mathschoolinternational.commathschoolinternational.com/Math-Books/Integral... · March 4, 2015 14:44 book-9x6 9571-Root page vii Contents Preface ix 1 Introductory Concepts 1 1.2

A FIRST COURSEIN

INTEGRAL EQUATIONS

Second Edition

Solutions Manual

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N E W J E R S E Y • L O N D O N • S I N G A P O R E • BE IJ ING • S H A N G H A I • H O N G K O N G • TA I P E I • C H E N N A I

World Scientific

A FIRST COURSEIN

INTEGRAL EQUATIONS

Second Edition

Abdul-Majid WazwazSaint Xavier University, USA

Solutions Manual

9571sc_9789814675154_tp.indd 2 13/2/15 9:42 am


Published by

World Scientific Publishing Co. Pte. Ltd.5 Toh Tuck Link, Singapore 596224USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication DataWazwaz, Abdul-Majid. A first course in integral equations. Solutions manual / by Abdul-Majid Wazwaz (Saint Xavier University, USA). -- Second edition. pages cm Includes bibliographical references and index. ISBN 978-9814675154 (pbk. : alk. paper) 1. Integral equations--Problems, exercises, etc. I. Title. QA431.W36 2015b 515'.45--dc23 2015008782

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THIS BOOK IS DEDICATED TO

My wife, our son, and our three daughtersfor supporting me in all my endeavors


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March 4, 2015 14:44 book-9x6 9571-Root page vii

Contents

Preface ix

1 Introductory Concepts 1

1.2 Classification of Linear Integral Equations . . . . . . . . . . 1

1.3 Solution of an Integral Equation . . . . . . . . . . . . . . . 2

1.4 Converting Volterra Equation to an ODE . . . . . . . . . . 4

1.5 Converting IVP to Volterra Equation . . . . . . . . . . . . . 7

1.6 Converting BVP to Fredholm Equation . . . . . . . . . . . 11

1.7 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Fredholm Integral Equations 15

2.2 Adomian Decomposition Method . . . . . . . . . . . . . . . 15

2.3 The Variational Iteration Method . . . . . . . . . . . . . . . 22

2.4 The Direct Computation Method . . . . . . . . . . . . . . . 25

2.5 Successive Approximations Method . . . . . . . . . . . . . . 29

2.6 Successive Substitutions Method . . . . . . . . . . . . . . . 33

2.8 Homogeneous Fredholm Equation . . . . . . . . . . . . . . . 35

2.9 Fredholm Integral Equation of the First Kind . . . . . . . . 39

3 Volterra Integral Equations 41

3.2 Adomian Decomposition Method . . . . . . . . . . . . . . . 41

3.3 The Variational Iteration Method . . . . . . . . . . . . . . . 54

3.4 The Series Solution Method . . . . . . . . . . . . . . . . . . 57

3.5 Converting Volterra Equation to IVP . . . . . . . . . . . . . 63

3.6 Successive Approximations Method . . . . . . . . . . . . . . 67

3.7 Successive Substitutions Method . . . . . . . . . . . . . . . 75

3.9 Volterra Equations of the First Kind . . . . . . . . . . . . . 79

vii


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viii Contents

4 Fredholm Integro-Differential Equations 854.3 The Direct Computation Method . . . . . . . . . . . . . . . 854.4 The Adomian Decomposition Method . . . . . . . . . . . . 904.5 The Variational Iteration Method . . . . . . . . . . . . . . . 944.6 Converting to Fredholm Integral Equations . . . . . . . . . 96

5 Volterra Integro-Differential Equations 1015.3 The Series Solution Method . . . . . . . . . . . . . . . . . . 1015.4 The Adomian Decomposition Method . . . . . . . . . . . . 1035.5 The Variational Iteration Method . . . . . . . . . . . . . . . 1055.6 Converting to Volterra Equations . . . . . . . . . . . . . . . 1075.7 Converting to Initial Value Problems . . . . . . . . . . . . . 1105.8 The Volterra Integro-Differential Equations of the First

Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6 Singular Integral Equations 1176.2 Abel’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . 1176.3 Generalized Abel’s Problem . . . . . . . . . . . . . . . . . . 1226.4 The Weakly Singular Volterra Equations . . . . . . . . . . . 1226.5 The Weakly Singular Fredholm Equations . . . . . . . . . . 130

