Solutions Manual Introduction Diﬀerential · First-Order Diﬀerential Equations and Their Applications 1.1 INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS There are no exercises

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

Copyright c� 2008 by Princeton University Press

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 v

Chapter 1. First-Order Differential Equations and Their Applications 1

1.1 Introduction to Ordinary Differential Equations 1

1.2 Definite Integral and the Initial Value Problem 1

1.3 First-Order Separable Differential Equations 3

1.4 Direction Fields 5

1.5 Euler’s Numerical Method (Optional) 7

1.6 First-Order Linear Differential Equations 10

1.7 Linear First-Order Differential Equations with Constant Coeffi

cients and Constant Input 15

1.8 Growth and Decay Problems 20

1.9 Mixture Problems 23

1.10 Electronic Circuits 25

1.11 Mechanics II: Including Air Resistance 26

1.12 Orthogonal Trajectories (optional) 27

Chapter 2. Linear Second and Higher-Order Differenial Equations 29

2.1 General Solution of Second-Order Linear Differential Equations 29

2.2 Initial Value Problem (For Homogeneous Equation) 30

2.3 Reduction of Order 32

2.4 Homogeneous Linear Constant Coefficient Differential Equations

(Second Order) 35

2.5 Mechanical Vibrations I: Formulation and Free Response 39

2.6 The Method of Undetermined Coefficients 45

2.7 Mechanical Vibrations II: Forced Response 58

2.8 Linear Electric Circuits 65

2.9 Euler Equation 68

2.10 Variation of Parameters (Second-Order) 70

2.11 Variation of Parameters (nth-Order) 75

Chapter 3. The Laplace Transform 82

3.1 Definition and Basic Properties 82

3.2 Inverse Laplace Transforms (Roots, Quadratics, & Partial Fractions) 86

3.3 Initial-Value Problems for Differential Equations 94

3.4 Discontinuous Forcing Functions 98

3.5 Periodic Functions 109

3.6 Integrals and the Convolution Theorem 114

3.7 Impulses and Distributions 118

iv CONTENTS

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

Their Phase Plane 121

4.1 Introduction 121

4.2 Introduction to Linear Systems of Differential Equations 121

4.3 Phase Plane for Linear Systems of Differential Equations 130

Chapter 5. Mostly Nonlinear First-Order Differential Equations 142

5.1 First-Order Differential Equations 142

5.2 Equilibria and Stability 142

5.3 One Dimensional Phase Lines 143

5.4 Application to Population Dynamics: The Logistic Equation 146

Chapter 6. Nonlinear Systems of Differential Equations in the Plane 150

6.1 Introduction 150

6.2 Equilibria of Nonlinear Systems, Linear Stability Analysis of Equi

librium, and Phase Plane 150

6.3 Population Models 161

6.4 Mechanical Systems 178

Preface

This Student Solutions Manual contains solutions to the odd-numbered exercises 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 problems 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 terminologies 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, constructive 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 publishing 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 1. x = 2e3t + 1. l.h.s. = dx = 6e3t .dt

r.h.s. = 3x − 3 = 3(2e3t + 1) − 3 = 6e3t . Hence l.h.s. = r.h.s. 3. x = t − 1. l.h.s. = dx = 1. r.h.s = x = t−1 = 1. Hence l.h.s. = r.h.s. dt t−1 t−1 5. x = et

2 . l.h.s. = dx = 2tet

2 . r.h.s = 2tx = 2tet

2 . Hence l.h.s. = r.h.s. dt

dx 7. x = e−2t . l.h.s. = dt = −2e−2t . r.h.s. = −2e2tx2 = −2e2t(e−2t)2 = −2e−2t . Hence l.h.s. = r.h.s. dx t t9. dt = 3e . Integrating we get, x = 3e + c.

11. dx = −5 cos 6t. Integrating we get, x = − 5 sin 6t + c.dt 6 dx 1

1

13. dt = 8 cos(t− 2 ). Use of definite integral gives x = 8

∫ 0 t cos t

− 2 dt + c. 15. dx = ln(4 + cos2 t). Use of definite integral gives

x dt

= ∫ 0 t ln(4 + cos2 t)dt + c

dx 117. dt = t4; x(2) = 3. Integrating we get x = 5 t

5 + c.

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

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

ln tx = 5 + ∫ t

dt. 2 4+cos2 t

21. dx = e t

; x(1) = 3. dx = e t

dt. Use of definite integral gives dt 1+t 1+t t t e e x − 3 =

∫ t dt = x = 3 +

∫ tdt.

1 1+t ⇒

1 1+t

d2 x23. dt2 = −15. Integrating we get

dx = −15t + c1 ( dx

dx dt dt = v0 at t = 0 =⇒ c1 = v0.)

= −15t + v0. Integrating again we getdt 15 t2x = − 2 + v0t + c2 (x = 0 at t = 0 = c2 = 0.)⇒

)

2 CHAPTER 1

v0Car stops when dx = 0 = v0 − 15t = 0 = t = (stopping time). dt 15 ⇒ ⇒So distance travelled is

v0 = 15√

10 m/sec. 20

20

2015 1 v

15 2 15 75 = 1v v( v0 )2 +15 x = − = = = 2 15 ⇒ 2 ⇒

25. d2 x = −2500. Integrating we get dx = −2500t + c1( dx = 60 at t = 0 dt2 dt dt = c1 = 60). So dx = −2500t + 60. Integrating again we get ⇒

2500 dt

x = − t2 + 60t + c2(x = 0 at t = 0 =⇒ c2 = 0.)2 Car stops when dx = 0 =⇒ −2500t + 60 = 0 dt = t = 60 (stopping time). So distance travelled is 2500 ⇒

2500 60 602 x = − 2 ( 2500 )2 + 2500 = 0.72 km. x27. d

2 = −2500. Integrating we get dx = −2500t + c1( dx = v0 at t = 0 dt2 dt dt

= c1 = v0). So dx = −2500t + v0. Integrating again we get ⇒ 2500 t2

dt x = − 2 + v0t + c2(x = 0 at t = 0 = c2 = 0) ⇒ Car stops when dx = 0 = v0 − 2500t = 0

v0 dt ⇒

= ⇒ t = 2500 (stopping time). So distance travelled is 20

202500 v0 v v)2( km. +x = − = 2 2500 2500 5000

29. d2 x = −6t. Integrating we get dx dt2 dt = −3t2 + c1( dx dt

= −3t2 + 62. Integrating again we get = 50 at t = 2

c1 = 62). So dx dt = ⇒ x = −t3 + 62t + c2(x = 0 at t = 2 =Car stops when dx = 0 =⇒ −3t2 + 62 = 0 dt

⇒ c2 = −116.) √

62 = t = (stopping time). So distance travelled is

x

⇒

= t(62 − 3

t2) − 116 = √

62 (62 − 62 ) − 116 3 3 √ 62

3 2( 2 3 − 116 km. = 3 )62 − 116 = 2

( 62 3

dV So dy 31. (a) V =volume, dt = Q m3/h. Let snow depth be y. 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

ΔV ΔxΔV = w(Δx)y = wctΔx Δt = wct Δt . Now taking limit as Δt 0 we get dV = Q

⇒ = wct dx dx = Q = 1 with

wc → dt dt ⇒ dt wct kt

k = .

(b) dx = Q

1

1 . Separating the variables we get, ∫

dx = 1 ∫

1 dt ⇒dt kt k t x = k ln t + a. At 11 A.M. t = 3 and x(3) = 0. So 0 = k

1 ln 3 + a ⇒1 a = − k 1 ln 3. Then at noon (t = 4),

x(4) = 1 ln 4 − ln 3 = 1 ln 4 .k k k 3 33. d

2 y = −g = −9.8 m/sec2 . Integrating we get dt2 dy = −9.8t + c1( dy dy

= v0 at t = 0 =⇒ c1 = v0.) So dt dt = −9.8t + v0. Integrating again we get dt

y = − 92 .8 t2 + v0t + c2(y = 0 at t = 0 = c2 = 0.)⇒ At maximum height dy = 0 = v0 − 9.8t = 0

v0 dt ⇒

= ⇒ t = 9.8 (time at maximum height). So maximum height is =

√1960 m/sec.

20

20

20y = − 9.8 2 (

v0 )2 +9.8 = 1 v 100 = 1 2

v v = = v09.8 2 9.8 9.8⇒ ⇒

FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS 3

35. d2 y = −g = −9.8 m/sec2 . Integrating we get dt2

dy = −9.8t + c1( dy = 0 at t = 0 = c1 = 0.)dt dt ⇒ So dy = −9.8t. Integrating again we get y = − 9.8 t2 + c2dt 2

(y = 200 at t = 0 = c2 = 200). y = − 9.8 t2 + 200. Now to fall, ⇒ 2 9.8 9.8

√ 400 20y = 0. So 0 = − 2 t2 + 200 ⇒ 2 t2 = 200 =⇒ t = 9.8 = √9.8 sec.

37. Since x(t0) = x0, the general solution x = ∫t

f (t) dt + c becomes

0

x0 = ∫t0

f (t) dt + c = c = x0 −

∫t0 f

(t) dt. Hence the solution is

0 ⇒

0

x = ∫t

f (t) dt + x0 −

∫t0 f

(t) dt = x0 +

∫tf

(t) dt.

0 0 t0

1.3 FIRST-ORDER SEPARABLE DIFFERENTIAL EQUATIONS

dx x+1 dx dt 1. dt = t . Separating variables gives ∫

x+1 = ∫

t . Integrating we get,

ln x + 1 = ln t + c1 ln |x+1| = c1

∣∣ x+1 c1 t|

x+1

| c1

| | ⇒ |t| ⇒ ∣∣ = e

dx ⇒ t

t = ±e = c ⇒ x = ct − 1.

3. dt = e . Separating variables gives ∫

dx = ∫

etdt. Integrating we get, x = et + c.

