Problems and Solutions - Arkansas Tech Faculty Web Sites

Problems and Solutions - Arkansas Tech Faculty Web Sites
A Second Course in Elementary Differential
Equations: Problems and Solutions
Marcel B. Finan
Arkansas Tech University
c
All
Rights Reserved
1
Contents
28 Calculus of Matrix-Valued Functions of a Real Variable
4
29 nth Order Linear Differential Equations: Exsitence and Uniqueness
13
30 The General Solution of nth Order Linear Homogeneous Equations
15
31 Fundamental Sets and Linear Independence
21
32 Higher Order Homogeneous Linear Equations with Constant
Coefficients
25
33 Non Homogeneous nth Order Linear Differential Equations 31
34 Existence and Uniqueness of Solution to Initial Value First
Order Linear Systems
45
35 Homogeneous First Order Linear Systems
50
36 First Order Linear Systems: Fundamental Sets and Linear
Independence
60
37 Homogeneous Systems with Constant Coefficients
64
38 Homogeneous Systems with Constant Coefficients: Complex
Eigenvalues
75
39 Homogeneous Systems with Constant Coefficients: Repeated
Eigenvalues
84
40 NonHomogeneous First Order Linear Systems
96
41 Solving First Order Linear Systems with Diagonalizable Constant Coefficients Matrix
110
42 Solving First Order Linear Systems Using Exponential Matrix
122
2
43 The Laplace Transform: Basic Definitions and Results
127
44 Further Studies of Laplace Transform
136
45 The Laplace Transform and the Method of Partial Fractions146
47 Laplace Transforms of Periodic Functions
156
47 Convolution Integrals
166
48 47
173
49 The Dirac Delta Function and Impulse Response
173
49 Solving Systems of Differential Equations Using Laplace Transform
179
50 Numerical Methods for Solving First Order Linear Systems:
Euler’s Method
188
3
28
Calculus of Matrix-Valued Functions of a
Real Variable
Problem 28.1
Consider the following matrices
t−1
t2
t −1
t+1
A(t) =
, B(t) =
, c(t) =
2
2t + 1
0 t+2
−1
(a) Find 2A(t) - 3tB(t)
(b) Find A(t)B(t) - B(t)A(t)
(c) Find A(t)c(t)
(d) Find det(B(t)A(t))
Solution.
(a)
2A(t) − 3tB(t) =
2
2t − 2 2t2
4
4t + 2
=
=
t−1
t2
2
2t + 1
− 3t
−
t −1
0 t+2
3t2
−3t
0 3t2 + 6t
2t − 2 − 3t2
2t2 + 3t
4
2 − 2t − 3t2
(b)
A(t)B(t) − B(t)A(t) =
t −1
0 t+2
t2 − t t3 + 2t2 − t + 1
2t
2t2 + 5t
t−1
t2
2
2t + 1
=
=
t −1
0 t+2
t2 − t − 2 t3 − 2t − 1
2t + 4
2t2 + 5t + 2
−
−
2 2t2 + t + 2
−4
−2
(t − 1)(t + 1) + t2 (−1)
2(t + 1) + (2t + 1)(−1)
t−1
t2
2
2t + 1
(c)
A(t)c(t) =
t−1
t2
2
2t + 1
t+1
−1
=
4
=
−1
1
(d)
2
t −1
t − t − 2 t3 − 2t − 1
t−1
t2
=
det(B(t)A(t)) = 0 t+2
2
2t + 1 2t + 4
2t2 + 5t + 2
3
2
=
−(t + 3t + 2t)
Problem 28.2
Determine all values t such that A(t) is invertible and, for those t-values,
find A−1 (t).
t+1
t
A(t) =
t
t+1
Solution.
We have det(A(t)) = 2t + 1 so that A is invertible for all t 6= − 12 . In this
case,
1
t + 1 −t
−1
A (t) =
−t t + 1
2t + 1
Problem 28.3
Determine all values t such that A(t) is invertible and, for those t-values,
find A−1 (t).
sin t − cos t
A(t) =
sin t cos t
Solution.
We have det(A(t)) = 2 sin t cos 2 = sin 2t so that A is invertible for all t 6=
where n is an integer. In this case,
1
cos t cos t
−1
A (t) =
sin 2t − sin t sin t
Problem 28.4
Find
lim
t→0
sin t
t
3t
e
3
t+1
2t
t2 −1
t cos t
sec t
Solution.
lim
t→0
sin t
t
3t
e
t cos t
sec t
3
t+1
2t
t2 −1
5
=
1 0 3
1 1 0
nπ
2
Problem 28.5
Find
lim
t→0
te−t tan t
t2 − 2 esin t
Solution.
lim
t→0
te−t tan t
t2 − 2 esin t
=
0 0
−2 1
Problem 28.6
Find A0 (t) and A00 (t) if
sin t 3t
2
t +2 5
A(t) =
Solution.
0
cos t 3
2t 0
− sin t 0
2
0
A (t) =
00
A (t) =
Problem 28.7
Express the system
y10 = t2 y1 + 3y2 + sec t
y20 = (sin t)y1 + ty2 − 5
in the matrix form
y0 (t) = A(t)y(t) + g(t)
Solution.
y(t) =
y1 (t)
y2 (t)
, A(t) =
t2 3
sin t t
, g(t) =
sec t
−5
Problem 28.8
Determine A(t) where
0
A (t) =
2t
1
cos t 3t2
, A(0) =
6
2 5
1 −2
Solution.
Integrating componentwise we find
2
t + c11
t + c12
A(t) =
sin t + c21 t3 + c22
Since
A(0) =
2 5
1 −2
=
c11 c12
c21 c22
by equating componentwise we find c11 = 2, c12 = 5, c21 = 1, and c22 = −2.
Hence,
2
t +2
t+5
A(t) =
sin t + 1 t3 + −2
Problem 28.9
Determine A(t) where
1 t
1 1
−1 2
00
A (t) =
, A(0) =
, A(1) =
0 0
−2 1
−2 3
Solution.
Integrating componentwise we find
t + c11
0
A (t) =
c21
Integrating again we find
A(t) =
t2
2
But
A(0) =
t2
2
+ c12
c22
+ c11 t + d11
c21 t + d21
t3
6
+ c12 t + d12
c22 t + d22
d11 d12
d21 d22
1 1
−2 1
=
so by equating componentwise we find d11 = 1, d12 = 1, d21 = −2, and
d22 = 1. Thus,
t2
t3
+
c
t
+
1
+
c
t
+
1
11
12
6
A(t) = 2
c21 t + −2
c22 t + 1
7
Since
A(1) =
−1 2
−2 3
=
3
2
+ c11 76 + c12
−2 + c21 1 + c22
we find c11 = − 25 , c12 = 56 , c21 = 0, c22 = 2. Hence,
t2 5
t3
5
−
t
+
1
+
t
+
1
2
6
6
A(t) = 2
−2
2t + 1
Problem 28.10 R
t
Calculate A(t) = 0 B(s)ds where
es
6s
B(s) =
cos 2πs sin 2πs
Solution.
Integrating componentwise we find
Rt s
Rt
t
e ds
6sds
e −1
0
0
Rt
= sin 2πt
A(t) = R t
cos 2πsds 0 sin 2πsds
2π
0
3t2
1−cos 2πt
2π
Problem 28.11
Construct a 2 × 2 nonconstant matrix function A(t) such that A2 (t) is a
constant matrix.
Solution.
Let
A(t) =
Then
2
A (t) =
0 t
0 0
0 t
0 0
0 t
0 0
=
0 0
0 0
Problem 28.12
(a) Construct a 2 × 2 differentiable matrix function A(t) such that
d 2
d
A (t) 6= 2A A(t)
dt
dt
That is, the power rule is not true for matrix functions.
(b) What is the correct formula relating A2 (t) to A(t) and A’(t)?
8
Solution.
(a) Let
A(t) =
Then
2
A (t) =
so that
d 2
A (t) =
dt
1 t
t2 0
1 + t3 t
t2
t3
3t2 1
2t 3t2
On the other hand,
d
2A(t) A(t) = 2
dt
1 t
t2 0
(b) (b) The correct formula is
2 + 4t3 2t + 2t4
2t2 + 2t5
2t3
d
A2 (t)
dt
=
1 t
t2 0
= A(t)A0 (t) + A0 (t)A(t)
Problem 28.13
Transform the following third-order equation
y 000 − 3ty 0 + (sin 2t)y = 7e−t
into a first order system of the form
x0 (t) = Ax(t) + b(t)
Solution.
Let x1 = y, x2 = y 0 , x3 = y 00 . Then by letting






x1
0
1 0
0
0 1 , b =  0 
x =  x2  , A =  0
x3
sin 2t 3t 0
7e−t
then the differential equation can be presented by the first order system
x0 (t) = Ax(t) + b(t)
Problem 28.14
By introducing new variables x1 and x2 , write y 00 − 2y + 1 = t as a system of
two first order linear equations of the form x0 + Ax = b
9
Solution.
By letting x1 = y and x2 = y 0 we have
x1
0 −1
x=
, A=
,
x2
−2 0
b=
0
t−1
Problem 28.15
Write the differential equation y 00 + 4y 0 + 4y = 0 as a first order system.
Solution.
By letting x1 = y and x2 = y 0 we have
x1
0 −1
x=
, A=
,
x2
4 4
b=
0
0
Problem 28.16
Write the differential equation y 00 + ky 0 + (t − 1)y = 0 as a first order system.
Solution.
By letting x1 = y and x2 = y 0 we have
x1
0
−1
x=
, A=
,
x2
t−1 k
b=
0
0
Problem 28.17
Change the following second-order equations to a first-order system.
y 00 − 5y 0 + ty = 3t2 , y(0) = 0, y 0 (0) = 1
Solution.
If we write the problem in the matrix form
x0 + Ax = b, x(0) = y0
then
0 −1
A=
−5 t
x1
y
0
0
x=
=
, b=
, y0 =
0
2
x2
y
3t
1
10
Problem 28.18
Consider the following system of first-order linear equations.
3 2
0
x =
·x
1 −1
Find the second-order linear differential equation that x satisfies.
Solution.
The system is
x01 = 3x1 + 2x2
x02 = x1 − x2
It follows that
x01 + 2x02 = 5x1 or x1 =
x01 +2x02
5
2
so we let w = x1 +2x
so that w0 = x1 . Thus, x01 = 3x1 +2x2 = 3x1 +(5w−x1 ) =
5
00
2x1 + 5w. Hence, x1 = 2x01 + 5w0 = 2x01 + 5x1 or x001 − 2x01 − 5x1 = 0
Problem 28.19
List all the permutations of S = {1, 2, 3, 4}.
Solution.
1
2
3
4
.
↓
&
.
↓
& .
↓
& .
↓
&
2
3
4
1
3
4 1
2
4 1
2
3
.& .& .& .& .& .& .& .& .& .& .& .&
3 4 2 4 2 3 3 4 1 4 1 3 2 4 1 4 1 2 2 3 1 3 1 2
Problem 28.20
List all elementary products from the matrices
(a)
a11 a12
,
a21 a22
(b)


a11 a12 a13
 a21 a22 a23 
a31 a32 a33
11
Solution.
(a) The only elementary products are a11 a22 , a12 a21 .
(b) An elementary product has the form a1∗ a2∗ a3∗ . Since no two factors come
from the same column, the column numbers have no repetitions; consequently
they must form a permutation of the set {1, 2, 3}. The 3! = 6 permutations
yield the following elementary products:
a11 a22 a33 , a11 a23 a32 , a12 a23 a31 , a12 a21 a33 , a13 a21 a32 , a13 a22 a31
Problem 28.21
Find det(A) if
(a)
A=
a11 a12
a21 a22
,
(b)


a11 a12 a13
A =  a21 a22 a23 
a31 a32 a33
Solution.
By using the definition of a determinant and Exercise ?? we obtain
(a) |A| = a11 a22 − a21 a12 .
(b) |A| = a11 a22 a33 − a11 a23 a32 + a12 a23 a31 − a12 a21 a33 + a13 a21 a32 − a13 a22 a31
12
29
nth Order Linear Differential Equations:
Exsitence and Uniqueness
For Problems 29.1 - 29.3, use Theorem 29.1 to find the largest interval
a < t < b in which a unique solution is guaranteed to exist.
Problem 29.1
y 000 −
1
y 00 + ln (t + 1) + (cos t)y = 0, y(0) = 1, y 0 (0) = 3, y 00 (0) = 0
t2 − 9
Solution.
The coefficient functions are all continuous for t 6= −3, −1, 3. Since t0 = 0,
the largest interval of existence is −1 < t < 3
Problem 29.2
y 000 +
1 0
y + (tan t)y = 0, y(0) = 0, y 0 (0) = 1, y 00 (0) = 2
t+1
Solution.
The coefficient functions are all continuous for t 6= −1 and t 6= (2n + 1) π2
where n is an integer. Since t0 = 0, the largest interval of existence is
−1 < t < π2
Problem 29.3
y 00 −
1
y 00 + ln (t2 + 1)y 0 + (cos t)y = 0, y(0) = 1, y 0 (0) = 3, y 00 (0) = 0
t2 + 9
Solution.
The coefficient functions are all continuous for t so that the interval of existence is −∞ < t < ∞
Problem 29.4
Determine the value(s) of r so that y(t) = ert is a solution to the differential
equation
y 000 − 2y 00 − y 0 + 2y = 0
13
Solution.
Inserting y and its derivatives into the equation we find
0 = y 000 − 2y 00 − y 0 + 2y
= (r3 − 2r2 − r + 2)ert
Since ert > 0, we must have 0 = r3 − 2r2 − r + 2 = r2 (r − 2) − (r − 2) =
(r − 2)(r2 − 1) = (r − 2)(r − 1)(r + 1). Hence, r = −1, 1, 2
Problem 29.5
Transform the following third-order equation
y 000 − 3ty 0 + (sin 2t)y = 7e−t
into a first order system of the form
x0 (t) = Ax(t) + b(t)
Solution.
Let x1 = y, x2 = y 0 , x3 = y 00 . Then by letting






x1
0
1 0
0
0 1 , b =  0 
x =  x2  , A =  0
x3
sin 2t 3t 0
7e−t
then the differential equation can be presented by the first order system
x0 (t) = Ax(t) + b(t)
14
30
The General Solution of nth Order Linear
Homogeneous Equations
In Problems 30.1 - 30.3, show that the given solutions form a fundamental
set for the differential equation by computing the Wronskian.
Problem 30.1
y 000 − y 0 = 0, y1 (t) = 1, y2 (t) = et , y3 (t) = e−t
Solution.
We have
1 et e−t
0 et −e−t
0 et e−t
W (t) =
t
e −e−t
= (1) t
e
e−t
−t
− et 0 −e−t
0 e
t
+ e−t 0 et
0 e
= 2 6= 0
Problem 30.2
y (4) + y 00 = 0, y1 (t) = 1, y2 (t) = t, y3 (t) = cos t, y4 (t) = sin t
Solution.
We have
W (t) = 1
0
0
0
t cos t
sin t
1 − sin t cos t
0 − cos t − sin t
0 sin t − cos t
= cos2 t + sin2 t = 1 6= 0
Problem 30.3
t2 y 000 + ty 00 − y 0 = 0, y1 (t) = 1, y2 (t) = ln t, y3 (t) = t2
15
Solution.
We have
W (t) =
−1
t
= (1) −t−2
1 ln t
0 1
t
0 − 12
t 0 2t 2t + t2 −
ln
t
0 2 2 t2
2t
2
0
0
t−1 = 3t−1 6= 0, t > 0
−t−2 Use the fact that the solutions given in Problems 30.1 - 30.3 for a fundamental
set of solutions to solve the following initial value problems.
Problem 30.4
y 000 − y 0 = 0, y(0) = 3, y 0 (0) = −3, y 00 (0) = 1
Solution.
The general solution is y(t) =
we have
y(0) = 3
y 0 (0) = −3
y 00 (0) = 1
c1 + c2 et + c3 e−t . With the initial conditions
=⇒ c1 + c2 + c3 =
3
=⇒
c2 − c3
= −3
=⇒
c2 + c3
=
1
Solving these simultaneous equations gives c1 = 2, c2 = −1 and c3 = 2 and
so the unique solution is
y(t) = 2 − et + 2e−t
Problem 30.5
y (4) + y 00 = 0, y( π2 ) = 2 + π, y 0 ( π2 ) = 3, y 00 ( π2 ) = −3, y 000 ( π2 ) = 1.
Solution.
The general solution is y(t) = c1 + c2 t + c3 cos t + c4 sin t. With the initial
conditions we have
y( π2 ) = 2 + π
y 0 ( π2 ) = 3
y 00 ( π2 ) = −3
y 000 ( π2 ) = 1
=⇒ c1 + π2 c2 + c4
=⇒
c2 − c3
=⇒
−c4
=⇒
c3
16
= 2+π
=
3
=
−3
=
1
Solving these simultaneous equations gives c1 = −(π + 1), c2 = 4, c3 = 1
and c4 = 3. Hence, the unique solution is
y(t) = −(π + 1) + 4t + cos t + 3 sin t
Problem 30.6
t2 y 000 + ty 00 − y 0 = 0, , y(1) = 1, y 0 (1) = 2, y 00 (1) = −6
Solution.
The general solution is y(t) = c1 + c2 ln t + c3 t2 .
we have
y(1) = 1 =⇒ c1 + c3
y 0 (1) = 2 =⇒ c2 + 2c3
y 00 (1) = −6 =⇒ −c2 + 2c3
With the initial conditions
=
1
=
2
= −6
Solving these simultaneous equations gives c1 = 2, c2 = 4 and c3 = −1 and
so the unique solution is
y(t) = 2 + 4 ln t − t2
Problem 30.7
In each question below, show that the Wronskian determinant W (t) behaves
as predicted by Abel’s Theorem. That is, for the given value of t0 , show that
W (t) = W (t0 )e
−
Rt
t0
pn−1 (s)ds
(a) W (t) found in Problem 30.1 and t0 = −1.
(b) W (t) found in Problem 30.2 and t0 = 1.
(c) W (t) found in Problem 30.3 and t0 = 2.
Solution.
(a) For the given differential equation pn−1 (t) = p2 (t) = 0 so that Abel’s theorem predict W (t) = W (t0 ). Now, for t0 = −1 we have W (t) = W (−1) =constant.
From Problem 28.1, we found that W (t) = 2.
(b) For the given differential equation pn−1 (t) = p3 (t) = 0 so that Abel’s theorem predict W (t) = W (t0 ). Now, for t0 = 1 we have W (t) = W (1) =constant.
17
From Problem 28.2, we found that W (t) = 1.
(c) For the given differential equation pn−1 (t) = p2 (t) = 1t so that Abel’s
R t ds
2
theorem predict W (t) = W (2)e− 2 s = W (2)eln ( t ) = 2t W (2). From Problem
28.3, we found that W (t) = 3t so that W (2) = 32
Problem 30.8
Determine W (t) for the differential equation y 000 +(sin t)y 00 +(cos t)y 0 +2y = 0
such that W (1) = 0.
Solution.
Here pn−1 (t) = p2 (t) = sin t. By Abel’s Theorem we have
W (t) = W (1)e−
Rt
1
sin sds
≡0
Problem 30.9
Determine W (t) for the differential equation t3 y 000 − 2y = 0 such that W (1) =
3.
Solution.
Here pn−1 (t) = p2 (t) = 0. By Abel’s Theorem we have
W (t) = W (1)e−
Rt
1
0ds
= W (1) = 3
Problem 30.10
Consider the initial value problem
y 000 − y 0 = 0, y(0) = α, y 0 (0) = β, y 00 (0) = 4.
The general solution of the differential equation is y(t) = c1 + c2 et + c3 e−t .
(a) For what values of α and β will limt→∞ y(t) = 0?
(b) For what values α and β will the solution y(t) be bounded for t ≥ 0, i.e.,
|y(t)| ≤ M for all t ≥ 0 and for some M > 0? Will any values of α and β
produce a solution y(t) that is bounded for all real number t?
Solution.
(a) Since y(0) = α, c1 + c2 + c3 = α. Since y 0 (t) = c2 et − c3 e−t and y 0 (0) = β,
c2 − c3 = β. Also, since y 00 (t) = c2 et + c3 e−t and y 00 (0) = 4 we have c2 + c3 = 4.
Solving these equations for c1 , c2 , and c3 we find c1 = α − 4, c2 = β/2 + 2
and c3 = −β/2 + 2. Thus,
y(t) = α − 4 + (β/2 + 2)et + (−β/2 + 2)e−t .
18
If α = 4 and β = −4 then
y(t) = 4e−t
and
lim 4e−t = 0.
t→∞
(b) In the expression of y(t) w know that e−t is bounded for t ≥ 0 whereas
et is unbounded for t ≥ 0. Thus, for y(t) to be bounded we must choose
β/2 + 2 = 0 or β = −4. The number α can be any number. Now, for the
solution y(t) to be bounded on −∞ < t < ∞ we must have simultaneously
β/2 + 2 = 0 and −β/2 + 2 = 0. But there is no β that satisfies these two
equations at the same time. Hence, y(t) is always unbounded for any choice
of α and β
Problem 30.11
Consider the differential equation y 000 + p2 (t)y 00 + p1 (t)y 0 = 0 on the interval
−1 < t < 1. Suppose it is known that the coefficient functions p2 (t) and p1 (t)
are both continuous on −1 < t < 1. Is it possible that y(t) = c1 + c2 t2 + c3 t4
is the general solution for some functions p1 (t) and p2 (t) continuous on −1 <
t < 1?
(a) Answer this question by considering only the Wronskian of the functions
1, t2 , t4 on the given interval.
(b) Explicitly determine functions p1 (t) and p2 (t) such that y(t) = c1 + c2 t2 +
c3 t4 is the general solution of the differential equation. Use this information,
in turn, to provide an alternative answer to the question.
Solution.
(a) The Wronskian of 1, t2 , t4 is
1 t2 t4
W (t) = 0 2t 4t3
0 2 12t2
= 16t3
Since 0 is in the interval −1 < t < 1 and W (0) = 0, {1, t2 , t4 } cannot
be a fundamental set and therefore the general solution cannot be a linear
combination of 1, t2 , t4 .
(b) First notice that y = 1 is a solution for any p1 and p2 . If y = t2 is a
solution then substitution into the differential equation leads to tp1 + p2 =
0. Since y = t4 is also a solution, substituting into the equation we find
t2 p1 + 3tp2 + 6 = 0. Solving for p1 and p2 we find p1 (t) = t32 and p2 (t) = − 3t .
Note that both functions are not continuous at t = 0
19
Problem 30.12
(a) Find the general solution to y 000 = 0.
(b) Using the general solution in part (a), construct a fundamental set
{y1 (t), y2 (t), y3 (t)} satisfying the following conditions
y1 (1) = 1, y10 (1) = 0, y100 (1) = 0.
y2 (1) = 0, y10 (1) = 1, y100 (1) = 0.
y1 (1) = 0, y10 (1) = 0, y100 (1) = 1.
Solution.
(a) Using antidifferentiation we find that y(t) = c1 + c2 t + c3 t2 .
(b) With the initial conditions of y1 we obtain the following system
c1 + c2 + c3 = 1
c2 + 2c3 = 0
2c3 = 0
Solving this system we find c1 = 1, c2 = 0, c3 = 0. Thus, y1 (t) = 1. Repeating
this argument for y2 we find the system
c1 + c2 + c3 = 0
c2 + 2c3 = 1
2c3 = 0
Solving this system we find c1 = −1, c2 = 1, c3 = 0. Thus, y2 (t) = t − 1.
Repeating this argument for y3 we find the system
c1 + c2 + c3 = 0
c2 + 2c3 = 0
2c3 = 1
Solving this system we find c1 = − 21 , c2 = −1, c3 =
1
(t − 1)2
2
20
1
.
2
Thus, y3 (t) =
31
Fundamental Sets and Linear Independence
Problem 31.1
Determine if the following functions are linearly independent
y1 (t) = e2t , y2 (t) = sin (3t), y3 (t) = cos t
Solution.
First take derivatives
y10 (t) = 2e2t y20 (t) = 3 cos (3t) y30 (t) = − sin t
y100 (t) = 4e2t y200 (t) = 9 sin (3t) y300 (t) = − cos t
The Wronskian is
2t
e
sin
3t
cos
t
2t
2e 3 cos (3t) − sin t W (t) =
2t
4e
9 sin (3t) − cos t = e2t (−3 cos 3t cos t − 9 sin 3t sin t) − sin 3t(−2et cos t + 4e2t sin t)
+
cos t(−18e2t sin 3t − 12e2t cos 3t)
Thus,W (0) = −15. Since this is a nonzero number, we can conclude that the
three functions are linearly independent
Problem 31.2
Determine whether the three functions : f (t) = 2, g(t) = sin2 t, h(t) = cos2 t,
are linearly dependent or independent on −∞ < t < ∞
Solution.
Computing the Wronskian
2
2 sin2 t
cos
t
0 sin 2t
− sin 2t W (t) =
0 2 cos 2t −2 cos 2t = 2[sin (2t)(−2 cos 2t) − (− sin 2t(2 cos 2t)) = 0
So the functions are linearly depedent
Problem 31.3
Determine whether the functions, y1 (t) = 1; y2 (t) = 1 + t; y3 (t) = 1 + t + t2 ;
are linearly dependent or independent. Show your work.
21
Solution.
0 =
ay1 + by2 + cy3
= a(1) + b(1 + t) + c(1 + t + t2 )
= (a + b + c) + (b + c)t + ct2
Equating coefficients we find
a+b+c = 0
b+c
= 0
c
= 0
Solving this system we find that a = b = c = 0 so that y1 , y2 , and y3 are
linearly independent
Problem 31.4
Consider the set of functions {y1 (t), y2 (t), y3 (t)} = {t2 + 2t, αt + 1, t + α}. For
what value(s) α is the given set linearly depedent on the interval −∞ < t <
∞?
Solution.
0 =
ay1 + by2 + cy3
2
= a(t + 2t) + b(αt + 1) + c(t + α)
=
b + αc + (2a + αb + c)t + at2
Equating coefficients we find
b + αc
= 0
2a + αb + c = 0
a
= 0
Solving this system we find that a = b = c = 0 provided that α 6= ±1. In
this case, y1 , y2 , and y3 are linearly independent
Problem 31.5
Determine whether the set {y1 (t), y2 (t), y3 (t)} = {t|t| + 1, t2 − 1, t} is linearly
independent or linearly dependent on the given interval
(a) 0 ≤ t < ∞.
(b) −∞ < t ≤ 0.
(c) −∞ < t < ∞.
22
Solution.
(a) If t ≥ 0 then y1 (t) = t|t| + 1 = t2 + 1. In this case, we have
0 =
ay1 + by2 + cy3
2
= a(t + 1) + b(t2 − 1) + c(t)
=
(a + b)t2 + ct + a − b
Equating coefficients we find
a+b = 0
c
= 0
a−b = 0
Solving this system we find that a = b = c = 0. This shows that y1 , y2 , and
y3 are linearly independent.
(b) If t ≤ 0 then y1 (t) = −t2 + 1 = −(t2 − 1) = −y2 (t) + 0y3 (t). Thus, y1 , y2 ,
and y3 are linearly dependent.
(c) Since y1 , y2 , and y3 are linearly independent on the interval 0 ≤ t < ∞,
they are linearly independent on the entire interval −∞ < t < ∞
In Problems 31.6 - 31.7, for each differential equation, the corresponding
set of functions {y1 (t), y2 (t), y3 (t)} is a fundamental set of solutions.
(a) Determine whether the given set {y 1 (t), y 2 (t), y 3 (t)} is a solution set to
the differential equation.
(b) If {y 1 (t), y 2 (t), y 3 (t)} is a solution set then find the coefficient matrix A
such that


 

y1
a11 a12 a13
y1
 y 2  =  a21 a22 a23   y2 
a31 a32 a33
y3
y3
(c) If {y 1 (t), y 2 (t), y 3 (t)} is a solution set, determine whether it is a fundamental set by calculating the determinant of A.
Problem 31.6
y 000 + y 00 = 0
{y1 (t), y2 (t), y3 (t)} = {1, t, e−t }
{y 1 (t), y 2 (t), y 3 (t)} = {1 − 2t, t + 2, e−(t+2) }
23
Solution.
000
00
(a) Since y 1 (t) = 1−2t, y 01 (t) = −2 and y 001 (t) = y 000
1 (t) = 0. Thus, y 1 +y 1 = 0.
000
Similarly, since y 2 (t) = t + 2, y 02 (t) = 1 and y 002 (t) = y 000
2 (t) = 0. Thus, y 2 +
y 002 = 0. Finally, since y 3 (t) = e−(t+2) , y 03 (t) = −e−(t+2) , y 003 (t) = e−(t+2) and
00
−(t+2)
y 000
. Thus, y 000
3 (t) = −e
3 + y 3 = 0. It follows, that {y 1 (t), y 2 (t), y 3 (t)} =
{1 − 2t, t + 2, e−(t+2) } is a solution set.
(b) Since y 1 = 1y1 −2y2 +0y3 , y 2 = 2y1 +1y2 +0y3 , and y 3 = 0y1 +0y2 +e−2 y3
we have