7 Nonlinear Fredholm Integral Equations 1337.2 Nonlinear Fredholm Integral Equations . . . . . . . . . . . . 133

7.2.1 The Direct Computation Method . . . . . . . . . . . 1337.2.2 The Adomian Decomposition Method . . . . . . . . 1417.2.3 The Variational Iteration Method . . . . . . . . . . . 148

7.3 Nonlinear Fredholm Integral Equations of the FirstKind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

7.4 Weakly-Singular Nonlinear Fredholm Integral Equations . . 153

8 Nonlinear Volterra Integral Equations 1578.2 Nonlinear Volterra Integral Equations . . . . . . . . . . . . 157

8.2.1 The Series Solution Method . . . . . . . . . . . . . . 1578.2.2 The Adomian Decomposition Method . . . . . . . . 1638.2.3 The Variational Iteration Method . . . . . . . . . . . 168

8.3 Nonlinear Volterra Integral Equations of the First Kind . . 1708.3.1 The Series Solution Method . . . . . . . . . . . . . . 1708.3.2 Conversion to a Volterra Equation of the Second

Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . 1728.4 Nonlinear Weakly-Singular Volterra Equation . . . . . . . . 173


March 4, 2015 14:44 book-9x6 9571-Root page ix

Preface

This Solutions Manual is provided as a supplement to accompany the textA First Course in Integral Equations, Second Edition. It is intended:• to develop readers’ skills for each linear or nonlinear exercise provided inthe textbook;• to illustrate the use of the newly developed methods along with the tra-ditional methods used in the text;• to help readers master integral equations concepts so they can solve reg-ular level and challenging exercises.

This Solutions Manual provides completely explained solutions to allexercises of the textbook. The manual can be used by a wide variety of au-dience in various fields of mathematics, physics, chemistry and engineering.Because the text A First Course in Integral Equations, Second Edition con-tains the newly developed methods and some of the classical methods, wehave aimed to make the Solutions Manual self-explanatory, instructional,and useful. The fully-worked solutions are explained in a systematic wayfocusing on the needs, suggestions and expectations of the readers. Thefollowing distinguishing features make this Solutions Manual significant:• it provides a fully worked solution for each problem in the text;• it applies the new methods along with some of the classic methods formost problems using the same strategies, notations, and terminologies usedin the text A First Course in Integral Equations, Second Edition;• it illustrates the solutions in a systematic fashion consistent with the ma-terial presented in the text;• it assists in helping readers gain knowledge and understanding of themathematical methods as they work through the problems and study thesolutions;• it addresses all questions and suggestions made by students and scholarswho used the first edition;• it complements the material and the worked examples of the textbook.

ix


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x Preface

The text A First Course in Integral Equations, Second Edition presentsa variety of valuable methods and applications which cannot be found inany other book. For this reason, and based upon a vast request by readers,this Solutions Manual has been made to provide detailed explanations andillustrations for solving each problem of the second edition.

I would like to acknowledge my wife for her genuine patience and I wishto thank her for sincerely encouraging me in all my endeavors, includingthe writing of the second edition of the textbook as well as the solutionsmanual. I would also like to acknowledge our son and three daughters fortheir support and encouragement.

Saint Xavier University Abdul-Majid WazwazChicago, IL 60655 e-mail: [email protected] 2015


March 4, 2015 14:44 book-9x6 9571-Root page 1

Chapter 1

Introductory Concepts

1.2 Classification of Linear Integral Equations

Exercises 1.2

1. Fredholm, linear, nonhomogeneous2. Volterra, linear, nonhomogeneous3. Volterra, nonlinear, nonhomogeneous4. Fredholm, linear, homogeneous5. Fredholm, linear, nonhomogeneous6. Fredholm, nonlinear, nonhomogeneous7. Fredholm, nonlinear, nonhomogeneous8. Fredholm, linear, nonhomogeneous9. Volterra, nonlinear, nonhomogeneous10. Volterra, linear, nonhomogeneous11. Volterra integro-differential equation, nonlinear12. Fredholm integro-differential equation, linear13. Volterra integro-differential equation, nonlinear14. Fredholm integro-differential equation, linear15. Volterra integro-differential equation, linear

16. u(x) = 1 +

∫ x

0

4u(t)dt

17. u(x) = 1 +

∫ x

0

3t2u(t)dt

18. u(x) = 4 +

∫ x

0

u2(t)dt

1



2 Chapter 1. Introductory Concepts

19. u′(x) = 1 +

∫ x

0

4tu2(t)dt, u(0) = 2

20. u′(x) = 1 +

∫ x

0

2tu(t)dt, u(0) = 0

21. Volterra–Fredholm integral equation, nonlinear, nonhomogeneous22. Volterra–Fredholm integro-differential equation, linear, nonhomoge-neous23. Volterra–Fredholm integro-differential equation, nonlinear, nonhomo-geneous24. Volterra (singular) integral equation, nonlinear, nonhomogeneous

1.3 Solution of an Integral Equation

Exercises 1.3

1. Substituting u(x) = x+ 124 in the right hand side yields

RHS = x+

∫ 14

0

(t+

1

24

)dt

= x+ [ 12 t2 + 1

24 t]140

= x+ 124

= u(x)= LHS.