5. dx = tx + 4x + 3t + 12 = (x + 3) (t + 4) . Separating variables gives dt

dx t2

∫ x+3 =

∫ (t + 4) dt. Integrating we get, ln x + 3 = 2 + 4t + c1

2

| |2

2 . dx ⇒ |x + 3| = ec1 e t2 +4t ⇒ x = ce t +4t − 3 where c = ±ec1

7. dt = 3. Separating variables gives ∫

dx = ∫

3dt. Integrating we get, x = 3t + c.

dx 59. dt = x . Separating variables gives ∫

x−5dx = ∫

dt. Integrating we get,

− x−4 = t + c. Using x(2) = 1 we get, − 1 = 2 + c ⇒ c = − 9 .4 4 4 Substituting c we get x−4 = 9 − 4t x = (9 − 4t)−1/4 .

dx⇒

11. dt = x2 cos(t2). Separating variables gives x−2dx = cos(t2)dt. Using

2 definite integrals we get,

∫xx−2dx =

∫tcos(t )dt

1 0

2 2⇒ − 1 + 1 = ∫ 0

t

cos(t )dt ⇒ 1 = 1 − ∫ 0

t

cos(t )dt x x 1 x =

2⇒

1−∫t

cos(t )dt 0

13. dx = t cos(x−1/2). Separating variables gives dx = tdt. Using dt cos(x−1/2)

dxdefinite integrals we get, ∫x

= ∫t

tdt cos(x−1/2)

2 1 2 ⇒

∫ 2

xdx = t − 1 = 1

(t2 − 1

) .

cos(x−1/2) 2 2 2

du t2+1 15. dt = u2+4 . Separating variables gives ∫ (

u2 + 4) du =

∫ (t2 + 1

) dt.

4 CHAPTER 1 3 3

Integrating we get, u 3 + 4u = t3 + t + c. Using u(0) = 1 we get,

1 13 + 4 = c c = . Substituting c we obtain the solution as u33 + 12u =

⇒t3 + 3t

3 + 13.

dx 2 217. dt = t2x + x + t2 +1 =

(x2 + 1

) (t2 + 1

) . Separating variables gives

dx 3∫

x2+1 = ∫ (

t2 + 1) dt. Integrating we get, tan−1 (x) = t3 + t + c

3 x = tan

( t3 + t + c

) . Using x(0) = 2 we get, c = tan−1(2). Hence ⇒ (

t)

the solution is x = tan 3 3

+ t + tan−1(2) . dx dx 19. dt = x (x − 1) . Separating variables gives

∫ x(x−1) =

∫ dt. Using

partial fractions to the integral on the left we get, 1 A B

x(x−1) = x + x−1 1 = A (x − 1) + Bx. Putting x = 0 and 1, ⇒respectively, we have, A = −1 and B = 1. Hence

dx dx dx ∫

x(x−1) = ∫

x−1 − ∫

x = ∫

dt ⇒ ln |x − 1| − c

ln |x| = t + c x−1ln

∣∣ x−1 ∣∣ = t + c = ket where k = ±e . Solving this x x⇒ ⇒1equation for x we obtain the general solution, x = 1−ket . 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. Hence the solutions are x = 1 and x = 0.1−ket dx 2 dx 21. dt = (x − 1) (x − 2) . Separating variables gives

∫ (x−1)(x−2)2 =

∫ dt.

Using partial fractions to the integral on the left we get, 1 A B C

(x−1)(x−2)2 = x−1 + x−2 + (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 the coefficients of x2 we have A + B = 0 B = −1. Hence

dx dx dx dx ⇒∫

(x−1)(x−2)2 = ∫

x−1 − ∫

x−2 + ∫

(x−2)2 = ∫

dt. Integrating both

sides we obtain the solution, ln − ln 1 = t + c.|x − 1| |x − 2| − x−2 Since x = 1 and x = 2 both satisfy the differential equation they are also solutions. Hence the solutions are ln − ln 1 = t + c, x = 1 and x = 2.|x − 1| |x − 2| − x−2

23. (tx + x) dt + (tx + t) dx = 0. Dividing by tx we get,(1 + 1 t

) dt +

(1 + x

1 ) dx = 0. Now integrating we have,

t + ln t + x + ln x = c ln tx| | | | x

| | = −t − x + c

c ⇒tx = ±ece−te−x tetxe

⇒ = c where c = ±e .

25.(t2 − 4

) dz +

(z2⇒− 9

) dt = 0. Dividing by

(t2 − 4

) (z2 − 9

) dz dt we get, (z2−9) + (t2−4) = 0. Now we use the formula

du 1 u−a∫ u2−a2 = 2a ln

∣∣∣ u+a ∣∣∣ (from integration table) to integrate and

1/4 get 1 ln

∣∣∣ ∣∣∣ + 1 ln

∣∣∣ ∣∣∣ = c ln

∣∣∣ ∣∣∣ 1/6 ∣∣∣

∣∣∣ = cz−3 t−2 z−3 t−2 6 z+3 4 t+2 z+3 t+2 )1/4(

z−3 )1/6 (

t−2 )1/4∣∣∣∣ c

( z−3

)1/6 ( t−2 c

∣∣∣∣ = e

⇒

= ±e

⇒

z+3 t+2 z+3 t+2 ⇒ ⇒

6 4 4 2(

z−3 ) 1

c (

t−2 )− 1

c (

t+2 ) 1

z−3 (

t+2 ) 3

z+3 = ±e t+2 = ±e t−2 ⇒ z+3 = c t−2


where c = (±ec)6 . 27. et+xdt + e2t−3xdx = 0. Dividing by exe2t we get,

e−tdt + e−4xdx = 0. Now integrating we have, −e−t e−4x = c4 e−4x = −4e−t − 4c ⇒ −4x = ln (−4e−t − 4c)

− ⇒

1⇒

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

dz = a + b dx dz = a + bf (z) which is a differential equationdt dt dt ⇒in z and t and can be solved by separation of variables as

dz ∫

= ∫

dt. a+bf (z) 31. dx = (t + 4x − 1)2 . Let z = t + 4x − 1. Then dx = z2 = f(z)dt dt

and dz = 1 + 4 dx = 1 + 4z2 . Separating the variables we get, dz 1

∫ 1+4

dt

z2 = ∫

dt. dt

We use the substitution u = 2z ⇒ dz = 2 du du to integrate the left hand side. This gives 12

∫ 1+u2 =

∫ dt ⇒

1 tan−1 (2z) = t + c 1 tan−1 (2t + 8x − 2) = t + c. dx z33.2

= et+x (t + x)−1 −⇒ 1.

2

Let z = t + x. Then dx = e z−1 dt dt − 1 z zand dz = 1 + dx = 1 + e z−1 − 1 = e z−1 . Separating the dt dt

variables we get, ∫

ze−z dz = ∫

dt. We use integration by parts to integrate the left hand side as u = z du = dz and dv = e−zdz.

v = −e−z . Then ∫

ze−z dz = ∫

udv = ⇒ uv −

∫ vdu = −ze−z ⇒

+ ∫

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.⇒

1.4 DIRECTION FIELDS

1.

−2 0 2

0

2

x

t

6

6 CHAPTER 1

3.

−2 0 2

0

2

t

x

5.

−2 0 2

0

2

x

t

1.4.1 Existence and Uniqueness

1. dx = x = f (t, x) and fx = 1 are continuous for all (t, x) . Sodt 1+t2 1+t2 unique solution exists for all (t0, x0) .

dx 2

)7/3 7 2)4/33. dt = (1 − t2 − x = f (t, x) and fx = 3

(1 − t2 − x are

continuous for all (t, x) . So unique solution exists for all (t0, x0) . dx 1/5 15. dt = (x + t) = f (t, x) is continuous for all (t, x) but fx = 5(x+t)4/5 is not continuous for x + t = 0.So unique solution exists for all (t0, x0) such that x0 + t0 = 0. dx cos t

6− cos t7. 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 = 1. dx 9. =

(1 − t2 − 2x2

)3/2 = f (t, x) and fx = −6x

(1 − t

62 − 2x2

)1/2 dt are continuous for 1 − t2 − 2x2 > 0. So unique solution exists for all (t0, x0) such that t20 + 2x0

2 < 1. dx 111. 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 = 0.

13. (a) Differentiating t2 + x2 = c wrt(with respect to) t we get,


2t + 2x dx = 0 dx t . Hence t2 + x2 = c definesdt ⇒ dt = − x

a solution.

(b) graph (c) At x0 = 0, as f (t, x) = − t and fx = t are discontinuous at x x2

this point. 2 dx 15. (a) Differentiating t + x = c wrt t we get, 1 + 2x dt = 0.

Hence t + x2 = c defines a solution.

(b) graph (c) Here dx 1 = f (t, x) and fx = 1 are discontinuous at 2dt = − 2x 2x

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 = c cos t = c cos t = cx = x cot t. dt x c sin t Hence x = c sin t defines a solution.

(b) graph (c) Here dx = x cos t = f (t, x) and fx = cos t are discontinuous dt sin t sin t

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, ...

19. (a) Differentiating x = 1 wrt t we get, dx 1 2 . t+c dt = − (t+c)2 = −xHence x = 1 defines a solution. t+c

(b) graph (c) Here dx = −x2 = f (t, x) and fx = −2x are continuous dt

everywhere. Hence there is NO point where the theorem fails to holds.

21. (a) For x = 1, l.h.s. is dx = 0 and r.h.s. = (x − 1)1/5 = 0.dt For x =

( 4 t + c

)5/4 + 1, l.h.s. is dx =

( 4 t + c

)1/4 and 5 dt 5

r.h.s. = (x − 1)1/5 = ((

4 t + c)5/4

+ 1 − 1)1/5

= (

4 t + c)1/4

.5 5

(b) By separating the variables we get, dx = dt which, on (x−1)1/5

integration, becomes x = (

45 t + c

)5/4 + 1. Then using the

initial condition x0 = 1 for any t0 we get c = − 4 t0 and thus 5 one solution is x =

( 45 t − 54 t0

)5/4 + 1. Another solution is

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 (d) Although f (t, x) = (x − 1)1/5 is continuous everywhere,

fx = 1 is not continuous at x0 = 1. As a result, (x−1)4/5 uniqueness does not hold and two solutions in part (b) is not a surprise.

1.5 EULER’S NUMERICAL METHOD (OPTIONAL)

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

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.