1 2 0
A =  −2 1 0 
0 0 e−2
(c) Since det(A) = 5e−2 6= 0, {y 1 (t), y 2 (t), y 3 (t)} = {1 − 2t, t + 2, e−(t+2) } is
a fundamental set of solutions
Problem 31.7
t2 y 000 + ty 00 − y 0 = 0, t > 0
{y1 (t), y2 (t), y3 (t)} = {t, ln t, t2 }
{y 1 (t), y 2 (t), y 3 (t)} = {2t2 − 1, 3, ln (t3 )}
Solution.
2 000
(a) Since y 1 (t) = 2t2 − 1, y 01 (t) = 4t, y 001 (t) = 4, and y 000
1 (t) = 0. Thus, t y 1 +
00
0
0
00
000
ty 1 − y 1 = 0. Similarly, since y 2 (t) = 3, y 2 (t) = y 2 (t) = y 2 (t) = 0. Thus,
3
3
00
0
0
00
3
t2 y 000
2 + ty 2 − y 2 = 0. Finally, since y 3 (t) = ln t , y 3 (t) = t , y 3 (t) = − t2 and
6
000
00
0
2 000
y 3 (t) = t3 . Thus, t y 3 + ty 3 − y 3 = 0. It follows, that {y 1 (t), y 2 (t), y 3 (t)} =
{2t2 − 1, 3, ln t3 } is a solution set.
(b) Since y 1 = −1y1 +0y2 +2y3 , y 2 = 3y1 +0y2 +0y3 , and y 3 = 0y1 +3y2 +0y3
we have


−1 3 0
A= 0 0 3 
2 0 0
(c) Since det(A) = 18 6= 0, {y 1 (t), y 2 (t), y 3 (t)} = {2t2 − 1, 3, ln t3 } is a
fundamental set of solutions
24
32
Higher Order Homogeneous Linear Equations with Constant Coefficients
Problem 32.1
Solve y 000 + y 00 − y 0 − y = 0
Solution.
The characteristic equation is
r3 + r2 − r − 1 = 0.
Factoring this equation using the method of grouping we find
r2 (r + 1) − (r + 1) = (r + 1)2 (r − 1) = 0
Hence, r = −1 is a root of multiplicity 2 and r = 1 is a of multiplicity 1.
Thus, the general solution is given by
y(t) = c1 e−t + c2 te−t + c3 et
Problem 32.2
Find the general solution of 16y (4) − 8y 00 + y = 0.
Solution.
The characteristic equation is
16r4 − 8r2 + 1 = 0.
This is a complete square
(4r2 − 1)2 = 0
Hence, r = − 21 and r =
solution is given by
1
2
are roots of multiplicity 2. Thus, the general
t
t
t
t
y(t) = c1 e− 2 + c2 te− 2 + c3 e 2 + c4 te 2
Problem 32.3
Solve the following constant coefficient differential equation :
y 000 − y = 0.
25
Solution.
In this case the characteristic equation is r3 − 1 = 0 or r3 = 1 = e2kπi . Thus,
2kπi
r = e 3 where k is an integer. Replacing k by 0,1, and 2 we find
r0 =
1 √
1
r1 = − 2 + i √23
r2 = − 21 − i 23
Thus, the general solution is
√
− 12 t
y(t) = c1 et + c2 e
√
1
3t
3t
+ c3 e− 2 t sin
cos
2
2
Problem 32.4
Solve y (4) − 16y = 0
Solution.
In this case the characteristic equation is r4 − 16 = 0 or r4 = 16 = 16e2kπi .
kπi
Thus, r = 2e 2 where k is an integer. Replacing k by 0,1,2 and 3 we find
r0
r1
r2
r3
= 2
= 2i
= −2
= −2i
Thus, the general solution is
y(t) = c1 e2t + c2 e−2t + c3 cos (2t) + c4 sin (2t)
Problem 32.5
Solve the initial-value problem
y 000 + 3y 00 + 3y 0 + y = 0, y(0) = 0, y 0 (0) = 1, y 00 (0) = 0.
Solution.
We have the characteristic equation
r3 + 3r2 + 3r + 1 = (r + 1)3 = 0
Which has a root of multiplicity 3 at r = −1. We use what we have learned
about repeated roots roots to get the general solution. Since the multiplicity
of the repeated root is 3, we have
y1 (t) = e−t , y2 (t) = te−t , y3 (t) = t2 e−t .
26
The general solution is
y(t) = c1 e−t + c2 te−t + c3 t2 e−t .
Now Find the first three derivatives
y 0 (t) =
−c1 e−t + c2 (1 − t)e−t + c3 (2t − t2 )e−t
00
y (t) = c1 e−t + c2 (−2 + t)e−t + c3 (t2 − 4t + 2)e−t
Next plug in the initial conditions to get
0 =
c1
1 =
c2
0 = −2 + 2c3
Solving these equations we find c1 = 0, c2 = 1, and c3 = 1. The unique
solution is then
y(t) = te−t + t2 e−t
Problem 32.6
Given that r = 1 is a solution of r3 + 3r2 − 4 = 0, find the general solution
to
y 000 + 3y 00 − 4y = 0
Solution.
Since r = 1 is a solution then using synthetic division of polynomials we can
write (r − 1)(r + 2)2 = 0. Thus, the general solution is given by
y(t) = c1 et + c2 e−2t + c3 te−2t
Problem 32.7
Given that y1 (t) = e2t is a solution to the homogeneous equation, find the
general solution to the differential equation
y 000 − 2y 00 + y 0 − 2y = 0
Solution.
The characteristic equation is given by
r3 − 2r2 + r − 2 = 0
27
Using the method of grouping we can factor into
(r − 2)(r2 + 1) = 0
The roots are r1 = 2, r2 = −i, and r3 = i. Thus, the general solution is
y(t) = c1 e2t + c3 cos t + c4 sin t
Problem 32.8
Suppose that y(t) = c1 cos t + c2 sin t + c3 cos (2t) + c4 sin (2t) is the general
solution to the equation
y (4) + a3 y 000 + a2 y 00 + a1 y 0 + a0 y = 0
Find the constants a0 , a1 , a2 , and a3 .
Solution.
The characteristic equation is of order 4. The roots are given by r1 = i, r2 =
−i, r3 = −2i, and r4 = 2i. Hence the characteristic equation is (r2 + 1)(r2 +
4) = 0 or r4 + 5r2 + 4 = 0. Comparing coefficients we find a0 = 4, a1 =
0, a2 = 5, and a3 = 0. Thus, the differential equation y (4) + 5y 00 + 4y = 0
Problem 32.9
Suppose that y(t) = c1 + c2 t + c3 cos 3t + c4 sin 3t is the general solution to
the homogeneous equation
y (4) + a3 y 000 + a2 y 00 + a1 y 0 + a0 y = 0
Determine the values of a0 , a1 , a2 , and a3 .
Solution.
The characteristic equation is of order 4. The roots are given by r = 0 (of
multiplicity 2), r = −3i and r = 3i. Hence the characteristic equation is
r2 (r2 + 9) = 0 or r4 + 9r2 = 0. Comparing coefficients we find a0 = a1 =
0, a2 = 9, and a3 = 0. Thus, the differential equation y (4) + 9y 00 = 0
Problem 32.10
Suppose that y(t) = c1 e−t sin t+c2 e−t cos t+c3 et sin t+c4 et cos t is the general
solution to the homogeneous equation
y (4) + a3 y 000 + a2 y 00 + a1 y 0 + a0 y = 0
Determine the values of a0 , a1 , a2 , and a3 .
28
Solution.
The characteristic equation is of order 4. The roots are given by r1 = −1 −
i, r2 = −1 + i, r3 = 1 − i, and r4 = 1 + i. Hence (r − (−1 − i))(r − (−1 + i)) =
r2 + 2r + 2 and (r − (1 − i))(r − (1 + i)) = r2 − 2r + 2 so that the characteristic
equation is (r2 + 2r + 2)(r2 − 2r + 2) = r4 + 4 = 0. Comparing coefficients we
find a0 = 4, a1 = a2 = a3 = 0. Thus, the differential equation y (4) + 4y = 0
Problem 32.11
Consider the homogeneous equation with constant coefficients
y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 = 0
Suppose that y1 (t) = t, y2 (t) = et , y3 (t) = cos t are several functions belonging to a fundamental set of solutions to this equation. What is the smallest
value for n for which the given functions can belong to such a fundamental
set? What is the fundamemtal set?
Solution.
The fundamental set must contain the functions y4 (t) = 1 and y5 (t) = sin t.
Thus, the smallest value of n is 5 and the fundamental set in this case is
{1, t, cos t, sin t, et }
Problem 32.12
Consider the homogeneous equation with constant coefficients
y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 = 0
Suppose that y1 (t) = t2 sin t, y2 (t) = et sin t are several functions belonging
to a fundamental set of solutions to this equation. What is the smallest value
for n for which the given functions can belong to such a fundamental set?
What is the fundamemtal set?
Solution.
The fundamental set must contain the functions y3 (t) = sin t, y4 (t) =
cos t, y5 (t) = t sin t, y6 (t) = t cos t, y7 (t) = t2 cos t, and y8 (t) = et cos t.
Thus, the smallest value of n is 8 and the fundamental set in this case is
{sin t, cos t, t sin t, t cos t, t2 sin t, t2 cos t, et sin t, et cos t}
29
Problem 32.13
Consider the homogeneous equation with constant coefficients
y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 = 0
Suppose that y1 (t) = t2 , y2 (t) = e2t are several functions belonging to a
fundamental set of solutions to this equation. What is the smallest value for
n for which the given functions can belong to such a fundamental set? What
is the fundamemtal set?
Solution.
The fundamental set must contain the functions y3 (t) = 1 and y4 (t) = t
Thus, the smallest value of n is 4 and the fundamental set in this case is
{1, t, t2 , e2t }
30
33
Non Homogeneous nth Order Linear Differential Equations
Problem 33.1
Consider the nonhomogeneous differential equation
t3 y 000 + at2 y 00 + bty 0 + cy = g(t), t > 0
Determine a, b, c, and g(t) if the general solution is given by y(t) = c1 t +
c2 t2 + c3 t4 + 2 ln t
Solution.
Since t, t2 , t4 are solutions to the homogeneous equation then
0 + 0 + bt + ct = 0
=⇒
b+c=0
0 + at2 (2) + bt(2t) + ct2 = 0
=⇒
2a + 2b + c = 0
t3 (24t) + at2 (12t2 ) + bt(4t3 ) + ct4 = 0 =⇒ 12a + 4b + c = −24
Solving the system of equations we find a = −4, b = 8, c = −8. Thus,
t3 y 000 − 4t2 y 00 + 8ty 0 − 8y = g(t)
But 2 ln t is a particular solution so that
g(t) = t3 (4t−3 − 4t2 (−2t−2 ) + 8t(2t−1 ) − 16 ln t = 28 − 16 ln t
Problem 33.2
Consider the nonhomogeneous differential equation
y 000 + ay 00 + by 0 + cy = g(t), t > 0
Determine a, b, c, and g(t) if the general solution is given by y(t) = c1 + c2 t +
c3 e2t + 4 sin 2t
Solution.
The characteristic equation is r2 (r − 2) = 0 so that the associated homogeneous equation is y 000 − 2y 00 = 0. Thus, a = −1, b = c = 0. The particular
solution is yp (t) = 4 sin 2t. Inserting into the equation we find
g(t) = −32 cos 2t − 2(−16 sin 2t) = −32 cos 2t + 32 sin 2t
31
Problem 33.3
Solve
y (4) + 4y 00 = 16 + 15et
Solution.
We first find the solution to the homogeneous differential equation. The
characteristic equations is
r4 + 4r2 = 0 or r2 (r2 + 4) = 0
The roots are
r = 0 (repeated twice), r = 2i, r = −2i
The homogeneous solution is
yh (t) = c1 + c2 t + c3 sin (2t) + c4 cos (2t)
Since g(t) is a sum of two terms, we can work each term separately. The trial
function for the function g(t) = 16 is yp = 1. Since this is a solution to the
homogeneous equation, we multiply by t to get
yp (t) = t
This is also a solution to the homogeneous equation, so multiply by t again
to get
yp (t) = t2
which is not a solution of the homogeneous equation. We write
yp1 = At2 , yp0 1 = 2At, yp001 = 2A, yp0001 = 2A, yp(4)
=0
1
Substituting back in, we get 0 + 4(2A) = 16 or A = 2. Hence
yp1 (t) = 2t2
Now we work on the second piece. The trial function for g(t) = 15et is et .
Since this in not a solution to the homogeneous equation, we get
yp2 = Aet , yp0 2 = Aet , yp002 = Aet , yp0002 = Aet , yp( 2 4) = Aet
Plugging back into the original equation gives Aet + 4Aet = 15et and this
implies A = 3. Hence
yp2 (t) = 3et
The general solution to the nonhomogeneous differential equation is
y(t) = c1 + c2 t + c3 sin (2t) + c4 cos (2t) + 2t2 + 3et
32
Problem 33.4
Solve: y (4) − 8y 00 + 16y = −64e2t
Solution.
The characteristic equation r4 − 8r2 + 16 = (r2 − 4)2 = 0 has two double
roots at r = −2 and r = 2. The homogeneous solution is then
yh (t) = c1 e−2t + c2 te−2t + c3 e2t + c4 te2t
Since the nonhomogeneous term is an exponential function, we use the method
of undetermined coefficients to find a particular solution of the form
yp (t) = At2 e2t
Plug this into the equation and get
(48A + 64At + 16At2 )e2t − 8(2A + 8At + 4At2 )e2t + 16At2 e2t = −64e2t
from which we see 32A = −64 or A = −2. The general solution is
y(t) = c1 e−2t + c2 te−2t + c3 e2t + c4 te2t − 2t2 e2t
Problem 33.5
Given that y1 (t) = e2t is a solution to the homogeneous equation, find the
general solution to the differential equation,
y 000 − 2y 00 + y 0 − 2y = 12 sin 2t
Solution.
The characteristic equation is r3 − 2r2 + r − 2 = 0. We know that r = 2 is a
root of this equation. Using synthetic division we can write
r3 − 2r2 + r − 2 = (r − 2)(r2 + 1) = 0
So the roots are r = 2, r = −i, r = i and the general solution is
yh (t) = c1 e2t + c2 cos t + c3 sin t
To find a particular solution we use the method of undetermined coefficients
by considering the trial function yp (t) = A cos 2t + B sin 2t. In this case, we
have
yp0 (t) = −2A sin 2t + 2B cos 2t
yp00 (t) = −4A cos 2t − 4b sin 2t
33
Inseting into the differential equation we find
12 sin 2t =
yp000 − 2yp00 + yp0 − 2yp
= 8A sin 2t − 8B cos 2t − 2(−4A cos 2t − 4B sin 2t)
+ (2B cos 2t − 2A sin 2t) − 2(A cos 2t + B sin 2t)
=
(6A − 6B) cos 2t + (6A + 6B) sin 2t
Equating coefficients we find 6A − 6B = 0 and 6A + 6B = 12. Solving we
find A = B = 1 and so the general solution is
y(t) = c1 e2t + c2 cos t + c3 sin t + cos 2t + sin 2t
Problem 33.6
Find the general solution of the equation
y 000 − 6y 00 + 12y 0 − 8y =
√ 2t
2te
Solution.
The characteristic equation r3 − 6r2 + 12r − 8 = (r − 2)3 = 0 has a triple
root at r = 2. Hence, the homogeneous solution is
yh (t) = c1 e2t + c2 te2t + c3 t2 e2t
We use the method of variation of paramemters to find the particular solution
yp (t) = u1 e2t + u2 te2t + r3 t2 e2t
The Wronskian is
2t
e
te2t
t2 e2t
2t
e2t + 2te2t
2te2t + 2t2 e2t
W (t) = 2e
2t
2t
2t
2t
4e 4e + 4te 2e + 8te2t + 4t2 e2t
= 2e6t
Also,
0
te2t
t2 e2t
= t2 e4t
2te2t + 2t2 e2t
W1 (t) = 0 e2t + 2te2t
1 4e2t + 4te2t 2e2t + 8te2t + 4t2 e2t 2t
2 2t
e
0
t
e
2t
2t
2
2t
= −2te4t
2te + 2t e
W2 (t) = 2e 0
4e2t 1 2e2t + 8te2t + 4t2 e2t 34
2t
2t
e
te
0
2t
2t
2t
e + 2te
0 = e4t
W3 (t) = 2e
4e2t 4e2t + 4te2t 1 Hence,
u2 (t) =
u3 (t) =
W1 (t)
g(t)dt
W (t)
W2 (t)
g(t)dt =
RW (t)
W3 (t)
g(t)dt
W (t)
R
u1 (t) =
R
√ 7
R √2 5
2 2
2 dt =
=
t
t
2
7 √
R √ 3
5
2
− 2t 2 dt = − 5 2 t 2
√ 3
R √2 1
2 2
2 dt =
=
t
t
2
3
Hence, the general solution is
√
7
√
7
2 2 2 2t
t e +
y(t) = c1 e2t + c2 te2t + c3 t2 e2t + 72 t 2 e2t − √
5
8 2 72 2t
2t
2t
2 2t
=
c1 e + c2 te + c3 t e + 105 t e
√
2 72 2t
t e
3
Problem 33.7
(a) Verify that {t, t2 , t4 } is a fundamental set of solutions of the differential
equation
t3 y 000 − 4t2 y 00 + 8ty 0 − 8y = 0
(b) Find the general solution of
√
t3 y 000 − 4t2 y 00 + 8ty 0 − 8y = 2 t, t > 0
Solution.
(a) Let y1 (t) = t, y2 (t) = t2 , y3 (t) = t4 . Then
t3 y1000 − 4t2 y100 + 8ty10 − 8y1
t3 y2000 − 4t2 y200 + 8ty20 − 8y2
t3 y3000 − 4t2 y300 + 8ty30 − 8y3
t
W (t) = 1
0
=
t3 (0) − 4t2 (0) + 8t(1) − 8t = 0
=
t3 (0) − 4t2 (2) + 8t(2t) − 8t2 = 0
3
= t (24t) − 4t2 (12t2 ) + 8t(4t3 ) − 8t4 = 0
t2 t4 2t 4t3 = 6t4 6= 0, t > 0
2 12t2 Since W (1) 6= 0 then {y1 , y2 , y3 } is a fundamental set of solutions.
(b) Using the method of variation of parameters we look for a solution of the
form
yp (t) = u1 t + u2 t2 + u3 t4
35
where
u01 (t) =
u02 (t) =
u03 (t) =
0 t2 t4
0 2t 4t3
1 2 12t2
t 0
1 0
0 1
− 52
2t
6t4
4
t 5
4t3 2t− 2
12t2 6t4
2
t t 0
1 2t 0
0 2 1
− 52
2t
6t4
3
= 23 t− 2
5
= −t− 2
9
= 13 t− 2
Thus,
R 2 −3
1
u1 (t) =
t 2 dt = − 43 t− 2
3
R −5
3
u2 (t) =
−t 2 dt = 23 t− 2
R 1 −9
2 − 72
u3 (t) =
t 2 dt = − 21
t
3
Thus, the general solution is
2 1
16 1
4 1 2 1
y(t) = c1 t + c2 t2 + c3 t4 − t 2 + t 2 − t 2 = c1 t + c2 t2 + c3 t4 − t 2
3
3
21
21
Problem 33.8
(a) Verify that {t, t2 , t3 } is a fundamental set of solutions of the differential
equation
t3 y 000 − 3t2 y 00 + 6ty 0 − 6y = 0
(b) Find the general solution of by using the method of variation of parameters
t3 y 000 − 3t2 y 00 + 6ty 0 − 6y = t, t > 0
Solution.
(a) Let y1 (t) = t, y2 (t) = t2 , y3 (t) = t3 . Then
t3 y1000 − 3t2 y100 + 6ty10 − 6y1
t3 y2000 − 3t2 y200 + 6ty20 − 6y2
t3 y3000 − 3t2 y300 + 6ty30 − 6y3
t
W (t) = 1
0
=
t3 (0) − 3t2 (0) + 6t(1) − 6t = 0
= t3 (0) − 3t2 (2) + 6t(2t) − 6t2 = 0
= t3 (6) − 3t2 (6t) + 6t(3t2 ) − 6t3 = 0
t2 t3 2t 3t2 = 2t3 6= 0, t > 0
2 6t 36
Since W (1) 6= 0 then {y1 , y2 , y3 } is a fundamental set of solutions for t > 0.
(b) Using the method of variation of parameters we look for a solution of the
form
yp (t) = u1 t + u2 t2 + u3 t3
where
u01 (t) =
u02 (t) =
u03 (t) =
0 t2 t3
0 2t 3t2
1 2 6t
2t3
3
t 3t2 t−2
6t t 0
1 0
0 1
−2
t
2t3
2
t t 0
1 2t 0
0 2 1
2t3
−2
t
=
1
2t
= − t12
=
1
2t3
Thus,
R
u1 (t) = R 2t1 dt = 12 ln t
u2 (t) = R − t12 dt = 1t
1
u3 (t) =
dt = − 4t12
2t3
Thus, the general solution is
y(t) = c1 t + c2 t2 + c3 t3 +
3
t
t
ln t + t = c1 t + c2 t2 + c3 t3 + ln t
2
4
2
since 34 t is a solution to the homogeneous equation
Problem 33.9
Solve using the method of undetermined coefficients: y 000 − y 0 = 4 + 2 cos t
Solution.
We first solve the homogeneous differential equation
y 000 − y 0 = 0
The characteristic equation is
r3 − r = 0
37
Factoring gives
r(r − 1)(r + 1) = 0
Solving we find r = 0, r = −1 and r = 1. The homogeneous solution is
yh (t) = c1 + c2 et + c3 e−t
The trial function generated by g(t) = 4 + 2 cos (2t) is
yp (t) = At + B cos (2t) + C sin (2t)
Then
yp0 = A − 2B sin (2t) + 2C cos (2t)
yp00 = −4B cos (2t) − 4C sin (2t)
8B sin (2t) − 8C cos (2t)
yp000 =
Plugging back into the original differential equation gives
[8B sin (2t) − 8C cos (2t)] − [A − 2B sin (2t) + 2C cos (2t)] = 4 + 2 cos (2t)
Combining like terms gives
−10C cos (2t) + 10B sin (2t) − A = 4 + 2 cos (2t)
Equating coefficients gives
−10C = 2
10B = 0
−A = 4
Solving we find A = −4, B = 0, and C = − 15 . The general solution is thus
y(t) = c1 + c2 et + c3 e−t − 4t −
1
sin (2t)
5
Problem 33.10
Solve using the method of undetermined coefficients: y 000 − y 0 = −4et
Solution.
We first solve the homogeneous differential equation
y 000 − y 0 = 0
38
The characteristic equation is
r3 − r = 0
Factoring gives
r(r − 1)(r + 1) = 0
Solving we find r = 0, r = −1 and r = 1. The homogeneous solution is
yh (t) = c1 + c2 et + c3 e−t
The trial function generated by g(t) = −4et is
yp (t) = Atet
Then
yp0 = Aet + Atet
yp00 = 2Aet + Atet
yp000 = 3Aet + Atet
Plugging back into the original differential equation gives
(3Aet + Atet ) − (Aet + Atet ) = −4et
Combining like terms gives
2Aet = −4et
Solving we find A = −2. The general solution is thus
y(t) = c1 + c2 et + c3 e−t − 2tet
Problem 33.11
Solve using the method of undetermined coefficients: y 000 − y 00 = 4e−2t
Solution.
The characteristic equation is
r3 − r2 = 0
Factoring gives
r2 (r − 1) = 0
39
Solving we find r = 0 (double root) and r = 1. The homogeneous solution is
yh (t) = c1 + c2 t + c3 et
The trial function generated by g(t) = 4e−2t is
yp (t) = Ae−2t
Then
yp0 = −2Ae−2t
yp00 = 4Ae−2t
yp000 = −8Ae−2t
Plugging back into the original differential equation gives
(−8e−2t ) − (4Ae−2t ) = 4e−2t
Combining like terms gives
−12Ae−2t = 4e−2t
Solving we find A = − 31 . The general solution is thus
1
y(t) = c1 + c2 et + c3 e−t − e−2t
3
Problem 33.12
Solve using the method of undetermined coefficients: y 000 −3y 00 +3y 0 −y = 12et .
Solution.
We first solve the homogeneous differential equation
y 000 − 3y 00 + 3y 0 − y = 0
The characteristic equation is
r3 − 3r2 + 3r − 1 = 0
Factoring gives
(r − 1)3 = 0
Solving we find r = 1 of multiplicity 3. The homogeneous solution is
yh (t) = c1 et + c2 tet + c3 t2 et
40
The trial function generated by g(t) = 12et is
yp (t) = At3 et
Then
yp0 =
3At2 et + At3 et
6Atet + 6At2 et + At3 et
yp00 =
t
000
yp = 6Ae + 18Atet + 9At2 et + At3 et
Plugging back into the original differential equation gives
(6Aet +18Atet +9At2 et +At3 et )−3(6Atet +6At2 et +At3 et )+3(3At2 et +At3 et )−At3 et = 12et
Combining like terms gives
6Aet = 12et
Solving we find A = 2. The general solution is thus
y(t) = c1 et + c2 tet + c3 t2 et + 2t3 et
Problem 33.13
Solve using the method of undetermined coefficients: y 000 + y = et + cos t.
Solution.
The characteristic equation is
r3 + 1 = 0
Factoring gives
(r + 1)(r2 − r + 1) = 0
√
1
2
−i
3
2
√
and r = 12 + i 23 . The homogeneous
√
√
t
t
3
3
−t
yh (t) = c1 e + c2 e 2 cos
t + c3 e 2 sin
t
2
2
The trial function generated by g(t) = et + cos t is
Solving we find r = −1, r =
solution is
yp (t) = Aet + B cos t + C sin t
Then
yp0 = Aet − B sin t + C cos t
yp00 = Aet − B cos t − C sin t
yp000 = Aet + B sin t − C cos t
41
Plugging back into the original differential equation gives
(Aet + B sin t − C cos t) + (Aet + B cos t + C sin t) = et + cos t
or
2Aet + (B + C) sin t + (B − C) cos t = et + cos t
Equating coefficients we find
2A
= 1
B+C = 0
B−C = 1
Solving we find A = B =
1
2
and C = − 12 . Thus, the general solution is
√
−t
y(t) = c1 e
√
t
3
3
1
+ c2 e cos
t + c3 e 2 sin
t + (et + cos t − sin t
2
2
2
t
2
In Problems 33.14 and 33.15, answer the following two questions.
(a) Find the homogeneous general solution.
(b) Formulate an appropriate for for the particular solution suggested by the
method of undetermined coefficients. You need not evaluate the undetermined coefficients.
Problem 33.14
y 000 − 3y 00 + 3y 0 − y = et + 4et cos 3t + 4
Solution.
(a) The characteristic equation is
r3 − 3r2 + 3r − 1 = (r − 1)3 = 0
So r = 1 is a root of multiplicity 3 so that the homogeneous general solution
is
yh (t) = c1 et + c2 tet + c3 t2 et .
(b) The trial function for the right-hand function g(t) = et + 4et cos 3t + 4 is
yp (t) = At3 et + Bet cos 3t + Cet sin 3t + D.
Problem 33.15
y (4) + 8y 00 + 16y = t cos 2t
42
Solution.
(a) The characteristic equation is
r4 + 8r2 + 16 = (r2 + 4)2 = 0
So r = −2i and r = 2i are roots of multiplicity 2 so that the homogeneous
general solution is
yh (t) = c1 cos 2t + c2 sin 2t + c3 t cos 2t + c4 t sin 2t.
(b) The trial function for the right-hand function g(t) = t cos 2t is yp (t) =
t2 (At + B) cos 2t + t2 (Ct + D) sin 2t.
Consider the nonhomogeneous differential equation
y 000 + ay 00 + by 0 + cy = g(t)
In Problems 33.16 - 33.17, the general solution of the differential equation is
given, where c1 , c2 , and c3 represent arbitrary constants. Use this information
to determine the constants a, b, c and the function g(t).
Problem 33.16
y(t) = c1 + c2 t + c3 e2t + 4 sin 2t.
Solution.
The roots of the characteristic equation are r = 0 of multiplicity 2 and r = 2.
Thus, the equation is
r2 (r − 2) = r3 − 2r2 = 0.
The corresponding differential equation is y 000 − 2y 00 = 0. Comparing coefficients we find a = −2, b = c = 0. The function yp (t) = 4 sin 2t is a
particular solution to the nonhomogeneous equation. Taking derivatives we
find yp0 (t) = 8 cos 2t, yp00 (t) = −16 sin 2t, yp000 (t) = −32 cos 2t. By plugging into
the equation we find −32 cos 2t + 32 sin 2t = g(t)
Problem 33.17
y(t) = c1 + c2 t + c3 t2 − 2t3
43
Solution.
The roots of the characteristic equation are r = 0 of multiplicity 3. Thus,
the equation is
r3 = 0.
The corresponding differential equation is y 000 = 0. Comparing coefficients we
find a = b = c = 0. The function yp (t) = −2t3 is a particular solution to the
nonhomogeneous equation. Taking derivatives we find yp0 (t) = −6t2 , yp00 (t) =
−12t, yp000 (t) = −12. By plugging into the equation we find −12 = g(t)
Problem 33.18
Consider the nonhomogeneous differential equation
t3 y 000 + at2 y 00 + bty 0 + cy = g(t), t > 0
Suppose that y(t) = c1 t + c2 t2 + c3 t4 + 2 ln t is the general solution to the
above equation. Determine the constants a, b, c and the function g(t)
Solution.
The functions y1 (t) = t, y2 (t) = t2 , and y3 (t) = t4 are solutions to the
homogeneous equation. Substituting into the equation we find
0 + 0 + bt + ct = 0
−→
b+c=0
0 + 2at2 + 2bt2 + ct2 = 0
−→
2a + 2b + c = 0
24t4 + 12at4 + 4bt4 + ct4 = 0 −→ 12a + 4b + c = −24
Solving this system of equations we find a = −4, b = 8, and c = −8. Thus,
t3 y 000 − 4t2 y 00 + 8ty 0 − 8y = g(t).
Since yp (t) = 2 ln t is a particular solution to the nonhomogeneous equation
then by substitution we find g(t) = t3 ( t43 ) − 4t2 (− t22 ) + 8t( 2t ) − 16 ln t =
28 − 16 ln t
44
34
Existence and Uniqueness of Solution to
Initial Value First Order Linear Systems
Problem 34.1
Consider the initial value problem
(t + 2)y10 = 3ty1 + 5y2 , y1 (1) = 0
(t − 2)y20 = 2y1 + 4ty2 , y2 (1) = 2
Determine the largest t-interval such that a unique solution is guaranteed to
exist.
Solution.
All the coefficient functions and the right-hand side functions are continuous
for all t 6= ±2. Since t0 = 1, the t−interval of existence is −2 < t < 2
Problem 34.2
Verify that the functions y1 (t) = c1 et cos t + c2 et sin t and y2 (t) = −c1 et sin t +
c2 et cos t are solutions to the linear system
y10 = y1 + y2
y20 = −y1 + y2
Solution.
Taking derivatives we find y10 (t) = (c1 + c2 )et cos t + (c2 − c1 )et sin t and
y20 (t) = −(c1 +c2 )et sin t+(c2 −c1 )et cos t. But y1 +y2 = (c1 +c2 )et cos t+(c2 −
c1 )et sin t = y10 (t) and −y1 + y2 = −(c1 + c2 )et sin t + (c2 − c1 )et cos t = y20 (t)
Problem 34.3
Consider the initial value problem
y0 (t) = Ay(t), y(0) = y0
where
3 2
−1
A=
, y0 =
4 1
8
1
−1
5t
−t
(a) Verify that y(t) = c1 e
+ c2 e
is a solution to the first
1
2
order linear system.
(b) Determine c1 and c2 such that y(t) solves the given initial value problem.
45
Solution.
(a) We have
0
y (t) = c1 e
and
5t
Ay(t) = c1 e
=
5t
3 2
4 1
5
5
1
1
5
5
5t
c1 e
−t
+ c2 e
+ c2 e
−t
+ c2 e
−t
1
−2
3 2
4 1
1
−2
−1
2
y0 (t)
=
(b) We need to solve the system
c1 − c2 = −1
c1 + 2c2 = 8
Solving this system we find c1 = 2 and c2 = 3. Therefore, the unique solution
to the system is
5t
2e − 3e−t
y(t) =
2e5t + 6e−t
Problem 34.4
√
Rewrite the differential equation (cos t)y 00 − 3ty 0 + ty = t2 + 1 in the matrix
form y(t) = P(t)y(t) + g(t).
Solution.
√
Rewriting the equation in the form y 00 − 3t sec ty 0 + t sec t = (t2 + 1) sec t
we find
0 0
1
y
y
0
√
y0 =
=
+
y0
y0
(t2 + 1) sec t
− t sec t 3t sec t
=
P(t)y + g(t)
Problem 34.5
Rewrite the differential equation 2y 00 + ty + e3t = y 000 + (cos t)y 0 in the matrix
form y(t) = P(t)y(t) + g(t).
46
Solution.
Rewriting the given equation in the form y 000 − 2y 00 + (cos t)y 0 − ty = e3t we
find

