# 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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