3. dx = −tx2 = f (t, x) , t0 = 0, x0 = 1, h = 1. We use recursive formula, dt xn+1 = xn + hf (tn, xn) = xn − tnx2 , where tn+1 = tn + hnto approximate

x1 = x0 − ht0x02 = 1 − 0 = 1 at t1 = 1

x2 = x1 − ht1x21 = 1 − 1 = 0 at t2 = 2

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

5. dx = 2x − 4t = f (t, x) , t0 = 0, x0 = 1, h = 0.5. We use recursive dt 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 = sin x = f (t, x) , t0 = 0, x0 = 0, h = 0.5. We use recursive dt 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 = −20x = f (t, x) , t0 = 0, x0 = 1, h = 0.2.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.2 (−20xn) = −3xn. Here xn = (−3)n 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 = −20x = f (t, x) , t0 = 0, x0 = 1, h = 0.01.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.01 (−20xn) = 0.8xn. Here xn = (0.8)

n 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.

9 FIRST-ORDER DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS

13. dx = x2 = f (t, x) , t0 = 0, x0 = 1, h = 0.2.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.2xn

2 , where tn+1 = tn + h to approximate x1 = x0 + 0.2x0

2 = 1 + 0.2 = 1.2 at t1 = 0.2 x2 = x1 + 0.2x1

2 = 1.2 + 0.288 = 1.488 at t2 = 0.4 x3 = x2 + 0.2x2

2 = 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 + 0.2x4

2 = 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.

15. dx = 1 − 2x + x2 = f (t, x) , t0 = 0, x0 = −5, h = 0.2.dt We use recursive formula,

xn+1 = xn + hf (tn, xn) = xn + 0.2

(1 − 2xn + x2

)

n

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

x4 = x3 + 0.

22 (x3 − 1) = 3.676 at t4 = 0.8

x5 = x4 + 0.2

2 (x4 − 1) = 5.109 at t5 = 1

2

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

x7 = x6 + 0.

22 (x6 − 1) = 19.694 at t7 = 1.4

x8 = x7 + 0.2

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.

17. dx = 1 − 2x + x2 = f (t, x) , t0 = 0, x0 = −5, h = 0.1.dt We use recursive formula, xn+1 = xn + hf (tn, xn) = xn + 0.1

(1 − 22xn + xn

)

= xn + 0.1 (xn − 1)2, where tn+1 = tn + h to approximate x1 = x0 + 0.1 (x0 − 1)2 = −1.4 at t1 = 0.1

2

x2 = x1 + 0.1 (x1 − 1) = −0.824 at t2 = 0.2 x3 = x2 + 0.1 (x

2

2 − 1) = −0.4913 at t3 = 0.3

2

x4 = x3 + 0.1 (x3 − 1) = −0.2689 at t4 = 0.4 x5 = x4 + 0.1 (x4 − 1)2 = −0.1079 at t5 = 0.5 x6 = x5 + 0.1 (x5 − 1)2 = 0.0148 at t6 = 0.6 x7 = x6 + 0.1 (x6 − 1)2 = 0.1119 at t7 = 0.7 x8 = x7 + 0.1 (x7 − 1)2 = 0.1908 at t8 = 0.8

( ∫ )

10 CHAPTER 1

2

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

x10 = x9 + 0.1 (x9 − 1)2 = 0.3116 at t10 = 1

x11 = x10 + 0.1 (x10 − 1)2 = 0.359 at t11 = 1.1

2

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

x13 = x12 + 0.1 (x1

2

2 − 1) = 0.436 at t13 = 1.3

x14 = x13 + 0.1 (x13 − 1)2 = 0.4678 at t14 = 1.4

x15 = x14 + 0.1 (x14 − 1)2 = 0.4961 at t15 = 1.5

2

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

x17 = x16 + 0.1 (x16 − 1)2 = 0.5444 at t17 = 1.7

2

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

x19 = x18 + 0.1 (x18 − 1)2 = 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

dx dx 1. dt = 3x. By separation we obtain ∫

x = 3 ∫

dt. Integration yields ln x = 3t + c1 x = ec1 e3t x = ce3t , where c = ±ec1 . dx | | ⇒ | | ⇒

dx 3. dt = 2tx. By separation we obtain ∫

x = ∫

2tdt. Integration yields

ln |x| = t2 + c1 ⇒ |x| = ec1 et2 ⇒ x = cet2 , where c = ±ec1 .

dx dx 1 ∫

dt 5. 2t dt + x = 0. By separation we obtain ∫

x = − 2 . Integration 1yields ln |x| = − ln |t| + c1 ⇒ ln |x| + ln

∣∣t1/2∣∣ = c1 ⇒

t

ln ∣∣xt1/2

∣∣ = c1 c1 c1 c1

∣∣xt1/2∣∣ = e

2

xt1/2 = ±e x = ct−1/2 where c = ±e . dx

⇒ 7.

⇒ + (cos t) x = 0

⇒ . By separation we obtain

∫ dx = −

∫ cos tdt. dt x

Integration yields ln |x| = − sin t + c1 ⇒ |x| = ec1 e− sin t x = ce− sin t , where c = ±ec1 .

dx 9. ⇒

+ (cos(t−1/2)

) x = 0. By separation we obtain

dx ∫dt

= − ∫

cos(t− 1

2 )dt. Definite integration yields

t t

ln

x

|x| = − ∫ 0

cos(s− 1 1

2 )ds + c1 ⇒ |x| = ec1 exp

( −

∫ 0

cos(s− 2 )ds

)

t1⇒ x = c exp − cos(s− 2 )ds , where c = ±ec1 .

0 dx dx 11. = −5x. By separation we obtain

∫ = −5

∫ dt. Integration dt x

yields ln |x| = −5t + c1 ⇒ x = ±ec1 e−5t = ce−5t . Using the initial condition x(0) = 9 we get, c = 9 and the solution is x = 9e−5t . dx dx 13. dt = 9x. By separation we obtain

∫ x = 9

∫ dt. Integration


yields ln x = 9t + c1 c1 e9t = ce9t . Using the initial | | ⇒ x = ±e27 condition x(3) = 7 we get, 7 = ce c = 7e−27 and the

solution is x = 7e9t−27 = 7e9(t−3). ⇒

dx sin t dx sin t15. dt + 4+et x = 0. By separation we obtain ∫

x = − ∫

4+et dt.

sin sDefinite integration yields ln |x| = − ∫ 5

t

4+es ds + c1 ⇒ (

sin s

) x = c exp −

∫ 5

t

4+es ds . Using the initial condition x(5) = 10 (

sin s

) we get, c = 10 and the solution is x = 10 exp −

∫ 5

t

4+es ds .

dx + t−2 dx 117. dt x = 0. By separation we obtain ∫

x = − ∫

t2 dt. Integration yields ln |x| = 1 t + c1 ⇒ x = ±ec1 e1/t = ce1/t. Using the initial condition x(1) = 3 we get, 3 = ce

c = 3e−1 and the solution is x = 3e( 1 t −1).⇒

1.6.3 Integrating Factors for First-Order Linear Differential Equations

dx dx e t 19. t dt + x = et

dt + xt = t , p(t) =

1 t . The integrating factor is ⇒

e ∫

p(t)dt = eln|t| = t. Multiplying by integrating factor we get dx t d t tt dt + x = e dt (xt) = e . Integration yields xt = e + c.⇒

Using the initial condition x(1) = 1 we get, 1 = e + c c = 1 − e. Hence the solution is xt = et + 1 − e x = e t+1−e .

dx t t21.⇒

= 3e . By integration we obtain ∫

dx = 3 ∫

etdt ⇒

x = 3et

+ c. dx 2t 1

⇒ 2t23.

(dt t2 + 1

) dx + 2tx = 1 + x = , p(t) = .dt ⇒ dt (t2+1) ∫ (t2+1) (t2+1)

2t dt ∫ p(t)dt (t2+1)The integrating factor e = e . Using the substitution 2tu = t2 + 1 we have

∫ (t2+1) dt = ln

∣∣t2 + 1∣∣ and then the integrating

ln t2+1factor is e | | = ∣∣t2 + 1

∣∣ = t2 + 1 (since positive for real t). Multiplying by integrating factor we get

d(t2 + 1

) dx + 2tx = 1

(x

(t2 + 1

)) = 1. Integration yields

x (t2 + 1

)dt = t + c x

⇒ =

dt t + c .

dx

⇒ t2+1 t2+1 ∫ p(t)dt 4t25. dt + 4x = t, p(t) = 4. The integrating factor is e = e .

Multiplying by integrating factor we get

4t dx 4t d e dt + 4e x = te

4t dt

(xe4t

) = te4t . Integration yields

4t ⇒

xe = ∫

te4tdt. We use integration by parts on the r.h.s. as u = t du = dt and dv = e4tdt v = 1 e4t . Then r.h.s.=

⇒∫ udv = uv −

∫ vdu = 1 te

⇒4t 1

∫4 e4tdt = 1 te4t 1 e4t .4 4 4 16

4t 1 1 4t 1−

1 −

So xe = 4 te4t − 16 e + c ⇒ x = 4 t − 16 + ce−4t

1 .

Using the initial condition x(0) = 0 we have, c = 16 .

1 1 1Hence the solution is x = 4 t − 16 + 16 e−4t .

27. t dx = 2x dx 2 x = 0, p(t) = − 2 . The integrating factor is e dt

= e−dt 2

∫ t= t−2

t

. ∫

p(t)dt

⇒ −1 dt = e−2 ln t t

Multiplying by integrating factor we get

12 CHAPTER 1

t−2 dx dt x

− 2 t3 x = 0 ⇒ d dt (

x t2

) = 0. Integration yields

= c. Using the initial condition x(1) = 4 we have, c = 4. t2 Hence the solution is x = 4t2 . This can also be done by separation.

29. t2 dx + tx = 1 dx + 1 x = t−2, p(t) = 1 . The integrating dt dt t t⇒∫ p(t)dt

∫ 1 dt ln tfactor is e = t.= e t = e

Multiplying by integrating factor we get t dx + x = 1 d (xt) = 1 . Integration yields xt

dt = ln t +

t c ⇒ dt

x = t−1 ln t t + ct−1 .

31. t dx + 3x = t ⇒

dx + 3 x = 1, p(t) = 3 . The integrating dt ⇒ dt 3 t∫ t∫

p(t)dt 1 dt 3 ln t = t3factor is e = e t = e . Multiplying by integrating factor we get t3 dx + 3t2x = t3 d

(t3x

) = t3 . Integration yields dt dt

4

⇒ t3x = t4 + c x =

14 t + ct

−3 .

One solution,

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

approaches ±∞ as t 0.

dx dx 1

→ 133. t − x = t∫

2

p(t)dt =

− e−

x ∫ = t, p(t) = − . The integrating 1 t

dt dt ⇒ t t dt = e− ln t = t−1factor is e .