0 


y
0
1
0
y
0
1   y 0  + 0 0 e3t
y0 =  y 0  =  0
y 00
t − cos t 2
y 00
=
P(t)y(t) + g(t)
Problem 34.6
The initial value problem
0 1
0
1
0
y (t) =
y+
, y(−1) =
−3 2
2 cos (2t)
4
was obtained from an initial value problem for a higher order differential
equation. What is the corresponding scalar initial value problem?
Solution.
Carrying the matrix arithmetic we find y 00 = −3y + 2y 0 + 2 cos 2t. Thus, the
initial value problem is
y 00 + 3y − 2y 0 = 2 cos 2t, y(−1) = 1, y 0 (−1) = 4
Problem 34.7
The initial value problem



y2



y3
 , y(1) = 
y0 (t) = 



y4
2
y2 + y3 sin y1 + y3

0
0 

−1 
2
was obtained from an initial value problem for a higher order differential
equation. What is the corresponding scalar initial value problem?
Solution.
Let

 
y1
y
 y2   y 0
 
y(t) = 
 y3  =  y 00
y4
y 000
47




Then
 
y0
y0
00
 y  
y 00


y0 = 
=
 y 000  
y 000
0
00
y + y sin y + (y 00 )2
y (4)





Equating components we find
y (4) = y 0 + y 00 sin y + (y 00 )2 .
Thus, the initial value problem is
y (4) = y 0 + y 00 sin y + (y 00 )2 , y(1) = y 0 (1) = 0, y 00 (1) = −1, y 000 (1) = 2
Problem 34.8
Consider the system of differential equations
y 00 = tz 0 + y 0 + z
z 00 = y 0 + z 0 + 2ty
Write the above system in the form
y0 = P(t)y + g(t)
where


y(t)
 y 0 (t) 

y(t) = 
 z(t) 
z 0 (t)
Identify P(t) and g(t).
Solution.
 
y0
0
 y 00   0
 
= 
 z0  =  0
z 00
2t

y0
=
1
1
0
1
0
1
0
0

0
y


t   y0
1  z
1
z0
P(t)y + G(t)
48

0
 0 
 
 0 
0

Problem 34.9
Consider the system of differential equations
y 00 = 7y 0 + 4y − 8z + 6z 0 + t2
z 00 = 5z 0 + 2z − 6y 0 + 3y − sin t
Write the above system in the form
y0 = P(t)y + g(t)
where


y(t)
 y 0 (t) 

y(t) = 
 z(t) 
z 0 (t)
Identify P(t) and g(t).
Solution.
 
0 1
0
0
y0
 y 00   4 7 −8 6
 
= 
 z0  =  0 0
0 1
00
z
3 −6 2
5

y0
=

0
y
  y 0   t2


 z 
0
0
− sin t
z

P(t)y + G(t)
49




35
Homogeneous First Order Linear Systems
In Problems 35.1 - 35.3 answer the following two questions.
(a) Rewrite the given system of linear homogeneous differential equations as
a homogeneous linear system of the form y0 (t) = P(t)y.
(b) Verify that the given function y(t) is a solution of y0 (t) = P(t)y.
Problem 35.1
y10 = −3y1 − 2y2
y20 = 4y1 + 3y2
and
et + e−t
−2et − e−t
−3 −2
4
3
et − e−t
−2et + e−t
y(t) =
Solution.
(a)
y1
y2
0
=
y1
y2
(b) We have
0
y =
and
P(t)y =
−3 −2
4
3
et + e−t
−2et − e−t
=
Problem 35.2
y10 =
y2
2
0
y2 = − t2 y1 + 2t y2
and
y(t) =
t2 + 3t
2t + 3
50
et − e−t
−2et + e−t
= y0
Solution.
(a)
y1
y2
0
0
− t22
=
1
2
t
y1
y2
(b) We have
0
y =
and
P(t)y =
0
− t22
1
2
t
2t + 3
2
t2 + 3t
2t + 3
=
2t + 3
2
= y0
Problem 35.3
y10 = 2y1 + y2 + y3
y20 = y1 + y2 + 2y3
y30 = y1 + 2y2 + y3
and

2et + e4t
y(t) =  −et + e4t 
−et + e4t

Solution.
(a)
0 


y1
2 1 1
y1
 y2  =  1 1 2   y2 
y3
1 1 2
y3

(b) We have


2et + 4e4t
y0 =  −et + 4e4t 
−et + 4e4t
and


 

2et + e4t
2et + 4e4t
2 1 1
P(t)y =  1 1 2   −et + e4t  =  −et + 4e4t  = y0
1 1 2
−et + e4t
−et + 4e4t
In Problems 35.4 - 35.7
(a) Verify the given functions are solutions of the homogeneous linear system.
51
(b) Compute the Wronskian of the solution set. On the basis of this calculation can you assert that the set of solutions forms a fundamental set?
(c) If the given solutions are shown in part(b) to form a fundamental set,
state the general solution of the linear homogeneous system. Express the
general solution as the product y(t) = Ψ(t)c, where Ψ(t) is a square matrix
whose columns are the solutions forming the fundamental set and c is a column vector of arbitrary constants.
(d) If the solutions are shown in part (b) to form a fundamental set, impose
the given initial condition and find the unique solution of the initial value
problem.
Problem 35.4
3t
4e + 2e−t
9 −4
0
2e3t − 4e−t
0
, y2 (t) =
y =
y, y(0) =
, y1 (t) =
6e3t + 5e−t
3e3t − 10e−t
15 −7
1
Solution.
(a) We have
y10
and
9 −4
15 −7
=
6e3t + 4e−t
9e3t + 10e−t
2e3t − 4e−t
3e3t − 10e−t
=
6e3t + 4e−t
9e3t + 10e−t
= y10
Similarly,
y20
and
9 −4
15 −7
=
12e3t − 2e−t
18e3t − 5e−t
4e3t + 2e−t
6e3t + 5e−t
=
12e3t − 2e−t
18e3t − 5e−t
(b) The Wronskian is given by
2e3t − 4e−t 4e3t + 2e−t
W (t) = 3t
3e − 10e−t 6e3t + 5e−t
= y20
= 20e2t
Since W (t) 6= 0, the set {y1 , y2 } forms a fundamental set of solutions.
(c) The general solution is
2e3t − 4e−t 4e3t + 2e−t
c1
y(t) = c2 y1 + c2 y2 =
3e3t − 10e−t 6e3t + 5e−t
c2
52
(d) We have
−2 6
−7 11
c1
c2
=
0
1
Solving this system we find c1 = −0.3, c2 = −0.1. Therefore the solution to
the initial value problem is
3t
2e3t − 4e−t
4e + 2e−t
e−3t + e−t
y(t) = −0.3
− 0.1
=
3e3t − 10e−t
6e3t + 5e−t
−1.5e3t + 2.5e−t
Problem 35.5
−3 −5
5
−5e−2t cos 3t
0
y =
y, y(0) =
, y1 (t) =
,
2 −1
2
e−2t (cos 3t − 3 sin 3t)
y2 (t) =
−5e−2t sin 3t
e−2t (3 cos 3t + sin 3t)
Solution.
(a) We have
y10
=
5e−2t (2 cos 3t + 3 sin 3t)
−11e−2t (cos 3t + sin 3t)
and
−2t
−3 −5
−5e−2t cos 3t
5e (2 cos 3t + 3 sin 3t)
=
= y10
2 −1
e−2t (cos 3t − 3 sin 3t)
−11e−2t (cos 3t + sin 3t)
Similarly,
y20
=
5e−2t (2 sin 3t − 3 cos 3t)
e−2t (−3 cos 3t − 11 sin 3t)
and
−3 −5
−5e−2t sin 3t
5e−2t (2 sin 3t − 3 cos 3t)
=
= y20
2 −1
e−2t (3 cos 3t + sin 3t)
e−2t (−3 cos 3t − 11 sin 3t)
(b) The Wronskian is given by
−5e−2t cos 3t
−5e−2t sin 3t
W (t) = −2t
−2t
e (cos 3t − 3 sin 3t) e (3 cos 3t − sin 3t)
53
= −15e−4t
Since W (t) 6= 0, the set {y1 , y2 } forms a fundamental set of solutions.
(c) The general solution is
−5e−2t cos 3t
−5e−2t sin 3t
c1
y(t) = c2 y1 + c2 y2 =
e−2t (cos 3t − 3 sin 3t) e−2t (3 cos 3t − sin 3t)
c2
(d) We have
−5 0
1 3
c1
c2
=
5
2
Solving this system we find c1 = −1, c2 = 1. Therefore the solution to the
initial value problem is
−5e−2t cos 3t
−5e−2t sin 3t
5e−2t (cos 3t − sin 3t)
y(t) =
− −2t
=
e−2t (cos 3t − 3 sin 3t)
e (3 cos 3t + sin 3t)
e−2t (2 cos 3t + 4 sin 3t)
Problem 35.6
1 −1
−2
1
e3t
0
y =
y, y(−1) =
, y1 (t) =
, y2 (t) =
−2e3t
−2 2
4
1
Solution.
(a) We have
y10
and
1 −1
−2 2
=
0
0
0
0
3e3t
−6e−3t
1
1
=
= y10
Similarly,
y20
and
1 −1
−2 2
=
e3t
−2e3t
=
3e3t
−6e−3t
(b) The Wronskian is given by
1 e3t
W (t) = 1 −2e3t
54
= −3e−3t
= y20
Since W (t) 6= 0, the set {y1 , y2 } forms a fundamental set of solutions.
(c) The general solution is
1 e3t
c1
y(t) = c2 y1 + c2 y2 =
1 −2e3t
c2
(d) We have
1 e−3
1 −2e−3
c1
c2
=
−2
4
Solving this system we find c1 = 0, c2 = −2e3 . Therefore the solution to the
initial value problem is
−2e3(t+1)
y(t) =
4e3(t+1)
Problem 35.7






 −2t 
0
−2 0 0
3
e
1 4  y, y(0) =  4  , y1 (t) =  0  , y2 (t) =  2et cos 2t 
y0 =  0
−et sin 2t
0
0 −1 1
−2


0
y3 (t) =  2et sin 2t 
et cos 2t
Solution.
(a) We have

−2e−2t

0
y10 = 
0

and
  −2t  

−2 0 0
e
−2e−2t
 0
 = y10
1 4  0  = 
0
0 −1 1
0
0

Similarly,


0
y20 =  2et (cos 2t − 2 sin 2t) 
−et (sin 2t + 2 cos 2t)
55
and


 

−2 0 0
0
0
 0
1 4   2et cos 2t  =  2et (cos 2t − 2 sin 2t)  = y20
0 −1 1
−et sin 2t
−et (sin 2t + 2 cos 2t)


0
y30 =  2et (sin 2t + 2 cos 2t) 
et (cos 2t − 2 sin 2t)
and


 

−2 0 0
0
0
 0
1 4   2et cos 2t  =  2et (sin 2t + 2 cos 2t)  = y30
0 −1 1
−et sin 2t
et (cos 2t − 2 sin 2t)
(b) The Wronskian is given by
−2t
e
0
0
t
t
2e cos 2t 2e sin 2t
W (t) = 0
0 −et sin 2t et cos 2t
= 2.
Since W (t) 6= 0, the set {y1 , y2 , y3 } forms a fundamental set of solutions.
(c) The general solution is
 −2t


c1
e
0
0
2et cos 2t 2et sin 2t   c2 
y(t) = c2 y1 + c2 y2 + c3 y3 =  0
c3
0 −et sin 2t et cos 2t
(d) We have


 

1 0 0
c1
3
 0 2 0   c2  =  4 
0 0 1
c3
−2
Solving this system using Cramer’s rule we find c1 = 3, c2 = 2, c3 = −2.
Therefore the solution to the initial value problem is
 −2t  
 
 

3e
0
0
3e−2t
y(t) =  0 + 4et cos 2t − 4et sin 2t  =  4et (cos 2t − sin 2t) 
0
−2et sin 2t
2et cos 2t
−2et (sin 2t + cos 2t)
In Problems 35.8 - 35.9, the given functions are solutions of the homogeneous
linear system.
56
(a) Compute the Wronskian of the solution set and verify the set is a fundamental set of solutions.
(b) Compute the trace of the coefficient matrix.
(c) Verify
Abel’s theorem by showing that, for the given point t0 , W (t) =
Rt
t0 tr(P(s))ds
W (t0 )e
.
Problem 35.8
t 6
5
5e−t
e
0
y =
y, y1 (t) =
, y2 (t) =
, t0 = −1, −∞ < t < ∞
−t
−7 −6
−7e
−et
Solution.
(a) The Wronskian is
5e−t
et
W (t) = −7e−t −et
= 2.
Since W (t) 6= 0, the set {y1 , y2 } forms a fundamental set of solutions.
(b) tr(P(t)) = 6 − 6 = 0. R
Rt
t
tr(P(s))ds
(c) W (t) = 2 and W (t0 )e t0
= 2e −1 0ds = 2
Problem 35.9
t 1
t
−1
e
0
y =
y, y1 (t) =
, y2 (t) =
, t0 = −1, t 6= 0, 0 < t < ∞
−1
−1
t
0
0 −t
Solution.
(a) The Wronskian is
−1 et
W (t) = −1
t
0
= −t−1 et .
Since W (t) 6= 0, the set {y1 , y2 } forms a fundamental set of solutions.
(b) tr(P(t)) = 1 − t−1
Rt
Rt
−1
tr(P(s))ds
(c) W (t) = −t−1 et and W (t0 )e t0
= −e · e 1 (1−s )ds = −t−1 et
Problem 35.10
The functions
y1 (t) =
5
1
,
y2 (t) =
57
2e3t
e3t
are known to be solutions of the homogeneous linear system y0 = Py, where
P is a real 2 × 2 constant matrix.
(a) Verify the two solutions form a fundamental set of solutions.
(b) What is tr(P)?
(c) Show that Ψ(t) satisfies the homogeneous differential equation Ψ0 = PΨ,
where
5 2e3t
Ψ(t) = [y1 (t) y2 (t)] =
1 e3t
(d) Use the observation of part (c) to determine the matrix P.[Hint: Compute
the matrix product Ψ0 (t)Ψ−1 (t). It follows from part (a) that Ψ−1 (t) exists.]
Are the results of parts (b) and (d) consistent?
Solution.
(a) The Wronskian is given by
5 2e3t
W (t) = 1 e3t
= 3e3t
Since
W (t) 6= 0, the set {y1 , y2 } forms a fundamental set of solutions.
(b) Since W 0 (t)−tr(P(t))W (t) = 0, 9e3t −3tr(P(t))e3t = 0. Thus, tr(P(t)) =
3.
(c) We have
Ψ0 (t) = [y10 y20 ] = [P(t)y1 P(t)y2 ] = P(t)[y1 y2 ] = P(t)Ψ(t)
(d) From part(c) we have
P(t) = Ψ0 (t)Ψ−1 (t) =
0 6e3t
0 3e3t
=
·
−2 10
−1 5
1
3e3t
e3t −2e3t
−1
5
The results in parts (b) and (d) are consistent since tr(P(t)) = −2 + 5 = 3
Problem 35.11
The homogeneous linear system
0
y =
3 1
−2 α
58
y
has a fundamental set of solutions whose Wronskian is constant, W (t) =
4, − ∞ < t < ∞. What is the value of α?
Solution.
We know that W (t) satisfies the equation W 0 (t) − tr(P(t))W (t) = 0. But
tr(P(t)) = 3 + α. Thus,R W 0 (t) − (3 + α)W (t) = 0. Solving Rthis equation
t
t
we find W (t) = W (0)e 0 (3+α)ds . Since W (t) = W (0) = 4, e 0 (3+α)ds = 1.
Evaluating the integral we find e(3+α)t = 1. This implies that 3 + α = 0 or
α = −3
59
36
First Order Linear Systems: Fundamental
Sets and Linear Independence
In Problems 36.1 - 36.4, determine whehter the given functions are linearly
dependent or linearly independent on the interval −∞ < t < ∞.
Problem 36.1
f1 (t) =
t
1
t2
1
0
0
, f2 (t) =
Solution.
Suppose
k1
t
1
+ k2
t2
1
=
Then k1 t+k2 t2 = 0 and k1 +k2 = 0 for all t. In particular, for t = −1 we have
−k1 + k2 = 0. But k1 + k2 = 0. These two equations imply that k1 = k2 = 0.
Hence, {f1 (t), f2 (t)} is a linearly independent set
Problem 36.2
et
1
f1 (t) =
, f2 (t) =
e−t
1
, f3 (t) =
Solution.
Note that
1
2
t
e
1
−
1
2
−t
e
1
−
et −e−t
2
0
et −e−t
2
0


0
= 0 
0
This shows that f1 (t), f2 (t), and f3 (t) are linearly dependent
Problem 36.3




 
1
0
0





f1 (t) = t , f2 (t) = 1 , f3 (t) = 0 
0
t2
0
60
Solution.
Note that




   
1
0
0
0







0 t +0 1 +1· 0 = 0 
0
t2
0
0
This shows that f1 (t), f2 (t), and f3 (t) are linearly dependent
Problem 36.4




 
1
0
1
2 
2




f1 (t) = sin t , f2 (t) = 2(1 − cos t) , f3 (t) = 0 
0
−2
1
Solution.
Note that

    

0
1
0
1
 sin2 t  − 1  2(1 − cos2 t)  −  0  =  0 
2
−2
1
0
0

This shows that f1 (t), f2 (t), and f3 (t) are linearly dependent
Problem 36.5
Consider the functions
f1 (t) =
t2
0
, f2 (t) =
2t
1
(a) Let Ψ(t) = [f1 (t) f2 (t)]. Determine det(Ψ(t)).
(b) Is it possible that the given functions form a fundamental set of solutions
for a linear system y0 = P(t)y where P(t) is continuous on a t-interval containing the point t = 0? Explain.
(c) Determine a matrix P(t) such that the given vector functions form a
fundamental set of solutions for y0 = P(t)y. On what t-interval(s) is the
coefficient matrix P(t) continuous?(Hint: The matrix Ψ(t) must satisfy
Ψ0 (t) = P(t)Ψ(t) and det(Ψ(t)) 6= 0.)
Solution.
(a) We have
2
t 2t = t2 .
F (t) = 0 1 61
(b) Since F (0) = 0, the given functions do not form a fundamental set for a
linear system y0 = P(t)y on any t−interval containing 0.
(c) For Ψ(t) to be a fundamental matrix it must satisfy the differential equation Ψ0 (t) = P(t)Ψ(t) and the condition det(Ψ(t)) 6= 0. But det(Ψ(t)) = t2
and this is not zero on any interval not containing zero. Thus, our coefficient
matrix P(t) must be continuous on either −∞ < t < 0 or 0 < t < ∞. Now,
from the equation Ψ0 (t) = P(t)Ψ(t) we can find P(t) = Ψ0 (t)Ψ−1 (t). That
is,
2t 2
1 −2t
1
0
−1
P(t) = Ψ (t)Ψ (t) = t2
0 0
0 t2
=
Problem
Let

1
0

y = 0
0
2t−1 −2
0
0
36.6


 t

 t
1 1
e + e−t 4e2t et + 4e2t
e
e−t
4e2t

e2t
−1 1  y, Ψ(t) =  0 −2e−t e2t  , Ψ(t) =  −2e−t e2t
2t
2t
2t
0
3e
3e
0 2
0
0
3e
(a) Verify that the matrix Ψ(t) is a fundamental matrix of the given linear
system.
(b) Determine a constant matrix A such that the given matrix Ψ(t) can be
represented as Ψ(t) = Ψ(t)A.
(c) Use your knowledge of the matrix A and assertion (b) of Theorem 36.4 to
determine whether Ψ(t) is also a fundamental matrix, or simply a solution
matrix.
Solution.
(a) Since

et −e−t 8e2t
Ψ0 (t) =  0 2e−t 2e2t 
0
0
6e2t

and
 t
  t

1 1 1
e
e−t
4e2t
e −e−t 8e2t
P(t)Ψ(t) =  0 −1 1   0 −2e−t e2t  =  0 2e−t 2e2t 
0 0 2
0
0
3e2t
0
0
6e2t

62
Thus, Ψ is a solution matrix. To show that Ψ(t) is a fundamental matrix
we need to verify that det(Ψ(t)) 6= 0. Since det(Ψ(t)) = −6e2t 6= 0, Ψ(t) is a
fundamental matrix.
(b) Note that
 t
  t


e + e−t 4e2t et + 4e2t
e
e−t
4e2t
1 0 1
 =  0 −2e−t e2t
 1 0 0 
e2t
Ψ(t) =  −2e−t e2t
2t
2t
2t
0
3e
3e
0
0
3e
0 1 1
Thus,


1 0 1
A= 1 0 0 
0 1 1
(c) Since det(A) = 1, Ψ(t) is a fundamental matrix
Problem 36.7
Let
0
y =
1 1
0 −2
y, Ψ(t) =
et
e−2t
0 −3e−2t
where the matrix Ψ(t) is a fundamental matrix of the given homogeneous
linear system.
Find a constant matrix A such that Ψ(t) = Ψ(t)A with
1 0
Ψ(0) =
.
0 1
Solution.
−1
We need to find
a matrix A such that Ψ(0) = Ψ(0)A or A = Ψ (0)Ψ(0) =
−3 −1
Ψ−1 (0) = − 13
0
1
63
37
Homogeneous Systems with Constant Coefficients
In Problems 37.1 - 37.3, a 2 × 2 matrix P and vectors x1 and x2 are given.
(a) Decide which, if any, of the given vectors is an eigenvector of P, and
determine the corresponding eigenvalue.
(b) For the eigenpair found in part (a), form a solution yk (t), where k = 1
or k = 2, of the first order system y0 = Py.
(c) If two solution are found in part (b), do they form a fundamental set of
solutions for y0 = Py.
Problem 37.1
P=
7 −3
16 −7
, x1 =
3
8
3
8
, x2 =
1
2
Solution.
(a) We have
Px1 =
7 −3
16 −7
= −1x1 .
Thus, x1 is an eigenvector corresponding to the eigenvalue r1 = −1. Similarly,
7 −3
1
Px2 =
= 1x2 .
16 −7
2
Thus, x2 is an eigenvector corresponding to the eigenvalue r1 = 1.
(b) Solutions to the system y0 = P(t)y are y1 (t) = e−t x1 and y2 (t) = et x2 .
(c) The Wronskian is
−t
t 3e
e
= −2 6= 0
W (t) = −t
8e
2et so that the set {y1 , y2 } forms a fundamental set of solutions
Problem 37.2
P=
−5 2
−18 7
, x1 =
64
1
3
, x2 =
1
2
Solution.
(a) We have
Px1 =
−5 2
−18 7
1
3
= 1x1 .
Thus, x1 is an eigenvector corresponding to the eigenvalue r1 = 1. Similarly,
−5 2
1
−1
Px2 =
=
.
−18 7
2
−4
Since the right-hand side cannot be a scalar multiple of x2 , x2 is not an
eigenvector of P.
(b) Solution to the system y0 = P(t)y is y1 (t) = et x1 .
(c) The Wronskian is not defined for this problem
Problem 37.3
P=
2 −1
−4 2
, x1 =
1
−2
1
−2
, x2 =
1
2
Solution.
(a) We have
Px1 =
2 −1
−4 2
= 4x1 .
Thus, x1 is an eigenvector corresponding to the eigenvalue r1 = 4. Similarly,
2 −1
1
Px2 =
= 0x2 .
−4 2
2
Thus, x2 is an eigenvector corresponding to the eigenvalue r1 = 0.
(b) Solutions to the system y0 = P(t)y are y1 (t) = e4t x1 and y2 (t) = x2 .
(c) The Wronskian is
e4t
1 W (t) = = 4e4t 6= 0
−2e4t 2 so that the set {y1 , y2 } forms a fundamental set of solutions
In Problems 37.4 - 37.6, an eigenvalue is given of the matrix P. Determine a
corresponding eigenvector.
65
Problem 37.4
P=
5
3
−4 −3
, r = −1
Solution.
We have
(P + I)x =
6
3
−4 −2
x1
x2
6x1 + 3x2
−4x1 − 2x2
=
0
0
Solving this system we find x2 = −2x1 . Letting x1 = 1 then x2 = −2 and an
eigenvector is
1
x=
−2
Problem 37.5


1 −7 3
P =  −1 −1 1  , r = −4
4 −4 0
Solution.
We have



  
5 −7 3
x1
5x1 − 7x2 + 3x3
0







−1
3
1
x
−x
+
3x
+
x
(P + 4I)x =
= 0 
2
1
2
3
4 −4 4
x3
4x1 − 4x2 + 4x3
0
Solving this system we find x1 = 2x2 and x3 = −x2 . Letting x2 = 1 then
x1 = 2 and x3 = −1. Thus, an eigenvector is


2
x= 1 
−1
Problem 37.6


1 3 1
P =  2 1 2 , r = 5
4 3 −2
66
Solution.
We have



  
−4 3
1
x1
−4x1 + 3x2 + x3
0







2 −4 2
x2
2x1 − 4x2 + 2x3
(P − 5I)x =
= 0 
4
3 −7
x3
4x1 + 3x2 − 7x3
0
Solving this system we find x1 = x2 = x3 . Letting x3 = 1 then x1 = 1 and
x2 = 1. Thus, an eigenvector is
 
1

x= 1 
1
In Problems 37.7 - 37.10, Find the eigenvalues of the matrix P.
Problem 37.7
P=
Solution.
The characteristic equation is
−5 − r
1
0
4−r
−5 1
0 4
= (r + 5)(r − 4) = 0
Thus, the eigenvalues are r = −5 and r = 4
Problem 37.8
P=
Solution.
The characteristic equation is
3 − r −3
−6 6 − r
3 −3
−6 6
= r(r − 9) = 0
Thus, the eigenvalues are r = 0 and r = 9
67
Problem 37.9


5 0 0
P= 0 1 3 
0 2 2
Solution.
The characteristic equation is
5−r
0
0
0
1−r
3
0
2
2−r
= (5 − r)(r + 1)(r − 4) = 0
Thus, the eigenvalues are r = 5, r = 4, and r = −1
Problem 37.10


