Solutions Manual to Introduction to Diffential Equations With Dynamical Systems

Solutions Manual to

Introduction to Differential Equations with Dynamical Systems

by Stephen L. Campbell and Richard Haberman

M. Ziaul Haque

PRINCETON UNIVERSITY PRESS PRINCETON AND OXFORD

c 2008 by Princeton University Press Copyright � Published by Princeton University Press 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press 6 Oxford Street, Woodstock, Oxfordshire, 0X20 1TW All Rights Reserved This book has been composed in LATEX press.princeton.edu

Contents

Preface Chapter 1. First-Order Differential Equations and Their Applications 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Introduction to Ordinary Differential Equations Definite Integral and the Initial Value Problem First-Order Separable Differential Equations Direction Fields Euler’s Numerical Method (Optional) First-Order Linear Differential Equations Linear First-Order Differential Equations with Constant Coeffi­

cients and Constant Input 1.8 Growth and Decay Problems 1.9 Mixture Problems 1.10 Electronic Circuits 1.11 Mechanics II: Including Air Resistance 1.12 Orthogonal Trajectories (optional) Chapter 2. Linear Second and Higher-Order Differenial Equations 2.1 2.2 2.3 2.4

General Solution of Second-Order Linear Differential Equations Initial Value Problem (For Homogeneous Equation) Reduction of Order Homogeneous Linear Constant Coefficient Differential Equations

(Second Order) 2.5 Mechanical Vibrations I: Formulation and Free Response 2.6 The Method of Undetermined Coefficients 2.7 Mechanical Vibrations II: Forced Response 2.8 Linear Electric Circuits 2.9 Euler Equation 2.10 Variation of Parameters (Second-Order) 2.11 Variation of Parameters (nth-Order)

Chapter 3. The Laplace Transform 3.1 3.2 3.3 3.4 3.5 3.6 3.7

Definition and Basic Properties Inverse Laplace Transforms (Roots, Quadratics, & Partial Fractions) Initial-Value Problems for Differential Equations Discontinuous Forcing Functions Periodic Functions Integrals and the Convolution Theorem Impulses and Distributions

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23

25

26

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29

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30

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45

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86

94

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109

114

118

iv

CONTENTS

Chapter 4. An Introduction to Linear Systems of Differential Equations and

Their Phase Plane 4.1 4.2 4.3

Introduction Introduction to Linear Systems of Differential Equations Phase Plane for Linear Systems of Differential Equations

Chapter 5. Mostly Nonlinear First-Order Differential Equations 5.1 5.2 5.3 5.4

First-Order Differential Equations Equilibria and Stability One Dimensional Phase Lines Application to Population Dynamics: The Logistic Equation

Chapter 6. Nonlinear Systems of Differential Equations in the Plane 6.1 Introduction 6.2 Equilibria of Nonlinear Systems, Linear Stability Analysis of Equi­

librium, and Phase Plane 6.3 Population Models 6.4 Mechanical Systems

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Preface This Student Solutions Manual contains solutions to the odd-numbered ex­ ercises in the text Introduction to Differential Equations with Dynamical Systems by Stephen L. Campbell and Richard Haberman. To master the concepts in a mathematics text the students must solve prob­ lems which sometimes may be challenging. This manual has been written focusing student’s needs and expectations. Instead of providing only the answer with very few steps, I include a reasonably detailed solution with a fair amount of detail when explaining the solution of the problem. The solutions are self-explanatory and consistent with the notations and termi­ nologies used in the text book. I hope this manual will help students build problem-solving skills. I would like to thank many people who have provided invaluable help, in many ways, in the preparation of this manual. First, I take this opportunity to thank Professor Richard Haberman for his generous expert help, construc­ tive comments and accuracy checking. I would also like to thank Professor Stephen L. Campbell for assembling the final manuscript, Professor Peter K. Moore for facilitating support process and Ms. Vickie Kearn of the pub­ lishing company for her patience and support. Finally, I must appreciate the patience of my wife, Rukshana, and my daughters, Zareen and Ehram for their understanding and compromise of summer time that was slighted because of my busy schedule. M. Ziaul Haque

Southern Methodist University

Dallas, TX, 75275, U.S.A.

July, 2007.

Chapter One

First-Order Differential Equations and Their Applications

1.1 INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS There are no exercises in this section.

1.2 DEFINITE INTEGRAL AND THE INITIAL VALUE PROBLEM 1-7. Substitute expression for x into the differential equation 3t 1. x = 2e3t + 1. l.h.s. = dx dt = 6e . 3t r.h.s. = 3x − 3 = 3(2e + 1) − 3 = 6e3t . Hence l.h.s. = r.h.s. x t−1 3. x = t − 1. l.h.s. = dx dt = 1. r.h.s = t−1 = t−1 = 1. Hence l.h.s. = r.h.s. 2

2

2

t t 5. x = et . l.h.s. = dx dt = 2te . r.h.s = 2tx = 2te . Hence l.h.s. = r.h.s. dx −2t −2t 7. x = e . l.h.s. = dt = −2e . r.h.s. = −2e2t x2 = −2e2t (e−2t )2 = −2e−2t . Hence l.h.s. = r.h.s. t t 9. dx dt = 3e . Integrating we get, x = 3e + c. dx 11. dt = −5 cos 6t. Integrating we get, x = − 56 sin 6t + c. Rt −1 − 12 13. dx ). Use of definite integral gives x = 8 0 cos t 2 dt + c. dt = 8 cos(t 2 15. dx dt =R ln(4 + cos t). Use of definite integral gives t x = 0 ln(4 + cos2 t)dt + c

dx 5 17. dt = t4 ; x(2) = 3. Integrating we get x = 1

5 t + c.

32 17 1 5 t = 2 =⇒ 3 = 5 + c =⇒ c = − 5 . So x = 5 t − 17

5 . ln t dx 19. dt = 4+cos2 t ; x(2) = 5. Use of definite integral gives R t ln t

x = 5 + 2 4+cos 2 t dt. et 1+t ;

t

21.

dx dt

e x(1) = 3. dx = 1+t dt. Use of definite integral gives R t e t R t et x − 3 = 1 1+t dt =⇒ x = 3 + 1 1+t

dt.

23.

d2 x

dt2 = −15. Integrating we get

dx dx

dt = −15t + c1 ( dt = v0 at t = 0 =⇒ c1 = v0 .)

dx dt = −15t + v0 . Integrating again we get

2 x = − 15 2 t + v0 t + c2 (x = 0 at t = 0 =⇒ c2 = 0.)

=

2

CHAPTER 1 v0 Car stops when dx dt = 0 =⇒ v0 − 15t = 0 =⇒ t = 15 (stopping time). So distance travelled is 2 √ 2 v02 v0 2 1 v0 1 v0 x = − 15 ( ) + = 2 15 15 2 15 =⇒ 75 = 2 15 =⇒ v0 = 15 10 m/sec. 2 dx 25. ddt2x = −2500. Integrating we get dx dt = −2500t + c1 ( dt = 60 at t = 0 dx =⇒ c1 = 60). So dt = −2500t + 60. Integrating again we get 2 x = − 2500 2 t + 60t + c2 (x = 0 at t = 0 =⇒ c2 = 0.) Car stops when dx dt = 0 =⇒ −2500t + 60 = 0 60 =⇒ t = 2500 (stopping time). So distance travelled is 60 2 602 x = − 2500 2 ( 2500 ) + 2500 = 0.72 km. 2 dx 27. ddt2x = −2500. Integrating we get dx dt = −2500t + c1 ( dt = v0 at t = 0 =⇒ c1 = v0 ). So dx dt = −2500t + v0 . Integrating again we get 2 t + v t + c2 (x = 0 at t = 0 =⇒ c2 = 0) x = − 2500 0 2 Car stops when dx dt = 0 =⇒ v0 − 2500t = 0 v0 =⇒ t = 2500 (stopping time). So distance travelled is v2

v2

v0 2 0 0 x = − 2500 2 ( 2500 ) + 2500 = 5000 km. 2 dx 2 29. ddt2x = −6t. Integrating we get dx dt = −3t + c1 ( dt = 50 at t = 2 dx 2 =⇒ c1 = 62). So dt = −3t + 62. Integrating again we get x = −t3 + 62t + c2 (x = 0 at t = 2 =⇒ c2 = −116.) 2 Car stops when dx dt = 0 =⇒ −3t + 62 = 0 q =⇒ t = 62 3 (stopping time). So distance travelled is q 62 x = t(62 − t2 ) − 116 = 62 3 (62 − 3 ) − 116 q ¡ 62 ¢ 32 2 − 116 km. = 62 3 ( 3 )62 − 116 = 2 3

dy 3 31. (a) V =volume, dV dt = Q m /h. Let snow depth be y. So dt = c ⇒ y = ct + c1 (y = 0 at t = 0 ⇒ c1 = 0). Thus y = ct. Now consider the snowplow has moved Δx over the time Δt and the approximate change in volume over this time is ΔV. Hence Δx ΔV = w(Δx)y = wctΔx ⇒ ΔV Δt = wct Δt . Now taking limit Q dx 1 dV as Δt → 0 we get dt = Q = wct dt ⇒ dx dt = wct = kt with wc k= Q. R R 1 dx = k1 1t dt ⇒ (b) dx dt = kt . Separating the variables we get, x = k1 ln t + a. At 11 A.M. t = 3 and x(3) = 0. So 0 = k1 ln 3 + a ⇒ a = − k1 ln 3. Then at noon (t = 4), x(4) = k1 ln 4 − k1 ln 3 = k1 ln 34 .

33.

d2 y 2 dt2 = −g = −9.8 m/sec . Integrating we get dy dy dt = −9.8t + c1 ( dt = v0 at t = 0 =⇒ c1 = v0 .) So dy dt = −9.8t + v0 . Integrating again we get y = − 92.8 t2 + v0 t + c2 (y = 0 at t = 0 =⇒ c2 = 0.) At maximum height dy dt = 0 =⇒ v0 − 9.8t = 0 v0 =⇒ t = 9.8 (time at maximum height). So maximum height is √ 2 2 v02 v0 2 1 v0 1 v0 1960 m/sec. y = − 9.8 2 ( 9.8 ) + 9.8 = 2 9.8 =⇒ 100 = 2 9.8 =⇒ v0 =

3

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 2

d y 2 dt2 = −g = −9.8 m/sec . Integrating we get dy dy dt = −9.8t + c1 ( dt = 0 at t = 0 =⇒ c1 = 0.) 9.8 2 So dy dt = −9.8t. Integrating again we get y = − 2 t + c2 9.8 2 (y = 200 at t = 0 =⇒ c2 = 200). y = − 2 t + 200. Now to

35.

2 − 9.8 2 t

q

fall,

400 9.8

sec. = 200 =⇒ t = = √20 9.8 t R ¡¢ 37. Since x(t0 ) = x0 , the general solution x = f t dt + c becomes y = 0. So 0 =

+ 200 ⇒

9.8 2 2 t

0

Rt0 ¡ ¢ Rt0 ¡ ¢ x0 = f t dt + c =⇒ c = x0 − f t dt. Hence the solution is 0

x=

Rt 0

0

¡¢ Rt ¡ ¢ Rt0 ¡ ¢ f t dt + x0 − f t dt = x0 + f t dt. t0

0

1.3 FIRST-ORDER SEPARABLE DIFFERENTIAL EQUATIONS R = dt . Integrating we get, ¯ ¯t |x+1| x+1 ¯ ¯ ln |x + 1| = ln |t| + c1 ⇒ ln |t| = c1 ⇒ t = ec1 c1 ⇒ x+1 = c ⇒ x = ct − 1. t = ±e R R dx t 3. dt = e . Separating variables gives dx = et dt. Integrating we get, x = et + c. 5. dx = tx + 4x + 3t + 12 = (x + 3) (t + 4) . Separating variables gives dt

R R dx 2

(t + 4) dt. Integrating we get, ln |x + 3| = t2 + 4t + c1 x+3 =

1.

dx dt

=

x+1 t .

Separating variables gives

t2

R

dx x+1

t2

⇒ |x + 3| = ec1 e 2 +4t ⇒ x = ce 2 R+4t − 3 R where c = ±ec1 . dx 7. dt = 3. Separating variables gives dx = 3dt. Integrating we get, x = 3t + c. R −5 R 5 9. dx x dx = dt. Integrating we get, dt = x . Separating variables gives −4 − x 4 = t + c. Using x(2) = 1 we get, − 14 = 2 + c ⇒ c = − 94 . −1/4 Substituting c we get x−4 = 9 − 4t ⇒ x = (9 − 4t) . 2 2 −2 dx = cos(t2 )dt. Using 11. dx dt = x cos(t ). Separating variables gives x Rx Rt 2 definite integrals we get, x−2 dx = cos(t )dt 1

⇒

1 − x +

⇒ x =

1=

Rt

0 1

0

2

cos(t )dt ⇒

1 x

Rt 2 1− cos(t )dt

=1−

Rt

2

cos(t )dt

0

0

13.

dx dt

= t cos(x−1/2 ). Separating variables gives Rx Rt definite integrals we get, cos(xdx−1/2 ) = tdt ⇒

15.

du dt

Rx 2

=

2

dx cos(x−1/2 ) t2 +1 u2 +4 .

=

t2 2

−

1 2

=

1 2

¡

¢

t2 − 1 .

Separating variables gives

dx cos(x−1/2 )

= tdt. Using

1

R¡

¢ ¢ R¡2 u2 + 4 du = t + 1 dt.

4

CHAPTER 1 3

3

Integrating we get, u3 + 4u = t3 + t + c. Using u(0) = 1 we get, 13 1 3 + 4 = c ⇒ c = 3 . Substituting c we obtain the solution as 3 3 u + 12u = t + 3t + 13. ¡ ¢¡ ¢ 17. dx = t2 x2 + x2 + t2 + 1 = x2 + 1 t2 + 1 . Separating variables gives dt ¢ R dx R¡2 3 −1 (x) = t3 + t + c x2 +1 = ³ t + 1 dt.´ Integrating we get, tan 3 ⇒ x = tan t3 + t + c . Using x(0) = 2 we get, c = tan−1 (2). Hence ³ 3 ´ the solution is x = tan t3 + t + tan−1 (2) . R dx R 19. dx dt. Using dt = x (x − 1) . Separating variables gives x(x−1) = partial fractions to the integral on the left we get, A B 1 x(x−1) = x + x−1 ⇒ 1 = A (x − 1) + Bx. Putting x = 0 and 1, respectively, have, A = −1 R we R and B = 1. Hence R dx R dx dx = − = dt ⇒ ln |x − 1| − ln |x| = t + c x(x−1) x ¯ x−1 ¯ x−1 x−1 ¯ ¯ ⇒ ln x = t + c ⇒ x = ket where k = ±ec . Solving this 1 equation for x we obtain the general solution, x = 1−ke t . Since x = 0 and x = 1 both satisfy the differential equation they are also solutions. The solution x = 1 corresponds to k = 0, however, x = 0 is not included in the general solution for any finite k. 1 Hence the solutions are x = 1−ke t and x = 0. R R 2 dx dx 21. dt = (x − 1) (x − 2) . Separating variables gives (x−1)(x−2) dt. 2 = Using partial fractions to the integral on the left we get, A B C

1 = x−1 + x−2 + (x−2) 2 ⇒

(x−1)(x−2)2 2

1 = A (x − 2) + B (x − 1) (x − 2) + C (x − 1) . Putting x = 1 and 2, respectively, we have, A = 1 and C = 1. Then equating 2 the of 0⇒B R xdx we have R = −1. Hence R coefficients R dx A +RB = dx dx − + = dt. Integrating both 2 = 2 x−1 x−2 (x−2) (x−1)(x−2)

1 sides we obtain the solution, ln |x − 1| − ln |x − 2| − x−2 = t + c. Since x = 1 and x = 2 both satisfy the differential equation they are also solutions. Hence the solutions are 1 = t + c, x = 1 and x = 2. ln |x − 1| − ln |x − 2| − x−2 23. ¡(tx + x) dt + (tx + t) dx = 0. Dividing by tx we get,

¢ ¡ ¢ 1 + 1t dt + 1 + x1 dx = 0. Now integrating we have,

t + ln |t| + x + ln |x| = c ⇒ ln |tx | = −t − x + c ⇒ c −t −x t x c tx ¡ 2= ±e¢ e e ¡ 2⇒ te¢ xe = c where c = ±e¡ 2. ¢¡ ¢ 25. t − 4 dz + z − 9 dt = 0. Dividing by t − 4 z 2 − 9 we get, (z2dz−9) + (t2dt = 0. Now we use the formula ¯ −4) ¯ R du ¯ u−a ¯ 1 u2 −a2 = 2a ln ¯ u+a ¯ (from integration table) to integrate and ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 ¯ t−2 ¯ ¯ z−3 ¯1/6 ¯ t−2 ¯1/4 get 61 ln ¯ z−3 + ln = c ⇒ ln =c⇒ ¯ ¯ ¯ ¯ ¯ ¯ z+3 4 z+3 t+2 ¯ ¯³ ¯ t+2 ³ ´1/6 ³ ´1/4 ¯ z−3 ´1/6 ³ t−2 ´1/4 ¯ t−2 ¯ ¯ = ec ⇒ z−3 = ±ec ⇒ t+2 z+3 t+2 ¯ z+3 ¯ ³ ³ ³ ´1 ´− 41 ´1 ´3 ³ t+2 4 t+2 2 z−3 6 = ±ec t−2 = ±ec t−2 ⇒ z−3 z+3 t+2 z+3 = c t−2

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

5

6

where c = (±ec ) . 27. et+x dt + e2t−3x dx = 0. Dividing by ex e2t we get,

−4x e−t dt + e−4x dx = 0. Now integrating we have, −e−t − e 4 = c

−4x −t −t ⇒e = −4e − 4c ⇒ −4x = ln (−4e − 4c) ⇒ x = − 41 ln (−4e−t + k) where k = −4c. 29. z = at + bx + c. Differentiating with respect to t we get,

dz dx dz dt = a + b dt ⇒ dt = a + bf (z) which is a differential equation

in and t and R z dz R can be solved by separation of variables as dt. = a+bf (z) 2

2 = (t + 4x − 1) . Let z = t + 4x − 1. Then dx dt = z = f (z) dx dz 2 and dt = 1 + 4 dt = 1 + 4z . Separating the variables we get, R R dz dt. We use the substitution u = 2z ⇒ dz = 12 du 1+4z 2 = R R du to integrate the left hand side. This gives 12 1+u dt ⇒ 2 = 1 1 −1 −1 (2z) = t + c ⇒ 2 tan (2t + 8x − 2) = t + c. 2 tan −1 t+x z −1 33. dx = e (t + x) − 1. Let z = t + x. Then dx −1 dt dt = e z dx dz z −1 z −1 1 + e z − 1 = e z . Separating the and dt = 1 + dt = R R variables we get, ze−z dz = dt. We use integration by parts to integrate the left Rhand side asR u = z ⇒ du = = e−z dz. R dz and dv −z −z −z ⇒Rv = −e . Then ze dz = udv = uv − vdu = −ze + e−z dz = −ze−z − e−z . This gives −e−z (z + 1) = t + c. Substituting z = t + x we get the solution as −e−(t+x) (t + x + 1) = t + c ⇒ e−t−x (t + x + 1) = −t + c.

31.

dx dt

1.4 DIRECTION FIELDS 1. 2

x 0

−2

0

t

2

6

CHAPTER 1

3. 2

x 0

−2

0

t

2

5. 2

x

0

−2

2

0 t

1.4.1 Existence and Uniqueness x 1 = 1+t 2 = f (t, x) and fx = 1+t2 are continuous for all (t, x) . So unique solution exists for all (t0 , x0 ) .

¡ ¢ ¡ ¢4/3

2 2 7/3 3. dx are = f (t, x) and fx = 73 1 − t2 − x2 dt = 1 − t − x continuous for all (t, x) . So unique solution exists for all (t0 , x0 ) . 1/5 1 5. dx = f (t, x) is continuous for all (t, x) but fx = 5(x+t) 4/5 dt = (x + t) is not continuous for x + t = 0.So unique solution exists for all (t0 , x0 ) such that x0 + t0 6= 0. cos t − cos t 7. dx dt = x−1 = f (t, x) and fx = (x−1)2 are not continuous at x = 1. So unique solution exists for all (t0 , x0 ) such that x0 6= 1. ¡ ¢ ¡ ¢1/2 2 2 3/2 9. dx = f (t, x) and fx = −6x 1 − t2 − 2x2 dt = 1 − t − 2x are continuous for 1 − t2 − 2x2 > 0. So unique solution exists for all (t0 , x0 ) such that t20 + 2x20 < 1. dx 1 11. dt = t1/3 = f (t, x) is not continuous at t = 0 although fx = 0 is continuous everywhere. So unique solution exists for all (t0 , x0 ) such that t0 = 6 0. 13. (a) Differentiating t2 + x2 = c wrt(with respect to) t we get,

1.

dx dt

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

7

dx t 2 2 2t + 2x dx dt = 0 ⇒ dt = − x . Hence t + x = c defines a solution.

(b) graph (c) At x0 = 0, as f (t, x) = − xt and fx = xt2 are discontinuous at this point. 15. (a) Differentiating t + x2 = c wrt t we get, 1 + 2x dx dt = 0.

Hence t + x2 = c defines a solution.

(b) graph 1 1 (c) Here dx dt = − 2x = f (t, x) and fx = 2x2 are discontinuous at x0 = 0. Hence the theorem fails to hold at this point. 17. (a) Differentiating x = c sin t wrt t we get, dx x cos t dt = c cos t = c x cos t = cx c sin t = x cot t. Hence x = c sin t defines a solution. (b) graph cos t cos t (c) Here dx dt = x sin t = f (t, x) and fx = sin t are discontinuous when sin t = 0 ⇒ t = nπ for n = 0, ±1, ±2, ...Hence the theorem fails to hold at the point t0 = nπ for n = 0, ±1, ±2, ... 1 1 2 19. (a) Differentiating x = t+c wrt t we get, dx dt = − (t+c)2 = −x . 1 defines a solution. Hence x = t+c (b) graph 2 (c) Here dx dt = −x = f (t, x) and fx = −2x are continuous everywhere. Hence there is NO point where the theorem fails to holds. 1/5 21. (a) For x = 1, l.h.s. is dx = 0. dt = 0 and r.h.s. = (x − 1) ¢5/4 ¢1/4 ¡ ¡4 4 + 1, l.h.s. is dx = t + c and For x = 5 t + c dt 5 ³¡ ´1/5 ¡ ¢ ¢1/4 5/4 1/5 r.h.s. = (x − 1) = 45 t + c +1−1 = 54 t + c . dx (b) By separating the variables we get, (x−1) 1/5 = dt which, on ¢5/4 ¡4 + 1. Then using the integration, becomes x = 5 t + c initial condition x0 = 1 for any t0 we get c = − 45 t0 and thus ¢5/4 ¡ + 1. Another solution is one solution is x = 54 t − 45 t0 clearly x = 1 because it satisfies the initial condition as well as the differential equation. So there are at least two solutions through the point (t0 , x0 ) with x0 = 1. (c) Graph 1/5 (d) Although f (t, x) = (x − 1) is continuous everywhere, 1 fx = (x−1)4/5 is not continuous at x0 = 1. As a result, uniqueness does not hold and two solutions in part (b) is not a surprise.

1.5 EULER’S NUMERICAL METHOD (OPTIONAL) 1.

dx dt

= x − t = f (t, x) , t0 = 0, x0 = 1, h = 0.5. We use recursive formula,

8

CHAPTER 1

xn+1 = xn + hf (tn , xn ) = xn + 0.5 (xn − tn ), where tn+1 = tn + h to approximate x1 = x0 + h (x0 − t0 ) = 1 + 0.5(1 − 0) = 1.5 at t1 = 0.5 x2 = x1 + h (x1 − t1 ) = 1.5 + 0.5(1.5 − 0.5) = 2 at t2 = 1 x3 = x2 + h (x2 − t2 ) = 2 + 0.5(2 − 1) = 2.5 at t3 = 1.5 x4 = x3 + h (x3 − t3 ) = 2.5 + 0.5(2.5 − 1.5) = 3 at t4 = 2 So estimate for x(2) is x4 = 3. 2 3. dx dt = −tx = f (t, x) , t0 = 0, x0 = 1, h = 1. We use recursive formula, xn+1 = xn + hf (tn , xn ) = xn − tn x2n , where tn+1 = tn + h to approximate

x1 = x0 − ht0 x20 = 1 − 0 = 1 at t1 = 1

x2 = x1 − ht1 x21 = 1 − 1 = 0 at t2 = 2

So estimate for x(2) is x2 = 0.

5. dx dt = 2x − 4t = f (t, x) , t0 = 0, x0 = 1, h = 0.5. We use recursive formula, xn+1 = xn + hf (tn , xn ) = xn + 0.5 (2xn − 4tn ) = 2 (xn − tn ) , where tn+1 = tn + h to approximate x1 = 2 (x0 − t0 ) = 2(1 − 0) = 2 at t1 = 0.5 x2 = 2 (x1 − t1 ) = 2(2 − 0.5) = 3 at t2 = 1 x3 = 2 (x2 − t2 ) = 2(3 − 1) = 4 at t3 = 1.5 x4 = 2 (x3 − t3 ) = 2(4 − 1.5) = 5 at t4 = 2 So estimate for x(2) is x4 = 5. 7. dx dt = sin x = f (t, x) , t0 = 0, x0 = 0, h = 0.5. We use recursive formula, xn+1 = xn + hf (tn , xn ) = xn + 0.5 (sin xn ) , where tn+1 = tn + h to approximate x1 = x0 + 0.5 sin x0 = 0 at t1 = 0.5 x2 = x1 + 0.5 sin x1 = 0 at t2 = 1 Similarly, xi = 0 at ti for all i = 0, 1, ..., 8. So estimate for x(4) is x8 = 0. 9. (a) dx dt = −20x = f (t, x) , t0 = 0, x0 = 1, h = 0.2. We use recursive formula, xn+1 = xn + hf (tn , xn ) = xn + 0.2 (−20xn ) = −3xn . n Here xn = (−3) at tn = 0.2n, n = 0, 1, 2, ... 10 So estimate for x(2) is x10 = (−3) = 59049. (b) xn oscillates wildly as n → ∞. x(2) = e−40 = 4.248 × 10−18 . So x10 = 59049 is not a very good approximation. 11. (a) dx dt = −20x = f (t, x) , t0 = 0, x0 = 1, h = 0.01. We use recursive formula, xn+1 = xn + hf (tn , xn ) = xn + 0.01 (−20xn ) = 0.8xn . n Here xn = (0.8) at tn = 0.01n, n = 0, 1, 2, ... 200 So estimate for x(2) is x200 = (0.8) = 4.1495 × 10−20 . (b) In this case, xn → 0 as n → ∞, so the numerical solution behaves like the actual solution x = e−20t and the statement x200 ≈ x(2) = e−40 = 4.248 × 10−18 is not too bad.

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS dx dt

= x2 = f (t, x) , t0 = 0, x0 = 1, h = 0.2. We use recursive formula, xn+1 = xn + hf (tn , xn ) = xn + 0.2x2n , where tn+1 = tn + h to approximate x1 = x0 + 0.2x20 = 1 + 0.2 = 1.2 at t1 = 0.2 x2 = x1 + 0.2x21 = 1.2 + 0.288 = 1.488 at t2 = 0.4 x3 = x2 + 0.2x22 = 1.488 + 0.443 = 1.9308 at t3 = 0.6 x4 = x3 + 0.2x23 = 1.9308 + 0.7456 = 2.6764 at t4 = 0.8 x5 = x4 + 0.2x24 = 2.6764 + 1.4326 = 4.109 at t5 = 1 So estimate for x(1) is x5 = 4.109. For h = 0.1 use the same formula and procedure as above. There are 11 points this time, that is 10 steps after initial step and the estimates for x(1) is x10 = 6.1289. For h = 0.01 the estimates for x(1) is 30.3897 using software and for h = 0.001 the estimates for x(1) is 193.1368 using software. 2 15. dx dt = 1 − 2x + x = f (t, x) , t0 = 0, x0 = −5, h = 0.2. We use recursive formula,

¡ ¢

xn+1 = xn + hf (tn , xn ) = xn + 0.2 1 − 2xn + x2n 2 = xn + 0.2 (xn − 1) , where tn+1 = tn + h to approximate

2 x1 = x0 + 0.2 (x0 − 1) = −5 + 7.2 = 2.2 at t1 = 0.2

2 x2 = x1 + 0.2 (x1 − 1) = 2.488 at t2 = 0.4

2 x3 = x2 + 0.2 (x2 − 1) = 2.931 at t3 = 0.6

2 x4 = x3 + 0.2 (x3 − 1) = 3.676 at t4 = 0.8

2 x5 = x4 + 0.2 (x4 − 1) = 5.109 at t5 = 1

2 x6 = x5 + 0.2 (x5 − 1) = 8.486 at t6 = 1.2

2 x7 = x6 + 0.2 (x6 − 1) = 19.694 at t7 = 1.4

2 x8 = x7 + 0.2 (x7 − 1) = 89.591 at t8 = 1.6

2 x9 = x8 + 0.2 (x8 − 1) = 1659.265 at t9 = 1.8

2 x10 = x9 + 0.2 (x9 − 1) = 551627.57 at t10 = 2 So estimate for x(2) is x10 = 551627.57 and it seems to tend to infinity. 2 17. dx dt = 1 − 2x + x = f (t, x) , t0 = 0, x0 = −5, h = 0.1. We use recursive formula, ¢ ¡ xn+1 = xn + hf (tn , xn ) = xn + 0.1 1 − 2xn + x2

n 2

= xn + 0.1 (xn − 1) , where tn+1 = tn + h to approximate 2

x1 = x0 + 0.1 (x0 − 1) = −1.4 at t1 = 0.1 2

x2 = x1 + 0.1 (x1 − 1) = −0.824 at t2 = 0.2 2

x3 = x2 + 0.1 (x2 − 1) = −0.4913 at t3 = 0.3 2

x4 = x3 + 0.1 (x3 − 1) = −0.2689 at t4 = 0.4 2

x5 = x4 + 0.1 (x4 − 1) = −0.1079 at t5 = 0.5 2

x6 = x5 + 0.1 (x5 − 1) = 0.0148 at t6 = 0.6 2

x7 = x6 + 0.1 (x6 − 1) = 0.1119 at t7 = 0.7 2

x8 = x7 + 0.1 (x7 − 1) = 0.1908 at t8 = 0.8 13.

9

10

CHAPTER 1

2

x9 = x8 + 0.1 (x8 − 1) = 0.2563 at t9 = 0.9

2

x10 = x9 + 0.1 (x9 − 1) = 0.3116 at t10 = 1

2

x11 = x10 + 0.1 (x10 − 1) = 0.359 at t11 = 1.1

2

x12 = x11 + 0.1 (x11 − 1) = 0.4 at t12 = 1.2

2

x13 = x12 + 0.1 (x12 − 1) = 0.436 at t13 = 1.3

2

x14 = x13 + 0.1 (x13 − 1) = 0.4678 at t14 = 1.4

2

x15 = x14 + 0.1 (x14 − 1) = 0.4961 at t15 = 1.5

2

x16 = x15 + 0.1 (x15 − 1) = 0.5215 at t16 = 1.6

2

x17 = x16 + 0.1 (x16 − 1) = 0.5444 at t17 = 1.7

2

x18 = x17 + 0.1 (x17 − 1) = 0.5652 at t18 = 1.8

2

x19 = x18 + 0.1 (x18 − 1) = 0.5841 at t19 = 1.9

2

x20 = x19 + 0.1 (x19 − 1) = 0.6014 at t20 = 2

So estimate for x(2) is x20 = 0.6014.

1.6 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 1.6.1 Form of the General Solution There are no exercises in this subsection. 1.6.2 Solutions of Homogeneous First-Order Linear Differential Equations R dx R 1. dx dt = 3x. By separation we obtain x = 3 dt. Integration yields ln |x| = 3t + c1 ⇒ |x| = ec1 e3t ⇒ x = ce3t , where c = ±ec1 . R dx R dx 3. dt = 2tx. By separation we obtain x = 2tdt. Integration yields 2 2 ln |x| = t2 + c1 ⇒ |x| = ec1 et ⇒ x = ceRt , where c R= ±ec1 . dx 1 dt 5. 2t dx dt + x = 0. By separation we obtain ¯ x = ¯ − 2 t . Integration ¯ 1/2 ¯ 1

1/2 ¯ ¯ ¯ ¯ yields ln |x| ¯ 1/2 ¯ =c− 2 ln |t|1/2+ c1 ⇒cln |x| + ln t−1/2 = c1 ⇒ ln xt c = c1 ⇒ ¯xt ¯ = e 1 ⇒ xt = ±e 1 ⇒ x = ct where c = ±e 1 . R dx R 7. dx dt + (cos t) x = 0. By separation we obtain x = − cos tdt. Integration yields ln |x| = − sin t + c1 ⇒ |x| = ec1 e− sin t − sin t ⇒x= c = ±ec1 . ¡ ce −1/2, where ¢ dx ) x = 0. By separation we obtain 9. dt + cos(t R dx R − 1

2 )dt. Definite integration yields

= − cos(t x µ t

¶ R Rt 1 1

ln |x| = − cos(s− 2 )ds + c1 ⇒ |x| = ec1 exp − cos(s− 2 )ds 0 0 µ ¶ Rt 1

⇒ x = c exp − cos(s− 2 )ds , where c = ±ec1 . 0 R R −5x. By separation we obtain dx 11. dx = dt x = −5 dt. Integration yields ln |x| = −5t + c1 ⇒ x = ±ec1 e−5t = ce−5t . Using the initial condition x(0) = 9 we get, c = 9 and is x = 9e−5t . R dxthe solution R dx 13. dt = 9x. By separation we obtain x = 9 dt. Integration

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

11

yields ln |x| = 9t + c1 ⇒ x = ±ec1 e9t = ce9t . Using the initial condition x(3) = 7 we get, 7 = ce27 ⇒ c = 7e−27 and the solution is x = 7e9t−27 = 7e9(t−3) . R dx R sin t

dx sin t 15. dt + 4+e t x = 0. By separation we obtain x =− 4+et dt. Rt sin s Definite integration yields ln |x| = − 4+es ds + c1 ⇒ 5 ¶ µ t R sin s x = c exp − 4+es ds . Using the initial condition x(5) = 10 5 ¶ µ t R sin s ds . we get, c = 10 and the solution is x = 10 exp − 4+e s 5R R 1 −2 x = 0. By separation we obtain dx 17. dx dt + t x =− t2 dt. Integration yields ln |x| = 1t + c1 ⇒ x = ±ec1 e1/t = ce1/t . Using the initial condition x(1) = 3 we get, 3 = ce 1 ⇒ c = 3e−1 and the solution is x = 3e( t −1) . 1.6.3 Integrating Factors for First-Order Linear Differential Equa­ tions t

x e 1 19. t dx x = et ⇒ dx dt + dt + t = t , p(t) = t . The integrating factor is R e p(t)dt = eln|t| = t. Multiplying by integrating factor we get d t t t t dx dt + x = e ⇒ dt (xt) = e . Integration yields xt = e + c. Using the initial condition x(1) = 1 we get, 1 = e + c t ⇒ c = 1 − e. Hence the solution is xt = et + 1 − e ⇒ x = e +1−e . t R R t t t = 3e . By integration we obtain dx = 3 e dt ⇒ x = 3e + c. 21. dx ¡dt ¢ dx 2t 1 2t 23. t2 + 1 dx dt + 2tx = 1 ⇒ dt + (t2 +1) x = (t2 +1) , p(t) = (t2 +1) . R

R

2t

dt

p(t)dt The integrating factor = e ¯ (t2 +1) ¯ . Using the substitution R e2t 2 u = t + 1 we have (t2 +1) dt = ln ¯t2 + 1¯ and then the integrating ¯ ¯ 2 factor is eln|t +1| = ¯t2 + 1¯ = t2 + 1 (since positive for real t).

Multiplying by integrating ¡factor get ¡2 ¡ 2 we ¢¢ ¢ d x t¡ + 1 ¢dx + 2tx = 1 ⇒ t + 1 = 1. Integration yields dt dt t + t2 c+1 . x t2 + 1 = t + c ⇒ x = t2 +1 R p(t)dt 25. dx = e4t .

dt + 4x = t, p(t) = 4. The integrating factor is e Multiplying by integrating¡ factor ¢ we get

d 4t 4t 4t = te4t . Integration yields e4t dx dt +R4e x = te ⇒ dt xe xe4t = te4t dt. We use integration by parts on the r.h.s. as u = t ⇒R du = dt and Rdv = e4t dt ⇒ v = R41 e4t . Then 1 4t r.h.s.= udv = uv − vdu = 41 te4t − 14 e4t dt = 41 te4t − 16 e . 1 1 1 1 So xe4t = 4 te4t − 16 e4t + c ⇒ x = 4 t − 16 + ce−4t . 1 Using the initial condition x(0) = 0 we have, c = 16 .

1 1 1 −4t Hence the solution is x = 4 t − 16 + 16 e . 2 2 = 2x ⇒ dx 27. t dx dt dt R− t x = 0, p(t) = − t . The integrating factor is R p(t)dt −2 1t dt −2 ln t −2 e =e =e =t . Multiplying by integrating factor we get

12

CHAPTER 1

¡x¢

2 d t−2 dx dt − t3 x = 0 ⇒ dt t2 = 0. Integration yields x t2 = c. Using the initial condition x(1) = 4 we have, c = 4. Hence the solution is x = 4t2 . This can also be done by separation. 1 29. t2 dx 1 ⇒ dx x = t−2 , p(t) = 1t . The integrating dt + tx = dt + R R t1 factor is e p(t)dt = e t dt = eln t = t. Multiplying by integrating factor we get 1 d 1 t dx dt + x = t ⇒ dt (xt) = t . Integration yields −1 xt = ln t + c ⇒ x = t ln t + ct−1 . dx 3 3 31. t dx dt + 3x =R t ⇒ dt + tRx = 1, p(t) = t . The integrating 1 factor is e p(t)dt = e3 t dt = e3 ln t = t3 . Multiplying by integrating we get ¡ 3 ¢factor d 2 3 3 t3 dx + 3t x = t ⇒ t x = t . Integration yields dt dt 4 t3 x = t4 + c ⇒ x = 41 t + ct−3 .

One solution, 14 t, is continuous at (0, 0) . All other solutions

approaches ±∞ as t → 0.

dx 1 2 33. t dx = t, p(t) = − 1t . The integrating dt − x = tR ⇒ dt − t x R 1 factor is e p(t)dt = e− t dt = e− ln t = t−1 . Multiplying by integrating we get ¡ −1factor ¢ d −2 t−1 dx + t x = 1 ⇒ t x = 1. Integration yields dt dt t−1 x = t + c ⇒ x = t2 + ct. All solutions are continuous and pass through (0, 0) .

x

t

35.

dx dtR

37.

dx dtR

+ 2tx = 1, p(t) = 2t. The integrating factor is R 2 e p(t)dt = e 2tdt = et . Multiplying factor ³ by ´integrating 2 2 d t dx t t2 t2 t2 we get e dt + 2te x = e ⇒ dt e x = e . Definite t Rt 2 2 2 R 2 2 integration yields et x = es ds + c ⇒ x = e−t es ds + ce−t . 0

0

+ t2 x = t, p(t) = t2 . The integrating factor is 1 3 e p(t)dt = e 3 t . Multiplying by integrating factor we get

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

13

´ 1 3 1 3 1 3 d 2 3t ⇒ 3t x e3 = te 3 t . dt + t x = te dt e Rt s3 1 3 Definite integration yields e 3 t x = se 3 ds + c 1 3 ¡ dx t

³

¢

0

⇒ x=e 39.

−t3 3

Rt

se

s3 3

ds + ce

−t3 3

.

0

dx dtR

+ et x = 3, p(t) = et . The integrating factor is t e p(t)dt = ee . Multiplying ³ by integrating factor we get ´ ¢ t ¡ d dx t et et et e e dt + e x = 3e ⇒ dt e x = 3e . Rt s t Definite integration yields ee x = 3 ee ds + c 0

−et

⇒ x = 3e

41.

Rt

es

−et

e ds + ce

.

0

dx dtR

1 + x = t+1 , x(2) = 1, p(t) = 1. The integrating factor is p(t)dt t e ¡ = ¢e . Multiplying by integrating factor we get et d et + x = t+1 ⇒ dt (et x) = t+1 . Definite integration yields et dx dt t R s e et x = s+1 ds + c. Using initial condition we get,

2

c = e2 and the solution is x = e−t

Rt

2

es s+1 ds

+ e2−t .

dx 1 1 1 43. 3t dx dt − x = t sin t ⇒ dt − 3tR x = 3 sin t, x(5) = 0, p(t) = − 3t . 1 The integrating factor is e p(t)dt =¡ e− 3 ln t =¢ t−1/3 . Multiplying 1 1 −1/3 by integrating factor we get t−1/3 dx sin t dt − 3t x = 3 t ¡ −1/3 ¢ 1 −1/3 d ⇒ dt t x = 3t sin t. Definite integration yields Rt t−1/3 x = 31 s−1/3 sin sds + c. Using initial condition we get, 5

c = 0 and the solution is x = 13 t1/3 t

Rt

s−1/3 sin sds.

5

dx 1 1e 1

t 45. 7t dx dt + x =R e ⇒ dt + 7t x = 7 t , p(t) = 7t . The integrating 1 factor is e ¡p(t)dt = e 7¢ln t = t1/7 . Multiplying factor ¡ 1/7 ¢ byeintegrating t d 1 et x = 7t6/7 . Definite we get t1/7 dx dt + 7t x = 7t6/7 ⇒ dt t Rt integration yields t1/7 x = 17 s−6/7 es ds + c ⇒

0

x=

47.

1 −1/7 7t

Rt

0

−6/7 s

s

e ds + ct−1/7 .

dt dt

1 ⇒ dx = x + t ⇒ dx − t = x, p(x) = −1. The integrating = x+t R p(x)dx −x factor¡ e by integrating factor we get ¢ = e−x . Multiplying dt d −x e−x dx − t = xe ⇒ (te ) = xe−x . Integration yields dx R −x −x te = xe dx + c. We use integration by parts on the r.h.s. as −x −x ⇒v= . Then u=x⇒ R du = dx andR dv = e dx −x R −e −x r.h.s.= udv = uv − vdu = −xe + e dx = −xe−x − e−x . So te−x = −xe−x − e−x + c ⇒ t = −x − 1 + cex .

dx dt

14

CHAPTER 1 R

49. Let u = e p(t)dt be the integrating factor of dx dt + p(t)xR = f (t). If we introduce an arbitrary constant Rc1 while we compute p(t)dtR then R new integrating factor is u1 = e p(t)dt+c1 = ec1 e p(t)dt = c2 e p(t)dt where c2 = ec1 > 0. Multiplying equation by this new ¡ the differential ¢ integrating factor we get, u1 dx + p(t)x = u f 1 (t) ⇒ dt d = u1 f (t) which, on Rintegration, becomes dt (u1 x) R R R u1 x = u1 f (t)dt + c3 ⇒ c2 e p(t)dt x = c e p(t)dt f (t)dt + c3 . R R 2 R R p(t)dt p(t)dt − p(t)dt Dividing by c e we get x = e f (t)dt e 2 R c3 c3 − p(t)dt . Now writing c2 = c we have the same general solution + c2 e R R R R x = e− p(t)dt f (t)e p(t)dt dt + ce− p(t)dt as (23) . R 51. Let u = e p(t)dt be the integrating factor of dx + p(t)x = f (t). So ¡ dxdt ¢ d dx du dx (ux) = u + x = u + xp(t)u = u + dt dt dt dt dt ¡ xp(t) . Now ¢ d

d since k is a constant dt (kux) = k dt (ux) = ku dx dt + xp(t) which implies that ku is also an integrating factor of (19) as the d (kux) of l.h.s. of (19) and ku is easily integrable with product dt respect to t. £ sin t ¤ sin t x = q(t) ⇒ dx 53. dx dt + t dt + p(t)x = q(t) where p(t) =R t . Multiplying this equation byR the integrating factor e p(t)dt we get R ¡ ¢ p(t)dt dx = q(t)e p(t)dt e ³ dt + p(t)x ´ d dt

R

R

= q(t)e p(t)dt . Definite integral yields R Rt ¡ ¢ R e p(t)dt x(t) = c + q t e p(t)dt dt. Particular solution with ⇒

xe

p(t)dt

0

xp (0) = 0 gives c = 0. Then xp = e−

p(t)dt

Rt ¡ ¢ R q t e

p(t)dt

0

R ¡¢ Rt R Rt ¡ ¢ ¡ ¢ xp = e p(t)dt e− p(t)dt q t dt = G t, t q t dt where R R R 0

R ¡ 0¢ G t, t = e p(t)dt e− p(t)dt = e p(t)dt− p(t)dt .

Using definite à integrals yields ! à ! ¡ ¢ Rt Rt Rt G t, t = exp p(s)ds − p(s)ds = exp p(s)ds t 0 à !0 à ! Rt sin s Rt = exp = exp − sins s ds . s ds t

55.

R

dx dt

dt ⇒

t

+ p(t)x = f (t). From the change of variable Rx = u(t)x1 (t) dx1 du − p(t)dt we have dx ,

dt = dt x1 + u dt . But since x1 = e dx1 dx du dt = −x1 p(t). Thus dt = dt x1 − ux1 p. Plugging in the original equation we get, du dt xR1 − ux1 p + ux1 p = f (t) ⇒ R du du − p(t)dt = f (t) ⇒ dt = e p(t)dtR f (t). Integrating we get, dt e R R R R u = f (t)e p(t)dt dt. Then x = e− p(t)dt f (t)e p(t)dt dt. Comparison: From³the original the integrating factor is ´ equation, R R R d p(t)dt p(t)dt p(t)dt x = f (t)e . Integrating we get, e and so dt e

15

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS R

e

p(t)dt

x=

R

R

f (t)e

p(t)dt

dt ⇒ x = e−

R

R p(t)dt

R

f (t)e

p(t)dt

dt.

1.7 LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS AND CONSTANT INPUT 1.

dx dt −

3. 5. 7. 9. 11.

13.

15.

17.

19.

21.

23.

8x = 0. Substituting x = ert we get, rert − 8ert = 0. Dividing by ert 6= 0, we have, r = 8. Thus the general solution is x = ce8t . dx rt we get, rert = −2ert . Dividing dt = −2x. Substituting x = e rt by e 6= 0, we have, r = −2. Thus the general solution is x = ce−2t . dx rt we get, rert + 7ert = 0. Dividing dt + 7x = 0. Substituting x = e rt by e 6= 0, we have, r = −7. Thus the general solution is x = ce−7t . dx rt we get, rert + ert = 0. Dividing dt + x = 0. Substituting x = e rt by e 6= 0, we have, r = −1. Thus the general solution is x = ce−t . dx rt we get, rert = 5ert . Dividing dt = 5x. Substituting x = e rt by e 6= 0, we have, r = 5. Thus the general solution is x = ce5t . dx 8 dt + 3x = 8. Substituting xp = A we get, 3A = 8 ⇒ A = 3 . Thus 8 xp = 3 . Let the associated homogeneous solution be x1 = ert . Then rert + 3ert = 0. Dividing by ert 6= 0, we have, r = −3. Thus the general solution is x = 38 + ce−3t . 9 dx dt − 4x = −9. Substituting xp = A we get, −4A = −9 ⇒ A = 4 . 9 Thus xp = 4 . Let the associated homogeneous solution be x1 = ert . Then rert − 4ert = 0. Dividing by ert 6= 0, we have, r = 4. Thus the general solution is x = 94 + ce4t . 4 15 dx 4 dt − 5 x = 3. Substituting xp = A we get, − 5 A = 3 ⇒ A = − 4 . 15 Thus xp = − 4 . Let the associated homogeneous solution be x1 = ert . 4 rt Then rert − 5 e = 0. Dividing by ert 6= 0, we have, r = 54 . 4 5t Thus the general solution is x = − 15 4 + ce . dx 2 4 2 4 dt + 3 x = − 3 . Substituting xp = A we get, 3 A = − 3 ⇒ A = −2. Thus xp = −2. Let the associated homogeneous solution be x1 = ert . Then rert + 23 ert = 0. Dividing by ert 6= 0, we have, r = − 23 . 2 Thus the general solution is x = −2 + ce− 3 t . dx dt = 2x + 18. Substituting xp = A we get, 2A + 18 = 0 ⇒ A = −9. Thus xp = −9. Let the associated homogeneous solution be x1 = ert . Then rert = 2ert . Dividing by ert 6= 0, we have, r = 2.

Thus the general solution is x = −9 + ce2t . dx 17 17 17 dt = −x − 3 . Substituting xp = A we get, −A − 3 = 0 ⇒ A = − 3 . rt Thus xp = − 17 . Let the associated homogeneous solution be x = e .

1 3 Then rert = −ert . Dividing by ert 6= 0, we have, r = −1.

−t Thus the general solution is x = − 17 3 + ce .

dx −4t −4t . Substituting xp = Ae we get, dt + 7x = 8e −4Ae−4t + 7Ae−4t = 8e−4t . Dividing by e−4t 6 = 0, we have, ⇒ −4A + 7A = 8 ⇒ A = 83 . Thus xp = 83 e−4t . Let the associated homogeneous solution be x1 = ert .

16

CHAPTER 1

25.

27.

29.

31.

33.

35.

37.

39.

41.

Then rert + 7ert = 0. Dividing by ert 6= 0, we have, r = −7.

Thus the general solution is x = 83 e−4t + ce−7t .

dx −5t . Substituting xp = Ae−5t we get,

dt − 2x = −3e −5t −5t −5Ae − 2Ae = −3e−5t . −5t Dividing by e 6= 0, we have, ⇒ −5A − 2A = −3 ⇒ A = 73 . Thus 3 −5t xp = 7 e . Let the associated homogeneous solution be x1 = ert .

Then rert − 2ert = 0. Dividing by ert 6= 0, we have, r = 2.

Thus the general solution is x = 73 e−5t + ce2t .

dx 4t 4t we get, 4Ae4t + 4Ae4t = 3e4t .

dt + 4x = 3e . Substituting xp = Ae Dividing by e4t 6= 0, we have, ⇒ 4A + 4A = 3 ⇒ A = 83 . Thus xp = 38 e4t . Let the associated homogeneous solution be x1 = ert .

Then rert + 4ert = 0. Dividing by ert 6= 0, we have, r = −4.

Thus the general solution is x = 83 e4t + ce−4t .

dx −2t . Substituting xp = Ae−2t we get,

dt + 3x = 3e −2t −2Ae + 3Ae−2t = e−2t . Dividing by e−2t 6= 0, we have, ⇒ −2A + 3A = 1 ⇒ A = 1. Thus xp = e−2t . Let the associated homogeneous solution be x1 = ert . Then rert + 3ert = 0. Dividing by ert 6= 0, we have, r = −3. Thus the general solution is x = e−2t + ce−3t . dx dt +7x = 3+5t. Substituting xp = At+B we get, A+7At+7B = 3+5t. Equating the coefficients of t and 1 we have 7A = 5 ⇒ A = 57 and 16 A + 7B = 3 ⇒ 7B = 3 − 75 = 16 7 ⇒ B = 49 5 16 Thus xp = 7 t + 49 . dx dt + 2x = 14t. Substituting xp = At + B we get, A + 2At + 2B = 14t. Equating the coefficients of t and 1 we have 2A = 14 ⇒ A = 7 and A + 2B = 0 ⇒ 2B = −A = −7 ⇒ B = − 27 . Thus xp = 7t − 27 .

dx 2 2 dt + x = 2t + 5t − 8. Substituting xp = At + Bt + C we get,

2 2 2At + B + At + Bt + C = 2t + 5t − 8.

Equating the coefficients of t2 , t and 1 we have A = 2,

2A+B = 5 ⇒ B = 5−2A = 1, and B +C = −8 ⇒ C = −8−B = −9

Thus xp = 2t2 + t − 9.

dx 3 3 2 dt + 3x = t . Substituting xp = At + Bt + Ct + D we get,

2 3 2 3At + 2Bt + C + 3At + 3Bt + 3Ct + 3D = t3 .

Equating the coefficients of t3 , t2 , t and 1 we have 3A = 1 ⇒ A = 13 ,

3A + 3B = 0 ⇒ B = −A = − 31 , 2B + 3C = 0 ⇒ C = − 23 B = 92 , 2 C + 3D = 0 ⇒ D = − 13 C = − 27 1 3 1 2 2 2 Thus xp = 3 t − 3 t + 9 t − 27 . dx dt + 5x = t. Substituting xp = At + B we get, A + 5B + 5At = t. Equating the coefficients of t and 1 we have 5A = 1 ⇒ A = 51 and 1 A + 5B = 0 ⇒ B = − 51 A = − 25 .

1 1

Thus xp = 5 t − 25 . dx dt + 2x = 3 sin 6t. Substituting xp = A sin 6t + B cos 6t we get, 6A cos 6t − 6B sin 6t + 2A sin 6t + 2B cos 6t = 3 sin 6t. Equating the coefficients of cos 6t and sin 6t we have

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

43.

45.

47.

49.

17

cos 6t : 6A + 2B = 0 sin 6t : 2A − 6B = 3 Solving these equations for A and B (multiplying first equation by 3 3 9 and adding with the second) we get A = 20 and B = − 20 . 3 9 Thus xp = 20 sin 6t − 20 cos 6t. dx dt − 5x = 2 cos t. Substituting xp = A cos t + B sin t we get, −A sin t + B cos t − 5A cos t − 5B sin t = 2 cos t. Equating the coefficients of sin t and cos t we have sin t : −A − 5B = 0 cos t : −5A + B = 2 Solving these equations for A and B (multiplying second equation 1 5 and B = 13 . by 5 and adding with the first) we get A = − 13 1 5 Thus xp = 13 sin t − 13 cos t. dx dt + 4x = 3 cos 2t + 5 sin 2t. Substituting xp = A cos 2t + B sin 2t we get, −2A sin 2t + 2B cos 2t + 4A cos 2t + 4B sin 2t = 3 cos 2t + 5 sin 2t. Equating the coefficients of sin 2t and cos 2t we have sin 2t : −2A + 4B = 5 cos 2t : 4A + 2B = 3 Solving these equations for A and B (multiplying first equation by 2 1 and adding with the second) we get B = 13 10 and A = 10 . 1 13 Thus xp = 10 cos 2t + 10 sin 2t. dx dt + 6x = cos t + sin 5t. Substituting xp = A cos t + B sin t + C sin 5t +D cos 5t we get, −A sin t + B cos t + 5C cos 5t − 5D sin 5t + 6A cos t +6B sin t + 6C sin 5t + 6D cos 5t = cos t + sin 5t. Equating the coefficients we have cos t : 6A + B = 1 sin t : −A + 6B = 0 cos 5t : 5C + 6D = 0 sin 5t : 6C − 5D = 1 Solving the first pair of equations for A and B (multiplying second 6 1 and A = 37 . equation by 6 and adding with the first) we get B = 37 Similarly, to solve the second pair of equations for C and D, 6 multiply the first by 5 and the second by 6 and add to get C = 61 5 and D = − 61 . 6 1 6 5 Thus xp = 37 cos t + 37 sin t + 61 sin 5t − 61 cos 5t. dx − 9x = 5 + 2 sin 3t. Substituting x = A + B sin 3t + C cos 3t p dt we get, 3B cos 3t − 3C sin 3t − 9A − 9B sin 3t − 9C cos 3t = 5 + 2 sin 3t. Equating the coefficients we have Non-t : −9A = 5 ⇒ A = − 59 cos 3t : 3B − 9C = 0 sin 3t : 9B − 3C = 2 Solving last pair of equations for B and C (multiplying the first equation by 3 and adding with the second) we get 1 1 and B = − 15 . Thus xp = − 95 − 15 sin 3t − 15 cos 3t. C = − 15

18

CHAPTER 1

51.

53.

55.

57.

59.

61.

dx dt

+ x = 2e3t + sin t. Substituting xp = Ae3t + B sin t + C cos t we get, 3Ae3t + B cos t − C sin t + Ae3t + B sin t + C cos t = 2e3t + sin t. Equating the coefficients we have e3t : 3A + A = 2 ⇒ A = 21 cos t : B+C =0 sin t : B−C =1 Solving last pair of equations for B and C ¡ (adding ) we get¢ B = 21 and C = −B = − 21 .Thus xp = 12 e3t + sin t − cos t . dx −3t . Substituting xp = Ae−3t we get, dt + 3x = 8e −3t −3Ae + 3Ae−3t = 8e−3t ⇒ 8e−3t = 0 which is impossible becasue e−3t 6= 0, ∀t. So a simple exponential, Ae−3t , does not work as a particular solution becasue e−3t is a solution of the associated homogeneous equation. Solve R by the integrating factor method : The integrating factor is e 3dt = ¡e3t . Multiplying the ¡equation by the integrating factor ¢ ¢ d 3t + 3x = 8 ⇒ xe = 8. Integrating we have we get e3t dx dt dt xe3t = 8t + c ⇒ x = 8te−3t + ce−3t . Conjecture: Notice that the particular solution part of this general solution is a constant multiple of t times the exponential. This indicates that we may find a particular solution by substituting Ate−3t , when simple exponential forcing e−3t is a solution of the associated homogeneous equation. dx 2t 2t we get, dt − 2x = 7e . Substituting xp = Ae 2t 2t 2t 2t 2Ae − 2Ae = 7e ⇒ 7e = 0 which is impossible becasue e2t 6= 0, ∀t. So a simple exponential, Ae2t , does not work as a particular solution becasue e2t is a solution of the associated homogeneous equation. 2t 2t 2t 2t Substituting x = ve2t we get, dv dt e + 2ve − 2ve = 7e dv 2t Dividing by e 6= 0 we have dt = 7 ⇒ v = 7t + c. Thus the general solution is x = 7te2t + ce2t . We can make the same conjecture as in #53. dx t rt be the solution of the associated dt − x = 4e . Let x1 = e homogeneous equation. On substitution we get, rert − ert = 0 ⇒ r = 1. So x1 = c1 et (similar to the forcing function). So let xp = Atet . Substituting this in the equation we get, Aet + Atet − Atet = 4et ⇒ A = 4. Thus the general solution is x = 4tet + cet . dx −t rt be the solution of the associated dt + x = 5e . Let x1 = e homogeneous equation. On substitution we get, rert + ert = 0 ⇒ r = −1. So x1 = c1 e−t (similar to the forcing function). So let xp = Ate−t . Substituting this in the equation we get, Ae−t − Ate−t + Ate−t = 5e−t ⇒ A = 5. Thus the general solution is x = 5te−t + ce−t . dx 7t rt be the solution of the associated dt − 7x = 8e . Let x1 = e

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

homogeneous equation. On substitution we get, rert − 7ert = 0 ⇒ r = 7. So x1 = c1 e7t (similar to the forcing function). So let xp = Ate7t . Substituting this in the equation we get, Ae7t + 7Ate7t − 7Ate7t = 8e7t ⇒ A = 8. Thus the general solution is x = 8te7t + ce7t . dx 63. dt + p(t)x = f (t), 0 ≤ t ≤ 2, f (t) = 0, x(0) = 2. For 0 ≤ t < 1, p(t) = 2 and so dx dt + 2x = 0 ⇒ x(t) = c1 e−2t . Using x(0) = 2 we get c1 = 2. So x(t) = 2e−2t for 0 ≤ t < 1. For 1 ≤ t ≤ 2, p(t) = 1 and so dx −t dt + x = 0 ⇒ x(t) = c2 e . In order for x to be continuous at t = 1 we must have lim− x(t) = x(1) ⇒ 2e−2 = c2 e−1 ⇒ t→1

c2 = 2e−1 . Thus the solution is ( 2e−2t , 0≤t<1 x(t) = −1 −t 2e e , 1 ≤ t ≤ 2 2 x 1

0

65.

1

t

2

dx dt

+ p(t)x = f (t), 0 ≤ t ≤ 2, p(t) = 0, x(0) = 0. For 0 ≤ t < 1, f (t) = 1 and so dx dt = 1 ⇒ x(t) = t + c1 .

Using x(0) = 0 we get c1 = 0. So x(t) = t for 0 ≤ t < 1.

For 1 ≤ t ≤ 2, f (t) = −1 and so dx dt = −1 ⇒ x(t) = −t + c2 .

In order for x to be continuous at t = 1 we must have lim x(t) = x(1) ⇒ 1 = −1 + c2 ⇒ c2 = 2. Thus the solution is t→1− ( t, 0≤t<1 x(t) = 2 − t, 1 ≤ t ≤ 2 x 1

0

67.

dx dt

1 t

+ p(t)x = f (t), 0 ≤ t ≤ 2, x(0) = 2. For 0 ≤ t < 1, −t p(t) = 1, f (t) = 0 and so dx dt + x = 0 ⇒ x(t) = c1 e .

Using x(0) = 2 we get c1 = 2. So x(t) = 2e−t for 0 ≤ t < 1. For 1 ≤ t ≤ 2, p(t) = 0, f (t) = 1 and so dx dt = 1 ⇒

19

20

CHAPTER 1

x(t) = t + c2 . In order for x to be continuous at t = 1 we must have lim x(t) = x(1) ⇒ 2e−1 = 1 + c2 ⇒ t→1−

c2 = 2e−1 − 1. Thus the solution is ( 2e−t , 0≤t<1 x(t) = −1 t + 2e − 1, 1 ≤ t ≤ 2 2 x 1

0

1

2

t

1.8 GROWTH AND DECAY PROBLEMS dP The growth rate k of a population P (t) is given by P1 dP dt = k ⇒ dt = kP kt whose solution with initial population P (0) is P (t) = P (0)e . The population will be doubled when P (t) = 2P (0). Then 2P (0) = P (0)ekt ⇒ ekt = 2 ⇒ kt = ln 2 ⇒ t = lnk 2 which is known as

doubling time, denoted by td as td = lnk2 , and we will use this formula throughout this section. 1. In this problem, the growth rate is k = 1.5% = 0.015 and so

ln 2 ≈ 46.2 years.

doubling time, td = lnk2 = 0.015 1 3. Using td = 8 hours = 3 day in the doubling time formula we get, 1 ln 2 3 = k ⇒ k = 3 ln 2 ≈ 2.08 = 208% per day. kt 5. Here, P (0) = 1500, t = 1 hour, P (1) ¡ 4=¢ 2000. So P (t) = P (0)e ⇒ 4 k k 2000 = 1500e ⇒ e = 3 ⇒ k = ln 3 . After 4 hour (i.e. t = 4), ¡ ¢4 4 P (t) = 1500e4k = 1500e4 ln( 3 ) = 1500 4 . 3

B

A

10

0.02

Q

c

5

0

7.

0.01

100

t

200

0

100

t

200

td = lnk2 = 3 ⇒ k = ln32 . Here, P (t) = 10P (0) and so 10P (0) = P (0)ekt ⇒ ekt = 10 ⇒ kt = ln 10 ⇒ ln 10 t = lnk10 = 3 ln 2 ≈ 9.97 years.

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

21

9. Let Q be the number of organisms.

dQ dQ Birth: dQ dt = k1 Q, Death: dt = −k2 Q, Addition: dt = k.

dQ ⇒ dt = k1 Q − k2 Q + k. 11. With the interest rate of 3% per year:

ln 2 Doubling time, td = lnk2 = 0.03 ≈ 23.10 years.