Multiplying by integrating factor we get t−1 dx + t−2x = 1 d

(t−1x

) = 1. Integration yields dt dt ⇒

t−1x = t + c x = t2 + ct. ⇒All solutions are continuous and pass through (0, 0) .

x

t

dx 35. dt + 2tx = 1, p(t) = 2t. The integrating factor is 2

∫ p(t)dt

∫ 2tdt te = e = e . Multiplying by integrating factor

2 2 2 2t2 dx t d (

t)

twe get e dt + 2tet x = e dt e x = e . Definite ⇒

2 2 2 2 2t s sintegration yields e x = ∫t

e ds + c x = e−t∫t

e ds + ce−t . 0

⇒ 0

37. dx + t2x = t, p(t) = t2 . The integrating factor is dt 1 3

∫ p(t)dt t3e = e . Multiplying by integrating factor we get


3 3 3 3e 1 t3

( dx + t2x

) = te

1 t3 d (e

1 t3 x)

= te 1 t3 .dt dt ⇒

3 3Definite integration yields e 1 t3 x =

∫tse

s 3

ds + c 0

s 3 3 3x = e

−t3 ∫tse

3

ds + ce −t3

.⇒ 0

39. dx + etx = 3, p(t) = et . The integrating factor is dt t

∫ p(t)dt ee = e . Multiplying by integrating factor we get

t t t t ee

( dx + etx

) = 3ee d

(ee x

) = 3ee .dt dt ⇒

t s e eDefinite integration yields e x = 3 ∫t

e ds + c 0

t s t ex = 3e−e ∫t

e ds + ce−e .⇒ 0

41. dx + x = 1 , x(2) = 1, p(t) = 1. The integrating factor is dt t+1 ∫ p(t)dt te = e . Multiplying by integrating factor we get

t t et

( dx + x

) = e d (etx) = e . Definite integration yields dt t+1 dt t+1 ⇒

st e e x = ∫t

s+1 ds + c. Using initial condition we get,

2

e s 2−tc = e2 and the solution is x = e−t ∫t

s+1 ds + e .

2

43. 3t dx − x = t sin t dx 1 x = 1 sin t, x(5) = 0, p(t) = − 1 .dt dt 3t 3 3t⇒ − 1

The integrating factor is e ∫

p(t)dt = e− 3 ln t = t−1/3 . Multiplying 1 1by integrating factor we get t−1/3

( dx − x

) = t−1/3 sin tdt 3t 3

d (t−1/3x

) = 1 t−1/3 sin t. Definite integration yields ⇒ dt

t3

1t−1/3x = ∫

3 s−1/3 sin sds + c. Using initial condition we get,

5

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

s−1/3 sin sds.

5

dx t dx 1 1 e t 145. 7t dt + x = 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 by integrating factor 1 e t d e t we get t1/7

( dx + x

) =

(t1/7x

) =

7t6/7 . Definite dt 7t 7t6/7 ⇒ dt

1integration yields t1/7x = ∫t

7 s−6/7esds + c

0 ⇒

x = 71 t−1/7

∫ts−6/7esds + ct−1/7 .

0

dx 1 dt dt47. = = x + t dx − t = x, p(x) = −1. The integrating dt x+t dx ⇒ ⇒ factor e

∫ p(x)dx = e−x . Multiplying by integrating factor we get

e−x (

dt d dx − t

) = xe−x ⇒ dx (te−x) = xe−x . Integration yields

te−x = ∫

xe−xdx + c. We use integration by parts on the r.h.s. as u = x du = dx and dv = e−xdx v = −e−x . Then ⇒ ⇒r.h.s.=

∫ udv = uv −

∫ vdu = −xe−x +

∫ e−xdx = −xe−x − e−x .

So te−x = −xe−x − e−x + c t = −x − 1 + cex .⇒

14 CHAPTER 1

49. Let u = e ∫

p(t)dt be the integrating factor of dx + p(t)x = f(t). If we dt introduce an arbitrary constant c1 while we compute

∫ p(t)dt then

new integrating factor is u1 = e ∫

p(t)dt+c1 = ec1 e ∫

p(t)dt = c2e ∫

p(t)dt

where c2 = ec1 > 0. Multiplying the differential equation by this new integrating factor we get, u1

( dx + p(t)x

) = u1f(t)

ddt ⇒

(u1x) = u1f(t) which, on integration, becomes udt

1x = ∫

u1f(t)dt + c3 c2e ∫

p(t)dtx = c2 ∫

e ∫

p(t)dtf(t)dt + c3. ∫ p(t)dt

⇒ Dividing by c2e we get x = e−

∫ p(t)dt

∫ e ∫

p(t)dtf(t)dt + cc2

3 e− ∫

p(t)dt . Now writing cc3

2 = c we have the same general solution

x = e− ∫

p(t)dt ∫

f(t)e ∫

p(t)dtdt + ce− ∫

p(t)dt as (23) . 51. Let u = e

∫ p(t)dt be the integrating factor of dx + p(t)x = f(t). So

d (ux) = u dx + x du = u dx + xp(t)u = u (

dx dt

+ xp(t)) . Now dt dt dt dt dt

dsince k is a constant (kux) = k d (ux) = ku (

dx + xp(t))

dt dt dt which implies that ku is also an integrating factor of (19) as the product d (kux) of l.h.s. of (19) and ku is easily integrable with dt respect to t.

53. dx + [

sin t ] x = q(t) dx + p(t)x = q(t) where p(t) = sin t .dt t dt t⇒ ∫

p(t)dt Multiplying this equation by the integrating factor e we get e ∫

p(t)dt (

dx + p(t)x)

= q(t)e ∫

p(t)dt

d (xe

∫ dt p(t)dt

) = q(t)e

∫ p(t)dt . Definite integral yields ⇒ dt

e ∫

p(t)dtx(t) = c + ∫t

q (t) e ∫

p(t)dtdt. Particular solution with 0

xp(0) = 0 gives c = 0. Then xp = e− ∫

p(t)dt ∫t

q (t) e ∫

p(t)dtdt 0

⇒

xp = ∫t

e ∫

p(t)dte− ∫

p(t)dtq (t) dt =

∫tG

(t, t

) q (t) dt where

0 0

G (t, t

) = e

∫ p(t)dte−

∫ p(t)dt = e

∫ p(t)dt−

∫ p(t)dt .

Using definite integrals yields ( t t

) ( t

) G

(t, t

) = exp

∫ p(s)ds −

∫ p(s)ds = exp

∫ p(s)ds

0 0 t ( t

) ( t

) sin s sin s = exp

∫ s ds = exp −

∫ s ds .

t t

55. dx + p(t)x = f(t). From the change of variable x = u(t)x1(t)dt

we have dx = du x1 + u

dx1 . But since x1 = e− ∫

p(t)dt ,dt dt dt dx1 = −x1p(t). Thus dx = du x1 − ux1p. Plugging in the dt dt dt original equation we get, du x1 − ux1p + ux1p = f(t)dt ⇒du e−

∫ p(t)dt = f(t) du = e

∫ p(t)dtf(t). Integrating we get, dt dt

u = ∫

f(t)e ∫

p(t)dtdt. ⇒

Then x = e− ∫

p(t)dt ∫

f(t)e ∫

p(t)dtdt. Comparison: From the original equation, the integrating factor is

d ( ∫

p(t)dt) ∫

p(t)dt e ∫

p(t)dt and so dt e x = f(t)e . Integrating we get,


e ∫

p(t)dtx = ∫

f(t)e ∫

p(t)dtdt x = e− ∫

p(t)dt ∫

f(t)e ∫

p(t)dtdt. ⇒

1.7 LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS

WITH CONSTANT COEFFICIENTS AND CONSTANT INPUT

dx rt rt 1. dt − 8x = 0. Substituting x = e we get, re − 8ert = 0. Dividing by ert = 0, we have, r = 8. Thus the general solution is x = ce8t . dx

6rt rt rt 3. = −2x. Substituting x = e we get, re = −2e . Dividing dt

by ert = 0, we have, r = −2. Thus the general solution is x = ce−2t . dx

6rt 5. dt + 7x = 0. Substituting x = e we get, re

rt + 7ert = 0. Dividing by ert = 0, we have, r = −7. Thus the general solution is x = ce−7t . dx

6rt rt 7. dt + x = 0. Substituting x = e we get, re + e

rt = 0. Dividing by ert = 0, we have, r = −1. Thus the general solution is x = ce−t . dx

6rt rt 9. dt = 5x. Substituting x = e we get, re

rt = 5e . Dividing by ert = 0, we have, r = 5. Thus the general solution is x = ce5t . dx

6811. dt + 3x = 8. Substituting xp = A we get, 3A = 8 A = 3 . Thus

8 ⇒

rt xp = 3 . Let the associated homogeneous solution be x1 = e . Then rert + 3ert = 0. Dividing by

8 ert = 06 , we have, r = −3.

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

13. dx − 4x = −9. Substituting xp = A we get, −4A = −9 A = 9 .dt 4 9

⇒ rt Thus xp = 4 . Let the associated homogeneous solution be x1 = e .

Then rert − 4ert = 0. Dividing by ert = 0, we have, r = 4. 9

64tThus the general solution is x = 4 + ce .

15. dx 4 x = 3. Substituting xp = A we get, − 4 A = 3 15 .dt − 5 5 ⇒ A = − 4 Thus xp 15 . Let the associated homogeneous solution be x1 = ert .= − 4 Then rert − 4 ert = 0. Dividing by e

15

rt = 06 , t

we have, r = 4 .5 5 4 5Thus the general solution is x = −

dx 2 4 + ce .4

2 417. dt + 3 x = − 3 . Substituting xp = A we get, Thus xp

A = − A = −2.3 3 ⇒ rt

2

= −2. Let the associated homogeneous solution be x1 Then rert + 3

2 ert = 0. Dividing by ert = 06 , we have, r = − 2 t

= e . .3

Thus the general solution is x = −2 + ce−dx

3 . 19. = 2x + 18. Substituting xp = A we get, 2A + 18 = 0 dt ⇒ A = −9.

rt = −9. Let the associated homogeneous solution be x1 Then rert = 2ert . Dividing by ert = 0, we have, r = 2.6

2tThus the general solution is x = −9 + ce . dx 17 17 17 21. dt = −x − 3 . Substituting xp = A we get, −A − 3 = 0 A = − 3 .