1 −7 3
P =  −1 −1 1 
4 −4 0
Solution.
The characteristic equation is
1−r
−7
3
−1 −1 − r 1
4
−4
−r
= −r(r − 4)(r + 4) = 0
Thus, the eigenvalues are r = 0, r = 4, and r = −4
In Problems 37.11 - 37.13, the matrix P has distinct eigenvalues. Using Theorem 35.4 determine a fundamental set of solutions of the system y0 = Py.
Problem 37.11
P=
−0.09 0.02
0.04 −0.07
68
Solution.
The characteristic equation is
−0.09 − r
0.02
0.04
−0.07 − r
= r2 + 0.16r + 0.0055 = 0
Solving this quadratic equation we find r = −0.11 and r = −0.05. Now,
0.02 0.02
x1
0.02x1 + 0.02x2
0
(P + 0.11I)x =
=
0.04 0.04
x2
0.04x1 + 0.04x2
0
Solving this system we find x1 = −x2 . Letting x2 = 1 then x1 = −1. Thus,
an eigenvector is
1
x1 =
−1
Similarly,
(P + 0.05I)x =
−0.04 0.02
0.04 −0.02
x1
x2
−0.04x1 + 0.02x2
0.04x1 − 0.02x2
=
0
0
Solving this system we find 2x1 = x2 . Letting x1 = 1 then x2 = 2. Thus, an
eigenvector is
1
x2 =
2
By Theorem 35.4, a fundamental set of solutions is given by {e−0.11t x1 , e−0.05t x2 }
Problem 37.12


1 2 0
P =  −4 7 0 
0 0 1
Solution.
The characteristic equation is
1−r
2
0
−4 7 − r
0
0
0
1−r
= (r − 1)(r − 3)(r − 5) = 0
69
Solving this equation we find

0 2

(P − I)x = −4 6
0 0
r1 = 1, r2 = 3, and r3 = 5. Now,


  
0
x1
2x2
0






0
x2
−4x1 + 6x2 = 0 
0
x3
0
0
Solving this system we find x1 = x2 = 0, and x3 is arbitrary. Letting x3 = 1,
an eigenvector is
 
0

x1 = 0 
1
Similarly,

  

0
−2x1 + 2x2
−2 2 0
x1







−4x1 + 4x2 = 0 
x2
(P − 3I)x = −4 4 0
0 0 −2
x3
0
−2x3

Solving this system we find x1 = x2 and x3 = 0. Letting x1 = x2 = 1, an
eigenvector is
 
1

x2 = 1 
0



  
−4 2 0
x1
−4x1 + 2x2
0
(P − 5I)x =  −4 2 0   x2   −4x1 + 2x2  =  0 
0 0 −4
x3
−4x3
0
Solving this system we find 2x1 = x2 , and x3 = 0. Letting x1 = 1, an
eigenvector is
 
1

x3 = 2 
0
By Theorem 35.4, a fundamental set of solutions is given by {et x1 , e3t x2 , e5t x3 }
Problem 37.13


3
1 0
P =  −8 −6 2 
−9 −9 4
70
Solution.
The characteristic equation is
3−r
1
0
−8 −6 − r
2
−9
−9
4−r
= (r − 1)(r − 2)(r + 2) = 0
Solving this equation we find r1 = −2, r2 = 1, and r3 = 2. Now,



  
5
1 0
x1
5x1 + x2
0







x2
−8x1 − 4x2 + 2x3 = 0 
(P + 2I)x = −8 −4 2
−9 −9 6
x3
−9x1 − 9x2 + 6x3
0
Solving this system we find x2 = −5x1 . Letting x1 = 1, we find x2 = −5 and
x3 = −6. An eigenvector is


1
x1 =  −5 
−6
Similarly,



  
2
1 0
x1
2x1 + x2
0







x2
−8x1 − 7x2 + 2x3 = 0 
(P − I)x = −8 −7 2
−9 −9 3
x3
−9x1 − 9x2 + 3x3
0
Solving this system we find x2 = −2x1 . Letting x1 = 1, we find x2 = −2 and
x3 = −3. An eigenvector is


1
x2 =  −2 
−3



  
1
1 0
x1
x1 + x2
0
(P − 2I)x =  −8 −8 2   x2   −8x1 − 8x2 + 2x3  =  0 
−9 −9 2
x3
−9x1 − 9x2 + 2x3
0
Solving this system we find −x1 = x2 . Letting x1 = 1, we find x2 = −1, x3 =
0. An eigenvector is


1
x3 =  −1 
0
By Theorem 35.4, a fundamental set of solutions is given by {e−2t x1 , et x2 , e2t x3 }
71
Problem 37.14
Solve the following initial value problem.
5
3
2
0
y =
y, y(1) =
−4 −3
0
Solution.
The characteristic equation is
5−r
3
−4 −3 − r
= (r + 1)(r − 3) = 0
Solving this quadratic equation we find r1 = −1 and r2 = 3. Now,
6
3
x1
6x1 + 3x2
0
(P + I)x =
=
−4 −2
x2
−4x1 − 2x2
0
Solving this system we find x2 = −2x1 . Letting x1 = 1 then x2 = −2. Thus,
an eigenvector is
1
x1 =
−2
Similarly,
(P − 3I)x =
2
3
−4 −6
x1
x2
2x1 + 3x2
−4x1 − 6x2
=
0
0
Solving this system we find 2x1 = −3x2 . Letting x1 = 3 then x2 = −2. Thus,
an eigenvector is
3
x2 =
−2
By Theorem 35.4, a fundamental set of solutions is given by {e−t x1 , e3t x2 }.
The general solution is then
y(t) = c1 e−t x1 + c2 e3t x2 .
Using the initial condtion we find c1 e−1 + 3c2 e3 = 2 and −2c1 e−1 − 2c2 e3 = 0.
Solving this system we find c1 = −e and c2 = e−3 . Hence, the unique solution
is given by
y(t) = −e1−t x1 + e3(t−1) x2
72
Problem 37.15
Solve the following initial value

4 2
y0 =  0 1
0 0
problem.



0
−1
3  y, y(0) =  0 
−2
3
Solution.
The characteristic equation is
4−r
2
0
0
1
−
r
3
0
0
−2 − r
Solving this equation we find

6 2
(P + 2I)x =  0 3
0 0
= (r + 2)(r − 1)(r − 4) = 0
r1 = −2, r2 = 1, and r3 = 4. Now,


  
0
x1
6x1 + 2x2
0
3   x2   3x2 + 3x3  =  0 
0
x3
0
0
Solving this system we find x2 = −3x1 . Letting x1 = 1, we find x2 = −3 and
x3 = 3. An eigenvector is


1
x1 =  −3 
3
Similarly,



  
3 2 0
x1
3x1 + 2x2
0







x2
3x3
(P − I)x = 0 0 3
= 0 
0 0 −3
x3
−3x3
0
Solving this system we find 3x1 + 2x2 = 0 and x3 = 0. Letting x1 = 2, we
find x2 = −3. An eigenvector is


2
x2 =  −3 
0



  
0 2
0
x1
2x2
0
(P − 4I)x =  0 −3 3   x2   −3x2 + 3x3  =  0 
0 0 −6
x3
−6x3
0
73
Solving this system we find x3 = x2 = 0 and x1 arbitrary. Letting x1 = 1,
an eigenvector is
 
1

x3 = 0 
0
By Theorem 35.4, a fundamental set of solutions is given by {e−2t x1 , et x2 , e4t x3 }.
The general solution is
y(t) = c1 e−2t x1 + c2 et x2 + c3 e4t x3 .
Using the initial condition we find c1 +2c2 +c3 = −1, −3c1 −3c2 = 0, 3c1 = 3.
Solving this system we find c1 = 1, c2 = −1, c3 = 0. Hence, the unique
solution to the initial value problem is
y(t) = e−2t x1 − et x2 + 0e4t x3
Problem 37.16
Find α so that the vector x is an eigenvector of P. What is the corresponding
eigenvalue?
2 α
1
P=
, u=
1 −5
−1
Solution.
We must have Pu = ru for some value r. That is
2 α
1
1
=r
1 −5
−1
−1
Equating components we find 2 − α = r and 1 + 5 = −r. Solving we find
r = −6 and α = 8
Problem 37.17
Find α and β so that the vector x is an eigenvector of P corresponding the
eigenvalue r = 1.
α β
−1
P=
, u=
2α β
1
Solution.
We must have Pu = u. That is
α β
−1
−1
=
2α −β
1
1
Equating components we find −α + β = −1 and 2α − β = 1. Solving we find
α = 0 and β = −1
74
38
Homogeneous Systems with Constant Coefficients: Complex Eigenvalues
Problem 38.1
Find the eigenvalues and the eigenvectors of the matrix
0 −9
P=
1 0
Solution.
The characteristic equation is
−r −9 2
1 −r = r + 9 = 0
Solving this quadratic equation we find r1 = −3i and r2 = 3i. Now,
3i −9
x1
3ix1 − 9x2
0
(P + 3iI)x =
=
1 3i
x2
x1 + 3ix2
0
Solving this system we find x1 = −3ix2 . Letting x2 = i then x1 = 3. Thus,
an eigenvector is
3
x1 =
i
An eigenvector corresponding to the eigenvalue 3i is then
3
x2 =
−i
Problem 38.2
Find the eigenvalues and the eigenvectors of the matrix
3 1
P=
−2 1
Solution.
The characteristic equation is
3−r
1
−2 1 − r
= r2 − 4r + 5 = 0
75
Solving this quadratic equation we find r1 = 2 − i and r2 = 2 + i. Now,
1+i
1
x1
(1 + i)x1 + x2
0
(P − (2 − i)I)x =
=
−2 −1 + i
x2
−2x1 − (1 − i)x2
0
Solving this system we find (1 + i)x1 = −x2 . Letting x1 = 1 − i then x2 = −2.
Thus, an eigenvector is
1−i
x1 =
−2
An eigenvector corresponding to the eigenvalue 2 − i is then
1+i
x2 =
−2
Problem 38.3
Find the eigenvalues and the eigenvectors

1 −4
P= 3 2
1 1
Solution.
The characteristic equation is
1 − r −4
−1
3
2−r
3
1
1
3−r
of the matrix

−1
3 
3
= −r3 + 6r2 − 21r = 26 = 0
Using the rational root test one finds that r = 2 is a solution so that the
characteristic equation is (r − 2)(r2 − 4r + 13) = 0. Solving this equation we
find r1 = 2, r2 = 2 − 3i, and r3 = 2 + 3i. Now,



  
−1 −4 −1
x1
−x1 − 4x2 − x3
0







3
0
3
x2
3x1 + 3x3
(P − 2I)x =
= 0 
1
1
1
x3
x1 + x2 + x3
0
Solving this system we find x1 = −x3 and x2 = 0. Letting x3 = −1 then
x1 = 1. Thus, an eigenvector is


1
x1 =  0 
−1
76
Next,



  
−1 + 3i −4 −1
x1
(−1 + 3i)x1 − 4x2 − x3
0







3
3i
3
x2
3x1 + 3ix2 + 3x3
(P−(2−3i)I)x =
= 0 
1
1 1 + 3i
x3
x1 + x2 + (1 + 3i)x3
0
Solving this system we find 3ix1 = (4 − i)x2 and x3 = (−1 + 3i)x1 − 4x2 .
Letting x2 = 3i then x1 = 4 − i and x3 = −1 + i. Thus, an eigenvector is


4−i
x2 =  3i 
−1 + i
An eigenvector corresponding to the eigenvalue 2 + 3i is then


4+i
x3 =  −3i 
−1 − i
In Problems 38.4 - 38.6, one or more eigenvalues and corresponding eigenvectors are given for a real matrix P. Determine a fundamental set of solutions
for y0 = Py, where the fundamental set consists entirely of real solutions.
Problem 38.4
P is a 2 × 2 matrix with an eigenvalue r = i and corresponding eigenvector
−2 + i
x=
5
Solution.
From the given information, a solution to the system is given by
−2 + i
(cos t + i sin t)(−2 + i)
it
y(t) = e
=
(cos t + i sin t)5
5
(−2 cos t − sin t) + (cos t − 2 sin t)i
5 cos t + 5i sin t
−2 cos t − sin t
5 cos t
=
=
+i
cos t − 2 sin t
5 sin t
Thus, a fundamental set of solution consists of the vectors
−2 cos t − sin t
cos t − 2 sin t
y1 (t) =
, y2 (t) =
5 cos t
5 sin t
77
Problem 38.5
P is a 2×2 matrix with an eigenvalue r = 1+i and corresponding eigenvector
−1 + i
x=
i
Solution.
From the given information, a solution to the system is given by
t
−1 + i
(e cos t + iet sin t)(−1 + i)
(1+i)t
y(t) = e
=
i
(et cos t + iet sin t)i
(−et cos t − et sin t) + (et cos t − et sin t)i
−et sin t + iet cos t
−et cos t − et sin t
−et sin t
=
=
+i
et cos t − et sin t
et cos t
Thus, a fundamental set of solution consists of the vectors
t
−et cos t − et sin t
e cos t − et sin t
y1 (t) =
, y2 (t) =
−et sin t
et cos t
Problem 38.6
P is a 4×4 matrix with eigenvalues r = 1+5i with corresponding eigenvector
 
i
 1 

x=
 0 
0
and eigenvalue r = 1 + 2i with corresponding eigenvector
 
0
 0 

x=
 i 
1
78
Solution.
From the given information, a solution to the system is given by
   t

(e cos 5t + iet sin 5t)i
i
 1   (et cos 5t + iet sin 5t) 
 

y(t) = e(1+5i)t 
 0 =

0
0
0


(−et sin 5t + iet cos 5t)
 et cos 5t + et sin 5t 




0
0
=

=

 t
−et sin 5t
e cos 5t
 et cos 5t 
 et sin 5t

 + i



0
0
0
0
This yields the two solutions

−et sin 5t
 et cos 5t
y1 (t) = 

0
0







et cos 5t

 t

 , y2 (t) =  e sin 5t 



0
0
Similarly,

 
0
0



0  
0
y(t) = e(1+2i)t 
 i  =  (et cos 2t + iet sin 2t)i
1
et cos 2t + iet sin 2t

0


0


t
t
 −e sin 2t + ie cos 2t 
et cos 2t + iet sin 2t

=


0
0



0
0



 −et sin 2t  + i  et cos 2t
et cos 2t
et sin 2t

=
79








This yields the two solutions



0
0



0
0


y3 (t) = 
 −et sin 2t  , y4 (t) =  et cos 2t
et cos 2t
et sin 2t




Thus, a fundamental set of solutions consists of the vectors y1 , y2 , y3 , y4
Problem 38.7
Solve the initial value problem
0 −9
6
0
y =
y, y(0) =
1 0
2
Solution.
By Problem 36.1, an eigenvector corresponding to the eigenvalue r = −3i is
3
x=
i
Thus, a solution corresponding to this eigenvector is
3
(cos 3t − i sin 3t)(3)
−3it
y(t) = e
=
i
(cos 3t − i sin 3t)i
=
3 cos 3t
sin 3t
+i
−3 sin 3t
cos 3t
This yields the two solutions
3 cos 3t
−3 sin 3t
y1 (t) =
, y2 (t) =
sin 3t
cos 3t
The general solution is then given by
3c1 cos 3t − 3c2 sin 3t
y(t) = c1 y1 + c2 y2 =
c1 sin 3t + c2 cos 3t
Using the initial condition we find c1 = 2 and c2 = 2. Hence, the unique
solution is
6 cos 3t − 6 sin 3t
y(t) =
2 sin 3t + 2 cos 3t
80
Problem 38.8
Solve the initial value problem
3 1
8
0
y =
y, y(0) =
−2 1
6
Solution.
By Problem 36.2, an eigenvector corresponding to the eigenvalue r = 2 − i is
1−i
x=
−2
Thus, a solution corresponding to this eigenvector is
2t
1−i
e (cos t − i sin t)(1 − i)
(2−i)t
y(t) = e
=
−2
e2t (cos t − i sin t)(−2)
=
e2t (cos t − sin t
−2e2t cos t
+i
−e2t (cos t + sin t)
2e2t sin t
This yields the two solutions
2t
e (cos t − sin t
−e2t (cos t + sin t)
y1 (t) =
, y2 (t) =
−2e2t cos t
2e2t sin t
The general solution is then given by
c1 e2t (cos t − sin t) − c2 e2t (cos t + sin t)
y(t) = c1 y1 + c2 y2 =
−2c1 e2t cos t + 2c2 e2t sin t
Using the initial condition we find c1 = −3 and c2 = −11. Hence, the unique
solution is
8 cos t + 14 sin t
2t
y(t) = e
6 cos t − 22 sin t
Problem 38.9
Solve the initial value problem




1 −4 −1
−1
3  y, y(0) =  9 
y0 =  3 2
1 1
3
4
81
Solution.
By Problem 36.3, an eigenvector corresponding to the eigenvalue r = 2 is


1
x1 =  0 
−1
Thus, a solution corresponding to this eigenvector is

  2t 
1
e
2t 


0
0 
y1 (t) = e
=
−1
−e2t
An eigenvector corresponding to the eigenvalue r = 2 − 3i is


4−i
x2 =  3i 
−1 + i
Thus, a solution corresponding to this eigenvector is

  2t

4−i
e (cos 3t − i sin 3t)(4 − i)
y(t) = e(2−3i)t  3i  =  e2t (cos 3t − i sin 3t)(3i) 
−1 + i
e2t (cos 3t − i sin 3t)(−1 + i)




e2t (4 cos 3t − sin 3t
−e2t (cos 3t + 4 sin 3t)
 + i

3e2t sin 3t
3e2t cos 3t
= 
2t
2t
e (− cos 3t + sin 3t
e (cos 3t + sin 3t)
This yields the two solutions
 2t



e (4 cos 3t − sin 3t
−e2t (cos 3t + 4 sin 3t)
 , y3 (t) = 

3e2t sin 3t
3e2t cos 3t
y2 (t) = 
2t
2t
e (− cos 3t + sin 3t)
e (cos 3t + sin 3t)
The general solution is then given by


c1 e2t + c2 e2t (4 cos 3t − sin 3t) − c3 e2t (cos 3t + 4 sin 3t)

3c2 e2t sin 3t + 3c3 e2t cos 3t
y(t) = c1 y1 (t)+c2 y2 (t)+c3 y3 (t) = 
2t
2t
2t
−c1 e + c2 e (− cos 3t + sin 3t) + c3 e (cos 3t + sin 3t)
82
Using the initial condition we find c1 = −2, c2 = 1, and c3 = 3. Thus, the
unique solution is


−2 + cos 3t − 13 sin 3t

3 sin 3t + 9 cos 3t
y(t) = e2t 
2 + 2 cos 3t + 4 sin 3t
83
39
Homogeneous Systems with Constant Coefficients: Repeated Eigenvalues
In Problems 39.1 - 39.4, we consider the initial value problem y0 = Py, y(0) =
y0 .
(a) Compute the eigenvalues and the eigenvectors of P.
(b) Construct a fundamental set of solutions for the given differential equation. Use this fundamental set to construct a fundamental matrix Ψ(t).
(c) Impose the initial condition to obtain the unique solution to the initial
value problem.
Problem 39.1
P=
3 2
0 3
Solution.
(a) The characteristic equation is
3−r
2
0
3−r
, y0 =
4
1
= (r − 3)2 = 0
and has a repeated root r = 3. We find an eigenvector as follows.
0 2
x1
2x2
0
=
=
0 0
x2
0
0
It follows that x2 = 0 and x1 is arbitrary. Letting x1 = 1 then an eigenvector
is
1
x1 =
0
(b) The above eigenvector yields the solution
3t e
y1 =
0
But we need two linearly independent solutions to form the general solution
of the given system and we only have one. We look for a solution of the form
x1
1
3t
3t
y2 (t) = e
+ te
x2
0
84
where
(P − 3I)
x1
x2
0 2
0 0
2x2
0
=
=
x1
x2
=
1
0
Solving this system we find x2 = 12 and x1 arbitrary. Let x1 = 0 then a
second solution is
0
1
3t
3t
y2 (t) = e
+ te
1
0
2
A fundamental matrix is
Ψ(t) =
e3t te3t
3t
0 e2
(c) Since y(t) = Ψ(t)c, Ψ(0)c = y0 or
1 0
c1
4
=
1
0 2
c2
1
Solving this system we find c1 = 4 and c2 = 2. Hence, the unique solution to
the initial value problem is
2t + 4
3t
y(t) = e
1
Problem 39.2
P=
3 0
1 3
Solution.
(a) The characteristic equation is
3−r
0
1
3−r
, y0 =
2
−3
= (r − 3)2 = 0
and has a repeated root r = 3. We find an eigenvector as follows.
0 0
x1
0
0
=
=
1 0
x2
x1
0
85
It follows that x1 = 0 and x2 is arbitrary. Letting x2 = 1 then an eigenvector
is
0
x1 =
1
(b) The above eigenvector yields the solution
0
y1 =
e3t
But we need two linearly independent solutions to form the general solution
of the given system and we only have one. We look for a solution of the form
x1
0
3t
t
y2 (t) = e
+ te
x2
1
where
(P − 3I)
x1
x2
0 0
1 0
0
x1
=
=
=
x1
x2
0
1
Solving this system we find x1 = 1 and x2 arbitrary. Let x2 = 0 then a
second solution is
1
0
3t
t
y2 (t) = e
+ te
0
1
A fundamental matrix is
Ψ(t) =
0 e3t
e3t te3t
(c) Since y(t) = Ψ(t)c, Ψ(0)c = y0 or
0 1
c1
2
=
1 0
c2
−3
Solving this system we find c1 = −3 and c2 = 2. Hence, the unique solution
to the initial value problem is
2
3t
y(t) = e
2t − 3
86
Problem 39.3
P=
−3 −36
1
9
Solution.
(a) The characteristic equation is
−3 − r −36
1
9−r
, y0 =
0
2
= (r − 3)2 = 0
and has a repeated root r = 3. We find an eigenvector as follows.
−6 −36
x1
−6x1 − 36x2
0
=
=
1
6
x2
x1 + 6x2
0
It follows that x1 = −6x2 . Letting x1 = 6 then an eigenvector is
6
x1 =
−1
(b) The above eigenvector yields the solution
6e3t
y1 =
−e3t
But we need two linearly independent solutions to form the general solution
of the given system and we only have one. We look for a solution of the form
x1
6
3t
t
y2 (t) = e
+ te
x2
−1
where
(P − 3I)
x1
x2
=
=
−6 −36
1
6
x1
x2
−6x1 − 36x2 x1 + 6x2
=
6
−1
Solving this system we find x1 + 6x2 = −1. Let x2 = 0 so that x1 = −1. Then
a second solution is
−1
6
3t
t
y2 (t) = e
+ te
0
−1
87
A fundamental matrix is
Ψ(t) =
6e3t (6t − 1)e3t
−e3t
−te3t
(c) Since y(t) = Ψ(t)c, Ψ(0)c = y0 or
6 −1
c1
0
=
−1 0
c2
2
Solving this system we find c1 = −2 and c2 = −12. Hence, the unique solution
to the initial value problem is
−72t
3t
y(t) = e
12t + 2
Problem 39.4
P=
6 1
−1 4
Solution.
(a) The characteristic equation is
6−r
1
−1 4 − r
, y0 =
4
−4
= (r − 5)2 = 0
and has a repeated root r = 5. We find an eigenvector as follows.
1
1
x1
x1 + x 2
0
=
=
−1 −1
x2
−x1 − x2
0
It follows that x2 = −x1 . Letting x1 = 1 then x2 = −1. An eigenvector is
1
x1 =
−1
(b) The above eigenvector yields the solution
5t e
y1 =
−e5t
88
But we need two linearly independent solutions to form the general solution
of the given system and we only have one. We look for a solution of the form
1
x1
t
5t
+ te
y2 (t) = e
−1
x2
where
(P − 5I)
x1
x2
=
=
1
1
−1 −1
x1 + x 2
−x1 − x2
x1
x2
=
1
−1
Solving this system we find x1 + x2 = 1. Let x2 = 0 then a second solution is
1
1
5t
t
y2 (t) = e
+ te
0
−1
A fundamental matrix is
Ψ(t) =
e5t (t + 1)e5t
−e5t
−te5t
(c) Since y(t) = Ψ(t)c, Ψ(0)c = y0 or
4
1 −1
c1
=
c2
−4
1 0
Solving this system we find c1 = 4 and c2 = 0. Hence, the unique solution to
the initial value problem is
4
5t
y(t) = e
−4
Problem 39.5
Consider the homogeneous linear system


2 1 0
y0 =  0 2 1  y
0 0 2
(a) Write the three component differential equations of y0 = Py and solve
these equations sequentially, first finding y3 (t), then y2 (t), and then y1 (t).
(b) Rewrite the component solutions obtained in part (a) as a single matrix
equation of the form y = Ψ(t)c. Show that Ψ(t) is a fundamental matrix.
89
Solution.
(a) We have
y10 = 2y1 + y2
y20 = 2y2 + y3
y30 =
2y3
Solving the last equation we find y3 (t) = c3 e2t . Substituting this into the
second equation we find
y20 − 2y2 = c3 e2t .
Solving this equation using the method of integrating factor we find
y2 (t) = c3 te2t + c2 e2t .
Substituting this into the first equation we find
y10 − 2y1 = c3 te2t + c2 e2t .
Solving this equation we find
y1 (t) = c1 e2t + c2 te2t + c3 t2 e2t .
(b)


c1
e2t te2t t2 e2t
y(t) =  0 e2t te2t   c2 
c3
0
0
e2t

Since
1 0 0
W (0) = det(Ψ(0)) = 0 1 0
0 0 1
=1
we have Ψ(t) is a fundamental matrix
In Problems 39.6 - 39.8, Find the eigenvalues and eigenvectors of P. Give the
geometric and algebraic multiplicity of each eigenvalue. Does P have a full
set of eigenvectors?
Problem 39.6


5 0 0
P= 1 5 0 
1 0 5
90
Solution.
The characteristic equation is
5−r
0
0
1
5−r
0
1
0
5−r
and has a repeated root r

0 0
 1 0
1 0
= (5 − r)3 = 0
= 5. We find an eigenvector as follows.

 
  
0
x1
0
0
0   x 2  =  x1  =  0 
0
x3
x1
0
It follows that x1 = 0, x2 and x3 are arbitrary. Thus,
 
 
0
0
x = x2  1  + x3  0 
0
1
so that two linearly independent eigenvectors are
 
 
0
0
x1 =  1  , x2  0 
0
1
Hence, r = 5 has algebraic multiplicity 3 and geometric multiplicity 2. Hence,
P is defective
Problem 39.7


5 0 0
P= 0 5 0 
0 0 5
Solution.
The characteristic equation is
5−r
0
0
0
5−r
0
0
0
5−r
91
= (5 − r)3 = 0
and has a repeated root r

0 0
 0 0
0 0
= 5. We find an eigenvector as follows.

    
0
x1
0
0
0   x2  =  0  =  0 
0
x3
0
0
It follows that x1 , x2 and x3 are arbitrary. Thus,
 
 
 
1
0
0





x = x1 0 + x2 1 + x3 0 
0
0
1
so that the three linearly independent eigenvectors are
 
 
 
0
1
0





x1 = 0 , x2 = 1 , x3 0 
1
0
0
Hence, r = 5 has algebraic multiplicity 3 and geometric multiplicity 3. Hence,
P has a full set of eigenvectors
Problem 39.8

2
 0
P=
 0
0
0
2
0
0
0
0
2
1
Solution.
The characteristic equation is
2−r
0
0
0
0
2−r
0
0
0
0
2−r
0
0
0
1
2−r
and has a repeated root

0 0
 0 0

 0 0
0 0

0
0 

0 
2
= (2 − r)4 = 0
r = 2. We find an eigenvector as

 
 
0 0
x1
0
0






0 0   x2   0   0
=
=
0 0   x3   0   0
1 0
x4
x3
0
92
follows.




It follows that x3 = 0, x1 , x2 and x4 are arbitrary. Thus,
 
 
 
1
0
0
 0 
 1 
 0 

 
 
x = x1 
 0  + x2  0  + x 4  0 
0
0
1
so that the three linearly independent eigenvectors are
 
 

1
0
0
 0 
 1 
 0

 

x1 = 
 0  , x2 =  0  , x3  0
0
0
1




Hence, r = 2 has algebraic multiplicity 4 and geometric multiplicity 3. Hence,
P defective
Problem 39.9
Let P be a 2 × 2 real matrix with an eigenvalue r1 = a + ib where b 6= 0. Can
P have a repeated eigenvalue? Can P be defective?
Solution.
P have the two distinct eigenvalues r1 = a + ib and r2 = a − ib. Thus, P has
a full set of eigenvectors
Problem 39.10
Dtermine the numbers x and y so that
symmetric.

0 1

P= y 2
6 2
the following matrix is real and

x
2 
7
Solution.
Since P is a real symmetric matrix, PT = P. That



0 y 6
0 1
T



P = 1 2 2 P= y 2
x 2 7
6 2
Equating entries we find x = 6 and y = 1
93
is,

x
2 
7
Problem 39.11
Dtermine the numbers x and y so that the following matrix is Hermitian.