With the 3% yield:

Yield = ek − 1 = 0.03 ⇒ ek = 1.03 ⇒ k = ln (1.03) . So, in this case

ln 2 Doubling time, td = lnk2 = ln(1.03) ≈ 23.45 years. 13. Here, the growth after one year is

ek − 1 = 0.064 ⇒ ek = 1.064 ⇒ k = ln (1.064) .

ln 2 ≈ 11.17 years. Then td = ln(1.064) 15. This is the case of exponential growth. So x(t) = x(0)ekt , k > 0. The cost of living rose from x(0) = 10, 000 to x(1) = 11, 000 in one year, that is, t = 1. So 11000 = 10000ek ⇒ ek = 1.1 ⇒ k = ln (1.1) ≈ 0.0953 = 9.53% per year. 17. Using the half-life formula T = lnk2 = 16 days, we get k = ln162 . At the end of t = 30 days, x(t) = 30 g. Then x(t) = x(0)e−kt ⇒ ln 2 ln 2 30 = x(0)e− 16 (30) ⇒ x(0) = 30e 16 (30) ≈ 110.04 g. dx 19. dt = −kx ⇒ x(t) = x(0)e−kt . When t = 10 years, x(t) = 0.80x(0). So 0.80x(0) = x(0)e−10k ⇒ e−10k = 0.8 ⇒ −10k = ln (0.8) ⇒ 1 k = − 10 ln (0.8) . ln 2 (a) Half-life: T = lnk2 = −10 ln(0.8) ≈ 31.063 years. (b) x(t) = 0.15x(0). So 0.15x(0) = x(0)e−kt ⇒ e−kt = 0.15 ⇒ −kt = ln (0.15) ⇒ t = − k1 ln (0.15) = 10 ln(0.15) ln(0.8) ≈ 85.018 years. Thus it will take 85.018 − 10 = 75.018 additional years. 21. Newton’s law of cooling dT dt = −k (T − Q0 ) has particular solution Q0 so the general solution is T (t) = Q0 + ce−kt . Here Q0 = 30. Using the initial temperature T (0) = 200, we get 200 = 30 + c ⇒ c = 170. Thus T (t) = 30 + 170e−kt . Again, using T (10) = 180, 15 ⇒ we get, 180 = 30 + 170e−10k ⇒ 170e−10k = 150 ⇒ e−10k = 17 ¡ 15 ¢ ¡ 15 ¢ 1 ln 15−ln 17 ln 17−ln 15 −10k = ln 17 ⇒ k = − 10 ln 17 = − = . 10 10 1 (a) T will be 400 C if 40 = 30 + 170e−kt ⇒ 170e−kt = 10 ⇒ e−kt = 17 ln(17) ⇒ −kt = − ln (17) ⇒ t = k1 ln (17) = ln1017−ln 15 ≈ 226.4 minutes. 15 (b) T (t) = 30 + ce−kt , where we previously had k = ln 17−ln . 10 At t = 0, T = 200, so c = 170 and T (t) = 30 + 170e−kt .

Now we solve this equation for t :

T (t) − 30 = 170e−kt ⇒ ln (T − 30) = ln (170)¡− kt ⇒ ¢ 10 T −30 1 . [ln (T − 30) − ln (170)] = ln 15−ln t = −k 17 ln 170 dT 23. Newton’s law of cooling dt = −k (T − Q0 ) has particular solution Q0 so the general solution is T (t) = Q0 + ce−kt . Here Q0 = 40. Using the initial temperature T (0) = 100, we get 100 = 40 + c ⇒ c = 60. Thus T (t) = 40 + 60e−kt . Again, using T (10) = 60, −10k we get, 60 = 40 ⇒ 60e−10k = 20 ⇒ e−10k = 13 ¢ 60e ¡ 1+ 1 ⇒ −10k = ln 3 ⇒ −10k = − ln 3 ⇒ k = 10 ln 3 ≈ 0.1099.

22

CHAPTER 1

25. (a) Newton’s law of cooling dT dt = −k (T − Q0 ) = −k [T − (20 + 10t)] . (b) Substituting k = 1 in the equation we get, dT dt = −T + 20 + 10t dT rt ⇒ dt + T = 20 + 10t. Let T1 = e be the solution of the homogeneous equation. Then rert + ert = 0 ⇒ r = −1. Thus T1 = ce−t . Substituting Tp = A + Bt we get, B+ A + Bt = 20 + 10t. Equating the coefficients of t and 1 we have B = 10 and A + B = 20 ⇒ A = 10. Thus the general solution is T (t) = 10 + 10t + ce−t . Using the initial temperature T (0) = 40 we get, 40 = 10 + c ⇒ c = 30. Hence the solution is T (t) = 10 + 10t + 30e−t . 27. (a) Let the amount invested be $A. The 6% interest rate will effect the amount in excess of $500. So the amount of interest per year will be $0.06 (A − 500) . Thus the rate of change of money can be written as the differential equation dA dt = 0.06 (A − 500) with initial condition A(0) = 2000. (b) dA dt − 0.06A = −30. Let Ap = B ⇒ −0.06B = −30 ⇒ B = 500 ⇒ Ap = 500 and Ah = ce0.06t . So the general solution is A(t) = 500 + ce0.06t . Using initial condition A(0) = 2000 we get,

c = 1500. Thus A(t) = 500 + 1500e0.06t .

After 10 years the amount will be A(10) = $3233.18.

(c) The full amount earns interest when dA dt = 0.06A ⇒ dA dt − 0.06A = 0, a homogeneous equation with general solution A(t) = ce0.06t which becomes A(t) = 2000e0.06t for the initial condition. After 10 years the amount will be A(10) = $3644.24 which is $411.06 more than that in part (b) . 29. Constant deposit rate is $B/day means $365B/year and 0.08P is the amount of interest per year with 8% interest rate. So dP dP dt = 0.08P + 365B ⇒ dt − 0.08P = 365B. Let Pp = A ⇒ 365 365 −0.08A = 365B ⇒ A = − 0.08 B ⇒ Pp = − 0.08 B and 365 0.08t Ph = ce . So the general solution is P (t) = − 0.08 B + ce0.08t . 365 Using initial condition P¡ (0) = 1000 we c = 1000 + 0.08 B. ¢ get, 365 365 0.08t Thus P (t) = − 0.08 B + 1000 + 0.08 B e . (a) After t = 5 years P = $10, ¡ 000. 365 ¢ 0.4 365 B + 1000 + 0.08 B e ⇒ So 10000 = − 0.08 ¡ 0.4 ¢ 80(10−e0.4 ) 0.4 800 = 365B e − 1 + 80e ⇒ B = 365(e0.4 −1) ≈ $3.79. ¡ ¢ 365 365 (b) Solve 10000¡= − 0.08 B(t) + ¢ 1000 + 0.08

B(t) e0.08t for B(t) : 800 = B(t) 365e0.08t −¢365 + 80e0.08t ¡ ⇒ B(t) 365e0.08t − 365 = 800 − 80e0.08t

¡ ¢¡ ¢−1 ⇒ B(t) = 800 − 80e0.08t 365e0.08t − 365 . dQ = 0.2Q + 400 cos (2πt) ⇒ − 0.2Q = 400 cos (2πt) . 31. dQ dt dt Substituting Qp = A cos (2πt) + B sin (2πt) we get

−2πA sin (2πt) + 2πB cos (2πt) − 0.2A cos (2πt) − 0.2B sin (2πt)

= 400 cos (2πt) .

Equating the coefficients of cos (2πt) and sin (2πt) we have

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

23

cos (2πt) : 2πB − 0.2A = 400 sin (2πt) : −0.2B − 2πA = 0 Solving these equations for A and B (Multiplying the first equation by 0.2 and the second equation by 2π and adding and simplifying) 200π we get, A = π2−20 +0.01 and B = π 2 +0.01 . Thus the particular solution is 1 Qp = π2 +0.01 (−20 cos (2πt) + 200π sin (2πt)) . Q1 = ce0.02t . So the general solution is

1 (−20 cos (2πt) + 200π sin (2πt)) + ce0.02t .

Q(t) = π2 +0.01 20 Using the initial condition Q(0) = 100 we get c = 100 + π2 +0.01 . ³ ´ 20 1 e0.02t . Hence Q(t) = π2 +0.01 (−20 cos (2πt) + 200π sin (2πt))+ 100 + π2 +0.01 1.9 MIXTURE PROBLEMS 1. (a) Let Q be the amount of salt in the tank of volume V = 300. This volume is constant as water is flowing in and out at the same Q . The rate (2 gal/min). So the concentration of salt is C = 300 Q inflow and outflow rate of salt are 2(0.4) and 2 300 , respectively. Thus the rate of change in salt can be written as dQ Q dQ Q dt = 2(0.4) − 2 300 ⇒ dt + 150 = 0.8. Substituting Qp = A t 1 ⇒ Q1 = e− 150 .

we get, A = 120 and Q1 = ert ⇒ r = − 150 t Thus Q(t) = 120 + ke− 150 . Using initial condition

Q(0) = 300(0.2) = 60 we have, k = −60. Hence

t t 2 1 − 150 . Q(t) = 120 − 60e− 150 and C(t) = Q(t) 300 = 5 − 5 e (b) When does C(t) = 0.3? t t t = ln 2 ⇒ 0.3 = 0.4 − 0.2e− 150 ⇒ e− 150 = 21 ⇒ 150 t = 150 ln 2 ≈ 104 minutes. 3. (a) Let Q be the amount of salt in the tank of volume V = 100. This

volume is constant as water is flowing in and out at the same

Q rate (5 L/h). So the concentration of salt is C = 100 . The

Q inflow and outflow rate of salt are 5(0.2) and 4 100 (evaporated

water contains no salt so that Q = 0), respectively.

Thus the rate of change in salt can be written as

dQ Q dQ Q dt = 5(0.2) − 4 100 ⇒ dt + 25 = 1. Substituting Qp = A t 1 we get, A = 25 and Q1 = ert ⇒ r = − 25 ⇒ Q1 = e− 25 .

t Thus Q(t) = 25 + ke− 25 . Using initial condition

Q(0) = 100(0.1) = 10 we have, k = −15. Hence

t t

− 25 Q(t) = 25 − 15e− 25 and C(t) = Q(t) . 100 = 0.25 − 0.15e (b) No. lim C(t) = 0.25 > 0.2. t→∞

5. (a) Let Q be the amount of iodine in the tank. Water is flowing in

and out at a different rate and so the volume is changing as

dV dt = 6 − 1 = 5 ⇒ V (t) = 5t + 500 since the initial half-volume is Q . The inflow and 500 gal. Concentration of iodine is C = 5t+500

24

CHAPTER 1

outflow rate of iodine are 0 (Pure water has no concentration) and Q 1 5t+500 , respectively. Thus the rate of change in iodine can be Q dQ dt written as dQ dt = − 5t+500 ⇒ Q = − 5t+500 ⇒ −1/5

ln Q = − 51 ln (5t + 500) + k1 ⇒ Q(t) = k2 (5t + 500) −1/5 = k (t + 100) . Using initial condition Q(0) = 10 we have, −1/5 1/5 10 = k (100) ⇒ k = 10 (100) . ³ ´1/5 100 Hence Q(t) = 10 t+100 . Since volume increases as t increases, tank will overflow when V = 1000. That is, tank overflows when 5t + 500 = 1000 means at t = 100. So for 0 ≤ t ≤ 100, ³ ´1/5 Q(t) 100 Q(t) = 10 t+100 and C(t) = 5t+500 . (b) Tank overflows when t = 100. After overflow, V = 1000 and Q . During overflow, mixture is leaving at the rate of 6 gal/min C = 1000 6Q (as pure water is entering at this rate) and so for t ≥ 100, dQ dt = − 1000 . 3 3 Since it has constant coefficients, Q(t) = c2 e− 500 t = c2 e− 500 (t−100) . In order for Q(t) to be continuous at t = 100 we must have ¡ ¢1/5 lim Q(t) = lim Q(t) ⇒ 10 100 = c2 200 t→100+

t→100−

−1/5

Q(t) So for t ≥ 100, Q(t) = 10 (2) e−3(t−100)/500 and C(t) = 1000 . 7. Let Q be the amount of pollutant in the lake. Water is flowing in and out at a different rate and so the volume is changing as dV dt = 5 − 2 = 3 ⇒ V (t) = 3t + 1000 since the initial volume is 1000 Q kL. Concentration of pollutant is C = 3t+1000 . The inflow and Q outflow rate of pollutant are 5 × 7 and 2 3t+1000 , respectively. Thus 2Q the rate of change in pollutant can be written as dQ dt = 35 − 3t+1000 ⇒ 2Q dQ Q + 3t+1000

R

= 35. The integrating factor is e

= (3t + 1000) ³

2/3

2 3t+1000 dt

= eln(3t+1000)

2/3

. Multiplying´by integrating factor and rearranging 2/3 2/3 Q (3t + 1000) = 35 (3t + 1000) ⇒ R 2/3 2/3 Q (3t + 1000) = 35 h(3t + 1000) dt + ki⇒ −2/3 5/3 Q(t) = (3t + 1000) 7 (3t + 1000) + k . Using initial condition h i −2 5 Q(0) = 2000 we have, 2000 = (10) 7 (10) + k ⇒ we have

d dt

5

7 (10) + k = 200000 ⇒hk = −500000. Thus i −2/3 5/3 Q(t) = (3t + 1000) 7 (3t + 1000) − 500000 −2/3

= 7 (3t + 1000) − 500000 (3t + 1000) and

−5/3 Q(t) = 7 − 500000 (3t + 1000) .

C(t) = 3t+1000 9. Let V1 and V2 be the volume of tank 1 and tank 2, respectively. V1 = (13 − 7) t + 150 = 6t + 150; V2 = (7 − 28) t + 250 = −21t + 250 Let S1 and S2 be the amount of salt in tank 1 and tank 2, respectively. S1 dS1 7S1 1 Then dS dt = 13 × 3 − 7 V1 ⇒ dt = 39 − 6t+150

7S1 S2 dS2 7S1 28S2 2 and dS dt = 6t+150 − 28 V2 ⇒ dt = 6t+150 − −21t+250 .

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

25

11. Let V1 and V2 be the volume of tank 1 and tank 2, respectively.

V1 = (21 − 18) t + 230 = 3t + 230; Tank 2 receives brine at the

rate 21 (18) = 9 gal/s. So V2 = (9 − 4) t + 275 = 5t + 275.

Let S1 and S2 be the amount of salt in tank 1 and tank 2, respectively. S1 dS1 18S1 1 Then dS dt = 21 × 5 − 18 V1 ⇒ dt = 105 − 3t+230 9S1 S2 9S1 4S2 dS2 dS2 . and dt = 3t+230 − 4 V2 ⇒ dt = 3t+230 − 5t+275 13. Let V1 , V2 and V3 be the volume of tank 1, tank 2 and tank 3, respectively. V1 = (11 − 18) t + 100 = −7t + 100; V2 = 200 (same rate of inflow and outflow). V3 = 300 (brine gets in and overflow so that volume remains the same). Let S1 , S2 and S3 be the amount of salt in tank 1, tank 2 and tank 3, respectively. Then S1 dS1 18S1 dS1 dt = 11 × 5 − 18 V1 ⇒ dt = 55 − −7t+100 18S1 S2 dS2 18S1 18S2 dS2 dt = −7t+100 − 18 V2 ⇒ dt = −7t+100 − 200 and dS3 18S2

dt = 200 . 1.10 ELECTRONIC CIRCUITS di 1. The differential equation is L dt + Ri = e. In this problem, Ri = v = 2i ⇒ R = 2, L = 1, e = 1. Substituting all these we get, di 1 dt + 2i = 1. Let ip = A. So 2A = 1 ⇒ A = 2 . 1 rt So ip = 2 . Taking i1 = e we have r = −2 ⇒ i1 = ce−2t . Thus i (t) = 21 + ce£ −2t . At t¤ = 0, i(0) = 12 + c ⇒ c = i(0) − 12 . Hence i (t) = 21 + i(0) − 21 e−2t . As t → ∞, i(t) → 21 . Alternatively, di = 0 so that 2i = 1 ⇒ i = 21 . for an equilibrium solution dt di 3. The differential equation is L dt + Ri = e. In this problem, Ri = v = i, L = 1, e = sin t. Substituting all these we get, di dt + i = sin t. Substituting ip = A sin t + B cos t in the differential equation we get A cos t − B sin t + A sin t + B cos t = sin t.

Equating the coefficients of cos t and sin t we get:

cos t : A + B = 0

sin t : A − B = 1

Solving these equations for A and B by adding we get,

A = 12 , B = − 21 . Thus ip = 12 sin t − 12 cos t. Taking

i1 = ert in the differential equation we have, r = −1 ⇒ i1 = ce−t . 1 Thus i (t) = 21 sin t − 12 cos t + ce−t . At t = 0, i(0) £ = − 21+ ¤c 1 1 1 ⇒ c = i(0) + 2 . Hence i (t) = 2 sin t − 2 cos t + i(0) + 2 e−t . di + Ri = e. For 0 ≤ t < 10, 5. The differential equation is L dt Ri = v = i ⇒ R = 1, L = 1, e = 9. Substituting all these we get, di rt dt + i = 9. Let ip = A. So A = 9 ⇒ ip = 9. Taking i1 = e −t −t we have r = −1 ⇒ i1 = c1 e . So i (t) = 9 + c1 e . Using the initial condition i(0) = 0 we get 0 = 9 + c1 ⇒ c1 = −9.

Thus for 0 ≤ t < 10, i (t) = 9 − 9e−t . For 10 ≤ t ≤ 20, Ri = v = i, L = 1, e = 0.

26

CHAPTER 1 di Substituting all these we get, dt + i = 0. This is a homogeneous equation and, in fact, the homogeneous part of the above equation. So i(t) = c2 e−t . In order for i(t) to be continuous at t = 10 we must have lim i(t) = i(10) ⇒ 9 − 9e−10 = c2 e−10 ⇒ t→10−

c2 = 9e10 − 9. Thus the solution is

( 9(1 − e−t ), 0 ≤ t < 10 i(t) = 9(e10 − 1)e−t , 10 ≤ t ≤ 20. 7. Capacitance, C = 0.5, Ri = v = 2i ⇒ R = 2, voltage source e = 6 sin t. In order to find charge q we can use: Cq + Ri = e dq q dq ⇒ Cq + R dq dt = e (since i = dt ) ⇒ 0.5 + 2 dt = 6 sin t ⇒ dq dt + q = 3 sin t. Substituting qp = A sin t + B cos t in this equation we get A cos t − B sin t + A sin t + B cos t = 3 sin t. Equating the coefficients of cos t and sin t we get: cos t : A+B =0 sin t : A−B =3 Solving these equations for A and B by adding we get, A = 23 , B = − 32 . Thus qp = 23 sin t − 32 cos t. Taking q1 = ert in the differential equation we have r = −1 ⇒ q1 = ce−t . Thus q (t) = 23 (sin t − cos t) + ce−t . At t = 0, q(0) = 1 = − 32 + c 3 ⇒ c = 1 + 2 = 52 . Hence q (t) = 32 (sin t − cos t) + 25 e−t . 1.11 MECHANICS II: INCLUDING AIR RESISTANCE 1. (a) The differential equation is m dv dt = −mg − kv ⇒ dv 1 20 dv = −20 (980) − 10v ⇒ dt dt + 2 v = −980. Substituting 1 vp = A we get, 2 A = −980 ⇒ A = −1960. So vp = −1960. t t v1 = ert ⇒ r = − 12 ⇒ v1 = ce− 2 . Thus v (t) = −1960 + ce− 2 . Using initial condition v(0) = 0 we³have, c =´ 1960. Hence t

t

v (t) = −1960 + 1960e− 2 = −1960 1 − e− 2 . ¡ ¢ (b) v (10) = 1960 e−5 − 1 = −1946.79. (c) lim v(t) = −1960. t→∞

3. The differential equation is m dv dt = −mg − kv ⇒ dv 1 70 dv = −70 (980) − 7v ⇒ dt dt + 10 v = −980. Substituting 1 vp = A we get, 10 A = −980 ⇒ A = −9800. So vp = −9800. t t 1 ⇒ v1 = ce− 10 . Thus v (t) = −9800 + ce− 10 . v1 = ert ⇒ r = − 10 Using the condition v(5) = 12600 we have, 1 12600 = −9800 + ce− 2 ⇒ c = 36931.36. So v (0) = −9800 + 36931.36 = 27131.36. 2 5. (a) The differential equation is m dv dt = −mg + v . Here, weight is 32 dv mg = 32 so that m = g = 1. Then dt = −32 + v 2 ⇒

27 ¡ ¢ R R dv dv 2 ⇒ 32−v dt. Integrating using tables gives 2 = − dt = − ¯ 32 − v¯ √ √ √ √ ¯ ¯ v+ v+ 32 32 1 √ ln ¯ v−√32 ¯ = −t + c ⇒ v−√32 = ke−2 32t , where k = ±e2 32c . 2 32

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

√

1000+√32 Using the initial condition v(0) = 1000 we have k = 1000− . 32 ´ ´ ³ ³ √ √ √ √ √ v+√32 −2 32t −2 32t −2 32t ⇒ v 1 − ke = − 32 1 + ke So v− 32 = ke √ √ ³ 1+ke−2√32t ´ 1000+√32 √ ⇒ v(t) = − 32 1−ke , where k = . −2 32t 1000− 32 √ (b) lim v(t) = − 32. t→∞

1.12 ORTHOGONAL TRAJECTORIES (OPTIONAL) 1. x = t + c ⇒ x − t = c. Differentiating with respect to (wrt) t: dx dx dt − 1 = 0 ⇒ dt = 1. The slope of the orthogonal family is thus dx dt = −1. Integrating we have, x(t) = −t + c2 . x

t dx x 3. xt = c. Differentiating wrt t: x + t dx dt = 0 ⇒ dt = − t . t dx The slope of the orthogonal family is thus dt = x ⇒ R R 2 2 xdx = tdt. Integrating we have, x2 = t2 + c ⇒ x2 − t2 = c2 , where c2 = 2c.

x

t

ln x ln t = c. Differentiating wrt t: dx ln x = 0 ⇒ dt = xt ln t . The slope of the orthogonal (ln t) R R dx t ln t family is thus dt = − x ln x ⇒ x ln xdx = − t ln tdt. Integration parts on both sides yields, 12 x2 ln |x| − 14 x2 = − 12 t2 ln |t| + 41 t2 + c

5. x = tc ⇒ ln x = c ln t ⇒ 1 1 dx x dt ln t−ln x t 2

by

28

CHAPTER 1

⇒ 2x2 ln |x| − x2 = t2 − 2t2 ln |t| + c2 , where c2 = 4c. 7. x = t2 + c ⇒ x − t2 = c. Differentiating wrt t: dx dx = 2t. The slope of the orthogonal family is thus dt − 2t = 0 ⇒ R dt R dx 1 = − ⇒ dx = − 12 1t dt. Integration yields, x = − 21 ln |t| + c2 . dt 2t √ √ 2 9. t = (x − c) ⇒ x − c = ± t ⇒ x ± t = c. Differentiating wrt t:

1 1 dx dx √ √ dt ± 2 t = 0 ⇒ dt = ± 2 t . The slope of the orthogonal family is √ R R√ thus dx tdt. Integrating we have, dt = ±2 t ⇒ dx = ±2 x = ± 43 t3/2 + c2 .

1 dx 11. x = tan (t + c) ⇒ tan−1 x = t + c. Differentiating wrt t: 1+x 2 dt = 1 dx 2 ⇒ dt = 1 + x . The slope of the orthogonal family is thus ¢ R¡ R dx 1 1 + x2 dx = − dt. Integrating we have, dt = − 1+x2 ⇒ 3

x + x3 = −t + c2 . x 13. x = c cos t. Differentiating wrt t: dx dt = −c sin t. Since c = cos t , x dx tan t. The slope of the orthogonal family is dt = − cos t sin t = −x R R 1 dx thus dt = x cot t ⇒ xdx = cot tdt. Integrating we have, x2 2 2 = ln |sin t| + c ⇒ x = 2 ln |sin t| + c2 , where c2 = 2c. x dx 2x 15. x = ct2 . Differentiating wrt t: dx dt = 2ct. Since c = t2R, dt = t . The R t dx slope of the orthogonal family is thus dt = − 2x ⇒ 2xdx = − tdt. 2 Integrating we have, x2 = − t2 + c ⇒ 2x2 = −t2 + c2 , where c2 = 2c. dx 2t 17. x3 + t2 = c. Differentiating wrt t: 3x2 dx dt + 2t = 0 ⇒ dt = − 3x2 . 2 3x The slope of the orthogonal family is thus dx dt = 2t ⇒ R dt R −2 1 3 3 x dx = 2 t . Integrating we have, − x = 2 ln |t| + c ⇒ ¢ ¡ ¡ 3 ¢−1 3 1 , where c2 = −c. x = − 2 ln |t| − c ⇒ x = c2 − 2 ln |t|

Chapter Two

Linear Second and Higher-Order Differenial Equations

2.1 GENERAL SOLUTION OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS 1. x = sin t ⇒ x′ = cos t ⇒ x′′ = − sin t. So x′′ + x = − sin t + sin t = 0. x = cos t ⇒ x′ = − sin t ⇒ x′′ = − cos t. So x′′ + x = 0. Thus {sin t, cos t} is a set of solutions for the associated homogeneous equation. Now xp = 1 ⇒ x′p = x′′p = 0 and so x′′p + xp = 1 =r.h.s. Thus the general solution is x(t) = 1 + c1 sin t + c2 cos t. Using the initial condition x(0) = 0 we have, 0 = 1 + c2 ⇒ c2 = −1. Differentiating the general solution we get, x′ (t) = c1 cos t − c2 sin t. The initial condition x′ (0) = 0 gives c1 = 0. Thus the solution of the initial value problem is x(t) = 1 − cos t. 3. x = et ⇒ x′ = et ⇒ x′′ = et . So x′′ − 3x′ + 2x = et − 3et + 2et 2t ′′ ′ = 0. For x = e2t , x′ = 2e2t and ©x′′ = 4e ª . So x − 3x + 2x 2t 2t 2t t 2t = 4e − 6e + 2e = 0. Thus e , e is a set of solutions for the associated homogeneous equation. Now xp = t + 23 ⇒ x′p = 1, x′′p = 0 and so x′′p − 3x′p + 2xp = −3 + 2t + 3 = 2t =r.h.s. Thus the general solution is x(t) = t + 32 + c1 et + c2 e2t .

Differentiating this we get, x′ (t) = 1 + c1 et + 2c2 e2t . Using the

initial conditions x(0) = 1 and x′ (0) = 0 we have, 1 = 23 + c1 + c2

and 0 = 1 + c1 + 2c2 . Solving these equations for c1 and c2 by

subtracting them we have c2 = − 21 and then c1 = 0.

Thus the solution of the initial value problem is x(t) = t + 23 − 21 e2t .

5. x = e−t cos t ⇒ x′ = −e−t cos t − e−t sin t ⇒ x′′ = e−t cos t + e−t sin t + e−t sin t − e−t cos t. So x′′ + 2x′ + 2x = e−t cos t + e−t sin t + e−t sin t − e−t cos t − 2e−t cos t −2e−t sin t + 2e−t cos t = 0. Similarly it can be shown that for x = e−t sin t, x′′ + 2x′ + 2x = 0. Thus {e−t cos t, e−t sin t} is a set of solutions for the associated homogeneous equation. Now xp = 3 ⇒ x′p = x′′p = 0 and so x′′p + 2x′p + 2xp = 6 =r.h.s. Thus the general solution is x(t) = 3+c1 e−t cos t+c2 e−t sin t. Using the initial condition x(0) = 1

30

CHAPTER 2

we have, 1 = 3 + c1 ⇒ c1 = −2. Differentiating the general solution we get, x′ (t) = −c1 e−t cos t−c1 e−t sin t−c2 e−t sin t+c2 e−t cos t. Using x′ (0) = 1 and c1 = −2 we have, 1 = 2 + c2 ⇒ c2 = −1. Thus the solution of the initial value problem is

x(t) = 3 − 2e−t cos t − e−t sin t.

7. x = sin t ⇒ x′ = cos t ⇒ x′′ = − sin t. So x′′ + x = − sin t + sin t = 0. x = cos t ⇒ x′ = − sin t ⇒ x′′ = − cos t. So x′′ + x = 0. Thus {sin t, cos t} is a set of solutions for the associated homogeneous equation. Now xp = t sin t, x′p = sin t + t cos t, x′′p = cos t + cos t − t sin t and so x′′p + xp = 2 cos t − t sin t + t sin t = 2 cos t =r.h.s. Thus the general solution is x(t) = t sin t + c1 cos t + c2 sin t. Using the initial condition x(0) = 1 we have, 1 = c1 . Differentiating the general solution we get, x′ (t) = sin t + t cos t − c1 sin t + c2 cos t. Using x′ (0) = −1 we have, −1 = c2 . Thus the solution of the initial value problem is x(t) = t sin t − sin t + cos t. 9. (a) x = et ⇒ x′ = et ⇒ x′′ = et . So x′′ − 3x′ + 2x = et − 3et + 2et 2t ′′ ′ = 0. For x = e2t , x′ = 2e2t and ©x′′ = 4e ª . So x − 3x + 2x 2t 2t 2t t 2t = 4e − 6e + 2e = 0. Thus e , e is a set of solutions for the associated homogeneous equation. xp = 2 cosh t ⇒ x′p = 2 sinh t, x′′p = 2 cosh t. So x′′p − 3x′p + 2xp = 2 cosh t − 6 sinh t + 4 cosh t = 6 (cosh t − sinh t) = 6e−t =r.h.s. The general solution is thus x = c1 et + c2 e2t + 2 cosh t. For xp = e−t , x′p = −e−t , x′′p = e−t . So x′′p − 3xp′ + 2xp = e−t + 3e−t + 2e−t = 6e−t =r.h.s. The general solution is thus x = c1 et + c2 e2t + e−t . (b) Using the initial conditions x(0) = 4, x′ (0) = 3 for the first solution we get, 4 = c1 + c2 + 2 and 3 = c1 + 2c2 . Solving these equations for for c1 and c2 we have c1 = c2 = 1. Thus the solution of the initial value problem is x = et + e2t + 2 cosh t = et + e2t + et + e−t = 2et + e2t + e−t . Using the same initial conditions for the second solution we get, 4 = c1 + c2 + 1 and 3 = c1 + 2c2 − 1. Solving these equations for for c1 and c2 we have c1 = 2, c2 = 1.Thus the solution of the initial value problem is x = 2et + e2t + e−t , which is the same as before.

2.2 INITIAL VALUE PROBLEM (FOR HOMOGENEOUS EQUATION) 1. x1 = sin t ⇒ x′1 = cos t ⇒ x′′1 = − sin t = −x1 ⇒ x′′1 + x1 = 0. x2 = cos t ⇒ x′2 = − sin t ⇒ x′′2 = − cos t = −x2 ⇒ x′′2 + x2 = 0. Thus x1 and x2 both are solutions·of x′′ + x ¸= 0. · ¸ x1 x2 sin t cos t The Wronskian is W [x1 , x2 ] = det = det cos t − sin t x′1 x′2

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

¡

¢

31

= − sin2 t + cos2 t = −1 6= 0. Hence {sin t, cos t} is a fundamental set of solutions. 3. Let x = tr . Then x′ = rtr−1 , x′′ = r (r − 1) tr−2 . So t2 x′′ − tx′ + x = 0 ⇒ tr [r (r − 1) − r + 1] = 0 ⇒ (r − 1) (r − 1) = 0 (since tr 6= 0) ⇒ r = 1(repeated). So we have two solutions: x =¸ t which are the · t t same. The Wronskian is W [x1 , x2 ] = det = 0. They don’t 1 1 form a fundamental set of solutions. · ¸ · ¸ x1 (1) x2 (1) 1 0 5. (a) The Wronskian is W (1) = det = det x′1 (1) x′2 (1) 1 −1 = −1 6= 0. Hence {x1 , x2 } is a fundamental set of solutions. (b) x3 = c1 x1 + c2 x2 ⇒ x′3 = c1 x′1 + c2 x′2 . Using all three sets of initial conditions we have, 2 = x3 (1) = c1 x1 (1) + c2 x2 (1) = c1 and 0 = x′3 (1) = c1 x′1 (1) + c2 x′2 (1) = c1 − c2 = 2 − c2 ⇒ c2 = 2. Thus x3 = 2x1 + 2x2 . ¸ · · ¸ 1 0 x1 (t0 ) x2 (t0 ) = det 7. The Wronskian is W (t0 ) = det 0 1 x′1 (t0 ) x′2 (t0 ) = 1 6= 0. Hence {x1 , x2 } is a fundamental set of solutions. Now x3 = c1 x1 + c2 x2 ⇒ x′3 = c1 x′1 + c2 x′2 . Using all three sets of initial conditions we have,

α = x3 (t0 ) = c1 x1 (t0 ) + c2 x2 (t0 ) = c1 × 1 + c2 × 0 = c1

β = x′3 (t0 ) = c1 x′1 (t0 ) + c2 x′2 (t0 ) = c1 × 0 + c2 × 1 = c2

Thus x3 = αx1 + βx2 .

9. Let x1 and x2 be two solutions of Airy’s equation x′′ + tx = 0,

so that x′′1 + tx1 = 0 and x′′2 + tx2 =· 0.

¸ x1 x2 Then the Wronskian is W (t) = det = x1 x′2 − x′1 x2 . x′1 x′2 dW ′ ′ ′′ ′ ′ ′′ ′′ ′′ dt = x1 x2 + x1 x2 − x1 x2 − x1 x2 = x1 x2 − x1 x2

= x1 (−tx2 ) − x2 (−tx1 )

= 0

So W (t) is constant.

11. x1 = sin t ⇒ x′1 = cos t ⇒ x′′1 = − sin t = −x1 ⇒ x′′1 + x1 = 0. x2 = cos t ⇒ x′2 = − sin t ⇒ x′′2 = − cos t = −x2 ⇒ x′′2 + x2 = 0. Thus x1 and x2 both are solutions·of x′′ + x ¸= 0. · ¸ x1 x2 sin t cos t = det The Wronskian is W [x1 , x2 ] = det x′1 x′2 cos t − sin t ¡ 2 ¢ 2 = − sin t + cos t = −1 6= 0. −

Rt

p(s)ds

According to (13), W (t) = W (t0 )e for the equation x′′ + p(t)x′ + q(t)x = 0. In this problem p(t) = 0 so that W (t) = W (t0 ) =constant. 13. x1 = et ⇒ x′1 = x′′1 = et = x1 . Then l.h.s. = x′′1 − 3x′1 + 2x1

et − 3et + 2et = 0 =r.h.s.

x2 = e2t ⇒ x′2 = 2e2t ⇒ x′′2 = 4e2t . Then

l.h.s.= x′′2 − 3x′2 + 2x2 = 4e2t − 6e2t + 2e2t = 0 =r.h.s.

Thus et and e2t both are solutions of x′′ − 3x′ + 2x = 0.

t0

32

CHAPTER 2

The Wronskian is W (t) = det

·

et e t

e2t 2e2t

−

Rt

¸

= 2e3t − e3t = e3t .

p(s)ds

According to (13), W (t) = W (t0 )e for the equation x′′ + p(t)x′ + q(t)x = 0. In this problem p(t) = −3 so that W (t) = W (t0 )e3t−3t0 = e3t0 e3t−3t0 = e3t . 15. x1 = t−1 ⇒ x′1 = −t−2 , x′′1 = 2t−3 . Then l.h.s. = t2 x′′1 + 4tx′1 + 2x1 = 2t−1 − 4t−1 + 2t−1 = 0 =r.h.s. x2 = t−2 ⇒ x′2 = −2t−3 , x′′2 = 6t−4 . Then l.h.s. = t2 x′′2 + 4tx′2 + 2x2 = 6t−2 − 8t−2 + 2t−2 = 0 =r.h.s. 2 ′′ ′ Thus t−1 and t−2 both are solutions · −1 of t x−2+ 4tx ¸ + 2x = 0. t t The Wronskian is W (t) = det −t−2 −2t−3 −4 −4 −4 = −2t + t = −t . −

t0

Rt

p(s)ds

According to (13), W (t) = W (t0 )e for the equation x′′ + p(t)x′ + q(t)x = 0. In this problem p(t) = 4t so that t0

ln

4 W (t) = W (t0 )e−4(ln t−ln t0 ) = −t− 0 e

³

t t0

´ −4

= −t−4 .

2.3 REDUCTION OF ORDER 1. x1 = t−1 ⇒ x′1 = −t−2 and x′′1 = 2t−3 so that 2t−1 − 3t−1 + t−1 = 0. −1 Let x = 1¢= vt ¡ and ¢substitute this in the original equation to ¡ vx ′ 2 −1 ′′ get¡ t vt + 3t vt−1 ¢+ vt−1¡ = 0 ⇒ ¢ t2 v ′′ t−1 − 2v ′ t−2 + 2vt−3 + 3t v ′ t−1 − vt−2 + vt−1 = 0 ⇒ v ′′ t + v ′ = 0, since the v terms cancel. Writing w =R v ′ and then R dt 1 dw dividing by t we get dw dt + t w = 0. By separation, w =− t , which, on integration, yields ln |w| = − ln t + c ⇒ w = c2 t−1 . But w = v ′ , so that v ′ = c2 t−1 . Now integrating again we get v = c2 ln t + c1 . Thus the general solution is −1 x = vx1 =©vt−1 = c2 t−1 ª ln t + c1 t . A fundamental set of solutions −1 −1 would be t , t ln t . 3. x1 = e−5t ⇒ x′1 = −5e−5t and x′′1 = 25e−5t so that 25e−5t − 50e−5t + 25e−5t = 0.¡Substitute x ¡= vx1 ¢= ve−5t ¢ ′ −5t ′′ in original equation to get¢ ve ¡ +10 ve−5t ¢+25ve−5t = 0 ⇒ ¡ ′′the −5t ′ −5t −5t ′ −5t −5t v e − 10v e + 25ve +10 v e − 5ve +25ve−5t = 0 ⇒ ′′ −5t ′ ′′ v e = 0, since the v (and v ) terms cancel ⇒ v = 0 since e−5t 6= 0. Now integrating twice we get v = c2 t + c1 . Thus the general solution is x = vx1©= ve−5t = ª c2 te−5t + c1 e−5t . A fundamental set of solutions −5t −5t . would be e , te 5. x1 = e−t ⇒ x′1 = −e−t and x′′1 = e−t , so te−t − (t − 1) e−t − e−t = 0. Let x = vx1 = ve−t and substitute this in the original equation to ′′ ′ get t (ve−t ) + (t − 1) (ve−t ) − ve−t = 0 ⇒ ′′ −t ′ −t −t t (v e − 2v e + ve ) + (t − 1) (v ′ e−t − ve−t ) − ve−t = 0 ⇒

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

33

tv ′′ e−t − (t + 1) v ′ e−t = ¡ 0,since ¢ the v terms cancel. Dividing by te−t 6= 0 we get, v ′′ − 1 + 1t v ′ = 0. Now w = v ′ yields ¢ ¢ ¡ R dw R ¡ dw 1 1 + 1t dt, which, dt − 1 + t w = 0. By separation, w = on integration, yields ln |w| = t + ln |t| + c ⇒ w = c2 tet . But w = v ′ ,

so that v ′ = c2 tet . Now using integration by parts we get

v = c2 (t − 1) et + c1 .

Thus the general solution is x = vx1 = ve−t = (c2 (t − 1) et + c1 ) e−t

= c2 (t − 1)+c1 e−t . A fundamental set of solutions would be {e−t , t − 1} .

7. x1 = t ⇒ x′1 = 1 and x′′1 = 0 so that x′′1 + tx′1 − x1 = t − t = 0. Let x = vx1 = vt and substitute this in the original equation to ′′ ′ ′′ ′ ′ get (vt) ¡ + t2 (vt) ¢ ′ − vt = 0 ⇒ (v t + 2v ) + t (v t + v) − vt = 0′ ⇒ ′′ v t + 2 + t v = 0, since the v terms cancel. Writing w = v and ¡ ¢ then dividing by t we get dw + 2t + t w = 0.

dt ¢ R R ¡2 By separation, dw

w =− t + t dt, which, on integration, yields 1 2 1 2 ln |w| = −2 ln |t| − 21 t2 + c ⇒ w = c2 e−2 ln|t|− 2 t = c2 t−2 e− 2 t . 1 2 But w = v ′ , so that v ′ = c2 t−2 e− 2 t . Now integrating by Rt 1 2 using definite integral we get, v = c2 s−2 e− 2 s ds + c1 . 1

Rt 1 2 Thus the general solution is x = vx1 = vt = c2 t s−2 e− 2 s ds + c1 t. 1 ½ ¾ Rt −2 − 1 s2 2 ds . A fundamental set of solutions would be t, t s e 1

9. Let x = tr . Then x′ = rtr−1 and x′′ = r (r − 1) tr−2 . Substituting these into t2 x′′ − 3tx′ + 4x = 0 we get, r (r − 1) tr − 3rtr + 4tr = 0 ⇒ 2 r2 − 4r + 4 = 0 (since tr 6= 0). Thus (r − 2) = 0 ⇒ r = 2. So x1 = t2 2 is a solution. Let x¡ = vx this in the original ¢ 1 = vt ¡ and ¢ substitute 2 2 ′′ 2 ′ 2 equation to get t vt − 3t vt + 4vt = 0 ⇒ ¡ ¡ ¢ ¢ t2 t2 v ′′ + 4tv ′ + 2v − 3t t2 v ′ + 2tv + 4vt2 = 0 ⇒ t4 v ′′ + t3 v ′ = 0, since the v terms cancel. v ′′ + 1t v ′ = 0. Writing w = v ′ we get R dw R 1 dw 1 dt + t w = 0. By separation, w =− t dt ⇒ ln |w | = − ln |t| + c ⇒ − ln|t| −1 ′ w = c2 e = c2 t . But w = v , so that v ′ = c2 1t . Now integrating again we get, v = c2 ln t + c1 . Thus the general solution is x = vt2 = (c2 ln t + c1 ) t2 = c2 t2 ln t + c1 t2 (i.e., the second solution is t2 ln t). 11. Let x = tr . Then x′ = rtr−1 and x′′ = r (r − 1) tr−2 . Substituting these into t2 x′′ + 7tx′ + 9x = 0 we get, r (r − 1) tr + 7rtr + 9tr = 0 2 ⇒ r2 + 6r + 9 = 0 (since tr 6= 0). Thus (r + 3) = 0 ⇒ r = −3. −3 −3 So x1 = t is a solution. Let x = substitute this 1 ¢= vt ¡ vx ¡and ¢ 2 −3 ′′ −3 ′ −3 in ¡the original equation to get t vt + 7t vt + 9vt =0⇒ ¡ ¢ ¢ t2 t−3 v ′′ − 6t−4 v ′ + 12t−5 v + 7t t−3 v ′ − 3t−4 v + 9vt−3 = 0, since the v terms cancel, t−1 v ′′ + t−2 v ′ = 0 ⇒ v ′′ + 1t v ′ = 0. R dw R 1 1 Writing w = v ′ we get dw dt + t w = 0. By separation, w =− t dt − ln|t| −1 ′ = c2 t . But w = v , so that ⇒ ln |w| = − ln |t| + c ⇒ w = c2 e

34

CHAPTER 2

v ′ = c2 1t . Now integrating again we get, v = c2 ln t + c1 . Thus the general solution is x = vt−3 = (c2 ln t + c1 ) t−3 = c2 t−3 ln t + c1 t−3 (i.e., the second solution is t−3 ln t). 13. Let x = sin (rt) . Then x′ = r cos (rt) and x′′ ¡= −r2 sin ¢ (rt) . Substituting these into x′′ + 4x = 0 we get, 4 − r2 sin (rt) = 0 ⇒ 4 − r 2 = 0 (since sin (rt) = 6 0 as nontrivial solution). Thus r = 2. So x1 = sin (2t) is a solution. Let x = vx1 = v sin (2t) and substitute this in the original equation to get ′′ (v sin (2t)) + 4v sin (2t) = 0 ⇒ ′′ (v sin (2t) + 4v ′ cos (2t) − 4v sin (2t)) + 4v sin (2t) = 0, since the v ′ terms cancel, v ′′ sin (2t) + 4v ′ cos (2t) = 0 ⇒ v ′′ + 4 cos(2t) sin(2t) v = 0. R

4 cos(2t) sin(2t) w = 0. By integrating factors, 1 = ce−4( 2 ln(sin(2t))) = c 21 = c csc2 (2t) .

dw dt + cos(2t) dt sin(2t)

Writing w = v ′ we get

we have w = ce−4 sin (2t) But w = v ′ , so that v ′ = c csc2 (2t) . Now using integration table we c get, v = − 2 cot (2t) + c1 = c2 cos(2t) + c1 . Thus the general solution sin(2t) ´ ³ cos(2t) is x = v sin (2t) = c2 sin(2t) + c1 sin (2t) = c1 sin (2t) + c2 cos (2t) .

15. Let x = ert ⇒ x′ = rert and¡ x′′ = r2 ert¢. Substituting these into x′′ − 4x′ + 4x = 0 we get, r2 − 4r + 4 ert = 0 ⇒ r2 − 4r + 4 = 0 2 (since ert 6= 0). Thus (r − 2) = 0 ⇒ r = 2. So x1 = e2t is a solution. 2t Let ¡x = vx this in the original equation to ¢ 1 = ve ¡ 2tand ¢′ substitute 2t ′′ get ve − 4 ve + 4ve2t = 0 ⇒ v ′′ e2t + 4v ′ e2t + 4ve2t − 4v ′ e2t − 8ve2t + 4ve2t = 0, since the v (and v ′ ) terms cancel, v ′′ e2t = 0 ⇒ v ′′ = 0(since e2t 6= 0). Integrating twice we have, v = c1 + c2 t. Thus the general solution is x = ve2t = (c1 + c2 t) e2t = c1 e2t + c2 te2t . ′′ 2 rt 17. Let x = ert ⇒ x′ = rert and Substituting these into ¡ 2x = r e ¢ . rt ′′ ′ x + 2x + x = 0 we get, r + 2r + 1 e = 0 2 ⇒ r2 + 2r + 1 = 0 (since ert 6= 0). Thus (r + 1) = 0 ⇒ r = −1. −t −t So x1 = e is a solution. Let x = vx1 = ve and substitute this in ′′ ′ the original equation to get (ve−t ) + 2 (ve−t ) + ve−t = 0 ⇒ ′′ −t ′ −t −t ′ −t −t −t v e − 2v e + ve + 2v e − 2ve + ve = 0, since the v (and v ′ ) terms cancel, v ′′ e−t = 0 ⇒ v ′′ = 0(since e−t 6= 0). Integrating twice we have, v = c1 + c2 t. Thus the general solution is x = ve−t = (c1 + c2 t) e−t = c1 e−t + c2 te−t . 19. Let x = ert ⇒ x′ = re¡rt and ¢x′′ = r2 ert . Substituting these into x′′ − 4x = 0 we get, r2 − 4 ert = 0 ⇒ r2 − 4 = 0 (since ert 6= 0). Thus r2 = 4 ⇒ r = ±2. So x1 = e2t is a solution. Let x = vx1 = ve2t and this in the original equation to get ¡ 2tsubstitute ¢′′ ve − 4ve2t = 0 ⇒ v ′′ e2t + 4v ′ e2t + 4ve2t − 4ve2t = 0, since the v terms cancel, v ′′ e2t + 4v ′ e2t = 0 ⇒ v ′′ + 4v ′ = 0(since e2t = 6 0). −4t Writing w = v ′ we get dw + 4w = 0. So w = ce . But w = v ′ , so dt 1 −4t ′ −4t that v = ce . Now integrating again we get, v = − 4 ce + c1 = c2 e−4t + c1 . Thus the general solution is

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

¡

35

¢

x = ve2t = c2 e−4t + c1 e2t = c1 e2t + c2 e−2t . 21. x2 = vx1 where v must be a non-constant function of t (otherwise x2 is a scalar multiple of x1 and so they dependent). · are linearly ¸ · ¸ x1 vx1 x1 x2 The Wronskian is W [x1 , x2 ] = det = det x′1 x′2 x′1 v ′ x1 + vx′1 2 2 = v ′ (x1 ) + vx1 x′1 − vx1 x′1 = v ′ (x1 ) 6= 0 (since v ′ 6= 0 as v is a non-constant function of t and x1 = 6 0 as a nontrivial solution). Hence {x1 , x2 } forms a fundamental set of solutions. 2.4 HOMOGENEOUS LINEAR CONSTANT COEFFICIENT DIFFERENTIAL EQUATIONS (SECOND ORDER) 1. Substituting x = ert into the equation x′′ + x′ − 6x = 0 we get r2 + r − 6 = 0 ⇒ (r + 3) (r − 2) = 0 ⇒ r = 2, −3.Thus the general solution is x = c1 e2t + c2 e−3t . 3. Substituting x = ert into the equation x′′ + x = 0 we get r2 +1 = 0 ⇒ r = ±i. Thus the general solution is x = c1 cos t+c2 sin t. 5. Substituting x = ert into the equation x′′ + 4x′ + 5x = 0 we get √ r2 + 4r + 5 = 0 ⇒ r = −4± 216−20 = −2 ± i.Thus the general solution is x = e−2t (c1 cos t + c2 sin t) . 7. Substituting x = ert into the equation x′′ − 3x′ + 2x = 0 we get r2 − 3r + 2 = 0 ⇒ (r − 1) (r − 2) ⇒ r = 1, 2.Thus the general solution is x = c1 et + c2 e2t . 9. Substituting x = ert into the equation x′′ − x′ = 0 we get r2 − r = 0 ⇒ r (r − 1) = 0 ⇒ r = 0, 1.Thus the general solution is x = c1 +c2 et . 11. Substituting x = ert into the equation 3x′′ = 0 we get 3r2 = 0

⇒ r = 0, a repeated root with multiplicity 2. Thus the general

solution is x = c1 + c2 t. 13. Substituting x = ert into the equation 3x′′ + 2x′ − x = 0 we get

1 . Thus

3r2 + 2r − 1 = 0 ⇒ (r + 1) (3r − 1) = 0 ⇒ r = −1, 3

1 −t t

the general solution is x = c1 e + c2 e 3 . 15. Substituting x = ert into the equation x′′ + x′ − 2x = 0 we get

r2 + r − 2 = 0 ⇒ (r + 2) (r − 1) = 0 ⇒ r = 1, −2. Thus

the general solution is x = c1 et + c2 e−2t . So x′ = c1 et − 2c2 e−2t .

Using the initial conditions x(0) = 0 and x′ (0) = 1 we get

c1 + c2 = 0 and c1 − 2c2 = 1. Solving these equations for

c1 and c2 , we have c1 = 31 and c2 = − 31 .

Hence x = 31 et − 13 e−2t .

17. Substituting x = ert into the equation 2x′′ + 4x = 0 we get √ 2 2 2r + 4 = 0 ⇒ √r = ±i 2. Thus the general solution √ r = −2 ⇒ is x = c1 cos 2t + c2 sin 2t. 19. Substituting x = ert into the equation 2x′′ + 8x′ + 6x = 0 we get

2r2 + 8r + 6 = 0 ⇒ r2 + 4r + 3 = 0 ⇒ (r + 3) (r + 1) = 0

36

CHAPTER 2

⇒ r = −1, −3.Thus the general solution is x = c1 e−t + c2 e−3t . So x′ = −c1 e−t − 3c2 e−3t . Using the initial conditions x(0) = 2 and x′ (0) = 0 we get c1 + c2 = 2 and −c1 − 3c2 = 0. Solving these equations for c1 and c2 , we have c1 = 3 and c2 = −1. Hence x = 3e−t − e−3t . 21. Substituting x = ert into the equation x′′ + 10x′ + 25x = 0 we

2 get, r2 + 10r + 25 = 0 ⇒ (r + 5) = 0 ⇒ r = −5, repeated

twice. Thus the general solution is x = c1 e−5t + c2 te−5t . 23. Substituting x = ert into the equation x′′ − 14x′ + 49x = 0 we

2 get, r2 − 14r + 49 = 0 ⇒ (r − 7) = 0 ⇒ r = 7, repeated

twice. Thus the general solution is x = c1 e7t + c2 te7t . 25. Substituting x = ert into the equation x′′ − 6x′ + 25x = 0 we get √ 6± 36−100 2 r − 6r + 25 = 0 ⇒ r = = 3 ± 4i.Thus the general 2 solution is x = e3t (c1 cos 4t + c2 sin 4t) . 27. Substituting x = ert √ into the equation x′′ − 12x = 0 we get 2 r − 12 √= 0 ⇒ r = √ ± 12. Thus the general solution is

x = c1 e 12t + c2 e− 12t . 29. Substituting x = ert into the equation x′′ + 4x′ + 8x = 0 we get

√ r2 + 4r + 8 = 0 ⇒ r = −4± 216−32 = −2 ± 2i.Thus the general

solution is x = e−2t (c1 cos 2t + c2 sin 2t) .

31. Substituting x = ert√into the √ equation x′′ + 8x = 0 we get

2 r +8=0 ⇒r =± √ −8 = ± 8i.Thus the general

√ solution is x = c1 cos 8t + c2 sin 8t.

33. Substituting x = ert into the equation x′′ + 6x′ + 7x = 0 we get

√ √ −6± 36−28 2 general ´ = −3 ± 2.Thus r + 6r + 7 = 0 ⇒ r = 2 ³ the √ √ √ √ (−3− 2)t −3t (−3+ 2)t 2t c1 e + c2 e− 2t . =e solution is x = c1 e + c2 e 35. The Wronskian W [eαt cos βt, eαt sin βt]

· ¸

eαt cos βt eαt sin βt = det eαt (α cos βt − β sin βt) eαt (α sin βt + β cos βt) ¡ ¢ 2αt α sin βt cos βt + β cos2 βt − α sin βt cos βt + β sin2 βt =e ¡ ¢ = βe2αt cos2 βt + sin2 βt = βe2αt 6= 0 (since β = 6 0). So {eαt cos βt, eαt sin βt} forms a fundamental set of solutions. 2 37. Repeated roots have b2 − 4ac = 0 ³⇒ c = 4ba , so that ´ ¢ ¡ b2 b2 b 2 ar2 + br + c = ar2 + br + 4a = a r2 + ab r + 4a = a r + 2a . 2

b . So repeated root is r = − 2a 39. Choose #21. This problem has a repeated root, r = −5. So one −5t solution is x1 = e−5t . Let x = vx = ve and this ¡ 1 −5t ¢′′ ¡ substitute ¢ −5t ′ −5t in the original equation to get ve + 10 ve + 25ve =0 ¡ ′′ −5t ¢ ′ −5t −5t ⇒ v e − 10v e + 25ve ¡ ¢ +10 v ′ e−5t − 5ve−5t + 25ve−5t = 0 ⇒ v ′′ e−5t = 0 ⇒ v ′′ = 0 since e−5t 6= 0. Now integrating twice we get v = c2 t + c1 . Thus the general solution is

x = vx1 = ve−5t = c2 te−5t + c1 e−5t .

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

37

41. c1 e−t + c2 e−2t will be the general solution if r = −1 and r = −2 are the roots of the characteristic equation. One such characteristic equation would be (r + 1) (r + 2) = r2 + 3r + 2 = 0. A corresponding differential equation is x′′ + 3x′ + 2x = 0. 43. c1 e3t + c2 te3t will be the general solution if r = 3 is a repeated (twice) root of the characteristic equation. One such characteristic 2 equation would be (r − 3) = r2 − 6r + 9 = 0. A corresponding ′′ differential equation is x − 6x′ + 9x = 0. 45. c1 sin 4t + c2 cos 4t will be the general solution if r = ±4i is a complex root of the characteristic equation. One such characteristic equation would be (r + 4i) (r − 4i) = r2 + 16 = 0. A corresponding differential equation is x′′ + 16x = 0. 47. c1 sin 3t + c2 cos 3t will be the general solution if r = ±3i is a complex root of the characteristic equation. One such characteristic equation would be (r + 3i) (r − 3i) = r2 + 9 = 0. A corresponding differential equation is x′′ + 9x = 0. 49. c1 + c2 t will be the general solution if r = 0 is a repeated root

of the characteristic equation. One such characteristic

equation would be r2 = 0. A corresponding

differential equation is x′′ = 0.

51. c1 et sin t + c2 et cos t will be the general solution if r = 1 ± i is a complex root of the characteristic equation. One such characteristic equation would be (r − 1 − i) (r − 1 + i) = r2 − 2r + 2 = 0. A corresponding differential equation is x′′ − 2x′ + 2x = 0. 2.4.1 Homogeneous Linear Constant Coefficient Differential Equa­ tions (nth-Order) 1. Substituting x = ert into the equation x′′′ − 6x′′ + 12x′ − 8x = 0 3 we get, r3 − 6r2 + 12r − 8 = 0 ⇒ (r − 2) = 0 ⇒ r = 2 is a repeated root of multiplicity 3.Thus the general solution is x = c1 e2t + c2 te2t + c3 t2 e2t . ′′′′ ′′ 3. Substituting x = ert ¡into the¢ ¡equation ¢ x − 5x + 4x = 0 we get, 4 2 2 2 r − 5r + 4 = 0 ⇒ r − 1 r − 4 = 0 ⇒ r = ±1, ±2. Thus the general solution is x = c1 et + c2 e−t + c3 e2t + c4 e−2t . 5. Substituting x = ert into the equation x′′′ + x′′ − 2x′ = 0 we get, r3 + r2 − 2r = 0 ⇒ r (r − 1) (r + 2) = 0 ⇒ r = 0, 1, −2. Thus the general solution is x = c1 + c2 et + c3 e−2t . 7. Substituting x = ert into the equation x′′′′ + 4x′′′ + 6x′′ + 4x′ + x = 0 4 we get, r4 + 4r3 + 6r2 + 4r + 1 = 0 ⇒ (r + 1) = 0 ⇒ r = −1 is a repeated root of multiplicity 4. Thus the general solution is x = c1 e−t + c2 te−t + c3 t2 e−t + c4 t3 e−t . 9. Substituting x = ert into the equation x′ − 3x = 0 we get, r − 3 = 0 ⇒ r = 3. Thus the general solution is x = c1 e3t .

38

CHAPTER 2

′′′ ′′ 11. Substituting x = ert into the ¡ 2equation ¢x + x − 2x = 0 we get, 3 2 r + r − 2 = 0 ⇒ (r − 1) r + 2r + 2 = 0 ⇒ r = 1, −1 ± i. Thus the general solution is x = c1 et + e−t (c2 cos t + c3 sin t) . 13. Substituting x = ert into the equation x′′′′ + 4x′′′ + 8x′′ + 8x′ + 4x = 0 we ¡get, r4 + 4r3 +¢ 8r2 + 8r + 4 = 0

⇒ r4 + 4r3 + 4r2 + 4r2 + 8r + 4 = 0

¡ ¢2 ¡ ¢ ⇒ r2 + 2r + 4 r2 + 2r + 4 = 0 ¡ 2 ¢2 ⇒ r + 2r + 2 = 0 ⇒ r = −1 ± i is a repeated root of multiplicity 2. Thus the general solution is x = e−t (c1 cos t + c2 sin t) + te−t (c3 cos t + c4 sin t) . 15. Substituting x = ert into the equation x(4) −4x(3) +6x(2) −4x(1) +x = 0 4 we get, r4 − 4r3 + 6r2 − 4r + 1 = 0 ⇒ (r − 1) = 0 ⇒ r = 1 is a repeated root of multiplicity 4. Thus the general solution is x = c1 et + c2 tet + c3 t2 et + c4 t3 et . 17. Substituting x = ert into the equation x(6) − 3x(4) + 3x(2) − x = 0 we get, r6 − 3r4 + 3r2 − 1 = 0

¡ ¢3 ¡ ¢2 ¡ ¢ ⇒ r2 − 3 r2 .1 + 3 r2 .12 − 13 = 0

¡ 2 ¢3 ⇒ r − 1 = 0 ⇒ r = ±1, each repeated with multiplicity 3. Thus the general solution is

x = c1 et + c2 tet + c3 t2 et + c4 e−t + c5 te−t + c6 t2 e−t .

rt (4) 4 19. Substituting ¡ 2 ¢ ¡x2= e ¢ into the 2equation x 2 − x = 0 we get, r − 1 = 0 ⇒ r − 1 r + 1 = 0 ⇒ r = 1 and r = −1 ⇒ r = ±1, ±i. Thus the general solution is x = c1 et + c2 e−t + c3 cos t + c4 sin t. 21. Substituting x = ert into the equation x′′′′ + 50x′′ + 625x = 0

¡ ¢2 we get, r4 + 50r2 + 625 = 0 ⇒ r2 + 25 = 0 ⇒ r = ±5i

repeated twice. Thus the general solution is

x = c1 cos 5t + c2 sin 5t + c3 t cos 5t + c4 t sin 5t.

′′′ ′′ ′ 23. Substituting x = ert into the equation ¡ 2 x + 3x¢ + x − 5x = 0 we 3 2 get, r + 3r + r − 5 = 0 ⇒ (r − 1) r + 4r + 5 = 0 ⇒ r = 1, −2 ± i. Thus the general solution is x = c1 et + e−2t (c2 cos t + c3 sin t) . 25. c1 e2t + c2 te2t + c3 will be the general solution if r = 2 is a root with multiplicity 2 and r = 0 is a simple root of the characteristic 2 equation. One such characteristic equation would be r (r − 2) = r3 − 4r2 + 4r = 0. A corresponding differential equation is x′′′ − 4x′′ + 4x′ = 0. 27. c1 sin 5t + c2 cos 5t + c3 t sin 5t + c4 t cos 5t will be the general solution if r = ±5i is a complex root with multiplicity 2 of the characteristic 2 equation. One such characteristic equation would be ((r − 5i) (r + 5i)) ¡ 2 ¢2 = r + 25 = r4 + 50r2 + 625 = 0. A corresponding differential equation is x′′′′ + 50x′′ + 625x = 0. 29. c1 e−2t + c2 te−2t + c3 t2 e−2t will be the general solution if r = −2 is a root with multiplicity 3 of the characteristic equation. One such 3 characteristic equation would be (r + 2) = r3 + 6r2 + 12r + 8 = 0. A corresponding differential equation is x′′′ + 6x′′ + 12x′ + 8x = 0.

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

39

31. c1 et sin t + c2 et cos t + c3 tet sin t + c4 tet cos t will be the general solution if r = 1 ± i is a complex root with multiplicity 2 of the characteristic equation. One such characteristic equation would be ¡ ¢2 2 ((r − 1 − i) (r − 1 + i)) = r2 − 2r + 2 =

r4 − 4r3 + 8r2 − 8r + 4 = 0. A corresponding differential

equation is x′′′′ − 4x′′′ + 8x′′ − 8x′ + 4x = 0.

33. c1 sin t + c2 cos t + c3 t sin t + c4 t cos t will be the general solution if the roots of the characteristic equation is r = ±i repeated twice. ¡ ¢2 2 One such characteristic equation would be ((r − i) (r + i)) = r2 + 1 = r4 + 2r2 + 1 = 0. A corresponding differential equation is x′′′′ + 2x′′ + x = 0. 35. Since there are 6 roots, four real (repeated), two complex, the general solution is a linear combination of 6 homogeneous solutions using 6 arbitrary ¡ constants: ¢ x = e8t c1 + c2 t + c3 t2 + c4 t3 + e8t (c5 cos 7t + c6 sin 7t) . 37. Since there are 9 roots, three real (repeated), one pair of complex

(repeated twice) and one pair of complex (simple), the general

solution is a linear combination of 9 homogeneous solutions

using 9 arbitrary constants:

x = c1 + c2 t + c3 t2 + e4t (c4 cos 5t + c5 sin 5t) + (c6 + c7 t) cos 5t

+ (c8 + c9 t) sin 5t. 39. Since there are six complex roots (one pair repeated three times), the general solution is a linear combination of 6 homogeneous solutions using 6 £¡ arbitrary constants: ¢ ¡ ¢ ¤ x = e2t c1 + c2 t + c3 t2 cos t + c4 + c5 t + c6 t2 sin t . 41. Since there are 8 complex roots (one pair repeated twice, other two pairs simple), the general solution is a linear combination of 8 homogeneous solutions using 8 arbitrary constants: x = e2t (c1 cos 6t + c2 sin 6t) + e−2t (c3 cos 6t + c4 sin 6t) + (c5 + c6 t) cos 6t + (c7 + c8 t) sin 6t. 2.5 MECHANICAL VIBRATIONS I: FORMULATION AND FREE RESPONSE p √ √ 2 9 + 49 = 58. The phase 1. The amplitude is R = c2 1 + c2 = angle φ is given by tan φ = cc21 = − 73 , fourth quadrant. So √ ¡ 7¢ = −1.1659 radians. Thus x (t) = 58 cos (5t + 1.1659) . φ = tan−1 − 3

p √ 3. The amplitude is R = c21 + c22 = 3 + 1 = 2. The phase angle

φ is given by tan φ = cc21 = √13 , first quadrant. So

³ ´ ¡ ¢ φ = tan−1 √13 = π6 radians. Thus x (t) = 2 cos 14t − π6 . p √ √ √ 5. The amplitude is R = c21 + c22 = 36 + 36 = 72 = 6 2. 6 The phase angle φ is given by tan φ = cc21 = −6 = −1,

second quadrant. So φ = tan−1 (−1) + π = − π4 + π = 3π

4 radians.

40

CHAPTER 2

√ ¢ ¡ Thus x (t) = 6 2 cos 5t − 3π 4 . √ p 2 7. The amplitude is R = c2 3 + 1 = 2. The phase angle 1 + c2 = −1 φ is given by tan φ = cc21 = √ , fourth quadrant. So 3 ´ ³ ¢ ¡ φ = tan−1 − √13 = − π6 radians. Thus x (t) = 2 cos 6t + π6 . p √ 2 9. The amplitude is R = c2 16 + 48 = 8. The phase angle 1 + c√2 = √ c2 4 3 φ is given by tan φ = c1 = −4 = − 3, second quadrant. So √ ¡ ¢ φ = tan−1 − 3 + −¢π3 + π = 2π 3 radians.