2. Substituting u(x) = x in the right hand side yields

RHS = 23x+

∫ 1

0

xt2dt

= 23x+ [ 13xt

3]10

= 23x+ 1

3x= u(x)= LHS.

3. Substituting u(x) = 2x in the right hand side yields

RHS = x+

∫ 1

0

4xt3dt

= x+ [xt4]10= 2x= u(x)= LHS

4. Substituting u(x) = sinx in the right hand side yields



1.3. Solution of an Integral Equation 3

RHS = x−∫ x

0

(x− t) sin tdt

= x− [−x cos t− sin t+ t cos t]x0= sinx= u(x)= LHS.

5. Substituting u(x) = coshx in the right hand side yields

RHS = 2 coshx− x sinhx− 1 +

∫ x

0

t cosh tdt

= 2 coshx− x sinhx− 1 + [t sinh t− cosh t]x0= coshx= u(x)= LHS.

6. Substituting u(x) = x in the right hand side yields

RHS = x+ 15x

5 −∫ x

0

t4dt

= x+ 15x

5 − [ 15 t5]x0

= x= u(x)= LHS.

7. Substituting u(x) = x2 in the right hand side yields

RHS = 2x− x4 +

∫ x

0

4t3dt

= 2x− x4 + [t4]x0= 2x= u

′(x)

= LHS.

8. Substituting u(x) = sinx in the right hand side yields

RHS = x cosx− 2 sinx+

∫ x

0

t sin tdt

= x cosx− 2 sinx+ [sin t− t cos t]x0= − sinx= u

′′(x)

= LHS.

9. Substituting u(x) = 3 in the left hand side yields

LHS =

∫ x

0

3 (x− t)2 dt

= [−(x− t)3]x0




= x3

= RHS.

10. Substituting u(x) = 32 in the left hand side yields

LHS =

∫ x

0

3

2(x− t)

12 dt

= [−(x− t) 32 ]x0

= x32

= RHS.

11. Substituting u(x) = 2 cosx− 1 into both sides yields

2 cosx− 1 =

∫ x

0

(x− t)(2 cos t− 1)dt

f(x) = x2

12. Substituting u(x) = ex into both sides yields

ex = f(x)

∫ x

0

(x− t)(et)dt

f(x) = 1 + x

13. Substituting u(x) = e−x3

into both sides yields

e−x3

= 1− α∫ x

0

3t2(e−)3

dt,

α = 1

14. Substituting u(x) = sinx into both sides yields

sinx = f(x)− 1

∫ π2

0

t sin t dt

f(x) = sinx

1.4 Converting Volterra Equation to an ODE

Exercises 1.4

1.

∫ x

0

3(x− t)2u(t)dt

2. 2xex3

− ex2

+

∫ x2

x

textdt

3.

∫ x

0

4(x− t)3u(t)dt

4. 4 sin 5x− sin 2x+

∫ 4x

x

cos(x+ t)dt



1.4. Converting Volterra Equation to an ODE 5

5. Differentiating both sides three times, we obtain

u′(x) = 1 +

∫ x

0

2(x− t)u(t)dt

u′′(x) =

∫ x

0

2u(t)dt

u′′′

(x) = 2u(x)dtSubstituting x = 0 in the original equation and the first two derivatives,we find u(0) = u

′(0) = 1, u

′′(0) = 0

Hence, the equivalent initial value problem is

u′′′

(x) = 2u(x), u(0) = u′(0) = 1, u

′′(0) = 0

6. Differentiating both sides two times, we obtain

u′(x) = ex −

∫ x

0

u(t)dt

u′′(x) = ex − u(x)

Substituting x = 0 in the original equation and the first derivative, we findu(0) = u

′(0) = 1

Hence, the equivalent initial value problem isu′′(x) + u(x) = ex, u(0) = u

′(0) = 1


u′(x) = 1 +

∫ x

0

u(t)dt

u′′(x) = u(x)