2
x + 3i
7
5
2 + yi 
P =  9 − 3i
7
2 + 5i
3
Solution.
T
Since P is a Hermitian matrix, P = P. That is,




2
9 + 3i
7
2
x + 3i
7
T
5
2 − 5i  P =  9 − 3i
5
2 + yi 
P =  x − 3i
7
2 − yi
3
7
2 + 5i
3
Equating entries we find x = 9 and y = −5
Problem 39.12
(a) Give an example of a 2 × 2 matrix P that is not invertible but have a full
set of eigenvectors.
(b) Give an example of a 2 × 2 matrix P that is invertible but does not have
a full set of eigenvectors.
Solution.
(a) Consider the matrix
P=
1 0
0 0
Then det(P) = 0 so that P is not invertible. The characterisitc equation of
this matrix is
1−r 0 = r(r − 1) = 0
0
−r so that the eigenvalues are r1 = 0 and r2 = 1. Hence, the set P has a full set
of eigenvectors.
(b) Consider the matrix
1 1
P=
0 1
Then det(P) = 1 so that P is invertible. The characterisitc equation of this
matrix is
1−r
1 = (r − 1)2 = 0
0
1−r 94
so that the r = 1 is a repeated eigenvalue. An eigenvector is found as follows.
0 1
x1
x2
0
=
=
0 0
x2
0
0
It follows that x2 = 0 and x1 is arbitrary. Thus, an eigenvector is given
1
x=
0
It follows that P is defective
95
40
NonHomogeneous First Order Linear Systems
In Problems 40.1 - 40.3, we consider the initial value problem y0 = Py +
g(t), y(t0 ) = y0 .
(a) Find the eigenpairs of the matrix P and form the general homogeneous
solution of the differential equation.
(b) Construct a particular solution by assuming a solution of the form suggested and solving for the undetermined constant vectors a,b, and c.
(c) Form the general solution of the nonhomogeneous differential equation.
(d) Find the unique solution to the initial value problem.
Problem 40.1
0
y =
−2 1
1 −2
y+
1
1
, y0 =
3
1
Try yp (t) = a.
Solution.
(a) The characteristic equation is
−2 − r
1
1
−2 − r
= (r + 3)(r + 1) = 0
Thus, the eigenvalues are r1 = −1 and r2 = −3. An eigenvector corresponding to r1 = −1 is found as follows
−1 1
x1
−x1 + x2
0
(P + I)x1 =
=
=
1 −1
x2
x1 − x2
0
Solving this system we find x2 = x1 . Letting x1 = 1 we find x2 = 1 and an
eigenvector is
1
x1 =
1
Similarly, for r2 = −3 we have
1 1
x1
x1 + x2
0
(P + 3I)x2 =
=
=
1 1
x2
x1 + x2
0
96
Solving this system we find x2 = −x1 . Letting x1 = 1 we find x2 = −1 and
an eigenvector is
1
x2 =
−1
Hence,
yh (t) = c1 e
−t
1
1
−3t
+ c2 e
(b) Inserting the suggested function yp =
a1
a2
1
−1
into the differential equation
leads to yp0 = Py + g(t) or
1
0
−2 1
a1
+
=
a2
1
0
1 −2
1
Solving this system we find a1 = a2 = 1. Thus, yp =
.
1
(c) The general solution is given by
1
1
1
c1 e−t + c2 e−3t + 1
−t
−3t
y(t) = c1 e
+ c2 e
+
=
1
−1
1
c1 e−t − c2 e−3t + 1
(d) Imposing the initial condition we find c1 + c2 + 1 = 3 and c1 − c2 + 1 = 1.
Solving this system we find c1 = c2 = 1. Hence, the unique solution to the
initial value problem is
−t
e + e−3t + 1
y(t) =
e−t − e−3t + 1
Problem 40.2
0
y =
0 1
1 0
y+
t
−1
, y0 =
2
−1
Try yp (t) = ta + b.
Solution.
(a) The characteristic equation is
−r 1 1 −r = (r − 1)(r + 1) = 0
97
Thus, the eigenvalues are r1 = −1 and r2 = 1. An eigenvector corresponding
to r1 = −1 is found as follows
1 1
x1
x1 + x2
0
(P + I)x1 =
=
=
1 1
x2
x1 + x2
0
Solving this system we find x2 = −x1 . Letting x1 = 1 we find x2 = −1 and
an eigenvector is
1
x1 =
−1
Similarly, for r2 = 1 we have
−1 1
x1
−x1 + x2
0
(P − I)x2 =
=
=
x2
x1 − x 2
1 −1
0
Solving this system we find x2 = x1 . Letting x1 = 1 we find x2 = 1 and an
eigenvector is
1
x2 =
1
Hence,
1
1
−t
t
yh (t) = c1 e
+ c2 e
−1
1
ta1 + b1
(b) Inserting the suggested function yp =
into the differential
ta2 + b2
equation leads to yp0 = Py + g(t) or
t
0
0 1
ta1 + b1
+
=
ta2 + b2
−1
0
1 0
0
Letting b1 = b2 = 0 we find a1 = 0 and a2 = −1. Thus, yp =
.
−t
(c) The general solution is given by
1
1
0
c1 e−t + c2 et
−t
t
y(t) = c1 e
+ c2 e
+
=
1
−1
−t
c1 e−t − c2 et − t
(d) Imposing the initial condition we find c1 + c2 = 2 and c1 − c2 = −1.
Solving this system we find c1 = 21 and c2 = 32 . Hence, the unique solution to
the initial value problem is
1 −t 3 t e + e
y(t) = 1 2−t 3 2t
e
− 2e − t
2
98
Problem 40.3
0
y =
−3 −2
4
3
y+
sin t
0
, y0 =
0
0
Try yp (t) = (sin t)a + (cos t)b.
Solution.
(a) The characteristic equation is
−3 − r −2
4
3−r
= (r − 1)(r + 1) = 0
Thus, the eigenvalues are r1 = −1 and r2 = 1. An eigenvector corresponding
to r1 = −1 is found as follows
−2 −2
x1
−2x1 − 2x2
0
(P + I)x1 =
=
=
4
4
x2
4x1 + 4x2
0
Solving this system we find x2 = −x1 . Letting x1 = 1 we find x2 = −1 and
an eigenvector is
1
x1 =
−1
Similarly, for r2 = 1 we have
0
−4x1 − 2x2
−4 −2
x1
=
=
(P − I)x2 =
4x1 + 2x2
0
x2
4
2
Solving this system we find x2 = −2x1 . Letting x1 = 1 we find x2 = −2 and
an eigenvector is
1
x2 =
−2
Hence,
yh (t) = c1 e
−t
1
−1
t
+ c2 e
(b) Inserting the suggested function yp =
1
−2
sin ta1 + cos tb1
sin ta2 + cos tb2
into the dif-
ferential equation leads to yp0 = Py + g(t) or
−3 −2
sin ta1 + cos tb1
sin t
cos ta1 − sin tb1
+
=
4
3
sin ta2 + cos tb2
0
cos ta2 − sin tb2
99
Multiplying and equating coefficients we find
a1
−b1
a2
−b2
=
−2b2 − 3b1
= −3a1 − 2a2 + 1
=
4b1 + 3b2
=
4a1 + 3a2
1
3
Solving
this system we find
a1 = 2 , a2 = −2, b1 = − 2 , b2 = 0. Thus,
3.5 sin t − 0.5 cos t
yp =
.
−2 sin t
(c) The general solution is given by
1
1
1.5 sin t − 0.5 cos t
−t
t
y(t) = c1 e
+ c2 e
+
−1
−2
−2 sin t
=
c1 e−t + c2 et + 1.5 sin t − 0.5 cos t
−c1 e−t − 2c2 et − 2 sin t
(d) Imposing the initial condition we find c1 + c2 = 0.5 and −c1 − 2c2 = 0.
Solving this system we find c1 = 1 and c2 = −0.5. Hence, the unique solution
to the initial value problem is
−t
e − 0.5et + 1.5 sin t − 0.5 cos t
y(t) =
−e−t + et − 2 sin t
Problem 40.4
Consider the initial value problem
π 0 2
0
= y0 .
y =
y + g(t), y
−2 0
2
Suppose we know that
y(t) =
1 + sin 2t
et + cos 2t
is the unique solution. Determine g(t) and y0 .
Solution.
We have
y0 = y
π 2
=
1 + sin π
π
e 2 + cos π
100
=
1
π
e2 − 1
Since y(t) is a solution, it satisfies the differential equation so that
2 cos 2t
0 2
1 + sin 2t
g(t) =
−
et − 2 sin 2t
−2 0
et + cos 2t
=
−2et
et + 2
Problem 40.5
Consider the initial value problem
1 t
2
0
y + g(t), y(1) =
.
y =
t2 1
−1
Suppose we know that
y(t) =
t+α
t2 + β
is the unique solution. Determine g(t) and the constants α and β.
Solution.
We have
1+α
1+β
= y(1) =
2
−1
Thus, 1 + α = 2 and 1 + β = −1. Solving these two equations we find α = 1
and β = −2. Now, inserting y into the differential equation we find
1
1 t
t+1
g(t) =
−
2t
t2 1
t2 − 2
=
−t3 + t
−t3 − 2t2 + 2t + 2
Problem 40.6
Let P(t) be a 2 × 2 matrix with continuous entries. Consider
the
differ1
ential equation y0 = P(t)y + g(t). Suppose that y1 (t) =
is the
−t
e
t −2
e
0
solution to y = P(t)y +
and y2 (t) =
is the solution to
0
−1
t e
y0 = P(t)y +
. Determine P(t). Hint: Form the matrix equation
−1
[y10 y20 ] = P[y1 y2 ] + [g1 g2 ].
101
Solution.
Following the hint we can write
1 et
−2 et
0
et
= P(t)
+
−e−t 0
e−t −1
0 −1
or
P(t)
1 et
e−t −1
=
2
0
−e−t 1
Solving for P(t) we find
−1
2
0
1 et
2
0
−1 −et
=
P(t) =
(−0.5)
−e−t 1
e−t −1
−e−t 1
−e−t 1
=
1 et
0 −1
Problem 40.7
Consider the linear system y0 = Py + b where P is a constant matrix and b
is a constant vector. An equilibrium solution, y(t), is a constant solution
of the differential equation.
(a) Show that y0 = Py + b has a unique equilibrium solution when P is
invertible.
(b) If the matrix P is not invertible, must the differential equation y0 =
Py + b possess an equilibrium solution? If an equilibrium solution does exist
in this case, is it unique?
Solution.
(a) If P is invertible and y is a constant solution then we must have Py+b =
0 or y = −A−1 b. This is the only equilibrium solution.
(b) Since Py + b = 0, Py = −b. This system has a unique solution only
when P is invertible. If P is not invertible then either this system has no
solutions or infinitely many solutions. That is, either no equilibrium solution
or infinitely many equilibrium solutions
Problem 40.8
Determine all the equilibrium solutions (if any).
2 −1
2
0
y =
y+
−1 1
−1
102
Solution.
Since det(P) = 1, the coefficient matrix is invertible and so there is a unique
equilibrium solution given by
1 1
−2
−1
−1
y = −P b = −
=
1 2
1
0
Problem 40.9
Determine all the equilibrium solutions (if any).


 
1 1 0
2
0



y = 0 −1 2 y + 3 
0 0 1
2
Solution.
Since det(P) = −1, the coefficient matrix is invertible and so there is a unique
equilibrium solution given by

−1   

1 1 0
2
−1
y = −P−1 b = −  0 −1 2   3  =  −1 
0 0 1
2
−2
Consider the homogeneous linear system y0 = Py. Recall that any associated fundamental matrix satisfies the matrix differential equation Ψ0 = PΨ.
In Problems 40.10 - 40.12, construct a fundamental matrix that solves the
matrix initial value problem Ψ0 = PΨ, Ψ(t0 ) = Ψ0 .
Problem 40.10
0
Ψ =
1 −1
−1 1
Ψ, Ψ(1) =
1 0
0 1
Solution.
0
We first find a fundamental matrix of the linear system y =
The characteristic equation is
1 − r −1
−1 1 − r
= r(r − 2) = 0
103
1 −1
−1 1
y.
and has eigenvalues r1 = 0 and r2 = 2. We find an eigenvector corresponding
to r1 = 0 as follows.
1 −1
x1
x1 − x2
0
=
=
−1 1
x2
−x1 + x2
0
It follows that x1 = x2 . Letting x1 = 1 then x2 = 1 and an eigenvector is
1
x1 =
1
An eigenvector corresponding to r2 = 2
−1 −1
x1
−x1 − x2
0
=
=
−1 −1
x2
−x1 − x2
0
Solving we find x1 = −x2 . Letting x1 = 1 we find x2 = −1 and an eigenvector
is
1
x2 =
−1
Thus, a fundamental matrix is
1 e2t
1 −e2t
Ψ=
.
But Ψ = ΨC. Using the initial condition we find I = Ψ(1) = Ψ(1)C and
therefore
−1
−1
1
1
1 e2
= 0.5
.
C=Ψ =
e−2 −e−2
1 −e2
Finally,
Ψ =
1 e2t
1 −e2t
=
0.5
(0.5)
1
e−2
1
−e−2
1 + e2(t−1) 1 − e2(t−1)
1 − e2(t−1) 1 + e2(t−1)
Problem 40.11
0
Ψ =
1 −1
−1 1
Ψ, Ψ(0) =
104
1 0
2 1
Solution.
From the previous problem, a fundamental matrix is
1 e2t
Ψ=
.
1 −e2t
But Ψ = ΨC. Using the initial condition we find Ψ(0) = Ψ(0)C and therefore
−1 1 1
1 0
3
1
C=
= 0.5
.
1 −1
2 1
−1 −1
Finally,
Ψ =
1 e2t
1 −e2t
=
(0.5)
3
1
−1 −1
3 − e2t 1 − e2t
3 + e2t 1 + e2t
0.5
Problem 40.12
0
Ψ =
1 4
−1 1
Ψ, Ψ
π 4
=
1 0
0 1
Solution.
We first find a fundamental matrix of the linear system y0 =
The characteristic equation is
1−r
4
−1 1 − r
1 4
−1 1
y.
= r2 − 2r + 5 = 0
and has eigenvalues r1 = 1 + 2i and r2 = 1 − 2i. We find an eigenvector
corresponding to r1 = 1 + 2i as follows.
−2i 4
x1
−2ix1 + 4x2
0
=
=
−1 −2i
x2
−x1 − 2ix2
0
It follows that x1 = −2ix2 . Letting x2 = 1 then x1 = −2i and an eigenvector
is
−2i
x=
.
1
105
Thus a solution to the system is
t
y(t) =
e (cos t + i sin t)
=
2et sin 2t
et cos 2t
+i
−2i
1
−2et cos 2t
et sin 2t
Thus, a fundamental matrix is
t
2e sin 2t −2et cos 2t
Ψ=
.
et cos 2t
et sin 2t
But Ψ = ΨC. Using the initial condition we find I = Ψ π4 = Ψ
therefore
π
−1 1 −π
−1
2e 4 0
e4
0
2
π
=
.
C=Ψ =
π
0 e4
0
e− 4
π
4
C and
Finally,
Ψ =
2et sin 2t −2et cos 2t
et cos 2t
et sin 2t
=
π
1 −π
e4
2
π
e− 4
0
π
et− 4 sin 2t −2et− 4 cos 2t
π
1 t− π4
cos 2t
et− 4 sin 2t
e
2
0
In Problems 40.13 - 40.14, use the method of variation of parameters to solve
the given initial value problem.
Problem 40.13
0
y =
9 −4
15 −7
y+
et
0
, y(0) =
2
5
Solution.
0
We first find a fundamental matrix of the linear system y =
The characteristic equation is
9−r
−4
15 −7 − r
= (r − 3)(r + 1) = 0
106
9 −4
15 −7
y.
and has eigenvalues r1 = −1 and r2 = 3. We find an eigenvector corresponding to r1 = 0 − 1 as follows.
10 −4
x1
10x1 − 4x2
0
=
=
15 −6
x2
15x1 − 6x2
0
It follows that x1 = 25 x2 . Letting x2 = 5 then x1 = 2 and an eigenvector is
2
x1 =
5
An eigenvector corresponding to r2 = 3
0
6x1 − 4x2
6 −4
x1
=
=
15x1 − 10x2
0
x2
15 −10
Solving we find x1 = 23 x2 . Letting x2 = 3 we find x1 = 2 and an eigenvector
is
2
x2 =
3
Thus, a fundamental matrix is
Ψ=
2e−t 2e3t
5e−t 3e3t
.
Therefore,
Ψ
−1
= −0.25
−3et
2et
−3t
5e
−2e−3t
.
But the variation of parameters formula is
Z
−1
y(t) = Ψ(t)Ψ (0)y(0) + Ψ(t)
t
Ψ−1 (s)g(s)ds.
0
Thus,
Z t
Z t
−1
Ψ (s)g(s)ds = 0.25
0
0
and
Z
Ψ(t)
t
−1
Ψ (s)g(s)ds =
0
−3e2s
5e−2s
ds = −0.125
3(e2t − 1)
−5(e−2t − 1)
0.75e−t − 2et + 1.25e3t
1.875e−t − 3.75et + 1.875e3t
107
.
Hence,
y(t) =
2e−t 2e3t
5e−t 3e3t
(−0.25)
=
−3 2
5 −2
2
5
+
0.75e−t − 2et + 1.25e3t
1.875e−t − 3.75et + 1.875e3t
2.75e−t − 2et + 1.25e3t
6.875e−t − 3.75et + 1.875e3t
Problem 40.14
0
y =
1 1
0 1
1
1
y+
, y(0) =
0
0
Solution.
0
We first find a fundamental matrix of the linear system y =
characteristic equation is
1−r
1
0
1−r
1 1
0 1
y. The
= (r − 1)2 = 0
and has repeated eigenvalue r = 1. We find an eigenvector corresponding to
r = 1 as follows.
0
x2
0 1
x1
=
=
x2
0
x2
0 1
It follows that x2 = 0 and x1 arbitrary. Letting x1 = 1 an eigenvector is
1
x=
0
A corresponding solution is
y1 (t) =
et
0
.
A second solution has the form
y2 (t) = te
t
et
0
t
+e
a1
a2
Inserting this into the differential equation and solving for a1 and a2 we find
a1 = 0 and a2 = 1. Thus,
t te
y2 (t) =
et
108
A fundamental matrix is
Ψ=
et tet
0 et
.
Therefore,
Ψ
−1
=
e−t −te−t
0
e−t
.
But the variation of parameters formula is
Z
−1
y(t) = Ψ(t)Ψ (0)y(0) + Ψ(t)
t
Ψ−1 (s)g(s)ds.
0
Thus,
y(t) =
=
et tet
0 et
et tet
0 et
Rt
0
e−s (1 − s)
e−s
te−t
1 − e−t
109
=
ds
tet
t
e −1
41
Solving First Order Linear Systems with
Diagonalizable Constant Coefficients Matrix
In Problems 41.1 - 41.4, the given matrix is diagonalizable. Find matrices T
and D such that T−1 PT = D.
Problem 41.1
P=
Solution.
The characteristic equation is
3−r
4
−2 −3 − r
3
4
−2 −3
= (r − 1)(r + 1) = 0
Thus, the eigenvalues are r1 = −1 and r2 = 1. An eigenvector corresponding
to r1 = −1 is found as follows
4
4
x1
4x1 + 4x2
0
(P + I)x1 =
=
=
−2 −2
x2
−2x1 − 2x2
0
Solving this system we find x2 = −x1 . Letting x1 = 1 we find x2 = −1 and
an eigenvector is
1
x1 =
−1
Similarly, for r2 = 1 we have
2
4
x1
2x1 + 4x2
0
(P − I)x2 =
=
=
−2 −4
x2
−2x1 − 4x2
0
Solving this system we find x1 = −2x2 . Letting x1 = 2 we find x2 = −1 and
an eigenvector is
2
x2 =
−1
110
Therefore
D=
−1 0
0 1
, T=
1
2
−1 −1
Problem 41.2
P=
Solution.
The characteristic equation is
2−r
3
2
3−r
2 3
2 3
= r(r − 5) = 0
Thus, the eigenvalues are r1 = 0 and r2 = 5. An eigenvector corresponding
to r1 = 0 is found as follows
2 3
x1
2x1 + 3x2
0
(P + 0I)x1 =
=
=
2 3
x2
2x1 + 3x2
0
Solving this system we find 2x1 + 3x2 = 0. Letting x1 = 3 we find x2 = −2
and an eigenvector is
3
x1 =
−2
Similarly, for r2 = 5 we have
−3 3
x1
−3x1 + 3x2
0
(P − 5I)x2 =
=
=
2 −2
x2
2x1 − 2x2
0
Solving this system we find x1 = x2 . Letting x1 = 1 we find x2 = 1 and an
eigenvector is
1
x2 =
1
Therefore
D=
0 0
0 5
, T=
3 1
−2 1
Problem 41.3
P=
1 2
2 1
111
Solution.
The characteristic equation is
1−r
2
2
1−r
= (r + 1)(r − 3) = 0
Thus, the eigenvalues are r1 = −1 and r2 = 3. An eigenvector corresponding
to r1 = −1 is found as follows
2 2
x1
2x1 + 2x2
0
(P + I)x1 =
=
=
2 2
x2
2x1 + 2x2
0
Solving this system we find x2 = −x1 . Letting x1 = 1 we find x2 = −1 and
an eigenvector is
1
x1 =
−1
Similarly, for r2 = 3 we have
−2 2
x1
−2x1 + 2x2
0
(P − 3I)x2 =
=
=
2 −2
x2
2x1 − 2x2
0
Solving this system we find x1 = x2 . Letting x1 = 1 we find x2 = 1 and an
eigenvector is
1
x2 =
1
Therefore
D=
−1 0
0 3
, T=
1 1
−1 1
Problem 41.4
P=
Solution.
The characteristic equation is
−2 − r
2
0
3−r
−2 2
0 3
= (r + 2)(r − 3) = 0
112
Thus, the eigenvalues are r1 = −2 and r2 = 3. An eigenvector corresponding
to r1 = −2 is found as follows
0 2
x1
2x2
0
(P + 2I)x1 =
=
=
0 5
x2
5x2
0
Solving this system we find x2 = 0. Letting x1 = 1 an eigenvector is
1
x1 =
0
Similarly, for r2 = 3 we have
0
−5x1 + 2x2
−5 2
x1
=
=
(P − 3I)x2 =
0
0
x2
0 0
Solving this system we find 5x1 − 2x2 = 0. Letting x1 = 2 we find x2 = 5 and
an eigenvector is
2
x2 =
5
Therefore
D=
−2 0
0 3
, T=
1 2
0 5
In Problems 41.5 - 41.6, you are given the characteristic polynomial for the
matrix P. Determine the geometric and algebraic multiplicities of each eigenvalue. If the matrix P is diagonalizable, find matrices T and D such that
T−1 PT = D.
Problem 41.5


7 −2 2
P =  8 −1 4  , p(r) = (r − 3)2 (r + 1).
−8 4 −1
Solution.
The algebraic multiplicity of r1 = −1 is 1. An eigenvector corresponding to
this eigenvalue is found as follows.


 

8 −2 2
x1
8x1 − 2x2 + 2x3
= 0
0 4   x2  = 
8x1 + 4x3
(P + I)x1 =  8
0
−8 4 0
x3
−8x1 + 4x2
113
Solving this system we find x2 = 2x1 and x3 = −2x1 . Letting x1 = 1 an
eigenvector is


1
x1 =  2 
−2
Hence, the geometric multiplicity of r1 = −1 is 1.
The algebraic multiplicity of r2 = 3 is 2. An eigenvector corresponding to
this eigenvalue is found as follows.


 

4 −2 2
x1
4x1 − 2x2 + 2x3
0
(P − 3I)x1 =  8 −4 4   x2  =  8x1 − 4x2 + 4x3  =
0
−8 4 −4
x3
−8x1 + 4x2 − 4x3
Solving this system we find 2x1 − x2 + x3 = 0. Letting x1 = 1 we find
 
 

 

0
1
x1
x1
 x2  =  2x1 + x3  = x1  2  + x3  1 
1
0
x3
x3
Thus, two linearly independent eigenvectors are
 
 
1
0



x2 = 2 , x3 = 1 
0
1
Hence, r2 = 3 has geometric multiplicity 2. It follows that P is diagonalizable
with




−1 0 0
1 0 1
D =  0 3 0 , T =  2 1 2 
0 0 3
−2 1 0
Problem 41.6


5 −1 1
P =  14 −3 6  , p(r) = (r − 2)2 (r − 3).
5 −2 5
Solution.
The algebraic multiplicity of r1 = 3 is 1. An eigenvector corresponding to
this eigenvalue is found as follows.


 

2 −1 1
x1
2x1 − x2 + x3
0
(P − 3I)x1 =  14 −6 6   x2  =  14x1 − 6x2 + 6x3  =
0
5 −2 2
x3
5x1 − 2x2 + 2x3
114
Solving this system we find x1 = 0 and x3 = x2 . Letting x2 = 1 an eigenvector
is
 
0

x1 = 1 
1
Hence, the geometric multiplicity of r1 = 3 is 1.
The algebraic multiplicity of r2 = 2 is 2. An eigenvector corresponding to
this eigenvalue is found as follows.


 

3 −1 1
x1
3x1 − x2 + x3
0
(P − 2I)x1 =  14 −5 4   x2  =  14x1 − 5x2 + 6x3  =
0
5 −2 3
x3
5x1 − 2x2 + 3x3
Solving this system we find x2 = 4x1 and x3 = x1 . Letting x1 = 1 we find
 
1

x2 = 4 
1
Hence, r2 = 2 has geometric multiplicity 1 so the matrix P is nondiagonalizable
Problem 41.7
At leat two (and possibly more) of the following four matrices are diagonalizable. You should be able to recognize two by inspection. Choose them and
give a reason for your choice.
5 6
3 6
3 0
1 3
(a)
, (b)
, (c)
, (d)
3 4
6 9
3 −4
−1 4
Solution.
The matrix (b) is diagonalizable since it is a real symmetric matrix. The
matrix (c) is diagonalizable since it is a lower triangular matrix with distinct
eigenvalues 3 and 4
Problem 41.8
Solve the following system by making the change of variables y = Tz.
2t
−4 −6
e − 2et
0
y =
y+
3
5
e−2t + et
115
Solution.
The characteristic equation is
−4 − r −6 = (r + 1)(r − 2) = 0
3
5−r Thus, the eigenvalues are r1 = −1 and r2 = 2. An eigenvector corresponding
to r1 = −1 is found as follows
−3 −6
x1
−3x1 − 6x2
0
(P + I)x1 =
=
=
3
6
x2
3x1 + 6x2
0
Solving this system we find x1 = −2x2 . Letting x2 = −1 we find x1 = 2 and
an eigenvector is
2
x1 =
−1
Similarly, for r2 = 2 we have
6 −6
x1
6x1 − 6x2
0
(P − 2I)x2 =
=
=
3 3
x2
3x1 + 3x2
0
Solving this system we find x1 = −x2 . Letting x2 = −1 we find x1 = 1 and
an eigenvector is
1
x2 =
−1
Therefore
T=
2
1
−1 −1
Thus,
−1
T
=
−1 0
0 2
1
1
−1 −2
−et
e2t
Letting y = Tz we obtain
0
z =
z+
That is,
z10 = −z1 − et
z20 = 2z2 + e2t
116
Solving this system we find
z(t) =
− 12 et + c1 e−t
te2t + c2 e2t
Thus, the general solution is
1 t
2
1
− 2 e + c1 e−t
2e−t e2t
c1
−et + te2t
y(t) = Tz(t) =
=
+ 1 t
e − te2t
−1 −1
te2t + c2 e2t
−e−t −e2t
c2
2
Problem 41.9
Solve the following system by making the change of variables y = Tz.
3 2
4t + 4
0
y =
y+
1 4
−2t + 1
Solution.
The characteristic equation is
3−r
2
1
4−r
= (r − 5)(r − 2) = 0
Thus, the eigenvalues are r1 = 2 and r2 = 5. An eigenvector corresponding
to r1 = 2 is found as follows
1 2
x1
x1 + 2x2
0
(P − 2I)x1 =
=
=
1 2
x2
x1 + 2x2
0
Solving this system we find x1 = −2x2 . Letting x2 = −1 we find x1 = 2 and
an eigenvector is
2
x1 =
−1
Similarly, for r2 = 5 we have
−2 2
x1
−2x1 + 2x2
0
(P − 5I)x2 =
=
=
1 −1
x2
x1 − x2
0
Solving this system we find x1 = x2 . Letting x2 = 1 we find x1 = 1 and an
eigenvector is
1
x2 =
1
117
Therefore
2 1
−1 1
T=
Thus,
T
−1
1
=
3
1 −1
1 2
Letting y = Tz we obtain
0
z =
2 0
0 5
z+
2t + 1
2
That is,
z10 = 2z1 + 2t + 1
5z2 + 2
z20 =
Solving this system we find
z(t) =
−t − 1 + c1 e2t
− 25 + c2 e5t
Thus, the general solution is
−2t − 12
c1
2 1
−t − 1 + c1 e2t
2e2t e5t
5
+
y(t) = Tz(t) =
=
c2
− 52 + c2 e5t
−e2t e5t
t + 53
−1 1
Problem 41.10
Solve the following system by making the change of variables x = Tz.
6
7
00
x =
x
−15 −16
Solution.
The characteristic equation is
6−r
7
−15 −16 − r
= (r + 1)(r + 9) = 0
Thus, the eigenvalues are r1 = −9 and r2 = −1. An eigenvector corresponding to r1 = −9 is found as follows
15
7
x1
15x1 + 7x2
0
(P + 9I)x1 =
=
=
−15 −7
x2
−15x1 − 7x2
0
118
Solving this system we find 15x1 = −7x2 . Letting x1 = 7 we find x2 = −15
and an eigenvector is
7
x1 =
−15
Similarly, for r2 = −1 we have
7
7
x1
7x1 + 7x2
0
(P + 1I)x2 =
=
=
−15 −15
x2
−15x1 − 15x2
0
Solving this system we find x1 = −x2 . Letting x2 = −1 we find x1 = 1 and
an eigenvector is
1
x2 =
−1
Therefore
T=
7
1
−15 −1
Letting x = Tz we obtain
00
z +
−9 0
0 −1
z=0
That is,
z100 = 9z1
z200 = z2
Solving we find
z(t) =
c1 e−3t + c2 e3t
k1 e−t + k2 et
.
The general solution is
x(t) =
Tz =
=
7
1
−15 −1
c1 e−3t + c2 e3t
k1 e−t + k2 et
7(c1 e−3t + c2 e3t ) + k1 e−t + k2 et
−15(c1 e−3t + c2 e3t ) − (k1 e−t + k2 et )
Problem 41.11
Solve the following system by making the change of variables x = Tz.
4 2
00
x =
x
2 1
119
Solution.
The characteristic equation is
4−r
2
2
1−r
= r(r − 5) = 0
Thus, the eigenvalues are r1 = 0 and r2 = 5. An eigenvector corresponding
to r1 = 0 is found as follows
4 2
x1
4x1 + 2x2
0
(P − 0I)x1 =
=
=
2 1
x2
2x1 + x2
0
Solving this system we find x2 = −2x1 . Letting x1 = 1 we find x2 = −2 and
an eigenvector is
1
x1 =
−2
Similarly, for r2 = 5 we have
−1 2
x1
−x1 + 2x2
0
(P − 5I)x2 =
=
=
2 −4
x2
2x1 − 4x2
0
Solving this system we find x1 = 2x2 . Letting x2 = 1 we find x1 = 2 and an
eigenvector is
2
x2 =
1
Therefore
T=
1 2
−2 1
Letting x = Tz we obtain
00
z +
0 0
0 5
z=0
That is,
z100 = 0
z200 = 5z2
Solving we find
z(t) =
√
√ c1 t + c2
k1 cos ( 5t) + k2 sin ( 5t)
120
.
The general solution is
x(t) = Tz =
=
1 2
−2 1
√
√ c1 t + c2
k1 cos ( 5t) + k2 sin ( 5t)
√
√
(c1 t + c2 ) + 2[k1 cos ( √5t) + k2 sin ( √5t)]
−2(c1 t + c2 ) + [k1 cos ( 5t) + k2 sin ( 5t)]
121
42
Solving First Order Linear Systems Using
Exponential Matrix
Problem 42.1 0
2
Find eP(t) if P =
.
−2 0
Solution.
We find
2
P =
−4 0
0 −4
= −4I.
Hence,
P4 = (−1)2 42 I, P6 = (−1)3 43 I, P2n = (−1)n 4n I.
It follows that
P2n+1 = P2n P = (−1)n 4n P.
Now we split the terms of the power series expansion of ePt into even powers
and odd powers of P. We get
P∞ (Pt)2n P∞ (Pt)2n+1
+ n=0 (2n+1)!
ePt =
P∞ t2n n=0 n(2n)!
P∞ t2n+1
n
n n
=
n=0 (2n)! (−1) 4 I +
n=0 (2n+1)! (−1) 4 P
P∞
P
2n
2n+1
n (2t)
=
I n=0 (−1)n (2t)
+ P2 ∞
n=0 (−1) (2n+1)!
(2n)!
=
I cos 2t + P2 sin 2t
=
cos 2t sin 2t
− sin 2t cos 2t
Problem 42.2
Consider the linear differential system
0
y = Py, P =
1
4
−1 −3
(a) Calculate ePt . Hint: Every square matrix satisfies its characteristic equation.
(b) Use the result from part (a) to find two independent solutions of the
differential system. Form the general solution.
122
Solution.
(a) Since the characteristic equation of P is p(r) = (r + 1)2 , (I + P)2 = 0.
But
ePt =
e(−I+(I+P))t
=
e−It e(I+P)t
2
+ ···)
= e−t I(I + t(I + P) + t2 (I+P)
2!
−t
=
e
(I
+
t(I
+
P))
1 0
2
4
−t
+t
= e
0 1
−1 −2
e−t
=
1 + 2t
4t
−t
1 − 2t
(b) Since y = ePt y(0) is the solution to y0 = Py, y(0) = y0 , we generate
two solutions with
1
1 + 2t
Pt
−t
y1 = e
=e
0
−t
y2 = e
Pt
0
1
=e
−t
4t
1 − 2t
Let Ψ(t) = [y1 y2 ]. Then det(Ψ(0)) = det(I) = 1 then {y1 , y2 } forms a
fundamental set of solutions. Thus, the general solution is y = c1 y1 + c2 y2
Problem 42.3
Show that if
then