¡ π = 2π

Thus x (t) = 8 cos 2t − 3 . 11. We determine the spring constant k from kΔL = mg where m = 30 gm, g = 980 cm/s2 and ΔL = 20 cm. So k = 980×30 = 1470. 20 The differential equation is mx′′ + kx = 0 ⇒ 30x′′ + 1470x = 0 ⇒ x′′ + 49x = 0. Substituting x = ert we have r2 + 49 = 0 ⇒ r = ±7i. Hence the general solution is x (t) = c1 cos 7t + c2 sin 7t ⇒ x′ (t) = −7c1 sin 7t + 7c2 cos 7t. Using the initial conditions x (0) = 10 and x′ (0) = 0 we get, c1 = 10 and c2 = 0. Thus the motion is x = 10 cos 7t. 13. We determine the spring constant k from kΔL = mg where

m = 8 slugs, g = 32 f t/s2 and ΔL = 2 f t. So k = 8×232 = 128.

For spring-mass system m = 2 slugs, the differential equation is

2x′′ + 128x = 0 ⇒ x′′ + 64x = 0. Substituting x = ert we have

r2 + 64 = 0 ⇒ r = ±8i. Hence the general solution

is x (t) = c1 cos 8t + c2 sin 8t ⇒ x′ (t) = −8c1 sin 8t + 8c2 cos 8t.

Using the initial conditions x (0) = 2 and x′ (0) = 1 we get, c1 = 2 and c2 = 81 . Thus the motion is x = 2 cos 8t + 18 sin 8t. 15. Here m = 10. Frequency 2ωπ = 5 cycles/sec⇒ ω = 10π ⇒ q k k 2 2 lb/ft. m = 10π ⇒ 10 = 100π ⇒ k = 1000π 17. Here m = 16 g, k = 64 gm/s2 . The differential equation is

16x′′ + 64x = 0 ⇒ x′′ + 4x = 0. Substituting x = ert we have

r2 + 4 = 0 ⇒ r = ±2i. Hence the general solution is

(0) = c1 and x′ (0) = 2c2 .

x (t) = c1 cos 2t + c2 sin 2t. So ³ x´ ¡ ¢ √ Now, phase, φ = π3 = tan−1 cc21 ⇒ cc21 = tan π3 = 3 ⇒ p √ c2 = 3c1 and amplitude, c21 + c22 = 2 ⇒ c21 + c22 = 4 ⇒ √ ¡√ ¢2 c21 + 3c1 = 4 ⇒ 4c2 3. 1 = 4 ⇒ c1 = 1 and then c2 = √ Thus the initial conditions must be x (0) = 1 and x′ (0) = 2 3. k x = 0. 19. (a) The differential equation is mx′′ + kx = 0 ⇒ x′′ + m q k k Substituting x = ert we have r2 + m =0⇒r=± m i. q q k k Hence the general solution is x (t) = c1 cos m t + c2 sin m t⇒ q q q q k k k k x′ (t) = − m c1 sin m t+ m c2 cos m t. Using the initial conditions x (0) = 0 and x′ (0) = q10 we get, pm pm k t. c1 = 0 and c2 = 10 k . Thus x (t) = 10 k sin m

41 p (b) The amplitude is R = c21 + c22 = 10 m k. (c) From part (b) the amplitude decreases as k increases. (d) From part (b) the amplitude increases as m increases. k 21. (a) The differential equation is mx′′ + kx = 0. ⇒ x′′ + m x = 0. q k k rt 2 Substituting x = e we have r + m = 0 ⇒ r = ± m i. q q k k Hence the general solution is x (t) = c1 cos m t + c2 sin m t q q q q k k k k ⇒ x′ (t) = − m c1 sin m t+ m c2 cos m t. Using the initial conditions x (0) = 1 and x′ (0) =1 q= 1 wepget, c1 q pm k m k and c2 = k . Thus the motion is x (t) = cos mt + k sin mt . p p (b) The amplitude is R = c21 + c22 = 1 + m k. (c) From part (b) the amplitude decreases to 1 as k increases and increases as m increases. 23. Multiplying mx′′ + kx = 0 by x′ we get, mx′ x′′ + kxx′ = 0.

Now integrating R this equation with respect to t we have

R mx′ x′′ dt + kxx′ dt =constant. ⇒ R d R d ′ 2 2 (x ) dt + k 12 dt (x) dt =constant

m 12 dt 2 1 1 ′ 2 ⇒ 2 m (x ) + 2 k (x) =constant

⇒ 12 mv 2 + 12 kx2 =constant. √ 2 25. ddt2x + 7x = 0. Substituting x = ert we have r2 + 7 = 0 ⇒ r = ± 7i. √ √ Hence the general solution is x (t) = c1 cos 7t + c2 sin 7t. √ d2 x 27. dt2 − 7x = 0. Substituting x = ert we have r2 − 7 = 0 ⇒ r = ± 7. √ √ Hence the general solution is x (t) = c1 e 7t + c2 e− 7t . √ 2 29. ddt2x + 5x = 0. Substituting x = ert we have r2 + 5 = 0 ⇒ r = ± 5i. √ √ Hence the √ general solution √ √is x (t) = √ c1 cos 5t + c2 sin 5t and dx sin 5c 5t + 5c cos 5t. Using the initial conditions (t) = − 1 2 dt x (0) = 2 and dx (0) = 3 we have, c = 2 and c2 = √35 . 1 dt √ √ Thus x (t) = 2 cos 5t + √35 sin 5t. The amplitude is q q p 9 29 2 = R = c2 4 + = and phase angle is

+ c 1 5 ³ 2´ ³ ´5 3

φ = tan−1 cc12 = tan−1 2√ = 0.59087. 5 q ¡√ ¢ 5t − 0.59087 . Hence x (t) = 29 5 cos

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

p

31. Suppose dθ dt = c1 (constant). The circle of radius r can be represented parametrically by x = r cos θ, y = r sin θ. Integrating the differential equation we have, θ = c1 t + c2 , where c2 is constant. Thus x component of the object satisfies a simple harmonic motion x = r cos (c1 t + c2 ) . Hence the number of cycles per second is c1 ω0 2π = 2π (here ω0 = c1 ). That is, c1 =cycles per 2π seconds, which is the circular frequency. 33. Here m = 10, k = 30, δ = 40. The differential equation is 10x′′ + 40x′ + 30x = 0 ⇒ x′′ + 4x′ + 3x = 0, x (0) = 3, x′ (0) = −5.

42

CHAPTER 2

Substituting x = ert we have r2 + 4r + 3 = 0 ⇒ (r + 1) (r + 3) = 0 ⇒ r = −1, −3. Hence the general solution is x (t) = c1 e−t + c2 e−3t . Also x′ (t) = −c1 e−t − 3c2 e−3t . Using initial conditions we get, c1 + c2 = 3

−c1 − 3c2 = −5

Adding these equations we get c2 = 1 and then c1 = 2 . Thus

x (t) = 2e−t + e−3t .

Here, the roots are real (and negative), so the system is overdamped.

35. Here m = 1, ΔL = 20, kΔL = mg = 980 ⇒ k = 980 20 = 49, δ = 0. The differential equation describing the motion is x′′ + 49x = 0, x (0) = 1, x′ (0) = 7. Substituting x = ert we have r2 + 49 = 0 ⇒ r = ±7i. Hence the general solution is x (t) = c1 cos 7t + c2 sin 7t. Also x′ (t) = −7c1 sin 7t + 7c2 cos 7t. Using initial conditions we get, c1 = 1 and c2 = 1. Thus x (t) = cos 7t + sin 7t. This motion is harmonic. 37. Here ΔL = 2 32 , k = 12, mg = kΔL = 32 ⇒ m = 1, δ = 7. The differential equation describing the motion is

x′′ + 7x′ + 12x = 0, x (0) = −1, x′ (0) = 1. Substituting x = ert

we have r2 + 7r + 12 = 0 ⇒ (r + 4) (r + 3) = 0 ⇒ r = −4, −3.

Hence the general solution is x (t) = c1 e−4t + c2 e−3t . Also

x′ (t) = −4c1 e−4t − 3c2 e−3t . Using initial conditions we get,

c1 + c2 = −1; −4c1 − 3c2 = 1 ⇒ c1 = 2 and c2 = −3. Thus

x (t) = 2e−4t − 3e−3t . Here, the roots are real (and negative), so the system is overdamped.

39. Here k = 5, m = 1, δ = 4. The differential equation describing the motion is x′′ + 4x′ + 5x = 0, x (0) = 2, x′ (0) = 0. Substituting x = ert we have r2 + 4r + 5 = 0 ⇒ r = −2 ± i. Hence the general solution is x (t) = e−2t (c1 cos t + c2 sin t) . Also x′ (t) = e−2t ((−2c1 + c2 ) cos t − (c1 + 2c2 ) sin t) . Using initial conditions we get, c1 = 2; −2c1 + c2 = 0 ⇒ c2 = 4. Thus x (t) = e−2t (2 cos t + 4 sin t) . Here, the roots are complex (with negative real part), so we have oscillation. The amplitude is a damped p √ √ R = c21 +³c22 ´= 4 + 16 = 20 and the phase angle is φ = tan−1

c2 c1

= tan−1 (2) = 1.107. √ Hence x (t) = e−2t 20 cos (t − 1.107) . 41. (Refer to the equation (36) of the text) The general solution for the δ δ critically damped motion is x (t) = c1 e− 2m t + c2 te− 2m t ⇒ £ ¤ δ δ x′ (t) = − 2m (c1 + c2 t) + c2 e− 2m t . The mass may change direction at the point where the derivative is zero. So we solve δ δ x′ (t) = 0 for t. Since e− 2m t = (c1 + c2 t) + c2 = 0 ⇒ 6 0, − 2m c1 2m t = δ − c2 , which is just one number. If t < 0, this does not happen. √ 43. If δ < √4mk, then damped oscillation.

If δ > 4mk, then overdamped.

43 √ And if δ = 4mk, then critically damped.

2 For small m (0 < m < 4δk ) motion will be overdamped. As m is

2 increased, motion will be critically damped at m = 4δk and motion 2 will be a damped oscillation if m is increased further (m > 4δk ). q √ k 45. Without damping: natural frequency, ω0 = m = k (since m = 1). √ 0 k = 10π ⇒ k = 100π 2 . Now ω 2π = 5 ⇒ √ √ 2 2 With damping: pseudo frequency, β = 4mk−δ = 4k−δ (as m = 1). 2m 2

√ β 4k−δ 2 2 2

So 2π = 4 ⇒ = 8π ⇒ 4k − δ = 256π ⇒ 2 δ 2 = 400π 2 − 256π 2 = 144π 2 ⇒ δ = 12π. 2 2 47. Substituting x = ert in 2 ddt2x + 4 dx dt + 3x = 0, we have 2r + 4r + 3 = 0 √ 2 ⇒ r = −1 ± ³ 2 i. Thus the general´solution is

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

√

√

x (t) = e−t c1 cos 22 t + c2 sin 22 t . Here, underdamped system since the roots are complex (with negative real part). 2 2 49. Substituting x = ert in 3 ddt2x + 5 dx dt + 4x = 0, we have 3r + 5r + 4 = 0 √ 23 ⇒ r = − 65 ± ³ 6 i. Thus the general´solution is x (t) = e−t c1 cos

√

23 6 t

+ c2 sin

√

23 6 t

.

Here, underdamped system since the roots are complex (with negative real part). 2 2 51. Substituting x = ert in 3 ddt2x + 8 dx dt + 4x = 0, we have 3r + 8r + 4 = 0 2 ⇒ (r + 2) (3r + 2) = 0 ⇒ r = −2, − 3 . Thus the general solution is 2 x (t) = x (t) = c1 e−2t + c2 e− 3 t . Here, overdamped system since the roots are real (and negative). 2 53. Substituting x = ert in 9 ddt2x + 12 dx dt + 4x = 0, we have 2 9r2 + 12r + 4 = 0 ⇒ (3r + 2) = 0 ⇒ r = − 23 repeated twice. 2 2 Thus the general solution is x (t) = c1 e− 3 t + c2 te− 3 t . Here, critically damped system since the roots are repeated negative real. 2 2 55. Substituting x = ert in 3 ddt2x + 4 dx dt + x = 0, we have 3r + 4r + 1 = 0 1 ⇒ (r + 1) (3r + 1) = 0 ⇒ r = −1, − 3 . Thus the general solution is 1 x (t) = c1 e−t + c2 e− 3 t . Here, overdamped system since the roots are real (and negative). 57. For an overdamped spring-mass system, (Refer to the equation (34) and (35) of the text) the general solution is x (t) = c1 er1 t + c2 er2 t . So x′ (t) = r1 c1 er1 t + r2 c2 er2 t . Using initial condition x (0) = 1 we get, c1 + c2 = 1 ⇒ c2 = 1 − c1 and x′ (0) = v0 we get, r1 c1 + r2 c2 = v0 ⇒ r1 c1 + r2 (1 − c1 ) = v0 ⇒ 0 c1 (r1 − r2 ) = v0 − r2 ⇒ c1 = − rr21 −v −r2 and then r2 −v0 r1 −v0 c2 = 1 − c1 = 1 + r1 −r2 = r1 −r2 . 0 r1 t 0 r2 t So x′ (t) = −r1 rr21 −v + r2 rr11 −v . Now local maximum or −r2 e −r2 e minimum (if exists) occurs at t such that x′ (t) = 0 ⇒

44

CHAPTER 2 er 1 t er 2 t

1 −v0 ) = rr21 (r (r2 −v0 ) = µ v0 ¶

v 1− r0 1− (r1 −r2 )t 1 e = 1− v0 ⇒ (r1 − r2 ) t = ln 1− vr10 ⇒

r2 r2 µ v0 ¶

1− 1 ln 1− vr10 .

t = (r1 −r 2) 0 r1 t 0 r2 t r1 rr21 −v = r2 rr11 −v ⇒ −r2 e −r2 e

v 1− r0 1 v0 1− r 2

⇒

r2

59. Substituting√x = ert in x′′ + δx′ + x = 0, we have r2 + δr + 1 = 0 2 ⇒ r = −δ± 2 δ −4 . r1 t r1 t r2 t Then x¡(t) = c√ + ¢c2 er2 t ; x′ (t) = 1e ¡ c1 r1 e√ + c2 r¢2 e where 1 1 2 2 r1 = 2 −δ + δ − 4 and r2 = 2 −δ − δ − 4 . For δ > 2 : Using the initial conditions x (0) = 1 and x′ (0) = −1 we get c1 + c2 = 1 ⇒ c2 = 1 − c1 and +1 r1 c1 + r2 c2 = −1 ⇒ c1 (r1 − r2 ) = −1 − r2 ⇒ c1 = − rr12−r 2 r2 +1 r1 +1 r2 +1 r1 t +1 r2 t and c2 = 1 + r1 −r2 = r1 −r2 . Thus xδ (t) = − r1 −r2 e + rr11−r e . 2 In order to find lim xδ (t) we must find limits of c1 , c2 , r1 , and r2 δ →2+

when δ → 2+ as they depend on δ.√ 2 +1 lim+ c1 = lim+ − rr12−r . Using L’Hospital’s Rule = lim+ δ+2√δδ2−4−2 2 −4

δ →2

δ →2

δ →2

and then simplifying we get lim+ δ+ δ →2

√

δ 2 −4 2δ

= 21 . Similar procedure

produces lim c2 = 21 . It is obvious that lim r1 = lim r2 = −1. δ →2+

Thus lim+ xδ (t) = 12 e−t + 21 e−t = e−t .

δ →2+

δ →2+

δ →2

For δ = 2 :

δ 2 − 4 = 0 ⇒ critically damped. Then r1 = r2 = −δ 2 = −1 and

x2 (t) = c1 e−t + c2 te−t . So x′2 (t) = −c1 e−t + c2 e−t − c2 te−t . Using initial conditions we get, c1 = 1 and −c1 + c2 = −1 ⇒ c2 = 0. Hence x2 (t) = e−t . Thus lim+ xδ (t) = x2 (t) = e−t for all t ≥ 0. δ →2

61. Substituting√x = ert in x′′ + δx′ + x = 0, we have r2 + δr + 1 = 0 2 ⇒ r = −δ± 2 δ −4 . r1 t r1 t r2 t Then x¡(t) = c√ + ¢c2 er2 t ; x′ (t) = 1e ¡ c1 r1 e√ + c2 r¢2 e where 1 1 2 2 r1 = 2 −δ + δ − 4 and r2 = 2 −δ − δ − 4 . For δ > 2 :

Using the initial conditions x (0) = 0 and x′ (0) = 1 we get

c1 + c2 = 0 ⇒ c2 = −c1 .

1 1 , c = r2 −r . r1 c1 + r2 c2 = 1 ⇒ c1 (r1 − r2 ) = 1 ⇒ c1 = r1 −r µ t2√ 2 ¶ √1 t 2 δ −4 − δ 2 −4 t −e 2 e2 1 1 r1 t r2 t −2δ √ Thus xδ (t) = r1 −r2 e + r2 −r1 e = e . δ 2 −4 µ t√ ¶ √ δ 2 −4 −t δ 2 −4 t √−e 2 lim+ xδ (t) = lim+ e− 2 δ lim+ e 2 . 2 −4

δ δ →2 δ →2 √ δ→2 , then Q → 0). Let Q = µ δ 2 − 4. (If δ → 2+¶ √ √ t tQ 2 − t Q

−t δ 2 −4 2 2 e 2 δ −4 −e √ So lim+ = lim e 2 −e Q

δ 2 −4 δ →2

Q→ 0

(This is of the form 00 . So use L’Hospital’s Rule)

45

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

= lim

Q→0

t

t t 2 Q t − 2 Q +2e 2e

1

= t. Then

t

lim+ xδ (t) = t lim+ e− 2 δ = te−t .

δ →2

δ →2

For δ = 2 : δ 2 − 4 = 0 ⇒ critically damped. Then r1 = r2 = −2δ = −1 and x2 (t) = c1 e−t + c2 te−t . So x′2 (t) = −c1 e−t + c2 e−t − c2 te−t . Using initial conditions we get, c1 = 0 and −c1 + c2 = 1 ⇒ c2 = 1. Hence x2 (t) = te−t . Thus lim+ xδ (t) = x2 (t) for all t ≥ 0. δ →2

2.6 THE METHOD OF UNDETERMINED COEFFICIENTS 1. Yes, because it is a constant coefficients problem with polynomial times sinusoidal forcing. 3. No, because forcing includes ln |t| which is not a polynomial. sin t 5. No, because cos t is not allowed as a sinusoidal forcing. 7. No, because it is not a constant coefficients problem. 9. No, because forcing includes t−1 which is not a polynomial. 11. Yes, because it is a constant coefficients problem and the second part of forcing.function sinh 3t is 12 e3t − 21 e−3t , a linear combination of exponential functions. 13. Substituting x = ert in the homogeneous part of x′′ + 9x = t3 + 6 we get, r2 + 9 = 0 ⇒ r = ±3i. Thus xh = c1 cos 3t + c2 sin 3t. The forcing function is a polynomial of degree 3, r = 0 is not a root ⇒ k = 0. So xp = A0 + A1 t + A2 t2 + A3 t3 , x′p = A1 + 2A2 t + 3A3 t2 , x′′p = 2A2 + 6A3 t. Substituting these in the original equation we have, ¡ ¢ 2A2 + 6A3 t + 9 A0 + A1 t + A2 t2 + A3 t3 = t3 + 6 ⇒ 9A0 + 2A2 + (6A3 + 9A1 ) t + 9A2 t2 + 9A3 t3 = t3 + 6 Equating the coefficients of like powers of t gives

t3 : 9A3 = 1 ⇒ A3 = 1

9 2 t : 9A2 = 0 ⇒ A2 = 0

6 2 t: 6A3 + 9A1 = 0 ⇒ A1 = − 69 A3 = − 81 = − 27 6 2 0 t : 9A0 + 2A2 = 6 ⇒ A0 = 9 = 3 6 Thus xp = 32 − 81 t + 19 t3 and the general solution is x = xp + xh ⇒ 2 x = 23 − 27 t + 19 t3 + c1 cos 3t + c2 sin 3t. 15. Substituting x = ert in the homogeneous part of x′′ + 8x′ = 7t + 11 we get, r2 + 8r = 0 ⇒ r (r + 8) = 0 ⇒ r = 0, −8. Thus xh = c1 + c2 e−8t . The forcing function is a polynomial of degree 1 and r = 0 is a root of multiplicity 1 ⇒ k = 1. So xp = t (A0 + A1 t) = A0 t + A1 t2 , x′p = A0 + 2A1 t, x′′p = 2A1 . Substituting these in the original equation we have, 2A1 + 8A0 + 16A1 t = 7t + 11. Equating the coefficients of like powers of t gives 7 t: 16A1 = 7 ⇒ A1 = 16 81 0 t : 2A1 + 8A0 = 11 ⇒ A0 = 64 81 7 2 Thus xp = 64 t + 16 t and the general solution is x = xp + xh ⇒ 7 2 −8t x = 81 . 64 t + 16 t + c1 + c2 e

46

CHAPTER 2

17. Substituting x = ert in the homogeneous part of x′′ − 7x′ + 12x = 5e2t we get r2 − 7r + 12 = 0 ⇒ (r − 3) (r − 4) = 0 ⇒ r = 3, 4. Thus xh = c1 e3t +c2 e4t . The forcing function is of the form eαt , α = 2 but r = 2 is not a root ⇒ k = 0. So xp = Ae2t , x′p = 2Ae2t , x′′p = 4Ae2t . Substituting these in the original equation we have, 4Ae2t − 14Ae2t + 12Ae2t = 5e2t ⇒ 2Ae2t = 5e2t ⇒ A = 52 . Thus xp = 52 e2t and the general solution is x = 52 e2t + c1 e3t + c2 e4t . x′ = 5e2t + 3c1 e3t + 4c2 e4t . Using the initial conditions we get, 1 = 52 + c1 + c2 0 = 5 + 3c1 + 4c2 Multiplying the first equation by 4 and then subtracting the second from it we have, c1 = −1. Then c2 = − 12 . Thus x = 52 e2t − e3t − 21 e4t . 19. Substituting x = ert in the homogeneous part of x′′ − 3x′ + 2x = 2et we get r2 − 3r + 2 = 0 ⇒ (r − 1) (r − 2) = 0 ⇒ r = 1, 2. Thus xh = c1 et + c2 e2t . The forcing function is of the form eαt , α = 1 and r = 1 is a root with multiplicity 1 ⇒ k = 1. So xp = Atet , x′p = Aet + Atet , x′′p = 2Aet + Atet .

Substituting these in the original equation we have,

2Aet + Atet − 3Aet − 3Atet + 2Atet = 2et , since tet terms cancel

−Aet = 2et ⇒ A = −2. Thus xp = −2tet and the general solution is

x = xp + xh = −2tet + c1 et + c2 e2t . 21. Substituting x = ert in the homogeneous part of x′′ − 2x′ + 5x = 4et √ 2± 4−20 2 ⇒ r = 1 ± 2i. we get r − 2r + 5 = 0 ⇒ r = 2 Thus xh = et (c1 cos 2t + c2 sin 2t) . The forcing function is of the form eαt , α = 1 but r = 1 is not a root ⇒ k = 0. So xp = Aet , x′p = Aet , x′′p = Aet . Substituting these in the original equation we have Aet − 2Aet + 5Aet = 4et ⇒ 4Aet = 4et ⇒ A = 1. Thus xp = et and the general solution is x = et +et (c1 cos 2t + c2 sin 2t) . 23. Substituting x = ert in the homogeneous part of x′′ − 9x = 5e−3t we get r2 − 9 = 0 ⇒ r = ±3. Thus xh = c1 e3t + c2 e−3t . The forcing function is of the form eαt , α = −3 and r = −3 is a root with multiplicity 1 ⇒ k = 1. So xp = Ate−3t , x′p = Ae−3t − 3Ate−3t , x′′p = −3Ae−3t − 3Ae−3t + 9Ate−3t . Substituting these in the original equation we have, −3Ae−3t − 3Ae−3t + 9Ate−3t − 9Ate−3t = 5e−3t ⇒ −6Ae−3t = 5e−3t (since te−3t terms cancel) ⇒ A = − 65 . Thus xp = − 65 te−3t and the general solution is

x = − 56 te−3t + c1 e3t + c2 e−3t .

25. Substituting x = ert in the homogeneous part of x′′ + 2x′ + 5x = 3 sin t √ 4−20 −2± ⇒ r = −1 ± 2i. we get, r2 + 2r + 5 = 0 ⇒ r = 2 Thus xh = e−t (c1 cos 2t + c2 sin 2t) . Since sin t is not a homogeneous solution, xp = A cos t + B sin t, x′p = −A sin t + B cos t, x′′p = −A cos t − B sin t. Substituting these in the original equation

we have,

−A cos t − B sin t − 2A sin t + 2B cos t + 5A cos t + 5B sin t = 3 sin t

Equating the coefficients of like terms we have,

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

47

cos t : 4A + 2B = 0 sin t : −2A + 4B = 3 Multiplying the second equation by 2 and then adding we get, 3 3 10B = 6 ⇒ B = 53 and then A = − 10 . Thus xp = − 10 cos t + 35 sin t and the general solution is 3 x = − 10 cos t + 53 sin t + e−t (c1 cos 2t + c2 sin 2t) .

3 ′ x = 10 sin t + 35 cos t + e−t (−2c1 sin 2t + 2c2 cos 2t)

−e−t (c1 cos 2t + c2 sin 2t)

Using the initial conditions we get, 3 17 3 + c1 ⇒ c1 = 13 1 = − 10 10 and 1 = 5¡ + 2c2 − c1 ⇒ c2 = ¢20 . 3 3 17 −t 13 Thus x = − 10 cos t + 5 sin t + e 10 cos 2t + 20 sin 2t . 27. Substituting x = ert in the homogeneous part of x′′ + 9x = 4 sin 3t we get r2 + 9 = 0 ⇒ r = ±3i. Thus xh = c1 cos 3t + c2 sin 3t. Since β = 3 and r = βi = 3i is a root of characteristic equation with multiplicity 1, k = 1. So xp = t (A cos 3t + B sin 3t) , x′p = A cos 3t + B sin 3t + t (−3A sin 3t + 3B cos 3t) , x′′p = −3A sin 3t + 3B cos 3t − 3A sin 3t + 3B cos 3t +t (−9A cos 3t − 9B sin 3t) = −6A sin 3t + 6B cos 3t − 9At cos 3t − 9Bt sin 3t. Substituting these in the original equation we have, −6A sin 3t + 6B cos 3t − 9At cos 3t − 9Bt sin 3t + 9At cos 3t +9Bt sin 3t = 4 sin 3t ⇒ −6A sin 3t + 6B cos 3t = 4 sin 3t (since last 4 terms cancel) Equating the coefficients of like terms we have,

cos 3t : 6B = 0 ⇒ B = 0

sin 3t : −6A = 4 ⇒ A = − 32 . Thus xp = − 32 t cos 3t and the general solution is

x = − 23 t cos 3t + c1 cos 3t + c2 sin 3t.

29. Substituting x = ert in the homogeneous part of x′′ + x = cos 2t we get r2 + 1 = 0 ⇒ r = ±i. Thus xh = c1 cos t + c2 sin t. Since β = 2 but r = βi = 2i is not a root of characteristic equation, k = 0. So xp = A cos 2t + B sin 2t, x′p = −2A sin 2t + 2B cos 2t, x′′p = −4A cos 2t − 4B sin 2t. Substituting these in the original equation we have, −4A cos 2t − 4B sin 2t + A cos 2t + B sin 2t = cos 2t ⇒ −3B sin 2t − 3A cos 2t = cos 2t. Equating the coefficients of like terms we have,

cos 2t : −3A = 1 ⇒ A = − 31

sin 3t : −3B = 0 ⇒ B = 0. Thus xp = − 1 3 cos 2t and the general solution is

x = − 13 cos 2t + c1 cos t + c2 sin t ⇒ x′ = 32 sin 2t − c1 sin t + c2 cos t.

Using the initial conditions we get, 0 = − 31 + c1 ⇒ c1 = 13 ; 2 = c2 . Thus x = − 31 cos 2t + 31 cos t + 2 sin t. 31. Substituting x = ert in the homogeneous part of x′′ − 4x = te3t we get, r2 − 4 = 0 ⇒ r = ±2 . Thus xh = c1 e2t + c2 e−2t . The forcing function is of p(t)eαt where p(t) is a first degree polynomial, α = 3 but r = 3 is not a root ⇒ k = 0. So xp = (A0 + A1 t) e3t ,

48

CHAPTER 2

x′p = A1 e3t + 3 (A0 + A1 t) e3t = (3A0 + A1 ) e3t + 3A1 te3t ,

x′′p = 3 (3A0 + A1 ) e3t +3A1 e3t +9A1 te3t = (9A0 + 6A1 ) e3t +9A1 te3t .

Substituting these in the original equation we have,

(9A0 + 6A1 ) e3t + 9A1 te3t − 4 (A0 + A1 t) e3t = te3t ⇒

(5A0 + 6A1 ) e3t + 5A1 te3t = te3t

Equating the coefficients of like terms we have,

te3t : 5A1 = 1 ⇒ A1 = 51 6 3t 0 + 6A1 = 0 ⇒ A0 = − 25 ¡ e6 : 1 ¢ 5A 3t Thus xp = − 25 + 5 t e and the general solution is

6 3t x = xp + xh = − 25 e + 51 te3t + c1 e2t + c2 e−2t . rt 33. Substituting x = e in the homogeneous part of x′′ − 4x′ + 3x = tet we get r2 − 4r + 3 = 0 ⇒ (r − 1) (r − 3) ⇒ r = 1, 3 . Thus xh = c1 et + c2 e3t . The forcing function is of the form p(t)eαt where p(t) is a first degree polynomial, α = 1 and r = 1 is a root of homogeneous equation with multiplicity 1 ⇒ k = 1. ¡ ¢ t 2 t e , So xp = t (A0 + A1 t) e¡t = A0 t + A 1 ¢ £ ¤ x′p = (A0 + 2A1 t) et + A0 t + A1 t2 et = A0 + (A0 + 2A1 ) t + A1 t2 et , £ ¤ x′′p = (A0 + 2A1 + 2A1 t) et + A0 + (A0 + 2A1 ) t + A1 t2 et £ ¤ = 2A0 + 2A1 + (A0 + 4A1 ) t + A1 t2 et .

Substituting these in the original equation £ ¤ t £ we have

¤ 2 2A¡0 + 2A1 + (A + 4A ) t + A t e −4 A0 + (A0 + 2A1 ) t + A1 t2 et 0 1 1 ¢ +3 A0 t + A1 t2 et = tet ⇒ (−2A0 + 2A1 ) et − 4A1 tet = tet

(since t2 et terms cancel). Equating the coefficients of like terms we have,

tet : −4A1 = 1 ⇒ A1 = − 41 1 t ¡ e1 : 1 2−2 ¢ tA0 + 2A1 = 0 ⇒ A0 = A1 = − 4 Thus xp = − 4 t − 4 t e and the general solution is x = xp + xh = − 41 tet − 41 t2 et + c1 et + c2 e3t . 35. Substituting x = ert in the homogeneous part of x′′ + 16x = 3 cos 4t we get r2 + 16 = 0 ⇒ r = ±4i. Thus xh = c1 cos 4t + c2 sin 4t. Since β = 4 and r = βi = 4i is a root of characteristic equation with multiplicity 1, k = 1. So xp = t (A cos 4t + B sin 4t) , x′p = A cos 4t + B sin 4t + t (−4A sin 4t + 4B cos 4t) , x′′p = −4A sin 4t + 4B cos 4t − 4A sin 4t + 4B cos 4t +t (−16A cos 4t − 16B sin 4t) = −8A sin 4t + 8B cos 4t + t (−16A cos 4t − 16B sin 4t) . Substituting these in the original equation we have, −8A sin 4t + 8B cos 4t + t (−16A cos 4t − 16B sin 4t) +16t (A cos 4t + B sin 4t) = 3 cos 4t ⇒ −8A sin 4t + 8B cos 4t = 3 cos 4t (since last 4 terms cancel) Equating the coefficients of like terms we have,

cos 4t : 8B = 3 ⇒ B = 83

sin 4t : −8A = 0 ⇒ A = 0. Thus xp = 83 t sin 4t and the general solution is

x = 83 t sin 4t + c1 cos 4t + c2 sin 4t.

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

49

37. Substituting x = ert in the homogeneous part of

√ x′′ − 2x′ + 5x = et cos 3t we get r2 − 2r + 5 = 0 ⇒ r = 2± 24−20

⇒ r = 1 ± 2i. Thus xh = et (c1 cos 2t + c2 sin 2t) . Since α = 1 and β = 3 but r = 1 ± 3i is not root of characteristic equation ⇒ k = 0. So xp = et (A cos 3t + B sin 3t) , x′p = et (A cos 3t + B sin 3t) + et (−3A sin 3t + 3B cos 3t) = [(A + 3B) cos 3t + (B − 3A) sin 3t] et ,

′′ xp = [(A + 3B) cos 3t + (B − 3A) sin 3t] et

+ [−3 (A + 3B) sin 3t + 3 (B − 3A) cos 3t] et = [(6B − 8A) cos 3t + (−6A − 8B) sin 3t] et . Substituting these in the original equation we have, [(6B − 8A) cos 3t + (−6A − 8B) sin 3t] et −2 [(A + 3B) cos 3t + (B − 3A) sin 3t] et +5et (A cos 3t + B sin 3t) = et cos 3t ⇒ [(6B − 8A) − 2 (A + 3B) + 5A] et cos 3t+[(−6A − 8B) − 2 (B − 3A) + 5B] = et cos 3t t ⇒ −5Ae cos 3t − 5Bet sin 3t = et cos 3t. Equating the coefficients of like terms we have,

et cos 3t : −5A = 1 ⇒ A = − 51

t e sin 3t : −5B = 0 ⇒ B = 0.

t Thus xp = − 1 e cos 3t and the general solution is

5 x = − 15 et cos 3t + et (c1 cos 2t + c2 sin 2t) .

39. Substituting x = ert in the homogeneous part of x′′ + 4x′ = 12t2 + et

we get r2 + 4r = 0 ⇒ r (r + 4) = 0 ⇒ r = 0, −4.

Thus xh = c1 + c2 e−4t . The first forcing term is a second degree

polynomial and r = 0 is a root with multiplicity 1 ⇒ k = 1.

αt The second forcing term of the form ¡ is ¢ e twith α = 1 and r = 1 is

2 not a root. So xp = t At + Bt + C + De ,

x′p = 3At2 + 2Bt + C + Det , x′′p = 6At + 2B + Det .

Substituting these in the original equation we have,

6At + 2B + Det + 12At2 + 8Bt + 4C + 4Det = 12t2 + et .

Equating the coefficients of like terms we have,

et : D + 4D = 1 ⇒ D = 15

2 t : 12A = 12 ⇒ A = 1

t: 6A + 8B = 0 ⇒ B = − 43 3

t0 : ¡ 2B + 4C¢= 0 ⇒ C = 8

3 3 1 t 2 Thus xp = t t − 4 t + 8 + 5 e and the general solution is

x = xp + xh = t3 − 43 t2 + 38 t + 15 et + c1 + c2 e−4t .

3 t + 38 + 51 et − 4c2 e−4t .

x′ = 3t2 − 2 Using the initial conditions we get,

1 = c1 + c2 + 51

3

+ 1 −1

17 1 = 83 + 15 − 4c2 ⇒ c2 = 8 45 = 15+8−40 = − 160 . Then from 160 1 4 17 29 the first equation we have, c1 = 1 − 5 − c2 = 5 + 160 = 145 160 = 32 . 3 2 3 1 t 29 17 −4t 3 Thus x = t − 4 t + 8 t + 5 e + 32 − 160 e .

41. Substituting x = ert in the homogeneous part of x′′ + 2x′ + x = 3e−t 2 we get r2 + 2r + 1 = 0 ⇒ (r + 1) = 0 ⇒ r = −1 with multiplicity 2.

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CHAPTER 2

Thus xh = c1 e−t + c2 te−t . The forcing term is of the form eαt with

α = −1 and r = −1 is a root with multiplicity 2 ⇒ k = 2.

So xp = At2 e−t , x′p = 2Ate−t − At2 e−t , x′′p = 2Ae−t − 2Ate−t − 2Ate−t + At2 e−t = 2Ae−t − 4Ate−t + At2 e−t .

Substituting these in the original equation we have,

2Ae−t − 4Ate−t + At2 e−t + 4Ate−t − 2At2 e−t + At2 e−t = 3e−t ⇒

2Ae−t = 3e−t (since te−t and t2 e−t terms cancel). Equating the coefficients of like terms we have,

e−t : 2A = 3 ⇒ A = 23 3 2 −t Thus xp = 2 t e and the general solution is

x = 23 t2 e−t + c1 e−t + c2 te−t .

43. Substituting x = ert in the homogeneous part of x′′ −4x′ +4x = tet −et +2e3t = (t − 1) et +2e3t we get, r2 −4r +4 = 0 2 ⇒ (r − 2) = 0 ⇒ r = 2 with multiplicity 2. Thus xh = c1 e2t + c2 te2t . The first forcing term is of the form p(t)eαt where p(t) is a first degree polynomial, α = 1 but r = 1 is not a root. The second forcing term is of the form eαt with α = 3 but r = 3 is not a root ⇒ k = 0. So xp = (A0 + A1 t) et + A2 e3t , x′p = A1 et + (A0 + A1 t) et + 3A2 e3t = (A0 + A1 + A1 t) et + 3A2 e3t , x′′p = A1 et + (A0 + A1 + A1 t) et + 9A2 e3t = (A0 + 2A1 + A1 t) et + 9A2 e3t . Substituting these in the original equation we have, (A0 + 2A1 + A1 t) et +9A2 e3t −4 (A0 + A1 + A1 t) et −12A2 e3t + 4 (A0 + A1 t) et + 4A2 e3t = tet − et + 2e3t ⇒ (A0 − 2A1 ) et + A1 tet + A2 e3t = tet − et + 2e3t Equating the coefficients of like terms we have e3t : A2 = 2 tet : A1 = 1 et : A0 − 2A1 = −1 ⇒ A0 = 2A1 − 1 = 1. Thus xp = (1 + t) et + 2e3t and the general solution is

x = xp + xh = (1 + t) et + 2e3t + c1 e2t + c2 te2t .

45. Substituting x = ert in the homogeneous part¢ of ¡ x′′ + 5x′ + 4x = 8t2 + 3 + 2 cos 2t = 8t2 + 3 + 2 cos 2t we get, r2 + 5r + 4 = 0 ⇒ (r + 1) (r + 4) = 0 ⇒ r = −1, −4. Thus xh = c1 e−t + c2 e−4t . Here first part of the forcing function is a second degree polynomial and r = 0 is not a root. In the second part β = 2 and r = βi = 2i is not a root ⇒ k = 0. So xp = A0 + A1 t + A2 t2 + B1 cos 2t + B2 sin 2t, x′p = A1 + 2A2 t − 2B1 sin 2t + 2B2 cos 2t, x′′p = 2A2 − 4B1 cos 2t − 4B2 sin 2t. Substituting these in the original equation we have, 2A2¡− 4B1 cos 2t − 4B2 sin 2t + 5 (A1 + 2A2 t ¢− 2B1 sin 2t + 2B2 cos 2t) +4 A0 + A1 t + A2 t2 + B1 cos 2t + B2 sin 2t = 8t2 + 3 + 2 cos 2t ⇒ (2A2 + 5A1 + 4A0 )+(10A2 + 4A1 ) t+4A2 t2 +10B2 cos 2t−10B1 sin 2t = 8t2 + 3 + 2 cos 2t Equating the coefficients of like terms we have,

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

cos 2t : 10B2 = 2 ⇒ B2 = 51 sin 2t : 10B1 = 0 ⇒ B1 = 0 t2 : 4A2 = 8 ⇒ A2 = 2 t: 10A2 + 4A1 = 0 ⇒ A1 = −5 t0 : 2A2 + 5A1 + 4A0 = 3 ⇒ A0 = 6 Thus xp = 6 − 5t + 2t2 + 15 sin 2t and the general solution is x = 6 − 5t + 2t2 + 51 sin 2t + c1 e−t + c2 e−4t .

47. Substituting x = ert in the homogeneous part of x′ + 3x = t2 + 1 we get r + 3 = 0 ⇒ r = −3. Thus xh = ce−3t . Here the forcing function is a second degree polynomial and r = 0 is not a root ⇒ k = 0. So xp = A0 + A1 t + A2 t2 , x′p = A1 + 2A2 t. Substituting these ¡ in the original¢ equation we have,

A1 + 2A2 t + 3 A0 + A1 t + A2 t2 = t2 + 1 ⇒

3A0 + A1 + (3A1 + 2A2 ) t + 3A2 t2 = t2 + 1 Equating the coefficients of like terms we have,

t2 : 3A2 = 1 ⇒ A2 = 31 t: 3A1 + 2A2 = 0 ⇒ A1 = − 29 0 t : 3A0 + A1 = 1 ⇒ A0 = 11 27 11 2 Thus xp = 27 − 9 t + 13 t2 and the general solution is

11 x = 27 − 92 t + 13 t2 + ce−3t . 49. Substituting x = ert in the homogeneous part of x′′ + 4x = sin 2t we get, r2 + 4 = 0 ⇒ r = ±2i. Thus xh = c1 cos 2t + c2 sin 2t. Since β = 2 and r = 2i = βi is a root with multiplicity 1, k = 1 and so xp = t (A cos 2t + B sin 2t) , x′p = A cos 2t + B sin 2t + t (−2A sin 2t + 2B cos 2t) , x′′p = −2A sin 2t + 2B cos 2t − 2A sin 2t + 2B cos 2t

+t (−4A cos 2t − 4B sin 2t)

= −4A sin 2t + 4B cos 2t + t (−4A cos 2t − 4B sin 2t) . Substituting these in the original equation we have, −4A sin 2t + 4B cos 2t + t (−4A cos 2t − 4B sin 2t) +4t (A cos 2t + B sin 2t) = sin 2t ⇒ −4A sin 2t + 4B cos 2t = sin 2t (since t cos 2t terms cancel) Equating the coefficients of like terms we have,

cos 2t : 4B = 0 ⇒ B = 0

sin 2t : −4A = 1 ⇒ A = − 41 Thus xp = − 41 t cos 2t and the general solution is x = − 14 t cos 2t + c1 cos 2t + c2 sin 2t. 51. Substituting x = ert in the homogeneous part of 3x′ − 2x = tet

2 we get, 3r − 2 = 0 ⇒ r = 32 . Thus xh = ce 3 t .

The forcing term is of the form p(t)eαt where p(t) is a first degree polynomial, α = 1 but r = 1 is not a root ⇒ k = 0. So xp = (A0 + A1 t) et , x′p = A1 et + (A0 + A1 t) et .

Substituting these in the original equation we have,

3A1 et + 3 (A0 + A1 t) et − 2 (A0 + A1 t) et = tet

⇒ (A0 + 3A1 ) et + A1 tet = tet Equating the coefficients of like terms we have,

51

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CHAPTER 2

tet : A1 = 1

et : A0 + 3A1 = 0 ⇒ A0 = −3A1 = −3.

Thus xp = (−3 + t) et and the general solution is 2 x = xp + xh = (−3 + t) et + ce 3 t . rt 53. Substituting x = e in the homogeneous part of x′′ − 5x′ + 6x = t5 + 7t3 + 4t we get r2 − 5r + 6 = 0 ⇒ (r − 2) (r − 3) = 0 ⇒ r = 2, 3. Thus e2t and e3t are independent homogeneous solutions. Here forcing function is a fifth degree

polynomial but r = 0 is not a root.

So the form is xp = A0 + A1 t + A2 t2 + A3 t3 + A4 t4 + A5 t5 .

55. Substituting x = ert in the homogeneous part of x′′ + 9x′ = t3 we get r2 + 9r = 0 ⇒ r (r + 9) = 0 ⇒ r = 0, −9. Thus 1 and e−9t are independent homogeneous solutions. Here forcing function is a third degree polynomial and 1 ⇒ k = 1. ¡ r = 0 is a root with multiplicity ¢ So the form is xp = t A0 + A1 t + A2 t2 + A3 t3 . 57. Substituting x = ert in the homogeneous part of x′′ + 5x′ + 6x = 5e4t we get r2 + 5r + 6 = 0 ⇒ (r + 2) (r + 3) = 0 ⇒ r = −2, −3. Thus e−2t and e−3t are independent homogeneous solutions. The forcing term is of the form eαt where α = 4 but r = 4 is not a root ⇒ k = 0. So the form is xp = Ae4t . 59. Substituting x = ert in the homogeneous part of x′′ − 4x = 5e2t we get, r2 − 4 = 0 ⇒ (r + 2) (r − 2) = 0 ⇒ r = 2, −2. Thus e2t and e−2t are independent homogeneous solutions. The forcing term is of the form eαt where α = 2 and r = 2 is a root with multiplicity 1 ⇒ k = 1. So the form is xp = Ate2t . 61. Substituting x = ert in the homogeneous part of x′′ + 6x′ + 9x = e−3t 2 we get r2 + 6r + 9 = 0 ⇒ (r + 3) = 0 ⇒ r = −3 with multiplicity 2. −3t −3t Thus e and te are independent homogeneous solutions. The forcing term is of the form eαt where α = −3 and r = −3 is a root with multiplicity 2 ⇒ k = 2. So the form is xp = At2 e−3t . 63. Substituting x = ert in the homogeneous part of x′′ + 2x′ + x = t3 e−t 2 we get r2 + 2r + 1 = 0 ⇒ (r + 1) = 0 ⇒ r = −1 with multiplicity 2. −t −t Thus e and te are independent homogeneous solutions. The forcing term is of the form p(t)eαt where p(t) is a third degree polynomial, α = −1 and r = −1 2⇒ ¡ is a root with multiplicity ¢ k = 2. So the form is xp = t2 A0 + A1 t + A2 t2 + A3 t3 e−t . 65. Substituting x = ert in the homogeneous part of x′′ −7x′ +12x = t5 e4t we get r2 − 7r + 12 = 0 ⇒ (r − 3) (r − 4) = 0 ⇒ r = 3, 4. Thus e3t and e4t are independent homogeneous solutions. The forcing term is of the form p(t)eαt where p(t) is a fifth degree polynomial, α = 4 and 1 ⇒ k = 1. r = 4 is a root with multiplicity ¡ ¢ So the form is xp = t A0 + A1 t + A2 t2 + A3 t3 + A4 t4 + A5 t5 e4t . 67. Substituting x = ert in the homogeneous part of x′′ − 6x′ + 9x = t4 e3t 2 we get r2 − 6r + 9 = 0 ⇒ (r − 3) = 0 ⇒ r = 3 with multiplicity 2. 3t 3t Thus e and te are independent homogeneous solutions. The

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

53

forcing term is of the form p(t)eαt where p(t) is a fourth degree polynomial, α = 3 and¡r = 3 is a root with multiplicity¢ 2 ⇒ k = 2. So the form is xp = t2 A0 + A1 t + A2 t2 + A3 t3 + A4 t4 e3t . 69. Substituting x = ert in the homogeneous part of x′′ + 3x′ − 10x = t2 e2t + e5t we get, r2 + 3r − 10 = 0 ⇒

(r − 2) (r + 5) = 0 ⇒ r = 2, −5. Thus e2t and e−5t are independent homogeneous solutions. Here the first forcing term is of the form p(t)eαt where p(t) is a second degree polynomial, α = 2 and r = 2 is a root¡ with multiplicity¢ 1 ⇒ k = 1. So the form is xp1 = t A0 + A1 t + A2 t2 e2t .

The second forcing term is not a homogeneous solution ⇒ k = 0.

So the ¡form is xp2 = Be5t ¢ . Combining them we get, the form to be xp = t A0 + A1 t + A2 t2 e2t + Be5t . 71. Substituting x = ert in the homogeneous part of x′′ + 5x′ = cos 5t we get, r2 + 5r = 0 ⇒ r (r + 5) = 0 ⇒ r = 0, −5. Thus 1 and e−5t are independent homogeneous solutions. Since the forcing function cos 5t is not a homogeneous solution, k = 0. So the form is xp = A cos 5t + B sin 5t. 73. Substituting x = ert in the homogeneous part of

x′′ − 7x′ + 12x = t2 sin 4t we get, r2 − 7r + 12 = 0 ⇒ (r − 3) (r − 4) = 0 ⇒ r = 3, 4.Thus e3t and e4t are independent homogeneous solutions. Here the forcing term is of the form p(t) sin βt where p(t) is a second degree polynomial, β = 4 but r = βi = 4i is not a root ⇒ k = 0.¡ ¢ ¡ ¢ So the form is xp = A0 + A1 t + A2 t2 cos 4t+ B0 + B1 t + B2 t2 sin 4t. 75. Substituting x = ert in the homogeneous part of x′′ + 25x = cos 5t we get r2 +25 = 0 ⇒ r = ±5i. Thus cos 5t and sin 5t are independent homogeneous solutions. Here β = 5 and r = βi = 5i is a root with multiplicity 1 ⇒ k = 1. So the form is xp = t (A cos 5t + B sin 5t) . 77. Substituting x = ert in the homogeneous part of

√ x′′ − 2x′ + 5x = 3et sin 2t we get, r2 − 2r + 5 = 0 ⇒ r = 2± 24−20

⇒ r = 1 ± 2i. Thus et cos 2t and et sin 2t are independent homogeneous solutions. Here the forcing term is of the form eαt sin βt with α = 1, β = 2 and r = α + βi = 1 + 2i is a root with multiplicity 1 ⇒ k = 1. So the form is xp = t (Aet cos 2t + Bet sin 2t) . 79. Substituting x = ert in the homogeneous part of x′′ + 9x = te−t sin 2t we get, r2 + 9 = 0 ⇒ r = ±3i. Thus cos 3t and sin 3t are independent homogeneous solutions. Here the forcing term is of the form p(t)eαt sin βt where p(t) is a first degree polynomial and α = −1, β = 2 but r = α + βi = −1 + 2i is not a root ⇒ k = 0. So the form is xp = (A0 + A1 t) e−t cos 2t + (B0 + B1 t) e−t sin 2t. 81. Substituting x = ert in the homogeneous part of

√ x′′ + 2x′ + 2x = t2 et sin t we get, r2 + 2r + 2 = 0 ⇒ r = −2±2 4−8 ⇒ r = −1 ± i. Thus e−t cos t and e−t sin t are independent homogeneous solutions. Here the forcing term is of the form p(t)eαt sin βt where

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CHAPTER 2

p(t) is a second degree polynomial and α = 1, β = 1 and

r = α + βi = 1 + i is ¡ not a root ⇒ k ¢= 0.

¡ ¢ So the form is xp = A0 + A1 t + A2 t2 et cos t+ B0 + B1 t + B2 t2 et sin t.

83. Substituting x = ert in the homogeneous part of

x′′ + 2x′√+ 2x = e−t cos t + et sin t we get, r2 + 2r + 2 = 0 ⇒

r = −2±2 4−8 ⇒ r = −1 ± i. Thus e−t cos t and e−t sin t are independent homogeneous solutions. Here the first forcing term is of the form eαt sin βt with α = −1, β = 1 and r = α + βi = −1 + i is a root with multiplicity 1 ⇒ k = 1. For the second forcing term, r = α + βi = 1 + i is not a root.

So the form is xp = A1 te−t cos t + B1 te−t sin t + A2 et cos t + B2 et sin t.

85. Substituting x = ert in the homogeneous part of

x′′ − x = e−t − et + et cos t we get, r2 − 1 = 0 ⇒ r = ±1.

Thus et and e−t are independent homogeneous solutions.

So for the first and second forcing terms k = 1.

The third forcing term is of the form eαt sin βt with α = 1, β = 1 but

r = α + βi = 1 + i is not a root ⇒ k = 0.

So the form is xp = Ate−t + Btet + Cet cos t + Det sin t. 87. Substituting x = ert in the homogeneous part of

x′′ + 16x = t cos 4t + e−t sin 4t + 3e−4t we get, r2 + 16 = 0 ⇒ r = ±4i.

Thus cos 4t and sin 4t are independent homogeneous solutions.

The first forcing term is of the form p(t) sin βt where p(t) is a first

degree polynomial, β = 4 and r = βi = 4i is a root with multiplicity

1 ⇒ k = 1. So the form is xp1 = t [(A0 + A1 t) cos 4t + (B0 + B1 t) sin 4t] .

The second forcing term is of the form eαt sin βt where α = −1, β = 4 but r = α + βi = −1 + 4i is not a root ⇒ k = 0. So the form is xp2 = A2 e−t cos 4t + B2 e−t sin 4t.

The third forcing term is of the form eαt with α = −4 but r = α = −4

is not a root ⇒ k = 0. So the form is xp3 = A3 e−4t .

Now combining them we have the form as

xp = t [(A0 + A1 t) cos 4t + (B0 + B1 t) sin 4t] + A2 e−t cos 4t

+B2 e−t sin 4t + A3 e−4t .

89. Substituting x = ert in the homogeneous part of

x′′ − 2x√′ + 2x = t3 e−t sin t + et cos t we get, r2 − 2r + 2 = 0 ⇒ r = 2± 24−8 ⇒ r = 1 ± i. Thus et cos t and et sin t are independent homogeneous solutions. The first forcing term is of the form p(t)eαt sin βt where p(t) is a third degree polynomial, α = −1, β = 1 and r = ¡ α + βi = −1 + i is not ¢a root ⇒ k¡= 0. So the form is ¢ xp1 = A0 + A1 t + A2 t2 + A3 t3 e−t cos t+ B0 + B1 t + B2 t2 + B3 t3 e−t sin t. The second forcing term is of the form eαt cos βt where α = 1, β = 1 and r = α + βi = 1 + i is a root with multiplicity 1 ⇒ k = 1. So the form is xp2 = t (Cet cos t + Det sin t) . Now combining them we have ¢the form as¡ ¡ ¢ xp = A0 + A1 t + A2 t2 + A3 t3 e−t cos t+ B0 + B1 t + B2 t2 + B3 t3 e−t sin t +Ctet cos t + Dtet sin t.

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

55

91. Substituting x = ert in the homogeneous part of

x′′ + 2x′ + x = tet sin 3t + et cos 3t we get, r2 + 2r + 1 = 0 ⇒ 2 (r + 1) = 0 ⇒ r = −1 with multiplicity 1. Thus e−t and te−t are independent homogeneous solutions. The forcing term is of the form

p(t)eαt cos βt + q(t)eαt sin βt where p(t) is a polynomial of degree

zero, q(t) is a polynomial of degree one, α = 1, β = 3 and r = α + βi

= 1 + 3i is not a root ⇒ k = 0. s = max (0, 1) = 1.

So the form is xp = (A0 + A1 t) et sin 3t + (B0 + B1 t) et cos 3t. 93. Substituting x = ert in the homogeneous part of

x′′ + 4x′ √+ 8x = t2 e−2t sin 2t + te−2t cos 2t we get, r2 + 4r + 8 = 0 ⇒

r = −4± 216−32 ⇒ r = −2 ± 2i. Thus e−2t cos 2t and e−2t sin 2t are independent homogeneous solutions. The forcing term is of the

form p(t)eαt cos βt + q(t)eαt sin βt where p(t) is a first degree

polynomial, q(t) is a second degree polynomial, α = −2, β = 2 and

r = α + βi = −2 + 2i is a root with multiplicity 1 ⇒ k = 1.

s = max £¡ (1, 2) = 2. So the¢ form is

¡ ¢ ¤ xp = t A0 + A1 t + A2 t2 e−2t sin 2t + B0 + B1 t + B2 t2 e−2t cos 2t .

95. Substituting x = ert in the homogeneous part of x′′ + 4x′ + √ 13x = t3 e−2t sin 3t we get r2 + 4r + 13 = 0

−4± 16−52

⇒r= 2 ⇒ r = −2 ± 3i. Thus e−2t cos 3t and e−2t sin 3t are independent

homogeneous solutions. The forcing term is of the form p(t)eαt sin βt

where p(t) is a third degree polynomial, α = −2, β = 3 and

r = α + βi = −2 + 3i is a root with multiplicity 1 ⇒ k = 1.

So the ¡form is

¢ ¡ ¢ xp = t A0 + A1 t + A2 t2 + A3 t3 e−2t cos 3t+t B0 + B1 t + B2 t2 + B3 t3 e−2t sin 3t.

97. (a) Substituting x = ert in the homogeneous part of x′′ = t3 + 7t − 2

we get, r2 = 0 ⇒ r = 0 with multiplicity 2. Thus 1 and t are

independent homogeneous solutions. The forcing term is a third

degree polynomial, and¡ r = 0 is a root with multiplicity 2 ⇒ k = 2.

¢ So the form is xp = t2 A0 + A1 t + A2 t2 + A3 t3 . (b) Integrating x′′ = t3 + 7t − 2 with respect to t we get, with respect to t we

x′ = 14 t4 + 27 t2 − 2t + c2 . Integrating this ¡ 1again ¢ 1 5 get, x = 20 t + 67 t3 − t2 + c2 t + c1 = t2 20 t3 + 67 t − 1 + c2 t + c1 . The last two terms correspond to the homogeneous and ¡ 1 3 7 solution ¢ t + 6t − 1 . hence the particular solution is xp = t2 20 99. Substituting x = ert in the homogeneous part of

x′′′ − 3x′′ + 3x′ − x = e2t we get, r3 − 3r2 + 3r − 1 = 0 ⇒

3 (r − 1) = 0 ⇒ r = 1 with multiplicity 3. Thus

xh = c1 et + c2 tet + c3 t2 et . The forcing term is of the form e αt with α = 2 but r = 2 is not a root⇒ k = 0. So xp = Ae2t ,

2t

x′p = 2Ae2t , xp′′ = 4Ae2t , x′′′ p = 8Ae . Substituting these in

the original equation we have,

8Ae2t − 12Ae2t + 6Ae2t − Ae2t = e2t ⇒ Ae2t = e2t ⇒ A = 1.

Thus xp = e2t and the general solution is x = e2t +c1 et +c2 tet +c3 t2 et .

56

CHAPTER 2

101. Substituting x = ert¡ in the¢ homogeneous part of x′′′ − x′ = sin t we get, r3 − r = 0 ⇒ r r2 − 1 = 0 ⇒ r = 0, ±1. Thus xh = c1 +c2 et +c3 e−t . Since sin t is not a homogeneous solution, xp = A cos t + B sin t, x′p = −A sin t + B cos t, x′′p = −A cos t − B sin t, x′′′ p = A sin t − B cos t. Substituting these in the original equation we have,

A sin t − B cos t + A sin t − B cos t = sin t ⇒ 2A sin t − 2B cos t = sin t.

Equating the coefficients of like terms

cos t : −2B = 0 ⇒ B = 0

sin t : 2A = 1 ⇒ A = 12 1 Thus xp = 2 cos t and the general solution is

x = 21 cos t + c1 + c2 et + c3 e−t .

103. Substituting x = ert in ¡the homogeneous part of x′′′ + x′ = 3 + 2 cos t ¢ 3 2 we get, r + r = 0 ⇒ r r + 1 = 0 ⇒ r = 0, ±i. Thus xh = c1 + c2 cos t + c3 sin t. The first forcing term is a constant and r = 0 is a root ⇒ k = 1. So xp1 = At. The second forcing term involves cos t which is also a homogeneous solution.

So xp2 = Bt cos t + Ct sin t.

Combining them we get, xp = At + Bt cos t + Ct sin t,

x′p = A + B cos t − Bt sin t + C sin t + Ct cos t,

x′′p = −B sin t − B sin t − Bt cos t + C cos t + C cos t − Ct sin t

= −2B sin t + 2C cos t − Bt cos t − Ct sin t, x′′′ p = −2B cos t − 2C sin t − B cos t + Bt sin t − C sin t − Ct cos t = −3B cos t − 3C sin t + Bt sin t − Ct cos t. Substituting these in the original equation we have, −3B cos t − 3C sin t + Bt sin t − Ct cos t + A + B cos t − Bt sin t +C sin t + Ct cos t = 3 + 2 cos t ⇒ A − 2B cos t − 2C sin t = 3 + 2 cos t

(since t cos t and t sin t terms cancel)

Equating the like terms we have,

cos t : −2B = 2 ⇒ B = −1 sin t : −2C = 0 ⇒ C = 0 t0 : A = 3

Thus xp = 3t − t cos t and the general solution is

x = 3t − t cos t + c1 + c2 cos t + c3 sin t

x′ = 3 − cos t + t sin t − c2 sin t + c3 cos t

x′′ = sin t + sin t + t cos t − c2 cos t − c3 sin t

Using the initial conditions we have,

0 = c1 + c2

0 = 3 − 1 + c3 ⇒ c3 = −2

0 = −c2 ⇒ c2 = 0. Then c1 = 0. Thus x = 3t − t cos t − 2 sin t. 105. Substituting x = ert in the homogeneous part of x′′′′ − 16x = 5tet we get r4 − 16 = 0 ⇒ r4 = 16 ⇒ r2 = ±4 ⇒ r = ±2, ±2i. Thus xh = c1 e2t + c2 e−2t + c3 cos 2t + c4 sin 2t. The forcing function is of the form p(t)eαt where p(t) is a first degree polynomial, α = 1,

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

57

and r = 1 is not a root ⇒ k = 0. So xp = (A0 + A1 t) et = A0 et +A1 tet

x′p = A0 et + A1 et + A1 tet , x′′p = A0 et + 2A1 et + A1 tet ,

t t t ′′′′ t t t x′′′ p = A0 e + 3A1 e + A1 te , xp = A0 e + 4A1 e + A1 te .

Substituting these in the original equation we have,

A0 et + 4A1 et + A1 tet − 16A0 et − 16A1 tet = 5tet ⇒

−15A0 et + 4A1 et − 15A1 tet = 5tet . Equating the like terms we have,

tet : −15A1 = 5 ⇒ A1 = − 31 4 4 t e : −15A0 + 4A1 = 0 ⇒ A0 = 15 A1 = − 45 . 4 t 1 t Thus xp = − 45 e − 3 te and the general solution is 4 t x = − 45 e − 13 tet + c1 e2t + c2 e−2t + c3 cos 2t + c4 sin 2t. 107. Substituting x = ert in the homogeneous part of

′′′′ ′′ 4x ¢= e2t − e3t we get r4 − 5r2 + 4 = 0 ⇒

¡x 2 − 5x ¢¡ + 2 r − 1 r − 4 = 0 ⇒ r2 = 1, 4 ⇒ r = ±1, ±2. Thus xh = c1 et + c2 e−t + c3 e2t + c4 e−2t . The first forcing function, e2t , is also a homogeneous solution ⇒ k = 1. So xp1 = Ate2t . The second forcing function e3t is not a homogeneous solution ⇒ k = 0. So xp2 = Be3t . Combining them we get, xp = Ate2t + Be3t .

x′p = Ae2t + 2Ate2t + 3Be3t , x′′p = 4Ae2t + 4Ate2t + 9Be3t ,

2t 2t 3t 2t 2t 3t ′′′′ x′′′ p = 12Ae + 8Ate + 27Be , xp = 32Ae + 16Ate + 81Be .

Substituting these in the original we have, ¡ equation ¢ 2t 3t 2t 2t 32Ae2t + 16Ate + 81Be − 5 4Ae + 4Ate + 9Be3t ¡ ¢ +4 Ate2t + Be3t = e2t − e3t ⇒ 12Ae2t + 40Be3t = e2t − e3t (since te2t terms cancel). Equating the like terms we have, 1 e2t : 12A = 1 ⇒ A = 12 1 3t e : 40B = −1 ⇒ B = − 40 . 1 1 3t 2t Thus xp = 12 te − 40 e and the general solution is

1 1 3t x = 12 te2t − 40 e + c1 et + c2 e−t + c3 e2t + c4 e−2t . 109. Substituting x = ert in the homogeneous ¡ ¢part of x′′′ − 3x′′ + 3x′ − x = t2 et − 3et = t2 − 3 et we get,

3 r3 − 3r2 + 3r − 1 = 0 ⇒ (r − 1) = 0 ⇒ r = 1 with multiplicity 3.

t t 2 t Thus e , te , and t e are independent homogeneous solutions.

The forcing term is of the form p(t)eαt where p(t) is a second degree

polynomial, α = 1 and¡r = 1 is a root with ¢ multiplicity 3 ⇒ k = 3.