17 ⇒

rt Thus xp = − 3 . Let the associated homogeneous solution be x1 = e .

Then rert = −ert . Dividing by ert = 06 , we have, r = −1.

Thus the general solution is x = − 17 + ce−t .3 dx 23. dt + 7x = 8e

−4t . Substituting xp = Ae−4t we get, −4Ae−4t + 7Ae−4t = 8e−4t . Dividing by e−4t = 0, we have, ⇒ −4A + 7A = 8 A = 38 . Thus

8 e−4t 6 ⇒

rt xp = 3 . Let the associated homogeneous solution be x1 = e .

Thus xp = e .

16

25.

27.

29.

31.

33.

35.

37.

39.

41.

CHAPTER 1

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

Thus the general solution is x = 83 e

−4t6+ ce−7t .

dx − 2x = −3e−5t . Substituting xp = Ae−5t we get,dt −5Ae−5t − 2Ae−5t = −3e−5t . Dividing by e−5t = 0, we have, ⇒ −5A − 2A = −3 A = 73 . Thus

3 6 ⇒

rt xp = 7 e−5t . Let the associated homogeneous solution be x1 = e .

Then rert − 2ert = 0. Dividing by ert = 06 , we have, r = 2.

Thus the general solution is x = 37 e

−5t + ce2t .

dx + 4x = 3e4t . Substituting xp = Ae4t we get, 4Ae4t + 4Ae4t = 3e4t .dt Dividing by e4t = 0, we have, 4A + 4A = 3 A = 8

3 . Thus 3 4t

6 ⇒ ⇒ rt xp = 8 e . Let the associated homogeneous solution be x1 = e .

Then rert + 4ert = 0. Dividing by 3 ert = 06 , we have, r = −4.

Thus the general solution is x = 8 e4t + ce−4t .

dx + 3x = 3e−2t . Substituting xp = Ae−2t we get,dt −2Ae−2t + 3Ae−2t = e−2t . Dividing by e−2t = 0, we have, ⇒ −2A + 3A = 1 A = 1. Thus

= e−2t 6 ⇒

rt xp . Let the associated homogeneous solution be x1 = e . Then rert + 3ert = 0. Dividing by ert = 06 , we have, r = −3. Thus the general solution is x = e−2t + ce−3t . dx +7x = 3+5t. Substituting xp = At+B we get, A+7At+7B = 3+5t. dt Equating the coefficients of t and 1 we have 7A = 5 A = 57 and

5 16 16 ⇒

A + 7B = 3 7B = 3 − = B = 5⇒

16 7 7 ⇒ 49

Thus xp = 7 t + 49 . dx + 2x = 14t. Substituting xp = At + B we get, A + 2At + 2B = 14t. dt Equating the coefficients of t and 1 we have 2A = 14 A = 7 and

7 ⇒

A + 2B = 0 2B = −A = −7 B = − 2 .⇒ 7

⇒Thus xp = 7t − 2 .

dx + x = 2t2 + 5t − 8. Substituting xp = At2 + Bt + C we get,dt 2At + B + At2 + Bt + C = 2t2 + 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 + 3x = t3 . Substituting xp = At3 + Bt2 + Ct + D we get,dt 3At2 + 2Bt + C + 3At3 + 3Bt2 + 3Ct + 3D = t3 .

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

1 2 ⇒

23

3A + 3B = 0 B = −A = − 3 , 2B + 3C = 0 C = − 3 B = 9 ,⇒ 1 2

⇒C + 3D = 0 D = − 3 C = − 27

1⇒

1 2 2Thus xp = t3 t2 + dx

3 − 3 9 t − 27 . + 5x = t. Substituting xp = At + B we get, A + 5B + 5At = t. dt

Equating the coefficients of t and 1 we have 5A = 1 A = 1 and 1 1

⇒ 5 A + 5B = 0 ⇒ B = − 5 A = − 25 .

1 1Thus xp = 5 t − 25 . dx + 2x = 3 sin 6t. Substituting xp = A sin 6t + B cos 6t we get, dt 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


cos 6t : 6A + 2B = 0 sin 6t : 2A − 6B = 3 Solving these equations for A and B (multiplying first equation by 3 and adding with the second) we get A = 3 and B 9 20 = − 20 . Thus xp = 3 sin 6t − 9 cos 6t. 20 20

43. dx − 5x = 2 cos t. Substituting xp = A cos t + B sin t we get, dt −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 by 5 and adding with the first) we get A = − 5 and B = 1 .13 13 Thus xp = 1 sin t − 5 cos t. 13 13

45. dx + 4x = 3 cos 2t + 5 sin 2t. Substituting xp = A cos 2t + B sin 2tdt 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

13 1and adding with the second) we get B = 10 and A = 10 . Thus xp = 1 cos 2t + 13 sin 2t. 10 10 dx 47. 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

1 6equation by 6 and adding with the first) we get B = 37 and A = 37 . Similarly, to solve the second pair of equations for C and D,

6multiply the first by 5 and the second by 6 and add to get C = 61 and D = − 5 .61 Thus xp = 6 cos t + 1 sin t + 6 sin 5t − 5 cos 5t. 37 37 61 61

49. dx − 9x = 5 + 2 sin 3t. Substituting xp = A + B sin 3t + C cos 3tdt 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 = − 95 ⇒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 5 1 1C = − 15 and B = − 5 . Thus xp = − 9 − 5 sin 3t − 15 cos 3t.

18

51.

53.

55.

57.

59.

61.

CHAPTER 1

dx + x = 2e3t + sin t. Substituting xp = Ae3t + B sin t + C cos tdt 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 = 1 ⇒ 2 cos t : B + C = 0 sin t : B − C = 1 Solving last pair of equations for B and C (adding ) we get

1 1 1B = 2 and C = −B = − 2 .Thus xp = 2 (e3t + sin t − cos t

) .

dx + 3x = 8e−3t . Substituting xp = Ae−3t we get, dt −3Ae−3t + 3Ae−3t = 8e−3t ⇒ 8e−3t = 0 which is impossible becasue e−3t = 06 , ∀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 by the integrating factor method : The integrating factor ∫

3dt 3tis e = e . Multiplying the equation by the integrating factor 3t

( dx dwe get e dt + 3x

) = 8 dt

(xe3t

) = 8. Integrating we have ⇒

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 − 2x = 7e2t . Substituting xp = Ae2t we get, dt 2Ae2t − 2Ae2t = 7e2t 7e2t = 0 which is impossible

2t ⇒

becasue e = 06 , ∀t. So a simple exponential, Ae2t , does not work as a particular solution becasue e2t is a solution of the associated homogeneous equation. Substituting x = ve2t we get, dv e2t + 2ve2t − 2ve2t = 7e2t dt Dividing by e2t = 0 we have dv = 7 v = 7t + c. Thus dt 6 ⇒the general solution is x = 7te2t + ce2t . We can make the same conjecture as in #53. dx t rt − x = 4e . Let x1 = e be the solution of the associated dt homogeneous equation. On substitution we get, rert − ert = 0

r = 1. So x1 = c1et (similar to the forcing function). ⇒So let xp = Atet . Substituting this in the equation we get, Aet + Atet − Atet = 4et

t⇒ A = 4. Thus the general

solution is x = 4tet + ce . rt dx + x = 5e−t . Let x1 = e be the solution of the associated dt

homogeneous equation. On substitution we get, rert + ert = 0 ⇒ r = −1. So x1 = c1e−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 − 7x = 8e . Let x1 = e be the solution of the associated dt


homogeneous equation. On substitution we get, rert − 7ert = 0 r = 7. So x1 = c1e7t (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 .

⇒

63. dx + p(t)x = f(t), 0 ≤ t ≤ 2, f(t) = 0, x(0) = 2.dt

For 0 ≤ t < 1, p(t) = 2 and so dx + 2x = 0dt x(t) = c1e−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 + x = 0 x(t) = c2e−t . In order for x to be continuous dt ⇒at t = 1 we must have lim x(t) = x(1) 2e−2 = c2e−1

t 1− ⇒ ⇒

→c2 = 2e

−1 . Thus the solution is {

2e−2t , 0 ≤ t < 1 x(t) =

2e−1e−t , 1 ≤ t ≤ 2

2

x

1

0 1 2 t

65. dx + p(t)x = f(t), 0 ≤ t ≤ 2, p(t) = 0, x(0) = 0.dt

For 0 ≤ t < 1, f(t) = 1 and so dx = 1 x(t) = t + c1.dt ⇒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 x(t) = −t + c2.dt = −1 ⇒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

t 0 1

1

67. dx + p(t)x = f(t), 0 ≤ t ≤ 2, x(0) = 2. For 0 ≤ t < 1,dt

p(t) = 1, f(t) = 0 and so dx + x = 0 x(t) = c1e−t .dt ⇒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 = 1 dt ⇒

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) = t + 2e−1 − 1, 1 ≤ t ≤ 2

x

0 1 2

1

2

t

1.8 GROWTH AND DECAY PROBLEMS

The growth rate k of a population P (t) is given by 1 dP = k dP = kP P dt dt ⇒whose solution with initial population P (0) is P (t) = P (0)ekt . The population will be doubled when P (t) = 2P (0). Then 2P (0) = P (0)ekt ekt = 2 kt = ln 2 t = ln 2 which is known as ⇒ ⇒ ⇒ k

doubling time, denoted by td as = ln 2 , and we will use this td k formula throughout this section. 1. In this problem, the growth rate is k = 1.5% = 0.015 and so

doubling time, td = ln 2 = ln 2 ≈ 46.2 years.k 0.015 3. Using td = 8 hours = 3

1 day in the doubling time formula we get, 1 ln 2 = k = 3 ln 2 ≈ 2.08 = 208% per day. 3 k ⇒

5. Here, P (0) = 1500, t = 1 hour, P (1) = 2000. So P (t) = P (0)ekt

2000 = 1500ek ek = 43 k = ln (

43

) . After 4 hour (i.e. t = 4),

⇒

3P (t) = 1500e4k ⇒ = 1500e 4 ln

⇒( 4 ) = 1500

( 4 )4

.3

A B

10 0.02

Q c

5 0.01

0 100 200 0 100 200 t t

7. td = ln 2 = 3 k = ln 2 . Here, P (t) = 10P (0) and so k 3⇒ kt 10P (0) = P (0)e ekt = 10 kt = ln 10

ln 10 3 ln 10 ⇒ ⇒ ⇒

t = = ≈ 9.97 years. k ln 2


9. Let Q be the number of organisms.

Birth: dQ = k1Q, Death:

dQ = −k2Q, Addition: dQ = k.dt dt dt dQ ⇒ dt = k1Q − k2Q + k.

