a1] x + 12a4 x 2 + [20a5 – a3] x 3 . . .
+ [(n + 2)(n + 1) an+2 – (n – 2)an ] x n = 0
Note that every power of x must be 0 for the equation to work, so the last term
in the previous equation gives you the recurrence relation for this solution:
(n + 2)(n + 1) an+2 – (n – 2)an = 0
After some simplifying, you get:
an+2 =
^ n - 2h a n
^ n + 2h^ n + 1h
That doesn’t look so bad. This equation gives you these values for the coefficients when you plug in for n:
a2 = –a0
a 3 = - 1 a1
6
a4 = 0
a5 = 1 a3
20
Substituting a3 in terms of a1 into the previous equation gives you the
following:
a 5 = 1 a 3 = - 1 a1
20
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Part III: The Power Stuff: Advanced Techniques
Here’s a6:
a6 = 2 a4
30
And substituting a4 into this equation gives you:
a 6 = 2 a 4 = 2 ^ 0h = 0
30
30
Keep going! Here’s a7:
a7 = 3 a5
42
Or, to simplify:
a7 = 1 a5
14
Substituting a5 in terms of a1 into this equation gives you:
1 ^ -1h
a7 = 1 a5 =
a
14
14 120 1
So:
a 7 = -1 a 1
1, 680
How about a8? Here’s what you get:
a8 = 4 a6
56
But you know that a6 = 0, so:
a8 = 4 a6 = 0
56
In fact, because a4 = 0, all subsequent even coefficients must be 0 by the
recurrence relation:
an+2 =
^ n - 2h a n
^ n + 2h^ n + 1h
Putting together the solution
Because all the even coefficients are 0 beyond a2, life is a little easier. You just
need to plug in the odd coefficients to get the entire solution to the differential equation. Here’s what you have for the solution:
y = a 0 + a1 x - a 0 x 2
- 1 a1 x 3 - 1 a1 x 5 - 1 a1 x 7 + . . .
6
120
1, 680
Chapter 9: Getting Serious with Power Series and Ordinary Points
Grouping terms together by their coefficients gives you:
y = a 0 ^1 - x h +
a 1d x - 1 a 1 x 3 - 1 a 1 x 5 - 1 a 1 x 7 + . . .n
6
120
1, 680
If you define the following:
y1 = (1 – x 2)
and:
y 2 = d x - 1 a 1 x 3 - 1 a 1 x 5 - 1 a 1 x 7 + . . .n
6
120
1, 680
then you can see that the general solution is:
y = a0 y1 + a1y2
A famous problem: Airy’s equation
Before you wrap up the chapter, take a look at one more differential equation.
Get ready, though. It’s a famous one! In fact, it even has its own name —
Airy’s equation. Here’s what it looks like:
y" – xy = 0
To solve this equation, you can assume, as you do earlier in this chapter, that
the solution is of the following form:
3
y=
!a
n
xn
n=0
I explain the rest of the process in the following sections.
Differentiating and substituting power series
into the differential equation
As always, be sure to start off by differentiating the equation. Doing so
gives you:
3
y l= ! na n x n - 1
n =1
After differentiating this equation one more time, you get:
y'' = ! n^ n - 1h a n x n - 2
3
n=2
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Part III: The Power Stuff: Advanced Techniques
Who was Airy?
Sir George Biddell Airy (1801–1892) was an
English mathematician and astronomer. In fact,
he was the Astronomer Royal from 1835 to 1881.
He was a man of many achievements. For
instance, he found planetary orbits, measured
the Earth’s density, and came up with a solution
of two-dimensional problems in mechanics.
He’s also the one responsible for establishing
Greenwich in Britain at the location of the prime
meridian.
Plugging the derivatives into the original differential equation gives you this
result:
! n^ n - 1h a
3
3
n
x n-2 - x !an x n = 0
n=2
n=0
With some simplifying you get:
! n^ n - 1h a
3
3
n
x n-2 -
n=2
!a
n
x n +1 = 0
n=0
You can also write this equation as:
! n^ n - 1h a
3
3
n
x n-2 -
n=2
!a
n
x n +1
n=0
Ensuring the same index value
Now you need to compare the coefficients of equal powers of x on the two
sides of this equation to find the recurrence relationship. To compare those
coefficients, it helps to make sure that the powers of x are the same in each
series.
Compare the coefficients in terms of x n. Shifting the series on the right (as I
show you earlier in this chapter) gives you this result:
! n^ n - 1h a
3
3
n
x n - 2 - ! an -1 x n
n=2
n =1
You can also shift the series on the left by substituting n + 2 for n, which
gives you:
! ^ n + 2h^ n + 1h a
3
n=0
3
n+2
x n = ! an -1 x n
n =1
Chapter 9: Getting Serious with Power Series and Ordinary Points
Using the recurrence relation to find coefficients
Now you’re ready to compare coefficients of powers of x. Note from the previous section that the series on the left starts at n = 0 and the series on the
right starts at n = 1; to compare the terms, it’s easier if the series starts at the
same value, say n = 1. You can break the first term out of the series on the left
to make the initial values of the indexes of the two series match, like this:
! ^ n + 2h^ n + 1h a
3
2a 2 +
n =1
3
n+2
x n = ! an -1 x n
n =1
Note that the left side of this equation has a constant term, 2a2, but the right
side doesn’t. So right off the bat you know that a2 = 0.
Because the two sides of this equation must be equal in their terms, you get
the following for the recurrence relation:
(n + 2)(n + 1)an + 2 = an – 1
n = 1, 2, 3, 4 . . .
The coefficients are determined in steps of three because if you know an – 1,
you can get an + 2. That is, when you have a0, you know you can get a3, a6, a9,
and so on. And when you have a1, you can get a4, a7, a10, and so on. Because
you already know that a2 = 0, you know that a5 = 0, a8 = 0, a11 = 0, and so on.
Start with the sequence a0, a3, a6, a9, and so on. And remember that a0 and a1
are set by the initial conditions. This means that you should start with a3.
Here’s a3 in terms of a0:
a0
^ 2h^ 3 h
a3 =
Here’s a6:
a6 =
a3
^ 5 h^ 6 h
Substituting a3 in terms of a0 into this equation gives you:
a6 =
a0
^ 5 h^ 6 h^ 2h^ 3 h
Here’s a9:
a9 =
a6
^ 8 h^ 9 h
Substituting a6 into this equation gives you:
a9 =
a0
^ 8 h^ 9 h^ 5 h^ 6 h^ 2h^ 3 h
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Part III: The Power Stuff: Advanced Techniques
So:
a0
^ 2h^ 3 h^ 5 h^ 6 h . . . ^ 3n - 4 h^ 3n - 3 h^ 3n - 1h^ 3nh
a 3n =
n = 1, 2, 3 . . .
Now look at the sequence a1, a4, a7, a10, and so on. Because a1 is set by the initial conditions, you should start with a4:
a1
^ 3 h^ 4 h
a4 =
Here’s a7:
a7 =
a4
^ 6 h^7 h
Substituting a4 in terms of a1 gives you:
a7 =
a1
^ 6 h^7 h^ 3 h^ 4 h
And here’s a10:
a 10 =
a7
^ 9 h^10h
Substituting a7 into this equation gives you:
a 10 =
a1
^ 9 h^10h^ 6 h^7 h^ 3 h^ 4 h
So you can say that in general:
a 3n + 1 =
a1
^ 3 h^ 4 h^ 6 h^7 h . . . ^ 3n - 3 h^ 3n - 2h^ 3nh^ 3n + 1h
n = 1, 2, 3 . . .
Putting together the solution
After you get through with all the previous steps, you’ll know that you can
write the general solution to Airy’s equation like this:
R
V
3
6
x 3n
W+
y = a 0 S1 + x + x + . . .
6
180
S
^ 2h^ 3 h^ 5 h^ 6 h . . . ^ 3n - 4 h^ 3n - 3 h^ 3n - 1h^ 3nh W
R T 4
V X
7
3n + 1
x
x
x
S
W
a1 x +
+
+...
12 504
S
^ 3 h^ 4 h^ 6 h^7 h . . . ^ 3n - 3 h^ 3n - 2h^ 3nh^ 3n + 1h W
T
X
You can write this in the following shorter form:
R
V
3
x 3n
W+
y = a o S1 + !
S
n = 0 ^ 2 h^ 3 h^ 5 h^ 6 h . . . ^ 3n - 4 h^ 3n - 3 h^ 3n - 1h^ 3n h W
X
R3 T
V
3n + 1
x
S
W
a1 !
S n = 0 ^ 3 h^ 4 h^ 6 h^7 h . . . ^ 3n - 3 h^ 3n - 2h^ 3nh^ 3n + 1h W
T
X
Chapter 9: Getting Serious with Power Series and Ordinary Points
So if you write:
3
y1= 1 +
!
n=0
x 3n
^ 2h^ 3 h^ 5 h^ 6 h . . . ^ 3n - 4 h^ 3n - 3 h^ 3n - 1h^ 3nh
and:
3
y2=
!
n=0
x 3n + 1
^ 3 h^ 4 h^ 6 h^7 h . . . ^ 3n - 3 h^ 3n - 2h^ 3nh^ 3n + 1h
you can see that the general solution to Airy’s equation is:
y = a0 y1 + a1 y2
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Chapter 10
Powering through Singular Points
In This Chapter
Perusing singular points
Examining Euler equations
Surveying series solutions near regular singular points
I
n Chapter 9, I briefly introduce you to the concept of singular points, but
this chapter is where the real action is. I cover a number of topics, including working with Euler equations, handling power series solutions near singular points, and dealing with a mix of the two — series solutions to Euler
equations near singular points.
Pointing Out the Basics of
Singular Points
In this chapter, you work with second order homogeneous differential equations of the following form:
P ^xh
d2 y
dy
+ Q^ x h
+ R^ x h y = 0
dx
dx 2
where P(x), Q(x), and R(x) have no common factors.
You can also divide each term in the equation by P(x) to get:
d2y
dy
+ p^ x h
+ q^ xh y = 0
dx
dx 2
where p(x) = Q(x)/P(x) and q(x) = R(x)/P(x).
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Part III: The Power Stuff: Advanced Techniques
So far, so good, right? Not so fast! Now I’m going to throw singular points into
the mix. Points where functions go to infinity are called singular points. At singular points, P(x0) = 0, and at least one of Q(x0) and R(x0) isn’t zero. So at least
one of p(x) or q(x) becomes unbounded (goes to infinity) as x → x0.
In the following sections, I introduce you to the fundamentals of singular
points, including how to find them, how they behave, and the difference
between regular and irregular points.
Finding singular points
Time to get your feet wet! What are the singular points of the following differential equation?
_4 - x 2i
d2y
dy
+ x3
+ ^1 + x h y = 0
dx
dx 2
The singular points are where P(x) = 0, so you have:
(4 – x 2) = 0
Simply use your excellent algebra skills to solve this equation, and you find
that the singular points are x = ±2.
As another example, determine the singular points of this differential equation:
x2
d2y
dy
+ _8 - x 3i
+ ^1 - 9x h y = 0
dx
dx 2
Here, P(x) is simply x 2. So the singular point is x = 0.
The behavior of singular points
When studied closely, singular points look like they have the potential to
become unruly. After all, singular points are where a solution may go to infinity or change rapidly in magnitude. If you’re asking whether you can simply
ignore them, the answer is no.
Why? Because solutions to differential equations with singular points vary so
much near those singular points that you need to take special care. In fact,
it’s often the case that the solution to a differential equation is the most interesting around its singular points. That’s where the most interesting physics
goes on. For example, an electrical circuit may reach resonance there. And
reaching resonance is often the whole point of amplifying circuits, so ignoring
the behavior at singular points just wouldn’t do.
Chapter 10: Powering through Singular Points
For example, take a look at this second order homogeneous equation:
x2
d2y
- 2y = 0
dx 2
By using your unbeatable differential equation solving skills, you can tell that
the two independent solutions to this differential equation are:
y1 = x 2
and
y2 = x –1
The y1 solution is fine — its behavior is well-defined around x = 0, for example. In fact, the y1 solution is still analytic (meaning that it has a working
Taylor expansion; see Chapter 9 for more about Taylor series) as x → 0, even
though the differential equation looks like this if you divide by P(x):
d 2 y 2y
- 2 =0
dx 2
x
This equation looks unbounded as x → 0, but everything is okay if you substitute y1 = x 2 here.
The situation is different with y2 = x–1 (also written as 1/x), however. This solution isn’t analytic at x = 0 (in other words, it has no Taylor expansion at x = 0).
So you can’t use the methods of Chapter 9 to solve this differential equation.
As you find out in the next section (and throughout this chapter), different
kinds of singular points exist — and all of them have varying levels of manageability. It is often the case that you can’t use the Taylor expansion used in
Chapter 9 to solve differential equations with singular points, because those
expansions become unbounded at the singular points. Because of this, you
have to use more general expansions from time to time.
Regular versus irregular singular points
An important concept when it comes to singular points is severity. A singular
point’s severity is an indication of how strong the singular point is — how
strongly it tends toward infinity. The severity of a singular point has a lot to
do with how you handle the solution; does the solution, for instance, include
terms like 1/x, or does it include terms like 1/x8? (It turns out that you can
handle 1/x in most cases, but 1/x8? Good luck with that one.) The idea is to
extend the techniques of Chapter 9, which allow you to use series expansions
near ordinary points, to help you use series expansions near singular points —
if they’re well-behaved enough.
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Part III: The Power Stuff: Advanced Techniques
A well-behaved singular point is called a regular singular point. Regular singular points are well-behaved enough that you can handle them using the techniques in this chapter.
What’s the definition of a regular singular point? They’re defined in terms of
the ratio Q(x)/P(x) and R(x)/P(x), where P(x), Q(x), R(x) are the polynomial
coefficients in the differential equation that you’re trying to solve:
P ^xh
d2y
dy
+ Q^ x h
+ R^ x h y = 0
dx
dx 2
In order for x 0 to be a regular singular point of this equation, these two relations have to be true:
Q^ x h
remains finite
P ^xh
lim _ x - x 0 i
x " x0
and
lim _ x - x 0 i
2
x " x0
R^ x h
remains finite
P ^xh
Another way of thinking about this is to say that x0 is a regular singular point
if (x – x0)Q(x)/P(x) and (x – x0)2Q(x)/P(x) have Taylor expansions (that is,
they’re analytic) around x0.
As you may have guessed, if singular points aren’t regular, they’re irregular.
And irregular singular points are much more — shall I say interesting? — to
handle.
Take a look at some example differential equations to see if their singular
points are regular or irregular.
Example 1
Start with this differential equation (which you just saw in the previous
section):
d 2 y 2y
- 2 =0
dx 2
x
The solutions for this equation are:
y1 = x 2
and
y2 = x–1
Chapter 10: Powering through Singular Points
There’s a singular point at x = 0 because y2 becomes infinite. Now take some
time to evaluate the relations:
Q^ x h
remains finite
P ^xh
lim _ x - x 0 i
x " x0
and
lim _ x - x 0 i
2
x " x0
R^ x h
remains finite
P ^xh
You test the first relation by plugging in some numbers, like so:
0
lim
^xh = 0
x"0
1
And because 0 is finite, you’re okay so far. Now for the second relation,
R(x) = –2/x 2 and P(x) = 1:
2 -2
= -2
lim
^xh
x"0
x2
Because –2 is finite, x = 0 is a regular singular point of your original differential equation. Cool, huh?
Example 2
Now try this one:
^ x - 4h x
2
d2y
dy
+ 4x
+ ^ x - 4h y = 0
dx
dx 2
What are the singular points in this equation? Are they regular? Here, P(x) =
(x – 4)2 x, and that’s 0 at x = 0 and x = 4. So x = 0 and x = 4 are your singular
points. Start with the x = 0 singular point first; here’s what the first relation
has to say about it:
lim
^xh
x"0
4x
=1
2
^ x - 4h x 4
Because 1⁄4 is finite, you’re good to go. Now for the second relation:
lim
^xh
x"0
2
^ x - 4h
^ x - 4h x
2
=0
Zero is finite, so the x = 0 singular point is a regular singular point.
So far, so good. Now, however, you have to try the other singular point, x = 4.
The first relation gives you:
lim
^ x - 4h
x"4
4x
2
^ x - 4h x
" ^ x -4 4h
is NOT FINITE
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Part III: The Power Stuff: Advanced Techniques
Whoops — this relation is unbounded as x → 4, so x = 4 isn’t a regular singular point. In other words, it’s an irregular singular point.
Example 3
Try your hand at this equation, which is a famous one, the Legendre equation
(see the nearby sidebar “Discovering the legacy of Legendre” for more on the
man who unearthed this famous equation):
_1 - x 2 i
d2y
dy
- 2x
+ α ^ α + 1h y = 0
dx
dx 2
where α is a constant. Because P(x) = (1 – x 2), the singular points are x = ±1.
Look at the x = 1 point first. From the first relation and because (1 + x)(1 – x) =
(1 – x 2):
lim
^ x - 1h
x "1
2x
_1 - x 2 i
= - 2x = -1
^1 + x h
As you know, –1 is, of course, finite.
Now for the second relation:
lim
^ x - 1h
x "1
α ^ α + 1h
_1 - x 2 i
2
This equation can be broken down to:
lim
^ x - 1h
x "1
α ^ α + 1h
^1 - x h^1 + x h
2
which is:
lim
- ^ x - 1h
x "1
α ^ α + 1h
=0
^1 + x h
Because 0 is finite, x = 1 is a regular singular point.
Now try the other singular point, x = –1. Here’s what the first relation gives you:
lim ^ x + 1h
2x
_1 - x 2 i
x " -1
=
2x = -1
^1 - x h
As you know, –1 is finite. Now check out the second relation, which gives you:
lim ^ x + 1h
x " -1
2
α ^ α + 1h
_1 - x 2 i
Chapter 10: Powering through Singular Points
Discovering the legacy of Legendre
Adrien-Marie Legendre (1752–1833) was the
French mathematician who discovered the everimportant Legendre equation. His work was done
in the fields of statistics, abstract algebra, mathematical analysis, and number theory. He was
born to a rich family, and studied physics in Paris.
Then he took up teaching at a military academy.
He took on this job because he enjoyed it, not
because he had to. He started publishing works
on physics — specifically on the motion of
cannonballs — but then moved on to math. In
1782, he became a member of the French
Academy of Sciences. He even got a crater on
the moon named after him. How many mathematicians can say that?
You can also write this as:
lim ^ x + 1h
x " -1
2
α ^ α + 1h
^1 - x h^1 + x h
or:
lim ^ x + 1h
x " -1
α ^ α + 1h
^1 - x h
The limit is:
lim ^ x + 1h
x " -1
α ^ α + 1h
=0
^1 - x h
Because 0 is finite, x = –1 is also a regular singular point.
Exploring Exciting Euler Equations
A good way to understand how to handle regular singular points is to see
how one of the most famous differential equations — which has a regular singular point — is handled. The equation I’m talking about is Euler’s equation:
x2
d2y
dy
+ αx
+ βy= 0
dx
dx 2
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Part III: The Power Stuff: Advanced Techniques
Here, α and β are real constants. If you assume that the solution to this equation has the following form:
y = xr
(where r is a constant), substituting the solution into the equation gives you:
[r(r – 1) + αr + β] x r = 0
Dividing both sides by x r gives you:
r(r – 1) + αr + β = 0
or:
r 2 – r + αr + β = 0
So:
r 2 + (α – 1) r + β = 0
The roots, r1 and r2, of this equation are:
- ^ α - 1h ! ^ α - 1h - 4β
2
2
r1, r 2 =
Because you’re considering the general Euler’s equation, you don’t know
what α and β are, so you have to consider three cases for these roots:
r1 and r2 are real and distinct
r1 and r2 are real and equal
r1 and r2 are complex conjugates
In the following section, you consider each of these cases separately.
Euler was a pretty busy guy; he also came up with Euler’s method, which is a
simple numerical method of solving differential equations. Flip to Chapter 4
for details.
Real and distinct roots
If the roots of r 2 + (α – 1) r + β = 0 are real and distinct, r1 ≠ r2. The general
solution to Euler’s equation is:
y = c1 x r + c 2 x r
1
2
Chapter 10: Powering through Singular Points
To see how to come up with this solution, try solving this differential equation:
4x 2
d2y
dy
+ 6x
- 2y = 0
dx
dx 2
This equation has a regular singular point at x = 0, and you can assume that
the solution is of the following form:
y = xr
Substituting this solution into the original equation gives you:
[4r (r – 1) + 6r – 2] x r = 0
or:
[4r (r – 1) + 6r – 2] = 0
After some simplifying, the equation looks like this:
4r 2 + 2r – 2 = 0
Factoring gives you:
2(2r – 1)(r + 1) = 0
So the roots are:
r1 = 1⁄2
and
r2 = –1
This means that the general solution to the original equation is
y = c1 x ⁄ + c2 x–1
1
2
Note how this technique is similar to the one in the situation where the differential equation you’re dealing with has constant coefficients (as in Chapter 5)
and you assume a solution of the form y = e r x. Here, there are powers of x 2 and
x already appearing in the coefficients of the differential equation, so you
assume that the solution is of the form y = x r.