So the form is xp = t3 A0 + A1 t + A2 t2 et . 111. Substituting x = ert in the homogeneous part of x′′′′ − 4x′′′ + 6x′′ − 4x′ + x = t3 et + t2 e−t we get, 4 r4 − 4r3 + 6r2 − 4r + 1 = 0 ⇒ (r − 1) = 0 ⇒ r = 1 with multiplicity t t 2 t 3 t 4. Thus e , te , t e and t e are independent homogeneous solutions. The first forcing term is of the form p(t)eαt , where p(t) is a third degree polynomial, α = 1 and ¡r = 1 is a root with multiplicity 4⇒ ¢ k = 4. So the form is xp1 = t4 A0 + A1 t + A2 t2 + A3 t3 et . The second forcing term is of the form p(t)eαt , where p(t) is a second

58

CHAPTER 2

degree polynomial, α ¡= −1 but r = −1¢is not a root ⇒ k = 0.

So the form is xp2 = B0 + B1 t + B2 t2 e−t . Now combining them we have the ¡ ¢ form¡ as

¢ xp = t4 A0 + A1 t + A2 t2 + A3 t3 et + B0 + B1 t + B2 t2 e−t . 113. Substituting x = ert in the homogeneous part of

x′′′ 2x′′ + 2x¢′ = 3e−t cos t we get, r3 + 2r2 + 2r = 0 ⇒ ¡ + 2 r r + 2r + 2 = 0 ⇒ r = 0, −1 ± i. Thus 1, e−t cos t, and e−t sin t are independent homogeneous solutions. The forcing term is of the form eαt cos βt where α = −1, β = 1 and r = α + βi is a root with multiplicity 1 ⇒ k = 1. So the form is xp = t (Ae−t cos t + Be−t sin t) . 115. Substituting x = ert in the homogeneous part of

x′′′′ + 4x′′′ + 8x′′ + 8x′ + 4x = 7e−t cos t we get

3 r4 + + 8r2 +¢8r + ¡ 4r ¡ 4 = 0 ¢ 4 ⇒ r + 4r2 + 4 + 4r3 + 8r + 4r2 = 0 (regrouping) ¡ ¢2 ¡ ¢ ⇒ r2 + 2 + 4r r2 + 2 + 4r2 = 0 (perfect square form) ¡ 2 ¢2 ⇒ r + 2r + 2 = 0 ⇒ r = −1±i with multiplicity 2.Thus e−t cos t, e−t sin t, te−t cos t and te−t sin t are independent homogeneous solutions.. The forcing term is of the form eαt cos βt where α = −1, β = 1 and r = α + βi is a root with multiplicity 2 ⇒ k = 2. So the form is xp = t2 (Ae−t cos t + Be−t sin t) .

2.7 MECHANICAL VIBRATIONS II: FORCED RESPONSE 2.7.1 Friction is Absent(δ = 0) 1. The characteristic equation of mx′′ + 4x = 13 cos ωt, by substituting q 4 x = ert , is mr2 + 4 = 0 which has pure imaginary roots r = ±i m . q 4 Resonance occurs if ω = m . If forcing function has a frequency q ω 1 4 of 20 Hz, then 2π = 20 ⇒ ω = 40π. So 40π = m ⇒ m = 400π 2. ′′ 3. The characteristic equation of 36x + kx = 4 cos ωt, by substituting q

k x = ert , is 36r2 + k = 0 which has pure imaginary roots r = ±i 36 . q k Resonance occurs if ω = 36 . If forcing function has a frequency ω of 22 Hz, then 2π = 22 ⇒ ω = 44π. q 2

k So 44π = 36 ⇒ k = 36 (44π) = 69696π 2 .

5. The characteristic equation of mx′′ + 10x = F cos ωt, by substituting q

x = ert , is mr2 + 10 = 0 which has pure imaginary roots r = ±i 10 m. q Resonance occurs if ω = 10 m . If forcing function has a frequency between 10 and 70 Hz, then 10 < 2ωπ < 70 ⇒ 20π < ω < 140π ⇒ q 1 m 1 20π < 10 m < 140π ⇒ (140π)2 < 10 < (20π)2

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

59

10 10 ⇒ (140π) . 2 < m < (20π)2 7. The characteristic equation of 15x′′ +8x = f (t), by substituting x = ert , q

8 is 15r2 + 8 = 0 which has pure imaginary roots r = ±i 15 . q 8 Resonance occurs if ω = 15 . q ω 1 8 Then the frequency of the forcing function is 2π = 2π 15 . 9. The characteristic equation of mx′′ + 15x = F sin ωt by substituting q

x = ert , is mr2 + 15 = 0 which has pure imaginary roots r = ±i 15 m. q Resonance occurs if ω = 15 m . The frequency of the forcing function q 1 ω 15 is 2π = 30 ⇒ ω = 60π. So 60π = 15 m ⇒ m = (60π)2 = 240π 2 gm.

β−ω 11. Let φ = β+ω 2 t, θ = 2 t. Then φ − θ = ωt and φ + θ = βt. Now using the given formula we have,

cos(φ − θ) = cos φ cos θ + sin φ sin θ and

cos(φ + θ) = cos φ cos θ − sin φ sin θ.

Subtracting we get, cos(φ +´θ) = 2 sin φ sin θ ⇒ ³ − θ) ´ −cos(φ ³ β+ω β−ω cos ωt − cos βt = 2 sin 2 t sin 2 t .

13. mg = kd, mx′′ = FT = −k (x + d) + mg = −kx, mx′′ + kx = 0. 15. The characteristic equation of mx′′ + kx = F cos ωt, by substituting q

k x = ert , is mr2 + k = 0 which has pure imaginary roots r = ±i m = q k ±iω0 where ω0 = m . Thus xh = c1 cos ω0 t + c2 sin ω0 t. Since the forcing term, cos ωt, is not a homogeneous solution (because ω 6= ω0 ), xp = A cos ωt + B sin ωt, x′p = −Aω sin ωt + Bω cos ωt, x′′p = −Aω 2 cos ωt − Bω 2 sin ωt. Substituting these in the original equation we have, 2 −Amω cos ωt −¢Bmω 2 sin¡ωt + Ak cos ωt ¡ ¢ + Bk sin ωt = F cos ωt ⇒ Ak − Amω 2 cos ωt + Bk − Bmω 2 sin ωt = F cos ωt. Equating the coefficients of like terms we have,

cos ωt : Ak − Amω 2 = F ⇒ A = k−Fmω2

0 sin ωt : Bk − Bmω 2 = 0 ⇒ B = k−mω 6 mω 2 ) 2 = 0 (since k = F So xp = k−mω2 cos ωt and thus the general solution is x = k−Fmω2 cos ωt + c1 cos ω0 t + c2 sin ω0 t Fω x′ = − k−mω 2 sin ωt − c1 ω0 sin ω0 t + c2 ω0 cos ω0 t

Using the initial conditions

x(0) = x0 = k−Fmω2 + c1 ⇒ c1 = x0 − k−Fmω2

and x′ (0) = 0 = c2 ω0 ⇒ c2³= 0 (since ω´0 6= 0).

Thus x =

F k−mω 2

F k−mω 2

cos ωt + x0 −

F k−mω 2

cos ω0 t ⇒

x= (cos ωt − cos ω0 t) + x0 cos ω0 t. 17. The characteristic equation of mx′′ + kx = F sin ωt, by substituting q k x = ert , is mr2 + k = 0 which has pure imaginary roots r = ±i m =

60

CHAPTER 2

q

k ±iω0 where ω0 = m . Thus xh = c1 cos ω0 t + c2 sin ω0 t. Since the forcing term, sin ωt, is not a homogeneous solution (because ω 6= ω0 ), xp = A cos ωt + B sin ωt, x′p = −Aω sin ωt + Bω cos ωt, x′′p = −Aω 2 cos ωt − Bω 2 sin ωt. Substituting these in the original equation we have, 2 −Amω cos ωt −¢Bmω 2 sin¡ωt + Ak cos ωt ¡ ¢ + Bk sin ωt = F sin ωt ⇒ Ak − Amω 2 cos ωt + Bk − Bmω 2 sin ωt = F sin ωt. Equating the coefficients of like terms we have, 0 cos ωt : Ak − Amω 2 = 0 ⇒ A = k−mω 6 mω 2 ) 2 = 0 (since k = F sin ωt : Bk − Bmω 2 = F ⇒ B = k−mω 2 F So xp = k−mω2 sin ωt and thus the general solution is x = k−Fmω2 sin ωt + c1 cos ω0 t + c2 sin ω0 t Fω x′ = k−mω 2 cos ωt − c1 ω0 sin ω0 t + c2 ω0 cos ω0 t Using the initial conditions x(0) = x0 = c1 and x′ (0) = 0 ω F F ω = k−mω 2 + c2 ω0 ⇒ c2 = − ω k−mω 2 . 0 Thus x = k−Fmω2 sin ωt + x0 cos ω0 t − ωω0 k−Fmω2 sin ω0 t ⇒ ³ ´ x = k−Fmω2 sin ωt − ωω0 sin ω0 t + x0 cos ω0 t.

19. The characteristic equation of mx′′ + kx = F cos ωt by substituting q k x = ert , is mr2 + k = 0 which has pure imaginary roots r = ±i m = q k ±iω0 where ω0 = m . Thus xh = c1 cos ω0 t + c2 sin ω0 t. Since the forcing term, cos ωt, is a homogeneous solution (because ω = ω0 ),

xp = t (A cos ωt + B sin ωt) ,

x′p = (A cos ωt + B sin ωt) + t (−Aω sin ωt + Bω cos ωt) ,

¡ ¢ x′′p = −2Aω sin ωt + 2Bω cos ωt − t Aω 2 cos ωt + Bω 2 sin ωt .

Substituting these in the original¡equation we have,

£ ¢¤

m −2Aω sin ωt + 2Bω cos ωt − t Aω 2 cos ωt + Bω 2 sin ωt +kt (A cos ωt + B sin ωt) = F cos ωt

Since k = mω 2 , t cos ωt and t sin ωt terms cancel,

−2Amω sin ωt + 2Bmω cos ωt = F cos ωt.

Equating the coefficients of like terms we have,

F cos ωt : 2Bmω = F ⇒ B = 2ωm

sin ωt : −2Amω = 0 ⇒ A = 0.

F t sin ωt and thus the general solution is

So xp = 2mω F

x = 2mω t sin ωt + c1 cos ω0 t + c2 sin ω0 t F F ′ x = 2mω sin ωt + 2m t cos ωt − c1 ω0 sin ω0 t + c2 ω0 cos ω0 t Using the initial conditions x(0) = 0 = c1 and x′ (0) = 0 = c2 ω0 F ⇒ c2 = 0 (since ω0 6= 0). Thus x = 2mω t sin ωt. rt 21. Substituting x = e in the homogeneous part of x′′ + 4x = 8 cos 5t we have, r2 + 4 = 0 ⇒ r = ±2i. Thus xh = c1 cos 2t + c2 sin 2t. Since the forcing term, cos 5t, is not a homogeneous solution (no resonance) xp = A cos 5t + B sin 5t, x′p = −5A sin 5t + 5B cos 5t, x′′p = −25A cos 5t − 25B sin 5t. Substituting these in the original

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

61

equation we have,

−25A cos 5t − 25B sin 5t + 4A cos 5t + 4B sin 5t = 8 cos 5t ⇒ −21A cos 5t − 21B sin 5t = 8 cos 5t

Equating the coefficients of like terms we have,

8 cos 5t : −21A = 8 ⇒ A = − 21 sin 5t : −21B = 0 ⇒ B = 0. 8 So xp = − 21 cos 5t and thus the general solution is 8 cos 5t + c1 cos 2t + c2 sin 2t. x = − 21 23. Substituting x = ert in the homogeneous part of x′′ − 4x = 8 cos 5t we have, r2 − 4 = 0 ⇒ r = ±2. Thus xh = c1 e2t + c2 e−2t . Since the forcing term, cos 5t, is not a homogeneous solution (no resonance) xp = A cos 5t + B sin 5t, x′p = −5A sin 5t + 5B cos 5t, x′′p = −25A cos 5t − 25B sin 5t. Substituting these in the original equation we have, −25A cos 5t − 25B sin 5t − 4A cos 5t − 4B sin 5t = 8 cos 5t ⇒ −29A cos 5t − 29B sin 5t = 8 cos 5t

Equating the coefficients of like terms we have,

8

cos 5t : −29A = 8 ⇒ A = − 29 sin 5t : −21B = 0 ⇒ B = 0. 8 So xp = − 29 cos 5t and thus the general solution is

8 x = − 29 cos 5t + c1 e2t + c2 e−2t .

25. Substituting x = ert in the homogeneous part of x′′ + 4x = 3 cos 2t we have, r2 + 4 = 0 ⇒ r = ±2i. Thus xh = c1 cos 2t + c2 sin 2t. Since the forcing term, cos 2t, is a homogeneous solution (resonance) xp = t (A cos 2t + B sin 2t) , x′p = A cos 2t + B sin 2t + t (−2A sin 2t + 2B cos 2t) , x′′p = −4A sin 2t + 4B cos 2t + t (−4A cos 2t − 4B sin 2t) . Substituting these in the original equation we have, −4A sin 2t + 4B cos 2t + t (−4A cos 2t − 4B sin 2t) +4t (A cos 2t + B sin 2t) = 3 cos 2t ⇒ Since t cos 2t and t sin 2t terms cancel, −4A sin 2t+4B cos 2t = 3 cos 2t. Equating the like coefficients of terms we have, cos 2t : 4B = 3 ⇒ B = 43 sin 2t : −4A = 0 ⇒ A = 0. So xp = 43 t sin 2t and thus the general solution is

x = 34 t sin 2t + c1 cos 2t + c2 sin 2t.

27. By substituting x = ert , the characteristic equation of mx′′ +q kx = F cos ωt is mr2 +q k = 0 which has pure imaginary roots k k . Thus xh = c1 cos ωt + c2 sin ωt. r = ±i m = ±iω where ω = m Since the forcing term, cos ωt, is a homogeneous solution

xp = t (A cos ωt + B sin ωt) ,

x′p = (A cos ωt + B sin ωt) + t (−Aω sin ωt + Bω cos ωt) ,

¡ ¢ x′′p = −2Aω sin ωt + 2Bω cos ωt − t Aω 2 cos ωt + Bω 2 sin ωt .

Substituting these in the original¡equation we have,

£ ¢¤

m −2Aω sin ωt + 2Bω cos ωt − t Aω 2 cos ωt + Bω 2 sin ωt

62

CHAPTER 2

+kt (A cos ωt + B sin ωt) = F cos ωt Since k = mω 2 , t cos ωt and t sin ωt terms cancel, −2Amω sin ωt + 2Bmω cos ωt = F cos ωt. Equating the coefficients of like terms we have, F cos ωt : 2Bmω = F ⇒ B = 2ωm

sin ωt : −2Amω = 0 ⇒ A = 0.

F So xp = 2mω t sin ωtand thus the general solution is

F

t sin ωt + c1 cos ωt + c2 sin ωt. x = 2mω 2.7.2 Friction is Present(δ > 0) (Damped Forced Oscillations) 29. x′′ + 6x′ + 25x = 3 cos 4t. The forcing term, cos 4t, cannot be a homogeneous solution because δ = 6 > 0 is present. So xp = A cos 4t + B sin 4t, x′p = −4A sin 4t + 4B cos 4t, x′′p = −16A cos 4t − 16B sin 4t. Substituting these in the original equation we have, −16A cos 4t − 16B sin 4t − 24A sin 4t + 24B cos 4t + 25A cos 4t +25B sin 4t = 3 cos 4t ⇒ (9A + 24B) cos 4t + (9B − 24A) sin 4t = 3 cos 4t Equating the coefficients of like terms we have,

cos 4t : 9A + 24B = 3 ⇒ 3A + 8B = 1

sin 4t : 9B − 24A = 0 ⇒ −24A + 9B = 0. Multiplying the first equation by 8 and then adding we get, 8 9 3 73B = 8 ⇒ B = 73 and then A = 24 B = 73. . 3 8 So xp = 73 cos 4t + 73 sin 4t. In order to find this in amplitude-phase q¡ ¢ √ ¡ 8 ¢2 3 2 form, we find R = = 7373 , 73 ³ +´ 73 8 ¡ ¢ = tan−1 83 = 1.212. φ = tan−1 73 3

√

73

Then xp = 7373 cos (4t − 1.212) . 31. x′′ + 4x′ + 13x = 5 cos 2t. The forcing term, cos 2t, cannot be a

homogeneous solution because δ = 4 > 0 is present. So

xp = A cos 2t + B sin 2t,

x′p = −2A sin 2t + 2B cos 2t,

x′′p = −4A cos 2t − 4B sin 2t.

Substituting these in the original equation we have, −4A cos 2t − 4B sin 2t − 8A sin 2t + 8B cos 2t + 13A cos 2t +13B sin 2t = 5 cos 2t ⇒ (9A + 8B) cos 2t + (9B − 8A) sin 2t = 5 cos 2t Equating the coefficients of like terms we have,

cos 2t : 9A + 8B = 5

sin 2t : −8A + 9B = 0

Multiplying the first equation by 8 and the second equation by 9 and then adding we get, 40 8 9 145B = 40 ⇒ B = 145 = 29 and then A = 98 B = 29 .

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

63

8 cos 2t + 29 sin 2t. In order to find this in amplitude-phase q¡ ¢ √ ¡ 8 ¢2 145 9 2 form, we find R = , + 29 = 29 29 ³ 8 ´ ¡ ¢ φ = tan−1 29

= tan−1 89 = 0.7266. 9

So xp =

9 29

√29

145 cos (2t − 0.7266) . Then xp = 29 ′′ ′ 33. x + 8x + 41x = 3 sin t. The forcing term, sin t, cannot be a

homogeneous solution because δ = 8 > 0 is present. So

xp = A cos t + B sin t,

x′p = −A sin t + B cos t,

x′′p = −A cos t − B sin t. Substituting these in the original equation we have,

−A cos t−B sin t−8A sin t+8B cos t+41A cos t+41B sin t = 3 sin t ⇒ (40A + 8B) cos t + (40B − 8A) sin t = 3 sin t Equating the coefficients of like terms we have, cos t : 40A + 8B = 0 ⇒ 5A + B = 0 ⇒ B = −5A sin t : −8A + 40B = 3 ⇒ −8A − 200A = 3 (substitution) 3 15 3 15 ⇒ A = − 208 . Then B = 208 . So xp = − 208 cos t + 208 sin t.

In order to find this in amplitude-phase form, we find

q¡ √ √ ¢ ¡ 15 ¢2 234 3 2 − 208 = 320826 ,

+ 208 = 208 R= ³ 15 ´ 3 φ = tan−1 −2083 + π (Since − 208 < 0) = tan−1 (−5) + π = 1.7682. √208

Then xp = 320826 cos (t − 1.7682) . 35. x′′ + 3x′ + 2x = sin t. The forcing term, sin t, cannot be a

homogeneous solution because δ = 3 > 0 is present. So

xp = A cos t + B sin t,

x′p = −A sin t + B cos t,

x′′p = −A cos t − B sin t.

Substituting these in the original equation we have,

−A cos t − B sin t − 3A sin t + 3B cos t + 2A cos t + 2B sin t = sin t ⇒ (A + 3B) cos t + (B − 3A) sin t = sin t Equating the coefficients of like terms we have,

cos t : A + 3B = 0

sin t : −3A + B = 1.

Multiplying the first equation by 3 and then adding we get, 1 3 10B = 1 ⇒ B = 10 . Then A = −3B = − 10 . 3 1 So xp = − 10 cos t + 10 sin t. In order to find this in amplitude-phase q¡ √ ¢ ¡ 1 ¢2 3 2 − 10 form, we find R = + 10 = 1010 , ³ 1 ´ ¡ ¢ 3 < 0) = tan−1 − 13 + π = 2.8198. φ = tan−1 −103 + π (Since − 10 √

10

Then xp = 1010 cos (t − 2.8198) . 37. x′′ + 2x′ + 2x = sin t. The forcing term, sin t, cannot be a

homogeneous solution because δ = 2 > 0 is present. So

xp = A cos t + B sin t,

x′p = −A sin t + B cos t,

64

CHAPTER 2

x′′p = −A cos t − B sin t.

Substituting these in the original equation we have,

−A cos t − B sin t − 2A sin t + 2B cos t + 2A cos t + 2B sin t = sin t ⇒ (A + 2B) cos t + (B − 2A) sin t = sin t Equating the coefficients of like terms we have,

cos t : A + 2B = 0

sin t : −2A + B = 1.

Multiplying the first equation by 2 and then adding we get, 5B = 1 ⇒ B = 51 . Then A = −2B = − 52 . So xp = − 25 cos t + 51 sin t. In order in amplitude-phase form, we find q¡ to ¢find this √ ¡ 1 ¢2 2 2 R= − 5 + 5 = 55 , ³ 1 ´ ¡ ¢ 2 < 0) = tan−1 − 21 + π = 2.6779. φ = tan−1 −52 + π (Since − 5 √

5

Then xp = 55 cos (t − 2.6779) . 39. x′′ + 2x′ + x = sin t. The forcing term, sin t, cannot be a

homogeneous solution because δ = 2 > 0 is present. So

xp = A cos t + B sin t,

x′p = −A sin t + B cos t,

x′′p = −A cos t − B sin t.

Substituting these in the original equation we have,

−A cos t − B sin t − 2A sin t + 2B cos t + A cos t + B sin t = sin t ⇒ 2B cos t − 2A sin t = sin t Equating the coefficients of like terms we have,

cos t : 2B = 0 ⇒ B = 0

sin t : −2A = 1 ⇒ A = − 12 .

So xp = − 12 cos t. In order to find this in amplitude-phase form, q¡ ¢ 2 we find R = − 12 = 12 , φ = tan−1 (0) + π (Since − 21 < 0) = π. Then xp = 21 cos (t − π) . ω2 1 41. y = s µ . Let z = ω 2 . ¶2 2 ω 2

2

ω0

2

ω0

Then y = √ ⇒ y ′ = − 12

0

+ 4γ

1− ω2

1 (1−z)2 +4γ 2 z −2(1−z)+4γ 2

h i− 21 2 = (1 − z) + 4γ 2 z . 1−z−2γ 2

3 . [(1−z) +4γ 2 z] [(1−z)2 +4γ 2 z] 2 The critical point is z such that y ′ = 0 ⇒ 1 − z − 2γ 2 = 0 ⇒ z = 1 − 2γ 2 .This will be a solution√ if the denominator is defined 2 (i.e. (1 − z) + 4γ 2 z > 0). If γ < 22 then z = 1 − 2γ 2 > 0 2 and thus (1 − z) + 4γ 2 z > 0. So z = 1 − 2γ 2 is a valid critical 2 = z = 1 − 2γ 2 ⇒ point. Hence maximum occurs at ω ω02 p ω = ω0 1 − 2γ 2 . If γ → 0 then ω → ω0 . 43. From equation (30) in the text we have, δ δ √ = √ δ 2 k = √ δ . γ = √2mk = k m

2m

=

2

3 2

m

4m

m

4mk

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

65

√ Since 4mk is the damping coefficient for the critically damped motion, γ is the ratio of the damping coefficient to the damping coefficient at the critical damping. 45. Graph

2.8 LINEAR ELECTRIC CIRCUITS 1. Substituting R = 3, L = 12 , C = 25 , e = 3 cos t in the equation 2

2

dq 1 1d q 5 L ddt2q + R dq dt + C q = e we have, 2 dt2 + 3 dt + 2 q = 3 cos t ⇒ ′′ ′ rt q + 6q + 5q = 6 cos t. Substituting q = e in the homogeneous part we get, r2 + 6r + 5 = 0 ⇒ (r + 5) (r + 1) = 0 ⇒ r = −1, −5. Thus qh = c1 e−t + c2 e−5t . Since cos t is not a homogeneous solution, qp = A cos t + B sin t, qp′ = −A sin t + B cos t, qp′′ = −A cos t − B sin t. Substituting these in the original equation

we have,

−A cos t − B sin t − 6A sin t + 6B cos t + 5A cos t + 5B sin t = 6 cos t

Equating the coefficients of like terms we have,

cos t : 4A + 6B = 6 sin t : −6A + 4B = 0 ⇒ −3A + 2B = 0 Multiplying the second equation by 3 and then subtracting from 6 9 the first we get, 13A = 6 ⇒ A = 13 and then B = 23 A = 13 . 6 9 Thus qp = 13 cos t + 13 sin t and the general solution is

6 9

q = 13 cos t + 13 sin t + c1 e−t + c2 e−5t .

6 9

i = q ′ = − 13 sin t + 13 cos t − c1 e−t − 5c2 e−5t .

Using the initial conditions q(0) = 0, i(0) = 1 we get,

6 0 = 13 + c1 + c2

9

1 = 13 − c1 − 5c2 . 1 1 Adding them we get, 1 = 15 13 − 4c2 ⇒ c2 = 26 and then c1 = − 2 . 6 9 1 −t 1 −5t Thus q = 13 cos t + 13 sin t − 2 e + 26 e and 9 5 −5t 6 sin t + 13 cos t + 12 e−t − 26 e .

i = − 13 3. Substituting R = 32 , L = 1, C = 2, e = 32 in the equation 2

2

d q 1 3 dq 1 3 L ddt2q + R dq dt + C q = e we have, dt2 + 2 dt + 2 q = 2 ⇒ ′′ ′ rt 2q + 3q + q = 3. Substituting q = e in the homogeneous part of we get, 2r2 + 3r + 1 = 0 ⇒ (2r + 1) (r + 1) = 0 ⇒ 1 r = −1, − 21 . Thus qh = c1 e−t + c2 e− 2 t . Since the constant forcing is not a homogeneous solution, qp = A, qp′ = 0, qp′′ = 0. Substituting these in the original equation we have, A = 3 Thus qp = 3 and the general solution is 1 q = 3 + c1 e−t + c2 e− 2 t .

1 i = q ′ = −c1 e−t − 12 c2 e− 2 t .

Using the initial conditions q(0) = 2, i(0) = 4 we get,

2 = 3 + c1 + c2 ⇒ −1 = c1 + c2

4 = −c1 − 12 c2 .

Adding them we get, 3 = 12 c2 ⇒ c2 = 6 and then c1 = −7.

66

CHAPTER 2 − 12 t

− 12 t

Thus q = 3 − 7e−t + 6e and i = 7e−t − 3e . 1 5. Substituting R = 0, C = 10 , e = sin ωt in the equation 2

2

d q

1 L ddt2q + R dq dt + C q = e we have, L dt2 + 10q = sin ωt. rt By substituting q = e q , we obtain the characteristic equation

Lr2 + 10 = 0 ⇒ r = ± 10 L. The forcing term has frequency between 20 and 30 i.e.

ω 20 < 2π < 30 ⇒ 40π < ω < 60π. Resonance occurs if

q q 10 10 10 10 ω= i.e. if 40π < L L < 60π ⇒ (60π)2 < L < (40π)2 . Hence resonance does not occur for L such that

L < (6010π)2 or L > (4010π)2 .

½ 1, t ≤ π 1 7. Substituting R = 2, L = 1, C = 2 , e = 0, t > π 2

d q dq 1 in the equation L ½ dt2 + R dt + C q = e we have, 1, t ≤ π q ′′ + 2q ′ + 2q = . 0, t > π Substituting q = ert in the homogeneous part, we get r2 + 2r + 2 = 0 √ −2± 4−8 ⇒r= = −1 ± i. Thus qh = e−t (c1 cos t + c2 sin t) . 2 For t ≤ π, constant forcing, 1, is not a homogeneous solution. So

qp = A, qp′ = qp′′ = 0. Substituting these in the original equation

we have, 2A = 1 ⇒ A = 21 . So qp = 12 and the general solution

is q = 12 + e−t (c1 cos t + c2 sin t) .

i = q ′ = −e−t (c1 cos t + c2 sin t) + e−t (−c1 sin t + c2 cos t) .

Using the initial conditions q(0) = i(0) = 0 we get,

0 = 21 + c1 ⇒ c1 = − 12 0 = −c1 + c2 ⇒ c2 = c1 = − 21 .

Thus q = 21 − 12 e−t (cos t + sin t) and i = q ′ = e−t sin t for t ≤ π.

For t > π, q ′′ + 2q ′ + 2q = 0 ⇒ q is just the homogeneous

solution. So q = e−t (c1 cos t + c2 sin t)

i = q ′ = −e−t (c1 cos t + c2 sin t) + e−t (−c1 sin t + c2 cos t) .

Using the initial conditions q(π− ) = q(π+ ) we get,

1 −π 1 = −c1 e−π ⇒ c1 = − 21 eπ − 12 = − 12 (eπ + 1) and

2 + 2e from i(π− ) = i(π+ ), we have 0 = e−π (c1 − c2 ) ⇒ c1 − c2 = 0 ⇒ c2 = c1 = − 21 (eπ + 1) .

Thus q = − 21 (eπ + 1) e−t½[cos t + sin t] for t > π.

1 1 −t (cos t + sin t) , t ≤ π; 2 − 2e Hence the solution is q = 1 π − 2 (e + 1) e−t [cos t + sin t] , π < t ≤ 2π.

67

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

q

0.5

t π

2π

2

9. Substituting R = 0, e = 0 in the equation L ddt2q + R dq dt +

1 Cq d q 1 L 2 1 2 we have, L dt2 + C q = 0. Given that E = 2 i + 2C q ⇒

dq dq d2 q dq dE L di 1 1 dq dt =h 2 2i dt + 2Ci2q dt = L dt dt2 + C q dt (since i = dt )

2 d q 1 = dq dt L dt2 + C q = 0 ⇒ E is constant.

=e

2

11. Substituting R = 0, L = 9, C = 1, e = 4 cos 2t in the equation 2 1 ′′ rt L ddt2q + R dq dt + C q = e, we have 9q + q = cos 2t. Substituting q = e i 2 in the homogeneous part, we get 9r + 1 = 0 ⇒ r = ± 3 . Thus qh = c1 cos 3t + c2 sin 3t . Since cos 2t is not a homogeneous solution, qp = A cos 2t + B sin 2t, qp′ = −2A sin 2t + 2B cos 2t, qp′′ = −4A cos 2t − 4B sin 2t. Substituting these in the original

equation we have,

−36A cos 2t − 36B sin 2t + A cos 2t + B sin 2t = 4 cos 2t

Equating the coefficients of like terms we have,

4 cos 2t : −35A = 4 ⇒ A = − 35 sin 2t : −35B = 0 ⇒ B = 0. 4 cos 2t + c1 cos 3t + c2 sin 3t . If the free response, q = − 35 4 8 c1 cos 3t + c2 sin 3t , is absent then q = − 35 cos 2t, i = q ′ = 35 sin 2t. 4 So the initial conditions are q(0) = − 35 and i(0) = 0. 1 13. (a) The differential equation is Lq ′′ + C1 q = 0 ⇒ q ′′ + LC q = 0. q 1 1 rt 2 Substituting q = e we have r + LC = 0 ⇒ r = ± LC i. q q 1 1 Hence the general solution is q (t) = c1 cos LC t + c2 sin LC t⇒ q q q q 1 1 1 1 q ′ (t) = − LC c1 sin LC t + LC c2 cos LC t. ′ Using the initial conditions q (0) = 0 andq q (0) = 10 q we get, q 1 1 Thus q (t) = 10 LC sin LC t. p √ 2 2 (b) The amplitude is R = c1 + c2 = 10 LC. (c) From part (b) the amplitude increases as C increases. 1 15. (a) The differential equation is Lq ′′ + C1 q = 0 ⇒ q ′′ + LC q = 0. q 1 1 rt 2 Substituting q = e we have r + LC = 0 ⇒ r = ± LC i. q q 1 1 t + c2 sin LC t⇒ Hence the general solution is q (t) = c1 cos LC

c1 = 0 and c2 = 10

1 LC .

68

.

CHAPTER 2

q q q q 1 1 1 1 q ′ (t) = − LC c1 sin LC t + LC c2 cos LC t. ′ Using the initial conditions q (0) = 1 and q (0) q= 1 we√get, c1 = q1 √ 1 1 t and c2 = LC. Thus the motion is q (t) = cos LC t+ LC sin LC

p √ (b) The amplitude is R = c21 + c22 = 1 + LC. (c) From part (b) the amplitude increases as C and/or L increases. 1 17. Given that E = 1, R = 6, ω = 3, L = 1, C = 13 . Eω to obtain Plug these in the formula I = q 1 2 ( C −Lω2 ) +R2 ω2 3 3 3 . = √340 = √16+324 I=√ 2 (13−9) +36×9

2.9 EULER EQUATION 1. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 . Substituting these in the equation t2 x′′ + tx′ − x = 0, we have r (r − 1) tr + rtr − tr = 0. Dividing by tr 6= 0 yields the indicial equation r (r − 1) + r − 1 = 0 ⇒ r2 − 1 = 0 ⇒ r = 1, −1. Thus the general solution is x = c1 t + c2 t−1 . 3. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 . Substituting these in the equation t2 x′′ + 3tx′ + x = 0, we have r (r − 1) tr + 3rtr + tr = 0. Dividing by tr 6= 0 yields the indicial equation r (r − 1)+3r+1 = 0 ⇒ 2 r2 + 2r + 1 = 0 ⇒ (r + 1) ⇒ r = −1, −1. Thus the general solution is x = c1 t−1 + c2 t−1 ln t. 5. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 . Substituting these in the equation 4t2 x′′ +8tx′ +x = 0, we have 4r (r − 1) tr +8rtr +tr = 0. Dividing by tr 6 = 0 yields the indicial equation 4r (r − 1) + 8r + 1 = 2 0 ⇒ 4r2 + 4r + 1 = 0 ⇒ (2r + 1) ⇒ r = − 12 , − 21 . Thus the general − 12 − 21 solution is x = c1 t + c2 t ln t. 7. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 . Substituting these in the equation t2 x′′ +4tx′ +2x = 0, we have r (r − 1) tr +4rtr +2tr = 0. Dividing by tr 6= 0 yields the indicial equation r (r − 1)+4r+2 = 0 ⇒ r2 + 3r + 2 = 0 ⇒ (r + 1) (r + 2) ⇒ r = −1, −2. Thus the general solution is x = c1 t−1 + c2 t−2 . x′ = −c1 t−2 − 2c2 t−3 . Using the initial

conditions we get,

1 = c1 + c2

0 = −c1 − 2c2 .

Adding them we have, −c2 = 1 ⇒ c2 = −1 and then c1 = 2.

Thus x = 2t−1 − t−2 . 9. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 . Substituting these in the equation t2 x′′ + tx′ + 4x = 0, we have r (r − 1) tr + rtr + 4tr = 0. Dividing by tr 6= 0 yields the indicial equation r (r − 1) + r + 4 = 0 ⇒ r2 + 4 = 0 ⇒ r = ±2i. Thus the general solution is x = c1 cos (2 ln t) + c2 sin (2 ln t) . 11. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 . Substituting these in the equation t2 x′′ +3tx′ +8x = 0, we have r (r − 1) tr +3rtr +8tr = 0.

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

69

Dividing by tr 6= 0 yields the indicial equation r (r − 1)+3r+8 = 0 ⇒ √ r2 + 2r + 8 = 0 ⇒ r = −1 ± i 7. Thus the general solution is £ ¡√ ¡√ ¢ ¢¤ x = t−1 c1 cos 7 ln t + c2 sin 7 ln t . 13. Choose #3.

Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 . Substituting these in

the equation t2 x′′ + 3tx′ + x = 0, we have r (r − 1) tr + 3rtr + tr = 0.

Dividing by tr 6= 0 yields the indicial equation r (r − 1)+3r+1 = 0 ⇒ 2 r2 + 2r + 1 = 0 ⇒ (r + 1) ⇒ r = −1, −1 (repeated roots). So one −1 solution is t . Let the other solution be x = vt−1 . Then substituting x, x′ = v ′ t−1 − vt−2 and x′′ = v ′′ t−1 − 2v ′ t−2 + 2vt−3 in the original equation we have, ¡ ¢ ¡ ¢ t2 v ′′ t−1 − 2v ′ t−2 + 2vt−3 + 3t v ′ t−1 − vt−2 + vt−1 = 0. After cancellation of v terms and simplification this yields tv ′′ + v ′ = 0. dt Let w = v ′ . Then tw′ + w = 0. By separation, dw w =− t ⇒ −1 ′ ′ −1 ln |w| = − ln t + c ⇒ w = c2 t . Since w = v , v = c2 t . Integration yields v = c2 ln t + c1 . Thus the general solution is x = vt−1 = (c2 ln t + c1 ) t−1 = c1 t−1 + c2 t−1 ln t. 1 dx dx ds dx 15. (i) t = ks ⇒ s = k1 t ⇒ ds dt = k . ¡Now t¢dt = ks ds dt = s ds and ¡ ¢ 2 2 d dx dx ds ds 2 2d x 1 2 d2 x 2 2 d t2 ddt2x = k 2 s2 dx dt = k s ds ds dt dt = k s ds2 k2 = s ds2 . 2 So the Euler equation at2 ddt2x + bt dx dt + cx = 0 becomes 2 dx 2 d x

as ds2 + bs ds + cx = 0. dx ds = k1 dx (ii) In this case, dx dt = ds dt ds and ¡ ¢ ¡ ¢ ¡ 1 dx ¢ 1 2 d x dx d dx ds ds d d 1 d2 x

dt2 = dx dt = ds ds dt dt = ds k ds k = k2 ds2 .

2 So the constant coefficient equation a ddt2x + b dx dt + cx = 0 becomes 2 2 d x dx 2 + cx = 0 ⇒ a + bk + ck x = 0.

a k12 ddsx2 + b k1 dx ds ds2 ds r r (iii) Substituting t = ks in c1 tr1 + c2 tr2 we get, c1 (ks) 1 + c2 (ks) 2 r1 r2 r1

r1 r1 r2 r2 = c1 k s + c2 k s = c1 s + c2 s , where c1 = c1 k and c2 = c2 k r2 are real constants.

r r (iv) Substituting t = ks in c1 tr1 +c2 tr2 ln t we get, c1 (ks) 1 +c2 (ks) 2 ln (ks) r2 r2 r1 r1 = c1 k s + c2 k s (ln k + ln s) . Note that the roots are repeated so r1 = r2 and thus c1 k r1 sr1 +c2 k r1 sr1 (ln k + ln s) = c1 sr1 +c2 sr1 ln s, where c1 = k r1 (c1 + c2 ln k) and c2 = c2 k r1 are real constants. (v) Substituting t = ks in c1 tα cos (β ln t) + c2 tα sin (β ln t) we get, α α c1 (ks) cos (β ln (ks)) + c2 (ks) sin (β ln (ks))

α α = c1 k s cos (β ln k + β ln s) + c2 k α sα sin (β ln k + β ln s)

= c1 k α sα [cos (β ln k) cos (β ln s) − sin (β ln k) sin (β ln s)]

+c2 k α sα [sin (β ln k) cos (β ln s) + cos (β ln k) sin (β ln s)] = c1 sα cos (β ln s) + c2 sα sin (β ln s) , where c1 = c1 k α cos (β ln k) + c2 k α sin (β ln k) and c2 = −c1 k α sin (β ln k) + c2 k α cos (β ln k) are real constants. 17. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 ⇒ x′′′ = r (r − 1) (r − 2) tr−3 ⇒ x′′′′ = r (r − 1) (r − 2) (r − 3) tr−4 . Substituting these in the equation t4 x′′′′ + 6t3 x′′′ + 7t2 x′′ + tx′ − x = 0, we have r (r − 1) (r − 2) (r − 3) tr + 6r (r − 1) (r − 2) tr + 7r (r − 1) tr

70

CHAPTER 2

+rtr − tr = 0.

Dividing by tr = 6 0 yields the indicial equation

r (r − 1) (r − 2) (r − 3)+6r (r − 1) (r − 2)+7r (r − 1)+r −1 = 0 ⇒ r4 − 1 = 0 ⇒ r2 = 1, −1 ⇒ r = 1, −1, ±i.

Thus the general solution is x = c1 t+c2 t−1 +c3 cos (ln t)+c4 sin (ln t) .

19. Let x = tr ⇒ x′ = rtr−1 ⇒ x′′ = r (r − 1) tr−2 ⇒ x′′′ = r (r − 1) (r − 2) tr−3 . Substituting these in the equation t3 x′′′ + tx′ − x = 0, we have r (r − 1) (r − 2) tr + rtr − tr = 0. Dividing by tr = 6 0 yields the indicial equation r (r − 1) (r − 2) + r − 1 = 0

3 ⇒ r3 − 3r2 + 3r − 1 = 0 ⇒ (r − 1) = 0 ⇒ r = 1, 1, 1.

2 Thus the general solution is x = c1 t + c2 t ln t + c3 t (ln t) .

2.10 VARIATION OF PARAMETERS (SECOND-ORDER) In problems 21 and 23 we derive the solution obtained by variation of parameters. In the other problems, the algebraic system of equations (obtained from the variation of parameters) v1′ x1 + v2′ x2 = 0 v1′ x′1 + v2′ x′2 = 0 x2 and will be solved for v1′ and v2′ by using the formulae v1′ = − fW f x1 ′ v2 = W where W [x1 , x2 ] is the Wronskian of the independent homogeneous solutions x1 and x2 and f is the forcing function. 1. Substituting x = ert in the homogeneous part of x′′ − x = e2t we have

the characteristic equation r2 − 1 = 0 ⇒ r = 1, −1. So x1 = et and

x2 = e−t are homogeneous solutions. The Wronskian,

· t ¸ e e−t t −t W [e , e ] = det = −2et e−t = −2 6= 0. Thus {et , e−t } et −e−t is a fundamental set of solutions and here f = e2t .

Variation of parameters: x = v1 x1 + v2 x2 .

2t −t 2t t x1 x2 = e 2e = 12 et and v2′ = fW = − e 2e = − 12 e3t . So v1′ = − fW Integrating¡ we have,¢ v1 = ¡12 et + c1 and¢ v2 = − 61 e3t + c2 .

Thus x = 12 et + c1 x1 + − 16 e3t + c2 x2 ⇒ x = 21 e2t + c1 et + − 16 e2t + c2 e−t ⇒ x = 13 e2t + c1 et + c2 e−t

is the general solution.

Yes, the method of undetermined coefficients could have been used

because of exponential forcing.

3. Substituting x = ert in the homogeneous part of x′′ + x = sin1 t we have

the characteristic equation r2 + 1 = 0 ⇒ r = ±i. So x1 = cos t and

x2 = sin t are homogeneous solutions.¸ The Wronskian,

· cos t sin t W [cos t, sin t] = det = cos2 t + sin2 t = 1 6= 0. − sin t cos t Thus {cos t, sin t} is a fundamental set of solutions and here f = sin1 t . Variation of parameters: x = v1 x1 + v2 x2 .

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

71

x2 x1 t So v1′ = − fW = − sin1 t sin t = −1 and v2′ = fW = cos sin t . Integrating we have, v1 = −t + c1 and v2 = ln |sin t| + c2 . Thus x = (−t + c1 ) x1 + (ln |sin t| + c2 ) x2 ⇒ x = (−t + c1 ) cos t + (ln |sin t| + c2 ) sin t ⇒ x = sin t ln |sin t| − t cos t + c1 cos t + c2 sin t is the general solution. No, sin1 t is not a right form of forcing function for using the method of undetermined coefficients. 5. Substituting x = ert in the homogeneous part of x′′ + x = tan t we have the characteristic equation r2 + 1 = 0 ⇒ r = ±i. So x1 = cos t and x2 = sin t are homogeneous solutions. The Wronskian, · ¸ cos t sin t W [cos t, sin t] = det = cos2 t + sin2 t = 1 6= 0. − sin t cos t Thus {cos t, sin t} is a fundamental set of solutions and here f = tan t.

Variation of parameters: x = v1 x1 + v2 x2 .

2 f x1 x2 t ′ = − tan t sin t = − sin t =

So v1′ = − fW cos t and v2 = WR = tan t cos R R 1−cos 2 t sin t. Integrating we have, v1 = − cos t dt = − sec tdt + cos tdt

= − ln |sec t + tan t| + sin t + c1 and v2 = − cos t + c2 .

Thus x = (− ln |sec t + tan t| + sin t + c1 ) x1 + (− cos t + c2 ) x2 ⇒

x = (− ln |sec t + tan t| + sin t + c1 ) cos t + (− cos t + c2 ) sin t

Since sin t cos t terms cancel

x = − (cos t) ln |sec t + tan t| + c1 cos t + c2 sin t.

is the general solution.

No, tan t is not a right form of forcing function for using the method

of undetermined coefficients.

7. Substituting x = ert in the homogeneous part of x′′ + 3x′ + 2x = 1+1e2t we have the characteristic equation r2 + 3r + 2 = 0 ⇒ (r + 1) (r + 2) = 0 ⇒ r = −1, −2. So x1 = e−t and x2 = e−2t ·are homogeneous ¸ £ ¤ e−t e−2t solutions. The Wronskian, W e−t , e−2t = det −t −e −2e−2t © −t −2t ª −3t −3t −3t = −2e +e = −e 6= 0. Thus e , e is a fundamental set of solutions and here f = 1+1e2t . Variation of parameters: x = v1 x1 + v2 x2 . f x1 x2 1 e−2t et 1 e−t ′ = − 1+e So v1′ = − fW 2t −e−3t = 1+e2t and v2 = W = 1+e2t −e−3t R et e2t t = − 1+e 2t . Integrating we have, v1 = 1+e2t dt (Substitution: u = e ) R 1 ⇒ v1 = 1+u2 du = tan−1 u + c1 = tan−1 (et ) + c1 and R e2t R du 1 2t v2 = − 1+e 2t dt (Substitution: u = 1 + e ) ⇒ v2 = − 2 u ¢ ¡ 1 2t 2t + c2 .

= − 21 ln u + c (since u = 1 + e > 0 ) ⇒ v = − ln 1 + e 2 2 ¢ 2 ¤ £ −1 t ¤ £ 1 ¡ 2t + e c x ⇒

Thus£ x = tan (e ¤) + c1 x£1 + −¡2 ln 1 + 2 2 ¢ ¤ 1 2t + c2 e−2t ⇒ x = tan−1 (et ) + c1 e−t + − 2 ln 1 ¢+ e ¡ x = e−t tan−1 (et ) − 21 e−2t ln 1 + e2t + c1 e−t + c2 e−2t is the general

solution.

No, 1+1e2t is not a right form of forcing function for using the method

of undetermined coefficients.

3 9. Substituting x = ert in the homogeneous part of x′′ − 6x′ + 9x = e3t t 2

72

CHAPTER 2 2

we have the characteristic equation r2 − 6r + 9 = 0 ⇒ (r − 3) = 0 ⇒ r = 3, 3. So x1 = e3t and x2 = te3t are homogeneous · 3t ¸ £ ¤ e te3t solutions. The Wronskian, W e3t , te3t = det 3t e3t + 3te3t © 3t 3t ª 3e 6t 6t 6t 6t = e + 3te − 3te = e 6= 0. Thus e , te is a fundamental set 3t 3

2

of solutions and here f = e t . Variation of parameters: x = v1 x1 + v2 x2 . 3t 3

5

3 e3t 3

x2 f x1 3t 2

′ 2

2

= −e3t t 2 te So v1′ = − fW e6tR = −t and v2 = W = e t e6t = t . 7

5

2 2

t + c1 and Integrating we have, v1 = −t 2 dt = − 7

R 3

5

2 2

v2 = t 2 dt ³ = 5 t + c´2 . ´ ³ 7

5

2 2

2 2

Thus x = − 7

t + c1 x1 + 5

t + c2 x2 ⇒ ³ ´ ³ ´ 5

2 7

2

x = − 7 t 2 + c1 e3t + 5 t 2 + c2 te3t ⇒ 7

7

2 2 3t 2 2 3t t e + 5

t e + c1 e3t + c2 te3t ⇒ x = − 7

4 7

3t x = 35 t 2 e + c1 e3t + c2 te3t is the general solution. 3

No, e3t t 2 is not a right form of forcing function for using the method of undetermined coefficients. t 11. Substituting x = ert in the homogeneous part of 4x′′ +4x′ +x = t−2 e− 2

2 we have the characteristic equation 4r2 + 4r + 1 = 0 ⇒ (2r + 1) = 0 t t − 2

⇒ r = − 21 , − 1

and x2 = te− 2 are homogeneous 2 . So x1 = e · ¸ t t i h t t e− 2

te− 2

− 2

− 2

= det solutions. The Wronskian, W e , te t t t 1 − 2

− 2

− 2

e e− 2 − 1

2 te n t o t 1 −t = e−t − 21 te−t + 2

te = e−t = 6 0. Thus e− 2 , te− 2 is a fundamental t

set of solutions and here f = t−2 e− 2 . Variation of parameters: x = v1 x1 + v2 x2 . t

−t

x2 = −t−2 e− 2 tee−t2 = −t−1 and So v1′ = − fW t

−t

x1 v2′ = fW = t−2 e− 2 ee−t2 = tR−2 . Integrating we have, v1 = −t−1 dt = − ln t + c1 and R v2 = t−2 dt = −t−1 + c2 . ¡ ¢ Thus x = (− ln t + c1 ) x1 + −t−1 + c2 x2 ⇒ ¡ ¢ t t x = (− ln t + c1 ) e− 2 + −t−1 + c2 te− 2 ⇒ t t t t x = −e− 2 ln t − t−1 te− 2 + c1 e− 2 + c2 te− 2 ⇒ t t t t − 2

− 2

− 2

− 2

x = −e ln t − e + c1 e + c2 te ⇒ t t t x = −e− 2 ln t + (c1 − 1) e− 2 + c2 te− 2 ⇒ t t t x = −e− 2 ln t + c1 e− 2 + c2 te− 2 (where c1 = c1 − 1) is the general solution. t No, t−2 e− 2 is not a right form of forcing function for using the method of undetermined coefficients. 13. Substituting x = ert in the homogeneous part of x′′ − x′ − 6x = e−2t we have the characteristic equation r2 − r − 6 = 0 ⇒ −2t (r − 3) (r + 2) = 0 ⇒ r = 3, −2. So x1 = e3t£ and x2 = are ¤ e 3t −2t = homogeneous solutions. The Wronskian, W e , e

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

·

¸

73

© ª e3t e−2t = −2et − 3et = −5et 6= 0. Thus e3t , e−2t is 3t −2t 3e −2e a fundamental set of solutions and here f = e−2t . Variation of parameters: x = v1 x1 + v2 x2 . x2 x1 e3t e−2t 1 −5t 1 So v1′ = − fW = −e−2t −5e and v2′ = fW = e−2t −5e t = 5e t = −5. R 1 1 −5t −5t + c1 and

Integrating R we have, v1 = 5 e dt = − 25 e v2 = − 15 ¡dt = − 51 t + c2¢.

¡ ¢ 1 −5t Thus¡ x = − 25 e ¢ + c1 ¡x1 + − 1

5 t¢+ c2 x2 ⇒ 1 −5t x = − 25 e + c1 e3t + − 15 t + c2 e−2t ⇒ ¡ ¢ −2t 1 1 −2t 1 −2t e x = − 25 e − 5 te +c1 e3t +c2 e−2t = − 51 te−2t +c1 e3t + c2 − 25

1 x = − 15 te−2t + c1 e3t + c2 e−2t , (where c2 = c2 − 25 )

is the general solution.

Yes, the method of undetermined coefficients could have been used

because of exponential © forcing.

ª 15. The fundamental set is t, t2 and dividing the equation by t2 we ¸ · £ 2¤ t t2 have the forcing function f = t. The Wronskian, W t, t = det 1 2t = 2t2 − t2 = t2 . Let x1 = t and x2 = t2 . Variation of parameters: x = v1 x1 + v2 x2 . x1 x2 = −t and v2′ = fW

= 1. Integrating we have, So v1′ = − fW 1 2 v1 = − 2 t ¡+ c1 and v¢2 = t + c2 .

¡ ¢ 2 2 Thus x = − 21 t2 + c1 x1 + (t + c2 ) x2 = − 1

2 t + c1 t + (t + c2 ) t ⇒ 1 3 1 3 3 2 2 x = − 2 t + t + c1 t + c2 t ⇒ x = 2 t + c1 t + c2 t is the general solution. © ª 17. The fundamental set is 1, t−1 and dividing the equation by t2 we

have the forcing function f =¸ t−3 . The Wronskian,

· −1 £ −1 ¤ 1 t = det W 1, t = −t−2 . Let x1 = 1 and x2 = t−1 . 0 −t−2 Variation of parameters: x = v1 x1 + v2 x2 . x2 x1 = t−2 and v2′ = fW

= −t−1 . Integrating we have, So v1′ = − fW −1 v1 = −t ¡+ c1 and v¢2 = − ln t + c2 .

Thus x = −t−1 + c1 x1 +(− ln t + c2 ) x2 = −t−1 +c1 +(− ln t + c2 ) t−1 ⇒ x = −t−1 − t−1 ln t + c1 + c2 t−1 is the general solution. 19. Let x1 (t) and x2 (t) be the homogeneous solutions of x′′ +px′ +qx = f. If the Wronskian is W [x1 , x2 ] (t) and the variation of parameters is x = v1 x1 + v2 x2 then definite integral yields Rt (t) (t)x2 (t) v1′ = − Wf[x ⇒ v1 = − W [xx2,x f (t)dt

1 ,x2 ](t) 1 2 ](t) 0

Rt (t)x1 (t) x1 (t) and v2′ = Wf[x ⇒ v = f (t)dt. Since we have 2 1 ,x2 ](t) W [x1 ,x2 ](t) 0 used definite integrals, we do not have any arbitrary constants of integration and the particular solution is thus xp = v1 x1 + v2 x2 ⇒ Rt Rt (t) (t) f (t)dt + x2 (t) W [xx1,x f (t)dt ⇒ xp = −x1 (t) W [xx2,x ] t 1 2 ( ) 1 2 ](t) 0 0 det

74

CHAPTER 2

Rt x2 (t)x1 (t)−x1 (t)x2 (t) f (t)dt. W [x1 ,x2 ](t) 0

xp =

2

21. Substituting x = ert in the homogeneous part of x′′ − x = e−t we have the characteristic equation r2 − 1 = 0 ⇒ (r − 1) (r + 1) = 0 ⇒ r = 1, −1. So x1 = et and x2 = e−t are homogeneous · t ¸ e e−t t −t solutions. The Wronskian, W [e , e ] = det et −e−t = −1 − 1 = −2 6= 0. Thus {et , e−t } is a fundamental set 2 of solutions and here f = e−t .

Variation of parameters: x = v1 et + v2 e−t .

So x′ = v1 et − v2 e−t + v1′ et + v2′ e−t . We set v1′ et + v2′ e−t = 0

such that x′ = v1 et − v2 e−t ⇒ x′′ = v1 et + v2 e−t + v1′ et − v2′ e−t .

Substituting this in the original equation, we have 2 v1 et + v2 e−t + v1′ et − v2′ e−t − v1 et − v2 e−t = e−t ⇒

2 v1′ et − v2′ e−t = e−t . Now we solve the system of equations

v1′ et + v2′ e−t = 0

2 v1′ et − v2′ e−t = e−t 2 2 for v1′ and v2′ . Adding them yields 2v1′ et = e−t ⇒ v1′ = 12 e−t e−t 2 2 and v2′ e−t = −v1′ et = − 12 e−t ⇒ v2′ = − 12 et e−t . Rt 2 Definite integral yields, v1 = 12 e−s e−s ds + c1 and

0

v2 =

Rt

− 21

s −s2

e e

0 ·

Thus x = x= x=

1 t 2e

Rt 0

Rt 0

¸ · ¸ Rt 2 2 e−s e−s ds + c1 et + − 12 es e−s ds + c2 e−t ⇒ 0

−s −s2

e

0

Rt £ 1 0

x=

Rt

1 2

ds + c2 .

t−s 2e

e

t

ds + c1 e −

1 −t 2e

Rt 0

s −s2

e e

ds + c2 e−t ⇒

¤ 2 − 12 e−(t−s) e−s ds + c1 et + c2 e−t ⇒ 2

sinh (t − s) e−s ds + c1 et + c2 e−t is the general solution.

1 23. Substituting x = ert in the homogeneous part of x′′ + 5x′ + 6x = t+1 we have the characteristic equation r2 + 5r + 6 = 0 ⇒ (r + 2) (r + 3) = 0 ⇒ r = −2, −3. So x1 = e−2t and x2 = e−3t are homogeneous solutions. The Wronskian, · ¸ £ −2t −3t ¤ e−2t e−3t = det W e ,e = −e−5t 6= 0. −2t −3t − 2 e − 3 e © ª 1 Thus e−2t , e−3t is a fundamental set of solutions and here f = t+1 .

−2t −3t Variation of parameters: x = v1 e + v2 e .

So x′ = −2v1 e−2t − 3v2 e−3t + v1′ e−2t + v2′ e−3t . We set

v1′ e−2t + v2′ e−3t = 0 such that x′ = −2v1 e−2t − 3v2 e−3t ⇒ x′′ = 4v1 e−2t + 9v2 e−3t − 2v1′ e−2t − 3v2′ e−3t .

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

75

Substituting this in the original equation, we have 4v1 e−2t + 9v2 e−3t − 2v1′ e−2t − 3v2′ e−3t − 10v1 e−2t − 15v2 e−3t 1 1 +6v1 e−2t + 6v2 e−3t = t+1 ⇒ −2v1′ e−2t − 3v2′ e−3t = t+1 . Now we solve the system of equations

v1′ e−2t + v2′ e−3t = 0

1

−2v1′ e−2t − 3v2′ e−3t = t+1 ′ ′ for v1 and v2 . Multiplying the first equation by 2 and adding with 1 1 3t ⇒ v2′ = − t+1

e and then the second yields −v2′ e−3t = t+1 1 1 2t ′ −2t ′ −3t ′ v1 e = −v2 e = t+1 ⇒ v1 = t+1 e . .

Rt e3s Rt e 2s ds + c1 and v2 = − s+1 ds + c2 . Definite integral yields, v1 = s+1 0 0 ¸ · t ¸ ·t R e3s R e 2s −2t ds + c e + − ds + c e−3t ⇒ Thus x = 1 2 s+1 s+1 0

0

Rt x = e−2t 0

x=

Rt £ 0

Rt e −2t −3t ds + c e − e 1 s+1 2s

0

e2(s−t) − e3(s−t)

¤

1 s+1 ds

e3s s+1 ds

+ c2 e−3t ⇒

+ c1 e−2t + c2 e−3t is the general

solution. 25. Substituting x = tr , x′ = rtr−1 , x′′ = r (r − 1) tr−2 in the homogeneous part of t2 x′′ + tx′ − x = et and then dividing by tr 6= 0 we have r (r − 1) + r − 1 = 0 ⇒ r2 − 1 = 0 ⇒ r = 1, −1. So x1 = t and x2·= t−1 are homogeneous solutions. The Wronskian, ¸ £ −1 ¤ t t−1 −1 −1 = det W t, t = −t − t = −2t−1 6= 0. −2 1 − t © ª Thus t, t−1 is a fundamental set of solutions and here f = et .

Variation of parameters: x = v1 x1 + v2 x2 .

x2 f x1 t−1 1 t t 1 2 t ′ t So v1′ = − fW = −et −2t −1 = 2 e and v2 = W = e −2t−1 = − 2 t e .

Rt Definite integral yields, v1 = 21 et + c1 and v2 = − 21 s2 es ds. 0 · ¸ £1 t ¤ Rt 2 s 1 −1 Thus x = 2 e + c1 t + − 2 s e ds + c2 t ⇒ 0

x = 21 tet + c1 t −

x = 12 tet −

1 2

Rt

Rt 1

2

0

t−1 s2 es ds + c2 t−1 ⇒

t−1 s2 es ds + c1 t + c2 t−1 is the general solution.

0

2.11 VARIATION OF PARAMETERS (N TH-ORDER) The algebraic system of equations in this section (obtained from the variation of parameters)

v1′ x1 + v2′ x2 + ... + vn′ xn = 0

v1′ x′1 + v2′ x′2 + ... + vn′ x′n = 0

... ... ... ...

76

CHAPTER 2 (n−1)

(n−1)

(n−1)

v1′ x1 + v2′ x2 + ... + vn′ xn =0 ′ ′ ′ will be solved for v1 , v2 , ..., vn by using the formulae n+i f Wi vi′ = (−1) an W [x1 ,x2 ,...,xn ] where W [x1 , x2 , ..., xn ] is the Wronskian of the independent homogeneous solutions x1 , x2 , ..., xn , 6 0 is the coefficient of the f is the forcing function and an = nth-derivative term. ′ 2t 1. Substituting x = ert in the homogeneous part of x′′′ − ¡ x2 = e¢ , 3 we have the characteristic equation r − r = 0 ⇒ r r − 1 = 0 ⇒ r = 0, 1, −1. So x1 = 1, x2 = et and x3 = e−t are homogeneous solutions. The Wronskian,   1 et e−t W [1, et , e−t ] = det  0 et −e−t  = 2 6= 0. Thus {1, et , e−t } 0 et e−t is a fundamental set of solutions. Here f = e2t , n = 3, a3 = 1. Variation of parameters:  v2 x2 + v3 x3 .  x = v1 x1 + et e−t   det ¡ ¢ et −e−t 3+1 e2t = −e2t So v1′ = (−1) = e2t −2 1 2 2 1 2t ⇒ v1 = − 2 e + c1 .  −t 1 e  det ³ −t ´ 0 −e−t 3+2 e2t v2′ = (−1) = −e2t −e2 = 21 et 1 2 ⇒ v2 = 12 et + c2 .



1 et 0 e t





3+3 e2t = (−1) = 12 e2t et = 12 e3t

1 2 1 3t ⇒ v3 = 6 e¡ + c3 .

¢ ¢ ¢ ¡ ¡ Thus x = − 21 e2t + c1 1 + 12 et + c2 et + 61 e3t + c3 e−t ⇒

x = − 12 e2t + c1 + 21 e2t + c2 et + 16 e2t + c3 e−t ⇒ x = 61 e2t + c1 + c2 et + c3 e−t is the general solution. rt ′′′′ ′′′

v3′

det

3. Substituting x = e in the homogeneous part of x − x = t, we have the characteristic equation r4 − r3 = 0 ⇒ r3 (r − 1) = 0 ⇒ r = 0, 0, 0, 1. So x1 = 1, x2 = t, x3 = t2 and x4 = et are homogeneous solutions.  The Wronskian,    1 t t2 e t 1 2t et t   £ ¤ 0 1 2 t e t    W 1, t, t2 , et = det   0 0 2 et  = det 0 2 et 0 0 e 0 0 0 et

t = 2e ©6= 0.

ª Thus 1, t, t2 , et is a fundamental set of solutions.

Here f = t, n = 4, a4 = 1.

Variation of parameters: x = v1 x1+ v2 x2 + v3 x3 + v4 x4 .

 t t2 et    1 2t et  det  t £ ¡ ¢¤ 0 2 e 4+1 t ′ So v1 = (−1) = − 2et t et t2 − 2t + 2 1 2et

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

¡ 1

¢

4

3

77

2

= − 2 t3 − 2t2 + 2t ⇒ v1 = − t8 + t3 − t2 + c1 .

  1 t2 et   t  det  0 2t e  t 0 2 e 4+2 t = 2ett 2et (t − 1) = t2 − t v2′ = (−1) 1 2et 3 2 ⇒ v2 = t3 − t2 + c2 .   1 t et  t   det  0 1 e  t 0 0 e 4+3 t = − 2et t et = − 2t v3′ = (−1) 1 2et 2 ⇒ v3 = − t4 + c3 .   1 t t2    det  0 1 2t  0 0 2 4+4 t = 2et t 2 = ett v4′ = (−1) 1 2et −t −t ⇒ v4 = −te ³ 4− e 3 + c42. ´ ³ 3 ´ ³ 2 ´ 2 Thus x = − t8 + t3 − t2 + c1 1 + t3 − t2 + c2 t + − t4 + c3 t2

+ (−te−t − e−t + c4 ) et ⇒ 3 t 4 − t6 + c1 + c2 t + c3 t2 + c4 et is the general solution, x = − 24 1 where c 1 = c1 − 1, c2 = c2 − 1,c3 = c3 − 2 .

at bt ct e e e 5. W = det  aeat bebt cect  . Factoring out eat , ebt and ect from a2 eat b2 ebt c2 ect the first, second and  third column,  respectively, we get 1 1 1 W = eat ebt ect det  a b c  . Subtracting second column a2 b2 c2 from the third and, next, first column fromthe second to get  1 0 0 b−a c − b  . Factoring out b − a, W = e(a+b+c)t det  a a2 b2 − a2 c2 − b2 and c − b from the second and third we get  column, respectively,  1 0 0 1 1  W = (b − a) (c − b) e(a+b+c)t det  a 2 a b+a c+b = (b − a) (c − b) e(a+b+c)t (c + b − b − a) = (b − a) (c − a) (c − b) e(a+b+c)t . 7. Substituting x = ert in the homogeneous part of x′′′ −2x′′ −x′ +2x = et , we have the characteristic equation r3 − 2r2 − r + 2 = 0 ⇒ (r − 1) (r + 1) (r − 2) = 0 ⇒ r = 1, −1, 2. So x1 = et , x2 = e−t and x3 = e2t£ are homogeneous solutions. The Wronskian, using Exercise ¤ t −t 2t (1−1+2)t = 5, is W e , e , e e (−1 − 1) (2 − 1) (2 + 1) = −6e2t 6 = 0. © t −t 2t ª Thus e , e , e is a fundamental set of solutions. Here f = et , n = 3, a3 = 1.