Substituting x = 0 in the original equation and the first derivative, we findu(0) = 0, u

′(0) = 1

Hence, the equivalent initial value problem isu′′(x)− u(x) = 0, u(0) = 0, u

′(0) = 1


u′(x) = 1 + sinx+

∫ x

0

u(t)dt

u′′(x) = cosx+ u(x)

Substituting x = 0 in the original equation and the first derivative, we findu(0) = −1, u

′(0) = 1

Hence, the equivalent initial value problem isu′′(x)− u(x) = cosx, u(0) = −1, u′(0) = 1


u′(x) = 3 + 10x+ u(x) +

∫ x

0

2u(t)dt

u′′(x) = 10 + u

′(x) + 2u(x)




Substituting x = 0 in the original equation and the first derivative, we find

u(0) = 2, u′(0) = 5

Hence, the equivalent initial value problem isu′′(x)− u

′(x)− 2u(x) = 10, u(0) = 2, u

′(0) = 5


u′(x) = 6 + 5u(x)−

∫ x

0

6u(t)dt

u′′(x) = 5u

′(x)− 6u(x)

Substituting x = 0 in the original equation and the first derivative, we findu(0) = −5, u

′(0) = −19

Hence, the equivalent initial value problem isu′′(x)− 5u

′(x) + 6u(x) = 0, u(0) = −5, u

′(0) = −19

11. Differentiating both sides we obtainu′(x) = sec2x− u(x)

Substituting x = 0 in the original equation we findu(0) = 0Hence, the equivalent initial value problem isu′(x) + u(x) = sec2x, u(0) = 0

12. Differentiating both sides three times, we obtain

u′(x) = 1 + 5x+ 3u(x) +

∫ x

0

[6− 5(x− t)]u(t)dt

u′′(x) = 5 + 3u

′(x) + 6u(x)− 5

∫ x

0

u(t)dt

u′′′

(x) = 3u′′(x) + 6u

′(x)− 5u(x)