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Real and equal roots
Sometimes the two roots of r 2 + (α – 1) r + β = 0 are equal. For example, take a
look at this differential equation:
x2
d2y
dy
+ 3x
+ 2y = 0
dx
dx 2
Substituting y = x r into this equation gives you:
[r (r – 1) + 3 r + 2] x r = 0
or:
r (r – 1) + 3 r + 2 = 0
After some simplifying, this becomes:
r2 + 2 r + 2 = 0
which in turn becomes:
(r + 1)(r + 1) = 0
So the roots here are –1 and –1. How do you handle a case like this? Clearly
y1 = x –1, but what’s y2?
As you find out in Chapter 5, when you have a second order differential equation with constant coefficients, you try a solution of the following form:
y = erx
After you substitute this solution into the differential equation, you may find
that the two roots, r1 and r2, are equal. In that case, you end up with these two
solutions:
y1 = er
1
x
and:
y 2 = x er
1
x
Is there an analog for Euler equations? You bet. If r1 = r2, then:
y1= x r
1
And in that case:
y 2 = ln^ x h x r
1
Chapter 10: Powering through Singular Points
So the general solution of Euler equations with equal roots of the characteristic equation is:
y = c 1 x r + c 2 ln^ x h x r
1
1
And that’s it. The solution to your original equation is:
y = c1 x–1 + c2 ln(x) x–1
Complex roots
In this section, you take a look at the case where the roots of a Euler equation
are complex and of the form a ± ib. In this case, a and b are constants, and i is
the square root of –1. Your solutions look like this:
y1 = x a+ib
and:
y2 = x a–ib
To deal with complex roots, you have to get into some trigonometry. Start by
noting that this is true:
x r = erln(x)
And note that as long as x > 0:
x a+ib = e(a+ib)ln(x)
x>0
and this equals:
e(a+ib)ln(x) = ealn(x)eibln(x)
x>0
which is:
ealnxeiblnx = xaeiblnx
x>0
Whew! At this point, you can use the following relation:
emx = cos mx + i sin mx
x>0
Here, m is a constant. So after using the relation, your equation becomes:
x aeiblnx = x a[cos(bln(x)) + i sin(bln(x))]
x>0
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Part III: The Power Stuff: Advanced Techniques
Absorbing the pesky factor of i into c2 gives you the general solution, which
looks like this:
y = c1 xacos(bln(x)) + c2 x asin(bln(x))
x>0
Now try putting all this trig to work with an example. Try solving this Euler
equation, where all the coefficients are 1:
x2
d2y
dy
+x
+y=0
dx
dx 2
Trying a solution of the following form:
y = xr
gives you:
[r (r – 1) + r + 1] x r = 0
or:
r (r – 1) + r + 1 = 0
After some simplifying, this equation becomes:
r2 + 1 = 0
Uh oh! The roots here are complex: ±i. This means that a = 0 and b = 1, so
you get:
y = c1 cos(ln(x)) + c2 sin(ln(x))
x>0
This equation is only valid for x > 0. Let me guess: You’re wondering whether
you can do anything for x < 0. Well, actually, it can be shown that this is the
way to handle the regions x < 0 and x > 0 (not x = 0):
y = c1 cos(ln |x|) + c2 sin(ln |x|)
x>0
Putting it all together with a theorem
The following formal theorem summarizes the previous sections:
If you have a Euler equation:
x2
d2 y
dy
+αx
+βy=0
dx
dx 2
Chapter 10: Powering through Singular Points
where α and β are real constants, then the solution is of this fundamental
form in any interval that doesn’t include the origin:
y = xr
where you can find r by solving the characteristic equation:
r (r + 1) + α r + β = 0
If the roots are real and distinct, then the general solution is of this form:
y = c1 x
r1
+ c2 x
r2
If the roots are real and equal, then the general solution is of this form:
y = c1 x
r1
+ c 2 ln x x
r1
And if the roots are complex, a ± ib, then the general solution is of this
form:
y = c1 x acos(b ln x ) + c2 x asin(b ln x )
Figuring Series Solutions Near
Regular Singular Points
In the following sections, you take a look at how to find the series solution
near regular singular points. Have fun!
Identifying the general solution
Take a look at the general second order differential equation that you want to
solve:
P ^xh
d2y
dy
+ Q^ x h
+ R^ x h y = 0
dx
dx 2
You can assume that the solution has a regular singular point. For convenience, you can also assume that the singular point is at x0 = 0. (If the singular
point isn’t at zero, you can make a substitution of variables, x → x – c, where
c is a constant, so that it is).
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Divide this equation by P(x) to get:
d2y
dy
+ p^ x h
+ q^ xh y = 0
dx
dx 2
where:
p^ x h =
Q^ x h
P ^xh
q^ xh =
R^ x h
P ^xh
and:
From here, with a few tricks up your sleeve, you can decipher a general solution near singular points, as you find out in the following sections.
Seeing the series of products
Because x = 0 is a regular singular point, it follows that x p(x) and x 2 q(x) both
have finite limits as x → 0. This means that both products have convergent
series of the following form (see Chapter 9 for an introduction to power series):
x p^ x h =
3
!p
n
xn
n=0
and:
x 2 q^ xh =
3
!q
n
xn
n=0
Both series converge for x < a for some interval, a > 0.
Substituting the series into the differential equation
Now multiply the original differential equation by x 2 to get the following
equation:
x2
d2y
dy
+ x 8 x p^ x hB
+ x 2 q^ xh y = 0
dx
dx 2
Substituting the two series from the previous section into this equation gives
you this result:
x2
d2y
dx 2
+ x 7 p 0 + p 1 x + p 2 x 2 + . . . + p n x n + . . .A
dy
dx
+ 7 q 0 + q 1 x + q 2 x 2 + . . . + q n x n + . . .A y = 0
Chapter 10: Powering through Singular Points
Recognizing a Euler equation
The previous equation looks a little difficult. But don’t worry. Here’s what to
do: Try tackling problems of this kind by assuming that all coefficients except
p0 and q0 are equal to 0. Doing so in the case of the example gives you:
x2
d2y
dy
+ x p0
+ q0 y = 0
dx
dx 2
Does this equation look familiar? It should — it’s a Euler equation, like the
ones I discuss earlier in this chapter. The fact that this looks like a Euler
equation helps you see how to tackle the more general differential equation
with a regular singular point.
Here’s the key: If not all the coefficients (besides p0 and q0) are equal to 0, you
have to assume that the solution is of this Euler-like form:
3
y = xr !qn x n
n=0
This series is the same as this one:
3
y=
!q
n
x n+r
n=0
The fundamental solution is a Euler solution, with the power series added in
to take care of any non-Euler coefficients. In other words, even if your differential equation has a regular singular point, you can sometimes find a valid
solution of the form of this power series in the region of the singular point.
The basics of solving equations
near singular points
Before I get to a numerical example, I want to walk you through the steps of
solving an equation near singular points. The following sections show you
everything you need to know. Fasten your seat belt!
Determining the solution’s form and differentiating
Say you start with a differential equation of the following form:
x2
d2y
dy
- x 8 x p^ x hB
+ 8 x 2 q ^ x hB y = 0
dx
dx 2
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Part III: The Power Stuff: Advanced Techniques
where:
x p^ x h =
3
!p
n
xn
n=0
and:
x 2 q^ xh =
3
!q
n
xn
n=0
Here’s the Euler equation that matches the differential equation:
x2
d2y
dy
- p0 x
+ q0 y = 0
dx
dx 2
As you find out in the earlier section “Recognizing a Euler equation,” you can
assume that a solution to this equation is of the following form:
3
y = xr !qn x n
n=0
This solution is the same as the following:
3
y=
!q
n
x n+r
n=0
Differentiating this series gives you:
3
dy
= ! a n ^ r + nh x n + r - 1
dx n = 0
And after differentiating again you get:
3
d2y
n+r -2
2 = ! a n ^ r + n h^ r + n - 1h x
dx
n=0
Substituting series into the original equation
Now it’s time for the heavy-lifting. Substituting the previous three series
into the original differential equation and collecting terms gives you this form
of the differential equation:
a 0 f ^r h x r +
!= f ^n + r h a
n -1
3
n =1
where:
f(r) = r(r + 1) + p0r + q0
and
p 0 = lim
x p^ x h
x"0
n
+
!a
m=0
m
8^ m + r h p n - m + q n - m BG x n + r = 0
Chapter 10: Powering through Singular Points
and
q 0 = lim x 2 q ^ x h
x"0
Wow. What do you do next? Read on to find out.
Finding the indicial equation
You know that in order for the terms on the left side of the previous equation
to equal zero, every power of x must equal zero. In particular, note that this
means:
a0 f(r) x r = 0
Because a0 and x r aren’t zero, you know that:
f(r) = 0
When this equation is expanded, you get the following:
r(r + 1) + p0r + q0 = 0
This equation is the indicial equation for the original differential equation;
in other words, it’s the same characteristic equation for the roots you’d get if
you solved the differential equation’s corresponding Euler equation.
Working with the roots
You use the two roots of the indicial equation to find the two solutions to the
differential equation, y1 and y2.
Next, you can set the coefficients of x n+r to zero in the previous section’s
equation. Doing so gives you this relation:
f ^n + r h a n +
n -1
!a
m
8^ m + r h p n - m + q n - m B = 0 n $ 1
m=0
which gives you the following recurrence relation (see Chapter 9 for details):
- ! a m8^ m + r h p n - m + q n - m B
n -1
an =
m=0
f ^n + r h
n$1
You can then find the coefficients, an, from this recurrence relation.
There are two solutions to the original differential equation, each of which
corresponds to the two roots, r1 and r2. Here’s the first solution, y1:
y 1 = x r <1 +
1
! a _r i x
3
n
n =1
1
n
F x!0
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Part III: The Power Stuff: Advanced Techniques
So how do you go about finding the second solution, y2? Well, how you do
that depends on the two roots of the indicial equation. (See the later section
“Taking a closer look at indicial equations” for details.)
A numerical example of solving an
equation near singular points
The stuff in the previous sections can be tough, so take a look at a numerical
example to make it all clear in your mind.
Uncovering the solution’s form
To start, take a look at this differential equation:
2x 2
d2y
dy
-x
+ ^1 + x h y = 0
dx
dx 2
This equation has a regular singular point at x = 0. How would you find a solution to this equation? Well, after you give it some thought, it looks a lot like
this Euler equation:
2x 2
d2y
dy
-x
+y=0
dx
dx 2
So you can try a solution of the following form:
3
y=
!a
n
x n+r
n=0
Differentiating this solution gives you:
3
dy
= ! a ^ r + nh x n + r - 1
dx n = 0 n
After differentiating again, you get:
3
d2y
n+r -2
2 = ! a n ^ r + n h^ r + n - 1h x
dx
n=0
Using series as substitutes in the original equation
Substituting the previous three series into your original differential equation
gives you this impressive result:
Chapter 10: Powering through Singular Points
2 ! a n ^ r + nh^ r + n - 1h x n + r - 2
3
n=0
- ! a n ^ r + nh x n + r
3
n=0
3
+ !an x n+r
n=0
3
+ ! an x n + r +1 = 0
n=0
This equation can also be written as:
a 0 8 2r ^ r - 1h - r + 1B x r
+ ! b 8 2 ^ r + nh^ r + n - 1h - ^ r + nh + 1B a n + a n - 1 l x r + n = 0
3
n =1
Wow. Can you decipher this? Keep reading to find out!
Unearthing the indicial equation
Because the coefficients of all powers of x in the previous equation must be
zero for the whole equation to be zero, you can set the coefficient of x r to
be zero, like so:
2r(r – 1) – r + 1 = 0
Multiplying this out gives you the following equation:
2r 2 – 3r + 1 = 0
This turns out to be the same characteristic equation you would have
received from the Euler equation that’s closest to the equation that you’re
trying to solve, namely:
2x 2
d2y
dy
-x
+y=0
dx
dx 2
You can factor 2r 2 – 3r + 1 = 0 into the following indicial equation:
(r – 1)(2r – 1) = 0
As you can see, the roots of this equation are:
r1 = 1
and
r2 = 1⁄2
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These two roots are given a big name: exponents at the singularity. They’re
given this name because they describe the behavior of the solution near the
singular point, x = 0. You put these roots to work in the next two sections.
Getting back to the original differential equation in impressive series form,
you now set the coefficients of x r+n equal to zero to get this:
[2(r + n)(r + n – 1) – (r + n) + 1]an + an–1 = 0
You can rewrite this equation as:
an =
-an -1
2 ^ r + nh - 3 ^ r + nh + 1
2
Or, with some regrouping you get:
an =
-an -1
8 ^ r + nh - 1B8 2 ^ r + nh - 1B
For each root of the initial equation, r1 and r2, you can use this recurrence
relation to find the coefficients an. Keep reading to find out what to do next!
Applying the first root
Plugging the first root, r1 = 1, into the recurrence relation gives you:
an =
-an -1
n ^ 2n + 1h
So when you plug in 1 for a, you get this:
a1 =
-a0
3
And when you plug 2 in for a you get this:
a2 =
- a1
^ 5 h^ 2h
Substituting a1 into this equation gives you:
a2 =
-a0
^ 5 h^ 2h^ 3 h
Now here’s the next coefficient:
a3 =
-a2
^7 h^ 3 h
Substituting a2 into this equation gives you:
a3 =
-a0
^7 h^ 3 h^ 5 h^ 2h^ 3 h
Chapter 10: Powering through Singular Points
You can rewrite this equation as:
a3 =
-a0
^ 3 h^ 5 h^7 h^1h^ 2h^ 3 h
In general, you get this relation:
^ -1h a 0
8 ^ 3 h^ 5 h . . . ^ 2n + 1h B n!
n
an =
which means that y1 of the general solution is:
R
V
n
3
S
W
^ -1h x n
y 1 = x S1 + !
W x>0
n = 1 8 ^ 3 h^ 5 h . . . ^ 2n + 1h B n!
S
W
T
X
Here’s the million-dollar question: Does this series converge? As you may
expect, you can use the ratio test to find out (see Chapter 9 for details):
lim
n"3
an +1 x n +1
an x n
which works out to be:
lim
n"3
x
^ 2n + 3 h^ n + 1h
And as n → ∞, this limit → 0, so the series expansion is valid for all values of x.
Plugging in the second root
Now how about r2 = 1⁄2? To work with the second root, turn back to the following recurrence relation:
an =
-an -1
8 ^ r + nh - 1B8 2 ^ r + nh - 1B
Plugging in r = 1⁄2 gives you:
an =
-an -1
1
; n - E6 2n @
2
which equals:
an =
-an -1
^ 2n - 1h n
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As you may know, a0 is arbitrary and is set to match the initial conditions of
the problem. So you start by finding a1:
a1 =
-a0
^1h^1h
So:
a1 = –a0
Now for a2:
a2 =
- a1
^ 3 h^ 2h
Substituting a1 into this equation gives you the following result:
a2 =
a0
^1h^ 3 h^1h^ 2h
And for a3:
a3 =
-a2
^ 3 h^ 5 h
Substituting a2 into this equation gives you:
a3 =
-a0
^1h^ 2h^ 3 h^1h^ 3 h^ 5 h
And for a4:
a4 =
-a3
^ 4 h^7 h
Substituting a3 into this equation gives you:
a4 =
-a0
^1h^ 2h^ 3 h^ 4 h^1h^ 3 h^ 5 h^7 h
So, in general:
^ -1h a 0
n! ^1h^ 3 h^ 5 h^7 h . . . ^ 2n - 1h
n
an =
You can give the second solution, y2, like this:
R
V
n
3
S
W
^ -1h x n
1/2
y 2 = x S1 + !
W
n = 1 8 ^1h^ 3 h^ 5 h . . . ^ 2n - 1h n! B
S
W
T
X
x>0
Chapter 10: Powering through Singular Points
How about the radius of convergence? To find out, you can use the ratio test
as you did in the previous section:
lim
"3
n
an +1 x n +1
an x n
After some simplification, this equation can be rewritten as:
lim
n"3
x
^ 2n - 1h n
Congratulations! Now you know that the series converges for all x.
Taking a closer look at indicial equations
In the numerical example in the previous section, there were two real and distinct roots to the indicial equation (the equation you would get if the differential equation were an Euler equation). So you got two distinct solutions, y1
and y2. The general solution is:
y = c1 y1 + c2 y2
What happens if the indicial equation’s roots are the same — that is, real and
equal? In that case, the first solution is of the following form:
3
y1=
!a
n
x n+r
n=0
The second solution usually involves a logarithmic term. If the roots of the
indicial equation are complex, on the other hand, the solution usually
involves sines and cosines.
In the following sections, I examine a bit more closely the several types of
roots you can have in indicial equations.
Distinct roots that don’t differ by a positive integer
If r2 ≠ r1, and r1 – r2 isn’t a positive integer, the second solution, y2, is given by
the following:
y 2 = x r <1 +
2
! a _r
3
n
n =1
2
i x nF
x!0
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Because these two series:
! a _r i x
3
1+
n
n
1
n =1
and:
! a _r
3
1+
n
2
i xn
n =1
are analytic at x = 0. The behavior of the solutions at x = 0, where there’s a
regular singular point, is entirely due to these terms:
xr
1
xr
2
and
That’s why r1 and r2 are called the exponents at the singularity, as I explain
in the earlier section “Finding the indicial equation” — they determine what
happens at the singular point.
Equal roots
What if the roots r1 and r2 are equal? In that case, y2 takes the following form:
3
y 2 = y 1 ln x + x r
!b
1
n
_ r 1i x n
x>0
n =1
You have to calculate the coefficients, bn, as usual: You substitute this equation into the differential equation, collect terms, set the coefficients of x equal
to zero, and get a recurrence relationship working.
Roots that differ by a positive integer
If the roots of the indicial equation differ by a positive integer, r1 – r2 = N, things
get complex pretty quickly. And you can’t use a solution, y2, of this form:
y 2 = x r <1 +
2
! a _r
3
n
2
i x nF
x!0
n =1
Why? Well, because if r1 – r2 = N, you would start overlapping with the y1
solution:
y 1 = x r <1 +
1
! a _r i x
3
n
n =1
1
n
F
x!0
Chapter 10: Powering through Singular Points
So what does the second solution look like when r1 – r2 = N? Here’s what it
turns out to be:
y 2 = a y 1 ln x + x r <1 +
2
3
!c
n
_r 2i x n F
n =1
Here’s what the constant a looks like (which may be zero):
a = lim _ r - r 2 i a N ^ r h
r " r2
Here, aN is the Nth coefficient , where r1 – r2 = N, and
c n _ r 2 i = d 9 _ r - r 2 i a n^ r h C
dr
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Chapter 11
Working with Laplace Transforms
In This Chapter
Looking at the elements of a Laplace transform
Determining whether a Laplace transform converges
Figuring some Laplace transforms
Using Laplace transforms to solve differential equations
Getting a grip on factoring, convolutions, and step functions
T
his chapter is all about a powerful new tool for handling particularly
tough differential equations. This tool is called a Laplace transform, which
is a type of integral transform; you use it to change a differential equation
into something simpler, solve the simpler equation, and then invert the transform to recover the solution to your original differential equation. Cool, huh?
Read on for all the details.
Breaking Down a Typical
Laplace Transform
To get to know Laplace transforms, start by taking a look at what a general
integral transform looks like:
β
F ^ sh =
# K _ s, t i f ^ t h dt
α
In this case, f(t) is the function that you’re taking an integral transform of, and
F(s) is the transformed function. The limits of integration, α and β, can be
anything you choose, but the most common limits are –∞ to +∞. And here’s
the key to the transform: K(s, t) is called the kernel of the transform, and you
choose your own kernel. The idea is that choosing your own kernel gives you
a chance to simplify your differential equation more easily.
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Part III: The Power Stuff: Advanced Techniques
Who was Laplace?
Pierre-Simon Laplace (March 23, 1749–March 5,
1827) was a French mathematician and
astronomer. One of his most important contributions was his five-volume work, Mécanique
Céleste (Celestial Mechanics), in which he laid
the groundwork of modern mathematical astronomy. This work was used by Sir Isaac Newton.
Laplace was responsible for what is now called
Laplace’s equation and for the Laplace transform
(which is important to mathematical physics).
The Laplacian operator, which is a differential
operator that’s central to many areas of physics,
also is named after Laplace.
When you restrict yourself to differential equations with constant coefficients, as you do in this chapter, a useful kernel is e–st. Why? Because when
you differentiate this kernel with respect to t, you end up with powers of s,
which you can equate to the constant coefficients. Here’s what this process
looks like:
3
F ^ sh =
#e
- st
f ^ t h dt
0
Note that besides using the kernel e–st, the limits of integration in the previous
transform are from 0 to ∞ because negative values of t would make the integral diverge.