Variation of parameters: x = v1 x1 + v2 x2 + v3 x3 .

78

CHAPTER 2

So ⇒ v2′ ⇒

det

e−t −e−t

e t et

e2t 2e2t

3+1 et = (−1) 1 v1 = − 2t + c1 . 

v1′



det

3+2 et = (−1) 1 1 2t v2 = 12 e + c2 .

−6e2t

−6e2t

et et

e−t −e−t





= 



1 6et

= − 61et [2et + et ] = − 12 £ 3t ¤ 2e − e3t = 16 e2t

 3+3 et = (−1) = − 6e1t [−1 − 1] = 31 e−t 1 −6e2t 1 −t ⇒ v3 = − ¡3 e + c3¢. ¡ 1 2t ¢ ¡ ¢ Thus x = − 2t + c1 et + 12 e + c2 e−t + − 31 e−t + c3 e2t 1 t x = − 2t et + c1 et + 12 e + c2 e−t − 13 et + c3 e2t ⇒ t t t x = − 2 e + c1 e + c2 e−t + c3 e2t is the general solution, 1 − 31 . where c1 = c1 + 12 rt ′′′ ′′ ′

v3′

det

e2t 2e2t



⇒

9. Substituting x = e in the homogeneous part of x +2x −x −2x = 1, we have the characteristic equation r3 + 2r2 − r − 2 = 0 ⇒ (r − 1) (r + 1) (r + 2) = 0 ⇒ r = 1, −1, −2. So x1 = et , x2 = e−t and x3 = e−2t are The Wronskian, using £ homogeneous ¤ solutions. (1−1−2)t Exercise 5, is W et , e©−t , e−2t = eª (−1 − 1) (−2 − 1) (−2 + 1) = −6e−2t 6= 0. Thus et , e−t , e−2t is a fundamental set of solutions.

Here f = 1, n = 3, a3 = 1.

Variation of parameters: x = v1 x1 + v2x2 + v3 x3 .

 −t e e−2t  det −t £ ¤ −e −2e−2t 3+1 1 1 = − 6e−2t

−2e−3t + e−3t So v1′ = (−1) −2t 1 −6e = 16 e−t ⇒ v1 = − 61 e−t + c1 . 

et e−2t  det t e −2e−2t 3+2 1 1 v2′ = (−1) = 6e−2t [−2e−t − e−t ] = − 21 et 1 −6e−2t 1 t ⇒ v2 = − 2 e + c2.  e t e−t   det et −e−t 3+3 1 1 v3′ = (−1) = − 6e−2t [−1 − 1] = 31 e2t 1 −6e−2t ⇒ v3 = 16 e¡2t + c3 . ¢ ¡ ¢ ¢ ¡ Thus x = − 61 e−t + c1 et + − 21 et + c2 e−t + 16 e2t + c3 e−2t ⇒ x = − 61 + c1 et − 21 + c2 e−t + 61 + c3 e−2t ⇒ x = − 12 + c1 et + c2 e−t + c3 e−2t is the general solution. 11. Substituting x = ert in the homogeneous part of x′′′ +3x′′ −x′ −3x = et , we have the characteristic equation r3 + 3r2 − r − 3 = 0 ⇒ (r − 1) (r + 1) (r + 3) = 0 ⇒ r = 1, −1, −3. So x1 = et , x2 = e−t and x3 = e−3t are The Wronskian, using £ homogeneous ¤ solutions. −3t (1−1−3)t = Exercise 5, is W et , e−t , e e (−1 − 1) (−3 − 1) (−3 + 1) © t −t −3t ª −3t = −16e 6= 0. Thus e , e , e is a fundamental set of solutions. Here f = et , n = 3, a3 = 1.

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

79

Variation of parameters: x = v1 x1 + v2 x 2 + v3 x3 .

 −t e e−3t  det −t £ ¤ −e −3e−3t 3+1 et So v1′ = (−1) = − 16e1−4t −3e−4t + e−4t −3t 1 −16e = 18 ⇒ v1 = 81 t + c1 . 

et e−3t   det

£ ¤ et −3e−3t 3+2 et v2′ = (−1) = 16e1−4t −3e−2t − e−2t = − 14 e2t 1 −16e−3t ⇒ v2 = − 81 e2t + c2.  et e−t  det t e −e−t 3+3 et v3′ = (−1) = − 16e1−4t [−1 − 1] = 81 e4t

1 −16e−3t 1 4t

⇒ v3 = 32¡e + c3 .¢ ¡ 1 2t ¢ ¢ ¡ 1 4t

e + c2 e−t + 32 e + c3 e−3t ⇒ Thus x = 18 t + c1 et + − 8 t 1 t x = te8 + c1 et − 18 et + c2 e−t + 32 e + c3 e−3t ⇒ t x = te8 + c1 et + c2 e−t + c3 e−3t is the general solution, where

1 c1 = c1 + 32 − 1

8 .



eαt teαt t2 eαt . eαt + αteαt 2teαt + αt2 eαt 13. W = det  αeαt 2 αt αt 2 αt αt αt 2 2 αt α e 2αe + α te 2e + 4αte + α t e Factoring out eαt from the first, second  and third column,  1 t t2  1 + αt 2t + αt2 respectively, we get W = eαt eαt eαt det  α 2 2 2 2 α 2α + α t 2 + 4αt + α t (expanding with respect to the second   column)    1 1 0 t2 t t2  .  + det  α 1 2t + αt2 2t + αt2 = e3αt det  α αt 2 2 2 2 2 2 2 α 2α 2 + 4αt + α t α α t 2 + 4αt + α t The first determinant is zero since its second column equals the first column after factoring out t. Then 

 1 0 t2 

1 2t + αt2 W = e3αt det  α 2 α 2α 2 + 4αt + α2 t2

(expanding  with  respect to2thelast column)

 

1 0 t 1 0 0 

1 αt2  + det  α 1 2t = e3αt det  α 2 α 2α α2 t2 α2 2α 2 + 4αt

The first determinant is zero since its third column equals the first

column after factoring out t. Then 

 1 0 0  = e3αt (2 + 4αt − 4αt) = 2e3αt .

1 2t W = e3αt det  α 2 α 2α 2 + 4αt

15. Substituting x = ert in the homogeneous part of x′′′ + 3x′′ + 3x′ + x

= t−3 e−t , we have the characteristic equation r3 + 3r2 + 3r + 1 = 0

3 ⇒ (r + 1) = 0 ⇒ r = −1, −1, −1. So x1 = e−t , x2 = te−t and

2 −t x3 = t e are homogeneous solutions. The Wronskian, using

80

CHAPTER 2

£ ¤ Exercise 13, is W e−t , te−t , t2 e−t = 2e−3t 6= 0. Here f = t−3 e−t , n = 3, a3 = 1. Variation of parameters: x = v1 x1 + v2 x2 + v3 x3 . te−t t2 e−t  det −t −t −t ¡ ¢ e − te 2te − t2 e−t 3+1 t−3 e−t ′ So v1 = (−1) = 2t3 e1−2t t2 e−2t 1 2e−3t = 21t ⇒ v1 = 21 ln t + c1.  e−t t2 e−t  det ¡ ¢ −e−t 2te−t − t2 e−t 3+2 t−3 e−t ′ v2 = (−1) = − 2t3 e1−2t 2te−2t 1 2e−3t = − t12 ⇒ v2 = 1t + c2 .   e−t te−t   det ¡ ¢ −e−t e−t − te−t 3+3 t−3 e−t v3′ = (−1) = 2t3 e1−2t e−2t = 2t13 1 2e−3t ⇒ v3 = − ¡4t12 + c3 . ¢ ¡ ¢ −t ¡ 1 ¢ Thus x = 12 ln t + c1 e−t + 1 + − 4t2 + c3 t2 e−t ⇒ t + c2 te x = 12 (ln t) e−t + c1 e−t + e−t + c2 te−t − 14 e−t + c3 t2 e−t ⇒ x = 21 (ln t) e−t + c1 e−t + c2 te−t + c3 t2 e−t is the general solution, where c1 = c1 + 1 − 41 . 17. Substituting x = ert in the homogeneous part of x′′′ − 6x′′ + 12x′ − 8x 7 = t 2 e2t , we have the characteristic equation r3 − 6r2 + 12r − 8 = 0 3 ⇒ (r − 2) = 0 ⇒ r = 2, 2, 2. So x1 = e2t , x2 = te2t and 2 2t x3 = t e are homogeneous solutions. The Wronskian, using

£ ¤ 7 Exercise 13, is W e2t , te2t , t2 e2t = 2e6t 6= 0. Here f = t 2 e2t , n = 3, a3 = 1. Variation of parameters: x = v1 x1 + v2 x2 + v3 x3 . te2t t2 e2t 

det 2t 2t 2t 7 2t e + 2te 2te + 2t2 e2t 3+1 t 2 e

′ So v1 = (−1) 1 2e6t 7 ¡ ¢ 1 11 13 1 t2 2 4t = 2e4t t e = 2 t 2 ⇒ v1 = 13 t 2 + c1 .



e2t t2 e2t   det 7 ¡ 7 ¢ 2e2t 2te2t + 2t2 e2t 3+2 t 2 e2t ′ v2 = (−1) = − 2te24t 2te4t 1 2e6t 9 2 11 = −t 2 ⇒ v2 = − 11 t 2 + c2 .  e2t te2t  det 7 ¡ 7 ¢ 1 7 2e2t e2t + 2te2t 3+3 t 2 e2t t2 4t ′ = 2t2 = v3 = (−1) 6t 1 2e 2e4t e 1 29 ⇒ v3 = 9 t³ + c3 . ´ ³ ´ ³ ´ 2 11 1 9 1 13 2t 2t 2 2t 2 + c 1 e + − 11 t 2 + c2 te + 9 t 2 + c3 t e ⇒ 13 t 13 1 13 2 13 x = 13 t 2 e2t + c1 e2t − 11 t 2 e2t + c2 te2t + 19 t 2 e2t + c3 t2 e2t ⇒ 13 8 t 2 e2t + c1 e2t + c2 te2t + c3 t2 e2t is the general solution. x = 1287 19. Let x1 = £et , x2 = tet¤and x3 = t2 et . The Wronskian, using Exercise

13, is W et , tet , t2 et = 2e3t 6= 0. Here f = et , n = 3, a3 = 2.

Thus x =

Variation of parameters: x = v1 x1 + v2 x2 + v

3 x3 .

 t 2 t te t e 

det t e + tet 2tet + t2 et 3+1 et

′ So v1 =¡ (−1)¢ 2 2e3t

1 3 = 4e12t t2 e2t = 41 t2 ⇒ v1 = 12 t + c1 .

81

LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

v2′ =



det

3+2 et = (−1) 2 − 21 t ⇒ v2 = − 41 t2

t2 et 2te + t2 et t

2e3t

+ c2 . et tet t t e e + tet



¡ ¢ = − 4e12t 2te2t



 ¡ ¢ 3+3 et = (−1) = 4e12t e2t = 41 2 2e3t ⇒ v3 = 41 t¡+ c3 . ¢ ¡ ¢ ¡ ¢ 1 3 Thus x = 12 t + c1 et + − 41 t2 + c2 tet + 41 t + c3 t2 et 1 3 t x = 12 t e + c1 et − 14 t3 et + c2 tet + 41 t3 et + c3 t2 et ⇒ 1 3 t x = 12 t e + c1 et + c2 tet + c3 t2 et is the general solution.

v3′

det

et et



1

21. Let x1 = 1, x2 = t and x3 = t 2 . The Wronskian is   1 1 t t2 h i 1 1 − 12  = − 1 t− 32 = W 1, t, t 2 =  0 1 6 0. 4 2t 3 1 −2 0 0 −4t Here f = t3 , n = 3, a3 = t2 . Variation of parameters: x = v1 x1+ v2 x2 + v3 x3 .  1 t t2  det 1 ³ ´ 1 12 t− 2 5 3+1 t3 1 12 2 t − = 2t3 So v1′ = (−1) = −4t 3 2 t 2 1 − ⇒ v1 =

v2′

=

1 4 2t

+ c1 .

3+2 t3 (−1) t2

=

3+3 t3 (−1) t2 7



det

⇒ v2 = 32 t3 + c2 . v3′

−4t



det

1

1 0

t2 1 − 12 2t

3 − 14 t− 2

1 0

2

t 1 3

− 14 t− 2

 

 

5

= 4t 2

³

1 − 12 2t

´

= 2t2

5

5

= −4t 2 (1) = −4t 2

⇒ v3 = − 78 t 2 + c3 . ³ ´ 1 ¢ ¢ ¡ ¡ 7 Thus x = 12 t4 + c1 1 + 32 t3 + c2 t + − 87 t 2 + c3 t 2 ⇒ 1

x = 21 t4 + c1 + 23 t4 + c2 t − 78 t4 et + c3 t 2 ⇒ 1 1 4 x = 42 t + c1 + c2 t + c3 t 2 is the general solution.

⇒

Chapter Three

The Laplace Transform

3.1 DEFINITION AND BASIC PROPERTIES 1. Graph. f (t) is not piecewise continuous because lim+ f (t) does not exist. t→2

3. Graph. lim f (t) = 1, lim f (t) = 0 and f (1) = 12 . So f (t) is not continuous t→1+

t→1−

at t = 1, however, it is piecewise continuous because lim− f (t) and t→1

lim f (t) exist. t→1+ ½ 1, 0 ≤ t ≤ 1, 5. f (t) = 0, t > 1. R∞ R1 R∞ −st L [f (t)] = e−st f (t) dt = e−st dt+ 0×e−st dt = e−s |10 = 1s (1 − e−s ) . 0 1  0  t, 0 ≤ t < 1, 2 − t, 1 ≤ t ≤ 2,

7. f (t) =  0, t > 2

R∞ −st R1 R2 R∞ L [f (t)] = e f (t) dt = te−st dt + (2 − t) e−st dt + 0 × e−st dt 0 1 ¯ ¯1 ¯ 0 ¯2 2 = ¯− 1s te−st − s12 e−st ¯0 + ¯− 2s e−st + 1s te−st + s12 e−st ¯1 = − 1s e¡−s − s12 e−s + s12 −¢ 2s e−2s + 2s e−2s + s12 e−2s + 2s e−s − 1s e−s − s12 e−s = s12 1 − 2e−s + e−2s . R∞ Ra 9. L [sin bt] = e−st sin btdt = lim e−st sin btdt. Using an integration

a→∞ 0 0 ¯ −st ¯a Ra −st ¯ ¯ table, we get e sin btdt = ¯ se2 +b2 (−s sin bt − b cos bt)¯ 0

e−sa s2 +b2

0

b (−s sin ba − b cos ba) + s2 +b 2. i h −sa b b e = s2 +b Then L [sin bt] = lim s2 +b2 (−s sin ba − b cos ba) + s2 +b 2 2. a→∞ µ ¶ R∞ R∞ R∞ −st at R∞ −st −at at −at 1 11. L [sinh at] = e−st sinh atdt = e−st e −e dt = e e dt − e e dt 2 2 0 0 ´ 0 ³ 0 ´ ³ 1 1 1 s+a−s+a 1 a 1 at −at = 2 (L [e ] − L [e ]) = 2 s−a − s+a = 2 (s−a)(s+a) = s2 −a2 . µ∞ ¶ R∞ R∞ R −st ibt R∞ −st −ibt ibt −ibt 1 13. L [sin bt] = e−st sin btdt = e−st e −e dt = e e dt − e e dt 2i 2i 0 0 ³ ´ ³0 ´ 0

¡ £ ¤ £ ¤¢ 1 1 1 1 s+ib−s+ib 1 L eibt − L e−ibt = 2i − = = 2i s−ib s+ib 2i (s−ib)(s+ib)

=

83

THE LAPLACE TRANSFORM

=

b s2 +b2 .

R∞

2t −2t e−st e +e dt 2

15. L [3 cosh 2t] = 3 0 ¡ £ 2t ¤ £ ¤¢ 3 = 2 L e − L e−2t =

3 2

³

=

1 s−2

3 2

+

µ∞ R

−st 2t

e

0 ´ 1 s+2

=

e dt + ³ 3 2

R∞

−st −2t

e

e

0 ´ s+2+s−2 (s−2)(s+2)

=

dt

¶

3s s2 −4 .

30 6 17. L [5 sin 6t] = 5 s2 +6 2 = s2 +36 . 19. L [−t + 3] . Linearity implies −L [t] + L [3] = − s 12 + 3s . s 21. L [2 + cos 5t] . Linearity implies L [2] + L [cos 5t] = 2s + s2 +25 . µ∞ ¶ ∞ ∞ R R R −st e3t −e−3t 1 −st 3t −st −3t dt = 2 e e dt − e e dt 23. L [sinh 3t] = e 2 0 0 ³ ´ ³ 0 ´ ¡ £ 3t ¤ £ −3t ¤¢ 1 1 1 +3−s+3 1 3 1 = 2 s−3 − s+3 = 2 (ss− = 2 L e −L e 3)(s+3) = s2 −9 . £ −t ¤ £ ¤ 6 2 + s−3 . 25. L 2e + 6e3t . Linearity implies 2L [e−t ] + 6L e3t = s+1 1 2t 1 −2t 27. L [3t − 1 + cosh 2t] .£ Note cosh 2t = e + e . Linearity implies 2 ¤ £ ¤2 1 1 3L [t] − L [1] + 12 L e2t + 12 L e−2t = s 32 − 1s + 21 s−2 + 21 s+2 1 s 3 = s2 − s + s2 −4 . £ ¤ £ ¤ 29. L 7t3 + 11t + 8 . Linearity implies 7L t3 + 11L [t] + L [8] 8 42 11 8 + 11 = 7(3!) s2¤ + s = s4 + s2 + s . £ s3+1 £ ¤ £ ¤ 4(3!) 31. L 3t4 + 4t3 . Linearity implies 3L t4 + 4L t3 = 3(4!) s4+1 + s3+1 72 24 = s 5 + s4 . 33. We apply shifting theorem L [ect f (t)] = F (s − c) with c = 3 and 5 . f (t) = sin 5t, so that F (s) = L [f (t)] = L [sin 5t] = s2 +25 5 So F (s − 3) = (s−3)2 +25 . 35. We apply shifting theorem L [ect f (t)] = F (s − c) with c = 4 and s . f (t) = cos 7t, so that F (s) = L [f (t)] = L [cos 7t] = s2 +49 s−4 So F (s − 4) = (s−4)2 +49 . 37. We apply shifting theorem L [ect f (t)] = F (s − c) with c = −3 and 5 . f (t) = sin 5t, so that F (s) = L [f (t)] = L [sin 5t] = s2 +25 5 So F (s + 3) = (s+3)2 +25 . 39. We apply shifting theorem L [ect f (t)] = F (s − c) with c = −4 and s . f (t) = cos 7t, so that F (s) = L [f (t)] = L [cos 7t] = s2 +49 s+4 So F (s + 4) = (s+4)2 +49 . 41. We apply shifting theorem L [ect f (t)] =£F ¤(s − c) with c = 2 and f (t) = t6 , so that F (s) = L [f (t)] = L t6 = s6!7 . So F (s − 2) = (s−6!2)7 . 43. We apply shifting theorem L [ect f (t)] =£F ¤(s − c) with c = −2 and f (t) = t6 , so that F (s) = L [f (t)] = L t6 = s6!7 .

6!

So F (s + 2) = (s+2) 7. 45. We apply shifting theorem L [ect f (t)] = F (s − c) £ with ¤ c £= 3¤ and f (t) = t5 + t3 + 1, so that F (s) = L [f (t)] = L t5 + L t3 + L [1] 1 . = s5!6 + s3!4 + 1s . So F (s − 3) = (s−5!3)6 + (s−63)4 + s−3 ¤ £ ¤ £ 1 1 47. f (t) = L−1 [F (s)] = L−1 s2 − L−1 s = t − 1.

84

CHAPTER 3

53.

55.

57.

59. 61. 63. 65.

i

h

i

1 1 1 −1 3 s2 +9 = 3 L s2 +32 = 3 sin 3t. £ ¤ £ ¤ £ £ ¤ ¤ = L−1 s12 + 1s = L−1 s12 + L−1 1s f (t) = L−1 [F (s)] = L−1 1+s s2 = t + 1. h i f (t) = L−1 [F (s)] = L−1 3s − s72 + s219 +1

h i

£ ¤ £ ¤ = L−1 3s − L−1 s72 + L−1 s219 +1 h i £ ¤ £ ¤ = 3L−1 1s − 7L−1 s1!2 + 19L−1 s21+1 = 3 − 7t + 19 sin t. h i i i h h 7 −1 f (t) = L−1 [F (s)] = L−1 s3s+7 = L−1 s23s + L 2 +16 2 +16 s +16 i i h h 7 −1 7 s 4 −1 = 3 cos 4t + 4 sin 4t. = 3L 2 2 s2 +42 + 4 L hs +4 i h i h i 5 7 5 7 −1 −1 −1 −1 f (t) = L [F (s)] = L s+3 + s−5 = L s+3 + L s−5 h i h i 1 1 = 5L−1 s+3 + 7L−1 s−5 = 5e−3t + 7e5t . £ ¤ £ ¤ £ ¤ 1 −1 5! 1 5 f (t) = L−1 [F (s)] = L−1 hs−6 i= L−1 s16h = 5! L s6 = 5! t .

i 3 = 32 sin 3t. f (t) = L−1 [F (s)] = L−1 s22+9 = 23 L−1 s2 +3 2 £ ¤ £ ¤ 3 −1 9! 3 9 f (t) = L−1 [F (s)] = L−1 s310 = 9! L ·s9+1 = 9! ¸t . i h h i 1 f (t) = L−1 [F (s)] = L−1 2s23+7 = L−1 2 s23+ 7 = 23 L−1 s2 + 7 ( 2 2) # " √ q q q 7 = 23 27 L−1 2 ³ √2 7 ´ 2 = 23 27 sin 72 t.

49. f (t) = L−1 [F (s)] = L−1

51.

h

s +

2

⇒ F (s) = s13 .

£ ¤ £ 2! ¤ 1 2 = 2t . So f (t) =hL−1 [Fi(s)] = L−1 s13 = 12 L−1 s2+1

67. Let F (s − 4) = Now L−1

1 (s−4)3

1 (s−4)3

= L−1 [F (s − 4)] and the shifting theorem (14)

with c = 4 yields L−1 [F (s − 4)] = e4t f (t) = 12 t2 e4t .

(s−5) s ⇒ F (s) = s2s+9 = s2 +3 69. Let F (s − 5) = (s−5) 2 2. +9 h i s −1 −1 So f (t) = L [F (s)] = L s2 +32 = cos 3t. h i (s−5) Now L−1 (s−5) = L−1 [F (s − 5)] and the shifting theorem 2 +9

(14) with c = 5 yields L−1 [F (s − 5)] = e5t f (t) = e5t cos 3t. 7 4 71. Let F (s − 7) = (s−7)72 +16 ⇒ F (s) = s2 +16 = 74 s2 +4 2. i h 4 = 47 sin 4t. So f (t) = L−1 [F (s)] = 47 L−1 s2 +4 2 h i Now L−1 (s−7)72 +16 = L−1 [F (s − 7)] and the shifting theorem

(14) with c = 7 yields L−1 [F (s − 7)] = e7t f (t) = 74 e7t sin 4t. (s+3) ⇒ F (s) = s2s+5 = 2 s√ 2 . 73. Let F (s + 3) = (s+3) 2 +5 s + ( 5) · ¸ √ So f (t) = L−1 [F (s)] = L−1 2 s√ 2 = cos 5t. s + ( 5) i h (s+3) = L−1 [F (s + 3)] and the shifting theorem Now L−1 (s+3) 2 +5 √ (14) with c = −3 yields L−1 [F (s + 3)] = e−3t f (t) = e−3t cos 5t.

85

THE LAPLACE TRANSFORM

½

2, t = 1; . t, t = 6 1. R1 R1 R∞ L [f (t)] = te−st dt + 2e−st dt + te−st dt 0 ¯ ¯1 ¯1 1 ¯b = ¯− 1s te−st − s12 e−st ¯0 + 0 + lim ¯− 1s te−st − s12 e−st ¯1 b→∞ ¡ = − 1s e−s − s12 e−s + s12 + lim − 1s be−sb − s12 e−sb + 1s e−s +

75. f (t) =

b→∞

1 s2

1 −s s2 e

¢

(since lim e−sb = 0 and first two terms cancel with last two). b→∞ ¯ ¯b R∞ Now L [t] = te−st dt = lim ¯− 1s te−st − s12 e−st ¯0

b→∞ 0 ¡ ¢ = lim − 1s be−sb − s12 e−sb + s12 = s12 .

=

b→∞

√

77. Yes, since e t ≤ et for t ≥ 1. R∞ 79. M eαt e−st dt converges for s > α, and |f (t) e−st | ≤ M eαt e−st by 0

assumption. ³ ´ ¡ £ 5t ¤¢ £ ¤ 1 1 d d = − ds L e 81. L te5t = − ds s−5 = (s−5)2 . ³ ´ d d s s2 −25 83. L [t cos 5t] = − ds (L [cos 5t]) = − ds s2 +25 = (s2 +25)2 .

³ ´ £ ¤ 4s(s2 −9) d2 d2 s −2s 85. L t2 cos 3t = ds 2 (L [cos 3t]) = ds2 2 s +9 = (s2 +9)2 + (s2 +9)3 . ´ ³ £ ¡ £ 5t ¤¢ ¤ d d 3 87. L te5t sin 3t = − ds L e sin 3t = − ds = 6(s−5) 2 2. 2 (s−5) +9 ´[(s−5) +9] 2 ³ ¡ £ ¤¢ £ −4t ¤ d s+4 d = (s+4)2 −25 2 . L e−4t cos 5t = − ds 89. L te cos 5t = − ds (s+4)2 +25 [(s+4) +25] n at 91. The derivative theorem for higher powers of t is L [t e ] = (s−an)!n+1 . h i So L−1 (s−an)!n+1 = tn eat . i i h h 1 5 −3t 1 −1 5! 1 = 5! = 5! L t e . Thus L−1 (s+3) 6 5+1 i h h i (s+3) h ³ ´i d 5s 5 −1 3 = 65 L−1 s26s 93. L−1 (s+9) = L − = 56 t sin 3t. 2 2 6 ds s +9 i h 2 h +9³ ´i −9 d s 95. L−1 (ss2 +9) = L−1 − ds = t cos 3t. 2 s2 +9

h i i h 2

5 −1 1 s −9 5 L − = 18 97. L−1 (s2 +9) 2 2

2 (s +9) (s2 +9) i

h 5 −1 1 3 s2 −9 = 18 L 3 (s2 +9) − (s2 +9)2

h h i h ³ ´ii

5 d 1 −1 3 s −1 = 18 − ds 3L (s2 +9) − L s2 +9

£ ¤ 5 1 = 18 h 3 sin 3t i − t cos h3t . i

2s 1 −1 99. L−1 (s22s+1 = L 2 2 + 2

2 2 (s +1) (s +1) i

h+1) 2s 2 + = L−1 21 (s2 +1) 2 (s2 +1)2

i h i

h 1 −1 2s 1 s2 −1 −1 = 2L + L − 2 2

2 2 2 +1) (s +1) ³ h i (s +1) h ³ (s ´i´ h ³ ´i 1 1 d s d 1 −1 −1 −1 =2 L − L − + L − 2 2 2 (s +1) ds s +1 ds s +1 =

1 2

(sin t − t cos t) + t sin t.

86

CHAPTER 3

R∞

101. L [f (ct)] =

0

e−st f (ct) dt. Let u = ct ⇒ 1c u = t and 1c du = dt.

Then this integral becomes L [f (t)] =

R∞

1 c

R∞ 0

s

e− c u f (u) du = 1c F

e−st f (t) dt = F (s) .

0

103. Using (T2), F (s) = L [f (t)] = L [eat ] = −2 1 F ′ (s) = − (s − a) = (−1) (s−1!a)2 F ′′ (s) = − (−2) (s − a)

−3

= (−1)

F ′′′ (s) = − (−2) (−3) (s − a)

−4

2

2! (s−a)3 3

= (−1) −5

1 s−a

¡s¢ c

, since

−1

= (s − a)

. So

3! (s−a)4 4

F ′′′′ (s) = − (−2) (−3) (−4) (s − a) = (−1) (s−4!a)5 ... ... ... ... −(n+1) n F (n) (s) = − (−2) (−3) (−4) ... (−n) (s − a) = (−1) n

n! . (s−a)(n+1)

Now using Exercise 102, L [tn f (t)] = (−1) F (n) (s) , we have n n L [tn eat ] = (−1) (−1) (s−an)(!n+1) = (s−an)(!n+1) .

3.2 INVERSE LAPLACE TRANSFORMS (ROOTS, QUADRATICS, & PARTIAL FRACTIONS) 1 A B 1. Using partial fractions, s21−9 = (s−3)(s+3) = s−3 + s+3 . Multiplying by the denominator gives 1 = A (s + 3) + B (s − 3) . Letting s = 3 and −3 1 1 in this equation i gives A h= 6³, B = − 6 . ´i ³ i i´ h h h 1 1 1 1 − s+3 = 61 L−1 s−3 − L−1 s−3 Thus L−1 s21−9 = L−1 16 s−3 ¡ ¢ = 16 e3t − e−3t . ¸ · √ h i h i h i h i 7 5s 1 s 1 −1 5s−1 −1 −1 −1 −1 √ √ 2 3. L s2 +7 = L s2 +7 −L s2 +7 = 5L s2 +7 − 7 L s 2 + ( 7) √ √ 1 = 5 cos 7t − √7 sin 7t. s+2 A B 5. Using partial fractions, ss+2 2 −1 = (s−1)(s+1) = s−1 + s+1 . Multiplying by the denominator gives s + 2 = A (s + 1) + B (s − 1) . Letting s = 1

1 3 and −1 in hthis equation gives h A h − 2i.

i i = 2, B =

Thus L−1

s+2 s2 −1

= 23 L−1

1 s−1

− 21 L−1

1 3 t 1 −t s+1 = 2 e − 2 e .

A B s + s−1 . Multiplying by

2s−1 7. Using partial fractions, 2s2s−1 −s = s(s−1) = the denominator gives 2s − 1 = A (s − 1) + Bs. Letting s = 0 and 1 in this equation gives A = 1, and B = 1. Thus 2s−1 1 1

.

s2 −s = hs + s−1 i h i £ ¤ 1 −1 2s−1 −1 1 t −1 + So L = L L s2 −s s s−1 = 1 + e .

−s−1 A B 9. Using partial fractions, s2−s−1 +s−2 = (s−1)(s+2) = s−1 + s+2 . Multiplying by the denominator gives −s − 1 = A (s + 2) + B (s − 1) . Letting

87

THE LAPLACE TRANSFORM

2 1 s = 1 and h−2 in this i equation hgivesi A = − 3 ,hB =i − 3 . 1 1 1 −2t 2 −1 1 −1 2 t Thus L−1 s2−s−1 . +s−2 = − 3 L s−1 − 3 L s+2 = − 3 e − 3 e

4 A B 4 = (s−2)(s+3) = s−2 + s+3 . Multiplying 11. Using partial fractions, s2 +s−6 by the denominator gives 4 = A (s + 3) + B (s − 2) . Letting s = 2 4 4 and −3 in hthis equation i ³gives hA = i5 , B = − h 5 . i´ ¡ ¢ 1 1 4 = 54 L−1 s−2 − L−1 s+3 = 54 e2t − e−3t . Thus L−1 s2 +s−6

13.

15.

s s2 −2s+26

= h

2s+5 s2 −6s+18

= h

that L−1 Thus by thehshifting theorem (T16) with c = 1, we have i ¡ ¢ s 1 −1 t f (t) = L s2 −2s+26 = e cos 5t + 5 sin 5t . that L−1

17.

(s−1)+1 s = (s−1) . Replacing s − 1 by s, we compute 2 2 (s−1)i +25 +5i2 h i h s 1 −1 1 5 s+1 −1 s2 +52 = L s2 +52 + 5 L s2 +52 = cos 5t + 5 sin 5t.

2s+5 (s−3)i2 +9 2s+11 s2 +32 =

2(s−3)+11 2 2 . Replacing s − 3 by s, we compute (s−3) h +3 i h i s 11 −1 3 2L−1 s2 +3 = 2 cos 3t+ 11 2 + 3 L 2 2 s +3 3 sin 3t.

=

Thus by thehshifting theorem (T16) with c = 3, we have i 2s+5 11 3t −1 3t f (t) = L s2 −6s+18 = 2e cos 3t + 3 e sin 3t.

3s−2 s2 +10s+26

3s−2 = 3(s+5)−17 2 2 2 . Replacing s + 5 by s, we compute (s+5) (s+5) i +1 h +1 i h i s 1 3s−17 −1 −1 = 3L − 17L s2 +12 s2 +12 s2 +12 = 3 cos t − 17 sin t.

= h

that L−1 Thus by thehshifting theorem (T16) with c = −5, we have i 2s+5 −1 −5t f (t) = L cos t − 17e−5t sin t. s2 −6s+18 = 3e

s s A B 19. Using partial fractions, s2 −5s+6 = (s−2)(s−3) = s−2 + s−3 . Multiplying by the denominator gives s = A (s − 3) + B (s − 2) . Letting s = 2 and 3 in this h equation i gives³ A =h−2, iB = 3. h i´ s 1 1 −1 −1 −1 Thus L = −2 L + 3L = 3e3t − 2e2t . 2 s −5s+6 s−2 s−3

1 B C = As + s−1 + s−2 . 21. Using partial fractions, s3 −3s12 +2s = s(s−1)(s−2) Multiplying by the denominator gives 1 = A (s − 1) (s − 2) + Bs (s − 2) + Cs (s − 1) . Letting s = 0, 1 1 1 and 2 in this h C= i gives³ A =£2 ,¤B = −1, i´2 . h i h equation 1 1 + 12 L−1 s−2 Thus L−1 s3 −3s12 +2s = 21 L−1 1s − L−1 s−1

= 12 − et + 21 e2t . s2 −2 s2 −2 A B C 23. Using partial fractions, s3 +8s 2 +7s = s(s+1)(s+7) = s + s+1 + s+7 . Multiplying by the denominator gives s2 − 2 = A (s + 1) (s + 7) + Bs (s + 7) + Cs (s + 1) . Letting s = 0, 1 47 −1 and −7h in this equation gives A = − 27 , B = h 6, C h i ³ i´= 42 . i £ ¤ 2 s −2 1 47 −1 1 Thus L−1 s3 +8s = − 72 L−1 1s + 16 L−1 s+1 + 42 L 2 +7s s+7

47 −7t e . = − 72 + 61 e−t + 42 s+3 A B C 25. Using partial fractions, ss+3 3 −s = s(s−1)(s+1) = s + s−1 + s+1 . Multiplying by the denominator gives s + 3 = A (s − 1) (s + 1) + Bs (s + 1) + Cs (s − 1) . Letting s = 0, 1

88

CHAPTER 3

and −1 in hthis equation C = 1. h i ³gives£ A¤ = −3, Bh= 2,i´ i 1 1 −1 s+3 −1 1 −1 Thus L + L−1 s+1 s3 −s = −3 L s + 2L s−1

= −3 + 2et + e−t . 2s+1 2s+1 A Bs+C 27. Using partial fractions, s3 +4s 2 +13s = s(s2 +4s+13) = s + s2 +4s+13 . Multiplying¡ by the denominator gives ¢ 2s+1 = A s2 + 4s + 13 +(Bs + C) s = As2 +4As+13A+Bs2 +Cs. 1 . Equating the coefficients: Letting s = 0 in this equation gives A = 13 1 2 s : 0 = A + B ⇒ B = −A = − 13 4 s: 2 = 4A + C ⇒ C = 2 − 13 = 22 13 . 2s+1 1 1 s−22 1 1 (s+2)−24 Thus s3 +4s2 +13s = 13s − 13 = − 2 2 s2 +4s+13 ³ ´ 13s 13 (s+2) +3 (s+2) 1 1 3 = 13s − 13 (s+2)2 +32 − 8 (s+2)2 +32 .

Now applying the shifting theorem (T16) with c = −2 in the second and hthird terms,i we have 2s+1 1 1 −2t L−1 s3 +4s = 13 − 13 e (cos 3t − 8 sin 3t) . 2 +13s 2

2

s −3 s −3 A Bs+C 29. Using partial fractions, s3 +2s 2 +26s = s(s2 +2s+26) = s + s2 +2s+26 . Multiplying gives ¡ by the denominator ¢ s2 −3 = A s2 + 2s + 26 +(Bs + C) s = As2 +2As+26A+Bs2 +Cs. 3 . Equating the Letting s = 0 in this equation gives A = − 26 coefficients: 3 s2 : 1 = A + B ⇒ B = 1 + 26 = 29 26 3 s: 0 = 2A + C ⇒ C = −2A = 13 . 3 1 29s+6 3 1 29(s+1)−23 s2 −3 Thus s3 +2s2 +26s = − 26s + 26 s2 +2s+26 = − 26s + 26 (s+1)2 +52 ´ ³ (s+1) 3 23 1 5 = − 26s + 26 − 29 (s+1) 2 2 5 (s+1) +52 .

+52

Now applying the shifting theorem (T16) with c = −1 in the second and hthird terms,i we have ¡ ¢ 3 1 −t s2 −3 = − 26 + 26 e 29 cos 5t − 23 L−1 s3 +2s 2 +26s 5 sin 5t .

A s−8 = s−5 + Bs+C 31. Using partial fractions, (s−5)(s 2 s2 +4 . Multiplying by the ¡ +4) ¢ 2 denominator gives s − 8 = A s + 4 + (Bs + C) (s − 5) ⇒ s − 8 = As2 + 4A + Bs2 + (C − 5B) s − 5C. 3 Letting s = 5 in this equation gives A = − 29 . Equating the coefficients:

3

s2 : 0 = A + B ⇒ B = −A = 29 15 s: 1 = C − 5B ⇒ C = 1 + 29 = 44 29 . s−8 3 1 1 3s+44 So (s−5)(s2 +4) = − 29 s−5 + 29 2 s +4 ´ ³ 1 s 2 3 1 + 3 + 22 = − 29 2 2 s2 +22 . Thus h s−5 29 i s +2 s−8 1 3 5t L−1 (s−5)(s e + 29 (3 cos 2t + 22 sin 2t) . = − 29 2 +4) 2

A 33. Using partial fractions, (s−3)(s2s−2s+26) = s−3 + s2Bs+C −2s+26 . Multiplying by the denominator gives ¡ ¢ s2 = A s2 − 2s + 26 + (Bs + C) (s − 3) ⇒ s2 = As2 − 2As + 26A + Bs2 + (C − 3B) s − 3C.

89

THE LAPLACE TRANSFORM

9 Letting s = 3 in this equation gives A = 29 . Equating the coefficients:

9 s2 : 1 = A + B ⇒ B = 1 − 29 = 20

29 26A 1: 0 = 26A − 3C ⇒ C = 3 = 78 29 . 9 1 1 20s+78 9 1 2 10(s−1)+49 s2 Thus (s−3)(s2 −2s+26) = 29 s−3 + 29 s2 −2s+26 = 29 s−3 + 29 (s−1)2 +52 ´ ³ (s−1) 9 1 49 2 5 = 29 s−3 + 29 10 (s−1)2 +52 + 5 (s−1)2 +52 .

Now applying the shifting theorem (T16) with c = 1 in the second and hthird terms, we ihave ¡ ¢ 2 9 3t 2 t L−1 (s−3)(s2s−2s+26) = 29 e + 29 e 10 cos 5t + 49 5 sin 5t . 3

3

s −1 s −1 As+B Cs+D 35. Using partial fractions, s4 +10s 2 +9 = (s2 +9)(s2 +1) = s2 +9 + s2 +1 . Multiplying by the¡denominator gives ¡ ¢ ¢

s3 − 1 = (As + B) s2 + 1 + (Cs + D) s2 + 9 ⇒ s3 − 1 = (A + C) s3 + (B + D) s2 + (A + 9C) s + B + 9D.

Equating the coefficients:

s3 : 1=A+C s2 : 0=B+D s: 0 = A + 9C 1: −1 = B + 9D Subtracting the first from the third we get, 8C = −1 ⇒ C = − 18 . Then from the third, A = −9C = 89 . Subtracting the second from the fourth we get, 8D = −1 ⇒ D = − 18 . Then from the second, B = −D = 81 . ³ ´

´ ³

So =

s3 −1 s4 +10s2 +9

1 8

L−1

³ h

= 18

9s+1 s2 +9

−

s+1 s2 +1

=

1 8

9s+1 2 +32

´s

−

s+1 s2 +12

s 1 3 s 1 . Thus 9 s2 +3 2 + 3 s2 +32 − s2 +12 − s2 +12

i ¡ ¢ 3

1 1 s −1 s4 +10s2 +9 = 8 9 cos 3t + 3 sin 3t − cos t − sin t . 3

3

3

s s s 37. Using partial fractions, s4 −5s 2 +4 = (s2 −1)(s2 −4) = (s−1)(s+1)(s−2)(s+2) A B C D = s−1 + s+1 + s−2 + s+2 . Multiplying by the denominator gives 3 s = A (s + 1) (s − 2) (s + 2) + B (s − 1) (s − 2) (s + 2) +C (s − 1) (s + 1) (s + 2) + D (s − 1) (s + 1) (s − 2) . Letting s = 1, −1, 2 and −2 in this equation gives A = − 61 , B = − 16 , C = 32 , D = 23 . ³ ´ ³ ´

s3 s4 −5s2 +4

1 1 1 1 = − 16 s−1 + s+1 + 32 s−2 + s+2 . Thus

i ¡ ¢ 3 s L−1 s4 −5s = − 16 (et + e−t ) + 32 e2t + e−2t . 2 +4 s s s = (s2 −4)(s 39. Using partial fractions, s4 −16 2 +4) = (s−2)(s+2)(s2 +4) B Cs+D A = s−2 + s+2 + s2 +4 . Multiplying by the denominator gives ¡ ¢ ¡ ¢ s = A (s + 2) s2 + 4 +B (s − 2) s2 + 4 +(Cs + D) (s − 2) (s + 2) ⇒

s = (A + B + C) s3 + (2A − 2B + D) s2 + (4A + 4B − 4C) s + 8A +8B − 4D.

1 1 Letting s = 2 and −2 in this equation gives A = 16 , B = 16 .

Equating the coefficients:

So

h

90

CHAPTER 3 1 0 = A + B + C ⇒ C = − 16 − 0 = ´2A − 2B + D ⇒ D = 0 s 1 1 1 1 s So s4 −16 = 16 s−2 + s+2 − 8 s2 +4 . Thus

h i ¡ 2t ¢ s 1 L−1 s4 −16 e + e−2t − 18 cos 2t. = 16

s3 : s³2 :

1 16

= − 81

41. The derivative theorem for higher powers of t is L [tn eat ] = h i So L−1 (s−an)!n+1 = tn eat . Thush using ia = −3 and h n = 5 we i get 4 −1 5! 1 5 −3t 4 4 5 −3t −1 = 5! L t e = 30 t e . L = 5! (s+3)6 (s+3)5+1

n! . (s−a)n+1

43. The derivative theorem for higher powers of t is L [tn eat ] = (s−an)!n+1 . i h So L−1 (s−an)!n+1 = tn eat . Thus using a = − 31 and n = 4 we get ¸ · · ¸ h i 4 1 4 − 13 t 4 4! 4 −1 −1 −1 L = 3!3 =L = 4!35 L . 5 5t e 1 4+1 (3s+1)5 35 (s+ 31 ) s+ ( 3) 2 A C B D + (s+3) . 45. Using partial fractions, (s+3)2s(s−3)2 = s+3 2 + s−3 + (s−3)2 Multiplying by the denominator gives 2 2 2 2 s2 = A (s + 3) (s − 3) + B (s − 3) + C (s − 3) (s + 3) + D (s + 3) Letting s = 3 and −3 in this equation gives D = 14 , B = 14 . Equating the coefficients: s3 : 0=A+C s2 : 1 = −3A + B + D + 3C ⇒ 1 − 41 − 14 = −3A + 3C ⇒ 16 = −A + C 1 1 Adding them we get, C = 12 and then A = −C = − 12 . s2 1 1 1 1 1 1 1 1 So (s+3)2 (s−3)2 = − 12 s+3 + 4 (s+3)2 + 12 s−3 + 4 (s−3)2 . Thus i h 2 1 −3t 1 3t e + 41 te−3t + 12 e + 41 te3t L−1 (s+3)2s(s−3)2 = − 12 ¡ ¢ ¡ ¢ 1 1 + e3t 14 t + 12 = e−3t 14 t − 12 . A Cs+D B 47. Using partial fractions, (s+1)2s(s2 +1) = s+1 + (s+1) 2 + s2 +1 . Multiplying by gives ¡ the denominator ¡ ¢ ¢ 2 s = A (s + 1) s2 + 1 + B s2 + 1 + (Cs + D) (s + 1)

1 Letting s = −1 in this equation gives B = − 2 .

Equating the coefficients:

s3 : 0=A+C s2 : 0 = A + B + 2C + D s: 1 = A + C + 2D 1: 0=A+B+D Substituting A + C = 0 (first equation) into the third equation we get, D = 12 and then the last equation gives A = 0 (since B = − 12 ). Finally, the first equation yields C = 0 (since A = 0). 1 1

1 So (s+1)2s(s2 +1) = − 21 (s+1) 2 + 2 s2 +1 .

h i Thus L−1 (s+1)2s(s2 +1) = − 12 te−t + 12 sin t. 49. Using partial fractions,

s+5 (s+1)(s−1)3

=

A s+1

+

B s−1

+

C (s−1)2

+

D . (s−1)3

91

THE LAPLACE TRANSFORM

Multiplying by the denominator gives 3 2 s + 5 = A (s − 1) + B (s − 1) (s + 1) + C (s − 1) (s + 1) + D (s + 1) Letting s = 1 and −1 in this equation gives D = 3, A = − 21 . Equating the coefficients: s3 : 0 = A + B ⇒ B = −A = 21 2 s : 0 = −3A − B + C ⇒ C = 21 − 32 = −1 s+5 1 1 So (s+1)(s−1)3 = − 21 s+1 + 12 s−1 − (s−11)2 + 3 (s−11)3 . Thus i h ¡ ¢ s+5 = − 12 e−t + 21 et −tet + 32 t2 et = − 21 e−t +et 21 − t + 32 t2 . L−1 (s+1)(s−1) 3 2

−s As+B Cs+D 51. Using partial fractions, (ss2 +4) .

2 = s2 +4 + (s2 +4)2 Multiplying by the¡ denominator gives

¢ s2 −s = (As + B) s2 + 4 +Cs+D = As3 +Bs2 +(4A + C) s+4B+D Equating the coefficients:

s3 : 0 = A

s2 : 1 = B

s: −1 = 4A + C ⇒ C = −1 1: 0 = 4B + D ⇒ D = −4B = −4 2 −s 1 1 s 4 s+4 So (ss2 +4) = 2 s2 +4 − (s2 +4)2 = s2 +4 − (s2 +4)2 − (s2 +4)2 1 1 2·2·s 2· 2 2 4 (s2 +22 )2 − 2 (s2 +22 )2 . ¡ 2 −s = 12 sin 2t − 14 t sin 2t − 21 12 sin 2t − Thus L−1 (ss2 +4) 2 = 14 sin 2t − 41 t sin 2t + 21 t cos 2t. 3 As+B Cs+D −1 . Using partial fractions, (ss2 +9) 2 = s2 +9 + (s2 +9)2

h

53.

=

1 2 2 s2 +22

i

−

¢

t cos 2t

Multiplying by the¡ denominator gives ¢ s3 −1 = (As + B) s2 + 9 +Cs+D = As3 +Bs2 +(9A + C) s+9B+D Equating the coefficients:

s3 : 1 = A

s2 : 0 = B

s: 0 = 9A + C ⇒ C = −9A = −9 1: −1 = 9B + D ⇒ D = −1 3 −1 s s 1 9s+1 9s So (ss2 +9) = s2 +3 − (s2 +3 2 = s2 +9 − 2 − 2 )2 (s2 +9)2 (s2 +32 )2 Thus L−1

h

=

s s2 +32

s 3 −1 (s2 +9)2

i

1 3 2·3·s 2· 3 2 2 (s2 +32 )2 − 18 (s2 +32 )2 . ¡1 1 cos 3t − 32 t sin 3t − 18 3 sin 3t s3 −s2 As+B Cs+D = s2 +36 + (s2 +36)2 . (s2 +36)2

−

=

¢ − t cos 3t .

55. Using partial fractions, Multiplying by the denominator gives ¡ ¢ s3 − s2 = (As + B) s2 + 36 + Cs + D = As3 + Bs2 + (36A + C) s +36B + D Equating the coefficients:

s3 : 1 = A

s2 : −1 = B

s: 0 = 36A + C ⇒ C = −36A = −36 1: 0 = 36B + D ⇒ D = −36B = 36 3 s−1 s 1 36 36s−36 36s −s2 = s2 +6 + (s2 +6 So (ss2 +36) 2 = s2 +36 − 2 − s2 +62 − 2 )2 (s2 +36)2 (s2 +62 )2

92

CHAPTER 3 1 2· 6 2 (s2 +62 )2 . ¡ ¢ 3 −s2 = cos 6t− 16 sin 6t−3t sin 6t+ 12 16 sin 6t − t cos 6t Thus L−1 (ss2 +36) 2 1 = cos 6t − 12 sin 6t − 3t sin 6t − 12 t cos 6t. 3(s+1) (3s+3)−4 4 3s−1 = 2 = 2 − 2 (s2 +2s+26)2 [(s+1)2 +52 ] [(s+1)2 +52 ] [(s+1)2 +52 ] 2·5(s+1) 3 2 2.52 = 10 2 − 25 2. [(s+1)2 +52 ] [(s+1)2 +52 ]

h

57.

s s2 +62

2

Thus L−1

h

=

i

−

i

1 6 6 s2 +62

·6·s − 3 (s22+6 2 )2 +

£

¤

3 2 1 −t −t sin 5t − 25 (sin 5t − 5t cos 5t) 10 te 5e ¡ ¢ 3 2 1 = 10 te−t sin 5t − 25 e−t 5 sin 5t − t cos 5t . s2 As+B Cs+D fractions, (s2 +6s+25) . 2 = s2 +6s+25 + (s2 +6s+25)2 3s−1

(s2 +2s+26)2

=

59. Using partial Multiplying by¡the denominator gives

¢ s2 = (As + B) s2 + 6s + 25 + Cs + D

= As3 + (6A + B) s2 + (25A + 6B + C) s + 25B + D

Equating the coefficients:

s3 : 0 = A

s2 : 1 = 6A + B ⇒ B = 1

s: 0 = 25A + 6B + C ⇒ C = −6 1: 0 = 25B + D ⇒ D = −25 1 s2 6s+25 So (s2 +6s+25) = 2 s2 +6s+25 − (s2 +6s+25)2

6s+18 7

2 − 2 [(s+3)2 +42 ] [(s+3)2 +42 ] 3 2·4·(s+3) 7 4 2· 4 2 1 = 4 (s+3)2 +42 − 4 2 − 32 2. [(s+3)2 +42 ] [(s+3)2 +42 ] i h £ ¤ 7 1 −3t s2 = 14 e−3t sin 4t− 34 te−3t sin 4t− 32 (sin 4t − 4t cos 4t) Thus L−1 (s2 +6s+25) 2 4e ¡ ¢ 7 −3t 1 = 41 e−3t sin 4t− 34 te−3t sin 4t− 32 e 4 sin 4t − t cos 4t . s+5 A D B C 61. Using partial fractions, (s+1)(s−1) + (s−1) 3 = s+1 + s−1 + 3. (s−1)2 Multiplying by the denominator gives 3 2 s + 5 = A (s − 1) + B (s − 1) (s + 1) + C (s − 1) (s + 1) + D (s + 1) Letting s = 1 and −1 in this equation gives D = 3, A = − 12 . Equating the coefficients: s3 : 0 = A + B ⇒ B = −A = 21 2 s : 0 = −3A − B + C ⇒ C = 21 − 32 = −1 s+5 1 1 So (s+1)(s−1)3 = − 12 s+1 + 21 s−1 − (s−11)2 + 3 (s−11)3 . Thus h i ¡ ¢ s+5 L−1 (s+1)(s−1) = − 12 e−t + 21 et −tet + 32 t2 et = − 21 e−t +et 12 − t + 23 t2 . 3 63. Using partial fractions, 12 D E F G

B C = As + s−8 + (s−8) + (s−8) + (s−8) 2 + 4 + 6 .

s(s−8)6 (s−8)3 (s−8)5 Multiplying by the denominator gives

6 5 4 3 12 = A (s − 8) + Bs (s − 8) + Cs (s − 8) + Ds (s − 8) 2 +Es (s − 8) + F s (s − 8) + Gs 3 Letting s = 0 and 8 in this equation gives A = 12 86 , G = 2 .

Equating the coefficients:

s6 : 0 = A + B ⇒ B = −A = − 12

86 5 s : 0 = −48A − 40B + C ⇒ C = 12

85

=

1 (s+3)2 +42

−

93

THE LAPLACE TRANSFORM

0 = 960A + 640B − 32C + D ⇒ D = − 12 84 0 = −10240A − 5120B + 384C − 24D + E ⇒ E = 12 83 0 = 61440A + 20480B − 2048C + 192D − 16E + F ⇒ F = − 12 82 12 1 12 1 12 1 12 1 12 1 1 = 86 s − 12 + 6 8 s−8 85 (s−8)2 − 84 (s−8)3 + 83 (s−8)4 − 82 (s−8)5

s4 : s3 : s2 : So

12 s(s−8)6

1 + 23 (s−8) 6

=

12 1 12 1 12 1! 2! 86 s−8 + 85 (s−8)1+1 − 84 2! (s−8)2+1 1 4! 5! 3 1 − 12 82 4! (s−8)4+1 + 2 5! (s−8)5+1 .

12 1 86 s

−

+

12 1 3! 83 3! (s−8)3+1

Thush i 12 = L−1 s(s−8) 6

65.

12 12 8t 12 8t 86 − 86 e + 85 te 6 2 3 8t 1 4 8t 3 5 8t 2 8t e 4096 t e + 512 t e − 128 t e + 240 t ¢ ¡ − 12 12 12 3 1 3 1 4 1 5 8t 2 = 86 + e − 86 + 85 t − 2048 t + 256 t − 128 t + 80 t . A C B = s+1 + (s+1) Using partial fractions, (s+1)s+5 2 2 + s−1 (s−1)2

+

D . (s−1)2

Multiplying by the denominator gives 2 2 2 2 s+5 = A (s + 1) (s − 1) +B (s − 1) +C (s − 1) (s + 1) +D (s + 1) Letting s = 1 and −1 in this equation gives D = 23 , B = 1. Equating the coefficients:

s3 : 0 = A + C

s2 : 0 = −A + B + C + D

5 = −A + C (since D = 32 , B = 1) ⇒ − 2 s: 1 = −A − 2B − C + 2D 1: 5=A+B−C +D Adding these equations we get, C = − 45 and then A = −C = 45 . 1 5 1 1 3 1 So (s+1)s+5 = 54 s+1 + (s+1) . Thus 2 2 − 4 s−1 + 2 (s−1)2 (s−1)2 i h 5 −t 5 t 3 t s+5 −t −1 = 4 e + te − 4 e + 2 te L (s+1)2 (s−1)2 ¡ ¡ ¢ ¢ = e−t 54 + t + et − 54 + 32 t . 67. Using partial fractions, C D E F

s = As+B s2 +1 + s−7 + (s−7)2 + (s−7)3 + (s−7)4 .

(s2 +1)(s−7)4 Multiplying by the denominator ¡ 2 gives ¡ ¢ ¢ 4 3 2 s = (As + B) (s − 7) + C s ¡ + 1 (s¢ − 7) + D s2 + 1 (s − 7) ¡ 2 ¢ +E s + 1 (s − 7) + F s2 + 1 7 Letting s = 7 in this equation gives F = 50 . Equating the coefficients: s5 : 0=A+C s4 : 0 = −28A + B − 21C + D s3 : 0 = 294A − 28B + 148C − 14D + E s2 : 0 = −1372A − 294B − 364C + 50D − 7E + F s: 1 = 2401A − 1372B + 147C − 14D + E 1: 0 = 2401B − 343C + 49D − 7E + F Note this is a system of 6 equations with 5 unknowns as we know F. So the system considering 5 equations (excluding the fourth one, for example) with 5 unknowns can be solved (by using Maple, for example) to get

94

CHAPTER 3 527 −84 −527 161 A = 1562500 , B = 390625 , C = 1562500 , D = 62500 , E = −12 625 . s 527s−336 1 12 527 1 161 1 1

So (s2 +1)(s−7)4 = 1562500 s2 +1 − 1562500 s−7 + 62500 (s−7) 2 − 625 (s−7)3

7 1

+ 50 (s−7)4

527 s 84 1 527 1 161 1 1562500 s2 +1 − 390625 s2 +1 − 1562500 s−7 + 62500 (s−7)2 2! 3! 7 12 − 625(2!) (s−7)2+1 + 50(3!) (s−7)3+1 . h i s 527 84 527 L−1 (s2 +1)(s−7) = 1562500 cos t − 390625 sin t − 1562500 e7t 4 6 2 7t 7 3 7t 161 7t h + 62500 te − 625 t e + ³ 300 t e 1 527 cos t − 336 sin t + e7t −527 + 4025t − 15000t2 = 1562500

=

Thus

3.3 INITIAL-VALUE PROBLEMS FOR DIFFERENTIAL EQUATIONS 1. Taking the Laplace transform of both sides of y ′′ − 4y = 1 we get, L [y ′′ ] − 4L [y] = L [1] ⇒ s2 Y¡ (s) − ¢sy (0) − y ′ (0) − 4Y (s) = 1s ⇒ s2 Y (s) − 1 − 4Y (s) = 1s ⇒ s2 − 4 Y (s) = 1s + 1 ⇒ Y (s) = s(s1+s 2 −4) . Using partial fractions, 1+s 1 B C Y (s) = s(s2 −4) = s(s−2)(s+2) = As + s−2 + s+2 . Multiplying by the denominator gives

1 + s = A (s + 2) (s − 2) + Bs (s + 2) + Cs (s − 2) . Letting s = 0, 2

and −2 in this equation gives A = − 14 , B = 38 and C = − 18 .

1 1 1 Thus Y (s) = − 14 1s + 83 s− 2 − 8 s+2 . h i h i

£ ¤ 1 1 So L−1 [Y (s)] = − 14 L−1 1s + 83 L−1 s−2 − 81 L−1 s+2

⇒

1 y (t) = − 4 + 38 e2t − 18 e−2t .

3. Taking the Laplace transform of both sides of y ′′ + 3y ′ − 4y = 0

we get, L [y ′′ ] + 3L [y ′ ] − 4L [y] = 0 ⇒ s2 Y (s) − sy (0) − y ′ (0) + 3sY (s) − 3y (0) − 4Y (s) = 0 ⇒

s¡2 Y (s) − s − 0⇒ ¢ 1 + 3sY (s) − 3 − 4Y (s) = s+4 s+4 1 = (s+4)(s−1) = s−1 . s2 + 3s − 4 Y (s) = s + 4 ⇒ Y (s) = s2 +3s−4 h i 1 −1 −1 t So L [Y (s)] = L s−1 ⇒ y (t) = e .

5. Taking the Laplace transform of both sides of y ′′ + 5y ′ + 6y = 1 we get,

L [y ′′ ] + 5L [y ′ ] + 6L [y] = L [1] ⇒ s2 Y (s) − sy (0) − y ′ (0) + 5sY (s) − 5y (0) + 6Y (s) = 1

s ⇒ s2 Y (s) − s − 1 + 5sY (s) − 5 + 6Y (s) = 1s ⇒ ¡ 2 ¢ s2 +6s+1 s2 +6s+1 s + 5s + 6 Y (s) = 1s + s + 6 ⇒ Y (s) = s(s 2 +5s+6) = s(s+2)(s+3) . 2

s +6s+1 B C Using partial fractions, Y (s) = s(s+2)(s+3) = As + s+2 + s+3 .

Multiplying by the denominator gives

s2 + 6s + 1 = A (s + 3) (s + 2) + Bs (s + 3) + Cs (s + 2) .

Letting s = 0, −2 and −3 in this equation gives A = 16 , B = 72 and

1 1 C = − 38 . Thus Y (s) = 16 1s + 27 s+2 − 38 s+3

+

7(15625) 3 t 3

´i

.

95

THE LAPLACE TRANSFORM

h i h i £ ¤ 1 1 So L−1 [Y (s)] = 16 L−1 1s + 72 L−1 s+2 − 83 L−1 s+3

1 + 72 e−2t − 83 e−3t . ⇒ y (t) = 6 7. Taking the Laplace transform of both sides of y ′′ − 3y ′ + 2y = 1

we get, L [y ′′ ] − 3L [y ′ ] + 2L [y] = L [1] ⇒ s2 Y (s) − sy (0) − y ′ (0) − 3sY (s) + 3y (0) + 2Y (s) = 1s ⇒ 2 (s) − 1 − 2Y (s) = 1s ⇒ ¡s Y ¢ 3sY (s) + s+1 1 2 s − 3s + 2 Y (s) = s + 1 ⇒ Y (s) = s(s2s+1 −3s+2) = s(s−2)(s−1) . s+1 B C Using partial fractions, Y (s) = s(s−2)(s−1) = As + s−2 + s−1 . Multiplying by the denominator gives s + 1 = A (s − 1) (s − 2) + Bs (s − 1) + Cs (s − 2) . Letting s = 0, 1 and 2 in this equation gives A = 12 , B = 32 and 1 1 C = −2. Thus Y (s) = 21 1s + 23 s−2 − 2 s−1 h i h i £ ¤ 1 1 − 2L−1 s−1 So L−1 [Y (s)] = 12 L−1 1s + 32 L−1 s−2

⇒ y (t) = 21 + 23 e2t − 2et . 9. Taking the Laplace transform of both sides of y ′′ + 9y = e−t we get, 1 L [y ′′ ] + 9L [y] = L [e−t ] ⇒ s2 Y (s) − sy (0) − y ′ (0) + 9Y (s) = s+1 ¡ ¢ 1 1 ⇒ s2 Y (s) − s − 2 + 9Y (s) = s+1 ⇒ s2 + 9 Y (s) = s+1 +s+2⇒ Y (s) =

s2 +3s+3 (s+1)(s2 +9) . Using partial fractions, s2 +3s+3 A Bs+C (s+1)(s2 +9) = s+1 + s2 +9 .

Y (s) = Multiplying by the gives ¡ denominator ¢ s2 + 3s + 3 = A s2 + 9 + (Bs + C) (s + 1) . Letting s = −1 in this 1 . equation gives A = 10 Equating the coefficients:

9 s2 : 1 = A + B ⇒ B = 1 − A = 10 .

9 s: 3 = B + C ⇒ C = 3 − B = 3 − 10 = 21 10 . 1 1 9 s 21 1 Thus Y (s) = 10 s+1 + 10 s2 +9 + 10 s2 +9 s 21 1 3 1 1 9 = 10 2 +32 + 10 3 s2 +32 . s+1 + 10 sh i i i h h 1 −1 7 −1 3 1 9 −1 s So L−1 [Y (s)] = 10 L s+1 + 10 L s2 +32 + 10 L s2 +32

9 7 1 −t e + 10 cos 3t + 10 sin 3t. ⇒ y (t) = 10 11. Taking the Laplace transform of both sides of y ′′ + 4y ′ + 13y = 2 we get, L [y ′′ ] + 4L [y ′ ] + 13L [y] = L [2] ⇒ s2 Y (s) − sy (0) − y ′ (0) + 4sY (s) − 4y (0) + 13Y (s) = 2s ⇒ s2 Y (s) − s + 4sY (s) − 4 + 13Y (s) = 2s ⇒ ¡ 2 ¢ 2 +4s+2 s + 4s + 13 Y (s) = 2s + s + 4 ⇒ Y (s) = s(ss2 +4s+13) . 2

+4s+2 = As + s2Bs+C Using partial fractions, Y (s) = s(ss2 +4s+13) +4s+13 . Multiplying by the denominator gives ¡ ¢ s2 + 4s + 2 = A s2 + 4s + 13 + (Bs + C) s.