11. With the interest rate of 3% per year:

Doubling time, td = ln 2 = ln 2 ≈ 23.10 years.k 0.03 With the 3% yield:

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

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

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

ln 2 ⇒ ⇒

Then td = ln(1.064) ≈ 11.17 years. 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 = ln 2 = 16 days, we get k = ln 2 .k 16 At the end of t = 30 days, x(t) = 30 g. Then x(t) = x(0)e−kt

ln 2 ln 2⇒

16 16 30 = x(0)e− (30) x(0) = 30e (30) ≈ 110.04 g. 19. dx = −kx x(t) =

⇒ x(0)e−kt . When t = 10 years, x(t) = 0.80x(0).dt ⇒

So 0.801 x(0) = x(0)e−10k ⇒ e−10k = 0.8 ⇒ −10k = ln (0.8) ⇒

k = − 10 ln (0.8) . ln 2 ln 2 (a) Half-life: T = = −10 ln(0.8) ≈ 31.063 years. k

(b) x(t) = 0.15x(0). So 0.15x(0) = x(0)e−kt e−kt = 0.15 ⇒ ⇒ −kt = ln (0.15) t = − 1 ln (0.15) = 10 ln(0.15) ≈ 85.018 years. ⇒ k ln(0.8)

Thus it will take 85.018 − 10 = 75.018 additional years. 21. Newton’s law of cooling dT = −k (T − Q0) has particular solution Q0dt

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

−10k = ln (

15 )

1 ln ⇒(

15 ln 15−ln 17 ⇒ = ln 17−ln 15 .

⇒ 17 ⇒ k = − 10 17

) = − 10 10

e−kt 1(a) T will be 400C if 40 = 30+170e−kt 170e−kt = 10 = 17 ⇒ ⇒ ⇒ −kt = − ln (17) t = 1 ln (17) = 10 ln(17) ≈ 226.4 minutes. ⇒ k ln 17−ln 15

(b) T (t) = 30 + ce−kt , where we previously had k = ln 17−ln 15 .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

1 ⇒

10 ⇒

t = [ln (T − 30) − ln (170)] = ln (

T −30 ) . −k ln 15−ln 17 170 23. Newton’s law of cooling dT = −k (T − Q0) has particular solution dt

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,⇒e−10k 1we get, 60 = 40 + 60e−10k 60e−10k = 20 = ⇒ ⇒ 3

1 ⇒ −10k = ln (

1 ) ⇒ −10k = − ln 3 k = ln 3 ≈ 0.1099.3 10 ⇒

22 CHAPTER 1

25. (a) Newton’s law of cooling dT = −k (T − Q0) = −k [T − (20 + 10t)] .dt (b) Substituting k = 1 in the equation we get, dT = −T + 20 + 10tdt

dT rt ⇒ dt + T = 20 + 10t. Let T1 = rt e be the solution of the

homogeneous equation. Then re + 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 = 0.06 (A − 500) with dt initial condition A(0) = 2000.

(b) dA − 0.06A = −30. Let Ap = −30 B = 500 dt = B ⇒ −0.06B ⇒Ap = 500 and Ah = ce0.06t . So the general solution is ⇒

0.06tA(t) = 500 + ce . 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 = 0.06A dA

dt ⇒ − 0.06A = 0, a homogeneous equation with general solution dt

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 = 0.08P + 365B − 0.08P = 365B. Let Pp = Adt dt ⇒

365 365 ⇒

−0.08A = 365B A = − 0.08 B Pp = − 0.08 B and 0.08t

⇒ ⇒ 365 0.08tPh = ce . So the general solution is P (t) = − B + ce .0.08

Using initial condition P (0) = 1000 we get, c = 1000 + 365 B. 365 0.08tThus P (t) = − B +

(1000 + 365 B

) e .

0.08

0.08 0.08 (a) After t = 5 years P = $10, 000.

365 0.4So 10000 = − B + (1000 + 365 B

) e80(10

⇒ −e 0.4)

0.08 0.08

0.4 0.4800 = 365B (e − 1

) + 80e B = −1) ≈ $3.79.⇒ 365(e0.4

365 0.08t(b) Solve 10000 = − B(t) + (1000 + 365 B(t)

) e for B(t) :

0.08t 0.08t800 = B(t) (365e

0.08 − 365

) + 80e

0.08

0.08t 0.08tB(t) (365e − 365

) = 800 − 80e⇒

0.08tB(t) = (800 − 80e0.08t

) (365e − 365

)−1 .

dQ ⇒

dQ 31. = 0.2Q + 400 cos (2πt) − 0.2Q = 400 cos (2πt) .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

http:$3233.18


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)

−20 200πwe get, A = π2+0.01 and B = π2+0.01 . Thus the particular solution is = 1 (−20 cos (2πt) + 200π sin (2πt)) .Qp π2+0.01

Q1 = ce0.02t . So the general solution is

Q(t) = 1 (−20 cos (2πt) + 200π sin (2πt)) + ce0.02t .π2+0.01

20 Using the initial condition Q(0) = 100 we get c = 100 + π2+0.01 .

Hence Q(t) = 1 (−20 cos (2πt) + 200π sin (2πt))+(100 + 20

) e0.02t .π2+0.01 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 rate (2 gal/min). So the concentration of salt is C = Q . The 300 inflow and outflow rate of salt are 2(0.4) and 2 Q 300 , respectively. Thus the rate of change in salt can be written as dQ dQ Q= 2(0.4) − 2 Q + = 0.8. Substituting Qp = Adt 300 dt 150 ⇒

rt 1 t we get, A = 120 and Q1 = e Q1 = e−

t

⇒ r = − 150 ⇒ 150 .

Thus Q(t) = 120 + ke− 150 . Using initial condition

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

t Q(t) 2 1 t 150 .Q(t) = 120 − 60e− 150 and C(t) = 300 = 5 − 5 e−

(b) When does C(t) = 0.3? t t

150 0.3 = 0.4 − 0.2e− 150 = 1 t = ln 2 e− 2 150 ⇒ ⇒ ⇒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

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

inflow and outflow rate of salt are 5(0.2) and 4 Q (evaporated100 water contains no salt so that Q = 0), respectively.

Thus the rate of change in salt can be written as

dQ dQ Q= 5(0.2) − 4 Q + = 1. Substituting Qp = Adt 100 dt 25 ⇒

t we get, A = 25 and Q1 = ert r = − 1 Q1 = e− 25 .

t

⇒ 25 ⇒

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

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

Q(t) = 25 − 15e− t and C(t) = Q(t) = 0.25 − 0.15e− t25 25 .100

(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 = 6 − 1 = 5 V (t) = 5t + 500 since the initial half-volume is dt ⇒

Q500 gal. Concentration of iodine is C = 5t+500 . The inflow and

24 CHAPTER 1

outflow rate of iodine are 0 (Pure water has no concentration) and Q1 5t+500 , respectively. Thus the rate of change in iodine can be

written as dQ Q dQ dt dt = − 5t+500 ⇒ Q = − 5t+500 ⇒ 1ln Q = − ln (5t + 500) + k1 Q(t) = k2 (5t + 500)−1/5 5 ⇒

= k (t + 100)−1/5 . Using initial condition Q(0) = 10 we have, 10 = k (100)−1/5 k = 10 (100)1/5 . (⇒

100 )1/5

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, (

100 )1/5

Q(t)Q(t) = 10 t+100 and C(t) = 5t+500 .

(b) Tank overflows when t = 100. After overflow, V = 1000 and C = Q . During overflow, mixture is leaving at the rate of 6 gal/min 1000

dQ 6Q(as pure water is entering at this rate) and so for t ≥ 100, dt = − 1000 . 3 3

500 500 Since it has constant coefficients, Q(t) = c2e− t = c2e− (t−100). In order for Q(t) to be continuous at t = 100 we must have

lim Q(t) = lim Q(t) 10 (

100 )1/5

= c2 t 100− t 100+

⇒ 200 → →Q(t)So for t ≥ 100, Q(t) = 10 (2)−1/5 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 = 5 − 2 = 3 V (t) = 3t + 1000 since the initial volume is 1000 dt ⇒

QkL. Concentration of pollutant is C = 3t+1000 . The inflow and

outflow rate of pollutant are 5 × 7 and 2 Q , respectively. Thus 3t+1000 the rate of change in pollutant can be written as dQ = 35 − 2Q

2 dt 3t+1000 ⇒

dQ + 2Q = 35. The integrating factor is e ∫

3t+1000 dt = eln(3t+1000)2/3

Q 3t+1000

= (3t + 1000)2/3 . Multiplying by integrating factor and rearranging d 2/3we have dt

(Q (3t + 1000)2/3

) = 35 (3t + 1000) ⇒

2/3 2/3Q (3t + 1000) = 35

∫ (3t + 1000) dt + k ⇒

Q(t) = (3t + 1000)−2/3 [7 (3t + 1000)5/3 + k

] . Using initial condition

Q(0) = 2000 we have, 2000 = (10)−2 [7 (10)5 + k

] ⇒

57 (10) + k = 200000 k = −500000. Thus ⇒

5/3Q(t) = (3t + 1000)−2/3

[7 (3t + 1000) − 500000

]

= 7 (3t + 1000) − 500000 (3t + 1000)−2/3 and

C(t) = Q(t) = 7 − 500000 (3t + 1000)−5/3 .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. Then dS1 = 13 × 3 − 7 S1 dS1 = 39 − 7S1 dt V1 ⇒ dt 6t+150

and dS2 = 7S1 dS2 = 7S1 28S2 .dt 6t+150 − 28

SV2

2 ⇒ dt 6t+150 − −21t+250


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 1 (18) = 9 gal/s. So V2 = (9 − 4) t + 275 = 5t + 275.2

Let S1 and S2 be the amount of salt in tank 1 and tank 2, respectively. Then dS1 = 21 × 5 − 18 S1 dS1 = 105 − 18S1 dt V1 ⇒ dt 3t+230 and dS2 = 9S1 dS2 = 9S1 4S2 .dt 3t+230 − 4