The symbol for Laplace transforms is j $ f ^ t h . , which is the Laplace transform of f(t):
3
j $ f ^ t h. = F ^ sh
#e
- st
f ^ t h dt
0
Deciding Whether a Laplace
Transform Converges
A potential difficulty exists with Laplace transforms: Sometimes an integral
won’t converge. Frequently, integrals with infinite ranges don’t converge, and
if the Laplace transform of a function won’t converge, it won’t be of any use
to you in solving differential equations.
Chapter 11: Working with Laplace Transforms
Take the following equation, for example: f(t) = et. The Laplace transform for
this equation is:
3
F ^ sh =
#e
- st
e t dt
0
So here’s what the new and improved transform looks like:
3
F ^ sh =
#e
-s
dt
0
Taking the e–s factor outside the integral leaves you with this equation:
3
F ^ sh = e - s
# dt
0
This equation is all fine and good, but integrating it gives you:
F ^ sh = t e - s
3
0
And for finite values of s, this equation tends toward infinity instead of converging, which is obviously a problem.
So how do you know if a Laplace transform exists in finite form? It’s time for a
theorem. But first, I start by defining what it means for the function f(t) to be
piecewise continuous. The function f(t) is piecewise continuous if both of the
following are true:
f(t) is continuous on each subinterval ti – 1 < t < ti, where i stands for the
number of the interval
f(t) stays finite as the endpoints of each subinterval are approached
(from inside the subinterval)
Now that you’re armed with that definition, here’s the theorem that helps you
determine whether a Laplace transform exists for your particular function:
If f(t) is piecewise continuous in the interval 0 ≤ t ≤ α for any positive α,
and | f(t) | ≤ Ceat where t ≥ K, where C, a, and K are real and positive,
then the Laplace transform F(x) = k $ f ^ t h . exists for s > a.
Calculating Basic Laplace Transforms
In this section, you can take a look at how to calculate some basic Laplace
transforms. Enjoy!
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The transform of 1
In this section, you start off by calculating just about the easiest Laplace
transform there is — the Laplace transform of 1. Here’s what that transform
looks like:
3
j "1, =
#e
- st
^1h dt
0
Integration gives you:
3
j "1, =
#
e - st ^1h dt = 1s
s>0
0
So the Laplace transform of 1 remains finite for all s > 0, and it depends on the
value you choose for s.
The transform of eat
Now move on to solving the transform of eat: f(t) = eat. Here’s what the Laplace
transform looks like:
3
j # e at - =
#e
- st
e at dt
0
With a little simplifying, this equation becomes:
3
j # e at - =
#e
- (s - a) t
dt
0
And this, in turn, becomes the following after integration:
3
j # e at - =
#
e - (s - a) t dt = s -1 a
s>a
0
Again, this result depends on the value you choose for s.
The transform of sin at
The Laplace transforms in the previous two sections don’t look so bad. How
about now trying your hand at the Laplace transform of some trig functions,
such as sin at? Here’s the Laplace transform of sin at, which I calculate in the
following sections:
3
j " sin at , =
# sin at
0
e - st dt
Chapter 11: Working with Laplace Transforms
Integrating by parts
So how do you go about tackling the integration to this tricky transform?
Well, you have to be crafty here. The best way to solve this transform is to
integrate by parts, giving you:
- st
at
j " sin at , = e cos
a
3
0
3
- as
# cos at
e at dt
0
Wow, did all that work buy you anything? Well, sure! It turns out that the first
term is 1/a (substituting in 0 for t), so this breaks down to:
3
1- s
j " sin at , = a
a
# cos at
e at dt
0
Here’s where the clever part comes in. Note that the second term has become
similar to the original integral, except that it uses cosine. Maybe if you integrate by parts again, you’ll get back to an integral that uses sine. And if that’s
the case, maybe you can actually factor j " sin at , onto the left side, which
would leave you with some reasonable expression on the right.
Integrating the previous transform by parts again gives you:
3
1 - s 2 sin at e at dt
j " sin at , = a
#
a2 0
Simplifying the result
After going through the previous section, you may be asking yourself: “Is this
just getting worse and worse?” Actually, no. Things truly are improving —
note that the second term is s2/a2 multiplied by j " sin at ,. So this transform
becomes:
1 - s 2 j " sin at ,
j " sin at , = a
a2
It turns out that you have been clever after all, because you can recast the
equation this way:
2
1
j " sin at , + s 2 j " sin at , = a
a
Or even simpler, you get:
s 2 + a 2 j " sin at , = 1
a
a2
which becomes:
j " sin at , =
a
s2 + a2
s>0
Can that be it? Yes, that’s j " sin at ,. Great work!
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Consulting a handy table for some relief
As you see in the previous sections, things can become complex pretty fast
when it comes to calculating Laplace transforms. Are you going to be expected
to jump through these kinds of hoops each time you need a Laplace transform?
If so, wouldn’t it just be easier to hit yourself on your head with a brick?
Thankfully, you usually don’t have to jump through those hoops very often.
Instead, you can use some handy tables of Laplace transforms. After all, why
reinvent the wheel? Check out Table 11-1, which is designed to save you a lot
of work.
Keep in mind that the Laplace transform of f(t) is defined this way:
3
j $ f ^ t h. = F ^ sh =
#e
- st
f ^ t h dt
0
So the Laplace transform depends on the value you choose for s.
Table 11-1
Laplace Transforms of Common Functions
Function
Laplace Transform
Restrictions
1
1
s
s>0
e at
1
s-a
n!
s n +1
s
s 2 + a2
a
s 2 + a2
s
s 2 - a2
a
s 2 - a2
s-a
_s 2 - a 2 i + b 2
s>a
b
tn
cos at
sin at
cosh at
sinh at
e at cos bt
s > 0, n an integer > 0
s>0
s>0
s > |a|
s > |a|
s>a
e at sin bt
_s 2 - a 2 i + b 2
s>a
t n e at
s > a, n an integer > 0
f (ct)
n!
n +1
^s - a h
1
c j $f ^s /c h .
f (n)(t)
s n j $f ^t h . - s n - 1 f ^ 0 h - . . . - s f (n - 2) ^ 0 h - f (n - 1) ^ 0 h
c>0
Chapter 11: Working with Laplace Transforms
Solving Differential Equations
with Laplace Transforms
Now here comes the fun stuff: using Laplace transforms to solve differential
equations. Take a look at this differential equation, for example:
y" + 3y' + 2y = 0
The initial conditions for this equation are:
y(0) = 2
and
y'(0) = –3
“Wait a minute,” you say. “I know how to solve this! You know that to solve
this equation you just assume a solution of the following form:
y = ceat
Then you plug into the original equation to get:
ca2eat + 3caeat + 2ceat = 0
Dividing by ceat gives you:
a2 + 3a + 2 = 0
which becomes:
(a + 1) (a + 2) = 0
So a = –1 and –2, which means that the solution is:
y = c1e–t + c2e–2t
Now, matching the initial condition y(0) = 2 means that:
y(0) = c1 + c2 = 2
Take the first derivative of the solution, which gives you:
y' = –c1e–t + –2c2e–2t
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So y'(0) = –3 gives you:
y'(0) = –c1 + –2c2 = –3
And solving y(0) = c1 + c2 = 2 and y'(0) = –c1 + –2c2 = –3 results in:
c1 = 1
and:
c2 = 1
So the general solution is:
y = e–t + e–2t
Excellent, you solved the problem with traditional techniques. Now try doing
the same thing using Laplace transforms in the following sections. Having
already solved the problem means that you can check the answer you get.
And beginning with this relatively simple problem will show how to use
Laplace transforms to solve differential equations.
A few theorems to send you on your way
To start solving the example problem, you need a few more theorems. You’ve
figured out how to do a number of individual Laplace transforms, but what
about doing the Laplace transform of an equation like this one:
y" + 3y' + 2y = 0
The Laplace transform is a linear operator
The first theorem I want to introduce you to says that the Laplace transform
is a linear operator:
Because the Laplace transform is a linear operator, this is a true statement:
k $ c 1 f 1^ t h + c 2 f 2 ^ t h . = c 1 k $ f 1^ t h . + c 2 k $ f 2 ^ t h .
In other words, the Laplace transform of the sum of two terms is the sum
of the Laplace transforms of those two terms.
Taking the Laplace transform of the differential equation you want to solve
gives you:
j {y'' + 3y' + 2y}
Chapter 11: Working with Laplace Transforms
which becomes:
j {y''} + 3j { y'}+ 2j { y}
So you’ve made some progress in solving differential equations already — as
you can see, you can break things up by terms, which is a great help.
The Laplace transform of a first derivative
Okay, so you already know something about the term j # y - — that’s the
Laplace transform of the function y(x). But what about the j # y l - term? That
brings me to the next theorem:
Say that f(t) is continuous and f'(t) is piecewise continuous in an interval
0 ≤ t ≤ α and there exist constants C, β, and δ such that |f(t)| ≤ Ceβt for t ≥ δ.
In that case, k $ f l^ t h . exists for s > β, and:
k $ f l^ t h . = s k $ f ^ t h . - f ^ 0 h
The Laplace transforms of higher derivatives
What about j {f'' (t)}, j {f''' (t)}, all the way up to j $ f
^ t h . ? It turns out that
you can apply the previous equation over and over to find the higher derivatives. Here’s the theorem that formalizes this application:
(n)
Say that f(t), f'(t), f"(t) . . . f (n–1)(t) are continuous and f (n)(t) is piecewise continuous in an interval 0 ≤ t ≤ α and there exist constants C, β, and δ such
that |f(t)| ≤ Ceβt, |f'(t)| ≤ Ceβt , |f"(t)| ≤ Ceβt . . . |f (n–1)(t)| ≤ Ceβt for t ≥ δ.
In that case, k $ f (n) ^ t h . exists for s > β, and:
k $ f (n) ^ t h . = s n k $ f ^ t h . - s n - 1 f ^ 0 h - . . . - sf (n - 2) ^ 0 h - f (n - 1) ^ 0 h
Solving a second order
homogeneous equation
Alright, now that you’re armed with the theorems from the previous section,
you’re ready to solve this differential equation in the following sections:
y" + 3y' + 2y = 0
In general, here’s how the process works:
1. Figure out the Laplace transform of the differential equation.
2. Solve the equation algebraically.
3. Try to find the inverse transform.
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Using Laplace transforms in this process has one major advantage: It changes
a differential equation to an algebraic equation. The only sticky part is finding
the transforms and inverse transforms of the various terms in the differential
equation (and that’s what the tables of Laplace transforms are for).
You can generalize the solution technique used in the upcoming problem to
the general second order differential equation ay" + by' + cy = f(t). It turns out
that the Laplace transform of the solution to this differential equation is:
j $ y ^ sh. =
j $ f ^ t h.
^ as + b h y ^ 0 h + ay l^ 0 h
+ 2
as 2 + bs + c
as + bs + c
Note that when you use this equation, you don’t have to find the solution to
the homogeneous version of the differential equation first — doing so isn’t
necessary when you use Laplace transforms.
Finding the Laplace transform of the equation’s unknown solution
Taking the Laplace transform of the differential equation (see the earlier section “The Laplace transform is a linear operator”) gives you:
j {y''} + 3j { y'}+ 2j { y}
Then, using the theorem from the earlier sections on the Laplace transforms
of derivatives, you get:
j {y''} = s 2 j # y - - sy ^ 0 h - y l ^ 0 h
and
j # y l - = sj # y - - y ^ 0 h
After plugging these two derivatives into the Laplace transform of the original
differential equation, you get this result:
s 2 j # y - - sy ^ 0 h - y l ^ 0 h + 3 9 sj # y - - y ^ 0 h C + 2j # y - = 0
Collecting terms gives you:
_ s 2 + 3s + 2 i j # y - - ^ 3 + s h y ^ 0 h - y l^ 0 h = 0
Now you can use the initial conditions:
y(0) = 2
and
y'(0) = –3
Chapter 11: Working with Laplace Transforms
in the collected version of the equation’s Laplace transform, which gives you:
_ s 2 + 3s + 2 i j # y - - ^ 6 + 2s h + 3 = 0
Or, more simply:
_ s 2 + 3s + 2 i j # y - - 2s - 3 = 0
Moving all the terms to the correct spots gives you this result:
j # y- =
2s + 3
_ s 2 + 3s + 2 i
Factoring the denominator leaves you with this equation for the Laplace
transform of the differential equation’s solution:
j # y- =
2s + 3
^ s + 1h^ s + 2h
Discovering a function to match the Laplace transform
Now you have to find a function whose Laplace transform is the same as the
previous solution. To do that, use the method of partial fractions, and then
write the solution as:
j # y- =
2s + 3
b
= a +
^ s + 1h^ s + 2h ^ s + 1h ^ s + 2h
And now you have to figure out what a and b are. To do so, write the
fractions as:
j # y- =
a ^ s + 2h + b ^ s + 1h
2s + 3
=
^ s + 1h^ s + 2h
^ s + 1h^ s + 2h
At this point, you can equate the numerators in the fractions to get:
2s + 3 = a (s + 2) + b(s + 1)
Because it’s up to you to choose s, you can first set it to –1 (to make it easy
on yourself), which gives you:
1=a
And now you can set s to –2 to get:
–1 = –b
or:
1=b
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Okay, so a = 1 = b. You have these fractions:
j # y- =
a +
b
^ s + 1h ^ s + 2h
Substituting for a and b gives you a more detailed Laplace transform of the
differential equation’s solution:
j # y- =
1
1
+
^ s + 1h ^ s + 2h
Uncovering the inverse Laplace transform
to get the equation’s solution
Whew. After you know what the Laplace transform of the solution looks like,
it’s time to find the inverse Laplace transform to find the actual solution to
the differential equation.
Your calculator isn’t going to have an inverse Laplace transform j - 1$ .
button on it (and if it does, I’d be glad to buy it from you). Your best bet,
then, is to use a table, like Table 11-1 earlier in this chapter, to find the form
of the transform that you’re dealing with. Then you simply have to find the
function whose transform that is.
From Table 11-1, you can see that the Laplace transform of eat is:
j # e at - = s -1 a
s>a
This transform looks promising. You can use this information to get the
inverse transform. Take a look at the first term:
1
^ s + 1h
Comparing this to the Laplace transform of eat tells you that in this case,
a = –1, so the first term in the differential equation’s solution is:
y1 = e–t
Now look at the second term:
1
^ s + 2h
Comparing this to the Laplace transform of eat tells you that a = –2, so the
second term in the differential equation’s solution is:
y2 = e–2t
Chapter 11: Working with Laplace Transforms
And the solution to the differential equation is y = y1 + y2, which equals this
final result:
y = y1 + y2 = e–t + e–2t
This solution is confirmed by your earlier solution (which you determined at
the very beginning of this section). But this time, you did it with Laplace
transforms. Not too bad, huh?
Solving a second order nonhomogeneous
equation
Ready for another example? How about this little gem:
y" + y = –15 sin 4t
where
y(0) = 2
and
y'(0) = 5
“Hmm,” you say, “I think I know how to solve this one as well.” Good! You
must have read Chapter 6! So, you likely know that you should first take a
look at the homogeneous equation:
y" + y = 0
And because it looks like sines and cosines would work, you can assume a
solution of the following form:
y = c1sin t + c2 sin t
Now you need a particular solution. The right side has a term in sin 4t, so try a
similar term here. Doing so would give you this form for the general solution:
y = c1sin t + c2 cos t + c3sin 4t
Plugging this into y" + y = –15 sin 4t gives you:
–c1sin t – c2cos t – 16c3sin 4t + c1sin t + c2 cos t + c3sin 4t = sin 4t
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Or, with some simplification you get:
–15 c3 sin t = –15 sin 4t
So c3 = 1, and so far, your general solution looks like this:
y = c1sin t + c2 cos t + sin 4t
Now you can use the initial conditions to determine c1 and c2. Here’s y(0):
y(0) = c1sin t + c2 cos t + sin 4t = c2 = 2
So c2 = 2. So far, your general solution looks like this:
y = c1sin t + 2 cos t + sin 4t
Now you can determine c1 with the last initial condition, which means that
you have to calculate y'(0). Here’s what y' looks like:
y' = c1cos t – 2 sin t + 4 cos 4t
And here’s y'(0):
y'(0) = c1 + 4 = 5
So c1 = 1. And the general solution is:
y = sin t + 2 cos t + sin 4t
Okay, very nice. You obviously have been paying attention! Now, in the following sections you can try the same thing using Laplace transforms.
Determining the Laplace transform
Here’s the differential equation you’re trying to solve with the Laplace transform, in case you forgot:
y" + y = –15 sin 4t
To begin, take the Laplace transform of it, using the relation from the earlier
section “The Laplace transforms of higher derivatives”:
j {y''} = s 2 j # y - - sy ^ 0 h - y l ^ 0 h
So here’s your differential equation:
s 2 j # y - - sy ^ 0 h - y l^ 0 h + j # y - =
-15
_ s 2 + 16 i
Chapter 11: Working with Laplace Transforms
where you’ve put in the Laplace transform of sin 4t, according to Table 11-1.
Here are the initial conditions for this problem:
y(0) = 2
and
y'(0) = 5
Substituting these initial conditions into the previous equation gives you this
result:
s 2 j # y - - 2s - 5 + j # y - =
-15
_ s 2 + 16 i
Or, with some simplifying you get:
_ s 2 + 1i j # y - - 2s - 5 =
-15
_ s 2 + 16 i
Now, taking 2s + 5 over to the right side gives you:
_ s 2 + 1i j # y - =
-15
_ s 2 + 16 i
+ 2s + 5
Be careful, because you’re into some heavy algebra now! After the calculations, the equation works out to be:
_ s 2 + 1i j # y - =
-15 + ^ 2s + 5 h_ s 2 + 16 i
_ s 2 + 16 i
Next, after expanding the terms you get the following equation:
_ s 2 + 1i j # y - =
-15 + 2s 3 + 32s + 5s 2 + 80
_ s 2 + 16 i
which works out to become:
3
2
s + 65
j # y - = 2s +2 5s + 32
_ s + 1i_ s 2 + 16 i
Matching a function to the Laplace transform
Wow. How the heck do you expand the previous equation using partial fractions so you can find a function to match the transform? It’s actually pretty
easy! You simply break it up and assume a form like this:
j # y - = as2 + b + cs2 + d
_ s + 1i _ s + 16 i
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which is:
j # y- =
^ as + b h_ s 2 + 16 i + ^ cs + d h_ s 2 + 1i
_ s 2 + 1i_ s 2 + 16 i
Now tackle the numerator. The first term is:
(as + b)(s2 + 16)
Multiplying this out gives you:
as3 + 16as + bs2 + 16b
The second term in the numerator is:
(cs + d)(s2 + 1)
After you multiply the term out, you get:
cs3 + cs + ds2 + d
Adding the two terms that you multiplied out gives you this form for the
numerator:
as3 + 16as + bs2 + 16b + cs3 + cs + ds2 + d
Or, after some simplifying:
(a + c)s3 + (b + d)s2 + (16a + c)s + (16b + d)
This numerator must equal the numerator in the other equation you have for
the same Laplace transform, 2s3 + 5s2 + 32s + 65. So equating the two numerators gives you:
2s3 + 5s2 + 32s + 65 = (a + c)s3 + (b + d)s2 + (16a + c)s + (16b + d)
Does this really buy you anything? Well, it still looks like a cubic equation.
And it still is a cubic equation, but the powers of s must be equal on the two
sides of the equation, so you have these relations:
a+c=2
b+d=5
16a + c = 32
16b + d = 65
Chapter 11: Working with Laplace Transforms
Solving these equations for a, b, c, and d gives you the following:
a=2
b=1
c=0
d=1
Here’s your equation for the Laplace transform:
j # y - = as2 + b + cs2 + d
_ s + 1i _ s + 16 i
Expanding this gives you:
j # y- =
as
_ s 2 + 1i
+
b
_ s 2 + 1i
+
cs
_ s 2 + 16 i
+
d
_ s 2 + 16 i
Finally, plugging in values for a, b, c, and d gives you:
j # y- =
2s
_ s 2 + 1i
+
1
_ s 2 + 1i
+
1
_ s 2 + 16 i
Using the handy table to find the inverse Laplace transform
So the general solution, after you check Table 11-1 and plug in all the right
numbers, is:
y = sin t + 2 cos t + sin 4t
As you may have noticed, this solution agrees with the one you got earlier in
this chapter using the traditional method. Great work.
Solving a higher order equation
Try out this ever-popular higher order differential equation in the land of
Laplace transforms:
y(4) – y = 0
where:
y(0) = 0
y'(0) = 1
y"(0) = 0
y'''(0) = 0
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So now you probably want to know how to go about solving y(4) – y = 0 using
Laplace transforms. The following sections will show you how.
Figuring out the equation’s Laplace transform
A fourth derivative may seem dreadful, but the fact that three of the four initial conditions are zero will definitely help. Here’s the Laplace transform of
y(4) – y = 0:
s 4 j # y - - s 3 y ^ 0 h - s 2 y l^ 0 h - s y''(0) - y'''(0) - j { y} = 0
This equation looks like it has the potential of being a tough nut to crack, but
substituting the initial conditions simplifies things, giving you this:
s4 j # y - - s2 - j # y - = 0
Looks a lot more manageable, doesn’t it? Solving for
j # y- =
^ as + b h_ s 2 + 16 i + ^ cs + d h_ s 2 + 1i
j{y} gives you:
_ s 2 + 1i_ s 2 + 16 i
j # y- =
s2
s4 - 1
Unearthing a function to match the Laplace transform
Now the plan is to get the transform from the previous section into a form
that’s recognizable in Table 11-1. You can use partial fractions to do just that.