2

Letting s = 0 in this equation gives A = 13 .

Equating the coefficients:

11 .

s2 : 1 = A + B ⇒ B = 1 − A = 13 8 s: 4 = 4A + C ⇒ C = 4 − 4A = 4 − 13 = 44 13 .

96

CHAPTER 3 1 11s+44 2 1 1 11(s+2)+22 = 13 2 s + 13 (s+2)2 +32 i h13 s +4s+13 (s+2) 2 1 22 1 3 = 13 s + 13 11 (s+2)2 +32 + 3 (s+2)2 +32 i £ ¤ 11 −1 h (s+2) i 22 −1 h 2 −1 1 3 So L−1 [Y (s)] = 13 + 39 L L s + 13 L (s+2)2 +32 (s+2)2 +32

Thus Y (s) =

2 1 13 s

+

22 −2t 2 −2t + 11 cos 3t + 39 e sin 3t. ⇒ y (t) = 13 13 e (4) 13. Taking £ (4)the ¤ Laplace transform of both sides of y − y = 1 we get, L y − L [y] = L [1] ⇒ s4 Y (s) − s3 y (0) − s2 y ′ (0) − sy ′′ (0) − y ′′′ (0) − Y (s) = 1s ⇒ s4 Y (s) − 3s3 − 5s2 − Y (s) = 1s ⇒ ¡ 4 ¢ 4 3 +1 3s4 +5s3 +1 s − 1 Y (s) = 1s +3s3 +5s2 ⇒ Y (s) = 3ss(s+5s = s(s+1)(s−1)(s 4 −1) 2 +1) . Using partial fractions, A B C Ds+E

3s4 +5s3 +1 Y (s) = s(s+1)(s−1)(s 2 +1) = s + s+1 + s−1 + (s2 +1) .

Multiplying by the¡ denominator gives ¡ ¡ ¢ ¢ ¢ 3s4 + 5s3 + 1 = A s4 − 1 + Bs (s − 1) s2 + 1 + Cs (s + 1) s2 + 1 + (Ds + E) s (s + 1) (s − 1) . Letting s = 0, −1 and 1 in this equation gives A = −1, B = − 14 and C = 94 . Equating the coefficients: s4 : 3 = A + B + C + D ⇒ D = 3 − A − B − C = 2. s: 0 = −B + C − E ⇒ E = −B + C = 14 + 94 = 25 . 1 1 1 Thus Y (s) = − s − 14 s+1 + 49 s−1 + 2 s2s+1 + 25 s21+1 h i h i £ ¤ 1 1 So L−1 [Y (s)] = −L−1 1s − 14 L−1 s+1 + 94 L−1 s−1 h i h i +2L−1 s2s+1 + 52 L−1 s21+1

⇒ y (t) = −1 − 41 e−t + 49 et + 2 cos t + 25 sin t. 15. Taking the Laplace transform of both sides of y ′ − y = 1 − t we get, L [y ′ ] − L [y] = L [1] − L [t] ⇒ sY (s) − y (0) − Y (s) = 1s − s12 ⇒ s−1 s−1−s2 ⇒ sY (s) + 1 − Y (s) = s−1 s2 ⇒ (s − 1) Y (s) = s2 − 1 = s2 2 2 s−1 s 1 1 s−1−s Y (s) = s2 (s−1) = s2 (s−1) − s2 (s−1) = s2 − s−1 . h i £ ¤ 1 So L−1 [Y (s)] = L−1 s12 − L−1 s−1 ⇒ y (t) = t − et .

17. Taking the Laplace transform of both sides of y ′′ + 2y ′ + y = e−t

we get, L [y ′′ ] + 2L [y ′ ] + L [y] = L [e−t ] ⇒ 1

⇒

s2 Y (s) − sy (0) − y ′ (0) + 2sY (s) − 2y (0) + Y (s) = s+1 1 2 s Y (s) − 1 + 2sY (s) + Y (s) = s+1 ⇒ ¡ 2 ¢ 1 s+2 s + 2s + 1 Y (s) = s+1 + 1 ⇒ Y (s) = (s+1)(ss+2 . 2 +2s+1) = (s+1)3 s+2 A Using partial fractions, Y (s) = (s+1) 3 = s+1 + Multiplying by the denominator gives 2 s + 2 = A (s + 1) + B (s + 1) + C.

Letting s = −1 in this equation gives C = 1.

Equating the coefficients:

s2 : 0=A s: 1 = 2A + B ⇒ B = 1.

B (s+1)2

+

C . (s+1)3

97

THE LAPLACE TRANSFORM 1 1! 2! = +1 (s+1)3 i (s+1)1+1h 2 (s+1)i2+1 1! 2! + 12 L−1 (s+1) (s)] = L−1 (s+1) 1+1 2+1 1 2 −t −t te + 2 t e . ′

Thus Y (s) = So L−1 [Y

1 (s+1)2

+ h

⇒ y (t) = 19. Taking the Laplace transform of both sides of y + 2y = e−t cos t

we get, L [y ′ ] + 2L [y] = L [e−t cos t] ⇒

s+1 sY (s) − y (0) + 2Y (s) = (s+1) ⇒

2 +1

s+1 s+1

⇒ (s + 2) Y (s) = s2 +2s+2 ⇒ sY (s) + 2Y (s) = s2 +2s+2 s+1 Y (s) = (s+2)(s2 +2s+2) . Using partial fractions,

A Bs+C

Y (s) = (s+2)(ss+1 2 +2s+2) = s+2 + s2 +2s+2 .

Multiplying gives

¡ by the denominator ¢ s + 1 = A s2 + 2s + 2 + (Bs + C) (s + 2) .

Letting s = −2 in this equation gives A = − 21 .

Equating the coefficients:

s2 : 0 = A + B ⇒ B = −A = 21 s: 1 = 2A + C + 2B ⇒ C = 1 − 2A − 2B = 1. 1 s+2 Thus Y (s) = − 12 s+2 + 21 s2 +2s+2 i h 1 s+1 1 = − 21 s+2 + + 21 (s+1) 2 2 h i +1 (s+1) h +1 i h i 1 −1 1 −1 1 1 −1 s+1 1 −1 So L [Y (s)] = − 2 L + L 2 2 2 2 s+2 + 2 L 2 (s+1) +1 (s+1) +1

⇒ y (t) = − 21 e−2t + 21 e−t cos t + 12 e−t sin t. 21. Taking the Laplace transform of both sides of y ′′ + y = sin t we get, L [y ′′ ] + L [y] = L [sin t] ⇒ s2 Y (s) − sy (0) − y ′ (0) + Y (s) = s21+1 ⇒ ¡ ¢ s2 Y (s) + Y (s) = s21+1 ⇒ s2 + 1 Y (s) = s21+1 ⇒ Y (s) =

So L−1 [Y

1 1 = 21 (s22·1 2 )2 . (s2 +1)2 +1 i h 1 −1 2·11 (s)] = 2 L (s2 +12 )2

⇒ y (t) =

1 2

[sin t − t cos t] .

23. Taking the Laplace transform of both sides of y ′′ + 9y = cos 3t

we get, L [y ′′ ] + 9L [y] = L [cos 3t]

⇒ s2 Y (s) − sy (0) − y ′ (0) + 9Y (s) = s2s+9 ⇒ ¡ ¢ 2 s2 Y (s) − 1 + 9Y (s) = s2s+9 ⇒ s2 + 9 Y (s) = s2s+9 + 1 = s s+s+9 2 +9 s2 +s+9 . Using partial fractions, (s2 +9)2 s2 +s+9 Cs+D

= As+B s2 +9 + (s2 +9)2 .

(s2 +9)2

⇒ Y (s) =

Y (s) = Multiplying by the denominator ¡ ¢ gives s2 + s + 9 = (As + B) s2 + 9 + Cs + D. Equating the coefficients:

s3 : 0 = A

s2 : 1 = B

s: 1=A+C ⇒C =1 1: 9 = 9B + D ⇒ D = 0. 1 s 3 1 2· 3· s Thus Y (s) = s21+9 + (s2 +9) 2 = 3 s2 +32 + 6 2 2 )2 .

i h h (s +3 i

3 So L−1 [Y (s)] = 31 L−1 s2 +3 + 61 L−1 (s22·3·s 2 +32 )2

98

CHAPTER 3

⇒ y (t) = 13 sin 3t + 16 t sin 3t. 25. Taking the Laplace transform of both sides of y ′′ + 4y = cos 2t we get, L [y ′′ ] + 4L [y] = L [cos 2t] ⇒ s2 Y (s) − sy (0) − y ′ (0) + 4Y (s) = s2s+4 ⇒ ¡ ¢ s2 Y (s) + 4Y (s) = s2s+4 ⇒ s2 + 4 Y (s) = s2s+4 1 2·2·s s ⇒ Y (s) = (s2 +4) 2 = 4 2 2 2. (s h +2 ) i 1 −1 2·2·s −1 So L [Y (s)] = 4 L ⇒ y (t) = 14 t sin 2t. (s2 +22 )2

27. Taking the Laplace transform of both sides of y ′′ + y = sin t + cos t we get, L [y ′′ ] + L [y] = L [sin t] + L [cos t] ⇒ s2 Y (s) − sy (0) − y ′ (0) + Y (s) = s21+1 + s2s+1 ⇒ ¡ ¢ s2 Y (s) + Y (s) = s21+1 + s2s+1 ⇒ s2 + 1 Y (s) = s21+1 + s2s+1 1 s 1 2·12 + (s2 +1) + 1 2·1·s 2 .

2 = 2 (s2 +1)2 (s2 +12 )2 h 2 (s2 +12 )i

h i 2 L−1 [Y (s)] = 21 L−1 (s22·1 + 12 L−1 (s22·1·s +12 )2 +12 )2 1 1 y (t) = 2 (sin t − t cos t) + 2 t sin t = − 21 t cos t + 12 sin t + 21 t sin t. R∞ −st ′ ′

⇒ Y (s) = So

⇒

29. L [f (t)] = −st

e

f (t) dt. We use integration-by-parts taking

0

⇒ du = −se−st dt and dv = f ′ (t) dt ⇒ v = f (t) . R∞ R∞ So L [f ′ (t)] = e−st f ′ (t) dt = uv|∞ vdu 0 −

u=e

0

−st

=e

f

(t) |∞ 0

+s

R∞

0

−st

e

f (t) dt

0

= lim [e−st f (t)] − lim [e−st f (t)] + sF (s) = sF (s) − f (0+ ) . t→∞

t→0+

3.4 DISCONTINUOUS FORCING FUNCTIONS   0; 0 ≤ t < 2, 3; 2 ≤ t < 5, 1. f (t) =  t; 5 ≤ t. Thus 3 [H (t − 2) − H (t − 5)] + t [H (t − 5)]

= 3H (t − 2) + (t − 3) H (t − 5) .

99

THE LAPLACE TRANSFORM

8 7 6 5 4 3 2 1 1

2

3

4

5

6

7

8

t

   

sin t; 0 ≤ t < π, 0; π ≤ t < 2π, 3. f (t) = sin t; 2π ≤ t < 3π,    0; 3π ≤ t. Thus sin t [H (t) − H (t − π)] + sin t [H (t − 2π) − H (t − 3π)] = sin t [1 − H (t − π)] + sin t [H (t − 2π) − H (t − 3π)] . y

1 t π

2π

3π

4π

 1; 0 ≤ t < 1,    0; 1 ≤ t < 2, 5. f (t) = 1; 2 ≤ t < 3,    0, t ≥ 3 The unit-step function is H (t) − H (t − 1) + H (t − 2) − H (t − 3) .  0, 0 ≤ t < 1,

     t − 1; 1 ≤ t < 2,

1; 2 ≤ t < 3,

7. f (t) =   − t; 3 ≤ t < 4,

4    0, t ≥ 4 The unit-step function is

(t − 1) [H (t − 1) − H (t − 2)]+1 [H (t − 2) − H (t − 3)]+(4 − t) [H (t − 3) − H (t − 4)] .

= (t − 1) H (t − 1)+(2 − t) H (t − 2)+(3 − t) H (t − 3)+(t − 4) H (t − 4) .

100

CHAPTER 3

   

0, 0 ≤ t < 1, 2 − t; 1 ≤ t < 3, 9. f (t) = t − 4; 3 ≤ t < 5,    0, t ≥ 5 The unit-step function is (2 − t) [H (t − 1) − H (t − 3)] + (t − 4) [H (t − 3) − H (t − 5)] . = (2 − t) H (t − 1) + 2 (t − 3) H (t − 3) − (t − 4) H (t − 5) . 11. y (t) = tH (t − 2) . Here c = 2 and f (t − 2) = t. Let τ = t − 2.

Then t = τ + 2 and f (τ ) = τ + 2 ⇒ f (t) = t + 2. Thus

F (s) = L [f (t)] = s 1 2 + 2s . So Y (s) = L [y (t)] = L [tH (t − 2)] = L [f (t − 2) H (t − 2)] . Then (T¡ 15) gives H ¡(t − 2)]¢= e−2s F (s) ¢ that L [f (t − 2) 2 2 −2s 1 −2s 1 = e ¡ s2 + ¢s . Thus Y (s) = e s2 + s . 13. y (t) = t3 + 1 H (t − 1) . Here c = 1 and f (t − 1) = t3 + 1.

3

Let τ = t − 1.Then t = τ + 1 and f (τ ) = (τ + 1) + 1, so that 3 3 2 f (t) = (t + 1) + 1 = t + 3t + 3t + 2. Thus 6 3 + 2s . F (s) = L [f (t)] = s64 + £¡ s3 + s2 ¢ ¤ 3 So Y (s) = L [y (t)] = L t + 1 H (t − 1) = L [f (t − 1) H (t − 1)] . −s Then (T that L¢[f (t − 1) H (t − 1)] = ¡ 15) gives ¡ e F6 (s) 3 ¢ 6 3 2 2 −s 6 −s 6 =e s4 + s3 + s2 + s . Thus Y (s) = e s4 + s3 + s2 + s . 15. y (t) = sin tH (t − π) . Here c = π and f (t − π) = sin t.

Let τ = t − π.Then t = τ + π and f (τ ) = sin (τ + π) = − sin τ.

⇒ f (t) = − sin t. Thus F (s) = L [f (t)] = −L [sin t] = − s21+1 . So Y (s) = L [y (t)] = L [sin tH (t − π)] = L [f (t − π) H (t − π)] . Then (T³15) gives = e−πs F (s) ´ that L [f (t − π) H (t ³ − π)] ´ = e−πs − s21+1 . Thus Y (s) = e−πs − s21+1 . 17. y (t) = cos tH (t − 2π) . Here c = 2π and f (t − 2π) = cos t. Let τ = t − 2π.Then t = τ + 2π and f (τ ) = cos (τ + 2π) = cos τ. ⇒ f (t) = cos t. Thus F (s) = L [f (t)] = L [cos t] = s2s+1 . So Y (s) = L [y (t)] = L [cos tH (t − 2π)] = L [f (t − 2π) H (t − 2π)] . −2πs Then (T ³ 15) gives F (s) ´ that L [f (t − 2π) H³(t − 2π)] ´ =e s s −2πs −2πs . Thus Y (s) = e . =e s2 +1 s2 +1 19. y (t) = e2t H (t − 3) . Here c = 3 and f (t − 3) = e2t .

Let τ = t − 3.Then t = τ + 3 and f (τ ) = e2τ +6 = e6 e2τ .

£ ¤ 6 ⇒ f (t) = e6 e2t . Thus F (s) = L [f (t)] = e6 L e2t = se−2 . £ 2t ¤ So Y (s) = L [y (t)] = L e H (t − 3) = L [f (t − 3) H (t − 3)] . −3s Then (T³15) ´ gives that L [f (t − 3) ³ H (t − ´ 3)] = e F (s) 6 6 = e−3s se−2 . Thus Y (s) = e−3s se−2 .

21. y (t) = te5t H (t − 2) . Here c = 2 and f (t − 2) = te5t .

Let τ = t − 2.Then t = τ + 2 and f (τ ) = (τ + 2) e5τ +10

5t = 2e10 e5τ + e10 τ e5τ ⇒ f (t) = 2e10 e5t + e10 te ³ .

´ 1 2 e10 2e10 . Thus F (s) = L [f (t)] = s−5 + (s−5)2 = e10 (s−5) 2 + s−5 £ 5t ¤ So Y (s) = L [y (t)] = L te H (t − 2) = L [f (t − 2) H (t − 2)] .

101

THE LAPLACE TRANSFORM −2s Then (T 15) ³ gives that L´[f (t − 2) H (t − 2)] = e ³ F (s) 1 2 1 . Thus Y (s) = e−2s e10 (s−5) = e−2s e10 (s−5) 2 + s−5 2 + 1 s2 +s

1 s(s+1)

A s

B s+1 .

2 s−5

´

.

= = + Multiplying 23. Using partial fractions by the denominator we have 1 = A (s + 1) + Bs. Letting s = 0, 1 and s = −1 gives A = 1 and B = −1. So F (s) = s21+s = 1s − s+1 ⇒ f (t) = L−1 [F (s)] = 1 − e−t . Now e−2s s21+s and e−3s s21+s are of the form e−cs F (s) with c = 2 and c = 3, respectively.

Thenh using (T 15) we have i

L−1 e−2s s21+s + e−3s s21+s = f (t − 2) H (t − 2) + f (t − 3) H (t − 3)

£ ¤ £ ¤ = 1 − e−(t−2) H (t − 2) + 1 − e−(t−3) H (t − 3) .

25. F (s) = s2 +21s+2 = (s+1)12 +12 ⇒ f (t) = L−1 [F (s)] = e−t sin t. Now e−3s s2 +21s+2 is of the form e−cs F (s) with c = 3. Then usingh (T 15) we i have e−3s −1 −(t−3) L sin (t − 3) H (t − 3) . s2 +2s+2 = f (t − 3) H (t − 3) = e −s

27. The first term, se2 +1 , is of the form e−cs F1 (s) with c = 1

where F1 (s) = s21+1 ⇒ f1 (t) = L−1 [F1 (s)] = sin t.

h −s i Then using (T 15) we have L−1 se2 +1 = L−1 [e−s F1 (s)] = f1 (t − 1) H (t − 1) = sin (t − 1) H (t − 1) . −2s The second term, se2 +4 , is of the form e−cs F2 (s) with c = 2 2 −1 [F2 (s)] = 21 sin 2t. where F2 (s) = s21+4 = 12 s2 +2 2 ⇒ f2 (t) = L h −2s i £ ¤ Then using (T 15) we have L−1 se2 +1 = L−1 e−2s F2 (s) = f2 (t − 2)hH (t − 2) = 21i sin (2t − 4) H (t − 2) .

−s −2s Hence L−1 se2 +1 − se2 +1

= sin (t − 1) H (t − 1) − 21 sin (2t − 4) H (t − 2) . 29. F1 (s) = 4s ⇒ f1 (t) = L−1 [F1 (s)] = 4 and s −1 F2 (s) = s26s [F2 (s)] = 6 cos 3t. +9 = 6 s2 +32 ⇒ f2 (t) = L For both terms, c = 1. Then using (T 15) we have

³ ´i h 6 −1 −s 4 = f1 (t − 1) H (t − 1) + f2 (t − 1) H (t − 1)

L e s + s2 +9

= [4 + 6 cos (3t − 3)] H (t − 1) . 31. g (t) = H (t − 2) − H (t − 5) and the differential eqution is y ′′ + y = H (t − 2) − H (t − 5) , y (0) = y ′ (0) = 0. Taking the Laplace transform of both sides gives −2s −5s s2 Y (s) − sy (0) − y ′ (0) + Y (s) = e s − e s (since, for example, H (t − 2) = f (t − 2) H (t − 2) with f (t − 2) = 1 ⇒ f (t) = 1 ⇒

F (s) = 1s and using (T 15) with c = 2, L [f (t − 2) H (t − 2)] = −2s e−2s F (s) = e s ). ¡ 2 ¢ −2s −5s −2s −5s ⇒ s + 1 Y (s) = e s − e s ⇒ Y (s) = s(se 2 +1) − s(se 2 +1) . Using partial fractions s(s21+1) = As + Bs+C . Multiplying by 2 ¡ ¢ s +1 denominator we have 1 = A s2 + 1 + (Bs + C) s. Letting s = 0

102

CHAPTER 3

gives A = 1. Equating the coefficients:

s2 : 0 = A + B ⇒ B = −A = −1

s: 0=C So F (s) = s(s21+1) = 1s − s2s+1 ⇒ f (t) = L−1 [F (s)] = 1 − cos t. −2s

−5s

Now s(se 2 +1) and s(se 2 +1) are of the form e−cs F (s) with c = 2 and c = 5, respectively.

h −2s i Thus L−1 s(se 2 +1) = f (t − 2) H (t − 2) = [1 − cos (t − 2)] H (t − 2)

h −5s i and L−1 s(se 2 +1) = f (t − 5) H (t − 5) = [1 − cos (t − 5)] H (t − 5) .

So y (t)= L−1 [Y (s)] = [1 − cos (t − 2)] H (t − 2)−[1 − cos (t − 5)] H (t − 5) .  t; 0 ≤ t < 1, 2 − t; 1 ≤ t < 2,

33. g (t) =  0; 2 ≤ t.

⇒ g (t) = t [H (t) − H (t − 1)] + (2 − t) [H (t − 1) − H (t − 2)] = t + (2 − 2t) H (t − 1) − (2 − t) H (t − 2)

and the differential eqution is

y ′ − 3y = t + (2 − 2t) H (t − 1) − (2 − t) H (t − 2) , y (0) = 1.

For the second forcing term, (2 − 2t) H (t − 1) , c = 1 and

f (t − 1) = 2 − 2t. Let τ = t − 1.Then t = τ + 1 and

f (τ ) = 2 − 2τ + 2 = 4 − 2τ ⇒ f (t) = 4 − 2t.

Thus F (s) = L [4 − 2t] = 4s − s2 2 .

£ ¤ So L [(2 − 2t) H (t − 1)] = L f (t − 1) H (t − 1) = e−cs F (s)

−s −s = 4es − 2es2 .

For the third forcing term, (2 − t) H (t − 2) , c = 2 and

fe(t − 2) = 2 − t. Let τ = t − 2.Then t = τ + 2 and

1 e e f (τ ) = 2 − τ − 2 = −τ ⇒ fe h (t) = −t. Thus F i(s) = L [−t] = − s2 .−2s So L [(2 − t) H (t − 2)] = L fe (t − 2) H (t − 2) = e−cs Fe (s) = − e 2 . s

Taking the Laplace transform of both sides differential eqution gives −s −s −2s sY (s) − y (0) − 3Y (s) = s12 + 4es − 2es2 + e s2 ⇒ −s −2s −s sY (s) − 1 − 3Y (s) = 4es − 2es2 + s12 + e s2 ⇒ −s −s −2s (s − 3) Y (s) = 4es − 2es2 + s12 + e s2 + 1 −s −s −2s 4e 1 1 ⇒ Y (s) = s(s−3) − 2 s2e(s−3) + s2 (s−3) + s2e(s−3) + (s−3) −s

1 e 1 ⇒ Y (s) = (4s−2)e s2 (s−3) + s2 (s−3) + s2 (s−3) + (s−3) To find the inverse Laplace transform of the first term, we use A B C partial fractions by letting F1 (s) = s24s−2 (s−3) = s + s2 + s−3 . Multiplying by denominator we have 4s − 2 = As (s − 3) + B (s − 3) + Cs2 . Letting s = 0 and 3 gives 2 B = 32 and C = 10 9 . Equating the coefficient of s we get 10 0 = A + C ⇒ A = −C = − 9 . 10 1 2 1 10 1

So F1 (s) = s24s−2 (s−3) = − 9 s + 3 s2 + 9 s−3

2 10 3t ⇒ f1 (t) = L−1 [F1 (s)] = − 9 + 3 t + 10 9 e .

The first term,

(4s−2)e−s s2 (s−3) ,

−2s

is of the form e−cs F1 (s) with c = 1.

103

THE LAPLACE TRANSFORM

h

−s

i

Then using (T 15) , L−1 (4s−2)e = L−1 [e−s F1 (s)] = f1 (t − 1) H (t − 1) s2 (s−3) ¤ £ 10 2 3(t−1) H (t − 1) . = − 9 + 3 (t − 1) + 10 9 e To find the inverse Laplace transform of the second term, we 1 E F use partial fractions by letting F2 (s) = s2 (s−3) =D s + s2 + s−3 . Multiplying by denominator we have 1 = Ds (s − 3) + E (s − 3) + F s2 . Letting s = 0 and 3 gives E = − 31 and F = 91 . Then equating the coefficient of s2 we have 0 = D + F ⇒ D = −F = − 91 . 1 1 So F2 (s) = s2 (s−3) = − 91 1s − 31 s12 + 91 s−3 1 1 3t 1 −1 ⇒ f2 (t) = L [F2 (s)] = − 9 − 3 t + 9 e . −2s The third term, s2e(s−3) , is of the form e−cs F2 (s) with c = 2.

h −2s i £ ¤ Then using (T 15) , L−1 s2e(s−3) = L−1 e−2s F2 (s) = f2 (t − 2) H (t − 2) ¤ £ = − 19 − 31 (t − 2) + 19 e3(t−2) H (t − 2) . i h 1 = e3t . The inverse Laplace transform of the last term is L−1 s−3

£ 1 1 ¤ 1 3t 3t Thus y (t) = e + − 9 − 3 t + 9 e ¤ £ 3(t−1) + 2 (t − 1) + 10 H (t − 1)

+ − 10 9 e £ 19 13 ¤ + − 9 − 3 (t − 2) + 91 e3(t−2) H (t − 2) . 35. g (t) = H (t) − H (t − 3) and the differential eqution is

y ′′ + 2y ′ + 10y = H (t) − H (t − 3) , y (0) = 1, y ′ (0) = 0.

Taking the Laplace transform of both sides gives

−3s

s2 Y (s) − sy (0) − y ′ (0) + 2sY (s) − 2y (0) + 10Y (s) = 1s − e s (since, for example, H (t − 3) = f (t − 3) H (t − 3) with f (t − 3) = 1 ⇒ f (t) = 1 ⇒ F (s) = 1s and using (T 15) with c = 3, −3s

L [f (t − 3) H (t − 3)] = e−3s F (s) = e s ).

−3s ⇒ s2 Y (s) − s + 2sY (s) − 2 + 10Y (s) = 1s − e s ⇒

¡ 2 ¢ −3s s + 2s + 10 Y (s) = 1s − e s + s + 2 1 e−3s s+2 ⇒ Y (s) = s(s2 +2s+10) − s(s2 +2s+10) + (s2 +2s+10) . First, we will find the inverse Laplace transform of the first term. 1 = As + s2Bs+C . Multiplying by Using partial fractions s(s2 +2s+10) ¡ 2 ¢ +2s+10 denominator we have 1 = A s + 2s + 10 + (Bs + C) s. Letting 1 . Equating the coefficients: s = 0 gives A = 10 1 2 s : 0 = A + B ⇒ B = −A = − 10 2 s: 0 = 2A + C ⇒ C = −2A = − 10 1 1 1 1 s+2

So F (s) = s(s2 +2s+10) = 10 s − 10 s2 +2s+10

=

1 1 10 s −1

(s+1) 1 3

1 10 (s+1)2 +32 − 30 (s+1)2 +32 1 1 −t 1 −t (s)] = 10 − 10 e cos 3t − 30 e −cs

−

⇒ f (t) = L [F sin 3t. The second term is ofh the form e i F (s) with c = 3. e−3s = f (t − 3) H (t − 3) So using (T 15) , L−1 s(s2 +2s+10)

£1 ¤ 1 −(t−3) 1 −(t−3) = 10 − 10 e cos 3 (t − 3) − 30 e sin 3 (t − 3) H (t − 3) . For the third term:

104

CHAPTER 3

h

i

h

s+2 s+1 −1 (s2 +2s+10) = L (s+1)2 +32 e−t cos 3t + 13 e−t sin 3t.

L−1

i

+ 13 L−1

h

3 (s+1)2 +32

i

= So using all three inverse Laplace transform we have

1 1 −t 1 −t y (t) = L−1 [Y (s)] = 10 − 10 e cos 3t − 30 e sin 3t

£1 ¤ 1 −(t−3) 1 −(t−3) − 10 − 10 e cos 3 (t − 3) − 30 e sin 3 (t − 3) H (t − 3) +e−t cos 3t + 31 e−t sin 3t 1 9 −t 3 −t = 10 + 10 e cos 3t + 10

e sin 3t £1 ¤ 1 −(t−3) 1 −(t−3) − 10 − 10 e cos 3 (t − 3) − 30 e sin 3 (t − 3) H (t − 3) . 37. g (t) = 1−H (t − 1)+H (t − 2)−H (t − 3) and the differential eqution

is y ′′ +4y = 1−H (t − 1)+H (t − 2)−H (t − 3) , y (0) = 0, y ′ (0) = 0.

Taking the Laplace transform of both sides gives

−s −2s −3s

s2 Y (s) − sy (0) − y ′ (0) + 4Y (s) = 1s − e s + e s − e s ⇒ ¡ 2 ¢ −s −2s −3s s + 4 Y (s) = 1s − e s + e s − e s ⇒ −s −2s −3s

Y (s) = s(s21+4) − s(se2 +4) + s(se 2 +4) − s(se 2 +4) .

Using partial fractions F (s) = s(s21+4) = As + Bs+C s2 +4 . Multiplying by ¡ 2 ¢ denominator we have 1 = A s + 4 + (Bs + C) s. Letting

s = 0 gives A = 14 . Equating the coefficients:

s2 : 0 = A + B ⇒ B = −A = − 1

4 s: 0=C s

So F (s) = s(s21+4) = 14 1s − 41 s2s+4 = 14 1s − 41 s2 +2 2

1 1 −1 ⇒ f (t) = L [F (s)] = 4 − 4 cos 2t. The second, third and fourth terms are of the form e−cs F (s) with and i3, respectively. So using (T 15) , we have c = 1, h 2 −s e 1 1 −1 L 2 +4) = f (t − 1) H (t − 1) = 4 − 4 cos 2 (t − 1) H (t − 1) i h s(s−2s L−1 s(se 2 +4) = f (t − 2) H (t − 2) = 14 − 14 cos 2 (t − 2) H (t − 2) h −3s i L−1 s(se 2 +4) = f (t − 3) H (t − 3) = 14 − 14 cos 2 (t − 3) H (t − 3) .

Thus y (t) = L−1 [Y (s)] = 41 − 14 cos 2t − 41 + 41 cos 2 (t − 1) H (t − 1) 1 + 14 − 4 cos 2 (t − 2) H (t − 2) − 14 + 14 cos 2 (t − 3) H (t − 3) ⇒ 1 y (t) = 4 [− cos 2t + cos 2 (t − 1) H (t − 1) − cos 2 (t − 2) H (t − 2) + cos 2 (t − 3) H (t − 3)]. 39. g (t) = H (t) − H (t − 2) + e−t H (t − 2) and the differential eqution is

y ′′ + 9y = H (t) − H (t − 2) + e−t H (t − 2) , y (0) = 0, y ′ (0) = 0.

For H (t − 2) = f (t − 2) H (t − 2) with f (t − 2) = 1 ⇒ f (t) = 1 ⇒

F (s) = 1s and using (T 15) with c = 2, L [f (t − 2) H (t − 2)] = −2s e−2s F (s) = e s .

For the last forcing term, e−t H (t − 2) , c = 2 and f (t − 2) = e−t .

Let τ = t − 2.Then t = τ + 2 and f (τ ) = e−(τ +2) = e12 e−τ

£ ¤ 1 ⇒ f (t) = e1 2 e−t . Thus F (s) = L e12 e−t = e12 s+1 . £ ¤ −2s −cs −t So L [e H (t − 2)] = L f (t − 2) H (t − 2) = e F (s) = e12 es+1 . Then the Laplace transform of both sides of differential eqution gives −2s −2s s2 Y (s) − sy (0) − y ′ (0) + 9Y (s) = 1s − e s + e12 es+1 ⇒

105

THE LAPLACE TRANSFORM −2s

−2s

s2 Y (s) + 9Y (s) = 1s − e s + e12 es+1 ⇒ ¡ 2 ¢ −2s −2s s + 9 Y (s) = 1s − e s + e12 es+1 −2s

−2s

e ⇒ Y (s) = s(s21+9) − s(se 2 +9) + e12 (s+1)(s 2 +9) .

First, we will find the inverse Laplace transform of the first term.

Using partial fractions let F1 (s) = s(s21+9) = As + Bs+C s¢2 +9 . ¡ 2 Multiplying by denominator we have 1 = A s + 9 + (Bs + C) s. Letting s = 0 gives A = 91 . Equating the coefficients: s2 : 0 = A + B ⇒ B = −A = − 91 s: 0=C s

So F1 (s) = s(s21+9) = 91 1s − 91 s2s+9 = 19 1s − 91 s2 +3 2

1 1 −1 ⇒ f1 (t) = L [F1 (s)] = 9 − 9 cos 3t. −cs The second h term i is of the form e F1 (s) with c = 2. −2s

So L−1 s(se 2 +9) = f1 (t − 2) H (t − 2) ¤ £ = 19 − 1

9 cos 3 (t − 2) H (t − 2) . For the third term, using partial fractions let 1 A F3 (s) = (s+1)(s = s+1 + Bs+C 2 s2 +9 . Multiplying by denominator ¡ +9) ¢ 2 we have 1 = A s + 9 + (Bs + C) (s + 1) . Letting s = −1 1 gives A = 10 . Equating the coefficients

1

2 s : 0 = A + B ⇒ B = −A = − 10 1 s: 0 = B + C ⇒ C = −B = 10 .

1 1 1 1 s 1 1

F3 (s) = (s+1)(s 2 +9) = 10 s+1 − 10 s2 +9 + 10 s2 +9

s 1 3

1 1 1 = 10 s+1 − 10 s2 +32 + 30 s2 +32

1 1 1 −t e − 10 cos 3t + 30 sin 3t ⇒ f3 (t) = L−1 [F3 (s)] = 10 −cs The third term is of the form e F (s) with c = 2. 3 h i e−2s 1 So L−1 e12 (s+1)(s f (t − 2) H (t − 2) = 2 e2 3 £ 1 −(t−2) +9) ¤ 1 1 e − 10 cos 3 (t − 2) + 30 sin 3 (t − 2) H (t −¤2) = e12 10 £1 1 1 1 −1 So y (t) = L [Y (s)] = 9 − 9 cos 3t − 9 − 9 cos 3 (t − 2) H (t − 2) £ 1 −(t−2) ¤ 1 1 + e12 10 e − 10 cos 3 (t − 2) + 30 sin 3 (t − 2) H (t − 2) . 41. g (t) = e3t [H (t) − H (t − 4)] = e3t − e3t H (t − 4) and the differential equation is y ′ − 5y = e3t − e3t H (t − 4) , y (0) = 0. For the last forcing term, e3t H (t − 4) , c = 4 and f (t − 4) = e3t . 3(τ +4) 12 3τ Let τ = t − 4.Then t = τ + 4 and£f (τ ) = ¤ e 12 1= e e 12 3t 12 3t ⇒ f (t) = e e . Thus F (s) = L e e = e s−3 . Then using £ £ ¤ ¤ −4s (T 15) , L e3t H (t − 4) = L f (t − 4) H (t − 4) = e−4s F (s) = e12 es−3 . Then Laplace transform of both sides of differential equation gives −4s 1 − e12 es−3 sY (s) − y (0) − 5Y (s) = s−3 ⇒ (s − 5) Y (s) =

1 s−3

−4s

− e12 es−3

−4s

1 e ⇒ Y (s) = (s−3)(s−5) − e12 (s−3)(s−5) . 1 A B Using partial fractions let F (s) = (s−3)(s−5) = (s−3) + (s−5) . Multiplying by denominator we have 1 = A (s − 5) + B (s − 3) . Letting s = 3 and 5 gives A = − 12 and B = 21 .

106

CHAPTER 3 1 1 2 (s−5) e−4s ⇒ f (t) = L [F (s)] = − 21 e3t + 12 e5t . Now (s−3)(s−5) is of the h i −4s e form e−cs F (s) with c = 4 and using (T 15) , L−1 (s−3)(s−5) = ¤ £ £ ¤ L−1 e−4s F (s) = f (t − 4) H (t − 4) = − 12 e3(t−4) + 12 e5(t−4) H −1

So F (s) =

1 (s−3)(s−5) −1

1 = − 21 (s−3) +

(t − 4) . Hence y (t) = L [Y (s)] £ ¤ = − 21 e3t ¡+ 12 e5t¢− e12 − 21 e3(t−4) + 12 e5(t−4) H (t − 4) . 43. g (t) = cos tH t −¡π2 and ¢ the differential equation is y ′ + 3y = cos tH t − π2 , y (0) = 0. ¡ ¡ ¢ ¢ For the forcing term, cos tH t − π2 , c = π2 and f t − π2 = cos t. ¡ ¢ π Let τ = t − π2 .Then t = τ + π 2 and f (τ ) = cos τ + 2 = − sin τ ⇒ f (t) = − sin t. Thus F (s) = L [− sin t] = − s21+1 . Then using £ ¡ ¢¤ £ ¡ ¢ ¡ ¢¤ π (T 15) , L cos tH t − π2 = L f t − π2 H t − π2 = e− 2 s F (s) − π s

2 . = − es2 +1 Then Laplace transform of both sides of differential equation gives − π s −πs 2 2 sY (s) − y (0) + 3Y (s) = − es2 +1 ⇒ (s + 3) Y (s) = − es2 +1 −πs

e 2 ⇒ Y (s) = − (s+3)(s 2 +1) . 1 A Bs+C Using partial fractions let F (s) = (s+3)(s 2 +1) = s+3 + s2 +1 . ¡ 2 ¢ Multiplying by denominator we have 1 = A s + 1 +(Bs + C) (s + 3) . 1 Letting s = −3 gives A = 10 . Equating the coefficients 1 2 s : 0 = A + B ⇒ B = −A = − 10 3 s: 0 = 3B + C ⇒ C = −3B = 10 .

1 1 1 s 3 1 So F (s) = (s+3)(s2 +1) = 10 s+3 − 10 s2 +1 + 10 s21+1

1 −3t 1 3 ⇒ f (t) = L−1 [F (s)] = 10 e − 10 cos t + 10 sin t. −πs

e 2 −cs F (s) with c = π2 and using (T 15) , Now (s+3)(s 2 +1) is of the form e h i π ¤ ¡ ¢ ¡ ¢ £ −πs e− 2 s −1 L−1 (s+3)(s e 2 F (s) = f t − π2 H t − π2 2 +1) = L h ¡ ¢ ¡ ¢i ¡ ¢ 3 1 −3(t− π π 2 ) − 1 cos t − π = 10 e H t − π2 10 2 + 10 sin t − 2 h i ¡ ¢ 1 −3(t− π 2 ) − 1 sin t − 3 cos t H t − π = 10 e 10 10 2 . −1 Hence y (t) h = L [Yπ (s)]

1 −3(t− 2 ) 1 e sin t − = − 10 − 10

3 10

i ¡ ¢ cos t H t − π2 .

45. g (t) = sin tH (t − 3) and the differential equation is y ′ + 2y = sin tH (t − 3) , y (0) = 0. For the forcing term, sin tH (t − 3) , c = 3 and f (t − 3) = sin t. Let τ = t − 3.Then t = τ + 3 and f (τ ) = sin (τ + 3) ⇒ f (t) = sin (t + 3) = cos 3 sin t + sin 3 cos t. Thus

F (s) = L [sin (t + 3)] = cos 3 s21+1 + sin 3 s2s+1 . Then using (T 15) ,

£ ¤ L [sin tH (t − 3)] = L f (t − 3) H (t − 3) = e−3s F (s) −3s −3s = cos 3 se2 +1 + sin 3 se s2 +1 . Then Laplace transform of both sides of differential equation gives −3s −3s sY (s) − y (0) + 2Y (s) = cos 3 se2 +1 + sin 3 se s2 +1

107

THE LAPLACE TRANSFORM −3s

−3s

⇒ (s + 2) Y (s) = cos 3 se2 +1 + sin 3 se s2 +1 −3s

−3s

e se ⇒ Y (s) = cos 3 (s+2)(s 2 +1) + sin 3 (s+2)(s2 +1) . 1 A Bs+C Using partial fractions let F1 (s) = (s+2)(s 2 +1) = s+2 + s2 +1 . ¡ 2 ¢ Multiplying by denominator we have 1 = A s + 1 +(Bs + C) (s + 2) . Letting s = −2 gives A = 51 . Equating the coefficients s2 : 0 = A + B ⇒ B = −A = − 51 2 s: 0 = 2B + C ⇒ C = −2B = 5

. 1 1 1 s 2 1 1 So F1 (s) = (s+2)(s2 +1) = 5 s+2 − 5 s2 +1 + 5 s2 +1

⇒ f1 (t) = L−1 [F1 (s)] = 51 e−2t − 15 cos t + 25 sin t. e−3s −cs F1 (s) with c = 3 and using (T 15) , Now (s+2)(s 2 +1) is of the form e h i £ ¤ −3s e −1 −1 −3s L e F1 (s) = f1 (t − 3) H (t − 3) (s+2)(s2 +1) = L ¤ £ = 51 e−2(t−3) − 51 cos (t − 3) + 25 sin (t − 3) H (t − 3) . s A Bs+C Using partial fractions let F2 (s) = (s+2)(s 2 +1) = s+2 + s2 +1 . ¡ 2 ¢ Multiplying by denominator we have s = A s + 1 +(Bs + C) (s + 2) . Letting s = −2 gives A = − 52 . Equating the coefficients s2 : 0 = A + B ⇒ B = −A = 52 1 s: 1 = 2B + C ⇒ C = 1 − 2B = 5

. s 2 1 2 s 1 1 So F2 (s) = (s+2)(s2 +1) = − 5 s+2 + 5 s2 +1 + 5 s2 +1

⇒ f2 (t) = L−1 [F2 (s)] = − 52 e−2t + 52 cos t + 15 sin t. se−3s −cs Now (s+2)(s F2 (s) with c = 3 and using (T 15) , 2 +1) is of the form e h i £ ¤ −3s se −1 L−1 (s+2)(s e−3s F2 (s) = f2 (t − 3) H (t − 3) 2 +1) = L £ ¤ 1 = − 25 e−2(t−3) + 2

5 cos (t − 3) + 5 sin (t − 3) H (t − 3) . −1 Hence y (t) = [Y (s)]

£ 1L−2(t−3) ¤ 2 = cos 3 5 e − 1

5 cos (t − 3) + 5 sin (t − 3) H (t ¤− 3) £ 2 −2(t−3) + sin 3 − 5 e + 25 cos (t − 3) + 51 sin (t − 3) H (t − 3) 1 = 5 H (t − 3) [(2 cos 3 + sin 3) sin (t − 3)+(2 sin 3 − cos 3) cos (t − 3) + (cos 3 − 2 sin 3) e−2(t−3) ]. 47. g (t) = sin t [H (t) − H (t − 4)] = sin t − sin tH (t − 4) and the

differential equation is

y ′′ + 9y = sin t − sin tH (t − 4) , y (0) = 0, y ′ (0) = 1. For the last forcing term, sin tH (t − 4) , c = 4 and f (t − 4) = sin t. Let τ = t − 4.Then t = τ + 4 and f (τ ) = sin (τ + 4) so that f (t) = sin (t + 4) = cos 4 sin t + sin 4 cos t. Thus F (s) = L [sin (t + 4)] = cos 4 s21+1 + sin 4 s2s+1 . Then using (T 15) , £ ¤ L [sin tH (t − 4)] = L f (t − 4) H (t − 4) = e−4s F (s) −4s −4s = cos 4 se2 +1 + sin 4 se s2 +1 . Then Laplace transform of both sides of differential equation gives −4s −4s s2 Y (s) − sy (0) − y ′ (0) + 9Y (s) = s21+1 − cos 4 se2 +1 − sin 4 se s2 +1 ⇒ −4s

−4s

s2 Y (s) − 1 + 9Y (s) = s21+1 − cos 4 se2 +1 − sin 4 se s2 +1 ⇒ ¡ 2 ¢ −4s 1 e−4s ⇒ s + 9 Y (s) = 1 + s2 +1 − cos 4 s2 +1 − sin 4 se s2 +1

108

CHAPTER 3 −4s

−4s

1 e se ⇒ Y (s) = s21+9 + (s2 +1)(s 2 +9) − cos 4 (s2 +1)(s2 +9) − sin 4 (s2 +1)(s2 +9) . Now we will find the inverse Laplace transform of all four terms.

3 −1 Let F1 (s) = s21+9 = 13 s2 +3 [F1 (s)] = 31 sin 3t.

2 ⇒ f1 (t) = L 1 As+B Cs+D Using partial fractions, let F2 (s) = (s2 +1)(s 2 +9) = s2 +1 + s2 +9 .

Multiplying by¡ denominator we have

¡ ¢ ¢ 1 = (As + B) s2 + 9 + (Cs + D) s2 + 1 .

Equating the coefficients

s3 : 0 = A + C ⇒ C = −A s2 : 0 = B + D ⇒ D = −B s: 0 = 9A + C ⇒ 9A − A = 0 ⇒ A = 0 and so C = 0. 1: 1 = 9B + D ⇒ 9B − B = 1 ⇒ B = 18 and so D = − 18 . 1 1 1 1 1 1 1 1 3

Thus F2 (s) = (s2 +1)(s 2 +9) = 8 s2 +1 − 8 s2 +9 = 8 s2 +1 − 24 s2 +32 .

1 sin 3t. ⇒ f2 (t) = L−1 [F2 (s)] = 18 sin t − 24 e−4s F3 (s) = (s2 +1)(s2 +9) is of the form e−cs F2 (s) with c = 4 and using i h £ ¤ e−4s = L−1 e−4s F2 (s) = f2 (t − 4) H (t − 4) (T 15) , L−1 [F3 (s)] = L−1 (s2 +1)(s 2 +9) ¤ £ 1 sin 3 (t − 4) H (t − 4) . = 18 sin (t − 4) − 24 As+B Cs+D s Using partial fractions let F4 (s) = (s2 +1)(s 2 +9) = s2 +1 + s2 +9 .

Multiplying by we have

¡ denominator ¡ ¢ ¢ s = (As + B) s2 + 9 + (Cs + D) s2 + 1 .

Equating the coefficients

s3 : 0 = A + C ⇒ C = −A

s2 : 0 = B + D ⇒ D = −B s: 1 = 9A + C ⇒ 9A − A = 1 ⇒ A = 18 and so C = − 81 . 1 : 0 = 9B + D ⇒ 9B − B = 0 ⇒ B = 0 and so D = 0. s 1 s 1 s 1 s 1 s Thus F4 (s) = (s2 +1)(s 2 +9) = 8 s2 +1 − 8 s2 +9 = 8 s2 +1 − 8 s2 +32 .

1 ⇒ f4 (t) = L−1 [F4 (s)] = 18 cos t − 8

cos 3t.

−4s se −cs Now (s2 +1)(s2 +9) is of the form e F4 (s) with c = 4 and using (T 15) ,

i h £ −4s ¤ se−4s −1 e F4 (s) = f4 (t − 4) H (t − 4) L−1 (s2 +1)(s 2 +9) = L ¤ £ = 18 cos (t − 4) − 1

8 cos 3 (t − 4) H (t − 4) . Hence y (t) = L−1 [Y (s)]

1 = 31 sin 3t +£18 sin t − 24 sin 3t ¤ 1 1 − cos 4 £8 sin (t − 4) − 24 sin 3 (t − 4) ¤ H (t − 4) − sin 4 81 cos (t − 4) − 18 cos 3 (t − 4) H (t − 4) 1 1 = 3 sin 3t + 18 sin t − 24 sin 3t

1 1 + 8 H (t − 4) [ 3 cos 4 sin 3 (t − 4) + sin 4 cos 3 (t − 4) − cos 4 sin (t − 4) − sin 4 cos (t − 4)]. n−1 49. g (t) = (−1) for n − 1 ≤ t < n, where n = 1, 2, ..., ∞ ⇒

g (t) = 1 [H (t) − H (t − 1)] − 1 [H (t − 1) − H (t − 2)]

+1 [H (t − 2) − H (t − 3)] − 1 [H (t − 3) − H (t − 4)] − ...∞ g (t) = 1 − 2H (t − 1) + 2H (t − 2) − 2H (t − 3) + 2H (t − 4) − ...∞ and the differential equation is y ′′ + 4y = 1 − 2H (t − 1) + 2H (t − 2) − 2H (t − 3)

109

THE LAPLACE TRANSFORM

+2H (t − 4) − ...∞, y (0) = 0, y ′ (0) = 0. Taking the Laplace transform of both sides of differential equation gives s2 Y (s) − sy (0) − y ′ (0) + 4Y (s) −s −2s −3s −4s = 1s − 2es + 2e s − 2e s + 2e s − ... ∞ ¡ 2 ¢ −s −2s −3s −4s ⇒ s + 4 Y (s) = 1s − 2es + 2e s − 2e s + 2e s − ... ∞ ∞ P n e−ns (−1) s(s ⇒ Y (s) = s(s21+4) + 2 2 +4) . n=1

Using partial fractions, let F (s) = s(s21+4) = As + Bs+C s¢2 +4 . ¡ 2 Multiplying by denominator we have 1 = A s + 4 + (Bs + C) s. Letting s = 0 gives A = 41 . Equating the coefficients s2 : 0 = A + B ⇒ B = −A = − 14 s: 0 = C. s

So F (s) = s(s21+4) = 41 1s − 41 s2s+4 = 41 1s − 41 s2 +2 2

1 1 −1 ⇒ f (t) = L [F (s)] = 4 − 4 cos 2t. e−ns −cs F (s) with c = n and using (T 15) , Now s(s 2 +4) is of the form e h −ns i e L−1 s(s = L−1 [e−ns F (s)] = f (t − n) H (t − n) 2 ¤ £ 1 1 +4)

= 4 − 4 cos 2 (t − n) H (t − n) . So y (t) = L−1 [Y (s)] = 14 − 41 cos 2t

∞ ¤ P n£ (−1) 41 − 14 cos 2 (t − n) H (t − n) +2 n=1

⇒ y (t) =

1 4

(1 − cos 2t) +

1 2

∞ P

n=1

n

(−1) H (t − n) [1 − cos 2 (t − n)] .

3.5 PERIODIC FUNCTIONS 1. g (t) = |sin t| is a periodic function with period T = π and |sin t| = sin t for 0 ≤ t < π.So G (s) = L [g (t)] = Using integration table we have that ¯π ¯ −st Rπ −st ¯ ¯e = e sin tdt = ¯ 1+s 2 (−s sin t − cos t)¯ 0

0

Thus G (s) =

1 1−e−πs

1+e−πs 1+s2 .

1+e−πs

(s2 +1)(1−e−πs ) .

y

1 t π

2π

3π

4π

Rπ 0

e−st sin tdt.

110

CHAPTER 3

R2

1 1−e−2s

3. L [g (t)] =

1 1−e−2s

e−st g (t) dt =

0

·1 R

t2 e−st dt +

0

¸ ¢ 2 − t2 e−st dt .

R2 ¡ 1

We the method of integration by parts twice to find R 2 use t e−st dt = − 1s t2 e−st − s22 te−st − s23 e−st . ·1 ¸ R 2 −st R2 −st R2 2 −st 1 t e dt + 2 e dt − t e dt Now L [g (t)] = 1−e−2s 0 1 1 ¯ ¯1 2 = 1−e1−2s [¯− 1s t2 e−st − s22 te−st − s23 e−st ¯0 − 2s |e−st |1 ¯ 1 2 −st ¯ 2 − ¯− t e − 22 te−st − 23 e−st ¯ ] =

= =

s s s 1 1 −s 1 − s22 e−s − s23 e−s + s23 − 2s e−2s + 2s e−s 1−e−2s [− s e 4 −2s + 4s e−2s + s23 e−2s − 1s e−s − s22 e−s − s23 e−s ] s2 e £ + ¤ 1 4 −s 4 −s 2 2 −2s + s42 e−2s + s23 e−2s 1−e−2s £− s2 e ¡ − s3 e ¢ + s3 +¡ s e ¢ ¤ 1 2 4 4 −s + s23 + s42 + 2s e−2s . 1−e−2s + s3 − s3 + s2 e

1 t −1

5. L [g (t)] = =

1 1−e−s

1 1−e−s

R1

e−st g (t) dt =

¯ (1−s)t0¯1

¯e ¯ ¯ 1−s ¯ = 0

1 1−e−s

e1−s −1 (1−s)(1−e−s ) .

R1

e−st et dt =

0

1 1−e−s

R1

e(1−s)t dt

0

7. gb (t) = 1 [H (t) − H (t − 1)] ½ − 1 [H (t − 1) − H (t − 2)] = 1 − 2H (t − 1) g (t) , 0 ≤ t ≤ 2, +H (t − 2) . So gb (t) = 0, t > 2. 2 R L [g (t)] = 1−e1−2s e−st g (t) dt = 1−e1−2s L [gb (t)] = 1−e1−2s L [1 − 2H (t − 1) + H (t − 2)] 0 £ ¤ = 1−e1−2s 1s − 2s e−s + 1s e−2s .

1

1

2

3

4

5

t

1 1 1 2 −1 [F (s)] s2 +hs3 . Then f (t) =iL h = t + 2 ti. ¡ ¢ = L−1 1−1e−s s12 + s13 = L−1 1−1e−s F (s) i ∞ h P 2 (t − n) + 21 (t − n) H (t − n) H (t − n) = n=0

9. Let F (s) = So g (t) ∞ P = f n=0

(t − n) .

111

THE LAPLACE TRANSFORM s s2 +4 h −1

s . s 2 +2 ³2

Then f (t) = L−1 [F (s)] = cos 2t. ´i h i

1 s 1 −1 So g (t) = L = L F (s) 1−e−s s2 +4 1−e−s ∞ ∞ P P [cos 2 (t − n)] H (t − n) . = f (t − n) H (t − n) =

11. Let F (s) =

=

n=0 2 1 e−s f (t) = ´i L−1 [F (s)]h = t+ 21 (t − i1) s2 +h s3 . Then ³ −s L−1 1−e1−2s s12 + es3 = L−1 1−e1−2s F (s)

n=0

13. Let F (s) =

H (t − 1) .

So g (t) = ∞ P = f (t − 2n) H (t − 2n) n=0

i ∞ Ph 2

= (t − 2n) + 21 (t − 2n − 1) H (t − 2n − 1) H (t − 2n) . n=0 ¡ ¢ ¡ ¢ − πs 2 15. Let F (s) = 1s + es2 +1 . Then f (t) = L−1 [F (s)] = 1+sin t − π2 H t − π2 . h ³ ´i h i − πs 2 So g (t) = L−1 1−e1−πs 1s + es2 +1 = L−1 1−e1−πs F (s) ∞ P = f (t − nπ) H (t − nπ) = =

n=0

∞ £ P

n=0

∞ P

n=0

¡ ¢ ¡ ¢¤ 1 + sin t − nπ − π2 H t − nπ − π

H (t − nπ) 2

¡ ¢ ¡ ¢ H (t − nπ) + sin t − nπ − π2 H t − nπ − π

2 .

1 e−2s = ´i L−1 [F (s)]h = 12 t2 + 16 (ti− s3 +h s4 . Then ³ f (t)−2s = L−1 1+e1−5s F (s) L−1 1+e1−5s s13 + e s4

17. Let F (s) =

3

2) H (t − 2) .

So g (t) = ∞ P n = (−1) f (t − 5n) H (t − 5n) n=0

h i ∞ P n 2 3

= (−1) 12 (t − 5n) + 16 (t − 5n − 2) H (t − 5n − 2) H (t − 5n) n=0 h i ∞ P 2 3 n = (−1) 12 (t − 5n) H (t − 5n) + 16 (t − 5n − 2) H (t − 5n − 2) . n=0

³ ´ ³ −s ´

2 19. G (s) = s12 csch(s) = s12 es −e = 1−e1−2s 2es2 . −s 2e−s = L−1 [F (s)] s2h . Then³f (t) ´i i H h = 2 (t − 1) −s 1 2e 1 −1 −1 L =L 1−e−2s s2 1−e−2s F (s)

Let F (s) =

So g (t) = ∞ P = f (t − 2n) H (t − 2n) =

=

n=0 ∞ P

n=0 ∞ P

n=0

(t − 1) .

[2 (t − 2n − 1) H (t − 2n − 1)] H (t − 2n) 2 (t − 2n − 1) H (t − 2n − 1) .

21. Graphing the step functions makes  result simple.

t > b,

 1, 0, a < t < b = H (t − b) , If b > a, H (t − a) H (t − b) =  0, t < a.

112

CHAPTER 3

 t > a,  1, 0, b < t < a = H (t − a) , If a > b, H (t − a) H (t − b) =  0, t < b. 23. First, we will find the Laplace transform of |sin 2t| which is a periodic function with period T = π2 and |sin 2t| = sin 2t for π R2 −st π 1 0 ≤ t < 2 .So L [|sin 2t|] = e sin 2tdt. Using −πs 2 1−e

π

0

R2

e−st sin 2tdt π ¡ π ¢ = s21+4 |e−st (−s sin 2t − 2 cos 2t)|02 = s22+4 1 + e− 2 s . ³ ´ ³ ´ π 1 Thus L [|sin 2t|] = s22+4 + s22+4 e− 2 s . π 1−e− 2 s Taking the Laplace transform of the differential equation y ′′ + y = |sin 2t| , y (0) = y ′ (0) = 0³we have ´³ integration table we have that

0

´ π 1 s2 Y (s) − sy (0) − y ′ (0) + Y (s) = s22+4 + s22+4 e− 2 s π 1−e− 2 s ´³ ´ ³ ¡ 2 ¢ 1 2 2 s −π ⇒ s + 1 Y (s) = s2 +4 + s2 +4 e 2 −πs 1−e i³2 h ´ π 1 2 2 −2s . ⇒ Y (s) = (s2 +4)(s 2 +1) + (s2 +4)(s2 +1) e −πs 1−e 2 Using partial fractions, 2 As+B Cs+D let F1 (s) = (s2 +4)(s 2 +1) = s2 +1 + s2 +4 . Multiplying by the ¡ 2 ¡ ¢ ¢ denominator we get 2 = (As + B) s + 4 + (Cs + D) s2 + 1 . Equating the coefficients: s3 : 0 = A + C ⇒ C = −A s2 : 0 = B + D ⇒ D = −B s: 0 = 4A + C ⇒ 4A − A = 0 ⇒ A = 0 and then C = 0 1: 2 = 4B + D ⇒ 4B − B = 2 ⇒ B = 23 and then D = − 32 . 2 2 1 2 1 2 1 1 2 Thus F1 (s) = (s2 +4)(s 2 +1) = 3 s2 +1 − 3 s2 +4 = 3 s2 +1 − 3 s2 +22 2 1 −1 ⇒ f1 (t) = L [F1 (s)] = 3 sin t − 3 sin 2t. π 2 −π 2 s is of the form e−cs F (s) with c = F2 (s) = (s2 +4)(s 2 +1) e 1 2. ¡ ¢ ¡ ¢ π π −1 So f2 (t) = L ¢ f1 t − 2 H t ¤− 2¡ ¢ £ [F¡2 (s)] = = 32 sin t − π2 − 13 sin (2t − π) H t − π2 . 2 2 −π 2 s. Let F (s) = F1 (s) + F2 (s) = (s2 +4)(s 2 +1) + (s2 +4)(s2 +1) e Then f (t) = L−1 [F (s)] ¡ ¢ ¢ £ ¤ ¡ t − π2 = 23 sin t − 31 hsin 2t + 23 sin t − π2 − 13 sin (2ti− ³ π) H ´ 2 (s2 +4)(s2 +1) 1 π F (s) . 1−e− 2 s

and Y (s) =

+

2 −π 2s (s2 +4)(s2 +1) e

= Hence the inverse Laplace htransform yields i 1 y (t) = L−1 [Y (s)] = L−1 F (s) π 1−e− 2 s ∞ ¢ ¡ ¡ ¢ P nπ nπ = f t− 2 H t− 2 =

n=0 ∞ P n=0

¡ { 32 sin t −

nπ 2

¢

−

1 3

¡ sin 2 t −

nπ 2

¢

1 π 1−e− 2 s

113

THE LAPLACE TRANSFORM

£2

¡

¢ π

¤

¡

¢ π

¡

1 nπ nπ sin t − nπ 2 − 2 − 3 sin (2t − nπ − π) H t − 2 − 2 }H t − 2 ∞ £ ¡ ¢ ¡ ¢ ¤ P = 13 { 2 sin t − nπ − sin (2t − nπ) H t − nπ 2 2 n=0 ¢ ¢ £ ¡ ¤ ¡ π nπ π + 2 sin t − 2 − 2 − sin (2t − nπ − π) H t − nπ 2 − 2 }. 25. From Exercise 5, g (t) = et and its Laplace transform is R1 R1 R1 L [g (t)] = 1−e1 −s e−st g (t) dt = 1−e1 −s e−st et dt = 1−e1 −s e(1−s)t dt 0 0 ¯ (1−s)t ¯1 ´ ³ ´³0 ¯e ¯ e1−s −1 1 e−s 1 1 . = 1−e−s ¯ 1−s ¯ = (1−s)(1−e−s ) = s−1 − e s−1 −s 1−e 0 Taking the Laplace transform of the differential equation

y ′ + y = g (t) , y (0) = 1, ³we have

´ ´³ 1 e−s 1 sY (s) − y (0) + Y (s) = s−1 − e s−1 1−e−s ³ ´ ´³ 1 e−s 1 ⇒ (s + 1) Y (s) = 1 + s−1 − e s−1 1−e−s ³ ´³ ´ 1 e−s 1 1 + (s−1)(s+1) − e (s−1)(s+1) .

⇒ Y (s) = s+1 −s 1−e Using partial fractions,

1 A B let F1 (s) = (s−1)(s+1) = s−1 + s+1 . Multiplying by the

denominator we get 1 = A (s + 1) + B (s − 1) . Letting s = 1 and 1 1 − 21 s+1 ⇒ s = −1 gives A = 21 and B = − 12 . So F1 (s) = 21 s−1

+

3

−s

e f1 (t) = L−1 [F1 (s)] = 12 et − 21 e−t = sinh t. Now F2 (s) = (s−1)(s+1) is of the form e−cs F1 (s) with c = 1. So f2 (t) = L−1 [e−s F1 (s)] = f1 (t − 1) H (t − 1) = sinh (t − 1)hH (t − 1) . i 1 e−s Then f (t) = f1 (t) − ef2 (t) = L−1 (s−1)(s+1) − e (s−1)(s+1)

= sinh th³ − e sinh (t − 1) H (t − 1) . ´ ³ ´i

e−s 1 1 −1 − e So L −s

(s−1)(s+1) (s−1)(s+1) 1−e ∞ P = f (t − n) H (t − n) =

=

n=0 ∞ P

n=0 ∞ P

n=0

[sinh (t − n) − e sinh (t − n − 1) H (t − n − 1)] H (t − n) sinh (t − n) H (t − n) − e sinh (t − n − 1) H (t − n − 1) .

Hence the inverse Laplaceh transform ³ yields ´³ ´i 1 e−s 1 1 −1 −1 + − e y (t) = L [Y (s)] = L −s s+1 (s−1)(s+1) (s−1)(s+1) 1−e ∞ P −t =e + sinh (t − n) H (t − n)−e sinh (t − n − 1) H (t − n − 1) . n=0 ½ 1, 0 ≤ t < 1, 27. From Exercise 7, g (t) = is a periodic −1, 1 ≤ t < 2. ·1 ¸ R −st R2 −st 1 function with period 2. So L [g (t)] = 1−e−2s e dt − e dt 0 1 ·¯ ¸ ¯ ¯ ¯ £ 1 2 −s 1 −2s ¤ ¯ e−st ¯1 ¯ e−st ¯2 1 1 . = 1−e−2s ¯− s ¯ + ¯ s ¯ = 1−e−2s s − s e + s e 0

1

Taking the Laplace transform of the differential equation

¢

114

CHAPTER 3

y ′′ − 4y = g (t) , y (0) = y ′ (0) = 0, we have £ ¤

s2 Y (s) − sy (0) − y ′ (0) − 4Y (s) = 1−e1−2s 1s − 2s e−s + 1s e−2s ¡ 2 £ ¤ ¢ −2s ⇒ s − 4 Y (s) = 1−e1−2s 1s − 2s e−s + 1

se h i 1 2 1 ⇒ Y (s) = 1−e1−2s s(s−2)(s+2) − s(s−2)(s+2) e−s + s(s−2)(s+2) e−2s . Using partial fractions, 1 B C let F1 (s) = s(s−2)(s+2) = As + s−2 + s+2 . Multiplying by the

denominator we get 1 = A (s − 2) (s + 2) + Bs (s + 2) + Cs (s − 2) .