d1 0 0
D =  0 d2 0 
0 0 d3


ed1 0
0
eD =  0 ed2 0 
0
0 ed3
Solution.
One can easily show by induction on n that
 n

d1 0 0
Dn =  0 dn2 0  .
0 0 dn3
123
Thus,

eD =
P∞
=
Dn
n=0 n!
=
 P
∞
 n=0
0

0
dn
1
n!
P∞
n=0

dn1 0 0
1 
0 dn2 0 
n!
0 0 dn3
0
P∞
0
0
0
P∞
dn
2
n=0 n!

dn
3


n=0 n!

0
ed1 0
 0 ed2 0 
0
0 ed3

=
Problem 42.4
Solve the initial value problem
3 0
0
y =
y, y(0) = y0
0 −1
Solution.
The solution is given by
Pt
y(t) = e y0 =
e3t 0
0 e−t
y0
Problem 42.5
Show that if r is an eigenvalue of P then er is an eigenvalue of eP .
Solution.
Since r is an eigenvalue of P, there is a nonzero vector x such that Px = rx.
In this case,
2
eP x =
I + P + P2! + · · · x
2
=
x + Px + P2!x + · · ·
2
=
x + rx + r2! x + · · ·
2
= (1 + r + r2! + · · · )x = er x
Problem 42.6
Show that det(eA ) = etr(A) . Hint: Recall that the determinant of a matrix
is equal to the product of its eigenvalues and the trace is the sume of the
eigenvalues. This follows from the expansion of the characteristic equation
into a polynomial.
124
Solution.
Suppose r and v are two eigenvalues of A. Then tr(A) = r + v. Hence,
er+v = er · ev . But er and ev are eigenvalues of eA . It follows that etr(A) is the
product of the eigenvalues of eA . But det(eA ) is the product of eigenvalues
of eA
Problem 42.7
Prove: For any invertible n × n matrix P and any n × n matrix A
−1 AP
eP
= P−1 eA P
(Thus, if A is similar to B; then eA is similar to eB ).
Solution.
One can easily show by induction on n that (P−1 AP)n = P−1 An P. Thus,
P∞ (P−1 AP)n
−1
eP AP =
n!
Pn=0
∞ (P−1 An P)
=
n=0
P∞ n!An = P−1
n=0 n! P
=
P−1 eA P
Problem 42.8
Prove: If AB = BA then eA+B = eA eB .
Solution.
Since A and B commute, the binomial formula is valid. That is
X n!
(A + B)n =
Ap Bq .
p!q!
p+q=n
Here the sum runs over non-negative integers p and q that sum to n. We
really need commutativity here, in order to put the As on one side and B s
on the other.
Now we can compute
P∞ (A+B)n
eA+B = P
n=0
n!
P
∞
1
n!
=
Ap Bq
n=0
p+q=n
n!
P∞ P∞ Ap!q!
p Bq
=
P p=0 p q=0
P p!q! q ∞ A
∞ B
=
p=0 p!
q=0 q!
=
eA eB
125
Problem 42.9
Prove: For any square matrix A, eA is invertible with (eA )−1 = e−A .
Solution.
For any t and s we have (tA)(sA) = (sA)(tA). From the previous problem
we can write etA+sA = etA esA . Now, let t = 1 and s = −1 to obtain I = e0 =
eA e−A . This shows that eA is invertible and (eA )−1 = e−A
Problem 42.10
Consider the two matrices
1 0
0 1
A=
, B=
0 −1
−1 0
Show that AB 6= BA and eA+B 6= eA eA .
Solution.
A simple calculation shows that
0 1
0 −1
AB =
, BA =
1 0
−1 0
Since A is diagonal, we have
A
e =
e 0
0 e−1
A simple algebra one finds
B
e =
cos 1 sin 1
− sin 1 cos 1
Since
A+B=
0 0
0 0
we have
eA+B = I
On the other hand,
A B
e e =
e cos 1
e sin 1
−1
−1
−e sin 1 e cos 1
126
43
The Laplace Transform: Basic Definitions
and Results
Problem 43.1
R∞
Determine whether the integral 0
verges, give its value.
1
dt
1+t2
converges. If the integral con-
Solution.
We have
R∞
0
1
dt
1+t2
RA 1
A
= limA→∞ 0 1+t
2 dt = limA→∞ [arctan t]0
=
limA→∞ arctan A = π2
So the integral is convergent
Problem 43.2
R∞
Determine whether the integral 0
verges, give its value.
Solution.
We have
R∞
0
t
dt
1+t2
=
=
1
2
limA→∞
RA
t
dt
1+t2
A
= 12 limA→∞ [ln (1 + t2 )]0
1
limA→∞ ln (1 + A2 ) = ∞
2
0
2t
dt
1+t2
converges. If the integral con-
Hence, the integral is divergent
Problem 43.3
R∞
Determine whether the integral 0 e−t cos (e−t )dt converges. If the integral
converges, give its value.
Solution.
Using substitution we find
R∞
0
R e−A
e−t cos (e−t )dt =
limA→∞ 1 − cos udu
−A
= limA→∞ [− sin u]1e = limA→∞ [sin 1 − sin (e−A )]
=
sin 1
Hence, the integral is convergent
127
Problem 43.4
Using the definition, find L[e3t ], if it exists. If the Laplace transform exists
then find the domain of F (s).
Solution.
We have
L[e3t = limA→∞
=
=
=
RA
e3t e−st dt = limA→∞ et(3−s) dt
h t(3−s) iA
limA→∞ e 3−s
0 i
h A(3−s)
e
1
limA→∞ 3−s − 3−s
1
, s>3
s−3
0
Problem 43.5
Using the definition, find L[t − 5], if it exists. If the Laplace transform exists
then find the domain of F (s).
Solution.
Using integration by parts we find
L[t − 5] = limA→∞
RA
0
(t − 5)e
−st
dt = limA→∞
=
limA→∞
−(A−5)e−sA +5
s
1
s2
=
h
−(t−5)e−st
s
−
h
e−st
s2
iA
0
+
1
s
RA
0
−st
e
dt
iA 0
− 5s , s > 0
Problem 43.6
2
Using the definition, find L[e(t−1) ], if it exists. If the Laplace transform
exists then find the domain of F (s).
Solution.
We have
Z
∞
Z
(t−1)2 −st
∞
2
dt =
e(t−1) −st dt.
0
0
1
+
=, for any fixed s we
Since limt→∞ (t − 1)2 − st = limt→∞ t2 1 − (2+s)
t
t2
e
e
2
can choose a positive CR such that (t − 1)2R− st ≥ 0. In this case, e(t−1) −st ≥ 1
2
∞
∞
and this implies that 0 e(t−1) −st dt ≥ C dt. The integral on the right is
divergent so that the integral on the left is also divergent by the comparison
2
theorem of improper integrals. Hence, f (t) = e(t−1) does not have a Laplace
transform
128
Problem 43.7
Using the definition, find L[(t − 2)2 ], if it exists. If the Laplace transform
exists then find the domain of F (s).
Solution.
We have
L[(t − 2)2 ] = lim (t − 2)2 e−st dt.
T →∞
0
Using integration by parts with u = e−st and v = (t − 2)2
T
Z
Z T
2 T
(t − 2)2 e−st
2 −st
+
(t−2)e−st dt =
(t−2) e dt = −
s
s 0
0
0
we find
4 (T − 2)2 e−sT 2
−
+
s
s
s
Thus,
Z
2 −st
(t − 2) e
lim
T →∞
T
0
0
4 2
dt = + lim
s s T →∞
Z
T
(t − 2)e−st dt
0
−st
Using by parts with u = e
and v = t − 2 we find
T
Z T
(t − 2)e−st
1 −st
−st
(t − 2)e dt = −
+ 2e
.
s
s
0
0
Letting T → ∞ in the above expression we find
Z T
2
1
lim
(t − 2)e−st dt = − + 2 , s > 0.
T →∞ 0
s s
Hence,
4 2
F (s) = +
s s
1
4
2
2
4
− + 2 = − 2 + 3, s > 0
s s
s s
s
Problem 43.8
Using the definition, find L[f (t)], if it exists. If the Laplace transform exists
then find the domain of F (s).
0,
0≤t<1
f (t) =
t − 1,
t≥1
Solution.
We have
Z
T
L[f (t)] = lim
T →∞
1
129
(t − 1)e−st dt.
Z
0
T
(t−2)e−st dt.
Using integration by parts with u0 = e−st and v = t − 1 we find
T
Z T
(t − 1)e−st
1 −st
e−s
−st
lim
(t − 1)e dt = lim −
− 2e
= 2 , s>0
T →∞ 1
T →∞
s
s
s
1
Problem 43.9
Using the definition, find L[f (t)], if it exists. If the Laplace transform exists
then find the domain of F (s).

0≤t<1
 0,
t − 1, 1 ≤ t < 2
f (t) =

0,
t ≥ 2.
Solution.
We have
2
Z 2
(t − 1)e−st
1 −st
e−2s 1 −s −2s
−st
(t−1)e dt = −
− 2e
+ 2 (e −e ), s 6= 0
L[f (t)] =
=−
s
s
s
s
1
1
Problem 43.10
Let n be a positive integer. Using integration by parts establish the reduction
formula
Z
Z
tn e−st n
n −st
+
tn−1 e−st dt, s > 0.
t e dt = −
s
s
Solution.
−st
Let u0 = e−st and v = tn . Then u = − e s and v 0 = ntn−1 . Hence,
Z
Z
tn e−st n
n −st
+
tn−1 e−st dt, s > 0
t e dt = −
s
s
Problem 43.11
For s > 0 and n a positive integer evaluate the limits
limt→0 tn e−st
(b) limt→∞ tn e−st
Solution.
n
(a) limt→0 tn e−st = limt→0 etst = 10 = 0.
(b) Using L’Hôpital’s rule repeatedly we find
n!
=0
t→∞ sn est
lim tn e−st = · · · = lim
t→∞
130
Problem 43.12
(a) Use the previous two problems to derive the reduction formula for the
Laplace transform of f (t) = tn ,
L[tn ] =
n n−1
L[t ], s > 0.
s
(b) Calculate L[tk ], for k = 1, 2, 3, 4, 5.
(c) Formulate a conjecture as to the Laplace transform of f (t), tn with n a
positive integer.
Solution.
(a) Using the two previous problems we find
h
iT
R T n −st
n −st
n
L[t ] = limT →∞ 0 t e dt = limT →∞ − t es
+
0
=
n
s
limT →∞
RT
L[t]
L[t2 ]
L[t3 ]
L[t4 ]
L[t5 ]
=
=
=
=
=
0
n
s
n−1 −st
R
0
Tt
e
dt
tn−1 e−st dt = ns L[tn−1 ], s > 0
(b) We have
1
s2
2
L[t] = s23
s
3
L[t2 ] = s64
s
4
L[t3 ] = 24
s
s5
5
120
4
L[t
]
=
s
s5
(c) By induction, one can easily shows that for n = 0, 1, 2, · · ·
L[tn ] =
n!
sn+1
, s>0
From a table of integrals,
R αu
sin βu
e sin βudu = eαu α sin βu−β
α2 +β 2
R αu
sin βu
e cos βudu = eαu α cos βu+β
α2 +β 2
Problem 43.13
Use the above integrals to find the Laplace transform of f (t) = cos ωt, if it
exists. If the Laplace transform exists, give the domain of F (s).
131
Solution.
We have
(
−st
L[cos ωt] = lim − e
T →∞
−s cos ωt + ω sin ωt
s2 + ω 2
T )
=
0
s2
s
, s>0
+ ω2
Problem 43.14
Use the above integrals to find the Laplace transform of f (t) = sin ωt, if it
exists. If the Laplace transform exists, give the domain of F (s).
Solution.
We have
(
L[sin ωt] = lim − e−st
T →∞
−s sin ωt + ω cos ωt
s2 + ω 2
T )
=
0
s2
ω
, s>0
+ ω2
Problem 43.15
Use the above integrals to find the Laplace transform of f (t) = cos ω(t − 2),
if it exists. If the Laplace transform exists, give the domain of F (s).
Solution.
Using a trigonometric identity we can write f (t) = cos ω(t − 2) = cos ωt cos 2ω+
sin ωt sin 2ω. Thus, using the previous two problems we find
L[cos ω(t − 2)] =
s cos 2ω + ω sin 2ω
, s>0
s2 + ω 2
Problem 43.16
Use the above integrals to find the Laplace transform of f (t) = e3t sin t, if it
exists. If the Laplace transform exists, give the domain of F (s).
Solution.
We have
L[e3t sin t] =
limT →∞
RT
0
e−(s−3)t sin tdt
h
iT −(s−3)t (s−3) sin t+cos t
= limT →∞ − e
(s−3)2 +1
0
=
1
,
(s−3)2 +1
132
s>3
Problem 43.17
Use the linearity property of Laplace transform to find L[5e−7t + t + 2e2t ].
Find the domain of F (s).
Solution.
We have L[e−7t ] =
Hence,
1
,
s+7
s > −7, L[t] =
1
,
s2
s > 0, and L[e2t ] =
L[5e−7t + t + 2e2t ] = 5L[e−7t ] + L[t] + 2L[e2t ] =
1
,
s−2
s > 2.
1
5
2
+ 2+
, s>2
s+7 s
s−2
Problem 43.18
Consider the function f (t) = tan t.
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on 0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
Solution.
sin t
and this function is discontinuous at t = (2n +
(a) Since f (t) = tan t = cos
t
π
1) 2 . Since this function has vertical asymptotes there it is not piecewise
continuous.
(b) The graph of the function does not show that it can be bounded by
exponential functions. Hence, no such numbers a and M
Problem 43.19
Consider the function f (t) = t2 e−t .
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on 0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
Solution.
(a) Since t2 and e−t are continuous everywhere, f (t) = t2 e−t is continuous on
0 ≤ t < ∞.
(b) By L’Hôpital’s rule one has
t2
=0
t→∞ et
Since f (0) = 0, f (t) is bounded. Since f 0 (t) = (2t − t2 )e−t , f (t) has a
maximum when t = 2. The value of this maximum is f (2) = 4e−2 . Hence,
M = 4e−2 and a = 0
lim
133
Problem 43.20
Consider the function f (t) =
2
et
e2t +1
.
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on 0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
Solution.
t2
(a) Since et2 and e2t +1 are continuous everywhere, f (t) = e2te +1 is continuous
on 0 ≤ t < ∞.
2
2
(b) Since e2t + 1 ≤ e2t + e2t = 2e2t , f (t) ≥ 12 et e−2t = 12 et −2t . But for t ≥ 4
we have t2 − 2t >
at infinity
t2
t2
.
2
Hence, f (t) > 12 e 2 . So f (t) is not of exponential order
Problem 43.21
Consider the floor function f (t) = btc, where for any integer n we have
btc = n for all n ≤ t < n + 1.
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on 0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
Solution.
(a) The floor function is a piecewise continuous function on 0 ≤ t < ∞.
(b) Since btc ≤ t < et for 0 ≥ t < ∞ we have M = 1 and a = 1
Problem 43.22
3
Find L−1 s−2
.
Solution. 1
Since L s−a
=
L
1
,
s−a
−1
Problem 43.23
Find L−1 − s22 +
s > a we have
1
s+1
3
s−2
= 3L
−1
1
s−2
.
134
= 3e2t , t ≥ 0
Solution.
Since L[t] =
1
,
s2
s > 0 and L
L−1 − s22 +
1
s+1
1
s−a
=
1
,
s−a
s > a we have
1
= −2L−1 s12 + L−1 s+1
=
−2t + e−t , t ≥ 0
Problem 43.24
2
2
Find L−1 s+2
+ s−2
.
Solution.
We have
2
1
1
2
−1
−1
−1
+
= 2L
+2L
= 2(e−2t +e2t ), t ≥ 0
L
s+2 s−2
s+2
s−2
135
44
Further Studies of Laplace Transform
Problem 44.1
Use Table L to find L[2et + 5].
Solution.
L[2et + 5] = 2L[et ] + 5L[1] =
5
2
+ , s>1
s−1 s
Problem 44.2
Use Table L to find L[e3t−3 h(t − 1)].
Solution.
L[e3t−3 h(t − 1)] = L[e3(t−1) h(t − 1)] = e−s L[e3t ] =
e−s
, s>3
s−3
Problem 44.3
Use Table L to find L[sin2 ωt].
Solution.
1 − cos 2ωt
1
1
L[sin ωt] = L[
] = (L[1]−L[cos 2ωt]) =
2
2
2
2
1
s2
− 2
s s + 4ω 2
Problem 44.4
Use Table L to find L[sin 3t cos 3t].
Solution.
L[sin 3t cos 3t] = L[
sin 6t
1
3
= L[sin 6t] = 2
, s>0
2
2
s + 26
Problem 44.5
Use Table L to find L[e2t cos 3t].
Solution.
L[e2t cos 3t] =
s−3
, s>3
(s − 3)2 + 9
136
, s>0
Problem 44.6
Use Table L to find L[e4t (t2 + 3t + 5)].
Solution.
L[e4t (t2 +3t+5)] = L[e4t t2 ]+3L[e4t t]+5L[1] =
Problem 44.7
Use Table L to find L−1 [ s210
+
+25
2
3
5
, s>4
+
+
3
2
(s − 4) (s − 4) s − 4
4
].
s−3
Solution.
L−1 [
s2
10
4
5
1
+
] = 2L−1 [ 2
]+4L−1 [
] = 2 sin 5t+4e3t , t ≥ 0
+ 25 s − 3
s + 25
s−3
Problem 44.8
5
Use Table L to find L−1 [ (s−3)
4 ].
Solution.
L−1 [
5
5
3!
5
] = L−1 [
] = e3t t3 , t ≥ 0
4
4
(s − 3)
6
(s − 3)
6
Problem 44.9
−2s
Use Table L to find L−1 [ es−9 ].
Solution.
e−2s
L [
] = e9(t−2) h(t − 2) =
s−9
−1
0,
9(t−2)
e
0≤t<2
, t≥2
Problem 44.10
−3s
Use Table L to find L−1 [ e s2(2s+7)
].
+16
Solution.
−3s (2s+7)
L−1 [ e
s2 +16
−3s
−3s
] =
2L−1 [ se2 +16s ] + 74 L−1 [ e 4 s2 + 16]
= 2 cos 4(t − 3)h(t − 3) + 74 sin 4(t − 3), t ≥ 0
137
Problem 44.11
Graph the function f (t) = h(t − 1) + h(t − 3) for t ≥ 0, where h(t) is the
Heaviside step function, and use Table L to find L[f (t)].
Solution.
Note that

 0, 0 ≤ t < 1
1, 1 ≤ t < 3
f (t) =

2,
t≥3
The graph of f (t) is shown below. Using Table L we find
L[f (t)] = L[h(t − 1)] + L[h(t − 3)] =
e−s e−3s
+
, s>0
s
s
Problem 44.12
Graph the function f (t) = t[h(t − 1) − h(t − 3)] for t ≥ 0, where h(t) is the
Heaviside step function, and use Table L to find L[f (t)].
Solution.
Note that

 0, 0 ≤ t < 1
t, 1 ≤ t < 3
f (t) =

0,
t≥3
The graph of f (t) is shown below. Using Table L we find
L[f (t)] =
L[(t − 1)h(t − 1) + h(t − 1) − (t − 3)h(t − 3) − 3h(t − 3)]
= L[(t − 1)h(t − 1)] + L[h(t − 1)] − L[(t − 3)h(t − 3)] − 3L[h(t − 3)]
−s
3s
−3s
e−s
=
+ e s − es2 − e s , s > 1
s2
138
Problem 44.13
Graph the function f (t) = 3[h(t − 1) − h(t − 4)] for t ≥ 0, where h(t) is the
Heaviside step function, and use Table L to find L[f (t)].
Solution.
Note that

 0, 0 ≤ t < 1
3, 1 ≤ t < 4
f (t) =

0,
t≥4
The graph of f (t) is shown below. Using Table L we find
L[f (t)] = 3L[h(t − 1)] − 3L[h(t − 4)] =
3e−s 3e−4s
−
, s>0
s
s
Problem 44.14
Graph the function f (t) = |2 − t|[h(t − 1) − h(t − 3)] for t ≥ 0, where h(t) is
the Heaviside step function, and use Table L to find L[f (t)].
Solution.
Note that
f (t) =


0,
0≤t<1
|2 − t|, 1 ≤ t < 3

0,
t≥3
139
The graph of f (t) is shown below. Using Table L we find
L[f (t)] = L[−(t − 1)h(t − 1) + h(t − 1) + (t − 2)h(t − 2) + (t − 2)h(t − 2) − (t − 3)h(t − 3) − h
=
−L[(t − 1)h(t − 1)] + L[h(t − 1)] + 2L[(t − 2)h(t − 2)] − L[(t − 3)h(t − 3)] − L[h(t
−s
−2s
−3s
−3s
−s
=
− es2 + e s + 2es2 − e s2 − e s , s > 1
Problem 44.15
Graph the function f (t) = h(2 − t) for t ≥ 0, where h(t) is the Heaviside step
function, and use Table L to find L[f (t)].
Solution.
Note that
f (t) =
1, 0 ≤ t ≤ 2
0,
t>2
The graph of f (t) is shown below. Using Table L we find
Z
L[f (t)] =
2
e
0
−st
e−st
dt =
−s
2
=
0
1 − e−2s
, s>0
s
Problem 44.16
Graph the function f (t) = h(t − 1) + h(4 − t) for t ≥ 0, where h(t) is the
Heaviside step function, and use Table L to find L[f (t)].
140
Solution.
Note that

 1, 0 ≤ t < 1
2, 1 ≤ t ≤ 4
f (t) =

1,
t≥4
The graph of f (t) is shown below. Using Table L we find
Z 4
1 + e−s − e−4s
e−s
e−st dt =
L[f (t)] = L[h(t−1)]+L[h(4−t)] =
+
, s>0
s
s
0
Problem 44.17
The graph of f (t) is given below. Represent f (t) as a combination of Heaviside step functions, and use Table L to calculate the Laplace transform of
f (t).
Solution.
From the graph we see that
f (t) =
(t − 2)[h(t − 2) − h(t − 3)] + [h(t − 3) − h(t − 4)]
= (t − 2)h(t − 2) − [(t − 3) + 1]h(t − 3) + h(t − 3) − h(t − 4)
=
(t − 2)h(t − 2) − (t − 3)h(t − 3) − h(t − 4)
Thus,
e−2s − e−3s e−4s
L[f (t)] = L[(t−2)h(t−2)]−L[(t−3)h(t−3)]−L[h(t−4)] =
−
, s>0
s2
s
141
Problem 44.18
The graph of f (t) is given below. Represent f (t) as a combination of Heaviside step functions, and use Table L to calculate the Laplace transform of
f (t).
Solution.
From the graph we see that
f (t) =
(t − 1)[h(t − 1) − h(t − 2)] + (3 − t)[h(t − 2) − h(t − 3)]
= (t − 1)h(t − 1) − [(t − 2) + 1]h(t − 2) + [−(t − 2) + 1]h(t − 2) + (t − 3)h(t − 3)
=
(t − 1)h(t − 1) − 2(t − 2)h(t − 2) + (t − 3)h(t − 3)
Thus,
L[f (t)] = L[(t−1)h(t−1)]−2L[(t−2)h(t−2)]+L[(t−3)h(t−3)] =
e−s − 2e−2s + e−3s
, s>0
s
Problem 44.19
i
h
12
−1
.
Using the partial fraction decomposition find L
(s−3)(s+1)
Solution.
Write
12
A
B
=
+
(s − 3)(s + 1)
s−3 s+1
Multiply both sides of this equation by s − 3 and cancel common factors to
obtain
12
B(s − 3)
=A+
.
s+1
s+1
Now, find A by setting s = 3 to obtain A = 3. Similarly, by multiplying both
sides by s + 1 and then setting s = −1 in the resulting equation leads to
B = −3. Hence,
12
1
1
=3
−
(s − 3)(s + 1)
s−3 s+1
142
Finally,
L−1
h
12
(s−3)(s+1)
i
1 1 = 3L−1 s−3
− 3L−1 s+1
=
3e3t − 3e−t , t ≥ 0
Problem 44.20
h −5s i
Using the partial fraction decomposition find L−1 24e
.
s2 −9
Solution.
Write
A
B
24
=
+
(s − 3)(s + 3)
s−3 s+3
Multiply both sides of this equation by s − 3 and cancel common factors to
obtain
24
B(s − 3)
=A+
.
s+3
s+3
Now, find A by setting s = 3 to obtain A = 4. Similarly, by multiplying both
sides by s + 3 and then setting s = −3 in the resulting equation leads to
B = −4. Hence,
24
1
1
=4
−
(s − 3)(s + 3)
s−3 s+3
Finally,
L−1
h
24e−5s
(s−3)(s+3)
i
=
4L−1
h
i
e−5s
−
s−3
−3(t−5)
= 4[e3(t−5) − e
4L−1
h
e−5s
s+3
i
]h(t − 5), t ≥ 0
Problem 44.21
Use Laplace transform technique to solve the initial value problem
y 0 + 4y = g(t), y(0) = 2
where