Because s4 – 1 = (s2 + 1) (s2 – 1), you can write the previous transform in the
following format:
j # y - = as2 + b + cs2 + d
s -1
s +1
Adding these terms together gives you:
j # y- =
^ as + b h_ s 2 + 1i + ^ cs + d h_ s 2 - 1i
s4 - 1
The numerator here must equal the numerator in the other equation you
have for j # y - . Equating the numerators gives you:
s2 = (as + b)(s2 + 1) + (cs + d)(s2 – 1)
Don’t forget that you can choose the value of s yourself. For example, if you
choose s to equal 1, the previous equation simplifies to:
1 = 2(a + b)
Chapter 11: Working with Laplace Transforms
How about if you choose s = –1? Then you get this result:
1 = 2(–a + b)
Adding these two equations together gives you:
2 = 4b
So:
b = 1⁄2
Substituting b = 1⁄2 into 1 = 2(a + b) gives you:
1 = 2(a + 1⁄2)
Or with some simplification:
1 = 1 + 2a
So:
a=0
Okay, so a = 0 and b = 1. How about c and d? You can set s = 0 in 2s2 =
(as + b)(s2 + 1) + (cs + d)(s2 – 1) to get:
0=b–d
Or:
b=d
And because b = 1⁄2, d = 1⁄2 also.
Alright, that gives you a, b, and d. But you still have to figure out c. Take a
look at your equation that equated the two numerators:
s2 = (as + b)(s2 + 1) + (cs + d)(s2 – 1)
Note that equating the cubic terms on the two sides gives you this equation:
0 = as3 + cs3
But because a = 0, this becomes:
0 = cs3
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So c = 0. That gives you:
a=0
b = 1⁄2
c=0
d = 1⁄2
Substituting these numbers into the Laplace transform gives you this more
detailed result:
1
1
2
j # y- = 2
+ 2
s - 1 s2 + 1
So, as you can see, having used the initial conditions in the problem has significantly simplified the form of the Laplace transform.
Getting the equation’s inverse Laplace transform
After you have the Laplace transform, check out Table 11-1 for the inverse
transform, and then plug in the correct numbers. Here’s the result you
should get:
y = 1 ^ sin h t + sin t h
2
Factoring Laplace Transforms
and Convolution Integrals
Sometimes, the Laplace transforms you end up with require a little extra
work to solve. But no worries; you just have to be a little creative. And in the
following sections, I explain how to do just that. For instance, I show you how
to factor a Laplace transform into a sum of fractions and how to work with
convolution integrals.
Factoring a Laplace transform
into fractions
When you face an especially funky-looking Laplace transform, factoring the
denominator and the numerator often helps. For example, imagine that you
end up with this Laplace transform when solving a differential equation:
j # y- =
s+4
s 2 + 4s + 8
Chapter 11: Working with Laplace Transforms
There’s no entry in Table 11-1 that matches this transform. But the form of
the denominator indicates that you may be able to factor it. Here’s what the
factoring may look like:
s2 + 4s + 8 = (s + 2)2 + 22
This equation looks somewhat more promising, don’t you think? You can now
convert your original transform into this new form:
j # y- =
s+4
2
^ s + 2h + 2 2
Hey, wait a minute. This expression isn’t in Table 11-1 either. But take a closer
look at the numerator; you can factor it as well. For instance, you can rewrite
the numerator like this:
s + 4 = (s + 2) + 2
This in turn gives you the following for the Laplace transform:
j # y- =
^ s + 2h + 2
^ s + 2h + 2 2
2
Or, with a little simplifying you get:
j # y- =
^ s + 2h
^ s + 2h + 2 2
2
+
2
2
^ s + 2h + 2 2
Doesn’t this look a whole lot better? According to Table 11-1, you can now
find the inverse Laplace transform, which is:
j # y - = e - 2t cos 2t + e - 2t sin 2t
Checking out convolution integrals
The example in the previous section factored a Laplace transform into a sum
of fractions. But what if, instead of a sum, you end up with a product of recognizable Laplace transforms? What do you do then?
For example, what if you had this Laplace transform:
j # y- =
a
2
s 2^ s + a h
Looking at Table 11-1, you can see that the inverse transform of this expression doesn’t appear. However, if you write this Laplace transform as:
a
j # y - = 12
s ^ s + ah2
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then you can see that this is the product of the Laplace transforms for t and
sin at. This product brings you to an important question: Can you write the
inverse Laplace transform this way:
j # y - = t sin at
Nope, it doesn’t work like that. The inverse of the sum of Laplace transforms
is the sum of the inverses of the individual Laplace transforms, but it doesn’t
work that way when you multiply Laplace transforms together. Instead, you
have to turn to convolution integrals, as defined in the following theorem:
If:
k $ f ^ t h . k $ g ^ t h . = k $ h^ t h .
then:
t
h^ t h =
# f ^ t - α h g ^ α h dα
0
which also equals:
t
h^ t h
# f ^ α h g ^ t - α h dα
0
where α is a new variable. The integrals in these transforms are called
convolution integrals.
Sometimes you may see the convolution operation denoted with an asterisk
(*), like this:
h(t) = (f * g)(t)
Using this new operator means that you can write the products of Laplace
transforms in much the same way that you would use the multiplication
operator:
f(t) * 0 = 0 * f(t) = 0
f(t) * g(t) = g(t) * f(t)
f(t) * (g(t) + i(t))= f(t) * g(t) + f(t) * i(t)
(f(t) * g(t)) * i(t) = f(t) * (g(t) * i(t))
Armed with this theorem, you can now tackle the example problem:
a
j # y - = 12
s ^ s + ah2
Chapter 11: Working with Laplace Transforms
These terms are the Laplace transforms of t and sin at. So you can write the
inverse Laplace transform like this:
t
y ^t h
# ^ t - α h sin ^ at h dα
0
which equals:
y ^ t h = at - sin2at
a
Surveying Step Functions
In the world of Laplace transforms, there’s an interesting function called the
step function, which looks like this: ua(t). In the following sections, I break
down the elements of the step function, and I show how it relates to Laplace
transforms.
Defining the step function
The function’s value is defined as 0 up to a certain point, a, but then as 1
thereafter. Here’s the definition of the step function in equation form:
0 t <a
u a^t h = '
1 t $a
As an example, if a = 0, you have that u0(t) equals:
0 t <0
u 0^ t h = '
1 t $0
Note that the subscript on the u indicates the point at which the step function “steps.”
And if a = 1, you have that u1(t) equals:
0 t <1
u 1^ t h = '
1 t $1
You can see what ua(t) looks like graphically in Figure 11-1.
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Figure 11-1: 1
The step
function in
graph form.
a
Figuring the Laplace transform
of the step function
Now that you’re familiar with the step function, go a bit further and take the
Laplace transform of the step function:
3
j $ u a^ t h. =
#e
- st
u a ^ t h dt
0
Substituting the step function into this transform gives you the following result:
3
j $ u a^ t h. =
#e
- st
dt
a
Note that the lower limit of the integral has become a. Why? Because the integral is zero up to that point. Now this equation becomes:
- as
j $ u a^ t h. = es
s>0
And that brings me to a new theorem:
If a is a positive constant, then:
k $ u a ^ t h f ^ t - ah . = e - as k $ f ^ t h .
This theorem can help you with Laplace transforms that have exponentials in
them. To see how, try finding the inverse Laplace transform of this equation:
j # y - = 1 - e2
s
-s
Begin by breaking it into the following:
-s
j # y - = 12 - e 2
s
s
Chapter 11: Working with Laplace Transforms
Using the theorem, you can find the inverse Laplace transform:
y(t) = t – u1(t)(t – 1)
That is, for t < 1:
y(t) = t
and for t ≥ 1:
y(t) = 1
So the step function helped you find a Laplace transform. Pretty cool, right?
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Chapter 12
Tackling Systems of First Order
Linear Differential Equations
In This Chapter
Brushing up on matrix basics
Handling matrix operations
Checking out linear independence, eigenvalues, and eigenvectors
Working out homogeneous and nonhomogeneous systems
T
his chapter is all about systems of differential equations, where a system
is a set of linear differential equations that share some common variables. In this chapter, I show you how to apply many of the techniques used
to handle systems of standard algebraic equations to solving systems of differential equations.
Being able to handle systems of differential equations is useful in two cases:
When you want to reduce the order of a differential equation to a system
of first order differential equations
When you have a problem that consists of interdependent differential
equations — such as when you have an electrical circuit with linked
loops in it that share the current (don’t worry; I won’t delve into such
science here)
This chapter is heavy in its use of matrices, which are usually used to solve systems of linear equations. After you brush up on the fundamentals of matrices,
you find out about some important concepts, such as linear independence,
eigenvalues, and eigenvectors. I wrap up the chapter by explaining how to
solve both homogeneous and nonhomogeneous systems.
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Introducing the Basics of Matrices
Take a look at the following system of three simultaneous equations:
x+y+z=6
x – y – z = –4
x+y–z=0
How do you solve this system of equations for x, y, and z? A fair bit of algebra
is involved here, so to keep things straight, you can use a matrix. You’re most
likely already familiar with matrices, but in the following sections, I go over
the fundamentals just in case you need to refresh your memory.
Setting up a matrix
Simply put, a matrix is a set of numbers arranged into columns and rows,
like this:
Ja a a N
2
3O
K 1
K b1 b 2 b 3O
O
K
K c1 c 2 c 3 O
P
L
When creating a matrix, you place the coefficients of each equation in a row.
For instance, a1, a2, and a3 are from one equation; b1, b2, and b3 are from a
second equation; and c1, c2, and c3 are from a third equation. Similarly, the first
coefficient of any equation is in the first column; the second coefficient of any
equation is in the second column; and the third coefficient of any equation is
in the third column. (I’m sure you get the idea!) A matrix can contain as many
rows and columns as you need.
Putting numbers into rows and columns like this helps you keep track of
them. (Neatniks rejoice!) For example, you can represent the coefficients in
these equations:
x+y+z=6
x – y – z = –4
x+y–z=0
like this:
J1 1 1N
O
K
K 1 -1 -1O
O
K
K 1 1 -1O
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Chapter 12: Tackling Systems of First Order Linear Differential Equations
And to keep track of the constants on the right side of these equations, you
can create what’s called an augmented matrix. With this type of matrix, a vertical line keeps the constants on the right sides of the equations separate
from the coefficients on the left sides of the equations, like this:
J1 1 1N J 6 N
OK O
K
K 1 -1 -1O K - 4 O
O
K
K 1 1 -1O KK 0 OO
PL P
L
Working through the algebra
After you set up a matrix, it’s easy to keep track of the algebraic manipulations involved in solving the system of simultaneous equations. Here, you’re
going to work on the individual rows.
You can see how this works using the matrix you set up in the previous section. Remember that your goal is to solve for x, y, and z. First, add the second
row to the first and place the result in the first row, making sure that you
leave the second row intact. Doing all this gives you:
J2 0 0 N J 2 N
OK O
K
K 1 -1 -1O K - 4 O
O
K
K 1 1 -1O KK 0 OO
PL P
L
Now divide the first row by 2 to simplify:
J1 0 0 N J 1N
OK O
K
K 1 -1 -1O K - 4 O
O
K
K 1 1 -1O KK 0 OO
PL P
L
Next, add –1 times the first row to the second row and place the result in the
second row:
J 1 0 0 N J 1N
OK O
K
K 0 -1 -1O K - 5 O
O
K
K 1 1 -1O KK 0 OO
PL P
L
Then multiply the second row by –1:
J 1 0 0 N J 1N
OK O
K
K 0 1 1O K 5 O
O
K
K 1 1 -1O KK 0 OO
PL P
L
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Add –1 times the first row to the third row and place the result in the third
row, like so:
J 1 0 0 N J 1N
OK O
K
K 0 1 1O K 5 O
OK O
K
K 0 1 -1O K -1O
PL P
L
Now add –1 times the second row to the third, placing the result in the
third row:
J1 0
0 NO JK 1NO
K
K0 1
1O K 5 O
OK O
K
K 0 0 - 2O K - 6 O
PL P
L
Finally, you divide the third row by –2:
J 1 0 0 N J 1N
OK O
K
K 0 1 1O K 5 O
OK O
KK
0 0 1O K 3 O
PL P
L
Note that you now have a triangular matrix on the bottom left; in other words,
the bottom left triangle’s elements all equal zero. This triangle simplifies matters considerably, because this new augmented matrix now represents this
system:
x=1
y+z=5
z=3
And from the latter two equations, finding the solution for y is pretty trivial:
y = 2. Now you’ve solved for all three variables. Nice work.
Examining matrices
Did you know that you can name matrices? Yup, that’s right. In matrix terms,
a name is denoted in bold. For instance, you can name the following matrix,
which contains both real and complex elements, A (imaginative, I know):
J 1
2 + 2i N
O
A = KK
3 + 3i
4 O
L
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Chapter 12: Tackling Systems of First Order Linear Differential Equations
The transpose of a matrix swaps its rows and columns; in other words, the
rows become the columns, and the columns become the rows. Here’s what
the transpose of A (called AT ) looks like:
J 1
3 + 3i N
O
A T = KK
2 + 2i
4 O
L
P
You can also define the complex conjugate of a matrix. To denote a complex
conjugate, you simply add a line above the name of a matrix. The complex
conjugate of a matrix is simply the complex conjugate element by element of
the matrix. Here’s the complex conjugate of A, where you flip the sign of the
imaginary part:
J 1
2 - 2i N
O
A = KK
3 - 3i
4 O
L
P
The adjoint of a matrix is the transpose of the complex conjugate of the
matrix. You denote an adjoint by adding an asterisk (*) to the name of
the matrix. Here’s what A* looks like:
J 1
3 - 3i N
O
A * = KK
2 - 2i
4 O
L
P
Mastering Matrix Operations
After you’re familiar with the basics of matrices, you can take the next step
and start working with a variety of matrix operations, such as addition, subtraction, multiplication, and other fun stuff. I explain what you need to know
in the following sections.
Equality
Two matrices are considered equal if every element in the first matrix is
equal to the corresponding element in the other matrix. For instance, if:
J1 2 N
O
A = KK
3 4O
L
P
And:
J1 2 N
O
B = KK
3 4O
L
P
then A = B. Easy enough, right?
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Addition
If you need to, you can add two matrices together. Doing so involves adding
the elements at corresponding positions in the two matrices. For example, if:
J1 2 N
J5 6 N
O and B = K
O
A = KK
O
K 7 8O
3 4
L
P
L
P
then:
J1 2 N J5 6 N J 6 8 N
O+ K
O= K
O
A + B = KK
3 4 O K 7 8 O K 10 12O
L
P L
P L
P
And just so you know: A + B = B + A.
Subtraction
Like matrix addition, matrix subtraction works as you’d expect: element by
element. For example, take a look:
J1 2 N J5 6 N J- 4 - 4 N
O
O- K
O= K
A - B = KK
3 4O K7 8O K - 4 - 4O
L
P L
P L
P
Note that A – B doesn’t equal B – A. In fact, as you’d expect, A – B = –(B – A),
as you can see here:
J5 6 N J1 2 N J 4 4 N
O- K
O= K
O
B - A = KK
7 8O K 3 4O K 4 4O
L
P L
P L
P
Multiplication of a matrix and a number
In some cases, you need to multiply a matrix by a number; to do so, you multiply each element in the matrix by that number. For example, 4A looks like this:
J1 2 N J 4 8 N
O= K
O
4A = 4 KK
3 4 O K 12 16 O
L
P L
P
Multiplication of two matrices
How about multiplying two matrices together, such as A and B? Unfortunately,
this is a little more involved than just adding them. It turns out that AB is
defined when the number of columns in A is the same as the number of rows in
B. That is, if A is an l × m matrix (that’s row × column notation, so A has l rows
Chapter 12: Tackling Systems of First Order Linear Differential Equations
and m columns) and B is an m × n matrix, the product AB exists — and the
product is an l × n matrix.
If AB = C, here’s how it works: the (i, j) (that’s row, column) element of C is
found by multiplying each element of the ith row of A by the matching element in the jth column of B. Then when you add the products you get:
m
C ij = ! A ik B kj = A i1 B 1j + A i2 B 2j + A i3 B 3j + . . . + A im B mj
k =1
Now put in the numbers:
J1 2 N J5 6 N
OK
O
AB = KK
3 4O K7 8O
L
PL
P
Applying the rules of multiplying matrices, you get:
J1 2 N J5 6 N J 5 + 14 6 + 16 N J19 22 N
O= K
OK
O= K
O
AB = KK
3 4 O K 7 8 O K 15 + 28 18 + 32O K 43 50 O
L
PL
P L
L
P
P
One thing to note is that in general, AB ≠ BA. Using the same example, here’s
what BA looks like:
J5 6 N J1 2 N J5 + 18 10 + 24 N J23 34 N
O= K
OK
O= K
O
BA = KK
7 8 O K 3 4 O K 7 + 24 14 + 32 O K 31 46 O
L
PL
P L
P
P L
Multiplication of a matrix and a vector
You’ll often see matrices that only have one set of rows or one set of
columns. These matrices are called vectors. For example, here’s a vector with
one column:
J1 N
K O
K 2O
KK OO
3
L P
Here’s a vector with one row:
_1 2 3 i
Sometimes you’ll see vectors that are used to hold variables, as in this one,
which I’ll call x:
JxN
K O
x = K yO
KK OO
z
L P
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With vectors, you can express a set of simultaneous equations like this:
J1 2 N J x N J 3 N
OK O K O
K
K 3 4O K y O = K 4O
OK O K O
KK
5 6O K z O K 5O
PL P L P
L
This set of simultaneous equations can be written as:
Ax = b
Identity
The identity matrix, which is labeled I, holds 1s along its upper-left to lowerright diagonal, and all 0s otherwise. Here’s a 2 × 2 identity matrix (with 2 rows
and 2 columns):
J1 0 N
O
I = KK
0 1O
L
P
And here’s a 3 × 3 identity matrix (with 3 rows and 3 columns):
J1 0 0 N
O
K
I = K 0 1 0O
O
KK
0 0 1O
P
L
Multiplying any matrix, A, by the identity matrix gives you A again. (I explain
how to multiply two matrices together earlier in this chapter.) For example,
take a look at the multiplication of a matrix called A and the 3 × 3 identity
matrix:
J 1 2 3 N J1 0 0 N J 1 2 3 N
O
O K
OK
K
AI = K 4 5 6 O K 0 1 0 O = K 4 5 6 O
O
O K
OK
KK
7 8 9O K 0 0 1O K 7 8 9O
P
P L
PL
L
The inverse of a matrix
If you have simultaneous equations such as these:
x+y+z=6
x – y – z = –4
x+y–z=0
Chapter 12: Tackling Systems of First Order Linear Differential Equations
you can write a system in matrix form like this:
J1 1 1N J x N J 6 N
OK O K O
K
K 1 -1 -1O K y O = K - 4 O
O
K
K 1 1 -1O KK z OO KK 0 OO
PL P L P
L
You can further simplify this so that it looks like:
Ax = b
where:
J1 1 1N
O
K
A = K 1 -1 -1O
O
K
K 1 1 -1O
P
L
JxN
K O
x = K yO
KK OO
z
L P
J 6N
K O
b = K - 4O
KK OO
0
L P
Wouldn’t it be nice if you could solve for x by some way dividing Ax = b by A
to get the following:
x = A–1b
Doing so would give you the solutions, x, y, and z, to your simultaneous equations. So can you find A–1? Yes, you sure can! This is called the inverse of A
(and A–1 A = I, where I is the identity matrix). So how do you find the inverse
of a matrix? You can choose from several ways, but the easiest is the one that
I present in the following sections.
Generally, to find the inverse of a matrix A, apply the steps to A in order to
reduce it to the identity matrix I (that is, 1s along the upper-left to lower-right
diagonal and 0s otherwise). Then apply those same steps, in the same order,
to the identity matrix. You should end up with A–1. To wrap up, simply solve
for x by multiplying A–1 and b.
Before I can explain how to find an inverse, I need to cover an important concept that’s helpful when you’re working with inverses: determinants. Read on
for details.
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Finding the determinant of a matrix
An important quantity when it comes to matrices is the determinant. The
determinant reduces a matrix to a single significant number; whether that
number is zero is important for inverses as well as other calculations. For
example, when checking to see whether the solutions you get to a system of
linear differential equations are linearly independent (and form a general
solution), you need to use determinants. (I discuss linear independence in
more detail later in this chapter.)