1 Letting s = 0, 2 and s = −2 gives A = − 14 , B = 18 and C = 8

.

11 1 1 1 1

So F1 (s) = − 4 s + 8 s−2 + 8 s+2 ⇒ f1 (t) = L−1 [F1 (s)] = − 14 + 81 e2t + 18 e−2t = 14 (cosh 2t − 1) . So f2 (t) = L−1 [e£−s F1 (s)] =¤ f1 (t − 1) H (t − 1) = 14 [cosh (2t − 2) − 1] H (t − 1) and f3 (t) = L−1 e−2s F1 (s) = f1 (t − 2) H (t − 2) = 14 [cosh (2t − 4) − 1] H (t − 2) Then fh(t) = f1 (t) − 2f2 (t) + f3 (t)

i

1

1 2 − s(s−2)(s+2) e−s + s(s−2)(s+2) e−2s = L−1 s(s−2)(s+2)

= 41 (cosh 2t − 1)− 21 [cosh (2t − 2) − 1] H (t − 1)+ 14 [cosh (2t − 4) − 1] H (t − 2) . Hence the inverse Laplaceh³ transform ´ ³yields ´i 1 2 1 −s −2s y (t) = L−1 [Y (s)] = L−1 1−e1−2s − e + e s(s−2)(s+2) s(s−2)(s+2) s(s−2)(s+2) ∞ P = f (t − 2n) H (t − 2n) =

n=0 ∞ P n=0

=

1 4

{ 41 (cosh (2t − 4n) − 1)− 21 [cosh (2t − 4n − 2) − 1] H (t − 2n − 1)

+ 14 [cosh (2t − 4n − 4) − 1] H (t − 2n − 2)}H (t − 2n) ∞ P {[cosh (2t − 4n) − 1] H (t − 2n)−2 [cosh (2t − 4n − 2) − 1] H (t − 2n − 1)

n=0

+ [cosh (2t − 4n − 4) − 1] H (t − 2n − 2)}. n 29. The Laplace transform of each term, tn! , of et is £ tn ¤ 1 n! 1 L n! = n! sn+1 = sn+1 . Applying the of Laplace transform we have ¸ property · ∞linearity ∞ ∞ ∞ ¡ ¢n £ tn ¤ P P P P tn 1 1 1 t L n! = . Now L [e ] = L n! = sn+1 = s s ∞ ¡ ¢n P 1

s n=0

¯1¯ ¯ ¯<1 s

n=0

n=0

n=0

n=0

is a geometric series which converges to

⇒ s > 1. Thus for s > 1, L [et ] =

1 s s s−1

=

1 1− 1s

=

s s−1

if

1

s−1 .

3.6 INTEGRALS AND THE CONVOLUTION THEOREM 1. f (t) = t, and g (t) = et . So f ∗ g =

Rt

τ et−τ dτ = et

0

Rt 0

τ e−τ dτ. Now using

integration by parts (u = τ, dv = e−τ dτ ) we get

t f ∗ g = et |−τ e−τ − e−τ |0 = et (−te−t − e−t + 1) = −t − 1 + et .

Rt 3. f (t) = 1, and g (t) = 1. So 1 ∗ 1 = 1 · 1dτ = t. 0

115

THE LAPLACE TRANSFORM 1 s2 −1

1 s2 +1 .

Then f (t) = L−1 [F (s)] = t and h i g (t) = L [G (s)] = sin t. So L−1 s12 · s21+1 = L−1 [F (s) · G (s)] Rt Rt = f (τ ) g (t − τ ) dτ = τ sin (t − τ ) dτ. Now using integration by 0 0 h i parts (u = τ, dv = sin (t − τ ) dτ ) we get L−1 s12 · s21+1

5. Let F (s) =

and G (s) =

t

= |τ cos (t − τ ) + sin (t − τ )|0 = t − sin t.

7. Taking the Laplace transform of both sides of the differential equation y ′′ − 5y ′ + 4y = f (t) , y (0) = y ′ (0) = 0, we have s2 Y¡ (s) − sy (0)¢− y ′ (0) − 5sY (s) + 5y (0) + 4Y (s) = F (s) ⇒ s2 − 5s + 4 Y (s) = F (s) ⇒ Y (s) = F (s) G (s) 1 1 where G (s) = s2 −5s+4 = (s−4)(s−1) . 1 A B Using partial fractions G (s) = s2 −5s+4 = s− 1 + s−4 . Multiplying by denominator we get 1 = A (s − 4)+B (s − 1) . Letting 1 1 s = 1 and 4 gives A = − 31 and B = 31 . So G (s) = 31 s−4 − 31 s−1 ⇒ 1 t 1 4t −1 g (t) = L [G (s)] = 3 e − 3 e . Now the inverse Laplace transform using the convolution theorem yields y (t) = L−1 [Y (s)] = L−1 [F (s) G (s)] ¤ £ Rt Rt = f (τ ) g (t − τ ) dτ = f (τ ) 13 e4(t−τ ) − 31 et−τ dτ. 0

0

0

0

9. Taking the Laplace transform of both sides of the differential equation y ′′ − 9y = f (t) , y (0) = y ′ (0) = 0, we have

′ s2 Y¡ (s) − sy ¢ (0) − y (0) − 9Y (s) = F (s)

2 ⇒ s − 9 Y (s) = F (s) ⇒ Y (s) = F (s) G (s) 1 where G (s) = s21−9 = (s+3)(s−3) . 1 A B Using partial fractions G (s) = (s+3)(s−3) = s+3 + s−3 . Multiplying by denominator we get 1 = A (s − 3)+B (s + 3) . Letting 1 1 − 61 s+3 ⇒ s = −3 and 3 gives A = − 61 and B = 61 . So G (s) = 16 s−3 1 1 g (t) = L−1 [G (s)] = 6 e3t − 6 e−3t . Now the inverse Laplace transform using the convolution theorem yields

y (t) = L−1 [Y (s)] = L−1 [F (s) G (s)]

¤ £ Rt Rt = f (τ ) g (t − τ ) dτ = f (τ ) 61 e3(t−τ ) − 61 e−3(t−τ ) dτ.

11. Taking the Laplace transform of both sides of the differential equation y ′′ + 4y = f (t) , y (0) = 0, y ′ (0) = 7, we have

′ s2 Y¡ (s) − sy ¢ (0) − y (0) + 4Y (s) = F (s)

2 ⇒ s + 4 Y (s) = F (s) + 7 ⇒ Y (s) = F (s) G (s) + s27+4 2 −1 [G (s)] = 21 sin 2t. where G (s) = s21+4 = 12 s2 +2 2 ⇒ g (t) = L Now the inverse Laplace transform using the convolution h itheorem 7 −1 −1 −1 yields y (t) = L [Y (s)] = L [F (s) G (s)] + L s2 +4 t Rt R = f (τ ) g (t − τ ) dτ + 72 sin 2t = 12 f (τ ) sin 2 (t − τ ) dτ + 72 sin 2t. 0

0

13. Taking the Laplace transform of both sides of the differential equation

116

CHAPTER 3

y ′ + 7y = f (t) , y (0) = 2, we have sY (s) − y (0) + 7Y (s) = F (s) 2 ⇒ (s + 7) Y (s) = F (s) + 2 ⇒ Y (s) = F (s) G (s) + s+7 1 −1 −7t where G (s) = s+7 ⇒ g (t) = L [G (s)] = e . Now the invrese Laplace transform using the convolution h i theorem 2 −1 −1 −1 yields y (t) = L [Y (s)] = L [F (s) G (s)] + L s+7 t Rt R = f (τ ) g (t − τ ) dτ + 2e−7t = f (τ ) e−7(t−τ ) dτ + 2e−7t . 0

0

15. Taking the Laplace transform of both sides of the differential equation y ′′ − 2y ′ + 10y = f (t) , y (0) = y ′ (0) = 0, we have ′ s2 Y¡ (s) − sy (0) − ¢ y (0) − 2sY (s) + 2y (0) + 10Y (s) = F (s) 2 ⇒ s − 2s + 10 Y (s) = F (s) ⇒ Y (s) = F (s) G (s) where G (s) = s2 −21s+10 = 13 (s−1)32 +32 ⇒ g (t) = L−1 [G (s)] = 31 et sin 3t. Now the inverse Laplace transform using the convolution theorem Rt yields y (t) = L−1 [Y (s)] = L−1 [F (s) G (s)] = f (τ ) g (t − τ ) dτ 0

= 31

Rt

t−τ

f (τ ) e

0

sin 3 (t − τ ) dτ.

17. Taking the Laplace transform of both sides of the differential equation y ′′ + 4y ′ + 13y = f (t) , y (0) = y ′ (0) = 0, we have ′ s2 Y¡ (s) − sy (0) − ¢ y (0) + 4sY (s) − 4y (0) + 13Y (s) = F (s) 2 ⇒ s + 4s + 13 Y (s) = F (s) ⇒ Y (s) = F (s) G (s) where G (s) = s2 +41s+13 = 13 (s+2)32 +32 ⇒ g (t) = L−1 [G (s)] = 13 e−2t sin 3t.

Now the inverse Laplace transform using the convolution theorem

Rt yields y (t) = L−1 [Y (s)] = L−1 [F (s) G (s)] = f (τ ) g (t − τ ) dτ 0

Rt 1

= 3

0

19. x (t) =

f (τ ) e−2(t−τ ) sin 3 (t − τ ) dτ. Rt 0

cos (t − τ ) x (τ ) dτ + sin t = (h ∗ x) (t) + sin t, where

h (t) = cos t. Now taking Laplace transform of both sides using the convolution theorem we have X (s) = L [h (t)] L [x (t)] + L [sin t] ⇒ X (s) = s2s+1 X (s) + s21+1 ⇒ √ ³ ´ 3 1 2 ³ √ ´2 1 − s2s+1 X (s) = s21 +1 ⇒ X (s) = s2 −s+1 = √23 1 2 (s− 2 ) + 23 Inverse Laplace transform then √yields 1 x (t) = L−1 [X (s)] = √23 e 2 t sin 23 t. Rt 21. x (t) = e−(t−τ ) x (τ ) dτ + 2 = (h ∗ x) (t) + 2, where 0

h (t) = e−t . Now taking Laplace transform of both sides using the convolution theorem we have 1 X (s) = L [h (t)] L [x (t)] + L [2] ⇒ X (s) = s+1 X (s) + 2s ⇒

117

THE LAPLACE TRANSFORM

³

´

1 1 − s+1 X (s) = 2s ⇒ X (s) = 2(s+1) = 2s + s22 . s2 Inverse Laplace transform then yields

x (t) = L−1 [X (s)] = 2 + 2t.

Rt 23. x (t) = 2 sin (2t − 2τ ) x (τ ) dτ + sin t = (h ∗ x) (t) + sin t, where 0

h (t) = 2 sin 2t. Now taking Laplace transform of both sides using the convolution theorem we have X (s) = L [h (t)] L [x (t)] + L [sin t] ⇒ X (s) = s24+4 X (s) + s21+1 ⇒ ³ ´ 2 1 − s24+4 X (s) = s21+1 ⇒ X (s) = s2s(s2+4 +1) . Using partial fraction s2 +4 s2 (s2 +1) 2

= As + sB2 + Cs+D . Multiplying by denominator we get 2 ¡ 2 ¢ s +1¡ 2 ¢ s + 4 = As s + 1 + B s + 1 + (Cs + D) s2 . Letting s = 0 in this equation gives B = 4 Equating the coefficients: s3 : 0 = A + C

s2 : 1 = B + D ⇒ D = 1 − B = −3

s: 0 = A. So C = 0 (from the first equation) s2 +4 So s2 (s2 +1) = s42 − s23+1 . Hence X (s) = s42 − s23+1 . Inverse Laplace transform then yields

x (t) = L−1 [X (s)] = 4t − 3 sin t.

Rt 25. x (t) = − sinh (t − τ ) x (τ ) dτ + 3 = (h ∗ x) (t) + 3, where 0

h (t) = − sinh t = − 21 et + 21 e−t . Now taking Laplace transform of both sides using the convolution theorem we³ have ´ 1 1 X (s) = L [h (t)] L [x (t)]+L [3] ⇒ X (s) = 21 s+1 − s−1 X (s)+ 3s ⇒ ³ ´ 2 1 1 1 − 12 s+1 + 12 s−1 X (s) = 3s ⇒ s2 s−1 X (s) = 3s ⇒ X (s) = 3s − s33 . Inverse Laplace transform then yields

x (t) = L−1 [X (s)] = 3 − 23 t2 .

R∞ 27. F (s) = L [f (t)] = e−st f (t) dt. ¯ ∞ ¯∞ 0 ¯ R ¯ R −st R∞ ¯ So |F (s)| = ¯ e f (t) dt¯¯ ≤ |e−st f (t)| dt = e−st |f (t)| dt 0

≤

R∞

0

−st

e

αt

M e dt = M

0

R∞ 0

0

−(s−α)t

e

dt =

M s−a

for s > α.

29. (i) Continuity at t = 1 :

lim g (t) = e and lim g (t) = −1 + e + 1 = e ⇒ lim g (t) exists

t→1

t→ 1+

t→1−

and lim g (t) = e = g (1) ⇒ g (t) is continuous at t = 1. t→1

Differentiability at t = 1 : 1+h h = lim −1+e h +e does not exist. g ′ (1) = lim g(1+h)−g(1) h h→0

h→0

Now, in terms of unit ¡ step tfunction, ¢ g (t) can be written as t−1 H (t − 1) et [1 − H (t − 1)] + −1 + e + e £ ¤ = e t + et−1 − 1 H (t − 1)

118

CHAPTER 3

Taking Laplace transform of both sides of the differential equation e−s y ′ − y = H (t − 1) , y (0) = 1, we have³ sY (s) − ´ y (0) − Y (s) = s ⇒ (s − 1) Y (s) − 1 =

e−s s

⇒ Y (s) = 1 +

e−s 1 s s−1 A B s + s−1 .

=

1 s−1

+

e− s s(s−1) .

1 = Multiplying by Using partial fraction let F (s) = s(s−1) denominator we get 1 = A (s − 1) + Bs. Letting s = 0 and 1 in this 1 equation gives A = −1 and B = 1. Hence F (s) = s−1 − 1s ⇒ h −s i f (t) = L−1 [F (s)] = et − 1. Thus L−1 s(es−1) = L−1 [e−s F (s)] £ ¤ = f (t − 1) H (t − 1) = eht−1 −i 1 H (t h− 1) . i

1 e−s Hence L−1 [Y (s)] = L−1 s−1 + L−1 s(s−1) ⇒

£ t−1 ¤ t y (t) = e + e − 1 H (t − 1) which is the same as unit step function of g (t) . dy (ii) From (i) dy dt − y = H (t − 1) ⇒ dt = y + H (t − 1) , y (0) = 1. Integrating from 0 to t we have Rt y (t) = y (0) + [y (τ ) + H (τ − 1)] dτ ⇒ 0

y (t) = 1 +

Rt 0

[y (τ ) + H (τ − 1)] dτ valid for all t including t = 1.

3.7 IMPULSES AND DISTRIBUTIONS 1.Taking the Laplace transform of both sides of the differential equation y ′ + 8y = δ (t − 1) + δ (t − 2) , y (0) = 0, we have sY (s) − y (0) + 8Y (s) = e−s + e−2s −2s e−s 1 ⇒ (s + 8) Y (s) = e−s + e−2s ⇒ Y (s) = s+8 + es+8 . Let F (s) = s+8 −1 −8t ⇒ f (t) = L [F (s)] = e . Then using L−1 [e−cs F (s)] = f (t − c) H (t −hc) the i invesehLaplace i transform −1 −1 e−s −1 e−2s yields y (t) = L [Y (s)] = L s+8 + L s+8

= e−8(t−1) H (t − 1) + e−8(t−2) H (t − 2) . 3. Taking the Laplace transform of both sides of the differential equation y ′′ + 6y ′ + 109y = δ (t − 1) + δ (t − 7) , y (0) = y ′ (0) = 0, we have s2 Y (s) − sy (0) − y ′ (0) + 6sY (s) − y (0) + 109Y (s) = e−s + e−7s ¡ ¢ e−7s e−s + s2 +6s+109 ⇒ s2 + 6s + 109 Y (s) = e−s +e−7s ⇒ Y (s) = s2 +6s+109 1 10 1 = 10 Let F (s) = s2 +6s+109 (s+3)2 +102 1 −3t ⇒ f (t) = L−1 [F (s)] = 10 e sin 10t. Then using −1 −cs L [e F (s)] = f (t − c) H (t − h transform h c) the inveseiLaplace

1 1 +L−1 e−7s s2 +6s+109 yields y (t) = L−1 [Y (s)] = L−1 e−s s2 +6s+109

=

1 −3(t−1) 10 e

i

1 −3(t−7) sin 10 (t − 1) H (t − 1)+ 10 e sin 10 (t − 7) H (t − 7) .

5. Taking the Laplace transform of both sides of the differential equation y ′′ + 4y ′ + 3y = 1 + δ (t − 3) , y (0) = 0, y ′ (0) = 1, we have

119

THE LAPLACE TRANSFORM

s2 Y (s) − sy (0) − y ′ (0) + 4sY (s) − 4y (0) + 3Y (s) = 1s + e−3s ¡ ¢ −3s 1 ⇒ s2 + 4s + 3 Y (s) − 1 = 1s + e−3s ⇒ Y (s) = s1+e 2 +4s+3 + s(s2 +4s+3) −3s

1 e 1 ⇒ Y (s) = (s+3)(s+1) + (s+3)(s+1) + s(s+3)(s+1)

. Using partial fractions 1 A B let F1 (s) = (s+3)(s+1) = s+3 + s+1 ⇒ 1 = A (s + 1) + B (s + 3) .

Letting s = −3 and s = −1 gives A = − 21 and B = 21 . Thus 1 1 F1 (s) = 21 s+1 − 12 s+3 ⇒ f1 (t) = L−1 [F1 (s)] = 21 e−t − 12 e−3t . Then using L£−1 [e−cs F1¤(s)]£= f1 (t − c) H (t − c) ¤we have f2 (t) = L−1 e−3s F1 (s) = 21 e−(t−3) − 12 e−3(t−3) H (t − 3) . 1 A B Using partial fractions let F3 (s) = s(s+3)(s+1) = s+3 + s+1 + Cs ⇒ 1 = As (s + 1) + Bs (s + 3) + C (s + 3) (s + 1) . Letting s = −3, s = −1 and s = 0 gives A = 16 , B = − 21 and C = 31 . 1 1 1 Thus F3 (s) = 61 s+3 − 12 s+1 + 3s 1 . ⇒ f3 (t) = L−1 [F3 (s)] = 61 e−3t − 21 e−t + 3

−1 Hence y (t) = L £[Y (s)] = f1 (t) + f2 (t) + f3 (t)

¤ = 21 e−t − 21 e−3t + 12 e−(t−3) − 12 e−3(t−3) H (t − 3) + 16 e−3t − 21 e−t + 31 £ ¤ = 13 − 31 e−3t + 12 e−(t−3) − 12 e−3(t−3) H (t − 3) ¡ ¢ £ ¤ = 31 1 − e−3t + 12 e−(t−3) − e−3(t−3) H (t − 3) .

7. Taking the Laplace transform of both sides of the differential equation y ′′ + y = 1 + δ (t − 2π) , y (0) = 1, y ′ (0) = 0, we have ′ −2πs s2 Y¡ (s) − sy = 1s + ¡ e2 ¢ (0) − y (0) 1+ Y (s) ¢ 2 −2πs ⇒ s + 1 Y (s) − s = s + e ⇒ s + 1 Y (s) = s + e−2πs + 1s e−2πs + 21 . ⇒ Y (s) = (s2s+1) + (s 2 h i +1) s(s +1) f1 (t) = L−1 (s2s+1) = cos t h i f2 (t) = L−1 (s21+1) e−2πs = sin (t − 2π) H (t − 2π)

Using partial fractions let F3 (s) = s(s21+1) = As + Bs+C s2 +1 ¡ 2 ¢ ⇒ 1 = A s + 1 + (Bs + C) s. Now s = 0 yields A = 1. Equating the coefficients:

s2 : 0 = A + B ⇒ B = −A = −1

s: 0 = C. Thus F3 (s) = 1s − s2s+1 ⇒ f3 (t) = L−1 [F3 (s)] = 1 − cos t.

Hence y (t) = L−1 [Y (s)] = f1 (t) + f2 (t) + f3 (t)

⇒ y (t) = cos t + sin (t − 2π) H (t − 2π) + 1 − cos t

= 1 + sin (t − 2π) H (t − 2π) = 1 + sin tH (t − 2π) .

9. (a) The following are graphs of the solutions of Exercises 1 through 4.

120

CHAPTER 3

y

1 t 1

2

y

1

t

5 y

0.05 t

y

1

t 5

£ ¤ R∞ 11. L δ (n) (t − a) = e−st δ (n) (t − a) dt 0

=

R∞

−∞

δ (n) (t − a) e−st dt −

R0

−∞

δ (n) (t − a) e−st dt

If t = 6 a ⇒ t − a 6= 0, then the property (i) gives δ (n) (t − a) = 0. In this case, everything on the right hand side is 0. But for t = a R∞ (n) n dn −st property (ii) gives δ (t − a) e−st dt = (−1) dt ) |t=a n (e −∞ £ ¤ n n = (−1) (−1) sn e−as = sn e−as . Thus L δ (n) (t − a) = sn e−as .

Chapter Four

An Introduction to Linear Systems of Differential Equations and Their Phase Plane

4.1 INTRODUCTION There are no exercises in this section.

4.2 INTRODUCTION TO LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS ·

¸ 0 1 . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ −4 0 · ¸ −λ 1 det = 0 ⇒ λ2 + 4 = 0 ⇒ λ = ±2i (complex). −4 −λ ¸ · 2 1 3. A = . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ ·1 2 ¸ 2−λ 1 2 det = 0 ⇒ (2 − λ) − 1 = 0 ⇒ 2 − λ = ±1 1 2−λ ⇒ λ = 3, 1. · ¸ · ¸· ¸ · ¸ u 2−λ u 0 1 Eigenvector, , satisfies = . v 1 v 0 2−λ · ¸ · ¸ ¸· u −1 1 0 Eigenvector for λ = 3 : = 1 −1 v 0· ¸ 1 ⇒ −u + v = 0 ⇒ u = 1, v = 1 and eigenvector is . · ¸· ¸ · ¸ 1 u 0 1 1 Eigenvector for λ = 1 : = ⇒u+v =0⇒ 1 1 v 0 · ¸ 1 u = 1, v = −1 and eigenvector is .

−1

· ¸ 3 −1 5. A = . The eigenvalue, λ, satisfies 1 2 ¸ · 3−λ −1 =0 det (A − λI) = 0 ⇒ det 1 2−λ ⇒ (3 − λ)√(2 − λ) + 1 = √ 0 ⇒ λ2 − 5λ + 7 = 0

5± 25−28 5 ⇒λ= = 2 ± 23 i (complex). 2

1. A =

122 7. A =

CHAPTER 4

·

1 1

0 −3

¸

. The eigenvalue, λ, satisfies · ¸ 1−λ 0 det (A − λI) = 0 ⇒ det =0 1

−3 − λ ⇒ (1 − λ) (−3· − λ) ¸· ¸ · ¸ · 1, −3.

¸=0⇒λ= u 0 1−λ u 0 = . Eigenvector, , satisfies 0 v 1¸ · −3 ¸ − ·λ ¸ v · u 0 0 0 Eigenvector for λ = 1 : = ⇒ u − 4v = 0 ⇒ 1 −4 ¸ v 0 · 4 v = 1, u = 4 and eigenvector is . 1¸ · ¸ · ¸ · 0 u 4 0 ⇒ 4u = 0 = Eigenvector for λ = −3 : 0 1 0 v · ¸ 0 u = 0, v is free to choose, say, v = 1 and eigenvector is . 1 · ¸ 4 −3 9. A = . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 1 0 · ¸ 4 − λ −3 det = 0 ⇒ −λ (4 − λ) + 3 = 0 1 −λ 2 ⇒ λ − 4λ + ·3 = 0¸ ⇒ (λ − 1)·(λ − 3) = 0 ⇒¸λ· = 1,¸ 3. · ¸ u 4 − λ −3 u 0 Eigenvector, , satisfies = . v 1 −λ v 0 ¸ ¸· ¸ · · 0 3 −3 u ⇒ 3u − 3v = 0 ⇒ = Eigenvector for λ = 1 : 1 −1 v 0 · ¸ 1 u = 1, v = 1 and eigenvector is . 1 ¸ ¸ · · ¸· 0 u 1 −3 ⇒ u − 3v = 0 = Eigenvector for λ = 3 : 1 −3 0 · ¸v 1 u = 1, v = 3 and eigenvector is . 3

¸

· 3 2 11. A = . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 0 4 · ¸ 3−λ 2 = 0 ⇒ (3 − λ) (4 − λ) = 0 ⇒ λ = 3, 4. det 0 ·4 − λ¸ · ¸· ¸ · ¸ u 3−λ u 0 2 Eigenvector, , satisfies = . v 0 4−λ v 0 · ¸· ¸ · ¸ u 0 0 2 ⇒ 2v = 0 ⇒ = Eigenvector for λ = 3 : 0 1 v 0 · ¸ 1 v = 0, u is free to choose, say, u = 1 and eigenvector is . 0 ¸ · ¸ ¸· · −1 2 0 u Eigenvector for λ = 4 : = 0 0 v 0 · ¸ 2 . ⇒ −u + 2v = 0 ⇒ v = 1, u = 2 and eigenvector is 1

Linear Systems of Differential Equations and Their Phase Plane

·

¸

0 1 . trA = 0 + 0 = 0 and det A = 0 + 4 = 4. −4 0

The·eigenvalue,¸λ, satisfies det (A − λI) = 0 ⇒ −λ 1 = 0 ⇒ λ2 + 4 = 0 ⇒ λ = ±2i (complex). det −4 −λ Let λ1 = 2i and λ2 = −2i. Then λ1 + λ2 = 0 = trA and 2 λ1 λ2·= 2i (−2i) ¸ = −4i = 4 = det A. 2 1 15. A = . trA = 2 + 2 = 4 and det A = 4 − 1 = 3. 1 2

The eigenvalue, λ, satisfies · ¸ det (A − λI) = 0

2−λ 1 ⇒ det =0 1 2−λ 2 ⇒ (2 − λ) − 1 = 0 ⇒ 2 − λ = ±1 ⇒ λ = 3, 1.

Let λ1 = 3 and λ2 = 1. Then λ1 + λ2 = 4 = trA and

λ2 = ·λ1dx ¸ 3 (1) · = 3 = det ¸ · A. ¸ · ¸ · ¸· ¸ x x x 0 1 0 1 d dt ⇒ 17. = = dy dt −4 0 y y −4 0 y dt · ¸ · ¸ ¸ · x u u is where λ is eigenvalue and = eλt ⇒ y v v · ¸ 0 1 corresponding eigenvector of A = . −4 0

The·eigenvalue,¸λ, satisfies det (A − λI) = 0 ⇒ −λ 1 det = 0 ⇒ λ2 + 4 = 0 ⇒ λ = ±2i (complex). −4 −λ · dx ¸ · ¸· ¸ · ¸ · ¸· ¸ x x x 2 1 2 1 d dt ⇒ 19. = = dy dt 1 2 y y 1 2 y dt · ¸ · ¸ · ¸ x u u λt ⇒ =e where λ is eigenvalue and is y v v · ¸ 2 1 corresponding eigenvector of A = . 1 2

The eigenvalue, λ, satisfies · ¸ det (A − λI) = 0

2−λ 1 2 ⇒ det = 0 ⇒ (2 − λ) − 1 = 0 1 2−λ ⇒ 2 − λ = ±1· ⇒ λ ¸· ¸ · ¸ ¸ = 3, 1. · u 0 u 1 2−λ = . Eigenvector, , satisfies 1 ¸ · 2 −¸λ · v ¸ 0 v · u 0 −1 1 Eigenvector for λ = 3 : = 1 −1 v 0· ¸ 1 ⇒ −u + v = 0 ⇒ u = 1, v = 1 and eigenvector is . 1 · ¸ 1 e3t . Solution of the differential equation is 1 ¸· ¸ · ¸ · u 0 1 1 Eigenvector for λ = 1 : = ⇒u+v =0⇒ 1 1 v 0 13. A =

123

124

CHAPTER 4

·

¸

1 u = 1, v = −1 and eigenvector is . Solution of the −1 · ¸ 1 differential equation is et .

−1

General · · ¸ equation is

¸ solution · ¸of the differential x 1 1 3t et . e + c2 = c1 −1 y 1 · dx ¸ · ¸· ¸ · ¸ · ¸· ¸ 3 −1 x x 3 −1 x d dt ⇒ 21. = = dy dt 1 2 y y 1 2 y dt · ¸ · ¸ · ¸ u u x is where λ is eigenvalue and = eλt ⇒ v v y · ¸ 3 −1 corresponding eigenvector of A = . 1 2

The·eigenvalue, λ, satisfies det (A − λI) = 0 ⇒

¸ 3−λ −1 det = 0 ⇒ (3 − λ) (2 − λ) + 1 = 0 1 2−λ √

√

3 5 ⇒ λ2 − 5λ + 7 = 0 ⇒ λ = 5± 25−28 2 · =¸ 2 ±· 2 i (complex).

· dx ¸ · ¸· ¸ ¸· ¸

x x x 1 0 1 0 d dt = = 23. ⇒ dy dt 1 −3 y y 1 −3 y dt · · ¸ ¸ · ¸ x u u ⇒ = eλt where λ is eigenvalue and is y v v · ¸ 1 0 corresponding eigenvector of A = . 1 −3

The eigenvalue, λ, satisfies · ¸ det (A − λI) = 0

1−λ 0 ⇒ det = 0 ⇒ (1 − λ) (−3 − λ) = 0 1 −3 − λ ⇒ λ = 1, −3.· ¸ · ¸· ¸ · ¸ u 1−λ u 0 0 Eigenvector, , satisfies = . v 1 −3 − λ v 0 · ¸· ¸ · ¸ u 0 0 0 Eigenvector for λ = 1 : = ⇒ u − 4v = 0 ⇒ 1 −4 v 0 · ¸ 4 v = 1, u = 4 and eigenvector is . Solution of the differential 1 · ¸ 4 equation is et . 1 · ¸ · ¸ ¸· 0 u 4 0 = ⇒ 4u = 0 Eigenvector for λ = −3 : 0 1 0 v · ¸ 0 u = 0, v is free to choose, say, v = 1 and eigenvector is . 1 · ¸ 0 e−3t . Solution of the differential equation is 1 General equation is · ¸ ¸ solution · ¸of the differential · 0 4 x −3t t e . e + c2 = c1 1 1 y

125

Linear Systems of Differential Equations and Their Phase Plane

·

dx dt dy dt ·

¸

·

¸

·

¸

·

¸·

¸

x x 4 −3 x −3 d ⇒ dt = 0 y y 1 0 y · ¸ · ¸ x u u ⇒ = eλt where λ is eigenvalue and is y v v · ¸ 4 −3 corresponding eigenvector of A = . 1 0

The eigenvalue, λ, satisfies det (A − λI) = 0

· ¸ 4 − λ −3 ⇒ det = 0 ⇒ −λ (4 − λ) + 3 = 0 ⇒ λ2 − 4λ + 3 = 0 −λ 1 ⇒ (λ − 1) (λ · − 3) ¸= 0 ⇒ λ = ·1, 3.

¸· ¸ · ¸ u 4 − λ −3 u 0 Eigenvector, , satisfies = . v 1 −λ v 0 · ¸· ¸ · ¸ 3 −3 u 0 Eigenvector for λ = 1 : = ⇒ 3u − 3v = 0 ⇒ 1 −1 v 0 · ¸ 1 u = 1, v = 1 and eigenvector is . Solution of the differential 1 · ¸ 1 et . equation is 1 ¸· ¸ · ¸ · u 0 1 −3 Eigenvector for λ = 3 : = ⇒ u − 3v = 0 1 −3 v 0 · ¸ 1 u = 1, v = 3 and eigenvector is . Solution of the differential 3 · ¸ 1 equation is e3t . 3

General equation is

· · ¸ ¸ solution · ¸of the differential x 1 1 t 3t

e . = c1 e + c2 y 1 3 · dx ¸ · ¸· ¸ ¸ · ¸· ¸ · 3 2 3 2 x x x d dt ⇒ 27. = = dy dt y 0 4 y y 0 4 dt · · ¸ ¸ · ¸ x u u λt ⇒ =e where λ is eigenvalue and is y v v · ¸ 3 2 corresponding eigenvector of A = . 0 4

The eigenvalue, λ, satisfies · ¸ det (A − λI) = 0

3−λ 2 ⇒ det = 0 ⇒ (3 − λ) (4 − λ) = 0 ⇒ λ = 3, 4. 0 · 4 ¸− λ · ¸· ¸ · ¸ u 3−λ u 0 2 Eigenvector, , satisfies = . v 0 4−λ v 0 · ¸· ¸ · ¸ 0 2 u 0 Eigenvector for λ = 3 : = ⇒ 2v = 0 ⇒ 0 1 v 0 · ¸ 1 v = 0, u is free to choose, say, u = 1 and eigenvector is . 0 · ¸ 1 Solution of the differential equation is e3t . 0 25.

= ¸

4 1

¸·

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CHAPTER 4

Eigenvector for λ = 4 :

·

−1 0

2 0

¸·

u v

¸

=

·

0 0

¸

· ¸ 2 . ⇒ −u + 2v = 0 ⇒ v = 1, u = 2 and eigenvector is 1 · ¸ 2 Solution of the differential equation is e4t . 1

General equation is

· · ¸ ¸ solution · ¸of the differential x 2 1 4t 3t e + c2 e . = c1 0 1

y ¸ · ¸· ¸

¾ · dx =x 1 0 x x d dt ⇒ dt 29. dy = y y 1 2 dt ·= x ¸+ 2y ¸ ¸ · · u u x λt is where λ is eigenvalue and =e ⇒ y v v · ¸ 1 0 corresponding eigenvector of A = . 1 2

The·eigenvalue, λ, satisfies det (A − λI) = 0 ⇒

¸ 1−λ 0 = 0 ⇒ (1 − λ) (2 − λ) = 0 ⇒ λ = 1, 2. det 1 ·2 − λ¸ · ¸· ¸ · ¸ u 1−λ u 0 0 Eigenvector, , satisfies = . v 1 2−λ v 0 · ¸· ¸ · ¸ 0 0 u 0 ⇒u+v =0⇒ Eigenvector for λ = 1 : = 1 1 · v ¸ 0 −1 v = 1, u = −1 and eigenvector is . Solution of the differential 1 · ¸ −1 equation is et . 1 · ¸· ¸ · ¸ u 0 −1 0 Eigenvector for λ = 2 : = ⇒ −u = 0 ⇒ 1 0 v 0 · ¸ 0 u = 0, v is free to choose, say, v = 1 and eigenvector is . 1 ¸ · 0 e2t . Solution of the differential equation is 1

General of ¸ the differential · · ¸ equation is

¸ solution · −1 x 0 e2t . et + c2 = c1 1 1

y · ¸· ¸

¾ · ¸ dx = −x − 5y x x −1 −5 d dt ⇒ dt 31. = dy y y 1 1 =x+y dt ¸ · ¸ ¸ · · u u x is where λ is eigenvalue and = eλt ⇒ y v v · ¸ −1 −5 corresponding eigenvector of A = . 1 1

The·eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ ¸ −1 − λ −5 = 0 ⇒ (−1 − λ) (1 − λ) + 5 = 0 det 1 1−λ

Linear Systems of Differential Equations and Their Phase Plane

⇒ λ2 + 4 ⇒ λ ¸ · ¸· ¸

¾ = ±2i.· dx =x−y x 1 −1 x d dt ⇒ dt 33. dy = y 1 1 y

dt·= x ¸+ y ¸ · ¸ · u u x λt where λ is eigenvalue and =e ⇒ y v v · ¸ 1 −1 is corresponding eigenvector of A = . 1 1

The· eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ ¸ 1−λ −1 2 det = 0 ⇒ (1 − λ) + 1 = 0 1 1−λ 2 ⇒ (1 − λ) = −1¾⇒ 1 − λ· = ±i ¸ ⇒·λ = 1 ± i. ¸ · ¸ dx = −5x − 4y x −5 −4 x d ⇒ dt 35. dtdy = y 2 1 y = 2x + y dt ¸ · ¸ · ¸ · u u x λt is where λ is eigenvalue and =e ⇒ y v v

¸

· −5 −4 corresponding eigenvector of A = . 2 1

The·eigenvalue, λ, satisfies det (A − λI) = 0 ⇒

¸ −5 − λ −4 det = 0 ⇒ (−5 − λ) (1 − λ) + 8 = 0 2 1−λ ⇒ λ2 + 4λ + ·3 = 0¸ ⇒ (λ + 1)·(λ + 3) ⇒ λ = −1,¸ · −3. ¸ u u −5 − λ −4 Eigenvector, , satisfies = 0. v 2 1 − λ v · ¸ · ¸ ¸· 0 u −4 −4 Eigenvector for λ = −1 : = 2 2 v 0 · ¸ −1 ⇒ −4u − 4v = 0 ⇒ v = 1, u = −1 and eigenvector is . 1 · ¸ −1 Solution of the differential equation is e−t . 1 ¸ ¸· ¸ · · 0 −2 −4 u = Eigenvector for λ = −3 : 2 4 v 0 · ¸ −2 ⇒ −2u − 4v = 0 ⇒ v = 1, u = −2, and eigenvector is . 1 ¸ · −2 Solution of the differential equation is e−3t . 1

General of · ¸ solution · · ¸equation is

¸ the differential x −2 −3t −1 −t e . e + c2 = c1 1 1

y ¾ · ¸ · ¸· ¸

dx =y x x 0 1 d dt ⇒ dt 37. dy = y 2 1 y +y dt ·= 2x ¸ · ¸ · ¸ u u x λt is where λ is eigenvalue and =e ⇒ y v v · ¸ 0 1 corresponding eigenvector of A = . 2 1

127

128

CHAPTER 4

The·eigenvalue, λ,¸satisfies det (A − λI) = 0 ⇒

−λ 1 = 0 ⇒ −λ (1 − λ) − 2 = 0 det 2 1−λ ⇒ λ2 − λ − 2·= 0 ¸⇒ (λ + 1) (λ · 2. ¸ · ¸ · − 2) ⇒ λ = ¸−1, u u 0 −λ 1 Eigenvector, , satisfies = . v 2 ¸ · 1 −¸λ · v ¸ 0 · 0 1 1 u Eigenvector for λ = −1 : = ⇒u+v =0⇒ 2 2 v 0 · ¸ 1 u = 1, v = −1 and eigenvector is . Solution of the differential −1 · ¸ 1 e−t . equation is −1 · ¸· ¸ · ¸ u 0 −2 1 Eigenvector for λ = 2 : = 2 −1 v 0 · ¸ 1 ⇒ −2u + v = 0 ⇒ u = 1, v = 2, and eigenvector is . 2 · ¸ 1 e2t . Solution of the differential equation is 2

General of · ¸ solution · ¸ the differential · ¸ equation is

x 1 1 e2t . = c1 e−t + c2 2 y −1 ¸ ¸· ¾ · ¸ · dx = −5x − y −5 −1 x x d ⇒ dt 39. dtdy = y 3 −1 y

= 3x dt ¸ −y · ¸ · ¸

· u u x ⇒ = eλt where λ is eigenvalue and is y v v

¸

· −5 −1 corresponding eigenvector of A = . 3 −1

The·eigenvalue, λ, satisfies ¸ det (A − λI) = 0 ⇒ −5 − λ −1 det = 0 ⇒ (−5 − λ) (−1 − λ) + 3 = 0 3 −1 − λ 2 ⇒ λ + 6λ + ·8 = 0¸ ⇒ (λ + 2)·(λ + 4) ⇒ λ = −2, −4.

¸· ¸ · ¸ −5 − λ −1 u 0 u = . Eigenvector, , satisfies 0 v 3 ¸ · −1 ¸− λ · · ¸v −3 −1 u 0 Eigenvector for λ = −2 : = 3 1 v 0 · ¸ 1 . ⇒ −3u − v = 0 ⇒ u = 1, v = −3 and eigenvector is −3 · ¸ 1 Solution of the differential equation is e−2t . − · ¸ · 3¸ · ¸ −1 −1 u 0 Eigenvector for λ = −4 : = 3 3 v 0 · ¸ 1 ⇒ −u − v = 0 ⇒ u = 1, v = −1, and eigenvector is . −1

129

Linear Systems of Differential Equations and Their Phase Plane

·

¸

1 e−4t .

−1

General of equation is

· ¸ solution · ¸ the differential · ¸

x 1 1 = c1 e−2t + c2 e−4t . y −3 −1 ¸· ¸ ¾ · ¸ · dx x x −1 4 d dt = −x + 4y ⇒ dt = 41. dy y y −4 −1 dt·= −4x ¸ −y · ¸ · ¸ x u u λt ⇒ =e where λ is eigenvalue and is y v v

¸

· −1 4 corresponding eigenvector of A = .

−4 −1

The·eigenvalue, λ, satisfies ¸ det (A − λI) = 0 ⇒

−1 − λ 4 2 det = 0 ⇒ (−1 − λ) + 16 = 0

−4 −1 − λ 2 ⇒ (1 + λ) = −16 ¸ ±4i · ⇒ λ =¸ −1 · ±¸ 4i.

¾ ⇒ 1 ·+ λ = dx = 4x + 3y x x 4 3 d dt ⇒ dt 43. dy = y y 3 4 = 3x + 4y dt· · ¸ · ¸ ¸ x u u λt ⇒ where λ is eigenvalue and is =e y v v · ¸ 4 3 corresponding eigenvector of A = . 3 4

The·eigenvalue, λ, satisfies det (A − λI) = 0 ⇒

¸ 4−λ 3 2 = 0 ⇒ (4 − λ) − 9 = 0 det 3 4−λ ⇒ λ2 − 8λ + ·7 = 0¸ ⇒ (λ − 1)·(λ − 7) ⇒ λ = 1,¸ · 7. ¸ · ¸ u 0 4−λ u 3 = . Eigenvector, , satisfies v 0 ¸ 3· ¸4 − ·λ ¸ v · u 3 3 0 Eigenvector for λ = 1 : = ⇒ 3u + 3v = 0 ⇒ 3 3 v ¸ 0 · −1 v = 1, u = −1 and eigenvector is . Solution of the differential 1 · ¸ −1 equation is et . 1 ¸· ¸ · ¸ · u 0 −3 3 Eigenvector for λ = 7 : = 3 −3 v 0 · ¸ 1 . ⇒ −3u + 3v = 0 ⇒ u = 1, v = 1, and eigenvector is 1 · ¸ 1 Solution of the differential equation is e7t . 1

General of · ¸ the differential · ¸ equation is

¸ solution · x 1 −1 t e + c2 e7t . = c1 1 · 1

y ¸ · ¸

2 0 ,b= . 45. α = 0, β = 3, a = 0 1 Solution of the differential equation is

130

CHAPTER 4

·

¸

µ·

¸

·

¸

¶ x 2 0 = c1 e0t cos 3t − sin 3t

y 0 1 µ· ¸ ¶

· ¸ 2 0 0t cos 3t sin 3t + +c2 e 0 ¸ · 1 ¸ · 2 sin 3t 2 cos 3t = c1 + c2 . − sin 3t cos 3t · ¸ · ¸ 1 1 ,b= . 47. α = 1, β = 2, a = 0 3

µ· ¸ · ¸ ¶

· ¸ 1 1 x t cos 2t − = c1 e sin 2t

y µ·0 ¸ ·3 ¸ ¶

1 1 +c2 et sin 2t + cos 2t 0 3 · ¸ · ¸ cos 2t + sin 2t cos 2t − sin 2t . + c2 et = c1 et 3 cos 2t · −¸3 sin 2t · ¸ 1 1 49. α = 0, β = 5, a = ,b= . 0 1

· ¸ ¶

¸ · ¸ µ· x 1 1 0t sin 5t

cos 5t − = c1 e y µ·0 ¸ ·1 ¸ ¶

1 1 +c2 e0t sin 5t + cos 5t 0 1 · ¸ · ¸ cos 5t + sin 5t cos 5t − sin 5t . + c2 = c1 cos 5t − sin 5t 4.3 PHASE PLANE FOR LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 1. 2

y0

−2

3.

0 x

2

131

Linear Systems of Differential Equations and Their Phase Plane

2

y 0

−2

0 x

2

0

2

5. 2

y0

−2

·

x

¸ 1 0 . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 1 2 · ¸ 1−λ 0 det = 0 ⇒ (1 − λ) (2 − λ) = 0

1 2−λ ⇒ λ = 1, 2, both positive ⇒Unstable node.

¸ · ¸ · ¸ · u x u λt so that eigenvector, , satisfies =e v¸· v

¸ · ¸

· y u 0 0 1−λ = . 1 2−λ v · 0 ¸ · ¸ ¸· 0 0 0 u ⇒u+v =0⇒ Eigenvector for λ = 1 : = 1 1 · v ¸ 0 −1 v = 1, u = −1, and eigenvector is . Solution of the 1 · ¸ −1 et . differential equation is ¸· ¸ · ¸ ·1 u 0 −1 0 Eigenvector for λ = 2 : = ⇒ −u = 0 ⇒ 1 0 v 0

7. A =

132

CHAPTER 4

u = 0, v is free to choose, say, v = 1 and eigenvector is · ¸ 0 2t Solution of the differential equation is e . 1 Phase plane:

·

¸ 0 . 1

·

¸ −1 −5 . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 1 · 1 ¸ −1 − λ −5 det = 0 ⇒ (−1 − λ) (1 − λ) + 5 = 0 1 1−λ ⇒ λ2 + 4 ⇒ λ = ±2i, complex with 0 real part ⇒Stable center.

9. A =

y 0

0 x

¸ 1 −1 . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ · 1 1 ¸ 1−λ −1 2 = 0 ⇒ (1 − λ) + 1 = 0 det 1 1−λ 2 ⇒ (1 − λ) = −1 ⇒ 1 − λ = ±i ⇒ λ = 1 ± i, complex with positive ¸ ⇒ Unstable spiral.

· real part −5 −4 . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 13. A = 1 · 2 ¸ −5 − λ −4 det = 0 ⇒ (−5 − λ) (1 − λ) + 8 = 0 2 1−λ ⇒ λ2 + 4λ + 3 =·0 ⇒¸(λ + 1)·(λ +¸3) ⇒ λ = −1, −3, both negative u x so that eigenvector, = eλt ⇒ Stable node. v y

11. A =

·

133

Linear Systems of Differential Equations and Their Phase Plane

·

u v

¸

·

¸·

¸

·

¸

u 0 −5 − λ −4 = . 2 1−λ v 0 ¸ ¸ · · ¸· 0 u −4 −4 = Eigenvector for λ = −1 : 2 2 v 0 · ¸ −1 ⇒ −4u − 4v = 0 ⇒ v = 1, u = −1 and eigenvector is . 1 · ¸ −1 e−t . Solution of the differential equation is · ¸· 1 ¸ · ¸ −2 −4 u 0 Eigenvector for λ = −3 : = 2 4 v 0 · ¸ −2 ⇒ −2u − 4v = 0 ⇒ v = 1, u = −2, and eigenvector is . 1 ¸ · −2 e−3t . Solution of the differential equation is 1

·

, satisfies

¸ 0 1 . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 15. A = · 2 1 ¸ −λ 1 det = 0 ⇒ −λ (1 − λ) − 2 = 0 2 1−λ ⇒ λ2 − λ − 2 = 0 ⇒ (λ + 1) (λ − 2) ⇒ λ = ·−1, 2, ¸ one positive · ¸ x u λt and one negative ⇒Unstable saddle point. =e so y ¸· · ¸ · ¸ v· ¸ u u 0 −λ 1 that eigenvector, , satisfies = . v 2 1−λ v 0 · ¸ · ¸ ¸· 1 1 0 u ⇒u+v =0⇒ = Eigenvector for λ = −1 : 0 2 2· v¸ 1 u = 1, v = −1 and eigenvector is . Solution of the differential −1 · ¸ 1 e−t .

equation is −1 · ¸· ¸ · ¸

u 0 −2 1 Eigenvector for λ = 2 : = 2 −1 v 0 · ¸ 1 ⇒ −2u + v = 0 ⇒ u = 1, v = 2, and eigenvector is . 2

134

CHAPTER 4

Solution of the differential equation is

¾

d dt

·

x y

·

1 2

¸

e2t .

¸· ¸ x −5 −1 = ⇒ 17. y 3 −1 ¸ · ¸ · u u x are eigenvalue and where λ, = eλt ⇒ y v v

· ¸ −5 −1 corresponding eigenvector, respectively, of A = .

3 −1 The·eigenvalue, λ, satisfies ¸ det (A − λI) = 0 ⇒ −5 − λ −1 det = 0 ⇒ (−5 − λ) (−1 − λ) + 3 = 0 3 −1 − λ ⇒ λ2 + 6λ + 8 = 0 ⇒ (λ + 2) (λ + 4) ⇒ λ = −2, −4, both negative ⇒ Stable node. ¸ · ¸ ¸ · ¸· · 0 u u −5 − λ −1 = . Eigenvector, , satisfies 0 v 3 ¸ · −1 ¸− λ · · ¸v −3 −1 u 0 Eigenvector for λ = −2 : = 3 1 v 0· ¸ 1 ⇒ −3u − v = 0 ⇒ u = 1, v = −3 and eigenvector is . −3 · ¸ 1 Solution of the differential equation is e−2t .

−3 · ¸· ¸ · ¸

−1 −1 u 0 Eigenvector for λ = −4 : = 3 3 v 0 · ¸ 1 ⇒ −u − v = 0 ⇒ u = 1, v = −1, and eigenvector is . −1 · ¸ 1 Solution of the differential equation is e−4t .

−1

dx dt = −5x − y dy −y dt ·= 3x ¸

¸

·

Linear Systems of Differential Equations and Their Phase Plane

135

¾ · ¸ · ¸ · ¸ −x + 4y x x −1 4 d ⇒ dt = 19. y −4 −1 y −4x − y ¸ · ¸ · ¸ x u u ⇒ = eλt where λ, are eigenvalue and y v v ¸ · −1 4 corresponding eigenvector, respectively, of A = . −4 −1 The·eigenvalue, λ, satisfies ¸ det (A − λI) = 0 ⇒

−1 − λ 4 2 det = 0 ⇒ (−1 − λ) + 16 = 0 −4 −1 − λ 2 ⇒ (1 + λ) = −16 ⇒ 1 + λ = ±4i ⇒ λ = −1 ± 4i, complex with negative ¸ ·⇒Stable¸ ·spiral.

¸

·real part ¾ dx = 4x + 3y x x 4 3 d dt ⇒ dt 21. dy = y y 3 4 dt·= 3x ¸ + 4y · ¸ · ¸ x u u ⇒ = eλt where λ, are eigenvalue and y v v · ¸ 4 3 corresponding eigenvector, respectively, of A = . 3 4

The·eigenvalue, λ, satisfies det (A − λI) = 0 ⇒

¸ 4−λ 3 2 det = 0 ⇒ (4 − λ) − 9 = 0

3 4−λ ⇒ λ2 − 8λ + 7 = 0 ⇒ (λ − 1) (λ − 7) ⇒ λ = 1, 7, both positive ⇒Unstable node. · ¸ · ¸· ¸ · ¸ 4−λ u 0 u 3 Eigenvector, , satisfies = . v 3 4−λ v 0 · ¸· ¸ · ¸ 3 3 u 0 ⇒ 3u + 3v = 0 ⇒ = Eigenvector for λ = 1 : 3 3 · v ¸ 0 −1 v = 1, u = −1 and eigenvector is . Solution of the differential 1 · ¸ −1 equation is et . 1 · ¸· ¸ · ¸ u 0 −3 3 Eigenvector for λ = 7 : = 3 −3 v 0 · ¸ 1 . ⇒ −3u + 3v = 0 ⇒ u = 1, v = 1, and eigenvector is 1 dx dt = dy dt·=

136

CHAPTER 4

Solution of the differential equation is

·

0 23. A = −4 · −λ det −4 0 real part

·

·

1 1

¸

e7t .

¸ 1 . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 0 ¸ 1 = 0 ⇒ λ2 + 4 = 0 ⇒ λ = ±2i, complex with −λ ⇒Stable center.

¸ 2 1 25. A = . The eigenvalue, λ, satisfies det (A − λI) = 0 1 ¸ · 2

1 2−λ 2

⇒ det = 0 ⇒ (2 − λ) − 1 = 0 1 2−λ ⇒ node. · 2− ¸ λ = ±·1 ⇒ ¸λ = 3, 1, both positive· ⇒ Unstable ¸ u x u so that eigenvector, = eλt , satisfies v v

· y ¸ · ¸

¸· 2−λ 0 u 1 = . 1 2−λ v · 0 ¸· ¸ · ¸ u 0 −1 1 Eigenvector for λ = 3 : = 1 −1 v 0· ¸ 1 ⇒ −u + v = 0 ⇒ u = 1, v = 1 and eigenvector is . 1 · ¸ 1 Solution of the differential equation is e3t . ¸· ¸ 1· ¸ · u 0 1 1 Eigenvector for λ = 1 : = ⇒u+v =0⇒ 1 1 v 0 · ¸ 1 u = 1, v = −1 and eigenvector is . Solution of the −1

137

Linear Systems of Differential Equations and Their Phase Plane

differential equation is

·

¸

1 et .

−1

·

¸ 3 −1 27. A = . The eigenvalue, λ, satisfies det (A − λI) = 0 ⇒ 1 2 · ¸ 3−λ −1 = 0 ⇒ (3 − λ) (2 − λ) + 1 = 0 det 1 2−λ √

⇒ λ2 − 5λ + 7 = 0 ⇒ λ = 5± 25−28 = 2 positive real part ⇒ Unstable spiral.

·

5 2

±

√

3 2 i,

complex with

¸ 1 0 . The eigenvalue, λ, satisfies det (A − λI) = 0 1 −3 · ¸ 1−λ 0 ⇒ det = 0 ⇒ (1 − λ) (−3 − λ) = 0

1 −3 − λ

⇒ λ = 1, −3, one positive ·and one ¸ negative · ¸ x u so that eigenvector, ⇒ Unstable saddle point. = eλt y v

· ¸ · ¸· ¸ · ¸

0 u 1−λ u 0 , satisfies = . v 1 · −3 − λ v 0

¸· ¸ · ¸

u 0 0 0 Eigenvector for λ = 1 : = ⇒ u − 4v = 0 ⇒ 1 −4 v 0 · ¸ 4 v = 1, u = 4 and eigenvector is . Solution of the differential 1

· ¸ 4 et . equation is 1 ¸· ¸ · ¸ · u 0 4 0 Eigenvector for λ = −3 : = ⇒ 4u = 0 1 0 v 0

29. A =

138

CHAPTER 4

u = 0, v is free to choose, say, v = 1 and eigenvector is · ¸ 0 Solution of the differential equation is e−3t . 1

·

·

0 1

¸

.

¸ 4 −3 . The eigenvalue, λ, satisfies det (A − λI) = 0 1 ¸

· 0

4 − λ −3 ⇒ det = 0 ⇒ −λ (4 − λ) + 3 = 0 1 −λ ⇒ λ2 − 4λ + 3 = 0·⇒ (λ · − 3) ¸ = 0 ⇒ λ = 1, 3, both positive ¸ − 1) (λ x u so that eigenvector, ⇒Unstable node. = eλt y v

¸ · ¸ ¸· · ¸ · u 0 4 − λ −3 u = .

, satisfies v 1 · −λ v 0 ¸· ¸ · ¸ 3 −3 u 0 Eigenvector for λ = 1 : = ⇒ 3u − 3v = 0 ⇒ 1 −1 0 · ¸v 1 u = 1, v = 1 and eigenvector is . Solution of the differential 1

· ¸ 1 et . equation is 1 ¸· ¸ · ¸ · u 0 1 −3 Eigenvector for λ = 3 : = ⇒ u − 3v = 0 1 −3 0 · ¸v 1 u = 1, v = 3 and eigenvector is . Solution of the differential 3 · ¸ 1 equation is e3t . 3

31. A =

Linear Systems of Differential Equations and Their Phase Plane

·

139

¸ 2 −1 33. A = . The eigenvalue, λ, satisfies det (A − λI) = 0 1 0

¸

· 2 − λ −1 = 0 ⇒ −λ (2 − λ) + 1 = 0

⇒ det 1 −λ

2 ⇒ λ2 − 2λ + 1 = 0 ⇒ (λ − 1) ⇒ λ = 1, 1. Not ·two distinct ¸ eigenvalues.

1 0 35. A = . The eigenvalue, λ, satisfies det (A − λI) = 0 0 −3 · ¸ 1−λ 0 ⇒ det = 0 ⇒ (1 − λ) (−3 − λ) = 0 0

−3 − λ ⇒ λ = 1, −3, one positive ·and one ¸ ¸ negative

· u x so that eigenvector, = eλt ⇒ Unstable saddle point. y ¸· ¸v · ¸

¸ · · u 0 1−λ 0 u , satisfies = . v 0 · −3 − λ ¸ · v ¸ · 0 ¸

u 0 0 0 Eigenvector for λ = 1 : = ⇒ −4v = 0 ⇒ 0 −4 v 0 · ¸ 1 v = 0, u is free to choose, say, u = 1 and eigenvector is . 0 · ¸ 1 Solution of the differential equation is et . 0 · ¸ · ¸ ¸· 0 u 4 0 = ⇒ 4u = 0 ⇒ Eigenvector for λ = −3 : 0 0 0 v · ¸ 0 u = 0, v is free to choose, say, v = 1 and eigenvector is . 1 · ¸ 0 Solution of the differential equation is e−3t . 1

140

CHAPTER 4

·

¸ 1 0 . det A = −3, trA = −2. 1 −3 2 Now 4 det A = −12 < 4 = (trA) ⇒ eigenvalues are real. Also det unstable saddle point. · A < 0 implies ¸ 4 −3 (31) A = . det A = 3, trA = 4. 1 0 2 Now 4 det A = 12 < 16 = (trA) ⇒ eigenvalues are real. Also det and trA > 0 imply unstable node. · A>0¸ 2 −1 (33) A = . det A = 1, trA = 2. 1 0 2 Now 4·det A = 4¸ = (trA) ⇒ eigenvalues are real (repeated). 1 0 (35) A = . det A = −3, trA = −2. 0 −3 2 Now 4 det A = −12 < 4 = (trA) ⇒ eigenvalues are real. Also det A < 0 implies unstable saddle point. 39. 37. (29) A =

41.

Linear Systems of Differential Equations and Their Phase Plane

141

Chapter Five

Mostly Nonlinear First-Order Differential Equations

5.1 FIRST-ORDER DIFFERENTIAL EQUATIONS There are no exercises in this section.

5.2 EQUILIBRIA AND STABILITY dx dt

= x (x − 1) = f (x) Equilibria: f = 0 ⇒ x (x − 1) = 0 ⇒ x = 0, 1. 2 3. dx = (x − 1) (x − 2) = f (x) dt 2 Equilibria: f = 0 ⇒ (x − 1) (x − 2) = 0 ⇒ x = 1, 2. 5. dx dt = sin x = f (x) Equilibria: f = 0 ⇒ sin x = 0 ⇒ x = nπ, for n = 0, ±1, ±2, ... = x (a − bx) = f (x) 7. dx dt Equilibria: f = 0 ⇒ x (a − bx) = 0 ⇒ x = 0, ab . 2 9. dx dt = x (x − 1) = f (x) 2 Equilibria: f = 0 ⇒ x (x − 1) = 0 ⇒ x = 0, 1. f ′ (x) = 3x2 − 4x + 1. For x = 0, f ′ (0) = 1 > 0

⇒ x = 0 is an unstable equilibrium.

Solution of the linearized differential equation

dy ′ ′ dt = f (xe ) y = f (0) y = y for this equilibrium is

′ y = x − 0 = cef (0)t ⇒ x = cet . Graph:

For x¡= 1, f ¢′ (1) ¡ = 0 ¢⇒ Linear analysis is inconclusive.

dx 11. dt = x2 − 1 x2 − 3 = f (x) . √ ¡ 2 ¢¡ ¢ Equilibria: ¡ f =¢0 ⇒ x −¢1 x2 − 3 = 0 ⇒ x = ±1, ± 3.

¡ f ′ (x) = 2x x2 − 3 + x2 − 1 2x. For x = 1, f ′ (1) = −4 < 0 ⇒ x = 1 is a stable equilibrium.

Solution of the linearized differential equation dy ′ dt = f (xe ) y = −4y for this equilibrium is ′ y = x − 1 = cef (1)t = ce−4t ⇒ x = ce−4t + 1. Graph:

For x = −1, f ′ (−1) = 4 > 0 ⇒ x = −1 is an unstable equilibrium.

Solution of the linearized differential equation for this equilibrium is ′ y = x + 1 = cef (−1)t = ce4t ⇒ x = ce4t − 1.

1.

143

MOSTLY NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS

Graph: √ √ √ ¡√ ¢ For x = 3, f ′ 3 = 4 3 > 0 ⇒ x = 3 is an unstable

equilibrium. Solution of the linearized differential equation for this equilibrium is √ √ √ √ √ ′ y = x − 3 = cef ( 3)t = ce4 3t ⇒ x = ce4 3t + 3.

Graph: √ √ √ ¡ √ ¢ For x = − 3, f ′ − 3 = −4 3 < 0 ⇒ x = − 3 is a stable

equilibrium. Solution of the linearized differential equation for this equilibrium is √ √ √ √ √ ′ y = x + 3 = cef ( 3)t = ce−4 3t ⇒ x = ce−4 3t − 3. Graph: 13. dx dt = tan x = f (x) .

Equilibria: f = 0 ⇒ tan x = 0 ⇒ x = nπ, for all integer n.

f ′ (x) = sec2 x and f ′ (nπ) = 1 > 0 for all integer n.

So all equilibria are unstable.

Solution of the linearized differential equation

dy ′ ′ dt = f (xe ) y = f (nπ) y = y is f ′ (nπ)t y = x − nπ = ce = cet ⇒ x = cet + nπ for all integer n.

dx −3x 15. dt = e − 5 = f (x) . Equilibria: f = 0 ⇒¡e−3x = ¢5 ⇒ x = − 13 ln 5. ′ −3x f (x) = −3e and f ′ − 13 ln 5 = −15 < 0 . 1 So x = − 3 ln 5 is a stable equilibrium. Solution of the linearized ¡ 1 ¢ ′ ′ differential equation dy dt = f (xe ) y = f − 3 ln 5 y = −15y is ′ 1 y = x + 1 ln 5 = cef (− 3 ln 5)t = ce−15t ⇒ x = ce−15t − 1 ln 5. 3

3

5.3 ONE DIMENSIONAL PHASE LINES 1.

dx dt

= x (x − 1) = f (x) Phase line: dx dt

x 1

Equilibria: f = 0 ⇒ x (x − 1) = 0 ⇒ x = 0, 1. From the graph above, x = 0 is stable and x = 1 is unstable.

144 3.

CHAPTER 5 dx dt

2

= (x − 1) (x − 2) = f (x) Phase line: dx dt 5

x 0

1

2

2

Equilibria: f = 0 ⇒ (x − 1) (x − 2) = 0 ⇒ x = 1, 2. From the graph above, x = 1 is unstable and x = 2 is unstable. 5. dx dt = sin x = f (x)

Phase line: see text page 411.

Equilibria: f = 0 ⇒ sin x = 0 ⇒ x = nπ, for n = 0, ±1, ±2, ... From the graph of the phase line, x = nπ is unstable if n is even integer 7.

and x = nπ is stable if n is odd integer. = x2 + 1 = f (x)

Phase line:

dx dt

dx dt

x

Equilibria: f = 0 ⇒ x2 + 1 = 0 has no real solution (there is no equilibrium in this problem). 3 9. dx dt = x = f (x) Phase line:

MOSTLY NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS

dx dt

x

Equilibria: f = 0 ⇒ x3 = 0 ⇒ x = 0. From the graph above, x = 0 is unstable. 4 11. dx dt = (x − 3) = f (x) . Phase line: dx dt

x 3 4

Equilibria: f = 0 ⇒ (x − 3) = 0 ⇒ x = 3. From the graph above, x = 3 is unstable. 13. dx dt = (x − 1) (x − 2) (x − 3) (x − 4) = f (x) . Phase line:

Equilibria: f = 0 ⇒ (x − 1) (x − 2) (x − 3) (x − 4) = 0 ⇒ x = 1, 2, 3, 4. From the graph above, x = 1 and 3 are stable but x = 2 and 4 are unstable. 15.