Substituting x = 0 in the original equation and the first two derivatives,

we find u(0) = 1, u′(0) = 4, u

′′(0) = 23

Hence, the equivalent initial value problem isu′′′

(x)− 3u′′(x)− 6u

′(x) + 5u(x) = 0,

u(0) = 1, u′(0) = 4, u

′′(0) = 23

13. u′′′

(x)− 4u(x) = 24x,u(0) = u

′(0) = 0, u

′′(0) = 2

14. uiv(x)− u(x) = 0,u(0) = u

′(0) = 0, u

′′(0) = 2, u

′′′(0) = 0



1.5. Converting IVP to Volterra Equation 7

1.5 Converting IVP to Volterra Equation

Exercises 1.5

1. We first sety′(x) = u(x)

Integrating both sides from 0 to x we findy(x)− 1 =

∫ x0u(t)dt

Substituting in the original equation we find

u(x) = −1−∫ x

0

u(t)dt


Integrating both sides from 0 to x, we findy(x) =

∫ x0u(t)dt


u(x) = x+

∫ x

0

u(t)dt


Integrating both sides from 0 to x we findy(x) =

∫ x0u(t)dt


u(x) = sec2(x)−∫ x

0

u(t)dt

4. We first sety′′(x) = u(x)

Integrating both sides from 0 to x twice givesy′(x)− y′(0) =

∫ x0u(t)dt

y(x)− y(0)− xy′(0) =∫ x0

∫ x0u(t)dtdt

or equivalentlyy(x) = 1 +

∫ x0

(x− t)u(t)dtSubstituting in the original equation we find

u(x) = −1−∫ x

0

(x− t)u(t)dt


Integrating both sides from 0 to x twice gives

y′(x)− y′(0) =

∫ x

0

u(t)dt




y(x)− y(0)− xy′(0) =

∫ x

0

∫ x

0

u(t)dtdt

or equivalentlyy(x) = 1 + x+

∫ x0


u(x) = 1 + x+

∫ x

0

(x− t)u(t)dt



y′(x)− y′(0) =

∫ x

0

u(t)dt

y(x)− y(0)− xy′(0) =

∫ x

0

∫ x

0

u(t)dtdt

or equivalentlyy(x) = 1 + x+

∫ x0


u(x) = −11− 6x−∫ x

1

[5 + 6(x− t)]u(t)dt



y′(x)− y′(1) =

∫ x

1

u(t)dt

or equivalentlyy′(x) = 1 +

∫ x1u(t)dt


u(x) = −1−∫ x

1

u(t)dt



y′(x)− y′(0) =

∫ x

0

u(t)dt

y(x)− y(0)− xy′(0) =

∫ x

0

∫ x

0

u(t)dtdt

or equivalentlyy(x) = x+

∫ x0


u(x) = −1 + 4x+

∫ x

0

[2(x− t)− 1]u(t)dt



1.5. Converting IVP to Volterra Equation 9



y′(x)− y′(0) =

∫ x

0

u(t)dt

y(x)− y(0)− xy′(0) =

∫ x

0

∫ x

0

u(t)dtdt

or equivalentlyy(x) =

∫ x0


u(x) = sinx−∫ x

0

(x− t)u(t)dt



y′(x)− y′(0) =

∫ x

0

u(t)dt

y(x)− y(0)− xy′(0) =

∫ x

0

∫ x

0

u(t)dtdt

or equivalentlyy(x) = 1− x+

∫ x0


u(x) = x− sinx+ xex − ex −∫ x

0

[(x− t)ex − sinx]u(t)dt

11. We first sety′′′

(x) = u(x)Integrating both sides from 0 to x three times gives

y′′(x)− y′′(0) =

∫ x

0

u(t)dt

y′(x)− y′(0)− xy′′(0) =

∫ x

0

∫ x

0

u(t)dtdt

y(x)− y(0)− xy′(0)− 12x

2y′′(0) =

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdt

or equivalentlyy(x) = 2 + x2 + 1

2

∫ x0

(x− t)2u(t)dtSubstituting in the original equation we find

u(x) = 2x− x2 +

∫ x

0

[1 + (x− t)− 1

2(x− t)2

]u(t)dt

12. We first sety′′′

(x) = u(x)Integrating both sides from 0 to x three times gives




y′′(x)− y′′(0) =

∫ x

0

u(t)dt

y′(x)− y′(0)− xy′′(0) =

∫ x

0

∫ x

0

u(t)dtdt

or equivalentlyy′(x) = x+

∫ x0


u(x) = −3x− 4

∫ x

0

(x− t)u(t)dt

13. We first setyiv(x) = u(x)Integrating both sides from 0 to x three times gives

y′′′

(x)− y′′′(0) =

∫ x

0

u(t)dt

y′′(x)− y′′(0)− xy′′′(0) =

∫ x

0

∫ x

0

u(t)dtdt

y′(x)− y′(0)− xy′′(0)− 1

2x2y′′′

(0) =

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdt

y(x)− y(0)− xy′(0)− 12x

2y′′(0)− 1

6x3y′′′

(0) =

∫ x

0

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdtdt

and by changing multiple integrals to single integrals and by substitutingin the original equation we find

u(x) = 2 + x− 1

2x2 − 1

6x3 −

∫ x

0

[2(x− t) +

1

6(x− t)3

]u(t)dt

14. We first setyiv(x) = u(x)Integrating both sides from 0 to x three times gives

y′′′

(x)− y′′′(0) =

∫ x

0

u(t)dt

y′′(x)− y′′(0)− xy′′′(0) =

∫ x

0

∫ x

0

u(t)dtdt

y′(x)− y′(0)− xy′′(0)− 1

2x2y′′′

(0) =

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdt

y(x)− y(0)− xy′(0)− 12x

2y′′(0)− 1

6x3y′′′

(0) =

∫ x

0

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdtdt


u(x) = 1− 1

2x2 +

1

3!

∫ x

0

(x− t)3u(t)dt

15. We first setyiv(x) = u(x)



1.6. Converting BVP to Fredholm Equation 11

Integrating both sides from 0 to x three times gives

y′′′

(x)− y′′′(0) =

∫ x

0

u(t)dt

y′′(x)− y′′(0)− xy′′′(0) =

∫ x

0

∫ x

0

u(t)dtdt

y′(x)− y′(0)− xy′′(0)− 1

2x2y′′′

(0) =

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdt

y(x)− y(0)− xy′(0)− 12x

2y′′(0)− 1

6x3y′′′

(0) =

∫ x

0

∫ x

0

∫ x

0

∫ x

0

u(t)dtdtdtdt


u(x) = 2ex − 1− x−∫ x

0

(x− t)u(t)dt

1.6 Converting BVP to Fredholm Equation

Exercises 1.6


Integrating both sides from 0 to x twice we findy′(x) = y

′(0) +

∫ x0u(t)dt

y(x) = y(0) + xy′(0) +

∫ x0

(x− t)u(t)dtSubstituting x = 1 in the last equation and using the boundary conditionat x = 1 givesy′(0) = −