So how do you find a matrix’s determinant? Here’s an example that can show
you how. Say you have this 2 × 2 matrix, called A:
Ja b N
O
A = KK
c dO
L
P
The determinant of A, written as det(A), for a 2 × 2 matrix is defined this way:
det(A) = ad – cb
For example, here’s a 2 × 2 matrix with some specific numbers:
J1 2 N
O
A = KK
3 4O
L
P
What’s its determinant? Just plug the numbers into ad – cb, like so:
det(A) = (1)(4) – (3)(2) = –2
How about a 3 × 3 determinant? Say you have this 3 × 3 matrix A:
Ja b c N
O
K
A = Kd e f O
O
KK
g h iO
P
L
Be careful here! Determinants rapidly become more complex beyond 2 × 2.
Here’s what the determinant of the 3 × 3 matrix is:
det(A) = aei – afh – bdi + cdh + bfg – ceg
Finding the inverse of a matrix
To find the inverse of a matrix A, apply the steps you need to A to reduce it
to the identity matrix, I (that is, 1s along the upper-left to lower right diagonal
and 0s otherwise). Then apply those same steps, in the same order, to an
identity matrix, and you’ll end up with A–1.
Chapter 12: Tackling Systems of First Order Linear Differential Equations
Take a look at an example. For this system of equations:
x+y+z=6
x – y – z = –4
x+y–z=0
matrix A looks like this:
J1 1 1N
O
K
A = K 1 -1 -1O
O
K
K 1 1 -1O
P
L
Find A–1 by first reducing A to I. To do so, add –1 times the first row to the
third row and place the result in the third row:
J1 1
1NO
K
K 1 -1 -1O
O
K
K 0 0 - 2O
P
L
Now divide the third row by –2 to simplify:
J 1 1 1N
O
K
K 1 -1 -1O
O
KK
0 0
1O
P
L
Add the second row to the first row and place the result in the first row:
J2 0 0 N
O
K
K 1 -1 -1O
O
KK
0 0
1O
P
L
Next, divide the first row by 2 to simplify:
J 1 0 0N
O
K
K 1 -1 -1O
O
KK
0 0
1O
P
L
You’re looking good. Now add –1 times the first row to the second row and
place the result in the second row:
J 1 0 0N
O
K
K 0 -1 -1O
O
KK
0 0
1O
P
L
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Almost there. Add the third row to the second row and place the result in the
second row:
J 1 0 0N
O
K
K 0 -1 0 O
O
KK
0 0 1O
P
L
Finally, multiply the second row by –1 to get that final 1:
J1 0 0 N
O
K
K 0 1 0O
O
KK
0 0 1O
P
L
And there you have it — the identity matrix.
After you reduce a system to the identity matrix, you have to apply the same
sequence of steps to the identity matrix to find A–1. Here’s the 3 × 3 identity
matrix:
J1 0 0 N
O
K
I = K 0 1 0O
O
KK
0 0 1O
P
L
Now follow the same steps that you followed earlier in this section. First, add
–1 times the first row to the third row and place the result in the third row:
J 1 0 0N
O
K
K 0 1 0O
O
K
K -1 0 1O
P
L
Divide the third row by –2:
J1 0 0 N
O
K
K0 1 0 O
O
K1
K 2 0 -1 2O
P
L
Next, add the second row to the first row and place the result in the first row:
J1 1 0 N
O
K
K0 1 0 O
O
K1
K 2 0 -1 2O
P
L
Divide the first row by 2:
J1 2
K
K0
K1
K 2
L
1
2
1
0
0 NO
0 O
O
-1 2O
P
Chapter 12: Tackling Systems of First Order Linear Differential Equations
And add –1 times the first row to the second row and place the result in the
second row:
J12
K
K -1 2
K 1
K 2
L
1
1
0 NO
0 O
O
-1 2O
P
2
2
0
Now add the third row to the second row, placing the result in the second row:
J1 2
K
K0
K1
K 2
L
1
1
2
2
0
0 NO
-1 2O
O
-1 2O
P
Finally, multiply the second row by –1:
J1 2 1 2
0 NO
K
K 0 -1 2 1 2 O
O
K1
K 2 0 -1 2O
P
L
And there you have it: A–1.
Solving for x with a little multiplication
Now you have to find x, the solution to your system of equations, by using the
equation x = A–1b. Your equation looks like this when you fill in the numbers:
J x N J1 2 1 2
0 NOJ 6 N
K O
K O K
K y O = K 0 - 1 2 1 2 OK - 4 O
OK O
KK OO KK 1
z
0 - 1 2 OK 0 O
2
L P L
PL P
Performing matrix multiplication (as I describe earlier in this chapter)
gives you:
J x N J1 N
K O K O
K y O = K 2O
KK OO KK OO
z
3
L P L P
Voila! It worked. You found the solution to your simultaneous equations by
finding the inverse of a matrix.
Does solving for x always work? The answer is a resounding no. Why?
Sometimes matrices can’t be inverted (in which case they’re called singular,
or non-invertable, matrices).
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How can you tell if a matrix is singular before attempting to find its inverse?
You can check to see whether its determinant is zero. Matrices whose determinants are zero are singular matrices without an inverse; matrices whose
determinants aren’t zero do have an inverse. In the original matrix in my
example, the determinant is:
det(A) = aei – afh – bdi + cdh + bfg – ceg
Having Fun with Eigenvectors ’n’ Things
You’re now ready for more insight into matrices using eigenvectors and
eigenvalues. What are those? Read on to find out more.
Linear independence
Determining whether a set of vectors is linearly dependent can be quite
important. It’s especially important when you’re solving systems of linear differential equations to see whether you have linearly independent solutions.
You need to get a linearly independent set of solutions to make sure you have
the complete solution.
Say, for example, that you have a set of vectors, x1 to xk. They’re said to be
linearly dependent if there are constants, c1 to ck, such that:
c1x1 + c2x2 + . . . ckxk = 0
where 0 is a vector whose elements are all zero. In other words, x1 to xk are
considered linearly dependent if a linear relation relates them. If there are no
constants, c1 to ck (not all zero), such that this equation holds, the vectors, x1
to xk, are linearly independent.
How do you determine whether a set of vectors, x1 to xk, is linearly independent? You can assemble all the vectors, x1 to xk, into a single matrix and then
check the determinant. If the determinant is zero, the vectors are linearly
dependent. If the determinant isn’t zero, the vectors are linearly independent.
I walk you through an example in the following sections.
Assembling the vectors into one matrix
Say that you have the following vectors:
J 2N
K O
x1 = K 4O
K O
K - 2O
L P
Chapter 12: Tackling Systems of First Order Linear Differential Equations
J2 N
K O
x 2 = K 1O
KK OO
3
L P
J - 8N
O
K
x 3 = K 2O
O
K
K - 22O
P
L
Now you can assemble these vectors into a 3 × 3 matrix called x:
J 2 2 - 8N
O
K
x=K 4 1
2O
O
K
K - 2 3 - 22O
P
L
Determining the determinant
Okay, so now you have to figure out whether the determinant of x is zero. As
discussed earlier in this chapter, the determinant of a 3 × 3 matrix A:
Ja b c N
O
K
A = Kd e f O
O
KK
g h iO
P
L
is the following:
det(A) = aei – afh – bdi + cdh + bfg – ceg
So the determinant of x in my example is:
(2)(1)(–22) – (2)(2)(3) – (2)(4)(–22) + (–8)(4)(3) + (2)(2)(–2) – (–8)(1)(–2)
which becomes:
–44 – 12 + 176 – 96 – 8 – 16 = 0
As you can see, the determinant in this case is zero. Uh-oh. That’s no good.
The determinant is zero, which means that x1, x2, and x3 are linearly dependent. What should you do next? Remember the following:
c1x1 + c2x2 + . . . ckxk = 0
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What are c1, c2, and c3? You can figure out those constants by solving the augmented matrix. Here’s your matrix:
J 2 2 - 8N
O
K
K 4 1
2O
O
K
K - 2 3 - 22O
P
L
Add the first row to the third row and place the result in the third row:
J 2 2 - 8 N J0 N
OK O
K
K4 1
2O K 0 O
O
K
K 0 5 - 30 O KK 0 OO
PL P
L
Next, divide the first row by 2 to simplify:
J 1 1 - 4 N J0 N
OK O
K
K4 1
2O K 0 O
O
K
K 0 5 - 30 O KK 0 OO
PL P
L
Now add –4 times the first row to the second row (to get a zero in the first
position of Row 2). Place the result in the second row:
J1
1 - 4 NO J0 N
K
K O
K0 -3
18 O K 0 O
OK O
K
K0
5 - 30 O K 0 O
PL P
L
Divide the second row by –3 to simplify:
J 1 1 - 4 N J0 N
OK O
K
K 0 1 - 6O K 0O
O
K
K 0 5 - 30 O KK 0 OO
PL P
L
Add –5 times the second row to the third, and then place the result in the
third row:
J 1 1 - 4 N J0 N
OK O
K
K 0 1 - 6O K 0O
OK O
KK
0 0
0O K 0O
PL P
L
This matrix corresponds to a system of these two equations:
c1 + c2 – 4c3 = 0
c2 – 6c3 = 0
Chapter 12: Tackling Systems of First Order Linear Differential Equations
You can choose one of these constants arbitrarily. So, just for kicks, set c3 = 1.
If you plug this constant into the previous two equations and do a little algebra, you’ll find that c2 = 6 and c1 = –2.
Putting it all together
Now it’s time to assemble your constants and your vectors so you can check
your work. So:
J - 8N
J 2N
J2 N
O
K
K O
K O
c 1K 4 O + c 2 K 1 O + c 3 K 2 O
O
K
K O
KK OO
K - 22O
K - 2O
3
P
L
P
L
L P
becomes:
J 2N
J2 N J - 8 N J0 N
O K O
K O
K O K
- 2 K 4 O + 6 K 1 O + K 2O = K 0 O
O K O
K O
KK OO KK
K - 2O
3
- 22O K 0 O
P
L
P L P
L
L P
As you can see, you’ve determined c1, c2, and c3 correctly.
Eigenvalues and eigenvectors
Take a look at this vector equation:
Ax = y
When facing problems like this, it sometimes helps to consider A in a form
where a linear transformation converts x into y. That is, you want to look for
solutions of the following form:
Ax = λx
In this case, λ stands for a constant. You can rewrite this equation as:
(A – λI)x = 0
This equation has a solution only if (A – λI)–1 exists. In other words, you can
be assured of a solution if:
det(A – λI) ≠ 0
Any values of λ that satisfy (A – λI) x = 0 are called eigenvalues of the original
equation. And the solutions x to (A – λI) x = 0 are called the eigenvectors.
You can see how eigenvalues and eigenvectors work in the example that I go
through in the following sections.
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Changing the matrix to the right form
Here’s an example you can wrap your brain around: Try finding the eigenvalues and eigenvectors of the following matrix:
J-1 -1N
O
A = KK
2 - 4O
L
P
First, convert the matrix into the A – λI form. Doing so gives you:
J-1 - λ -1N
O
A - λ I = KK
2 -4 -λO
L
P
Figuring out the eigenvalues
Next, it’s time to find the determinant:
det(A – λI) = (–1 – λ)(–4 – λ) + 2
or:
det(A – λI) = λ2 + 5λ + 6
which can be factored into the following:
det(A – λI) = λ2 +5λ +6 = (λ + 2)(λ + 3)
So, by equating this equation to 0, the eigenvalues of A are λ1 = –2 and λ2 = –3.
Calculating the eigenvectors
How about finding the eigenvectors of A? To find the eigenvector corresponding to λ1, you have to substitute λ1 (–2) into the A – λI matrix, like so:
J1 -1N
O
A - λ I = KK
2 - 2O
L
P
Because (A – λI)x = 0, you have:
J 1 -1N J x 1 N J0 N
K
OK O K O
K 2 - 2O K x 2 O = K 0 O
L
PL P L P
And, because every row of this matrix equation must be true, you can
assume that x1 = x2. So, up to an arbitrary constant, the eigenvector corresponding to λ1 is:
J1N
c KK OO
1
L P
Chapter 12: Tackling Systems of First Order Linear Differential Equations
You can also drop the arbitrary constant, and write this eigenvector as:
J1N
K O
K 1O
L P
How about the eigenvector corresponding to λ2? By plugging λ2 (–3) into the
A – λI matrix, you get:
J2 -1N
O
A - λ I = KK
2 -1O
L
P
Then you have the following:
J2 -1N J x 1 N J0 N
K
OK O K O
K 2 -1O K x 2 O = K 0 O
L
PL P L P
So 2x1 – x2 = 0 and x1 = x2/2. This means that up to an arbitrary constant, the
eigenvector corresponding to λ2 is:
J1 N
c KK OO
2
L P
Again, you can drop the arbitrary constant, and simply write this
eigenvector as:
J1 N
K O
K 2O
L P
Solving Systems of First-Order Linear
Homogeneous Differential Equations
You’ve likely discovered how to work with systems of linear equations in
algebra classes. If so, you’re in luck because working with systems of linear
differential equations follows the same techniques. In the following sections,
I give you a good look at systems of first order linear homogeneous differential equations with constant coefficients like this:
x1' = x1 + x2
x2' = 4x1 + x2
These differential equations are linked — that is, both contain x1 and x2. As
such, they have to be solved together. (Flip to Chapter 5 for an introduction
to working with constant coefficients in homogeneous equations.) I show you
how in the following sections.
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Understanding the basics
You can write a set of equations such as x1' = x1 + x2 and x2' = 4x1 + x2 in this
matrix form:
J x1l
K
K x l2
L
N J 1 1N J x 1 N
O= K
OK O
O K 4 1O K x 2 O
PL P
P L
Or, shorter still, you can write the matrix like this:
x' = Ax
Here, x', A, and x are all matrices:
J x1l N
x l= KK OO
xl
L 2P
J 1 1N
O
A = KK
4 1O
L
P
J x 1N
x = KK OO
x
L 2P
If A is a matrix of constant coefficients, you can assume a solution of the following form:
x = ξert
Okay, you say, what’s that funny squiggle (ξ) doing there? Actually, it’s the
Greek letter xi in lowercase form. It couldn’t possibly stand for eigenvector,
could it? You’re right! You’re way ahead of me.
Substituting your supposed form of solution, x = ξert, into the system of differential equations, x' = Ax, gives you:
rξert = Aξert
where r is a constant. Do you see how using matrix notation makes working
with systems of differential equations easier? Now you can subtract Aξert
from both sides of the equation and do a little rearranging to get:
(A – r I)ξert = 0
or:
(A – r I)ξ = 0
Chapter 12: Tackling Systems of First Order Linear Differential Equations
As you may recall from the earlier section on eigenvalues and eigenvectors,
this is exactly the equation that specifies the eigenvalues and eigenvectors of
the matrix A. So the solution to the system of differential equations, x' = Ax,
is x = ξert (provided that r is an eigenvalue of A and ξ is the associated
eigenvector).
Making your way through an example
Now take a closer look at the example I give in the previous section:
J x1l
K
K x l2
L
N=
O
O=
P
J 1 1N J x 1 N
K
OK O
K 4 1O K x 2 O
L
PL P
Because this matrix has constant coefficients, you can try a solution of the
following form:
x = ξert
I walk you through the steps of solving this system in the following sections.
Getting the right matrix form
The next step you have to tackle is substituting x = ξert into the matrix, which
gives you:
J r ξ e rt N =
K 1 rt O
Kr ξ2 e O =
L
P
J 1 1N J ξ 1 e rt N
O
K
OK
K 4 1O K ξ e rt O
2
L
PL
P
By subtracting the left side from both sides of the equation, you can rewrite
this as:
J0 N =
K O
K 0O =
L P
J1 - r
1 N J ξ 1 e rt N
O
K
OK
K 4
1 - r O K ξ 2 e rt O
L
PL
P
Dividing by ert gives you:
J0 N =
K O
K 0O =
L P
J1 - r
1 N Jξ1N
K
OK O
K 4
1 - r O K ξ 2O
L
PL P
Finding the eigenvalues
This system of linear equations has a solution only if the determinant of the
2 × 2 matrix is zero (see the earlier section “Linear independence” for
details). So:
J1 - r
1 N
O= 0
det KK
4
1 - rO
L
P
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After expanding, your equation looks like this:
(1 – r)(1 – r) – 4 = 0
which becomes:
r 2 – 2r + 1 – 4 = 0
Or, with some simplification:
r 2 – 2r – 3 = 0
Finally, you can factor this into:
(r – 3)(r + 1) = 0
So the eigenvalues of your 2 × 2 matrix are:
r1 = 3
and
r2 = –1
Coming up with the eigenvectors
After you find the eigenvalues, you need to find the two eigenvectors.
Plugging the first eigenvalue from the previous section, r1 = 3, into the matrix
yields this result:
J0 N =
K O
K 0O =
L P
J- 2 1 N J ξ 1 N
K
OK O
K 4 - 2O K ξ 2 O
L
PL P
With some simplification you get:
–2ξ1 + ξ2 = 0
and:
4ξ1 – 2ξ2 = 0
These equations are the same up to a factor of –2. So you get:
2ξ1 = ξ2
Chapter 12: Tackling Systems of First Order Linear Differential Equations
The first eigenvector is (up to, as usual, an arbitrary nonzero constant that
doesn’t matter):
J ξ 1 N J1 N
K O= K O
K ξ 2 O K 2O
L P L P
How about the second eigenvector? It corresponds to the eigenvalue r2 = –1.
So plugging that into the matrix gives you:
J0 N =
K O
K 0O =
L P
J2 1N J ξ 1 N
K
OK O
K 4 2O K ξ 2 O
L
PL P
After simplification you get:
2ξ1 + ξ2 = 0
and:
4ξ1 + 2ξ2 = 0
These equations give you the same information as with the first eigenvector:
2ξ1 = – ξ2
So the second eigenvector becomes:
J ξ 1 N J-1N
K O= K O
K ξ 2 O K 2O
L P L P
Summing up the solution
The first solution to the original system of differential equations, based on
the first eigenvector and eigenvalue, is:
J1 N
x 1 = KK OO e 3t
2
L P
The second solution, based on the second eigenvector and eigenvalue, is:
J-1N
x 2 = KK OO e - t
2
L P
The general solution to this system is a linear combination of these two solutions (c1x1 + c2x2), which in this example looks like:
J1 N
J-1N
x = c 1KK OO e 3t + c 2 KK OO e - t
2
2
L P
L P
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The solution can also be written as:
J x 1N
J1 N
J-1N
K O = c 1K O e 3t + c 2 K O e - t
K 2O
K 2O
K x 2O
L P
L P
L P
By splitting up this equation, the solutions to the system of differential equations become:
x1 = c1 e3t – c2 e–t
and
x2 = 2c1 e3t + 2c2 e–t
That example wasn’t so bad, was it?
Solving Systems of First Order Linear
Nonhomogeneous Equations
Now I want you to turn your attention to systems of first order linear nonhomogeneous differential equations. I show you how to solve this system of
homogeneous differential equations in the previous section:
J x1l
K
K x l2
L
N=
O
O=
P
J 1 1N J x 1 N
K
OK O
K 4 1O K x 2 O
L
PL P
But what if the situation was changed to this:
J x1l
K
K x l2
L
N=
O
O=
P
J 1 1N J x 1 N J- 2e - t N
O
K
OK O K
K 4 1O K x 2 O + K 3t O
L
PL P L
P
The solution to the nonhomogeneous system is the general solution of the
homogeneous version of the system (which you already have from the previous section) plus a particular solution.
In the following sections, I show you how to assume the form of the solution
and determine the missing coefficients — all in a way entirely analogous to
the method of undetermined coefficients in linear nonhomogeneous differential equations (see Chapter 6 for an introduction to this method).
Chapter 12: Tackling Systems of First Order Linear Differential Equations
Assuming the correct form
of the particular solution
So how can you find a particular solution to a nonhomogeneous system?
Easy: By using the method of undetermined coefficients after it has been generalized to work with matrices. To do that, rewrite the system like this:
J x1l
K
K x l2
L
N=
O
O=
P
J 1 1N J x 1 N J- 2e - t N J 0 N
O K O
K
OK O K
K 4 1O K x 2 O + K 0 O + K 3t O
L
PL P L
P L P
It’s clear that the last two terms in this system involve terms in e–t and t, so
can you assume a solution of the following form?
x = ae–t + bt + c
Well, not so fast. As you may recall from the previous section, the eigenvalue
of the general solution to the homogenous equation here was –1. So there’s
already a e–t term in the general solution.