SV2

2 ⇒ 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 dS1 dS1 18S1 dt = 11 × 5 − 18

SV1

1 dt = 55 − −7t+100 ⇒

dS2 18S1 dS2 18S1 18S2= = and dt −7t+100 − 18SV2

2 ⇒ dt −7t+100 − 200

dS3 18S2= .dt 200

1.10 ELECTRONIC CIRCUITS

1. The differential equation is L di + Ri = e. In this problem, dt Ri = v = 2i R = 2, L = 1, e = 1. Substituting all these we get, di

⇒1+ 2i = 1. Let ip = A. So 2A = 1 A = .dt 2

1 rt ⇒

So ip = 2 . Taking i1 = e we have r = −2 i1 = ce−2t . Thus i (t) = 1 + ce−2t . At t = 0, i(0) = 1 +

⇒ c c = i(0) − 1 .2 2 2

Hence i (t) = 1 + [i(0) − 1

] e−2t

⇒ 1 . Alternatively, 2 2 . As t → ∞, i(t) → 2

1for an equilibrium solution di = 0 so that 2i = 1 i = .dt 2⇒ 3. The differential equation is L di + Ri = e. In this problem, dt

Ri = v = i, L = 1, e = sin t. Substituting all these we get, di + i = sin t. Substituting ip = A sin t + B cos t in the differential dt 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 = 1 , B = − 1 . Thus ip = 1 sin t − 1 cos t. Taking2 2 2 2 i1 = ert in the differential equation we have, r = −1 i1 = ce−t . Thus i (t) = 1 sin t − 1 cos t + ce−t . At t = 0, i(0) = −

⇒1 + c

1 c = i(0) + 21 . Hence

2 i (t) = 1 sin t − cos t +

[i(0) +

21 ] e−t .⇒ 2 2 2 2

5. The differential equation is L di + Ri = e. For 0 ≤ t < 10, dt Ri = v = i R = 1, L = 1, e = 9. Substituting all these we get, di

⇒rt + i = 9. Let ip = A. So A = 9 ip = 9. Taking i1 = edt

we have r = −1 i1 = c1e−t . So⇒i (t) = 9 + c1e−t .⇒

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

Substituting all these we get, di + i = 0. This is a homogeneous dt equation and, in fact, the homogeneous part of the above equation. So i(t) = c2e−t . In order for i(t) to be continuous at t = 10 we must have lim i(t) = i(10) 9 − 9e−10 = c2e−10

t 10− ⇒ ⇒

10 →

c2 = 9e − 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 ⇒ qe = 6 sin t. In order to find charge q we can use: C + Ri = e

q + R dq = e (since i = dq ) q + 2 dq = 6 sin t⇒dq

C dt dt ⇒ 0.5 dt ⇒ + q = 3 sin t. Substituting qp = A sin t + B cos t in this dt

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 = 32 , B = − 23 . Thus qp = 32 sin t − 32 cos t. Taking q1 = ert in the differential equation we have r = −1 q1 = ce−t . Thus q (t) = 3 (sin t − cos t) + ce−t . At t = 0

⇒ , q(0) = 1 = − 3 + c2 2

c = 1 + 3 = 25 . Hence q (t) = 3 (sin t − cos t) + 5 e−t .⇒ 2 2 2

1.11 MECHANICS II: INCLUDING AIR RESISTANCE

1. (a) The differential equation is m dv = −mg − kv dv

dt 1

⇒ 20 dv = −20 (980) − 10v + v = −980. Substituting dt dt 2

1 ⇒

vp = A we get, A = −980 A = −1960. So vp = −1960.2 ⇒1 = ce−

t . Thus v (t) = −1960 + ce−

Using initial condition v(0) = 0 we have, c = 1960. Hence

trt r = − 2 2v1 = e v1 .2⇒ ⇒ (1 − e−t t v (t) = −1960 + 1960e−

(b) v (10) = 1960 (e−5

= −1960 − 1

) = −1946.79. 2 2 .

(c) lim v(t) = −1960. t→∞

dv 3. The differential equation is m = −mg − kv dt ⇒ 70 dv = −70 (980)

1 − 7v ⇒ dv + 1 v = −980. Substituting dt dt 10

vp = A we get, A = −980 A = −9800. So vp = −9800.10 ⇒1 = ce−

t r = − 10 . Thus v (t) = −9800 + ce−

Using the condition v(5) = 12600 we have,

t 10 . rt v1 = e v110 ⇒ ⇒

1 212600 = −9800 + ce− c = 36931.36. So⇒

v (0) = −9800 + 36931.36 = 27131.36. 5. (a) The differential equation is m dv = −mg + v2 . Here, weight is dt

mg = 32 so that m = 32 = 1. Then dv = −32 + v2 g dt ⇒

http:�1946.79http:36931.36http:27131.36


dv = − (32 − v2

) ∫ dv = −

∫ dt. Integrating using tables gives dt 32−v2

1 v+√

32

⇒ v+

√32 2

√32c

2√

32 ln

∣∣∣ v−

√32

∣∣∣ = −t + c ⇒ v−

√32

= ke−2√

32t , where k = ±e . 1000+

√32 Using the initial condition v(0) = 1000 we have k = .

v+√

32 = ke−2√

32t (1 − ke−2

√32t

) 1000−√

32

So v−

√32

v = −√

32 (1 + ke−2

√32t

) ⇒

( 1+ke−2

√32t

) 1000+

√32 v(t) = −

√32

1−ke−2√

32t , where k = 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 − 1 = 0 = 1. The slope of the orthogonal family is thus dt dt

dx ⇒

= −1. Integrating we have, x(t) = −t + c2.dt

t

x

dx dx x3. xt = c. Differentiating wrt t: x + t dt = 0 dt = − t . The slope of the orthogonal family is thus

⇒dx = t

2 2 dt x ⇒

x t 2∫

xdx = ∫

tdt. Integrating we have, 2 = 2 + c ⇒ x − t2 = c2, where c2 = 2c.

t

x

x = tc ln x5. ln x = c ln t ln t = c. Differentiating wrt t: 1 dx

⇒ 1

⇒ x dt ln t−ln x t dx x ln x

(ln t)2 = 0 dt = t ln t . The slope of the orthogonal

t ln tfamily is thus dx ⇒

∫ x ln xdx = −

∫ t ln tdt. Integration by dt = − x ln x ⇒

1 2 1 1parts on both sides yields, 12 x2 ln |x| − 4 x = − 2 t2 ln |t| + 4 t2 + c

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 = 0 = 2t. The slope of the orthogonal family is thus dt dt dx 1

⇒∫ dx = − 1

∫ 1 dt. Integration yields, x = − 1 ln tdt = − 2t

2

⇒ 2 t 2 | | + c2. 9. t = (x − c) x − c = ±

√t x ±

√t = c. Differentiating wrt t:

dx 1 ⇒

dx 1 ⇒

dt 2√

t = 0 dt = ± 2√t . The slope of the orthogonal family is

thus

± dx = ±2

√⇒ t

∫ dx = ±2

∫ √tdt. Integrating we have, dt

4 t3/2 ⇒

x = ± 3 + c2.

11. x = tan (t + c) tan−1 x = t + c. Differentiating wrt t: 1 dx = 1 ⇒ 1+x2 dt

dx = 1 + x2 . The slope of the orthogonal family is thus ⇒ dt 1dx

dt = − 1+x∫ (

1 + x2) dx = −

∫ dt. Integrating we have,

3

2 ⇒ x + x = −t + c2.3

13. x = c cos t. Differentiating wrt t: dx = −c sin t. Since c = x ,dt cos t dx x dt = − cos t sin t = −x tan t. The slope of the orthogonal family is

1thus dx = cot t ∫

xdx = ∫

cot tdt. Integrating we have, 2

dt x ⇒ x 2 = ln sin t + c x

2 = 2 ln sin t + c2, where c2 = 2c.| | ⇒ |dx

| x dx 2x15. x = ct2 . Differentiating wrt t: dt = 2ct. Since c = t2 , dt = t . The

tslope of the orthogonal family is thus dx ∫

2xdx = − ∫

tdt. 2

dt = − 2x ⇒ Integrating we have, x2 = − t2 + c ⇒ 2x2 = −t2 + c2, where c2 = 2c.

17. x3 + t2 = c. Differentiating wrt t: 3x2 dx + 2t = 0 dx 2t 2 .dt ⇒2

dt = − 3xThe slope of the orthogonal family is thus dx = 3x

3 dt 1 dt

32t ⇒∫

x−2dx = ∫

. Integrating we have, − = ln t + c2 t x 2 1 3 3

)−1 | | ⇒

x = (− 2 ln |t| − c

) ⇒ x =

(c2 − 2 ln |t| , where c2 = −c.

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 xp

′ = xp′′ = 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 t + 2et ⇒ 2t⇒

2t 2t − 3e

= 0. For x = e , x′ = 2e and x′′ = 4e . So x′′ − 3x′ + 2x 2t t= 4e − 6e2t + 2e2t = 0. Thus

{e , e2t

} is a set of solutions for

the associated homogeneous equation. Now xp = t + 23

x′ p = 1, x′′ = 0 and so x′′ p − 3x′ p + 2xp = −3 + 2t + 3 = 2

⇒t =r.h.s. p

Thus the general solution is x(t) = t + 32 + c1et + c2e2t .