∫ 1

0(1− t)u(t)dt

Accordingly, we obtainy(x) = −x

∫ 1

0(1− t)u(t)dt+

∫ x0

(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = sinx+ 4x

∫ x0

(1− t)u(t)dt+ 4x∫ 1

x(1− t)u(t)dt− 4

∫ x0

(x− t)u(t)dtThis gives

u(x) = sinx+

∫ 1

0

K(x, t)u(t)dt

K(x, t) =

{4t(1− x) 0 ≤ t ≤ x4x(1− t) x ≤ t ≤ 1



′(0) +

∫ x0u(t)dt

y(x) = y(0) + xy′(0) +

∫ x0

(x− t)u(t)dtSubstituting x = 1 in the last equation and using the boundary condition




at x = 1 we obtainy′(0) = −

∫ 1

0(1− t)u(t)dt

Accordingly, we obtainy(x) = −x

∫ 1

0(1− t)u(t)dt+

∫ x0

(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = 1 + 2x2

∫ x0

(1− t)u(t)dt+ 2x2∫ 1

x(1− t)u(t)dt− 2x

∫ x0


u(x) = 1 +

∫ 1

0

K(x, t)u(t)dt,

K(x, t) =

{2xt(1− x) 0 ≤ t ≤ x2x2(1− t) x ≤ t ≤ 1



′(0) +

∫ x0u(t)dt

y(x) = y(0) + xy′(0) +

∫ x0

(x− t)u(t)dtSubstituting x = 1 in the last equation and using the boundary conditionat x = 1 we obtainy′(0) = −1−

∫ 1

0(1− t)u(t)dt

Accordingly, we obtainy(x) = 1− x− x

∫ 1

0(1− t)u(t)dt+

∫ x0

(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = (2x− 1) + x

∫ x0

(1− t)u(t)dt+ x∫ 1

x(1− t)u(t)dt+

∫ x0


u(x) = (2x− 1) +

∫ 1

0

K(x, t)u(t)dt,

K(x, t) =

{t(1− x) 0 ≤ t ≤ xx(1− t) x ≤ t ≤ 1



′(0) +

∫ x0u(t)dt

y(x) = y(0) + xy′(0) +

∫ x0

(x− t)u(t)dtSubstituting x = 1 in the first equation and using the boundary conditionx = 1 we obtainy′(0) = −

∫ 1

0(1− t)u(t)dt

Accordingly, we obtainy(x) = 1− x

∫ 1

0u(t)dt+

∫ x0

(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = x− 1 + x

∫ x0u(t)dt+ x

∫ 1

xu(t)dt−

∫ x0

(x− t)u(t)dt



1.7. Taylor Series 13

This gives u(x) = (x− 1) +

∫ 1

0

K(x, t)u(t)dt,

K(x, t) =

{t 0 ≤ t ≤ xx x ≤ t ≤ 1

1.7 Taylor Series

Exercises 1.7

1. f(x) = e2x 2. f(x) = e−3x

3. f(x) = ex − 1 4. f(x) = cos(2x)5. f(x) = sin(3x) 6. f(x) = sinh(2x)7. f(x) = cosh(2x) 8. f(x) = cosh(3x)− 19. f(x) = 1 + cos(2x) 10. f(x) = 1 + sinx



Chapter 2

Fredholm IntegralEquations

2.2 Adomian Decomposition Method

Exercises 2.2

1. We set

u0(x) =13

3x

Hence, we find

u1(x) = −1

4

∫ 1

0

13

3xt2dt

u1(x) = − 13

3× 12x

and

u2(x) =13

3× 144x

and so on. Accordingly, we find

u(x) =13

4x

(1− 1

12+

1

144+ · · ·

)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = 4x

2. We set

u0(x) = x3 − 1

5x

Hence, we find

15



16 Chapter 2. Fredholm Integral Equations

u1(x) =

∫ 1

0

x

(t4 − 1

5t2)dt =

2

15x

and

u2(x) =2

45x


u(x) = x3 − 15x+

2

15x

(1 +

1

3+

1

9+ · · ·

)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = x3

3. We setu0(x) = x2

Hence, we find

u1(x) =

∫ 1

0

xt3dt =1

4x

and

u2(x) =1

12x


u(x) = x2 +1

4x

(1 +

1

3+

1

9+ · · ·

)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = x2 + 3

8x

4. We setu0(x) = ex

Hence, we find

u1(x) = e−1∫ 1

0

etdt

u1(x) = 1− e−1andu2(x) = e−1 − e−2u3(x) = e−2 − e−3and so on. Accordingly, we findu(x) = ex + 1−