What do you do in cases like this? You do just what you would do for a single
differential equation — you have to add a term in te–t. That makes your
assumed solution look like this:
x = ate–t + be–t + ct + d
Here’s what your system of differential equations looks like currently:
J- 2e - t N J 0 N
O+ K O
x = A x + KK
0 O K 3t O
P L P
L
Plugging x = ate–t + be–t + ct + d into this system gives you:
J- 2e - t N J 0 N
O+ K O
a e - t - a te - t - be - t + c = Aa te - t + Abe - t + Act + Ad + KK
0 O K 3t O
L
P L P
And equating coefficients on the two sides gives you these results:
Aa = –a
J- 2 N
Ab = a - b - KK OO
0
J0 N L P
Ac = - KK OO
3
L P
Ad = c
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Part III: The Power Stuff: Advanced Techniques
Crunching the numbers
After assuming the correct form of your solution, you’re ready to do the
math. In the following sections, I explain how to calculate all your missing
coefficients. Remember that A is the following:
J 1 1N
O
A = KK
4 1O
L
P
Finding a
When you start calculating the missing coefficients, why not start with finding a? In this case, Aa = –a becomes:
Ja1N
J 1 1N J a 1 N
K
OK O = - K O
K 4 1OK a 2 O
K a2O
L
PL P
L P
Or, in simpler terms:
a1 + a2 = – a1
and:
4a1 + a2 = – a2
These can be rewritten even more simply as:
2a1 = –a2
and:
4a1 = –2a2
These equations differ by a factor of 2. Because you’re only interested in a
single particular solution, go with a1 = –1 and a2 = 2, which gives you:
J-1N
a = KK OO
2
L P
Finding b
Now find b. You already know the following:
J- 2 N
Ab = a - b - KK OO
0
L P
Chapter 12: Tackling Systems of First Order Linear Differential Equations
So after plugging in your solution for a, you get:
J b 1 + b 2 N J-1N J b 1 N J- 2 N
K
O K O K O K O
K 4b 1 + b 2 O = K 2 O - K b 2 O - K 0 O
L
P L P L P L P
Or, more simply:
2b1 + b2 = 1
and
4b1 + 2b2 = 2
A solution to these equations (which only differ by a factor of 2) is:
J 1N
b = KK OO
-1
L P
Finding c
Ready to find c? You already know the following:
J0 N
Ac = - KK OO
3
L P
So in other words, you get:
J 1 1N J c 1 N
J0 N
K
O K O = -K O
K 4 1O K c 2 O
K 3O
L
PL P
L P
Here’s what you get after simplifying:
c1 + c2 = 0
and:
4c1 + c2 = –3
The solution to these equations, after a little number crunching, is:
J-1N
c = KK OO
1
L P
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Part III: The Power Stuff: Advanced Techniques
Finding d
Finally you just have to find d. You know that Ad = c. In other words, by plugging in your solution for c you get:
d1 + d2 = –1
and:
4d1 + d2 = 1
So the solution to this pair of equations is:
J 23 N
d = KK 5 OO
- 3
L
P
Winding up your work
After going through the previous sections and assuming a form of the solution and crunching the numbers, you just have to put the work together to
get a particular solution to your original system. That particular solution is:
J x 1 N J-1N
J 1N
J-1N J 2 3 N
x = KK OO = KK OO te - t + KK OO e - t + KK OO t + KK 5 OO
-1
- 3
x
2
1
L P L
L P
L 2P L P
P
Chapter 13
Discovering Three Fail-Proof
Numerical Methods
In This Chapter
Brushing up on the Euler method
Taking it to the next level with the improved Euler method
Getting great results with the Runge-Kutta method
A
group of elite computer scientists shuffles into your office with a problem: “We’re not comfortable writing software that solves differential
equations. Isn’t there an easier, more computer-friendly way of handling
differential equations?”
“Well,” you say, “you can use the method of undetermined coefficients and
then solve for . . .”
“Solve for?” the group asks. “You mean using variables? No, no — we want
something numeric.”
“Ah,” you tell them. “You want methods like Euler’s method and the RungeKutta method.” You show them some programming code.
“That’s it!” the group cries, grabbing your code and running off.
“My bill,” you call after them, “will be in the mail.”
“Better use e-mail,” cries a junior member, disappearing around the corner.
Differential equations can stump even the best and brightest, but I have a
secret weapon for you: numerical methods. This chapter is all about the
computer-based methods that you can use to solve differential equations
when everything else fails. Do remember that you’ll get numbers out of these
techniques, not elegant, finished formulas. But sometimes, numbers are just
what you want, as is often the case in engineering.
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Part III: The Power Stuff: Advanced Techniques
In this chapter, I use the Java programming language, so if you want to follow
along, you need to install Java on your computer. It’s free when you visit
java.sun.com. When you’re at the Web site, simply click the Java SE link and
download and install the Java Development Kit, or JDK.
Number Crunching with Euler’s Method
Euler’s method, which I introduce in Chapter 4, allows you to handle differential equations in a numeric way. In the following sections, I explain the basics
of the method, and then I show you how to enter code into your computer so
you can see the method in action.
The fundamentals of the method
Take a look at this differential equation:
dy
= f _ x, y i
dx
The standard Euler’s method notes that you may not have the actual function
that represents the solution to your differential equation. However, when
you’re armed with the preceding equation, you do have the slope of that
curve everywhere. That is, the rate of change of the curve is the derivative.
(See Chapter 1 for a refresher on derivatives.)
Say that you have a point, (x0, y0), that’s on the solution curve. Because of
the preceding equation, you know that the slope of the solution curve at that
point is f(x0, y0). Suppose that you also want to find the numeric solution at
a point, (x, y), a short distance, h, away. Here’s how Euler’s method says that
you may find y:
y = y0 + ∆y
In other words, y is equal to y at an initial point, plus the change in y.
Because the slope, m, is defined as ∆y/∆x (a change in y divided by a change
in x), this equation also can be written like this:
y = y0 + m ∆x
And because ∆ x = x – x0, this equation is also:
y = y0 + m (x – x0)
Chapter 13: Discovering Three Fail-Proof Numerical Methods
Here’s the key: The slope m is equal to the derivative at (x0, y0), and because
of your original differential equation, you know that m = f (x0, y0). So you have:
y = y0 + f(x0, y0) (x – x0)
Now convert (x – x0) to the symbol that it usually goes by when you discuss
Euler’s method: h, which gives you this equation:
y1 = y0 + f(x0, y0) h
This equation can be generalized to any point, (xn, yn), along the solution
curve like this:
yn + 1 = yn + f(xn, yn) h
And there you have it — you’ve discovered the recurrence relation, which ties
one term to the next, for Euler’s equation. (Flip to Chapter 9 for an introduction to recurrence relations.)
Using code to see the method in action
Now try testing the basics from the previous section, using this differential
equation:
dy
=x
dx
where y(0) = 0.
I’ll spare you the details of solving this equation with traditional methods
(but you can head to Chapter 4 to find out how to do so). Without further
ado, here’s the exact solution:
2
y= x
2
Because you know the exact solution, you can check the accuracy of Euler’s
method in Java code. The following sections show you how.
Typing in the code
In Chapter 4, I develop a short Java program, e.java, to display the results of
Euler’s method versus the exact solution. The code starts at (x0, y0) = (0, 0),
which you know is on the solution curve because of the initial condition,
y(0) = 0. The code calculates 100 steps, using a step size of h = 0.1.
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Part III: The Power Stuff: Advanced Techniques
If you want to follow along, start by using the Java compiler, javac.exe, to
compile e.java:
C:\>javac e.java
If javac.exe isn’t in your computer’s path, you have to specify that path this
way to run javac.exe:
C:\>C:\jdk\bin\javac e.java
After compiling e.java, you should get a new file, e.class. Now you can execute the compiled code, e.class, using java.exe like this:
C:\>java e
Here’s the code that you’re going to modify in this chapter (note that the
sections of the code that you have to change when you’re solving your own
differential equations are given in bold):
public class e
{
double
double
double
double
x0 = 0.0;
y0 = 0.0;
h = 0.1;
n = 100;
public e()
{
}
public double f(double x, double y)
{
return x;
}
public double exact(double x, double y)
{
return x * x / 2;
}
public static void main(String [] argv)
{
e de = new e();
de.calculate();
}
public void calculate()
{
Chapter 13: Discovering Three Fail-Proof Numerical Methods
double x = x0;
double y = y0;
double k;
System.out.println(“x\t\tEuler\t\tExact”);
for (int i = 1; i < n; i++){
k = f(x, y);
y = y + h * k;
x = x + h;
System.out.println(round(x) + “\t\t” + round(y) +
“\t\t” + round(exact(x, 0)));
}
}
public double round(double val)
{
double divider = 100;
val = val * divider;
double temp = Math.round(val);
return (double)temp / divider;
}
}
Surveying the results
After being run, the code will display the current x value, the Euler approximation of the solution at that value, and the exact solution, like this:
C:\>java e
x
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Euler
0.0
0.01
0.03
0.06
0.1
0.15
0.21
0.28
0.36
0.45
0.55
0.66
0.78
0.91
1.05
1.2
1.36
1.53
Exact
0.01
0.02
0.05
0.08
0.13
0.18
0.24
0.32
0.4
0.5
0.6
0.72
0.85
0.98
1.13
1.28
1.45
1.62
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Part III: The Power Stuff: Advanced Techniques
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
1.71
1.9
2.1
2.31
2.53
2.76
3.0
3.25
3.51
3.78
4.06
4.35
4.65
4.96
5.28
5.61
5.95
6.3
6.66
7.03
7.41
7.8
8.2
8.61
9.03
9.46
9.9
10.35
10.81
11.28
11.76
12.25
12.75
13.26
13.78
14.31
14.85
15.4
15.96
16.53
17.11
17.7
18.3
18.91
19.53
20.16
20.8
21.45
22.11
22.78
23.46
1.81
2.0
2.21
2.42
2.65
2.88
3.13
3.38
3.65
3.92
4.21
4.5
4.81
5.12
5.45
5.78
6.13
6.48
6.85
7.22
7.61
8.0
8.41
8.82
9.25
9.68
10.13
10.58
11.04
11.52
12.0
12.5
13.0
13.52
14.04
14.58
15.12
15.68
16.24
16.82
17.4
18.0
18.6
19.22
19.84
20.48
21.12
21.78
22.44
23.12
23.8
Chapter 13: Discovering Three Fail-Proof Numerical Methods
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
24.15
24.85
25.56
26.28
27.01
27.75
28.5
29.26
30.03
30.81
31.6
32.4
33.21
34.03
34.86
35.7
36.55
37.41
38.28
39.16
40.05
40.95
41.86
42.78
43.71
44.65
45.6
46.56
47.53
48.51
24.5
25.2
25.92
26.64
27.38
28.12
28.88
29.64
30.42
31.2
32.0
32.8
33.62
34.44
35.28
36.12
36.98
37.84
38.72
39.6
40.5
41.4
42.32
43.24
44.18
45.12
46.08
47.04
48.02
49.0
Not bad. Euler’s method came pretty close to the exact solution. In fact, at a
value of x = 9.9, Euler’s method is still within 1% of the exact solution.
Moving On Up with the Improved
Euler’s Method
As I explain in the earlier section “The fundamentals of the method,” the
recurrence relation for the Euler method is given by:
yn + 1 = yn + f(xn, yn) h
where the differential equation you’re trying to solve is:
dy
= f _ x, y i
dx
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Part III: The Power Stuff: Advanced Techniques
This is a fairly simplistic method; it just extrapolates from the current point
to the next point using the known slope. What if the actual solution varies
faster than the Euler method takes into account?
Well, it turns out that an improved Euler method exists. This method, which
is also called the Heun formula, can be more accurate. I explain everything
you need to know in the following sections.
Understanding the improvements
The standard Euler method assumes that the difference between yn+1 and yn is:
∆y = f(xn) h
For simplicity’s sake, assume that f(x, y) is only a function of x, f(x).
This method doesn’t take into account possible steep increases — or
decreases — in the slope. A better solution would not only take the current
slope, f(xn), into account, but it also would take the slope at the next point
along the curve, f(xn + h), into account. You could then take the average of
those two slopes, like this:
m=
f _ x n i + f _ x n + hi
2
This is a better approximation than simply using the slope at xn, which the
standard Euler’s method does. Now you can multiply the average slope by
the interval length, h, to find the change in y:
∆y = h 9 f _ x n i + f _ x n + h i C
2
So you can express the recurrence relation for the (new and) improved
Euler’s method like this:
y n + 1 = y n + h 9 f _ x n i + f _ x n + hi C
2
Coming up with new code
In this section, I show you how to create a new program, e2.java, that will put
the improved Euler method to work. The program will display the traditional
Euler result, the improved Euler result, and the exact result. To illustrate the
new code, I use the same equation from the earlier section “Using code to see
the method in action” in the following sections.
Chapter 13: Discovering Three Fail-Proof Numerical Methods
Changes in the new code
The code for the improved Euler’s method is different from the one in the earlier
section “Using code to see the method in action.” Here, you create a new variable, yimp, in which you store the improved Euler results, and then you make
these changes to the calculate method (the bold indicates changes in the code):
public void calculate()
{
double x = x0;
double y = y0;
double yimp = y0;
double k;
System.out.println(“x\t\tEuler\t\tImproved\t\tExact”);
for (int i = 1; i < n; i++){
k = f(x, y);
y = y + h * k;
yimp = y + (f(x, y) + f(x + h, y))* (h/2);
x = x + h;
System.out.println(round(x) + “\t\t” + round(y) +
“\t\t” + round(yimp) + “\t\t”
+ round(exact(x, 0)));
}
}
Here’s what the program looks like now:
public class e2
{
double
double
double
double
x0 = 0.0;
y0 = 0.0;
h = 0.1;
n = 100;
public e2()
{
}
public double f(double x, double y)
{
return x;
}
public double exact(double x, double y)
{
return x * x / 2;
}
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Part III: The Power Stuff: Advanced Techniques
public static void main(String [] argv)
{
e2 de = new e2();
de.calculate();
}
public void calculate()
{
double x = x0;
double y = y0;
double yimp = y0;
double k;
System.out.println(“x\t\tEuler\t\tImproved\t\tExact”);
for (int i = 1; i < n; i++){
k = f(x, y);
y = y + h * k;
yimp = y + (f(x, y) + f(x + h, y))* (h/2);
x = x + h;
System.out.println(round(x) + “\t\t” + round(y) +
“\t\t” + round(yimp) + “\t\t”
+ round(exact(x, 0)));
}
}
public double round(double val)
{
double divider = 100;
val = val * divider;
double temp = Math.round(val);
return (double)temp / divider;
}
}
The results of the new code
So how does the code work out? Take a look:
C:\>java e2
x
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Euler
0.0
0.01
0.03
0.06
0.1
0.15
0.21
0.28
0.36
0.45
Improved
0.01
0.03
0.06
0.1
0.15
0.21
0.28
0.36
0.45
0.55
Exact
0.01
0.02
0.05
0.08
0.13
0.18
0.24
0.32
0.4
0.5
Chapter 13: Discovering Three Fail-Proof Numerical Methods
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
6.2
0.55
0.66
0.78
0.91
1.05
1.2
1.36
1.53
1.71
1.9
2.1
2.31
2.53
2.76
3.0
3.25
3.51
3.78
4.06
4.35
4.65
4.96
5.28
5.61
5.95
6.3
6.66
7.03
7.41
7.8
8.2
8.61
9.03
9.46
9.9
10.35
10.81
11.28
11.76
12.25
12.75
13.26
13.78
14.31
14.85
15.4
15.96
16.53
17.11
17.7
18.3
18.91
0.66
0.78
0.91
1.05
1.2
1.36
1.53
1.71
1.9
2.1
2.31
2.53
2.76
3.0
3.25
3.51
3.78
4.06
4.35
4.65
4.96
5.28
5.61
5.95
6.3
6.66
7.03
7.41
7.8
8.2
8.61
9.03
9.46
9.9
10.35
10.81
11.28
11.76
12.25
12.75
13.26
13.78
14.31
14.85
15.4
15.96
16.53
17.11
17.7
18.29
18.9
19.52
0.6
0.72
0.85
0.98
1.13
1.28
1.45
1.62
1.81
2.0
2.21
2.42
2.65
2.88
3.13
3.38
3.65
3.92
4.21
4.5
4.81
5.12
5.45
5.78
6.13
6.48
6.85
7.22
7.61
8.0
8.41
8.82
9.25
9.68
10.13
10.58
11.04
11.52
12.0
12.5
13.0
13.52
14.04
14.58
15.12
15.68
16.24
16.82
17.4
18.0
18.6
19.22
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Part III: The Power Stuff: Advanced Techniques
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
19.53
20.16
20.8
21.45
22.11
22.78
23.46
24.15
24.85
25.56
26.28
27.01
27.75
28.5
29.26
30.03
30.81
31.6
32.4
33.21
34.03
34.86
35.7
36.55
37.41
38.28
39.16
40.05
40.95
41.86
42.78
43.71
44.65
45.6
46.56
47.53
48.51
20.15
20.79
21.44
22.1
22.77
23.45
24.14
24.84
25.55
26.27
27.0
27.74
28.49
29.25
30.02
30.8
31.59
32.39
33.2
34.02
34.85
35.69
36.54
37.4
38.27
39.15
40.04
40.94
41.85
42.77
43.7
44.64
45.59
46.55
47.52
48.5
49.49
19.84
20.48
21.12
21.78
22.44
23.12
23.8
24.5
25.2
25.92
26.64
27.38
28.12
28.88
29.64
30.42
31.2
32.0
32.8
33.62
34.44
35.28
36.12
36.98
37.84
38.72
39.6
40.5
41.4
42.32
43.24
44.18
45.12
46.08
47.04
48.02
49.0
As you can see, Euler’s method and the improved Euler’s method come out
about the same here. They’re both off by 1% at x = 9.9.
Plugging a steep slope into the new code
The improvement in the improved Euler’s method isn’t often visible until you
have steep or quickly varying slopes. For example, take a look at this differential equation:
dy
= x6
dx
where y(0) = 0.
Chapter 13: Discovering Three Fail-Proof Numerical Methods
Here’s the exact solution (I’ll spare you the calculations so you can get to the
coding faster):
7
y= x
7
Okay, so if the improved Euler method is truly improved, you should be able
to see difference in this example, where the slope changes faster than in the
differential equation in the previous section. Keep reading to find out what
happens.
Checking out the results
As you can see, when you use the steeply sloping differential equation, there’s
a difference between the Euler method and the improved Euler method:
x
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
9.0
9.1
9.2
Euler
.
.
.
37696.93
42362.53
47514.57
53194.59
59446.94
66318.89
73860.78
82126.18
91172.01
101058.76
111850.58
123615.48
136425.51
150356.91
165490.34
181910.99
199708.84
218978.83
239821.07
262341.03
286649.77
312864.17
341107.13
371507.8
404201.83
439331.64
477046.59
517503.31
560865.93
607306.34
657004.47
710148.57
766935.49
Improved
Exact
42138.76
47271.35
52930.6
59160.78
66009.09
73525.81
81764.42
90781.79
100638.31
111398.05
123128.94
135902.94
149796.23
164889.33
181267.37
199020.24
218242.76
239034.95
261502.17
285755.38
311911.35
340092.85
370428.94
403055.15
438113.75
475754.01
516132.42
559412.98
605767.45
655375.61
708425.58
765114.08
825646.71
39990.86
44896.33
50308.78
56271.15
62829.24
70031.83
77930.87
86581.59
96042.7
106376.48
117649.0
129930.29
143294.47
157819.98
173589.72
190691.27
209217.07
229264.62
250936.7
274341.56
299593.14
326811.32
356122.1
387657.87
421557.64
457967.27
497039.75
538935.42
583822.28
631876.21
683281.29
738230.03
796923.72
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Part III: The Power Stuff: Advanced Techniques
9.3
9.4
9.5
9.6
9.7
9.8
9.9
827570.99
892270.01
961256.99
1034766.18
1113041.96
1196339.16
1284923.4
890238.25
959113.01
1032505.07
1110658.66
1193828.45
1282279.88
1376289.52
859572.67
926396.56
997624.71
1073496.4
1154261.21
1240179.33
1331521.93
Even though there is indeed a difference here (at x = 9.9, the Euler method
is off by 3.5% and the improved Euler method is off by 3.4%), the difference
isn’t huge.
Increasing your accuracy with a decreased step size
One way to increase your accuracy is to decrease your step size. For instance,
decrease the step size from h = 0.1 to h = 0.01. Also increase the number of
steps from n = 100 to n = 1,000. Here are the changes to make in the code in
e2.java:
public class e2
{
double
double
double
double
x0 = 0.0;
y0 = 0.0;
h = 0.01;
n = 1000;
.
.
.
And here’s what the results look like:
x
9.0
9.01
9.02
9.03
9.04
9.05
9.06
9.07
9.08
9.09
9.1
9.11
9.12
9.13
Euler
.
.
.