 0, 0 ≤ t < 1
12, 1 ≤ t < 3
g(t) =

0,
t≥3
143
Solution.
Note first that g(t) = 12[h(t − 1) − h(t − 3)] so that
L[g(t)] = 12L[h(t − 1)] − 12L[h(t − 3)] =
12(e−s − e−3s
, s > 0.
s
Now taking the Laplace transform of the DE and using linearity we find
L[y 0 ] + 4L[y] = L[g(t)].
But L[y 0 ] = sL[y] − y(0) = sL[y] − 2. Letting L[y] = Y (s) we obtain
sY (s) − 2 + 4Y (s) = 12
e−s − e−3s
.
s
Solving for Y (s) we find
Y (s) =
2
e−s − e−3s
+ 12
.
s+4
s(s + 4)
But
−1
L
2
= 2e−4t
s+4
and
h −s −3s i
−e
=
L−1 12 e s(s+4)
=
1
3L−1 (e−s − e−3s ) 1s − s+4
h −3s i
h −s i
h −3s i
h −s i
e
+ 3L−1 es+4
3L−1 e s − 3L−1 e s − 3L−1 s+4
= 3h(t − 1) − 3h(t − 3) − 3e−4(t−1) h(t − 1) + 3e−4(t−3) h(t − 3)
Hence,
y(t) = 2e−4t +3[h(t−1)−h(t−3)]−3[e−4(t−1) h(t−1)−e−4(t−3) h(t−3)], t ≥ 0
Problem 44.22
Use Laplace transform technique to solve the initial value problem
y 00 − 4y = e3t , y(0) = 0, y 0 (0) = 0.
144
Solution.
Taking the Laplace transform of the DE and using linearity we find
L[y 00 ] − 4L[y] = L[e3t ].
But L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 L[y]. Letting L[y] = Y (s) we obtain
s2 Y (s) − 4Y (s) =
1
.
s−3
Solving for Y (s) we find
Y (s) =
1
.
(s − 3)(s − 2)(s + 2)
Using partial fraction decomposition
A
B
C
1
=
+
+
(s − 3)(s − 2)(s + 2)
s−3 s+2 s−2
we find A = 51 , B =
1
,
20
and C = − 14 . Thus,
1 1 −1 1 1 −1 1 1
= 51 L−1 s−3
+ 20 L
− 4L
y(t) = L−1 [ (s−3)(s−2)(s+2)
s+2
s−2
1 3t
1 −2t
1 2t
=
e
+
e
−
e
,
t
≥
0
5
20
4
Problem 44.23
Rt
Obtain the Laplace transform of the function 2 f (λ)dλ in terms of L[f (t)] =
R2
F (s) given that 0 f (λ)dλ = 3.
Solution.
We have
L
hR
t
2
i
hR
i
R2
t
f (λ)dλ = L 0 f (λ)dλ − 0 f (λ)dλ
=
=
F (s)
−
s
F (s)
3
− s,
s
145
L[3]
s>0
45
The Laplace Transform and the Method
of Partial Fractions
In Problems 45.1 - 45.4, give the form of the partial fraction expansion for
F (s). You need not evaluate the constants in the expansion. However, if the
denominator has an irreducible quadratic expression then use the completing
the square process to write it as the sum/difference of two squares.
Problem 45.1
F (s) =
s3 + 3s + 1
.
(s − 1)3 (s − 2)2
Solution.
F (s) =
A1
A2
A3
B1
B2
+
+
+
+
3
2
2
(s − 1)
(s − 1)
s − 1 (s − 2)
s − 2)
Problem 45.2
s2 + 5s − 3
.
F (s) = 2
(s + 16)(s − 2)
Solution.
F (s) =
A1 s + A2
B1
+
2
s + 16
s−2
Problem 45.3
F (s) =
s3 − 1
.
(s2 + 1)2 (s + 4)2
Solution.
F (s) =
A1 s + A2 A3 s + A4
B1
B2
+ 2
+
+
2
2
2
(s + 1)
s +1
(s + 4)
s+4
Problem 45.4
F (s) =
s4 + 5s2 + 2s − 9
.
(s2 + 8s + 17)(s − 2)2
146
Solution.
F (s) =
A1
B1 s + B2
B3 s + B4
A2
+
+
+
2
1
(s − 2)
s − 2 (s + 4) + 1 (s + 4)2 + 1
Problemh45.5 i
1
Find L−1 (s+1)
.
3
Solution.
i
h
1
= 12 e−t t2 , t ≥ 0
Using Table L we find L−1 (s+1)
3
Problem45.6 Find L−1 s22s−3
.
−3s+2
Solution.
We factor the denominator and split the rational function into partial fractions:
2s − 3
A
B
=
+
.
(s − 1)(s − 2)
s−1 s−2
Multiplying both sides by (s − 1)(s − 2) and simplifying to obtain
2s − 3 = A(s − 2) + B(s − 1)
= (A + B)s − 2A − B.
Equating coefficients of like powers of s we obtain the system
A+B
= 2
−2A − B = −3
Solving this system by elimination we find A = 1 and B = 1. Now finding
the inverse Laplace transform to obtain
2s − 3
1
1
−1
−1
−1
L
=L
+L
= et + e2t , t ≥ 0.
(s − 1)(s − 2)
s−1
s−2
Problemh45.7 i
2
Find L−1 4ss3+s+1
.
+s
147
Solution.
We factor the denominator and split the rational function into partial fractions:
A Bs + C
4s2 + s + 1
=
+ 2
.
s(s2 + 1
s
s +1
Multiplying both sides by s(s2 + 1) and simplifying to obtain
4s2 + s + 1 = A(s2 + 1) + (Bs + C)s
= (A + B)s2 + Cs + A.
Equating coefficients of like powers of s we obtain A + B = 4, C = 1, A = 1.
Thus, B = 3. Now finding the inverse Laplace transform to obtain
2
s
1
−1 1
−1
−1
−1 4s + s + 1
=L
+3L
+L
= 1+3 cos t+sin t, t ≥ 0.
L
s(s2 + 1)
s
s2 + 1
s2 + 1
Problemh45.8
i
s2 +6s+8
−1
.
Find L
s4 +8s2 +16
Solution.
We factor the denominator and split the rational function into partial fractions:
B1 s + C1 B2 s + C2
s2 + 6s + 8
=
+ 2
.
2
2
(s + 4)
s2 + 4
s +4
Multiplying both sides by (s2 + 4)2 and simplifying to obtain
s2 + 6s + 8 =
(B1 s + C1 )(s2 + 4) + B2 s + C2
= B1 s3 + C1 s2 + (4B1 + B2 )s + 4C1 + C2 .
Equating coefficients of like powers of s we obtain B1 = 0, C1 = 1, B2 = 6,
and C2 = 4. Now finding the inverse Laplace transform to obtain
h 2
i
h
i
h
i
1 s
1
−1
−1
−1
L−1 s(s+6s+8
=
L
+
6L
+
4L
2 +4)2
2
2
s2 +4
(s2 +4)2
(s +4) 1
1
t
= 2 sin 2t + 6 4 sin 2t + 4 16 [sin 2t − 2t cos 2t]
3
=
t sin 2t + 43 sin 2t − 12 t cos 2t, t ≥ 0
2
148
Problem 45.9
Use Laplace transform to solve the initial value problem
y 0 + 2y = 26 sin 3t, y(0) = 3.
Solution.
Taking the Laplace of both sides to obtain
L[y 0 ] + 2L[y] = 26L[sin 3t].
Using Table L the last equation reduces to
sY (s) − y(0) + 2Y (s) = 26
3
2
s +9
.
Solving this equation for Y (s) we find
Y (s) =
3
78
+
.
s + 2 (s + 2)(s2 + 9)
Using the partial fraction decomposition we can write
A
Bs + C
1 2
s +9=
+ 2
.
s+2
s+2
s +9
Multipliying both sides by (s + 2)(s2 + 9) to obtain
1 =
A(s2 + 9) + (Bs + C)(s + 2)
= (A + B)s2 + (2B + C)s + 9A + 2C
Equating coefficients of like powers of s we find A + B = 0, 2B + C = 0, and
1
1
2
9A + 2C = 1. Solving this system we find A = 13
, B = − 13
, and C = 13
.
Thus,
3
9
s
Y (s) =
−6 2
+4 2
.
s+2
s +9
s +9
Finally,
y(t) =
L−1 [Y (s)]
= 9e−2t − 6 cos 3t + 4 sin 3t, t ≥ 0
149
= 9L−1
1
s+2
− 6L−1
s
s2 +9
+ 4L−1
3
s2 +9
Problem 45.10
Use Laplace transform to solve the initial value problem
y 0 + 2y = 4t, y(0) = 3.
Solution.
Taking the Laplace of both sides to obtain
L[y 0 ] + 2L[y] = 4L[t].
Using Table L the last equation reduces to
sY (s) − y(0) + 2Y (s) =
4
.
s2
Solving this equation for Y (s) we find
Y (s) =
3
4
+
.
s + 2 (s + 2)s2
Using the partial fraction decomposition we can write
A
Bs + C
1
=
+
.
2
(s + 2)s
s+2
s2
Multipliying both sides by (s + 2)s2 to obtain
1 =
As2 + (Bs + C)(s + 2)
= (A + B)s2 + (2B + C)s + 2C
Equating coefficients of like powers of s we find A + B = 0, 2B + C = 0, and
2C = 1. Solving this system we find A = 4, B = −1, and C = 2. Thus,
Y (s) =
4
1
1
− + 2 2.
s+2 s
s
Finally,
y(t) =
L−1 [Y (s)]
= 4e−2t − 1 + 2t, t ≥ 0
= 4L−1
1
s+2
− L−1
1
Problem 45.11
Use Laplace transform to solve the initial value problem
y 00 + 3y 0 + 2y = 6e−t , y(0) = 1, y 0 (0) = 2.
150
s
+ 2L−1
1
s2
Solution.
Taking the Laplace of both sides to obtain
L[y 00 ] + 3L[y 0 ] + 2L[y] = 6L[e−t ].
Using Table L the last equation reduces to
s2 Y (s) − sy(0) − y 0 (0) + 3(sY (s) − y(0))) + 2Y (s) =
6
.
s+1
Solving this equation for Y (s) we find
Y (s) =
s+5
6
s2 + 6s + 11
+
=
.
(s + 1)(s + 2) (s + 2)(s + 1)2
(s + 1)2 (s + 2)
Using the partial fraction decomposition we can write
s2 + 6s + 11
A
B
C
=
+
+
.
2
(s + 2)(s + 1)
s + 2 s + 1 (s + 1)2
Multipliying both sides by (s + 2)(s + 1)2 to obtain
s2 + 6s + 11 =
A(s + 1)2 + B(s + 1) + C
= As2 + (2A + B)s + A + B + C
Equating coefficients of like powers of s we find A = 1, 2A + B = 6, and
A + B + C = 11. Solving this system we find A = 3, B = −2, and C = 6.
Thus,
2
6
3
−
+
.
Y (s) =
s + 2 s + 1 (s + 1)2
Finally,
y(t) =
L−1 [Y (s)]
= 3e−2t − 2e−t + 6te−t
= 3L−1
1
s+2
− 2L−1
Problem 45.12
Use Laplace transform to solve the initial value problem
y 00 + 4y = cos 2t, y(0) = 1, y 0 (0) = 1.
151
1
s+1
+ 6L−1
h
1
(s+1)2
i
Solution.
Taking the Laplace of both sides to obtain
L[y 00 ] + 4L[y] = L[cos 2t].
Using Table L the last equation reduces to
s2 Y (s) − sy(0) − y 0 (0) + 4Y (s) =
s2
s
.
+4
Solving this equation for Y (s) we find
Y (s) =
s+1
s
+
.
s2 + 4 (s2 + 4)2
Using Table L again we find
i
h
s
y(t) = L−1 s2s+4 + 21 L−1 s22+4 + L−1 (s2 +4)
2
1
t
=
cos 2t + 2 sin 2t + 4 sin 2t, t ≥ 0
Problem 45.13
Use Laplace transform to solve the initial value problem
y 00 − 2y 0 + y = e2t , y(0) = 0, y 0 (0) = 0.
Solution.
Taking the Laplace of both sides to obtain
L[y 00 ] − 2L[y 0 ] + L[y] = L[e2t ].
Using Table L the last equation reduces to
s2 Y (s) − sy(0) − y 0 (0) − 2sY (s) + 2y(0) + Y (s) =
Solving this equation for Y (s) we find
Y (s) =
1
.
(s − 1)2 (s − 2)
Using the partial fraction decomposition, we can write
Y (s) =
A
B
C
+
+
.
2
s − 1 (s − 1)
s−2
152
1
.
s−2
Multipliying both sides by (s − 2)(s − 1)2 to obtain
1 =
A(s − 1)(s − 2) + B(s − 2) + C(s − 1)2
= (A + C)s2 + (−3A + B − 2C)s + 2A − 2B + C
Equating coefficients of like powers of s we find A+C = 0, −3A+B−2C = 0,
and 2A − 2B + C = 1. Solving this system we find A = −1, B = −1, and
C = 1. Thus,
1
1
1
−
.
Y (s) = −
+
s − 1 (s − 1)2 s − 2
Finally,
i
h
1 1 1
−1
−1
−1
−1
+
L
y(t) = L [Y (s)] = −L
−
L
2
s−1
(s−1)
s−2
=
−et − tet + e2t , t ≥ 0
Problem 45.14
Use Laplace transform to solve the initial value problem
y 00 + 9y = g(t), y(0) = 1, y 0 (0) = 0
where
g(t) =
6, 0 ≤ t < π
0, π ≤ t < ∞
Solution.
Taking the Laplace of both sides to obtain
L[y 00 ] + 9L[y] = L[g(t)] = 6L[h(t) − h(t − π)].
Using Table L the last equation reduces to
s2 Y (s) − sy(0) − y 0 (0) + 9Y (s) =
6
e−πs
−6
.
s
s
Solving this equation for Y (s) we find
Y (s) =
s+3
6
+
(1 − e−πs ).
2
2
s + 9 s(s + 9)
Using the partial fraction decomposition, we can write
6
A Bs + C
= + 2
.
+ 9)
s
s +9
s(s2
153
Multipliying both sides by s(s2 + 9) to obtain
6 = A(s2 + 9) + (Bs + C)s
= (A + B)s2 + Cs + 9A
Equating coefficients of like powers of s we find A + B = 0, C = 0, and
9A = 6. Solving this system we find A = 32 , B = − 32 , and C = 0. Thus,
s
3
21 2 s
−πs
Y (s) = 2
+
+ (1 − e )
−
.
s + 9 s2 + 9
3 s 3 s2 + 9
Finally,
y(t) = L−1 [Y (s)] = cos 3t + sin 3t + 23 (1 − cos 3t) − 23 (1 − cos 3(t − π))h(t − π)
=
cos 3t + sin 3t + 32 (1 − cos 3t) − 23 (1 + cos 3t)h(t − π), t ≥ 0
Problem 45.15
Determine the constants α, β, y0 , and y00 so that Y (s) =
transform of the solution to the initial value problem
2s−1
s2 +s+2
is the Laplace
y 00 + αy 0 + βy = 0, y(0) = y0 , y 0 (0) = y00 .
Solution.
Taking the Laplace transform of both sides we find
s2 Y (s) − sy0 − y00 + αsY (s) − αy0 + βY (s) = 0.
Solving for Y (s) we find
Y (s) =
sy0 + (y00 + αy0 )
2s − 1
= 2
.
2
s + αs + β
s +s+2
By identification we find α = 1, β = 2, y0 = 2, and y00 = −3
Problem 45.16
Determine the constants α, β, y0 , and y00 so that Y (s) =
transform of the solution to the initial value problem
s
(s+1)2
y 00 + αy 0 + βy = 0, y(0) = y0 , y 0 (0) = y00 .
154
is the Laplace
Solution.
Taking the Laplace transform of both sides we find
s2 Y (s) − sy0 − y00 + αsY (s) − αy0 + βY (s) = 0.
Solving for Y (s) we find
Y (s) =
sy0 + (y00 + αy0 )
s
= 2
.
2
s + αs + β
s + 2s + 1
By identification we find α = 2, β = 1, y0 = 1, and y00 = −2
155
47
Laplace Transforms of Periodic Functions
Problem 47.1
Find the Laplace transform of the periodic function whose graph is shown.
Solution.
The function is of period T = 2. Thus,
1 −st 2
Z 1
Z 2
3 −st
e
1
−st
−st
3
e dt +
e dt = − e
−
= (3 − 2e−s − e−2s ).
s
s 1 s
0
1
0
Hence,
L[f (t)] =
3 − 2e−s − e−2s
s(1 − e−2s
Problem 47.2
Find the Laplace transform of the periodic function whose graph is shown.
Solution.
The function is of period T = 4. Thus,
1 −st 3
Z 1
Z 3
2 −st
e
1
−st
−st
2
e dt +
e dt = − e
−
= (2 − e−s − e−3s ).
s
s 1 s
0
1
0
156
Hence,
L[f (t)] =
2 − e−s − e−3s
s(1 − e−4s
Problem 47.3
Find the Laplace transform of the periodic function whose graph is shown.
Solution.
The function is of period T = 2. Thus,
−s
1
Z 1
Z 2
e
e−s
(1+u)s
−st
−su
ue
du = − 2 (su + 1)e
(t−1)e dt =
= − 2 [(s+1)e−s −1].
s
s
0
1
0
Hence,
L[f (t)] =
e−s
[1 − (s + 1)e−s ]
2
−2s
s (1 − e
Problem 47.4
Find the Laplace transform of the periodic function whose graph is shown.
157
Solution.
The function is of period T = 2. Thus,
2
Z 2
1
1
−st
−st
te dt = − 2 (st + 1)e
= − 2 [(2s + 1)e−2s − 1].
s
s
0
0
Hence,
L[f (t)] =
s2 (1
1
[1 − (2s + 1)e−2s ]
− e−2s
Problem 47.5
State the period of the function f (t) and find its Laplace transform where

 sin t, 0 ≤ t < π
f (t + 2π) = f (t), t ≥ 0.
f (t) =

0, π ≤ t < 2π
Solution.
The graph of f (t) is shown below.
The function f (t) is of period T = 2π. The Laplace transform of f (t) is
Rπ
sin te−st dt
L[f (t)] = 0
1 − e−2πs
Using integration by parts twice we find
Z
e−st
sin te−st dt = −
(cos t + s sin t)
1 + s2
158
Thus,
Rπ
0
sin te−st dt =
h
iπ
e−st
(cos
t
+
s
sin
t)
− 1+s
2
e−πs
1
+ 1+s
2
1+s2
−πs
1+e
1+s2
=
=
Hence,
0
1 + e−πs
L[f (t)] =
(1 + s2 )(1 − e−2πs )
Problem 47.6
State the period of the function f (t) = 1 − e−t , 0 ≤ t < 2, f (t + 2) = f (t),
and find its Laplace transform.
Solution.
The graph of f (t) is shown below
The function is periodic of period T = 2. Its Laplace transform is
R2
(1 − e−t )e−st dt
0
L[f (t)] =
.
1 − e−2s
But
−st 2 (s+1)t 2
Z 2
e
e
1
1
−t −st
(1 − e )e dt =
+
= (1 − e−2s ) −
(1 − e−2(s+1) ).
−s 0
s+1 0 s
s+1
0
Hence,
L[f (t)] =
1
1 − e−2(s+1)
−
s (s + 1)(1 − e−2s
159
Problem 47.7
Using Example 44.3 find
−1
L
s2 − s
e−s
.
+
s3
s(1 − e−s )
Solution.
Note first that
s2 − s
1
e−s
= −
+
3
−s
s
s(1 − e )
s
1
se−s
−
s2 s2 (1 − e−s )
.
Using Example 44.3, we find
f (t) = 1 − g(t)
where g(t) is the sawtooth function shown below
Problem 47.8
An object having mass m is initially at rest on a frictionless horizontal surface.
At time t = 0, a periodic force is applied horizontally to the object, causing
it to move in the positive x-direction. The force, in newtons, is given by

 f0 , 0 ≤ t ≤ T2
f (t + T ) = f (t), t ≥ 0.
f (t) =

T
0, 2 < t < T
160
The initial value problem for the horizontal position, x(t), of the object is
mx00 (t) = f (t), x(0) = x0 (0) = 0.
(a) Use Laplace transforms to determine the velocity, v(t) = x0 (t), and the
position, x(t), of the object.
(b) Let m = 1 kg, f0 = 1 N, and T = 1 sec. What is the velocity, v, and
position, x, of the object at t = 1.25 sec?
Solution.
2
(a) Taking Laplace transform of both sides we find ms X(s) =
−s T f0
1−e 2
. Solving for X(s) we find
s
1−e−sT
X(s) =
f0
R
T
2
0
e−st dt
1−e−sT
=
f0 1
1
· 3·
.
m s 1 + e−s T2
Also,
V (s) = L[v(t)] = sX(s) =
1
1 1
1
f0 1
· 2·
· ·
.
T =
m s 1 + e−s 2
m s s(1 + e−s T2 )
Hence, by Example 44.1 and Table L we can write
Z
1 t
v(t) =
f (u)du.
m 0
Since X(s) =
f0
1 1
m s2 s(1+e−s T2 )
=
1
L[t]L[f (t)]
m
=
1
L[t
m
∗ f (t)] we have
Z
1
1 t
x(t) = (t ∗ f (t)) =
(t − u)f (u)du.
m
m 0
R1
R5
(b) We have x(1.25) = 02 ( 54 −u)du+ 14 ( 45 −w)dw = 17
meters and v(1.25) =
32
R5
R1
R5
3
4
f (u)du = 02 dt + 14 dt = 4 m/sec
0
Problem 47.9
Consider the initial value problem
ay 00 + by 0 + cy = f (t), y(0) = y 0 (0) = 0, t > 0
Suppose that the transfer function of this system is given by Φ(s) =
(a) What are the constants a, b, and c?
(b) If f (t) = e−t , determine F (s), Y (s), and y(t).
161
1
.
2s2 +5s+2
Solution.
(a) Taking the Laplace transform of both sides we find as2 Y (s) + bsY (s) +
cY (s) = F (s) or
Φ(s) =
1
1
Y (s)
= 2
= 2
.
F (s)
as + bs + c
2s + 5s + 2
By identification we find a = 2, b = 5, and c = 2.
1
(b) If f (t) = e−t then F (s) = L[e−t ] = s+1
. Thus,
Y (s) = Φ(s)F (s) =
1
.
(s + 1)(2s2 + 5s + 2)
Using partial fraction decomposition
1
A
B
C
=
+
+
(s + 1)(2s + 1)(s + 2)
s + 1 2s + 1 s + 2
Multiplying both sides by s + 1 and setting s = −1 we find A = −1. Next,
multiplying both sides by 2s+1 and setting s = − 12 we find B = 34 . Similarly,
multiplying both sides by s + 2 and setting s = −2 we find C = 13 . Thus,
1 2 −1 h 1 i 1 −1 1 −1
y(t) = −L
+ 3L
+ 3L
s+1
s+2
s+ 1
=
t
2
−e−t + 32 e− 2 + 13 e−2t
Problem 47.10
Consider the initial value problem
ay 00 + by 0 + cy = f (t), y(0) = y 0 (0) = 0, t > 0
Suppose that an input f (t) = t, when applied to the above system produces
the output y(t) = 2(e−t − 1) + t(e−t + 1), t ≥ 0.
(a) What is the system transfer function?
(b) What will be the output if the Heaviside unit step function f (t) = h(t)
is applied to the system?
Solution.
(a) Since f (t) = t, we have F (s) = s12 . Aslo, Y (s) = L[y(t)] = L[2e−t − 2 +
Y (s)
2
1
1
1
1
te−t + t] = s+2
− 2s + (s+1)
2 + s2 = s2 (s+1)2 . But Φ(s) = F (s) = (s+1)2 .
162
(b) If f (t) = h(t) then F (s) = 1s and Y (s) = Φ(s)F (s) =
partial fraction decomposition we find
1
s(s+1)2
1
1
1
1
1
−
−
s s + 1 (s + 1)2
Y (s) =
and
y(t) = L−1 [Y (s)] = 1 − e−t − te−t
Problem 47.11
Consider the initial value problem
y 00 + y 0 + y = f (t), y(0) = y 0 (0) = 0,
f (t) =