145

146

CHAPTER 5

x

1

t

0

17. x 3 1 0

t

−1 −

19.

dx dt

21.

dx dt

3

= x2 . Separation of variables gives dx x2 = dt which, on integration, yields − x1 = t + c ⇒ x = − t+1c . Note that c > 0 if x (0) < 0 and c < 0 if x (0) > 0. If x (0) > 0 and x (0) is close to zero, then x (t) increases since 1 dx dt = (t+c)2 > 0 and thus the solution is not stable.

= x2 . Separation of variables gives dx x2 = dt which, 1 on integration, yields − x1 = t + c ⇒ x (t) = − t+c for t ≥ 0. 1 Initial condition 0 > x (0) = − c (thus c > 0). Then from x (t) = − t+1c , we see that x (t) has a vertical asymptote at t = −c which is less than zero (since c > 0). Thus there are no vertical asymptotes for t ≥ 0. From x (t) = − t+1c , as t → ∞, x (t) → 0, but x (t) = 6 0 for finite t.

5.4 APPLICATION TO POPULATION DYNAMICS: THE LO­ GISTIC EQUATION 1.

¡ ¢ = ax − bx3 = a − bx2 x = kx where k = a − bx2 is the growth rate with a > 0, b > 0, x ≥ 0. p (a) k = a − bx2 > 0 ⇒ bx2 < a ⇒ x < ab . That is, growth rate is

pa positive if the population x < b .

dx dt

MOSTLY NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS

pa

k = a − bx2 < 0 ⇒ bx2 > a ⇒ x > p negative if the population x > ab . (b)

b.

147

That is, growth rate is

dx dt

x a b

(c) x

a b

t

0

3.

dx dt

= x (a − bx) = ax − bx2 . Using partial fraction we write 1 A B x(a−bx) = x + a−bx . Multiplying by the denominator gives 1 = A (a − bx) + Bx. Letting x = 0 and x = ab we have

R R dx

A = a1 and B = ab . Then x(a−bx) = dt ⇒

R R R dx b dx 1 dt ⇒ a1 ln |x| − a1 ln |a − bx| = t + c

a x + a a−bx = ¯ ¯ ³ ´ a1 ¯ x ¯ x x =e cet ⇒ a−bx = ceat ⇒ a1 ln ¯ a−bx ¯ = t + c ⇒ a−bx at at ⇒ x = ace − bce x a ac aceat b . ⇒ x (t) = 1+bce at = bc+e−at = 1 1+[ bc ]e−at a x(0) ac So x (0) = 1+b 1 = 1+bc ⇒ c = a−bx(0) . bc

Thus x (t) = 1+

5.

dx dt

"

a b

1 bx(0) a−bx(0)

#

= e−at

1+[

a

b a−bx(0) bx(0)

]e−at

.

= x (a − bx) . There are three parameters here: a, b, x (0) . b Let x = a

b y (Note y (0) = a x (0)). Then differentiation gives a dy dx a a b dt = dt = x (a − bx) = b y (a − ay) = b ya (1 − y) dy ⇒ dt = ay (1 − y) which has two parameters, a and y (0) = ab x (0) .

148

CHAPTER 5 dx dt

= ax + bx2 = (a + bx) x = kx where k = a + bx is the growth rate with a > 0, b > 0, x ≥ 0. (a) growth rate k = a + bx is an increasing function because it is linear with positive slope b. So growth rate increases as the population x increases. (b) 7.

dx dt

x

(c) x Finite time explosion (see d)

t

0

(d)

dx dt R

= x (a + bx) = ax + bx2 . Separation of variables gives R dx dt. Using integration table we get, x(a+bx) = ¯ ¯ ´ a1 ³ ¯ x ¯ 1 x ln = t + =e cet ⇒ a+xbx = ceat c ⇒ ¯ ¯ a a+bx a+bx

⇒ x = aceat + bceat x ⇒ x (t) = x(0) ac 1−bc ⇒ c = a+bx(0) . aceat = 1−bce at = ∞ when

aceat 1−bceat .

So x (0) =

Now x (t)

eat =

1 bc .

For this to occur

when t > 0, we need bc < 1. However, bc = since x (0) ≥ 0. a (t−t ) − a (t−t0 ) 0 9. (a) x (t) = α + β e a22 (t−t0 ) −e− a22 (t−t0 ) =

bx(0) a+bx(0)

e +e i h a i h a a a α e 2 (t−t0 ) +e− 2 (t−t0 ) +β e 2 (t−t0 ) −e− 2 (t−t0 ) a

a

e 2 (t−t0 ) +e− 2 (t−t0 )

< 1,

. a

Multiplying the numerator and denominator by e− 2 (t−t0 ) we have α[1+e−a(t−t0 ) ]+β [1−e−a(t−t0 ) ] x (t) = and comparing this with 1+e−a(t−t0 ) equation (2), x (t) =

1+

h

a b a−bx0 bx0

i

e−at

, we obtain

149

MOSTLY NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS α[1+e

−a(t−t0 )

]+β [1−e 1+e−a(t−t0 )

−a(t−t0 )

]

=

1+

Equating Numerator:

h

a b a−bx0 bx0

i

e−at

.

a 0) α + β + (α − β) e−a(t−t =

h ib 0 Denominator: 1 + e−a(t−t0 ) = 1 + a−bx e−at . bx0 From the first equation we equate exponential and nonexponential terms to have α + β = ab and α − β = 0. Adding these, we get α = 2ab and then β = 2ab . h i 0 From the second equation, 1 + e−at eat0 = 1 + a−bx e−at ⇒ bx0 ¯ ¯ ¯ ¯ ¯ a−bx0 ¯ ¯ a−bx0 ¯ 1 0 eat0 = a−bx bx0 ⇒ at0 = ln ¯ bx0 ¯ ⇒ t0 = a ln ¯ bx0 ¯ . a (t−t ) 0

− a (t−t0 )

2 sinh

a

(t−t )

(b) x (t) = α + β e a22 (t−t0 ) −e− a22 (t−t0 ) = α + β 2 cosh 2a (t−t00 ) e +e 2 = α + β tanh a2 (t − t0 ) . (c) tanh t 1

t −1

Asymptotic behavior: lim tanh t = 1 and lim tanh t = −1. t→∞

(d)

t→−∞

x(t) a b

α+β= a b α−β=0

(e) See part (a). 11. f (Q) = a + bQ + cQ2 . (i) f (0) = 0 ⇒ a = 0 (ii) f ′ (Q) = b + 2cQ ≈ b for Q small. So b > 0 for f ′ (Q) > 0. (iii) f ′ (Q) = b + 2cQ ≈ 2cQ for Q large. So c < 0 for f ′ (Q) < 0. Hence f (Q) = bQ + cQ2 with b > 0, c < 0 and thus f (Q) is a logistic equation.

Chapter Six

Nonlinear Systems of Differential Equations in the Plane

6.1 INTRODUCTION There are no exercises in this section.

6.2 EQUILIBRIA OF NONLINEAR SYSTEMS, LINEAR STABILITY ANALYSIS OF EQUILIBRIUM, AND PHASE PLANE 1.

dx dt = x (1 + y) = x + xy = f dy

dt = y (2 − 4x) = 2y − 4xy

=g (a) Equilibria: f = 0 and g = 0 ⇒ x (1 + y) = 0 ⇒Either x = 0 or y = −1 y (2 − 4x) = 0 ⇒If x = 0, then¡ y = 0.¢ If y = −1, then x = 12 . Thus"equilibria are 0) and 21 , −1 . # (0, · ¸ ∂f ∂f 1+y x ∂x ∂y = . (b) A = ∂g ∂g −4y 2 − 4x ∂x ∂y · ¸ 1 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 2 · ¸ 1−λ 0 det (A − λI) = 0 ⇒ det =0 0 2−λ ⇒ (1 − λ) (2 − λ) = 0 ⇒ λ = 1, 2 (both positive). So the equilibrium · ¸ · ¸ ¸ (0, 0) is an· unstable node. ¸ · 0 u 0 1−λ u = . Eigenvector, , satisfies v 0 ¸ 0· ¸2 − ·λ ¸ v · u 0 0 0 Eigenvector for λ = 1 : = ⇒ v = 0. 0 1 · v ¸ · 0¸ u 1 u is free to choose, say, u = 1 ⇒ = . v 0 · ¸ · ¸ ¸· −1 0 0 u ⇒ u = 0. Eigenvector for λ = 2 : = 0 0· ¸v · ¸0 u 0 v is free to choose, say, v = 1 ⇒ = . v 1

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

For equilibrium

(c)

¡1

¢

dy dx

=

dy dt dx dt

=

,A=

0 4 ·

¸

1 2

151

. The eigenvalue, λ, 0 ¸ −λ 21 satisfies det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 − 2 = 0 4 −λ √ ¡ ¢ ⇒ λ = ± 2 (one positive, one negative). So the equilibrium 21 , −1 is a saddle point. ¸· ¸ · ¸ · ¸ · u 0 u −λ 12 = . Eigenvector, , satisfies v 0 v 4 −λ · √ ¸ · ¸ ¸· 1 √ 0 u − 2 2 √ Eigenvector for λ = 2 : = v 0 4 − 2 √ √ 1 ⇒ − 2u + 2 v = 0. Let u = 1. Then v = 2 2. ¸ · ¸ · 1 u √ So = . v 2 2 · √ ¸· ¸ · ¸ √ u 0 2 √12 ⇒ = Eigenvector for λ = − 2 : v 0 2 4 √ √ 2u + 21 v = 0. Let u = 1. Then v = −2 2. · ¸ · ¸ 1√ u So = . v −2 2

(e) Nullclines:

2 , −1

·

y(2−4x) x(1+y) .

dy So dx = 0 ⇒ y (2 − 4x) = 0 ⇒ y = 0, x = 12 dy and dx = ∞ ⇒ x (1 + y) = 0 ⇒ x = 0, y = −1.

y

x y = −1

1 x=— 2

152 3.

CHAPTER 6

dx dt = 2x − 2xy = 2x (1 − y) dy

dt = y − xy = y (1 − x) = g

=f

(a) Equilibria: f = 0 and g = 0 ⇒ 2x (1 − y) = 0 ⇒Either x = 0 or y = 1 y (1 − x) = 0 ⇒If x = 0, then y = 0. If y = 1, then x = 1. Thus"equilibria are 0) and (1, 1) . # (0, · ¸

∂f ∂f

2 − 2y −2x ∂x ∂y = .

(b) A = ∂g ∂g −y 1 − x

∂x ∂y · ¸ 2 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 1 ¸ · 2−λ 0 =0 det (A − λI) = 0 ⇒ det 0 1−λ ⇒ (1 − λ) (2 − λ) = 0 ⇒ λ = 1, 2 (both positive). So the equilibrium · ¸ (0, 0) is an· unstable node. ¸ · ¸ · ¸ u 2−λ u 0 0 Eigenvector, , satisfies = . v 0 1−λ v 0 · ¸· ¸ · ¸ 1 0 u 0 ⇒ u = 0. Eigenvector for λ = 1 : = 0 0 · v¸ · 0¸ u 0 v is free to choose, say, v = 1 ⇒ = . v 1 · ¸ · ¸ ¸· 0 u 0 0 ⇒ v = 0. Eigenvector for λ = 2 : = 0 −1· ¸v · ¸0 u 1 u is free to choose, say, u = 1 ⇒ = . v 0 ¸ · 0 −2 For equilibrium (1, 1) , A = . The eigenvalue, λ, satisfies −1 0 · ¸ −λ −2 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 − 2 = 0 −1 −λ √ ⇒ λ = ± 2 (one positive, one negative).

So the equilibrium saddle point.¸ · · ¸ (1, 1) is a · ¸ · ¸ u u 0 −λ −2 Eigenvector, , satisfies = . v −1 −λ v 0 · √ ¸· ¸ · ¸ √ − 2 −√2 u 0 = Eigenvector for λ = 2 : v 0 −1 − 2 √ √ ⇒− = 0.¸ Let u = 2. Then v = −1.

· 2u¸ − 2v · √ u 2 is unstable direction. So = v −1 · √ ¸· ¸ · ¸ √ 2 √ −2 u 0 ⇒ Eigenvector for λ = − 2 : = v 0 −1 2 ¸ · √ ¸ · √ √ u 2 = 2u − 2v = 0. Let u = 2. Then v = 1. So v 1 is stable direction.

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

153

(c)

(e) Nullclines:

dy dx

=

dy dt dx dt

=

y(1−x) 2x(1−y) .

dy So dx = 0 ⇒ y (1 − x) = 0 ⇒ y = 0, x = 1 dy and dx = ∞ ⇒ 2x (1 − y) = 0 ⇒ x = 0, y = 1.

y

y=1

x

x=1

5.

dx dt = x − y = f dy

dt = −2x + 2xy

= −2x (1 − y) = g (a) Equilibria: f = 0 and g = 0 ⇒ x−y =0⇒ x=y −2x (1 − y) = 0 ⇒ −2x (1 − x) = 0 ⇒ x = 0 or x = 1. Then"y = 0 or y#= 1. Thus equilibria are (0, 0) and (1, 1) . ¸

· ∂f ∂f

1 −1 ∂y .

= (b) A = ∂x ∂g ∂g −2 + 2y 2x ∂x ∂y

· ¸ 1 −1 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies −2 0 ¸ · 1 − λ −1 = 0 ⇒ −λ (1 − λ) − 2 = 0 det (A − λI) = 0 ⇒ det −2 −λ 2 ⇒ λ − λ − 2 = 0 ⇒ (λ + 1) (λ − 2) ⇒ λ = −1, 2 (one positive, one negative). So the equilibrium saddle point. ¸ · · ¸ (0, 0) is a · ¸ · ¸ u 1 − λ −1 u 0 Eigenvector, , satisfies = . v −2 −λ v 0

154

CHAPTER 6

2 −2

−1 1

¸·

¸

·

¸

u 0 = ⇒ v 0 ¸ · ¸ · 1 u = is stable 2u − v = 0. Let u = 1. Then v = 2 ⇒ v 2 direction. ¸ · ¸ ¸· · u 0 −1 −1 = ⇒ Eigenvector for λ = 2 : v· −2 −2 ¸ 0· ¸ u 1 −u − v = 0. Let u = 1. Then v = −1 ⇒ = v −1 is unstable direction. ¸ · 1 −1 . The eigenvalue, λ, satisfies For equilibrium (1, 1) , A = 0 2 · ¸ 1−λ −1 det (A − λI) = 0 ⇒ det =0 0 2−λ ⇒ (1 − λ) (2 − λ) = 0 ⇒ λ = 1, 2 (both positive). So the equilibrium · ¸ · ¸ ¸ (1, 1) is an· unstable node. ¸ · −1 u u 0 1−λ = . Eigenvector, , satisfies 0¸ · 2 ¸− λ · 0 v ¸v · u 0 0 −1 Eigenvector for λ = 1 : = ⇒ v = 0. 0 1 v 0 · ¸ · ¸ u 1 u is free to choose, say, u = 1. So = . v 0 ¸· ¸ · ¸ · u 0 −1 −1 Eigenvector for λ = 2 : = ⇒ 0 0 v· ¸ 0· ¸ u 1 = . −u − v = 0. Let u = 1. Then v = −1. So −1 v

Eigenvector for λ = −1 :

(c)

·

(e) Nullclines:

dy dx

=

dy dt dx dt

=

−2x(1−y) . x−y

dy So dx = 0 ⇒ −2x (1 − y) = 0 ⇒ x = 0, y = 1 dy and dx = ∞ ⇒ x − y = 0 ⇒ x = y.

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

155

y y=x

y=1

x

7.

dx 2 dt = 1 − y = f dy

2 dt = 1 − x = g

(a) Equilibria: f = 0 and g = 0 ⇒

1 − y 2 = 0 ⇒ y = ±1 1 − x2 = 0 ⇒ x = ±1.

Thus"equilibria are 1) , (−1, −1) , (1, −1) and (−1, 1) .

# (1, · ¸

∂f ∂f

0 −2y

∂x ∂y . (b) A = = ∂g ∂g −2x 0

∂x ∂y · ¸ 0 −2 For equilibrium (1, 1) , A = . The eigenvalue, λ, satisfies −2 ¸0 · −λ −2 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 − 4 = 0 ⇒ λ2 = 4 −2 −λ ⇒ λ = −2, 2 (one positive, one negative). So the equilibrium saddle point.

· ¸· ¸ · ¸

¸ (1, 1) is a · u u 0 −λ −2 = . Eigenvector, , satisfies v −2 −λ v 0 · ¸ · ¸ ¸· 0 u −2 −2 ⇒ Eigenvector for λ = 2 : = −2 −2 v · 0 ¸ · ¸ u 1 −2u − 2v = 0. Let u = 1. Then v = −1 ⇒ = v −1 is unstable direction.

· ¸· ¸ · ¸

2 −2 u 0 Eigenvector for λ = −2 : = ⇒ −2 2 v 0 ¸ · ¸ · u 1 2u − 2v = 0. Let u = 1. Then v = 1 ⇒ = v 1 is stable direction.

· ¸

0 2 For equilibrium (−1, −1) , A = . The eigenvalue, λ, satisfies 2 0 · ¸ −λ 2 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 − 4 = 0 ⇒ λ2 = 4 2 −λ ⇒ λ = −2, 2 (one positive, one negative). So the equilibrium (−1, −1) is a saddle point.

156

CHAPTER 6

·

u v

¸

·

¸·

¸

·

¸

u 0 −λ 2 = . 2 −λ v 0 ¸ · ¸ ¸· 0 u 2 = −2 v 0 ¸ · ¸ · 1 u ⇒ −2u + 2v = 0. Let u = 1. Then v = 1 ⇒ = v 1 is unstable direction.

· ¸· ¸ · ¸

u 0 2 2 Eigenvector for λ = −2 : = ⇒ 2 2 v · ¸0 · ¸ u 1 2u + 2v = 0. Let u = 1. Then v = −1 ⇒ = v −1 is stable direction.

· ¸

0 2 For equilibrium (1, −1) , A = . The eigenvalue, λ, satisfies −2 0 · ¸ −λ 2 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 + 4 = 0 −2 −λ ⇒ λ = ±2i (complex).

may be different).

So the equilibrium (1, −1) is a· center (nonlinear ¸ 0 −2 For equilibrium (−1, 1) , A = . The eigenvalue, λ, satisfies

2 0 ¸ · −λ −2 = 0 ⇒ λ2 + 4 = 0 det (A − λI) = 0 ⇒ det 2 −λ ⇒ λ = ±2i (complex).

So the equilibrium (−1, 1) is a center (nonlinear may be different).

Eigenvector,

, satisfies · −2 Eigenvector for λ = 2 : 2

(c)

(e) Nullclines:

dy dx

=

dy dt dx dt

=

1−x2 1−y 2 .

dy So dx = 0 ⇒ 1 − x2 = 0 ⇒ x = ±1 dy and dx = ∞ ⇒ 1 − y 2 = 0 ⇒ y = ±1.

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

157

y

x

9.

dx 2 dt = x − y + x dy

dt = x + y = g

=f

(a) Equilibria: g = 0 ⇒ x + y = 0 ⇒ y = −x

f = 0 ⇒ x − y + x2 = 0 ⇒ 2x + x2 = 0 ⇒ x (2 + x) = 0 ⇒ x = 0 or −2. So y = 0 or 2. Thus"equilibria are 0) and (−2, 2) . # (0, · ¸ ∂f ∂f 1 + 2x −1 ∂x ∂y (b) A = = . ∂g ∂g 1 1 ∂x ∂y · ¸ 1 −1 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 1 1 · ¸ 1−λ −1 2 = 0 ⇒ (1 − λ) + 1 = 0 det (A − λI) = 0 ⇒ det 1 1−λ √

⇒ λ2 − 2λ + 2 = 0 ⇒ λ = 2± 24−8 = 1 ± i (complex with positive real part). So the equilibrium ·(0, 0) is unstable spiral. ¸ −3 −1 For equilibrium (−2, 2) , A = . The eigenvalue, λ, 1 1

· ¸ −3 − λ −1 satisfies det (A − λI) = 0 ⇒ det = 0

1 1−λ ⇒ (1 − λ) (−3 − λ) + 1 = 0 ⇒ λ2 + 2λ − 2 = 0

√ √ −2± 4+8 ⇒λ= = −1 ± 3 (one positive, one negative). 2 So the equilibrium · ¸ (−2, 2) is a· saddle point. ¸· ¸ · ¸ u u 0 −3 − λ −1 Eigenvector, , satisfies = . v 1 1−λ v 0 √ Eigenvector 3: · √ for λ = −1 · ¸ ·+ ¸ ¸ √ ¢ ¡ −2 − 3 −1√ u 0 = ⇒ −2 − 3 u − v = 0. v 0

1 2− 3 ¸

· ¸ · √ 1√ u Let u = 1. Then v = −2 − 3 ⇒ = v −2 − 3 is unstable direction.

√ Eigenvector 3 : · √ for λ = −1 ¸ ·− ¸ ¸ · √ ¢ ¡ u 0 −2 + 3 −1√ = ⇒ −2 + 3 u − v = 0.

v 0 1 2+ 3

158

CHAPTER 6

Let u = 1. Then v = −2 +

√

3⇒

is stable direction.

·

u v

¸

=

·

1√ −2 + 3

¸

(c)

(e) Nullclines:

dy dx

=

dy dt dx dt

=

x+y x−y+x2 .

dy = 0 ⇒ x + y = 0 ⇒ y = −x

So dx dy and dx = ∞ ⇒ x − y + x2 = 0 ⇒ y = x + x2 .

Note intersections at equilibria.

y

x

11.

dx dt dy dt

= −x − 2y = f = 2x − y + xy 2 = g (a) Equilibria: f = 0 ⇒ −x − 2y = 0 ⇒ x = −2y g =¡0 ⇒ 2x ¢− y + xy 2 = 0 ⇒ −4y − y − 2y 3 = 0 ⇒ −y 2y 2 + 5 = 0 ⇒ y = 0 (real only). So x = 0.

Thus"equilibria is # (0,·0) only.

¸ ∂f ∂f −1 −2 ∂y . = (b) A = ∂x ∂g ∂g 2 + y 2 −1 + 2xy ∂x ∂y · ¸ −1 −2 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 2 −1 ¸ · −1 − λ −2 =0 det (A − λI) = 0 ⇒ det 2 −1 − λ 2 ⇒ (−1 − λ)√ + 4 = 0 ⇒ λ2 + 2λ + 5 = 0

⇒ λ = −2± 2 4−20 = −1 ± 2i (complex with negative real part).

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

159

So the equilibrium (0, 0) is a stable spiral. (c)

(e) Nullclines: So

dy dx

and

dy dx

=

dy dt dx dt

=

2x−y+xy 2 −x−2y . 2

¡ ¢ = 0 ⇒ 2x − y + xy = 0 ⇒ x 2 + y 2 = y y ⇒ x = 2+y 2

dy dx

= ∞ ⇒ −x − 2y = 0 ⇒ y = − 12 x. y

x

dx dt

dy

dt

= x − xy + γx2 = x − xy − 8x2 = f = −y + xy = g (a) Equilibria: g = 0 ⇒ −y (1 − x) = 0 ⇒ y = 0 or x = 1.

f = 0 ⇒ x − xy − 8x2 = 0. If y = 0, then x − 8x2 = 0

⇒ x (1 − 8x) = 0 ⇒ x = 0, 18 . If x = 1, then 1 − y − 8 =¡ 0 ⇒¢ y = −7. 1 Thus"equilibria is # (0,·0) , 8 , 0 and (1, −7). ¸ ∂f ∂f 1 − y − 16x −x ∂y . (b) A = ∂x = ∂g ∂g y −1 +x ∂x ∂y · ¸ 1 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 −1 · ¸ 1−λ 0 det (A − λI) = 0 ⇒ det =0 0 −1 − λ ⇒ − (1 − λ) (1 + λ) = 0 ⇒ λ = 1, −1 (one positive, one negative). So the equilibrium (0, 0) is a saddle point.

13.

160

CHAPTER 6

¡1

¢

·

¸

−1 − 18 . The eigenvalue, λ, 0 − 78 · ¸ −1 − λ − 81 satisfies det (A − λI) = 0 ⇒ det =0 0 − 87 − λ

¡ 7 ¢ 7 ⇒ (−1 − λ) − 8 − λ ¡ 1 =¢0 ⇒ λ = −1, − 8 (both negative).

So the equilibrium 8 , 0 is a ·stable node.¸ −8 −1 . The eigenvalue, λ, For equilibrium (1, −7) , A = −7 0 · ¸ −8 − λ −1 satisfies det (A − λI) = 0 ⇒ det =0 −7 −λ For equilibrium

8, 0

,A=

√

⇒ λ2 + 8λ − 7 = 0 ⇒ λ = −8± 264+28 √ ⇒ λ = −4 ± 23 (one positive, one negative).

So the equilibrium (1, −7) is a saddle point. 1 2 2 15. dx dt = x − xy + γx = x − xy + 3 x = f dy dt = −y + xy = g (a) Equilibria: g = 0 ⇒ −y (1 − x) = 0 ⇒ y = 0 or x = 1. f = 0¡ ⇒ x −¢xy + 13 x2 = 0. If y = 0, then x + 31 x2 = 0 ⇒ x 1 + 13 x = 0 ⇒ x = 0, −3. If x = 1, then 1 − y + 31 = 0 ⇒ y = 43 ¡. ¢ 4 Thus"equilibria is # (0,·0) , (−3, 0) and 1, 3 . ¸ ∂f ∂f −x 1 − y + 23 x ∂y . = (b) A = ∂x ∂g ∂g y −1 +x ∂x ∂y · ¸ 1 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 −1 · ¸ 1−λ 0 det (A − λI) = 0 ⇒ det =0 0 −1 − λ ⇒ − (1 − λ) (1 + λ) = 0 ⇒ λ = 1, −1 (one positive, one negative). So the equilibrium (0, 0) is a saddle point.¸ · −1 3 . The eigenvalue, λ, For equilibrium (−3, 0) , A = 0 −4 · ¸ −1 − λ 3 satisfies det (A − λI) = 0 ⇒ det =0 0 −4 − λ ⇒ (−1 − λ) (−4 − λ) = 0 ⇒ λ = −1, −4 (both negative). So the equilibrium (−3, 0) is ·a stable node. ¸ 1 ¡ ¢ −1 For equilibrium 1, 43 , A = 34 . The eigenvalue, λ, satisfies 0 3 · 1 ¸ − λ −1 det (A − λI) = 0 ⇒ det 3 4 = 0 ⇒ λ2 − 13 λ + 34 = 0 −λ 3 √

⇒ 3λ2 − λ + 4 = 0 ⇒ λ = 1± 61−48 ⇒ λ = 16 ± with positive real part). ¡ ¢ So the equilibrium 1, 43 is an unstable spiral. 2 2 17. dx dt = x − xy + γx = x − xy + 8x = f dy dt = −y + xy = g

√

47 6 i

(complex

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

161

(a) Equilibria: g = 0 ⇒ −y (1 − x) = 0 ⇒ y = 0 or x = 1.

f = 0 ⇒ x − xy + 8x2 = 0. If y = 0, then x + 8x2 = 0

⇒ x (1 + 8x) = 0 ⇒ x = 0, − 18 . If x = 1, then 1 − y + 8 =¡ 0 ⇒ y¢ = 9. 1 Thus"equilibria is # (0,·0) , − 8 , 0 and (1, 9). ¸ ∂f ∂f 1 − y + 16x −x ∂y = . (b) A = ∂x ∂g ∂g y −1 +x ∂x ∂y · ¸ 1 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 −1 · ¸ 1−λ 0 =0 det (A − λI) = 0 ⇒ det 0 −1 − λ ⇒ − (1 − λ) (1 + λ) = 0 ⇒ λ = 1, −1 (one positive, one negative). So the equilibrium (0, 0) is a saddle point. ¸ · 1 ¡ 1 ¢ −1 8 For equilibrium − 8 , 0 , A = . The eigenvalue, λ, 0 − 98 · ¸ 1 −1 − λ 8 satisfies det (A − λI) = 0 ⇒ det =0 0 − 98 − λ ¡ 9 ¢ ⇒ (−1 − λ) − 8 − λ 0¢⇒ λ = −1, − 89 (both negative). ¡ = 1 So the equilibrium − 8 , 0 is · a stable ¸node. 8 −1 . The eigenvalue, λ, satisfies For equilibrium (1, 9) , A = 9 0 ¸ · 8 − λ −1 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 − 8λ + 9 = 0 −λ 9 √ √ ⇒ λ = 4 ± 7 (both positive). ⇒ λ = 8± 64−36 2 So the equilibrium (1, 9) is an unstable node.

6.3 POPULATION MODELS 1. Two Competing Species Models dx 2 dt = x (a − by − cx) = ax − bxy − cx dy 2 dt = y (q − rx − sy) = qy − rxy − sy . a = 2, b = c = 1, and q = r = 6, s = 0. So dx 2 dt = x (2 − y − x) = 2x − xy − x = f dy dt = y (6 − 6x) = 6y − 6xy = g (a) Equilibria: g = 0 ⇒ y (6 − 6x) = 0 ⇒ y = 0 and x = 1. f = 0 ⇒ x (2 − y − x) = 0. If y = 0, then x = 0, 2. If x = 1, then y = 1. Thus equilibria are (0,"0) , (2, 0) and 1) . # (1, · ¸ ∂f ∂f 2 − y − 2x −x ∂x ∂y Jacobian matrix A = = . ∂g ∂g −6y 6 − 6x ∂x · ∂y ¸ 2 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 6

162

CHAPTER 6

·

¸

2−λ 0 =0 0 6−λ ⇒ (2 − λ) (6 − λ) = 0 ⇒ λ = 2, 6 (both positive). So the equilibrium · ¸ · ¸ ¸ (0, 0) is an· unstable node. ¸ · 0 u 0 u 2−λ Eigenvector, , satisfies = . 0 v ¸ 0· ¸6 − ·λ ¸ v · u 0 0 0 Eigenvector for λ = 2 : = ⇒ v = 0. 0 4 v 0 · ¸ · ¸ u 1 . u is free to choose, say, u = 1 ⇒ = v 0 ¸· ¸ · ¸ · u 0 −4 0 Eigenvector for λ = 6 : = ⇒ u = 0. 0 0 v 0 · ¸ · ¸ u 0 v is free to choose, say, v = 1 ⇒ = . v 1 · ¸ −2 −2 For equilibrium (2, 0) , A = . The eigenvalue, λ, satisfies 0 −6 · ¸ −2 − λ −2 det (A − λI) = 0 ⇒ det =0 0 −6 − λ ⇒ (−2 − λ) (−6 − λ) = 0 ⇒ λ = −2, −6 (both negative). So the equilibrium stable node. · ¸ · ¸ ¸ (2, 0) is a · ¸· −2 − λ −2 u 0 u = . Eigenvector, , satisfies v 0 ¸ · −6 0 ¸ − ·λ ¸ v · u 0 0 −2 Eigenvector for λ = −2 : = ⇒ −2v = 0 0 −4 v · 0 ¸ · ¸ u 1 . ⇒ v = 0. u is free to choose, say, u = 1. So = v 0 · ¸· ¸ · ¸ 4 −2 u 0 Eigenvector for λ = −6 : = ⇒ 0 0 v· 0 ¸ · ¸ u 1 4u − 2v = 0. Let u = 1. Then v = 2. So = . v 2 · ¸ −1 −1 For equilibrium (1, 1) , A = . The eigenvalue, λ, satisfies −6 0 ¸ · −1 − λ −1 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 + λ − 6 = 0 −6 −λ ⇒ (λ + 3) (λ − 2) = 0 ⇒ λ = 2, −3 (one positive, one negative). So the equilibrium saddle point. ¸ · · ¸ · ¸ ¸ (1, 1) is a · u 0 u −1 − λ −1 = . Eigenvector, , satisfies 0 v v −6 − λ · ¸· ¸ · ¸ −3 −1 u 0 Eigenvector for λ = 2 : = −6 −2 v 0 · ¸ · ¸ u 1 ⇒ −3u − v = 0. Let u = 1. Then v = −3. So = −3 v is unstable direction. · ¸ · ¸ ¸· 2 −1 0 u = ⇒ Eigenvector for λ = −3 : v 0 −6 3 det (A − λI) = 0 ⇒ det

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

2u − v = 0. Let u = 1. Then v = 2. So

·

u v

¸

=

is stable direction. (b) Graph of phase portrait using linear systems.

·

1 2

¸

0

(c) Method of nullclines:

dy dx

=

dy dt dx dt

=

y(6−6x) x(2−y−x) .

dy So dx = 0 ⇒ y (6 − 6x) = 0 ⇒ y = 0, x = 1 dy and dx = ∞ ⇒ x (2 − y − x) = 0 ⇒ x = 0, y = 2 − x.

x=1

y

1

2

(d) Graph improved by using eigenvectors.

x

163

164

CHAPTER 6

x = 1

y

1

2

x

3. Two Competing Species Models dx 2

dt = x (a − by − cx) = ax − bxy − cx dy 2

dt = y (q − rx − sy) = qy − rxy − sy . 3

a = 2

, b = 2, c = 4, and q = r = s = 1. So ¡ 3

¢ 3

dx 2

dt = x 2 − 2y − 4x = 2 x − 2xy − 4x = f dy 2

dt = y (1 − x − y) = y − ¡ 3 xy − y = g¢ 3

(a) Equilibria: f = 0 ⇒ x 2

− 2y − 4x = 0 ⇒ x = 0 or 2x + y = 4

. g = 0 ⇒ y (1 − x − y) = 0 ⇒ y = 0 or x + y = 1. If x = 0, then considering both equations obtained from g = 0

we have y = 0, 1. Equilibria are then (0, 0) and (0, 1) . Again, if 2x + y = 3

4 , then considering both equations obtained from g = 0, we solve: ¡ 3 ¢ 2x + y = 3

4 and y = 0 to get 8 , 0 ; and ¡ 1 5 ¢ 2x + y = 3

4 and x + y = 1 to get, by subtracting, − 4 , 4 , which is out of the domain as we are considering ¡ 3 ¢ x ≥ 0, y ≥ 0. Thus equilibria are (0,"0) , (0, 1) and # ·8 , 0 . ¸ ∂f ∂f 3

− 2y − 8x −2x ∂x ∂y 2

Jacobian matrix A = . = ∂g ∂g −y 1 − x − 2y ∂x · ∂y3

¸ 0 For equilibrium (0, 0) , A = 2

. The eigenvalue, λ, satisfies 0 1 · 3

¸ −λ 0 2

det (A − λI) = 0 ⇒ det =0 1 − λ

0 ¡ 3

¢ 3

⇒ 2 − λ (1 − λ) = 0 ⇒ λ = 2

, 1 (both positive).

So the equilibrium node. ¸ · · ¸ (0, 0) is an· unstable ¸ · ¸ 3

u u 0 −λ 0 2

Eigenvector, , satisfies = . v

v

0 0 1−λ ¸· ¸ · ¸ · 1

u 0 2 0 = Eigenvector for λ = 1 : ⇒ u = 0. v

0 0 0

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

·

¸

·

¸

165

u 0 v is free to choose, say, v = 1 ⇒ = . 1

v ¸· ¸ · ¸ · 0 0 u 0 3

= Eigenvector for λ = 2

⇒ v = 0. : 1

v 0 0 − 2

· ¸ · ¸ 1 u u is free to choose, say, u = 1 ⇒ = . v 0

· 1

¸ − 2 0 . The eigenvalue, λ, satisfies For equilibrium (0, 1) , A = −1 −1 ¸ · 1

0 − 2 − λ

det (A − λI) = 0 ⇒ det =0 −1

−1 − λ ¡ 1

¢ 1

⇒ − 2 − λ (−1 − λ) = 0 ⇒ λ = − 2

, −1 (both negative). So the equilibrium (0, 1) is a stable node.

· ¸ · 1

¸ · ¸ ¸· u 0 u − 2 − λ 0 = , satisfies . Eigenvector, v 0 v −1

−1 − λ · 1

¸ ¸ · ¸· u 0 0 2 Eigenvector for λ = −1 :

= ⇒ 1

2 u = 0 v 0 −1 0 · ¸ · ¸ u 0 ⇒ u = 0. v is free to choose, say, v = 1. So . = v 1

· ¸ · ¸

¸· 0 0 0 u = ⇒ Eigenvector for λ = − 21 : v 0 −1 − 21 · ¸ ¸ · u 1 = . −u − 21 v = 0. Let u = 1. Then v = −2. So v −2 ¸ · 3 ¡ ¢ − 2 − 43 . The eigenvalue, λ, For equilibrium 38 , 0 , A = 5 0 · 38 ¸ −2 − λ − 43 satisfies det (A − λI) = 0 ⇒ det =0 5 0 8 −λ ¡ 3 ¢ ¡5 ¢ 3 5 ⇒ − 2 − λ 8 − λ¡ = 0¢ ⇒ λ = − 2 , 8 (one positive, one negative). 3 So the equilibrium point. ¸ · ¸ ¸· ¸ 8 , 0 is a ·saddle · 0 u u − 43 − 23 − λ = . Eigenvector, , satisfies 5 0 v v 0

− λ 8¸ · ¸ · · ¸ 0 0 − 43 u Eigenvector for λ = − 23 : = ⇒ − 43 v = 0 v 0 0 17 8 ¸ · ¸ · 1 u ⇒ v = 0. u is free to choose, say, u = 1. So = v 0 is stable direction. · 17 ¸ · ¸ ¸· 0 u − 8 − 43 Eigenvector for λ = 58 : ⇒ = 0 v 0

0 · ¸ · ¸ u 6 3 − 17

= u − v = 0. Let u = 6. Then v = −17. So 8 4 v −17 is unstable direction. (b) Graph of phase portrait using linear systems.

166

CHAPTER 6

1

0

3/8

(c) Method of nullclines:

dy dx

=

dy dt dx dt

=

y(1−x−y) . x( 32 −2y−4x)

dy So dx = 0 ⇒ y (1 − x − y) = 0 ⇒ y = 0, y = 1 − x ¡ ¢ dy and dx = ∞ ⇒ x 23 − 2y − 4x = 0 ⇒ x = 0, y = Lines do not intersect for x ≥ 0, y ≥ 0.

3 4

− 2x.

Nullcline graph in combined figure in answer to part (d).

(d) Graph improved by using eigenvectors. y

1

3/8

x

5. Predator-prey models: dx 2 dt = x (a − by − cx) = ax − bxy − cx dy 2 dt = y (−q + rx − sy) = −qy + rxy − sy . a = 3, b = c = 1, q = r = 1 and s = 0. So dx 2 dt = x (3 − y − x) = 3x − xy − x = f dy

dt = y (−1 + x) = −y + xy = g (a) Equilibria: g = 0 ⇒ y (−1 + x) = 0 ⇒ y = 0 or x = 1. f = 0 ⇒ x (3 − y − x) = 0. If y = 0, then x = 0, 3 and if x = 1, then y = 2.

Thus equilibria are (0,"0) , (3, 0) and 2) .

# (1, · ¸ ∂f ∂f 3 − y − 2x −x ∂y Jacobian matrix A = ∂x = . ∂g ∂g y −1 + x ∂x ∂y

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

·

¸

167

3 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 −1 · ¸ 3−λ 0 det (A − λI) = 0 ⇒ det =0 0 −1 − λ ⇒ (3 − λ) (−1 − λ) = 0 ⇒ λ = 3, −1 (one positive, one negative). So the equilibrium saddle point. · ¸ (0, 0) is a · ¸· ¸ · ¸ u 3−λ u 0 0 Eigenvector, , satisfies = . v 0 −1 − λ v 0 · ¸ · ¸ ¸· 0 u 0 0 ⇒ v = 0. Eigenvector for λ = 3 : = 0 −4· v 0 ¸ · ¸ u 1 u is free to choose, say, u = 1 ⇒ = is unstable v 0 direction.

· ¸· ¸ · ¸

u 0 4 0 Eigenvector for λ = −1 : = ⇒ u = 0. 0 0· ¸v · ¸0 0 u v is free to choose, say, v = 1 ⇒ = v 1

is stable direction.

· ¸ −3 −3 For equilibrium (3, 0) , A = . The eigenvalue, λ, satisfies 0 2 ¸ · −3 − λ −3 =0 det (A − λI) = 0 ⇒ det 0 2−λ ⇒ (−3 − λ) (2 − λ) = 0 ⇒ λ = −3, 2 (one positive, one negative). So the equilibrium saddle point. · ¸ (3, 0) is a · ¸· ¸ · ¸ u u 0 −3 − λ −3 Eigenvector, , satisfies = . v 0 ¸ · 2 ¸− λ · 0 ¸v · u 0 0 −3 Eigenvector for λ = −3 : = ⇒ v = 0. 0 5 v 0 ¸ · ¸ · 1 u is stable u is free to choose, say, u = 1. So = v 0 direction.

¸ · ¸

· ¸· −5 −3 0 u ⇒ Eigenvector for λ = 2 : = 0 0 v · ¸0 · ¸ u 3 −5u − 3v = 0. Let u = 3. Then v = −5. So = v −5 is unstable direction.

· ¸

−1 −1 For equilibrium (1, 2) , A = . The eigenvalue, λ, satisfies 2 0 ¸ · −1 − λ −1 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 + λ + 2 = 0 −λ 2 √

⇒ λ = −1±2 1−8 = 0 √ ⇒ λ = − 21 ± 27 i (complex with negative real part).

So the equilibrium (1, 2) is a stable spiral.

(b) Graph of phase portrait using linear systems.

168

CHAPTER 6

0

(c) Method of nullclines:

dy dx

=

dy dt dx dt

=

y(−1+x) x(3−y−x) .

dy So dx = 0 ⇒ y (−1 + x) = 0 ⇒ y = 0, x = 1. dy and dx = ∞ ⇒ x (3 − y − x) = 0 ⇒ x = 0, y = 3 − x.

y

3

x

7. Predator-prey models: dx 2 dt = x (a − by − cx) = ax − bxy − cx dy 2

dt = y (−q + rx − sy) = −qy + rxy − sy . a = 3, c = 0, q = r = s = b = 1. So dx dt = x (3 − y) = 3x − xy = f dy 2

dt = y (−1 + x − y) = −y + xy − y = g (a) Equilibria: f = 0 ⇒ x (3 − y) = 0 ⇒ x = 0 or y = 3. g = 0 ⇒ y (−1 + x − y) = 0. If x = 0, then y = 0, −1 and if y = 3, then x = 4. However, (0, −1) is not in the domain

since x ≥ 0, y ≥ 0.

Thus equilibria are (0,"0) , and (4,#3) .

· ¸ ∂f ∂f 3−y −x ∂x ∂y = . Jacobian matrix A = ∂g ∂g y −1 + x − 2y ∂x · ∂y ¸ 3 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 −1 · ¸ 3−λ 0 det (A − λI) = 0 ⇒ det =0 0 −1 − λ

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

169

⇒ (3 − λ) (−1 − λ) = 0 ⇒ λ = 3, −1 (one positive, one negative). So the equilibrium saddle point. · ¸ (0, 0) is a · ¸· ¸ · ¸ u 3−λ u 0 0 Eigenvector, , satisfies = . v 0 v 0 −1 − λ · ¸· ¸ · ¸ u 0 0 0 Eigenvector for λ = 3 : = ⇒ v = 0. 0 −4 v 0 · ¸ · ¸ u 1 u is free to choose, say, u = 1 ⇒ = is unstable v 0 direction.

· ¸· ¸ · ¸

u 0 4 0 ⇒ u = 0. Eigenvector for λ = −1 : = 0 0· ¸v · ¸0 u 0 v is free to choose, say, v = 1 ⇒ = is stable v 1 direction.

· ¸

0 −4 For equilibrium (4, 3) , A = . The eigenvalue, λ, satisfies 3 −3 · ¸ −λ −4 det (A − λI) = 0 ⇒ det = 0 ⇒ λ2 + 3λ + 12 = 0 3 −3 − λ √

⇒ λ = −3± 2 9−48 = 0 √ ⇒ λ = − 23 ± 239 i (complex with negative real part).

So the equilibrium (4, 3) is a stable spiral.

(b) Graph of phase portrait using linear systems. (c) Method of nullclines:

dy dx

=

dy dt dx dt

=

y(−1+x−y) x(3−y) .

dy = 0 ⇒ y (−1 + x − y) = 0 ⇒ y = 0, y = x − 1. So dx dy and dx = ∞ ⇒ x (3 − y) = 0 ⇒ x = 0, y = 3.

0

(c)

170

CHAPTER 6

y

x

(d) Graph improved by using eigenvectors. y

x

9. Predator-prey models: dx 2 dt = x (a − by − cx) = ax − bxy − cx dy 2 dt = y (−q + rx − sy) = −qy + rxy − sy . a = 3, b = 6, c = 2, q = r = s = 1. So dx 2 dt = x (3 − 6y − 2x) = 3x − 6xy − 2x = f dy 2 dt = y (−1 + x − y) = −y + xy − y = g (a) Equilibria: f = 0 ⇒ x (3 − 6y − 2x) = 0 ⇒ x = 0 or x + 3y = 23 . g = 0 ⇒ y (−1 + x − y) = 0 ⇒ y = 0 or x − y = 1. We solve x = 0 and g = 0 ⇒ y = 0, y = −1 ⇒ (0, 0) , (0, −1) . However, (0, −1) is not in the domain since x ≥ 0, y ≥ 0. Again we solve x + 3y = 23 and g =¡ 0, i.e ¢ or x + 3y = 32 and y = 0 ⇒ x = 23 ⇒ 23 , 0 x + 3y = 32 and¡ x −¢y = 1. Subtracting we have, y = 81 and then x = 89 ⇒ 89 , 1

8 . ¡ ¢ ¡ ¢ Thus equilibria are (0, 0) , 32 , 0 and 89 , 18 .

171 ¸ 3 − 6y − 4x −6x . y −1 + x − 2y

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

Jacobian matrix A =

"

∂f ∂x ∂g ∂x

∂f ∂y ∂g ∂y

·

#

=

·

¸ 3 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 −1 · ¸ 3−λ 0 det (A − λI) = 0 ⇒ det =0 0 −1 − λ ⇒ (3 − λ) (−1 − λ) = 0 ⇒ λ = 3, −1 (one positive, one negative). So the equilibrium saddle point. · ¸· ¸ · ¸ ¸ (0, 0) is a · 0 u u 0 3−λ = . Eigenvector, , satisfies 0 v 0¸ · −1 ¸ − ·λ ¸ v · u 0 0 0 Eigenvector for λ = 3 : = ⇒ v = 0. 0 −4 v 0 · ¸ · ¸ u 1 is unstable u is free to choose, say, u = 1 ⇒ = v 0 direction.

· ¸· ¸ · ¸

u 0 4 0 Eigenvector for λ = −1 : = ⇒ u = 0. 0 0 v 0 · ¸ · ¸ u 0 v is free to choose, say, v = 1 ⇒ = is stable v 1

direction.

¸ · ¡ ¢ −3 −9 For equilibrium 23 , 0 , A = . The eigenvalue, λ, 1 0 2 ¸ · −3 − λ −9 satisfies det (A − λI) = 0 ⇒ det =0 1 0 2 −λ ¡1 ¢ ⇒ (−3 − λ) + 2 − ¡λ =¢ 0 ⇒ λ = −3, 12 (one positive, one negative). 3 So the equilibrium · ¸ 2 , 0 is a ·saddle point. ¸· ¸ · ¸ −3 − λ −9 u u 0 Eigenvector, , satisfies = . 1 v v 0 0 −λ ¸ · ¸ · 2¸ · 0 −9 u 0 = ⇒ v = 0. Eigenvector for λ = −3 : v 0 0 72 ¸ · ¸ · 1 u is stable direction. u is free to choose, say, u = 1 ⇒ = v 0 ¸· ¸ · ¸ · 7 u 0 − 2 −9 1 Eigenvector for λ = 2 : = ⇒ v 0 0 0 · ¸ · ¸ u 18 7 − 2 u − 9v = 0. Let u = 18. Then v = −7 ⇒ = v −7 is unstable direction. ¸ · 9 ¢ ¡ − 4 − 27 4 . The eigenvalue, λ, For equilibrium 89 , 18 , A = 1 −1 ·8 9 8 ¸ −4 − λ − 27 4 satisfies det (A − λI) = 0 ⇒ det =0 1 − 18 − λ 8 ¡ 9 ¢¡ 1 ¢ 27 ⇒ − 4 − λ − 8 − λ + 32 = 0 ⇒ 8λ2 + 19λ + 9 = 0 √ √ 361−288 −19± 73 ⇒ λ = −19± 16 = (both negative real roots). ¡ ¢ 16 So the equilibrium 89 , 18 is a stable node.

172

CHAPTER 6

Eigenvector,

·

u v

¸

, satisfies

Eigenvector for λ =

√ −19+ 73 16

· :

− 49 − λ "

− 27 4 − 18 − λ

1 8 √ −17− 73 16 1 8

27 −√ 4

¸·

17− 73 16

u v #·

¸

u v

= ¸

· =

0 0 ·

¸

0 0

. ¸

⇒

√ − 27 v = 0. Let u = 108. Then v = −17 − 73

¸

·4 108√ u ⇒ = . v −17 − 73 " #· √ ¸ · ¸ −17+ 73 27 √ −√ u 0 −19− 73 16 4 Eigenvector for λ = : = ⇒ 16 17+ 73 1 v 0 8 16 √ √ −17+ 73 u − 27 v = 0. Let u = 108. Then v = −17 + 73

·16 ¸ · 4 ¸

108√ u = ⇒ . v −17 + 73 (b) Graph phase portrait using linear systems. √ −17− 73 u ·16 ¸

0

(c) Method of nullclines:

3/2 dy dx

=

dy dt dx dt

=

y(−1+x−y) x(3−6y−2x) .

dy So dx = 0 ⇒ y (−1 + x − y) = 0 ⇒ y = 0, y = x − 1. dy and dx = ∞ ⇒ x (3 − 6y − 2x) = 0 ⇒ x = 0, y = 12 − 13 x.

y

3/2

11. a = b = ³1, c = 2, ´q = 21 , r = 1. So dx dt

=x 1−

y 1+x

=x−

x 1+x y

=f

x

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE dy dt

³

= y − 12 +

x 1+x

´

173

x = − 12 y + 1+x y=g ³ ´ x (a) Equilibria: g = 0 ⇒ y − 21 + 1+x = 0 ⇒ y = 0 or x = 1. ´ ³ y = 0. Substituting y = 0 we have x = 0. f = 0 ⇒ x 1 − 1+x Again substituting x = 1 in f = 0,we have y = 2. Thus equilibria are (0,"0) and (1, 2) # # . " y x ∂f ∂f − − 1 2 1+x (1+x) ∂y Jacobian matrix A = ∂x = . y x ∂g ∂g − 12 + 1+x (1+x)2 ∂x ∂y ¸ · 1 0 For equilibrium (0, 0) , A = . The eigenvalue, λ, satisfies 0 − 21 ¸ · 1−λ 0 det (A − λI) = 0 ⇒ det =0 0 − 21 − λ ¡ 1 ¢ ⇒ (1 − λ) − 2 − λ = 0 ⇒ λ = 1, − 21 (one positive, one negative). So the equilibrium saddle point. · ¸ (0, 0) is a · ¸· ¸ · ¸ 0 1−λ u u 0 Eigenvector, , satisfies = . v 0 v − 21 − λ 0 ¸ · ¸ · ¸· 0 0 u 0 = ⇒ v = 0. Eigenvector for λ = 1 : v 0 0 − 32· ¸ · ¸ 1 u u is free to choose, say, u = 1 ⇒ = is unstable v 0 direction. ¸· ¸ · ¸ · 3 u 0 0 1 2 Eigenvector for λ = − 2 : = ⇒ u = 0. v 0 0 0 · ¸ · ¸ u 0 is stable direction. v is free to choose, say, v = 1 ⇒ = v 1 · 1 ¸ − 12 For equilibrium (1, 2) , A = 21 . The eigenvalue, λ, satisfies 0 · 1 2 ¸ − λ − 12 det (A − λI) = 0 ⇒ det 2 1 = 0 ⇒ λ2 − 21 λ + 14 = 0 −λ 2 ⇒ 4λ2 − 2λ √ +1=0 √ ⇒ λ = 2± 84−16 = 14 ± 14 3i (complex with positive real part). So the equilibrium (1, 2) is an unstable spiral. (b) Graph phase portrait using linear systems.

0

(c) Method of nullclines:

dy dx

=

dy dt dx dt

=

x y (− 12 + 1+x ) y x(1− 1+x )

.

174

CHAPTER 6

³

dy dx

´

x = 0 ⇒ y − 21 + 1+x = 0 ⇒ y = 0, x = 1. ³ ´ dy y and dx = ∞ ⇒ x 1 − 1+x = 0 ⇒ x = 0, y = 1 + x.

So

Combined figure for (c) and (d) y

x

(e) Graph by software. 3

2 y 1

0

13.

dT ∗ ∗ dt = kT0 V − bT dV

∗ dt = N T − cV kT0 = 43 , b = 1, N = 1 3 dT ∗ ∗ dt = −T + 4 V = f dV ∗ dt = T − 2V = g

1

x

2

3

and c = 2. So

Equilibria: f = 0 and g = 0 ⇒ −T ∗ + 34 V = 0 T ∗ − 2V = 0 Adding these we get V = 0 and then T ∗ = 0. Thus·equilibria is ¸(0, 0) ¸ ·. ∂f ∂f −1 34 ∗ ∂T ∂V A= = . The eigenvalue, λ, satisfies ∂g ∂g 1 −2 ∂T ∗ ∂V

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

·

¸

175

3 −1 − λ 4 =0 1 −2 − λ ⇒ (−1 − λ) (−2 − λ) − 43 = 0 ⇒ 4λ2 + 12λ + 5 = 0 ⇒ (2λ + 1) (2λ + 5) = 0 ⇒ λ = − 21 , − 25 (both negative). So the equilibrium stable node. · ¸ (0, 0) is a · ¸· ¸ · ¸ 3 u u 0 −1 − λ 4 Eigenvector, , satisfies = . v v 0 1 −2 − λ · 1 ¸ · ¸ ¸ · 3 −2 u 0 4 = ⇒ Eigenvector for λ = − 21 : v 0 1 − 23 ¸ · ¸ · u 3 . − 12 u + 43 v = 0. Let u = 3. Then v = 2 ⇒ = v 2 · 3 3 ¸· ¸ · ¸ 0 u 2 4 Eigenvector for λ = − 25 : = ⇒ v 0 1 12 ¸ · ¸ · u 1 3 3 u + v = 0. Let u = 1. Then v = −2 ⇒ = . 2 4 v −2 (b) See text page 423 for the graph of the phase portrait.

det (A − λI) = 0 ⇒ det

(c) Method of nullclines:

dV dT ∗

=

dV ∗ So dT − 2V = 0 ∗ = 0 ⇒ T dV

and dT ∗ = ∞ ⇒ −T ∗ + 34 V

dV dt dT ∗ dt

T ∗ −2V . −T ∗ + 43 V 1 ∗ ⇒ V = 2T . = 0 ⇒ V = 34 T ∗ .

=

v

T*

(d) Graph by software.

176

CHAPTER 6

2

y

2 –2

x

–2

15.

dT ∗ ∗ dt = kT0 V − bT dV ∗ dt = N T − cV 9 kT0 = 4 , b = 1, N = 1 dT ∗ 9 ∗ dt = −T + 4 V = f dV ∗ dt = T − V = g

and c = 1. So

(a) Equilibria: f = 0 and g = 0 ⇒ −T ∗ + 94 V = 0 T∗ − V = 0 Adding these we get V = 0 and then T ∗ = 0. Thus·equilibria is ¸(0, 0) ·. ¸ ∂f ∂f −1 94 ∗ ∂T ∂V . The eigenvalue, λ, satisfies = A= ∂g ∂g 1 −1 ∂T ∗ ∂V · ¸ 9 −1 − λ 4 det (A − λI) = 0 ⇒ det =0 1 −1 − λ 2 ⇒ (−1 − λ) − 49 = 0 ⇒ 4λ2 + 8λ − 5 = 0 ⇒ (2λ − 1) (2λ + 5) = 0 ⇒ λ = 21 , − 52 (one positive, one negative). So the equilibrium saddle point. · ¸ (0, 0) is a · ¸· ¸ · ¸ 9 u u 0 −1 − λ 4 Eigenvector, , satisfies = . v 0 v 1 −1 − λ ¸ · ¸ · · 3 ¸ 9 0 u −2 4 = ⇒ Eigenvector for λ = 12 : v 0 1 − 32 · ¸ · ¸ u 3 − 23 u + 49 v = 0. Let u = 3. Then v = 2 ⇒ = v 2 is unstable direction. ¸ · ¸ · 3 9 ¸· 0 u 2 4 = ⇒ Eigenvector for λ = − 25 : v 0 1 32 ¸ · ¸ · u 3 3 9 = 2 u + 4 v = 0. Let u = 3. Then v = −2 ⇒ v −2 is stable direction. (b) Graph phase portrait using linear systems.

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

v

T*

(c) Method of nullclines:

dV dT ∗

dV ∗ −V So dT ∗ = 0 ⇒ T dV and dT ∗ = ∞ ⇒ −T ∗

=

dV dt dT ∗ dt

=

T ∗ −V −T ∗ + 94 V ∗

.

=0⇒V =T . + 94 V = 0 ⇒ V = 94 T ∗ . y=v

T* = x

(d) Graph by software. 2 y

2 x

–2

–2

177

178

CHAPTER 6

6.4 MECHANICAL SYSTEMS 1.

saddle

Local maximum of potential is an unstable saddle point. 3.

saddle center

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

179

Local maximum of potential is an unstable saddle point, and local minimum of potential is a stable center. 5.

saddle center

center

Local maximum of potential is an unstable saddle point, and two local minimum of potential are stable centers. ¡ ¢ d2 x 7. dt2 = x 1 + x2 = f (x) ¡ ¢ Equilibria: f = 0 ⇒ Rx 1 + x2 = 0 R⇒¡ x = 0. ¢ Potential: V (x) = − f (x) dx = − x + x3 dx = − 21 x2 − 14 x4 .

180

CHAPTER 6

dx

dt

From the graph of the potential we see that the equlibrium x = 0 is a local maximum and thus it is an unstable saddle point. ¡ ¢ 2 9. ddt2x = x 1 − x2 = f (x) ¡ ¢ Equilibria: f = 0 ⇒ x 1 − x2 = 0 ⇒ x = 0 or x2 = 1 ⇒ x = 0, ±1. ¢ R R¡ Potential: V (x) = − f (x) dx = − x − x3 dx = − 21 x2 + 14 x4 .

x= 0 x= -1

x= 1

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

181

From the graph of the potential we see that the equlibrium x = 0 is a local maximum and x = ±1 are local minimum. Thus x = 0 is an unstable saddle point and x = ±1 are stable centers. 2 11. ddt2x = −x (6 − 3x) = −6x + 3x2 = f (x) Equilibria: f = 0 ⇒ −x (6 − 3x) = 0 ⇒ x = 0 or 3x = 6 ⇒ x = 0, 2. ¢ R R¡ Potential: V (x) = − f (x) dx = − −6x + 3x2 dx = 3x2 − x3 .

x=2

x=0

From the graph of the potential we see that the equlibrium x = 0 is a local minimum and x = 2 is a local maximum. Thus x = 0 is a stable center and x = 2 is an unstable saddle point. 13. Pendulum (nonlinear) without damping is given by 2 2 ml ddt2x + mg sin x = 0 ⇒ ddt2x = −β sin x where β = gl . dy d2 x Let y = dx dt . Then dt = dt2 = −β sin x.

The slope field equation is then

dy dx

=

dy dt dx dt

1 2 2y

=

−β sin x y

⇒

ydy = −β sin xdx. Integration yields = β cos x + E, where E is a constant of integration, known as energy. Thus the conservation of energy for the pendulum without damping is 1 2 y − β cos x E = 2

¡ ¢2 ⇒ E = 21 dx − β cos x.

dt 2 15. The general conservative system is given by m ddt2x = f (x) . dx d2 x dx Multiplying this by dx dt we have m dt dt2 = f (x) dt . Now we ¡ ¢ d dx 2 d2 x = dx can rewrite this equation using the fact 12 dt dt dt dt2 ¡ ¢ d dx 2 as 21 m dt = f (x) dx dt dt .

182

CHAPTER 6

Integrating this with respect to t we have

R R ¡ dx ¢2 1 d dt = f (x) dx 2 m dt dt dt dt

¡ dx ¢2 R 1 ⇒ 2 m dt = f (x) dx + E ¡ ¢2 R ⇒ E = 12 m dx − f (x) dx dt which is the conservation of energy. 17. Pendulum with resistive force is given by 2 ml ddt2x + mg sin x = −δl dx dt . In this problem m = δ = 1, l = 32. Also g = 32. So

2 d2 x dx 32 ddt2x + 32 sin x = −32 dx dt ⇒ dt2 = − dt − sin x.

2 dy d x dx dx Let y = dt . Then dt = dt2 = − dt − sin x = −y − sin x. Thus the diffrential equation can be written as a system: dx dt = y = f dy

dt = −y − sin x = g Equilibria: f = 0 ⇒ y = 0 and then g = 0 gives sin x = 0 ⇒ x = nπ, n is an integer. Since the original equation is for x only, the "equilibrium# is x = nπ, n is an integer.

¸

· ∂f ∂f 0 1 ∂x ∂y

Jacobian A = = . ∂g ∂g − cos x −1 ∂x ∂y Since cos x has different values at even and odd π we separate even and odd integers to have two sets of equilibria, x = 2nπ and x = (2n + 1) π, n is an integer. · ¸ 0 1 For the equlibria x = 2nπ, A = . The eigenvalue, −1 −1 · ¸ −λ 1 λ, satisfies det (A − λI) = det = 0 ⇒

−1 −1 − λ λ2 + λ + 1 = √ √ 0

⇒ λ = −1±2 1−4 = − 12 ± 2 3 i (complex with negative real part). Thus the equlibria x = 2nπ are stable ·spirals. ¸

0 1 For the equlibria x = (2n + 1) π, A = . 1 −1

The eigenvalue, λ, · satisfies

¸ −λ 1

=0⇒ det (A − λI) = det 1 −1 − λ √

√

λ2 +λ−1 = 0 ⇒ λ = −1±2 1+4 = −1±2 5 (one positive, one negative). Thus the equlibria x = (2n + 1) π are unstable saddle points. 19. Pendulum with resistive force is given by 2 ml ddt2x + mg sin x = −δl dx dt . In this problem m = 31 , δ = 1, l = 32. Also g = 32. So 32 dx d2 x dx 32 d2 x 3 dt2 + 3 sin x = −32 dt ⇒ dt2 = −3 dt − sin x. 2 dy d x dx Let y = dx dt . Then dt = dt2 = −3 dt − sin x = −3y − sin x. Thus the diffrential equation can be written as a system: dx dt = y = f dy

dt = −3y − sin x = g

183

NONLINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS IN THE PLANE

Equilibria: f = 0 ⇒ y = 0 and then g = 0 gives sin x = 0 ⇒ x = nπ, n is an integer. Since the original equation is for x only, the "equilibrium# is x = nπ, n is an integer.

· ¸

∂f ∂f 0 1 ∂y Jacobian A = ∂x = . ∂g ∂g − cos x −3 ∂x ∂y Since cos x has different values at even and odd π we separate even and odd integers to have two sets of equilibria, x = 2nπ and x = (2n + 1) π, n is an integer. · ¸ 0 1 For the equlibria x = 2nπ, A = . The eigenvalue, −1 −3 · ¸ −λ 1 =0⇒ λ, satisfies det (A − λI) = det −1 −3 − λ √

√

λ2 + 3λ + 1 = 0 ⇒ λ = −3±2 9−4 = −3±2 5 (both negative). Thus the equlibria x = 2nπ are stable ·nodes. ¸

0 1 For the equlibria x = (2n + 1) π, A = . 1 −3

The eigenvalue, λ, · satisfies

¸

−λ 1 det (A − λI) = det =0⇒ 1 −3 − λ √

√

λ2 + 3λ − 1 = 0 ⇒ λ = −3±2 9+4 = −3±2 13 (one positive, one negative). Thus the equlibria x = (2n + 1) π are unstable saddle points. 21. Graph using software. Same phase plane as in Figure 6.4.5. 23. Graph using software.

2

0

-2

−π

0

π

2π

3π

4π

5π

184

CHAPTER 6

25. Graph using software.

2

0

-2

−π

0

π

2π

3π

4π

5π