680627.03
685941.44
691291.38
696677.04
702098.63
707556.34
713050.38
718580.94
724148.23
729752.45
735393.8
741072.49
746788.73
752542.72
Improved
685923.78
691273.62
696659.18
702080.67
707538.28
713032.22
718562.68
724129.87
729733.98
735375.23
741053.82
746769.96
752523.84
758315.69
Exact
683281.29
688613.44
693981.23
699384.84
704824.47
710300.33
715812.61
721361.52
726947.25
732570.02
738230.03
743927.48
749662.57
755435.52
Chapter 13: Discovering Three Fail-Proof Numerical Methods
9.14
9.15
9.16
9.17
9.18
9.19
9.2
9.21
9.22
9.23
9.24
9.25
9.26
9.27
9.28
9.29
9.3
9.31
9.32
9.33
9.34
9.35
9.36
9.37
9.38
9.39
9.4
9.41
9.42
9.43
9.44
9.45
9.46
9.47
9.48
9.49
9.5
9.51
9.52
9.53
9.54
9.55
9.56
9.57
9.58
9.59
9.6
9.61
9.62
9.63
9.64
758334.67
764164.78
770033.28
775940.36
781886.24
787871.12
793895.24
799958.79
806061.99
812205.06
818388.22
824611.67
830875.66
837180.38
843526.06
849912.93
856341.2
862811.1
869322.86
875876.69
882472.83
889111.5
895792.94
902517.36
909285.01
916096.1
922950.88
929849.58
936792.43
943779.67
950811.53
957888.25
965010.06
972177.22
979389.95
986648.5
993953.12
1001304.04
1008701.51
1016145.77
1023637.07
1031175.66
1038761.79
1046395.71
1054077.66
1061807.9
1069586.69
1077414.26
1085290.89
1093216.82
1101192.32
764145.7
770014.09
775921.06
781866.84
787851.62
793875.62
799939.07
806042.16
812185.13
818368.17
824591.52
830855.39
837160.01
843505.58
849892.34
856320.5
862790.29
869301.93
875855.65
882451.68
889090.24
895771.56
902495.87
909263.4
916074.38
922929.04
929827.62
936770.36
943757.47
950789.21
957865.82
964987.51
972154.55
979367.16
986625.59
993930.09
1001280.88
1008678.23
1016122.37
1023613.55
1031152.02
1038738.02
1046371.82
1054053.64
1061783.76
1069562.41
1077389.87
1085266.37
1093192.17
1101167.54
1109192.73
761246.54
767095.82
772983.59
778910.05
784875.42
790879.9
796923.72
803007.07
809130.19
815293.29
821496.58
827740.28
834024.61
840349.8
846716.05
853123.6
859572.67
866063.48
872596.26
879171.23
885788.61
892448.65
899151.56
905897.57
912686.92
919519.84
926396.56
933317.32
940282.34
947291.87
954346.14
961445.39
968589.85
975779.78
983015.4
990296.96
997624.71
1004998.88
1012419.73
1019887.49
1027402.42
1034964.76
1042574.76
1050232.67
1057938.75
1065693.24
1073496.4
1081348.48
1089249.74
1097200.43
1105200.82
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Part III: The Power Stuff: Advanced Techniques
9.65
9.66
9.67
9.68
9.69
9.7
9.71
9.72
9.73
9.74
9.75
9.76
9.77
9.78
9.79
9.8
9.81
9.82
9.83
9.84
9.85
9.86
9.87
9.88
9.89
9.9
1109217.64
1117293.04
1125418.77
1133595.11
1141822.31
1150100.64
1158430.36
1166811.74
1175245.04
1183730.53
1192268.48
1200859.17
1209502.85
1218199.81
1226950.31
1235754.64
1244613.06
1253525.86
1262493.31
1271515.69
1280593.28
1289726.36
1298915.22
1308160.14
1317461.39
1326819.28
1117268.0
1125393.6
1133569.81
1141796.88
1150075.08
1158404.66
1166785.91
1175219.08
1183704.43
1192242.25
1200832.8
1209476.35
1218173.17
1226923.54
1235727.73
1244586.02
1253498.68
1262465.99
1271488.23
1280565.68
1289698.62
1298887.33
1308132.11
1317433.22
1326790.97
1336205.62
1113251.15
1121351.7
1129502.71
1137704.46
1145957.2
1154261.21
1162616.73
1171024.05
1179483.42
1187995.12
1196559.42
1205176.58
1213846.88
1222570.59
1231347.98
1240179.33
1249064.92
1258005.02
1266999.91
1276049.87
1285155.19
1294316.13
1303533.0
1312806.06
1322135.61
1331521.93
When using the smaller step size, the Euler method and the improved Euler
method are only off by about 0.35% at x = 9.9. Much better.
Adding Even More Precision with
the Runge-Kutta Method
If you don’t want to use either of the Euler methods, you’re in luck. There’s
another numerical method that you can use to solve differential equations:
the Runge-Kutta method. This method gives excellent results that are even
more accurate than the Euler or improved Euler methods.
The method’s recurrence relation
In the Runge-Kutta method, the recurrence relation is a weighted average of
terms:
y n +1 = y n + h 7c1 + c 2 + c 3 + c 4A
6
Chapter 13: Discovering Three Fail-Proof Numerical Methods
The story of Runge and Kutta
Carl David Tolmé Runge was a German mathematician. He was born in Havana, Cuba, on
August 30, 1856, where his father was the Danish
consul. The family later moved to Bremen,
Germany. Runge died on January 3, 1927.
Martin Wilhelm Kutta was also a German mathematician. He was born in Pitschen, Germany
(which today is part of Poland), on November 3,
1867. He went to the University of Breslau and
continued to work in Munich. He also spent a
year at the University of Cambridge. He became
a professor in Stuttgart, Germany, in 1911. Kutta
died on December 25, 1944.
In 1901, Runge and Kutta co-developed the
Runge-Kutta method, which is the powerful
method that’s used to solve ordinary differential
equations numerically.
where:
c1 = f(xn yn)
c 2 = f c x n + h, y n + c1 h m
2
2
c3 = f c x n + h, y n + c2 h m
2
2
c4 = f(xn + h, yn + c3h)
This method can easily be adapted to the code you use earlier in the chapter,
because f(x, y) doesn’t depend on y — it’s just f(x). In this case, the recurrence relation becomes:
y n + 1 = y n + h = f _ x n i + f c x n + h m + f c x n + h m + f _ x n + hi G
6
2
2
Working with the method in code
Here, you put the Runge-Kutta technique to work, solving the same differential equation that you saw earlier in the chapter:
dy
=x
dx
Inputting the code
The recurrence relation of the Runge-Kutta method looks easy enough to
implement in a new Java program, e3.java. This new program will put the
Runge-Kutta method to work. The new code is shown in bold.
309
310
Part III: The Power Stuff: Advanced Techniques
public class e3
{
double
double
double
double
x0 = 0.0;
y0 = 0.0;
h = 0.1;
n = 100;
public e3()
{
}
public double f(double x, double y)
{
return x;
}
public double exact(double x, double y)
{
return x * x / 2;
}
public static void main(String [] argv)
{
e3 de = new e3();
de.calculate();
}
public void calculate()
{
double x = x0;
double y = y0;
double yrk = y0;
double k;
System.out.println(“x\t\tEuler\t\tRungeKutta\t\tExact”);
for (int i = 1; i < n; i++){
k = f(x, y);
y = y + h * k;
yrk = y + (f(x, y) + f(x + h/2, y)/2 + f(x + h/2, y)/2
+ f(x + h, y))* (h/6);
x = x + h;
System.out.println(round(x) + “\t\t” + round(y) +
“\t\t” + round(yrk) + “\t\t”
+ round(exact(x, 0)));
}
}
public double round(double val)
Chapter 13: Discovering Three Fail-Proof Numerical Methods
{
double divider = 100;
val = val * divider;
double temp = Math.round(val);
return (double)temp / divider;
}
}
Examining the results
Here are the results of e3.java:
C:\>java e3
x
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Euler
0.0
0.01
0.03
0.06
0.1
0.15
0.21
0.28
0.36
0.45
0.55
0.66
0.78
0.91
1.05
1.2
1.36
1.53
1.71
1.9
2.1
2.31
2.53
2.76
3.0
3.25
3.51
3.78
4.06
4.35
4.65
4.96
5.28
5.61
5.95
6.3
6.66
7.03
Runge-Kutta
0.0
0.02
0.04
0.08
0.12
0.18
0.24
0.32
0.4
0.5
0.6
0.72
0.84
0.98
1.12
1.28
1.44
1.62
1.8
2.0
2.2
2.42
2.64
2.88
3.12
3.38
3.64
3.92
4.2
4.5
4.8
5.12
5.44
5.78
6.12
6.48
6.84
7.22
Exact
0.01
0.02
0.05
0.08
0.13
0.18
0.24
0.32
0.4
0.5
0.6
0.72
0.85
0.98
1.13
1.28
1.45
1.62
1.81
2.0
2.21
2.42
2.65
2.88
3.13
3.38
3.65
3.92
4.21
4.5
4.81
5.12
5.45
5.78
6.13
6.48
6.85
7.22
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Part III: The Power Stuff: Advanced Techniques
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
7.41
7.8
8.2
8.61
9.03
9.46
9.9
10.35
10.81
11.28
11.76
12.25
12.75
13.26
13.78
14.31
14.85
15.4
15.96
16.53
17.11
17.7
18.3
18.91
19.53
20.16
20.8
21.45
22.11
22.78
23.46
24.15
24.85
25.56
26.28
27.01
27.75
28.5
29.26
30.03
30.81
31.6
32.4
33.21
34.03
34.86
35.7
36.55
37.41
38.28
7.6
8.0
8.4
8.82
9.24
9.68
10.12
10.58
11.04
11.52
12.0
12.5
13.0
13.52
14.04
14.58
15.12
15.68
16.24
16.82
17.4
18.0
18.6
19.22
19.84
20.48
21.12
21.78
22.44
23.12
23.8
24.5
25.2
25.92
26.64
27.38
28.12
28.88
29.64
30.42
31.2
32.0
32.8
33.62
34.44
35.28
36.12
36.98
37.84
38.72
7.61
8.0
8.41
8.82
9.25
9.68
10.13
10.58
11.04
11.52
12.0
12.5
13.0
13.52
14.04
14.58
15.12
15.68
16.24
16.82
17.4
18.0
18.6
19.22
19.84
20.48
21.12
21.78
22.44
23.12
23.8
24.5
25.2
25.92
26.64
27.38
28.12
28.88
29.64
30.42
31.2
32.0
32.8
33.62
34.44
35.28
36.12
36.98
37.84
38.72
Chapter 13: Discovering Three Fail-Proof Numerical Methods
8.9
9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
39.16
40.05
40.95
41.86
42.78
43.71
44.65
45.6
46.56
47.53
48.51
39.6
40.5
41.4
42.32
43.24
44.18
45.12
46.08
47.04
48.02
49.0
39.6
40.5
41.4
42.32
43.24
44.18
45.12
46.08
47.04
48.02
49.0
When you look at the results, you can see that the Runge-Kutta method beat
the Euler method hands down. In fact, to two decimal places, the Runge-Kutta
method nearly always nailed the correct answer. Very cool.
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Part III: The Power Stuff: Advanced Techniques
Part IV
The Part of Tens
E
In this part . . .
very For Dummies book features a Part of Tens, and
who am I to break with tradition? This part is full of
cool stuff. You discover the ten top differential equation
online tutorials as well as the ten top online tools for solving differential equations. If you want to get more info on a
specific aspect of differential equations, try the tutorials.
If you want to work with challenging differential equations
and need a little help, take a look at the tools available
online.
Chapter 14
Ten Super-Helpful Online
Differential Equation Tutorials
In This Chapter
Understanding different aspects of differential equations with tutorials
Checking out helpful notes and videos
E
ven if you think of yourself as a math whiz, you still may find yourself
scratching your head at certain aspects of differential equations. Fear not:
A number of differential equation tutorials are available on the Internet — and
this chapter lists ten of my favorites. Each site deals with various aspects of
differential equations.
AnalyzeMath.com’s Introduction
to Differential Equations
If you’re new to differential equations, you may be looking for a brief
overview of the topic. The Web site AnalyzeMath.com offers just that. If
you visit www.analyzemath.com/calculus/Differential_Equations/
introduction.html, you can read an introductory tutorial on differential
equations. The site is run by Abdelkader Dendane, PhD, a lecturer in mathematics at United Arab Emirates University.
318
Part IV: The Part of Tens
Harvey Mudd College Mathematics
Online Tutorial
If you’re looking for a tutorial that gives good coverage of solving first order
ordinary differential equations, be sure to check out the Harvey Mudd
College Mathematics Online Tutorial at www.math.hmc.edu/calculus/
tutorials/odes.
John Appleby’s Introduction
to Differential Equations
John Appleby, a professor in the School of Mathematical Sciences at Dublin
City University in Ireland, maintains a helpful tutorial at webpages.dcu.ie/
~applebyj/ms225/ms225.html. Be sure to click the links under the Tutorial
Sheets and the Supplementary Notes headers. On his site, Appleby covers
separable and homogeneous equations, first order linear differential equations, second order linear differential equations, variation of parameters,
and more.
Kardi Teknomo’s Page
Dr. Kardi Teknomo is a research fellow at Human Centered Mobility
Technologies in Arsenal Research (which is located in Austria). Teknomo’s
tutorial focuses on solving differential equations using numerical methods.
You can find this great tutorial at people.revoledu.com/kardi/tutorial/
ODE/index.html.
Martin J. Osborne’s Differential
Equation Tutorial
Martin J. Osborne, a professor of economics at the University of Toronto, has
created a superb, multipart differential equation tutorial at www.economics.
utoronto.ca/osborne/MathTutorial/IDEF.HTM.
Chapter 14: Ten Super-Helpful Online Differential Equation Tutorials
Midnight Tutor’s Video Tutorial
If you’re looking for something fun and different, check out the Midnight
Tutor’s Video Tutorial, which is a video tutorial solving y' – ycos(t) = cos(t)
using separation of variables. To view the video, visit www.midnighttutor.
com/de_xprime-xcost-xcost.html.
The Ohio State University Physics
Department’s Introduction
to Differential Equations
If you struggle with homogeneous and nonhomogeneous linear differential
equations, there’s still hope! Check out the Ohio State University Physics
Department tutorial on these types of differential equations. You can find
the site at www.physics.ohio-state.edu/~physedu/mapletutorial/
tutorials/diff_eqs/intro.html.
Paul’s Online Math Notes
Paul’s Online Math Notes is an extensive set of online explanations of differential equations. You can visit this site at tutorial.math.lamar.edu/
Classes/DE/DE.aspx. (Who’s Paul? He’s Paul Dawkins, who teaches at
Lamar University in Beaumont, Texas.)
S.O.S. Math
S.O.S. Math (www.sosmath.com/diffeq/diffeq.html) is an all-purpose
resource for all of mathematics, but it has an entire section on differential
equations. S.O.S Math is a great, multipart tutorial — it’s as complete a tutorial as you’ll find online.
319
320
Part IV: The Part of Tens
University of Surrey Tutorial
England’s University of Surrey provides an excellent tutorial on first and
second order differential equations. You can view this site at www.maths.
surrey.ac.uk/explore/vithyaspages. Vithya Nanthakumaar, who studies
at the University of Surrey, created the site.
Chapter 15
Ten Really Cool Online Differential
Equation Solving Tools
In This Chapter
Solving equations with handy online tools
Plotting direction fields and graphing functions
S
uppose you’re working on a particularly hairy differential equation that
requires some busywork. Do you need to sit and do pages of calculations? Heck no! That’s what differential equation solving tools are for! And
tons are available. Some programs cost money — and sometimes, they cost a
great deal of money. However, a bunch of good tools also are available for
free online. This chapter has a sampling of those great freebies.
AnalyzeMath.com’s Runge-Kutta
Method Applet
The AnalyzeMath.com Runge-Kutta Applet solves a number of representative
differential equations using the Runge-Kutta method that I explain in
Chapter 13. You can find this applet at www.analyzemath.com/calculus/
RungeKutta/RungeKutta.html. A couple of cool features of this applet:
You can increase the number of points used in the calculation, and you can
zoom in and out.
Coolmath.com’s Graphing Calculator
The Coolmath.com graphing calculator is a fabulous online graphing applet
that draws functions for you. You can find this calculator at www.coolmath.
com/graphit/index.html.
322
Part IV: The Part of Tens
Direction Field Plotter
Want to see what the direction field for a differential equation’s solution looks
like? You’re in luck! You can find a great direction field plotter at www.math.
ubc.ca/~israel/applet/dfplotter/dfplotter.html. (Flip to Chapter 1
for an introduction to direction fields.)
An Equation Solver from QuickMath
Automatic Math Solutions
The solver from QuickMath Automatic Math Solutions is an equation solver,
not a differential equation solver. But it’s still a great tool when you need to
factor an equation, such as a characteristic equation, to find the roots. Here’s
the Web site where you can find the solver: www.hostsrv.com/webmab/app1/
MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=
solve&s3=basic.
First Order Differential Equation Solver
Here’s a cool one: the First Order Differential Equation Solver at www.cs.
gordon.edu/~senning/desolver/index.html. With this solver you can
use the Euler, Improved Euler, and Runge-Kutta methods to solve differential
equations. Just select the method you want to use from the drop-down box,
enter your equation, and then click the Submit button. You couldn’t find a
niftier math tool.
GCalc Online Graphing Calculator
If you’re working on a tough problem and don’t have a graphing
calculator handy, you can find a good graphing calculator applet at www.
calculator.com/calcs/GCalc.html. Just enter the function you want to
graph and press Enter on your keyboard. It doesn’t get easier than that!
Chapter 15: Ten Really Cool Online Differential Equation Solving Tools
JavaView Ode Solver
The JavaView Ode Solver at www.javaview.de/services/odeSolver/
index.html is a numerical differential equation solver that uses the
Runge-Kutta method in Chapter 13. (“Ode” stands for “ordinary differential
equation.”)
Math @ CowPi’s System Solver
The Math @ CowPi’s System Solver lets you solve simultaneous equations —
from 2 × 2 to 5 × 5. Just fill in all the blanks, and presto! You have your
answer. This great tool is available at math.cowpi.com/systemsolver.
This tool also can be useful when you break a higher-order differential equation into a system of lower-order ones.
A Matrix Inverter from QuickMath
Automatic Math Solutions
When working with systems of differential equations, you work with equations like Ax = b. The solution to an equation like this is x = A–1b, where A–1 is
the inverse of the matrix A. Don’t feel like solving all that? Here’s a tool that
finds the inverse of matrices for you in a snap: www.hostsrv.com/webmab/
app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=matrices&
s2=inverse&s3=basic. All you have to do is enter the matrix and click the
Inverse button.
Visual Differential Equation
Solving Applet
The Visual Differential Equation Solving Applet at www.falstad.com/diffeq
(which runs in browsers) lets you solve some common differential equations
and adjust numeric parameters.
323
324
Part IV: The Part of Tens
Index
• Symbols & Numerics •
* operator, 260
1 function, 244
•A•
absolute convergence of series, 192
adjoints of matrices, 269
advanced techniques, 21
Airy, George Biddell, 208
Airy’s equation
differentiating and substituting power
series into differential equation,
207–208
ensuring same index value, 208
overview, 207
putting together solution, 210–211
using recurrence relation to find
coefficients, 209–210
AnalyzeMath.com
introduction to differential equations, 317
Runge-Kutta applet, 321
angles
of pendulum equations, 18
phase, 149
Appleby, John, 318
augmented matrices, 267
•B•
bank interest rate problem, 55–59
body falling through air example, 13–14
boldfaced text in book, 2
•C•
calculator, graphing
Coolmath.com, 321
GCalc, 322
chain rule, 44, 65
classifying differential equations
linear versus nonlinear equations, 18–19
by order, 17
ordinary versus partial equations, 17–18
overview, 17
coefficients, undetermined. See method of
undetermined coefficients
complex conjugates of matrices, 269
complex roots, 100, 164–166, 223–224
compound interest rate problem, 55–59
conjugates of matrices, complex, 269
connecting slopes into integral curve,
14–15
constant coefficients, second order linear
homogeneous differential equations
with
elementary solutions, 94–95
linear equations, 95–96
overview, 94
constants, 11–12, 26
continuous functions, 35, 93, 195
conventions used in this book, 1–2
convergence, radius of, 193
convolution integrals, 259–261
Coolmath.com graphing calculator, 321
cos at function, 244
cosh at function, 244
cosine, finding particular solutions in
cosine form, 185
curve, 14–15. See also Euler’s method
•D•
damping, 148
Dawkins, Paul, 319
Dendane, Abdelkader, 317
derivatives
of cos(x), 12
involving multiple functions, 12–13
involving trigonometry, 12
Laplace transform of, 247
overview, 8
326
Differential Equations For Dummies
derivatives (continued)
partial, 17–18
of product of two functions, 13
of quotient of two functions, 13
of sin(x), 12
of sum (or difference) of two functions, 12
that are constants, 11–12
that are powers, 12
determinants, of matrices, 274
difference equations
equilibrium solutions
overview, 85
working with constant, 86–87
working without constant, 86
iterative solutions, 84–85
overview, 83
terminology related to, 84
direction fields
connecting slopes into integral curve,
14–15
equilibrium value in, 16
and first derivative, 20
of flow problem solution, 54
graphs
of advanced solution, 34
equilibrium value in, 16
of flow problem solution, 54
of nonlinear separable solution, 45
of separable equation with hard-to-find
explicit solution, 49
of separable equation with initial
conditions, 47
solution in, 15
of nonlinear separable equation, 45
overview, 13
plotter for, 322
plotting, 13–14
recognizing equilibrium value, 16
of separable equations, 47
distinct roots, 220–221. See also real and
distinct roots
drag coefficient, 14
drag force, mass with, 148–150
duplicate roots
fifth order equation with identical real
roots, 167–168
fourth order equation with identical real
roots, 166–167
identical complex roots, 170–171
identical imaginary roots, 169–170
overview, 166
•E•
e, raising to certain power, 12
ea, inverse of, 12
eat cos bt function, 244
eat function, 242, 244
eat sin bt function, 244
e.class, 81
eigenvalues and eigenvectors
calculating eigenvectors, 282–283,
286–287
changing matrix to right form, 282
figuring out eigenvalues, 282
overview, 281
equal roots, 222–223
equilibrium value, 16
erx, g(x) in form of, 127–128, 176–179
Euler equations
complex roots, 223–224
overview, 219–220
real and distinct roots, 220–221
real and equal roots, 222–223
recognizing, 227
Euler, Leonhard Paul, 76
Euler’s method
checking method’s accuracy on computer
defining initial conditions and functions,
78–79
examining entire code, 79
example at work, 80–83
overview, 77–78
fundamentals of method, 294–295
graph, 77
improved
coming up with new code, 300–304
overview, 299–300
plugging steep slope into new code,
304–308
understanding improvements, 300
overview, 75, 294
understanding method, 76–77
using code to see method in action
overview, 295
Index
surveying results, 297–299
typing in code, 295–297
exact differential equations. See also
nonexact differential equations
defining, 64
determining whether equation is exact,
66–70
solving, 74–75
typical, working out, 65–66
existence and uniqueness theorem,
35, 37–40
explicit solution, 44
exponents at the singularity, 232
•F•
F = ma (Newton’s second law), 14
falling body example, 13–14
f(ct) function, 244
fifth order equation with identical real
roots, 167–168
first order difference equations, 84
first order differential equations, 19–20.