Differentiating this we get, x′(t) = 1 + c1et + 2c2e2t . Using the

initial conditions x(0) = 1 and x′(0) = 0 we have, 1 = 32 + 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 + 32 − 1 e2t .2

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 xp

′ = x ′′ p = 0 and so ⇒ x′′ p + 2xp

′ + 2xp = 6 =r.h.s. Thus the general solution is x(t) = 3+c1e−t cos t+c2e−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) = −c1e−t cos t−c1e−t sin t−c2e−t sin t+c2e−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⇒

2t 2t= 0. For x = e , x ′ = 2e and x ′′ = 4e . So x ′′ − 3x ′ + 2x 2t t= 4e − 6e2t + 2e2t = 0. Thus

{e , e2t

} is a set of solutions for

the associated homogeneous equation. xp = 2 cosh t x′ = 2 sinh t, ⇒ p x′′ p = 2 cosh t. So x

′′ p − 3x′ + 2xp = 2 cosh t − 6 sinh t + 4 cosh t = p

6 (cosh t − sinh t) = 6e−t =r.h.s. The general solution is thus x = c1et + c2e2t + 2 cosh t. For xp = e−t , x ′ p = −e−t , x ′′ = e−t . So x ′′ p − 3x ′ + 2xp = p p e−t + 3e−t + 2e−t = 6e−t =r.h.s. The general solution is thus x = c1et + c2e2t + 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 x1′ = cos t x′′ 1 = − sin t = −x1 x′′ 1 + x1 = 0.⇒ ⇒ ⇒

x2 = cos t x2′ = − sin t x′′ 2 = − cos t = −x2 x2′′ + 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 x ′ 1 x2

′ cos t − sin t

31 LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

= − (sin2 t + cos2 t

) = −1 = 06 . 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 t2x′′ − tx′ + x = 0

tr [r (r − 1) − r + 1] = 0 (r − 1) (r − 1) = 0 (since tr = 0) ⇒ ⇒ 6r = 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) x2

′ (1) 1 −1 = 0. Hence {x1, x2} is a fundamental set of solutions. = −1 6

(b) x3 = c1x1 + c2x2 x3′ = c1x1

′ + c2x′ 2. Using all three sets of ⇒initial conditions we have, 2 = x3(1) = c1x1(1) + c2x2(1) = c1 and 0 = x′ 3(1) = c1x1

′ (1) + c2x′ 2(1) = c1 − c2 = 2 − c2 c2 = 2.⇒Thus x3 = 2x1 + 2x2. [

x1(t0) x2(t0) ] [

1 0 ]

7. The Wronskian is W (t0) = det = det x′ 1(t0) x2′ (t0) 0 1

= 1 = 06 . Hence {x1, x2} is a fundamental set of solutions. Now x3 = c1x1 + c2x2 x ′ 3 = c1x

′ 1 + c2x

′ 2. Using all three sets ⇒

of initial conditions we have,

α = x3(t0) = c1x1(t0) + c2x2(t0) = c1 × 1 + c2 × 0 = c1

β = x3

′ (t0) = c1x′ 1(t0) + c2x2′ (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.

Then the Wronskian is W (t) = det

[ xx

1 ′ 1

xx

2 ′ 2

] = x1x ′ 2 − x1′ x2.

dW dt = x

′ 1x

′ 2 + x1x2

′′ − x′ 1x′ 2 − x1′′ x2 = x1x2′′ − x′′ 1 x2

= x1 (−tx2) − x2 (−tx1)

= 0

So W (t) is constant.

11. x1 = sin t x′ 1 = cos t x

′′ 1 = − sin t = −x1 x1′′ + x1 = 0.⇒ ⇒ ⇒

x2 = cos t x2′ = − sin t x′′ 2 = − cos t = −x2 x2′′ + 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

= − (sin2 t + cos2 t

) = −1 = 06 .

x′ 1 x2′ cos t − sin t

−t

∫

0

t

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 t = et ⇒ x′ 1 = x′′ 1 = et = x1. Then l.h.s. = x′′ 1 − 3x1′ + 2x1

e − 3et + 2et = 0 =r.h.s.

x2 = e2t x′ 2 = 2e

2t x′′ 2 = 4e2t . Then

l.h.s.= x ′′ 2

⇒− 3x2′ + 2x2

⇒= 4e2t − 6e2t + 2e2t = 0 =r.h.s.

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

32 CHAPTER 2

t 2t The Wronskian is W (t) = det

[ et

e2t

] = 2e3t − e3t = e3t .

e 2e∫t

p(s)ds −According to (13), W (t) = W (t0)e t0 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. = t2x′′ 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

⇒ x2′′ + 4tx′ 2 + 2x2 = 6t

−2 − 8t−2 + 2t−2 = 0 =r.h.s. Thus t−1 and t−2 both are solutions of t2x′′ + 4tx′ + 2x = 0.[

t−1 t−2 ]

The Wronskian is W (t) = det −t−2 −2t−3 = −2t−4 + t−4 = −t−4 .

∫tp(s)ds −

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

t )−4

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

2.3 REDUCTION OF ORDER

1. x1 = t−1 ⇒ x1′ = −t−2 and x1′′ = 2t−3 so that 2t−1 − 3t−1 + t−1 = 0. Let x = vx1 = vt−1 and substitute this in the original equation to get t2

(vt−1

)′′ + 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 = v′ and then dw 1 dw dt dividing by t we get dt + t w = 0. By separation,

∫ w = −

∫ t ,

which, on integration, yields ln |w| = − ln t + c ⇒ w = c2t−1 . But w = v′ , so that v′ = c2t−1 . Now integrating again we get v = c2 ln t + c1. Thus the general solution is x = vx1 = vt−1 = c2t−1 ln t + c1t−1 . A fundamental set of solutions would be

{t−1, t−1 ln t

} .

3. x1 = e−5t ⇒ x1′ = −5e−5t and x′′ 1 = 25e−5t so that 25e−5t − 50e−5t + 25e−5t = 0. Substitute x = vx1 = ve−5t in the original equation to get

(ve−5t

)′′ +10

(ve−5t

)′ +25ve−5t = 0 (

v ′′ e−5t − 10v ′ e−5t + 25ve−5t)+10

(v ′ e−5t − 5ve−5t

)+25ve−5t = 0

⇒

v′′e−5t = 0, since the v (and v′) terms cancel ⇒ v′′ = 0 since e−5t = 06⇒ .

Now integrating twice we get v = c2t + c1. Thus the general solution is x = vx1 = ve−5t = c2te−5t + c1e−5t . A fundamental set of solutions would be

{e−5t, te−5t

} .

5. x1 = e−t x1′ = −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 (v′′e−t − 2v′e−t + ve−t) + (t − 1) (v′e−t − ve−t) − ve−t = 0 ⇒

{ }

33 LINEAR SECOND AND HIGHER-ORDER DIFFERENIAL EQUATIONS

tv′′e−t − (t + 1) v′e−t = 0,since the v terms cancel. Dividing by te−t 6= 0 we get, v′′ −

(1 + 1

) v′ = 0. Now w = v′ yields t

dw − (1 + 1

) w = 0. By separation,

∫ dw =

∫ (1 + 1

) dt, which, dt t w t

on integration, yields ln |w| = t + ln |t| + c ⇒ w = c2tet . But w = v ′ ,

so that v′ = c2tet . 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)+c1e−t . A fundamental set of solutions would be {e−t, t − 1} .

7. x1 = t x1′ = 1 and x′′ 1 = 0 so that x1

′′ + tx′ 1 − x1 = t − t = 0.⇒Let x = vx1 = vt and substitute this in the original equation to get (vt)′′ + t (vt)′ − vt = 0 ⇒ (v′′t + 2v′) + t (v′t + v) − vt = 0 ⇒ v ′′t +

(2 + t2

) v ′ = 0, since the v terms cancel. Writing w = v ′ and

dw + (

2then dividing by t we get dt t + t) w = 0.

dwBy separation, ∫

= − ∫ (

2 + t) dt, which, on integration, yields w t

2 21 11 w = c2e−2 ln|t|−t2 + c2 = c2t−2e−t tln |w| = −2 ln |t

But w = v′ , so that v′| − 2 2 .⇒

= c2t−2e−1 2 t

2 . Now integrating by ∫t 21

2s−2e− susing definite integral we get, v = c2 ds + c1. 1

∫ts−2e−

2 2

2

1 sThus the general solution is x = vx1 ds + c1t. = vt = c2t 1

∫ts−2e−

1 2 sA fundamental set of solutions would be ds .t, t

1

9. Let x = tr . Then x′ = rtr−1 and x′′ = r (r − 1) tr−2 . Substituting these into t2x ′′ − 3tx′ + 4x = 0 we get, r (r − 1) tr − 3rtr + 4tr = 0 ⇒ r2 − 4r + 4 = 0 (since tr = 0). 6 Thus (r − 2)2 = 0 ⇒ r = 2. So x1 = t2 is a solution. Let x = vx1 = vt2 and substitute this in the original equation to get t2

(vt2

)′′ − 3t (vt2

)′ + 4vt2 = 0

t2 (t2v′′ + 4tv′ + 2v

) − 3t

(t2v′ + 2tv

) + 4vt2 = 0

⇒ t4v′′ + t3v′ = 0,

since the v terms cancel. v ′′ + 1 t v ′ = 0. Writing w

⇒ = v ′ we get

dw + 1 w = 0. By separation, ∫

dw 1 dt ln w = − ln t + cdt t w = − ∫

t ⇒ | | | | ⇒ w = c2e− ln|t| = c2t−1 . But w = v′ , so that v′ = c2 1 t . Now integrating again we get, v = c2 ln t + c1. Thus the general solution is x = vt2 = (c2 ln t + c1) t2 = c2t2 ln t + c1t2

(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 t2x′′ + 7tx′ + 9x = 0 we get, r (r − 1) tr + 7rtr + 9tr = 0 r2 + 6r + 9 = 0 (since tr = 0). Thus (r + 3)2 = 0 r = −3.⇒ 6 ⇒

So x1 = t−3 is a solution. Let x = vx1 = vt−3 and substitute this in the original equation to get t2

(vt−3

)′′ + 7t

(vt−3

)′ + 9vt−3 = 0

t2 (t−3v′′ − 6t−4v′ + 12t−5v

) + 7t

(t−3v′ − 3t−4v

) + 9vt−3 = 0,

⇒

since the v terms cancel, t−1v′′ + t−2v′ = 0 v′′ + 1 v′ = 0. dw 1

⇒ t dw 1Writing w = v′ we get dt + t w = 0. By separation,

∫ w = −

∫ t dt

⇒ ln |w| = − ln |t| + c ⇒ w = c2e− ln|t| = c2t−1 . But w = v′ , so that

34 CHAPTER 2

v′ = c2 1 t . Now integrating again we get, v = c2 ln t + c1. Thus the general solution is x = vt−3 = (c2 ln t + c1) t−3 = c2t−3 ln t + c1t−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

2 = 0 (since sin (rt) = 0 as nontrivial solution). Thus ⇒ 4 − r 6r = 2. So x1 = sin (2t) is a solution. Let x = vx1 = v sin (2t) and substitute this in the original equation

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Solutions Manual Introduction Diﬀerential · First-Order Diﬀerential Equations and Their Applications 1.1 INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS There are no exercises