(e−1 − e−1 + e−2 − e−2 + · · ·

)u(x) = ex + 1

5. We setu0(x) = x+ sinxHence, we find



2.2. Adomian Decomposition Method 17

u1(x) = −x∫ π

2

0

(t+ sin t)dt

u1(x) = −(1 +π2

8)x

and

u2(x) = (1 +π2

8)π2

8x


u(x) = x+ sinx− (1 +π2

8)

(1− π2

8+π4

64+ · · ·

)x

And by evaluating the infinite geometric series at the right hand side, wefindu(x) = sinx

6. We setu0(x) = x+ cosxHence, we find

u1(x) = −2x

∫ π6

0

(t+ cos t)dt

u1(x) = −(1 +π2

36)x

and

u2(x) = (1 +π2

36)π2

36x


u(x) = x+ cosx− (1 +π2

36)

(1− π2

36+

π4

362+ · · ·

)x

And by evaluating the infinite geometric series at the right hand side, wefindu(x) = cosx

7. We setu0(x) = cos(4x) + 1

4xHence, we find

u1(x) = −x∫ π

8

0

(cos 4t+1

4t)dt

u1(x) = −1

4(1 +

π2

2× 82)x

and

u2(x) =1

4(1 +

π2

2× 82)

π2

2× 82x


u(x) = cos 4x+1

4x− 1

4

(1 +

π2

2× 82

)(1− π2

2× 82+

π4

4× 84+ · · ·

)x




Hence, we findu(x) = cos 4x

8. We setu0(x) = sinhx− e−1xHence, we find

u1(x) =

∫ 1

0

x t (sinh t− e−1t)dt

u1(x) =2

3e−1x

and

u2(x) =2

9e−1x


u(x) = sinhx− e−1x+2

3e−1x

(1 +

1

3+

1

9+ · · ·

)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = sinhx

9. We setu0(x) = 2e2x + (1− e2)xHence, we find

u1(x) = x

∫ 1

0

(2e2t + (1− e2)t

)dt

u1(x) = −1

2(1− e2)x

and

u2(x) = −1

4(1− e2)x


u(x) = 2e2x + (1− e2)x− 1

2(1− e2)x

(1 +

1

2+

1

4+ · · ·

)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = 2e2x

10. We setu0(x) = 1 + sec2xHence, we find

u1(x) = −∫ π

4

0

(1 + sec2t

)dt

u1(x) = −(1 +π

4)

and




u2(x) = (1 +π

4)π

4and so on. Accordingly, we find

u(x) = 1 + sec2x− (1 +π

4)

(1− π

4+π2

16+ · · ·

)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = sec2x

11. We setu0(x) = sinxHence, we find

u1(x) = esin−1x

∫ 1

−1sin tdt

u1(x) = 0andun(x) = 0, for n ≥ 1Accordingly, we findu(x) = sinx

12. We setu0(x) = tanxHence, we find

u1(x) = −etan−1x

∫ π3

−π3tan t dt

u1(x) = 0andun(x) = 0, for n ≥ 1Accordingly, we findu(x) = tanx

13. We setu0(x) = tan−1xHence, we find

u1(x) =1

2(ln 2− π

2)x+ x

∫ 1

0

tan−1dt

u1(x) =1

2(ln 2− π

2)x+ x[t(tan−1t)− 1

2ln(1 + t2)]10

and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = tan−1x

14. We set




u0(x) = coshxHence, we findu1(x) = (sinh 1)x+ (e−1 − 1)−

∫ 1

0(x− t) cosh tdt

u1(x) = (sinh 1)x+ (e−1 − 1)− [x sinh t− t sinh t+ cosh t]10and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = coshx