See also systems of first order linear
differential equations
linear, 23–40
basics of solving, 24–26
determining if solution exists for, 35–38
solving with integrating factors, 26–34
nonlinear, 38–40
online solver for, 322
separable, 41–61
finding explicit solutions from implicit
solutions, 45–47
implicit solutions, 43–45
sample flow problem, 52–55
sample monetary problem, 55–59
turning nonlinear separable equations
into linear separable equations, 49–51
using partial fractions in, 59–61
when can’t find explicit solution, 48–49
flow problem solution, 52–55
direction field of, 54
graph, 55
f (n)(t) function, 244
force
drag force, mass with, 148–150
periodic, 145, 146
restorative, 144
fourth order equation with identical real
roots, 166–167
fractions
factoring Laplace transforms into,
258–259
partial, using in first order differential
equations, 59–61
friction, mass without, 144–148
functions
continuous, 93, 195
cos at, 244
cosh at, 244
eat, 244
eat cos bt, 244
eat sin bt, 244
f(ct), 244
f (n)(t), 244
multiple, derivatives involving, 12–13
piecewise continuous, 241
product of two, derivatives of, 13
quotient of two, derivatives of, 13
sin at, 244
sinh at, 244
solving differential equations involving,
25–26
step, 261–263
sum (or difference) of two,
derivatives of, 12
t n, 244
t n eat, 244
f '(x), 11
•G•
GCalc online graphing calculator, 322
general solutions, 36
graphing calculator
Coolmath.com, 321
GCalc, 322
graphs
direction fields
of advanced solution, 34
equilibrium value in, 16
327
328
Differential Equations For Dummies
graphs (continued)
of flow problem solution, 54
of nonlinear separable solution, 45
of separable equation with hard-to-find
explicit solution, 49
of separable equation with initial
conditions, 47
solution in, 15
Euler’s method, 77
of flow problem solution, 55
math behind mass
with drag force, 150
without friction, 148
solution to equation, 105
with complex roots, 105
with real and distinct roots, 100
with real and imaginary roots, 164
step function, 262
g(x)
as combination of sines and cosines,
131–133, 182–185
in form of erx, 127–128, 176–179
as polynomial of order n, 128–131,
179–182
as product of two different forms,
133–134
•H•
Harvey Mudd College Mathematics Online
Tutorial, 318
heat conduction equation, 18
higher order differential equations
overview, 20–21
solving with Laplace transforms
figuring out equation’s Laplace
transform, 256
getting equation’s inverse Laplace
transform, 258
overview, 255
unearthing function to match Laplace
transform, 256–258
higher order linear homogeneous
differential equations
basics of
format, solutions, and initial conditions,
153–154
general solution of, 155
overview, 153
solutions as related to Wronskian, 156
different types of
complex roots, 164–166
duplicate roots, 166–171
overview, 156
real and distinct roots, 156–161
real and imaginary roots, 161–164
notation of, 152–153
overview, 151
higher order linear nonhomogeneous
differential equations. See also second
order linear nonhomogeneous
differential equations
method of undetermined coefficients for
overview, 174–175
when g(x) is combination of sines and
cosines, 182–185
when g(x) is in form erx, 176–179
when g(x) is polynomial of order n,
179–182
overview, 173
solving with variation of parameters
basics of method, 185–186
overview, 185
working through example, 186–188
homogeneous differential equations. See
higher order linear homogeneous
differential equations; linear
homogeneous differential equations;
second order homogeneous equations;
second order linear homogeneous
differential equations; systems of first
order linear homogeneous differential
equations
•I•
icons used in this book, 4
identical complex roots, 170–171
identical imaginary roots, 169–170
identity matrices, 272
imaginary roots, 161–164
indicial equations
distinct roots that don’t differ by positive
integer, 235–236
equal roots, 236
Index
overview, 235
roots that differ by positive integer,
236–237
initial conditions, separable equations
with, 47
integers, positive
distinct roots that don’t
differ by, 235–236
roots that differ by, 236–237
integral curve, connecting slopes into,
14–15
integrals, convolution, 259–261
integrating factors
solving linear first order differential
equations with
overview, 26
solving advanced example, 32–34
solving for integrating factor, 27–28
trying special shortcut, 30–32
using integrating factor to solve
differential equation, 28–29
using integrating factors in differential
equations with functions, 29
using with nonexact differential equations
completing process, 72–73
multiplying by factor you want to find,
71–72
overview, 70
integration by parts, 30–31
interest rate problem, 55–59
italics in book, 2
iterates of difference equation’s
solution, 85
•J•
Java programming language, downloading,
78, 294
JavaView Ode Solver, 323
John Appleby’s introduction to differential
equations, 318
juice flow problem, 52–55
•K•
Kardi Teknomo’s page, 318
kernel of Laplace transforms, 239
Kutta, Martin Wilhelm, 309
•L•
Laplace, Pierre-Simon, 240
Laplace transforms
breaking down, 239–240
calculating, 241–244
transform of 1, 242
transform of eat, 242
transform of sin at, 242–243
of common functions, 244
convolution integrals, 259–261
deciding whether converges, 240–241
factoring into fractions, 258–259
kernel of, 239
overview, 239
solving differential equations with
overview, 245–246
solving higher order equation, 255–258
solving second order homogeneous
equation, 247–251
solving second order nonhomogeneous
equation, 251–255
step functions, 261–263
Legendre, Adrien-Marie, 219
Legendre equation, 218, 219
limiting expressions, 11
line, slope of, 9
linear equations, versus nonlinear
equations, 18–19
linear first order differential equations.
See also nonlinear first order
differential equations
basics of solving
adding constants, 26
applying initial conditions from start,
24–25
overview, 24
solving differential equations involving
functions, 25–26
329
330
Differential Equations For Dummies
linear first order differential equations
(continued)
determining if solution exists for, 35–38
overview, 23
solving with integrating factors
overview, 26
solving advanced example, 32–34
solving for integrating factor, 27–28
trying special shortcut, 30–32
using integrating factor to solve
differential equation, 28–29
using integrating factors in differential
equations with functions, 29
linear homogeneous differential equations,
second order
characteristic equations, 96–109
with constant coefficients, 94–96
getting second solution by reduction of
order, 109–113
overview, 91–93
theorems, 114–122
linear independence
assembling vectors into one matrix,
278–279
determining determinant, 279–281
overview, 115–117, 278
linear nonhomogeneous differential
equations. See higher order linear
nonhomogeneous differential
equations; second order linear
nonhomogeneous differential equations
linear separable equations, 49–51
•M•
Martin J. Osborne’s differential equation
tutorial, 318
mass
with drag force, 148–150
without friction, 144–148
Math @ CowPi’s System Solver, 323
matrices
adjoints of, 269
assembling vectors into one matrix,
278–279
augmented, 267
basics of
overview, 266
setting up matrix, 266–267
working through algebra, 267–268
complex conjugates of, 269
identity matrices, 272
inverter from QuickMath Automatic Math
Solutions, 323
non-invertable, 277
operations
addition, 270
equality, 269
identity, 272
inverse of a matrix, 272–278
multiplication of matrix and number, 270
multiplication of matrix and vector,
271–272
multiplication of two matrices, 270–271
overview, 269
subtraction, 270
singular, 277
transpose of, 269
triangular, 268
method of undetermined coefficients. See
undetermined coefficients, method of
methods. See also partial fractions;
reduction of order method
Euler’s
checking method’s accuracy on
computer, 77–83
fundamentals of method, 294–295
graph, 77
improved, 299–308
overview, 75, 294
understanding method, 76–77
using code to see method in action,
295–299
Runge-Kutta
method’s recurrence relation, 308–309
overview, 308
working with method in code, 309–313
Midnight Tutor’s Video Tutorial, 319
monetary problem, first order differential
equations, 55–59
monofont text in book, 2
µ(t), 26, 27
multiple functions, derivatives involving,
12–13
Index
•N•
•O•
n, g(x) as polynomial of order n, 128–131,
179–182
Nanthakumaar, Vithya, 320
Newton’s second law (F = ma), 14
nonexact differential equations, using
integrating factors with. See also exact
differential equations
completing process, 72–73
multiplying by factor you want to find,
71–72
overview, 70
nonhomogeneous equation. See second
order nonhomogeneous equations;
systems of first order linear
nonhomogeneous equations
non-invertable matrices, 277
nonlinear equations, versus linear
equations, classifying by, 18–19
nonlinear first order differential equations,
determining if solution exists for,
38–40. See also linear first order
differential equations
nonlinear separable equations
direction field of, 45
turning into linear separable equations,
49–51
non-partial (ordinary) derivatives, 17–18
numerical methods
Euler’s method
checking method’s accuracy on
computer, 77–83
fundamentals of method, 294–295
improved, 299–308
overview, 75, 294
understanding method, 76–77
using code to see method in action,
295–299
overview, 293
Runge-Kutta method
method’s recurrence relation, 308–309
overview, 308
working with method in code, 309–313
Ohio State University Physics
Department’s introduction to
differential equations, 319
online tutorials
AnalyzeMath.com’s introduction to
differential equations, 317
Harvey Mudd College Mathematics
Online Tutorial, 318
John Appleby’s introduction to
differential equations, 318
Kardi Teknomo’s page, 318
Martin J. Osborne’s differential equation
tutorial, 318
Midnight Tutor’s Video Tutorial, 319
Ohio State University Physics
Department’s introduction to
differential equations, 319
overview, 317
Paul’s Online Math Notes, 319
S.O.S. Math, 319
University of Surrey Tutorial, 320
order, classifying by, 17
ordinary (non-partial) derivatives, 17–18
ordinary equations, versus partial
equations, classifying by, 17–18
ordinary points, 196
Osborne, Martin J., 318
•P•
parameters method
breaking down equations with variation of
applying method to any linear equation,
138–142
basics of method, 136
overview, 135
solving typical example, 137–138
variation of parameters method and
Wronskian, 142–143
331
332
Differential Equations For Dummies
parameters method (continued)
solving higher order linear
nonhomogeneous differential
equations, solving with variation of
basics of method, 185–186
overview, 185
working through example, 186–188
partial derivatives, 17–18
partial equations, versus ordinary
equations, classifying by, 17–18
partial fractions, using in first order
differential equations, 59–61
particular solution, 124
parts, integration by, 30–31
Paul’s Online Math Notes, 319
peanut butter price increase example, 8–10
pendulum angle equation, 18
periodic force, 145, 146
phase angle, 149
piecewise continuous functions, 241
pitcher of water problem, 52–55
plotting direction fields, 13–14
points, singular
behavior of, 214–215
Euler equations
complex roots, 223–224
overview, 219
real and distinct roots, 220–221
real and equal roots, 222–223
finding, 214
overview, 197, 213
regular
defined, 216
figuring series solutions near, 227–237
versus irregular, 215–219
severity of, 215
positive integers
distinct roots that don’t differ by, 235–236
roots that differ by, 236–237
power series
basics of, 191–192
determining whether converges with
ratio test
fundamentals of ratio test, 192–193
overview, 192
plugging in numbers, 193–195
solving second order differential
equations with
Airy’s equation, 207–211
overview, 196
when already know solution, 198–204
when don’t know solution beforehand,
204–207
powering through singular points
behavior of singular points, 214–215
Euler equations
complex roots, 223–224
overview, 219
real and distinct roots, 220–221
real and equal roots, 222–223
finding singular points, 214
overview, 213
regular singular points
defined, 216
figuring series solutions near, 225–237
versus irregular, 215–219
powers, derivatives that are, 12
product of two functions, derivatives of, 13
•Q•
quadratic equations, 45
QuickMath Automatic Math Solutions, 322
quotient of two functions, derivatives of, 13
•R•
radius of convergence, 193
ratio test, determining whether power
series converges with
fundamentals of ratio test, 192–193
overview, 192
plugging in numbers, 193–195
real and distinct roots
fourth order equation, 159–161
overview, 156
third order equation, 156–159
real and imaginary roots, 161–164
recurrence relation, 201, 295
reduction of order method, 108
Remember icon, 4
Index
resources
AnalyzeMath.com’s introduction to
differential equations, 317
Harvey Mudd College Mathematics
Online Tutorial, 318
John Appleby’s introduction to
differential equations, 318
Kardi Teknomo’s page, 318
Martin J. Osborne’s differential equation
tutorial, 318
Midnight Tutor’s Video Tutorial, 319
Ohio State University Physics
Department’s introduction to
differential equations, 319
overview, 317
Paul’s Online Math Notes, 319
S.O.S. Math, 319
University of Surrey Tutorial, 320
restorative force, 144
rocket wobbling problem, 41
roots
complex, 164–166
distinct, 156–161
duplicate, 166–171
Euler equations
complex roots, 223–224
distinct roots, 220–221
equal roots, 222–223
real roots, 220–223
imaginary, 161–164
indicial equations
distinct roots that don’t differ by
positive integer, 235–236
equal roots, 236
roots that differ by positive integer,
236–237
real and distinct
fourth order equation, 159–161
third order equation, 156–159
and solving equations near singular
points
applying first root, 232–233
overview, 229
plugging in second root, 233–235
Runge, Carl David Tolmé, 309
Runge-Kutta method
method’s recurrence relation, 308–309
overview, 308
working with method in code
examining results, 311–313
inputting code, 309–311
overview, 309
•S•
second order differential equations
overview, 20–21
solving with power series
Airy’s equation, 207–211
overview, 196–197
when already know solution, 198–204
when don’t know solution beforehand,
204–207
second order homogeneous equations,
solving with Laplace transforms
finding Laplace transform of equation’s
unknown solution, 249–250
overview, 247
uncovering inverse Laplace transform to
get equation’s solution, 250–251
second order linear homogeneous
differential equations. See also second
order linear nonhomogeneous
differential equations
basics of
homogeneous equations, 93
linear equations, 92–93
overview, 91
characteristic equations
complex roots, 100–105
identical real roots, 106–109
overview, 96
real and distinct roots, 97–100
with constant coefficients
elementary solutions, 94–95
linear equations, 95–96
overview, 94
333
334
Differential Equations For Dummies
second order linear homogeneous
differential equations (continued)
getting second solution by reduction of
order, 109–113
theorems
linear independence, 115–117
overview, 114
superposition of solutions, 114–115
Wronskian, 117–122
second order linear nonhomogeneous
differential equations
breaking down equations with variation
of parameters method
applying method to any linear equation,
138–142
basics of method, 136
overview, 135
solving typical example, 137–138
variation of parameters method and
Wronskian, 142–143
finding particular solutions with method
of undetermined coefficients
overview, 127
when g(x) is combination of sines and
cosines, 131–133
when g(x) is in form of erx, 127–128
when g(x) is polynomial of order n,
128–131
when g(x) is product of two different
forms, 133–134
general solution of, 124–126
mass with drag force, 148–150
mass without friction, 144–148
overview, 123
second order nonhomogeneous equation,
solving with Laplace transforms
determining Laplace transform, 252–253
matching function to Laplace transform,
253–255
overview, 251
using table to find inverse Laplace
transform, 255
separable equations
with hard-to-find explicit solution,
direction field of, 47, 49
with initial conditions, direction field of, 47
separable first order differential equations
basics of
finding explicit solutions from implicit
solutions, 45–47
implicit solutions, 43–45
turning nonlinear separable equations
into linear separable equations, 49–51
when can’t find explicit solution, 48–49
sample flow problem
determining basic numbers, 52–53
solving equation, 53–55
sample monetary problem
adding set amount of money, 58–59
compounding interest at set intervals,
56–57
general solution, 55–56
using partial fractions in, 59–61
series index, shifting, 195
severity of singular points, 215
shifting the series index, 195
sin at function, 242–243
sine, 185
singular matrices, 277
singular points, 197, 213–219
sinh at function, 244
sinusoidal, 149
sin(x), 12
slopes
connecting into, 14–15
of curves. See Euler’s method
of lines, 9
S.O.S. Math, 319
step functions, Laplace transforms,
261–263
sum (or difference) of two functions,
derivatives of, 12
superposition of solutions theorem,
114–115
systems of first order linear differential
equations. See also systems of first
order linear nonhomogeneous
equations; systems of first order linear
homogeneous differential equations
eigenvalues and eigenvectors
calculating eigenvectors, 282–283
changing matrix to right form, 282
Index
figuring out eigenvalues, 282
overview, 281
linear independence, 278–281
assembling vectors into one matrix,
278–279
determining determinant, 279–281
matrices, basics of
overview, 266
setting up matrix, 266–267
working through algebra, 267–268
matrix operations
addition, 270
equality, 269
identity, 272
inverse of a matrix, 272–278
multiplication of matrix and number, 270
multiplication of matrix and vector,
271–272
multiplication of two matrices, 270–271
overview, 269
subtraction, 270
overview, 265
systems of first order linear
nonhomogeneous equations
assuming correct form of particular
solution, 289
crunching numbers, 290–292
overview, 288
winding up work, 292
systems of first order linear homogeneous
differential equations
coming up with eigenvectors, 286–287
finding eigenvalues, 285–286
getting right matrix form, 285
overview, 283
summing up solution, 287–288
understanding basics, 284–285
•T•
Taylor expansion, 215
Taylor series, 195–196
Technical Stuff icon, 4
Teknomo, Kardi, 318
Tip icon, 4
t n eat function, 244
t n function, 244
tools for solving differential equations
AnalyzeMath.com’s Runge-Kutta
Applet, 321
Coolmath.com graphing calculator, 321
direction field plotter, 322
equation solver from QuickMath
Automatic Math Solutions, 322
First Order Differential Equation Solver, 322
GCalc online graphing calculator, 322
JavaView Ode Solver, 323
Math @ CowPi’s System Solver, 323
matrix inverter from QuickMath
Automatic Math Solutions, 323
overview, 321
Visual Differential Equation Solving
Applet, 323
transpose of matrices, 269
triangular matrices, 268
trigonometry, derivatives involving, 12
tutorials, online
AnalyzeMath.com’s introduction to
differential equations, 317
Harvey Mudd College Mathematics
Online Tutorial, 318
John Appleby’s introduction to
differential equations, 318
Kardi Teknomo’s page, 318
Martin J. Osborne’s differential equation
tutorial, 318
Midnight Tutor’s Video Tutorial, 319
Ohio State University Physics
Department’s introduction to
differential equations, 319
overview, 317
Paul’s Online Math Notes, 319
S.O.S. Math, 319
University of Surrey Tutorial, 320
•U•
undetermined coefficients, method of
finding particular solutions with
overview, 127
when g(x) is combination of sines and
cosines, 131–133
335
336
Differential Equations For Dummies
undetermined coefficients, method of
(continued)
when g(x) is in form of erx, 127–128
when g(x) is polynomial of order n,
128–131
when g(x) is product of two different
forms, 133–134
for higher order linear nonhomogeneous
differential equations
overview, 174–175
when g(x) is combination of sines and
cosines, 182–185
when g(x) is in form erx, 176–179
when g(x) is polynomial of order n,
179–182
University of Surrey Tutorial, 320
•V•
vectors, assembling into one matrix,
278–279
Visual Differential Equation Solving
Applet, 323
•W•
Warning icon, 4
Wronski, Jósef Maria Hoëné, 119
Wronskian
arrays and determinants, 117–119
formal theorem, 119–122
and higher order linear homogeneous
differential equation solutions, 156
overview, 117
and variation of parameters method,
142–143
•X•
x, raising to power of n, 12
Notes
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