 1,

0≤t≤1
f (t + 2) = f (t)
−1, 1 < t < 2
(a) Determine the system transfer function Φ(s).
(b) Determine Y (s).
Solution.
(a) Taking the Laplace transform of both sides we find
s2 Y (s) + sY (s) + Y (s) = F (s)
so that
Φ(s) =
1
Y (s)
= 2
.
F (s)
s +s+1
(b) But
R2
0
Using
B
C
A
+ s+1
+ (s+1)
=
2
s
=
A(s + 1)2 + Bs(s + 1) + Cs
= (A + B)s2 + (2A + B + C)s + A
Equating coefficients of like powers of s we find A = 1,
C = −1. Therefore,
where
1
.
s(s+1)2
f (t)e−st dt =
=
=
=
R 1 −st
R2
e dt − 1 e−st dt
0h
i1 h −st i2
e−st
− e−s
−s
1
(1
s
0
1
− e−s ) + 1s (e−2s − e−s )
163
(1−e−s )2
s
B = −1, and
Hence,
F (s) =
(1 − e−s )
(1 − e−s )2
=
s(1 − e−2s )
s(1 + e−s )
and
Y (s) = Φ(s)F (s) =
(1 − e−s )
s(1 + e−s )(s2 + s + 1)
Problem 47.12
Consider the initial value problem
y 000 − 4y = et + t, y(0) = y 0 (0) = y 00 (0) = 0.
(a) Determine the system transfer function Φ(s).
(b) Determine Y (s).
Solution.
(a) Taking Laplace transform of both sides we find
s3 Y (s) − 4Y (s) = F (s).
Thus,
Φ(s) =
1
Y (s)
= 3
.
F (s)
s −4
(b) We have
F (s) = L[et + t] =
1
1
s2 + s − 1
+ 2 =
.
s−1 s
(s − 1)s2
Hence,
s2 + s − 1
Y (s) = 2
s (s − 1)(s3 − 4)
Problem 47.13
Consider the initial value problem
y 00 + by 0 + cy = h(t), y(0) = y0 , y 0 (0) = y00 , t > 0.
Suppose that L[y(t)] = Y (s) =
and y00 .
s2 +2s+1
.
s3 +3s2 +2s
164
Determine the constants b, c, y0 ,
Solution.
Take the Laplace transform of both sides to obtain
1
s2 Y (s) − sy0 − y00 + bsY (s) − by0 + cY (s) = .
s
Solve to find
Y (s) =
=
=
sy +y00 +by0
1
+ s02 +bs+c
s3 +bs2 +cs)
s2 y0 +s(y00 +by0 )+1
s3 +bs2 +cs
s2 +2s+1
.
s3 +3s2 +2s
By comparison we find b = 3, c = 2, y0 = 1, and y00 + by0 = 2 or y00 = −1
165
47
Convolution Integrals
Problem 47.1
Consider the functions f (t) = g(t) = h(t), t ≥ 0 where h(t) is the Heaviside
unit step function. Compute f ∗ g in two different ways.
(a) By directly evaluating the integral.
(b) By computing L−1 [F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].
Solution.
(a) We have
Z
t
Z
f (t − s)g(s)ds =
(f ∗ g)(t) =
0
t
Z
h(t − s)h(s)ds =
0
t
ds = t
0
(b) Since F (s) = G(s) = L[h(t)] = 1s , (f ∗g)(t) = L−1 [F (s)G(s)] = L−1 [ s12 ] =
t
Problem 47.2
Consider the functions f (t) = et and g(t) = e−2t , t ≥ 0. Compute f ∗ g in
two different ways.
(a) By directly evaluating the integral.
(b) By computing L−1 [F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].
Solution.
(a) We have
(f ∗ g)(t) =
=
Rt
0
Rt
f (t − s)g(s)ds = 0 e(t−s) e−2s ds
h (t−3s) it
R
t t −3s
e 0 e ds = e −3
et −e−2t
3
=
0
1
1
(b) Since F (s) = L[et ] = s−1
and G(s) = L[e−2t ] = s+2
we have (f ∗ g)(t) =
1
−1
−1
L [F (s)G(s)] = L [ (s−1)(s−2) ]. Using partial fractions decomposition we
find
1
1 1
1
= (
−
).
(s − 1)(s + 2)
3 s−1 s+2
Thus,
1
1
1
et − e−2t
(f ∗ g)(t) = L−1 [F (s)G(s)] = (L−1 [
] − L−1 [
]=
3
s−1
s+2
3
166
Problem 47.3
Consider the functions f (t) = sin t and g(t) = cos t, t ≥ 0. Compute f ∗ g in
two different ways.
(a) By directly evaluating the integral.
(b) By computing L−1 [F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].
Solution.
(a) Using the trigonometric identity 2 sin p cos q = sin (p + q) + sin (p − q) we
find that 2 sin (t − s) cos s = sin t + sin (t − 2s). Hence,
Rt
Rt
(f ∗ g)(t) = 0 f (t − s)g(s)ds = 0 sin (t − s) cos sds
Rt
Rt
1
[ sin tds + 0 sin (t − 2s)ds]
=
2 0
Rt
t sin t
=
+ 14 −t sin udu
2
t sin t
=
2
(b) Since F (s) = L[sin t] =
1
s2 +1
s
s2 +1
and G(s) = L[cos t] =
(f ∗ g)(t) = L−1 [F (s)G(s)] = L−1 [
(s2
we have
t
s
] = sin t
2
+ 1)
2
Problem 47.4
Use Laplace
transform to
compute
the convolution P ∗ y, where |bf P (t) =
t
h(t)
h(t) e
and y(t) =
.
0
t
e−t
Solution.
We have
(P ∗ y)(t) =
L
−1
("
= L−1
But
and
1
s
0
1
s
1
2
s (s+1)
1
1
=
(s − 1)(s + 1)
2
1
s−1
1
s2
1
s2
+
1
s
1
s+1
1
(s−1)(s+1)
1
1
−
s−1 s+1
1
1
1
1
= 2− +
.
+ 1)
s
s s+1
s2 (s
167
#)
Hence,
(P ∗ y)(t) =
t
−t
t + e2 − e 2
t − 1 + e−t
Problem 47.5
Compute and graph f ∗ g where f (t) = h(t) and g(t) = t[h(t) − h(t − 2)].
Solution.
Since f (t) = h(t), F (s) = 1s . Similarly, since g(t) = th(t) − th(t − 2) =
−2s
−2s
th(t) − (t − 2)h(t − 2) − 2h(t − 2), G(s) = s12 − e s2 − 2e s . Thus, F (s)G(s) =
−2s
−2s
1
− e s3 − 2es2 . It follows that
s3
(f ∗ g)(t) =
t2 (t − 2)2
−
h(t − 2) − 2(t − 2)h(t − 2).
2
2
The graph of (f ∗ g)(t) is given below
Problem 47.6
Compute and graph f ∗ g where f (t) = h(t) − h(t − 1) and g(t) = h(t − 1) −
2h(t − 2)].
Solution.
−s
Since f (t) = h(t) − h(t − 1), F (s) = 1s − e s . Similarly, since g(t) = h(t − 1) −
−s
−2s
−s
−2s
−3s
2h(t − 2), G(s) = e s − 2e s . Thus, F (s)G(s) = e −3e s2 +2e . It follows
that
(f ∗ g)(t) = (t − 1)h(t − 1) − 3(t − 2)h(t − 2) + 2(t − 3)h(t − 3).
The graph of (f ∗ g)(t) is given below
168
Problem 47.7
Compute t ∗ t ∗ t.
Solution.
By the convolution theorem we have L[t ∗ t ∗ t] = (L[t])3 =
5
t5
t ∗ t ∗ t = L−1 s16 = t5! = 120
1 3
s2
=
1
.
s6
Hence,
Problem 47.8
Compute h(t) ∗ e−t ∗ e−2t .
Solution.
By the convolution theorem we have L[h(t)∗e−t ∗e−2t ] = L[h(t)]L[e−t ]L[e−2t ] =
1
· 1 · 1 . Using the partial fractions decomposition we can write
s s+1 s+2
1
1
1
1
1
=
−
+ ·
.
s(s + 1)(s + 2)
2s s + 1 2 s + 2
Hence,
h(t) ∗ e−t ∗ e−2t =
1
1
− e−t + e−2t
2
2
Problem 47.9
Compute t ∗ e−t ∗ et .
Solution.
By the convolution theorem we have L[t ∗ e−t ∗ et ] = L[t]L[e−t ]L[et ] =
1
· 1 . Using the partial fractions decomposition we can write
s+1 s−1
s2 (s
1
1
1
1
1
1
=− 2 − ·
− ·
.
+ 1)(s − 1)
s
2 s−1 2 s+1
169
1
s2
·
Hence,
t ∗ e−t ∗ et = −t +
Problem 47.10
et e−t
−
2
2
n f unctions
z
}|
{
Suppose it is known that h(t) ∗ h(t) ∗ · · · ∗ h(t) = Ct8 . Determine the constants C and the poisitive integer n.
Solution.
n f unctions
}|
{
z
We know that L[h(t) ∗ h(t) ∗ · · · ∗ h(t)] = (L[h(t)])n =
tn−1
= Ct8 . It follows that n = 9 and C = 8!1
(n−1)!
1
sn
so that L−1 [ s1n =
Problem 47.11
Use Laplace transform to solve for y(t) :
Z t
sin (t − λ)y(λ)dλ = t2 .
0
Solution.
Note that the given equation reduces to sin t ∗ y(t) = t2 . Taking Laplace
2
transform of both sides we find sY2(s)
= s23 . This implies Y (s) = 2(ss3+1) =
+1
2
+ s23 . Hence, y(t) = L−1 [ 2s + s23 ] = 2 + t2
s
Problem 47.12
Use Laplace transform to solve for y(t) :
Z t
y(t) −
e(t−λ) y(λ)dλ = t.
0
Solution.
Note that the given equation reduces to et ∗ y(t) = y(t) − t. Taking Laplace
(s)
transform of both sides we find Ys−1
= Y (s) − s12 . Solving for Y (s) we find
Y (s) = s2s−1
. Using partial fractions decomposition we can write
(s−2)
1
1
− 41
s−1
2
4
=
+
+
.
2
2
s (s − 2)
s
s
(s − 2)
Hence,
1 t 1
y(t) = − + + e2t
4 2 4
170
Problem 47.13
Use Laplace transform to solve for y(t) :
t ∗ y(t) = t2 (1 − e−t ).
Solution.
Taking Laplace transform of both sides we find
Y (s) =
2
s
−
2s2
.
(s+1)3
Y (s)
s2
=
2
2
− (s+1)
3.
s3
This implies
Using partial fractions decomposition we can write
1
2
1
s2
=
−
+
.
(s + 1)3
s + 1 (s + 1)2 (s + 1)3
Hence,
−t
y(t) = 2 − 2(e
− 2te
−t
t2 −t
t2 −t
+ e = 2 1 − (1 − 2t + )e
2
2
Problem 47.14
Use Laplace transform to solve for y(t) :
0
y = h(t) ∗ y, y(0) =
1
2
.
Solution.
Taking Laplace transform of both sides we find sY − y(0) = 1s Y. Solving for
Y we find
1
1
s
1
1
1
Y(s) = 2
+
=
.
2
s −1 2
2 s−1 s+1
Hence,
1
y(t) = (et + e−t )
2
1
2
= cosh t
1
2
Problem 47.15
Solve the following initial value problem.
Z t
0
y −y =
(t − λ)eλ dλ, y(0) = −1.
0
171
Solution.
Note that y 0 − y = t ∗ et . Taking Lalplace transform of both sides we find
1
1
1
. This implies that Y (s) = − s−1
+ s2 (s−1)
sY − (−1) − Y = s12 · s−1
2 . Using
partial fractions decomposition we can write
s2 (s
1
1
1
2
2
+
= + 2−
.
2
− 1)
s s
s − 1 (s − 1)2
Thus,
Y (s) = −
2
1
2
1
2
1
3
1
1
+ + 2−
+
=
+
−
+
.
s−1 s s
s − 1 (s − 1)2
s s2 s − 1 (s − 1)2
Finally,
y(t) = 2 + t − 3et + tet
172
48
47
49
The Dirac Delta Function and Impulse Response
Problem 49.1
Evaluate
R3
(a) 0 (1 + e−t )δ(t − 2)dt.
R1
(b) −2 (1 + e−t )δ(t − 2)dt.
R2
cos 2t
(c) −1
δ(t)dt.
te−t 

δ(t + 2)
R 2 2t
(d) −1 (e + t)  δ(t − 1)  dt.
δ(t − 3)
Solution.
R3
(a) 0 (1 + e−t )δ(t − 2)dt = 1 + e−2 .
R1
(b) −2 (1 + e−t )δ(t − 2)dt = 0 since 2 lies outside the integration interval.
(c)
#
" R
2
R2
cos
2tδ(t)dt
cos 2t
R−12 −t
δ(t)dt =
−1
te−t
te δ(t)dt
−1
cos 0
=
0 × t0
1
=
0
(d)


δ(t
+
2)
R 2 2t
(e + t)  δ(t − 1)  dt
−1
δ(t − 3)
 R2

2t
(e
+
t)δ(t
+
2)dt
 R−1

2
=  −1 (e2t + t)δ(t − 1)dt 
R 2 2t
(e + t)δ(t− 3)dt
−1
e−4 − 2
 e2 + 1 
=
0
173
Problem 49.2
Let f (t) be a function defined and continuous on 0 ≤ t < ∞. Determine
Z t
f (t − s)δ(s)ds.
(f ∗ δ)(t) =
0
Solution.
Let g(s) = f (t − s). Then
(f ∗ δ)(t) =
=
Rt
0
Rt
f (t − s)δ(s)ds = 0 g(s)δ(s)ds
g(0) = f (t)
Problem 49.3
R1
Determine a value of the constant t0 such that 0 sin2 [π(t − t0 )]δ(t− 21 )dt = 34 .
Solution.
We have
R1
3
sin2 [π(t
− t0 )]δ(t
− 21 )dt =
4
3
sin2 π ( 12 − t0
=
4√
sin π ( 12 − t0
= ± 23
Thus, a possible value is when π ( 12 − t0 = π3 . Solving for t0 we find t0 =
0
1
6
Problem
49.4
R5 n
If 1 t δ(t − 2)dt = 8, what is the exponent n?
Solution.R
5
We have 1 tn δ(t − 2)dt = 2n = 8. Thus, n = 3
Problem 49.5
RtRs
Sketch the graph of the function g(t) which is defined by g(t) = 0 0 δ(u −
1)duds, 0 ≤ t < ∞.
Solution.
Rs
Note first that 0 δ(u − 1)du = 1 if s > 1 and 0 otherwise. Hence,
g(t) =
Rt
1
0,
if t ≤ 1
h(s − 1)ds = t − 1, if t > 1
174
Problem 49.6
Rt
The graph of the function g(t) = 0 eαt δ(t − t0 )dt, 0 ≤ t < ∞ is shown.
Determine the constants α and t0 .
Solution.
Note that
g(t) =
0,
0 ≤ t ≤ t0
eαt0 , t0 < t < ∞
It follows that t0 = 2 and α = −1
Problem 49.7
(a) Use the method of integarting factor to solve the initial value problem
y 0 − y = h(t), y(0) = 0.
(b) Use the Laplace transform to solve the initial value problem φ0 − φ =
δ(t), φ(0) = 0.
(c) Evaluate the convolution φ ∗ h(t) and compare the resulting function with
the solution obtained in part(a).
175
Solution.
(a) Using the method of integrating factor we find, for t ≥ 0,
y0 − y
(e−t y)0
e−t y
y
y
=
h(t)
=
e−t
−t
= −e + C
= −1 + Cet
= −1 + et
(b) Taking Laplace of both sides we find sΦ − Φ = 1 or Φ(s) =
φ(t) = et .
(c) We have
Z t
Z t
(t−s)
(φ ∗ h)(t) =
e
h(s)ds =
e(t−s) ds = −1 + et
0
1
.
s−1
Thus,
0
Problem 49.8
Solve the initial value problem
y 0 + y = 2 + δ(t − 1), y(0) = 0, 0 ≤ t ≤ 6.
Graph the solution on the indicated interval.
Solution.
Taking Laplace of both sides to obtain sY + Y =
2
e−s
2
e−s
+ s+1
= 2s − s+1
+ s+1
. Hence,
s(s+1)
y(t) =
2
s
2 − 2e−t ,
t<1
−t
2 + (e − 2)e , t ≥ 1
176
+ e−s . Thus, Y (s) =
Problem 49.9
Solve the initial value problem
y 00 = δ(t − 1) − δ(t − 3), y(0) = 0, y 0 (0) = 0, 0 ≤ t ≤ 6.
Graph the solution on the indicated interval.
Solution. Taking Laplace of both sides to obtain s2 Y = e−s − e−3s . Thus,
−3s
−s
Y (s) = es2 − e s2 . Hence,
y(t) = (t − 1)h(t − 1) − (t − 3)h(t − 3).
Problem 49.10
Solve the initial value problem
y 00 − 2y 0 = δ(t − 1), y(0) = 1, y 0 (0) = 0, 0 ≤ t ≤ 2.
Graph the solution on the indicated interval.
Solution.
Taking Laplace transform of both sides and using the initial conditions we
find
s2 Y − s − 2(sY − 1) = e−s .
Solving for s we find Y (s) =
1
s
+
e−s
s(s−2)
=
1
s
−
e−s
2s
+
e−s
.
s−2
1
1
y(t) = 1 − h(t − 1) + e2(t−1) h(t − 1)
2
2
177
Hence,
Problem 49.11
Solve the initial value problem
y 00 + 2y 0 + y = δ(t − 2), y(0) = 0, y 0 (0) = 1, 0 ≤ t ≤ 6.
Graph the solution on the indicated interval.
Solution.
Taking Laplace transform of both sides to obtain s2 Y − 1 + 2sY + Y = e−2s .
e−2s
1
−t
+ (t −
Solving for Y (s) we find Y (s) = (s+1)
2 + (s+1)2 . Therefore, y(t) = te
2)e−(t−2) h(t − 2)
178
49
Solving Systems of Differential Equations
Using Laplace Transform
Problem 49.1
Find L[y(t)] where

e−t cos 2t
d 

0
y(t) =
dt
t
t+e

Solution.


 −e−t cos 2t + 2e−t sin 2t 

0
L[y(t)] = L 


1 + et


s+1
4
− (s+1)
2 +4 + (s+1)2 +4


=
0
1
1
+
 s 3−s s−1 
(s+1)2 +4
=

1
s

0
1
+ s−1
Problem 49.2
Find L[y(t)] where
t
Z
y(t) =
0


1
 u  du
e−u
Solution.


t


t2

L[y(t)] = L 
2


−t
−e
+1

1
=

1
s
179
s2
1
s3
−
1
s+1

Problem 49.3
Find L−1 [Y(s)] where

Y(s) = 
Solution.
We have
1
s
2
2
s +2s+2
1
s2 +s


L−1 [ 1s ]
=
1
2
2
−1
−1
L [ s2 +2s+2 ] = L [ (s+1)2 +12 ] = 2e−t sin t
1
L−1 [ s21+s ] = L−1 [ 1s − s+1
] = 1 − e−t
Thus,

1
L−1 [Y(s)] =  2e−t sin t 
1 − e−t

Problem 49.4
Find L−1 [Y(s)] where



1 −1 2
L[t3 ]
Y(s) =  2 0 3   L[e2t ] 
1 −2 1
L[sin t]
Solution.
We have

Y(s) =


1 −1 2
L[t3 ]
 2 0 3   L[e2t ] 
1 −2 1
L[sin t]

  3 
t

 1 −1 2
= L  2 0 3   e2t 


sin t
1 −2 1
=
 3

 t − e2t + 2 sin t 
L  2t3 + 3 sin t 


t3 − 2e2t + sin t
Thus,


t3 − e2t + 2 sin t
L−1 [Y(s)] =  2t3 + 3 sin t 
t3 − 2e2t + sin t
180
Problem 49.5
Use the Laplace transform to solve the initial value problem
5 −4
0
0
0
y =
y+
, y(0) =
5 −4
1
0
Solution.
Taking Laplace of both sides and using the initial condition we find
5 −4
sY =
Y + 0 1s
5 −4
Solving this matrix equation for Y we find
"
#
4
− s2 (s−1)
Y(s) =
.
s−5
s2 (s−1)
Using partial fractions decomposition we find
−
and
4
4
4
4
= 2+ −
− 1)
s
s s−1
s2 (s
s−5
5
4
4
= 2+ −
.
− 1)
s
s s−1
s2 (s
Hence,
Y(s) =
4
s2
5
s2
+ 4s −
+ 4s −
4
s−1
4
s−1
.
Finally,
y(t) =
4t + 4 − 4et
5t + 4 − 4et
Problem 49.6
Use the Laplace transform to solve the initial value problem
5 −4
3
0
y =
y, y(0) =
3 −2
2
181
Solution.
Taking Laplace of both sides and using the initial condition we find
3
5 −4
sY −
=
Y
2
3 −2
Solving this matrix equation for Y we find
3
s + 2 −4
1
Y(s) = (s−1)(s−2)
3
s−5
2
1
(s−1)(s−2)
=
3s − 2
2s − 1
Using partial fractions decomposition we find
−1
4
3s − 2
=
+
(s − 1)(s − 2)
s−1 s−2
and
2s − 1
−1
3
=
+
.
(s − 1)(s − 2)
s−1 s−2
Hence,
−1
s−1
−1
s−1
Y(s) =
+
+
4
s−2
3
s−2
.
Finally,
y(t) =
−et + 4e2t
−et + 3e2t
Problem 49.7
Use the Laplace transform to solve the initial value problem
1 4
0
3
0
y =
y+
, y(0) =
t
−1 1
3e
0
Solution.
Taking Laplace of both sides and using the initial condition we find
0
3
1 4
sY −
=
Y+
3
0
−1 1
s−1
s − 1 −4
1
s−1
Y =
182
3
3
s−1
Solving this matrix equation for Y we find
3
s−1
4
1
Y(s) = (s−1)2 +4
3
−1 s − 1
s−1
=
1
(s−1)2 +4
3(s − 1) +
0
12
s−1
Using partial fractions decomposition we find
12
3(s − 1) + s−1
s−1
12
=3
+
.
2
2
(s − 1) + 4
(s − 1) + 4 (s − 1)[(s − 1)2 + 4]
But
12
3
s−1
=
−
3
.
(s − 1)[(s − 1)2 + 4]
s−1
(s − 1)2 + 4
Hence,
3
s−1
Y(s) =
0
.
Finally,
y(t) =
3et
0
Problem 49.8
Use the Laplace transform to solve the initial value problem
−3 −2
1
0
00
0
y =
y, y(0) =
, y (0) =
4
3
0
1
Solution.
Taking Laplace of both sides and using the initial condition we find
1
0
−3 −2
2
s Y−s
−
=
Y
0
1
4
3
s2 + 3
2
−4 s2 − 3
Y
183
=
s
1
Solving this matrix equation for Y we find
2
s − 3 −2
s
1
Y(s) = s4 −1
2
4
s +3
1
1
s4 −1
=
s3 − 3s − 2
s2 + 4s + 3
Using partial fractions decomposition we find
s3 − 3s − 2
1
s
1
=
−
+
2
+
s4 − 1
s−1
s2 + 1 s2 + 1
and
s2 + 4s + 3
s+3
2
s
1
=
=
− 2
− 2
.
2
2
(s − 1)(s + 1)(s + 1)
(s − 1)(s + 1)
s−1 s +1 s +1
Hence,
Y(s) =
1
− s−1
+ 2 s2s+1 + s21+1
2
− s2s+1 − s21+1
s−1
.
Finally,
y(t) =
−et + 2 cos t + sin t
2et − cos t − sin t
Problem 49.9
Use the Laplace transform to solve the initial value problem
1 −1
2
0
0
00
0
y =
y+
, y(0) =
, y (0) =
1 −1
1
1
0
Solution.
Taking Laplace of both sides and using the initial condition we find
2 0
1 −1
2
s Y−s
=
Y + 1s
1
1 −1
s
s2 − 1
1
−1 s2 + 1
Y =
184
1
s
2
s
+s
Solving this matrix equation for Y we find
2
s + 1 −1
1
Y(s) = s4
1
s2 − 1
1
s3
1
s
=
+ s15
+ s15
3
t2
2
+ t4!
4
1 + t4!
y(t) =
Problem 49.10
Use the Laplace transform to solve the



1 0 0
y0 =  0 −1 1  y + 
0 0 2
1
s
2
s
+s
initial value problem

 
et
0


1
, y(0) = 0 
−2t
0
Solution.
Taking Laplace of both sides and using the initial condition we find


 1 
1 0 0
s−1
sY
=  0 −1 1  Y +  1s 
0 0 2
− s22


s−1
0
0
 0
s + 1 −1  Y =
0
0
s−2


1
s−1
1
s
− s22


Solving this matrix equation for Y we find
 1
 1 
0
0
s−1
s−1
1
1
1



Y(s) =  0 s+1
(s+1)(s−2)
s
2
1
− s2
0
0
s−2

=


1
(s−1)2
s(s−2)−2
s2 (s+1)(s−2)
−2
s2 (s−2)
185



Using partial fractions decomposition we find
s(s − 2) − 2
1
1
1
1
1
1
= 2+
− ·
− ·
+ 1)(s − 2)
s
2s 3 s + 1 6 s − 2
s2 (s
and
−2
1
1
1
1
=
−
·
.
+
s2 (s − 2)
s2 2s 2 s − 2
Hence,

Y(s) = 
1
s2
+
1
(s−1)2
1
1
1
− 13 · s+1
− 16 · s−2
2s
1
1
1
+ 2s
− 12 · s−2
s2

.
Finally,

tet
y(t) =  t + 21 − 13 e−t − 61 e2t 
t + 12 − 12 e2t

Problem 49.11
0
The Laplace transform
was
=
applied to the initial value problem y = Ay,
y(0)
y1 (t)
y1,0
.
y0 , where y(t) =
, A is a 2 × 2 constant matrix, and y0 =
y2,0
y2 (t)
The following transform domain solution was obtained
1
s − 2 −1
y1,0
L[y(t)] = Y(s) = 2
.
4
s−7
y2,0
s − 9s + 18
(a) what are the eigenvalues of A?
(b) Find A.
Solution.
(a) det(sI − A) = s2 − 9s + 18 = (s − 3)(s − 6) = 0. Hence, the eigenvalues
of A are r1 = 3 and r2 = 6.
(b) Taking Laplace transform of both sides of the differential equation we
find
sY − y0 =
AY
Y
= (sI − A)−1 y0
Letting s = 0 we find
Y(0) = −A
−1
1
=
18
186
−2 −1
4 −7
Hence,
A
−1
1
=
18
2 1
−4 7
and
det(A−1 ) =
It follows that
−1 −1
A = (A )
=
187
1
.
18
7 −1
4 2
50
Numerical Methods for Solving First Order Linear Systems: Euler’s Method
In Problems 50.1 - 50.3 answer the following questions:
(a) Solve the differential equation analytically using the appropriate method
of solution.
(b) Write the Euler’s iterates: yk+1 = yk + hf (tk , yk ).
(c) Using step size h = 0.1, compute the Euler approximations yk , k = 1, 2, 3
at times tk = a + kh.
(d) For k = 1, 2, 3 compute the error y(tk ) − yk where y(tk ) is the exact value
of y at tk .
Problem 50.1
y 0 = 2t − 1, y(1) = 0.
Solution.
(a) y = t2 − t
(b) yk+1 = yk + h(2tk − 1)
(c)
y1 =y0 + 0.1(2t0 − 1) = 0 + 0.1(2(1) − 1) = 0.1
y2 =y1 + 0.1(2t1 − 1) = 0.1 + 0.1(2(1.1) − 1) = 0.22
y3 =y2 + 0.1(2t2 − 1) = 0.22 + 0.1(2(1.2) − 1) = 0.36
(d)
y1err =0.11 − 0.1 = 0.01
y2err =0.24 − 0.22 = 0.02
y3err =0.39 − 0.36 = 0.03
Problem 50.2
y 0 = −ty, y(0) = 1.
Solution.
t2
(a) Using the method of integrating factor we find y = Ce− 2 . Since y(0) = 1,
t2
we find c = 1 so that y = e− 2 .
188
(b) yk+1 = yk − h(tk yk )
(c)
y1 =y0 − 0.1(t0 y0 ) = 1 − 0.1(0 × 1) = 1
y2 =y1 − 0.1(t1 y1 ) = 1 − 0.1(0.1)(1) = 0.99
y3 =y2 − 0.1(t2 y2 ) = 0.99 − 0.1(0.2 × 0.99) = 0.9702
(d)
y1err =0.99501 − 1 = −0.00499
y2err =0.98109 − 0.99 = −0.00891
y3err =0.95599 − 0.9702 = −0.01421
Problem 50.3
y 0 = y 2 , y(0) = 1.
Solution.
(a) Using the method of integrating factor we find − y1 = t+C. Since y(0) = 1,
1
we find c = −1 so that y = 1−t
.
(b) yk+1 = yk + hyk2
(c)
y1 =y0 + 0.1y02 = 1 + 0.1(12 ) = 1.1
y2 =y1 + 0.1y12 = 1.1 + 0.1(1.1)2 = 1.221
y3 =y2 + 0.1y22 = 1.221 + 0.1(1.221)2 = 1.370084
(d)
y1err =1.1111 − 1.1 = 0.0111
y2err =1.25 − 1.221 = 0.029
y3err =1.4286 − 1.370084 = 0.058516
In Problems 50.4 - 50.6 answer the following questions:
(a) Write the Euler’s method algorithm in explicit form. Specify the starting
values t0 and y0 .
(b) Give a formula for the kth t−value, tk . What is the range of the index k
if we choose h = 0.01?
(c) Use a calculator to carry out two steps of Euler’s method, finding y1 and
y2 .
189
Problem 50.4
2
−t2 t
1
0
y =
y+
, y(1) =
, 1 ≤ t ≤ 4.
2−t 0
t
0
Solution.
(a)
yk+1 = yk + h
−t2k tk
2 − tk 0
where t0 = 1 and
y0 =
2
0
yk +
1
tk
(b) We have tk = 1 + kh. Now, h = b−a
→ 0.01 = N3 → N = 300. Thus, the
N
range of the index k is 0 ≤ k ≤ 300.
(c) We have
1
1.99
−t20 t0
1
2
−1 1
2
y1 =y0 + h
y0 +
=
+ 0.01
+
=
2 − t0 0
tk
0
1 0
0
1
0.03
1
1.99
−(1.01)2 1.01
1.99
1
−t21 t1
y1 +
=
+ 0.01
+
y2 =y1 + h
2 − t1 0
t1
0.03
0.99
0
0.03
1.01
Problem 50.5


 
 
1 0 1
0
1
0





y = 3 2 1 y + 2 , y(−1) = 0  ,
1 2 0
t
1
− 1 ≤ t ≤ 0.
Solution.
(a)

yk+1
where t0 = 1 and



1 0 1
0
= yk + h  3 2 0  yk +  2 
1 2 0
tk


1
y0 =  0 
1
190
→ 0.01 = N1 → N = 100. Thus,
(b) We have tk = −1 + kh. Now, h = b−a
N
the range of the index k is 0 ≤ k ≤ 100.
(c) We have



  

 
 
1 0 1
0
1
1 0 1
1
0
=
y1 =y0 + h  3 2 0  y0 +  2  =  0  + 0.01  3 2 0   0  +
2 −1
1 2 0
t0
1
1 2 0
1



 



 

1 0 1
0
1.02
1 0 1
1.02
0
y2 =y1 + h  3 2 0  y1 +  2  =  0.06  + 0.01  3 2 0   0.06  +  2 
1 2 0
t1
1.00
1 2 0
1.00
−0.99
Problem 50.6
0
y =
1
t
sin t
1−t 1
y+
0
t2
1
tk
, y(1) =
0
0
, 1 ≤ t ≤ 6.
Solution.
(a)
yk+1 = yk + h
1 − tk
where t0 = 1 and
y0 =
sin tk
1
0
0
yk +
0
t2k
(b) We have tk = 1 + kh. Now, h = b−a
→ 0.01 = N5 → N = 500. Thus, the
N
range of the index k is 0 ≤ k ≤ 500.
(c) We have
1
sin t0
0
0
1 sin 1
0
0
0
t
0
y1 =y0 + h
y0 +
=
+ 0.01
+
=
2
t0
0
0
1
0
1
0.01
1 − t0
1
1
1
sin t1
sin 1.01
0
0
0
t1
1.01
y2 =y1 + h
y1 +
=
+ 0.01
+
0.01
1
0.01
(1.
t21
0.01
1 − t1
1
In Problems 50.7 - 50.8 answer the following questions.
(a) Rewrite the given initial value problem as an equivalent initial value
problem for a first order system, using the substitution z1 = y, z2 = y 0 , z3 =
y 00 , · · · .
(b) Write the Euler’s method algorithm zk+1 = zk + h[P (tk )zk + g(tk )], in
explicit form. Specify the starting values t0 and z0 .
191
(c) Using a calculator with step size h = 0.01, carry out two steps of Euler’s
method, finding z1 and z2 What are the corresponding numerical approximations to the solution y(t) at times t = 0.01 and t = 0.02?
Problem 50.7
y 00 + y 0 + t2 y = 2, y(1) = 1, y 0 (1) = 1.
Solution.
(a) Let z1 = y and z2 = y 0 so that z10 = z2 and z20 = y 00 . Thus,
z10 =z2
z20 =y 00 = 2 − t2 y − y 0 = 2 − t2 z1 − z10
= − z2 − t2 z1 + 2
Thus,
z2
=
Z (t) =
−t2 z1 − z2 + 2
0
1
z1
0
=
+
2
−t −1
z2
2
0
z1
z2
with
Z(1) =
1
1
.
(b) We have
Zk+1 = Zk + h
0
1
2
−tk −1
with t0 = 1 and
Z0 =
192
1
1
Zk +
0
2
(c) We have
0
1
Z1 =Z0 + h
Z0 +
−t20 −1
1
0
1
=
+ 0.01
1
−1 −1
1.01
=
1
0
1
Z2 =Z1 + h
Z1 +
−t21 −1
1.01
0
=
+ 0.01
1
−(1.01)2
1.02
=
0.99969
0
2
1
1
0
2
+
1
−1
0
2
1.01
1
+
0
2
Problem 50.8
y 000 + 2y 0 + ty = t + 1, y(0) = 1, y 0 (0) = −1, y 00 (0) = 0.
Solution.
(a) Let z1 = y, z2 = y 0 and z3 = y 00 . Then
z10 =z2
z20 =y 00 = z3
z30 =y 000 = t + 1 − ty − 2y 0 = t + 1 − tz1 − 2z2
Thus,


z1
z
2
0
Z (t) =  z2  =
−t2 z1 − z2 + 2
z3
0
1
z1
0
=
+
−t2 −1
z2
2
with
Z(1) =
193
1
1
.
(b) We have
Zk+1 = Zk + h
0
1
2
−tk −1
with t0 = 1 and
Zk +
1
1
0
1
Z0 +
Z1 =Z0 + h
−t20 −1
1
0
1
=
+ 0.01
1
−1 −1
1.01
=
1
0
1
Z1 +
Z2 =Z1 + h
−t21 −1
1.01
0
=
+ 0.01
1
−(1.01)2
1.02
=
0.99969
0
2
1
1
0
2
Z0 =
0
2
(c) We have
194
+
1
−1
0
2
1.01
1
+
0
2
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