15. We set

u0(x) =1

1 + x2Hence, we find

u1(x) = (eπ4 − e−π4 )x− x

∫ 1

−1

1

1 + t2etan

−1tdt

Using integration by substitution, we set v = tan−1t, we obtainu1(x) = (e

π4 − e−π4 )x− x[etan

−1t]1−1and, henceun(x) = 0, for n ≥ 1Accordingly, we find

u(x) =1

1 + x2

16. We set

u0(x) =1√

1− x2Hence, we find

u1(x) = (eπ6 − 1)x− x

∫ 12

0

1√1− t2

esin−1tdt

Using integration by substitution, we set v = sin−1t, we obtain u1(x) =

(eπ6 − 1)x− x[esin

−1t]120

and, henceun(x) = 0, for n ≥ 1accordingly, we find

u(x) =1√

1− x2

17. We set

u0(x) =1

1 + x2Hence, we find

u1(x) =π2

32x− x

∫ 1

0

1

1 + t2tan−1t dt

Using integration by substitution, we set v = tan−1t, we obtain




u1(x) =π2

32x− x[

(tan−1t)2

2]10

and, henceun(x) = 0, for n ≥ 1Accordingly, we find

u(x) =1

1 + x2

18. We setu0(x) = cos−1xHence, we find

u1(x) = −πx+ x

∫ 1

−1cos−1tdt

u1(x) = −πx+ x[t(cos−1t)−√

1− t2]1−1and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = cos−1x

19. We setu0(x) = xtan−1xHence, we find

u1(x) = (π

4− 1

2)x− x

∫ 1

0

t tan−1dt

u1(x) = (π

4− 1

2)x− 1

2x[(1 + t2)tan−1t− t]10

and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = xtan−1x

20. We setu0(x) = xsin−1x+ 1Hence, we find

u1(x) = −(π

8+ 1)x+ x

∫ 1

0

(tsin−1t+ 1)dt

u1(x) = (π

8+ 1)x+

1

4x[(2t2 − 1)sin−1t+ t

√1− t2]10

and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = xsin−1x+ 1

21. We set




u0(x) = sin x1+sin x + x− π

2x

u1(x) =∫ π

2

0xu0(t) dt = −x+ π

2x+ π2

8 x−π3

16xBy cancelling the noise terms, we findu(x) = sin x

1+sin x

22. We setu0(x) = sin x

1+cos x − x ln 2

u1(x) =∫ π

2

0xu0(t) dt = x ln 2− π2

8 x ln 2By cancelling the noise terms, we findu(x) = sin x

1+cos x

23. We setu0(x) = sec2 x

1+tan x − x ln 2

u1(x) =∫ π

4

0xu0(t) dt = x ln 2− π2

32x ln 2By cancelling the noise terms, we findu(x) = sec2 x

1+tan x

24. We setu0(x) = 1 + sinx− x− π2

8 x

u1(x) =∫ π

2

0xtu0(t) dt = x+ π2

8 x+ · · ·By cancelling the noise terms, we findu(x) = 1 + sinx

25. We setu0(x) = 1 + sinx+ cosx− 2x− π

2x+ π2 + π2

8

u1(x) =∫ π

2

0(x− t)u0(t) dt = 2x+ π

2x−π2 −

π2

8 + · · ·By cancelling the noise terms, we findu(x) = 1 + sinx+ cosx

26. We setu0(x) = x sinx− x− 2 + π

u1(x) =∫ π

2

0(x− t)u0(t) dt = x+ 2− π + · · ·

By cancelling the noise terms, we findu(x) = x sinx

2.3 The Variational Iteration Method

Exercises 2.3

1. Differentiating both sides givesu′(x) = 3x2 − 1

5 +∫ 1

0tu(t) dt, u(0) = 0

u0(x) = 0



2.3. The Variational Iteration Method 23

u1(x) = x3 − 15x

u2(x) = x3 − 115x

u3(x) = x3 − 145x

...un(x) = x3 − 1

5···3n−1x, n ≥ 1u(x) = x3

2. Differentiating both sides givesu′(x) = ex − 1 +

∫ 1

0tu(t) dt, u(0) = 1

u0(x) = 1u1(x) = ex − 1

2x

u2(x) = ex − 16x

u3(x) = ex − 118x

...un(x) = ex − 1

2···3n−1x, n ≥ 1u(x) = ex

3. Differentiating both sides givesu′(x) = 2

3 +∫ 1

0tu(t) dt, u(0) = 0

u0(x) = 0u1(x) = 2

3x

u2(x) = 89x

u3(x) = 2627x

...un(x) = 3n−1

3n x, n ≥ 1u(x) = x

4. Differentiating both sides givesu′(x) = 4x3 + 2x− 5

12 +∫ 1

0tu(t) dt, u(0) = 0

u0(x) = 0u1(x) = x2 + x4 − 5

12x

u2(x) = x2 + x4 − 536x

u3(x) = x2 + x4 − 5108x

...un(x) = x2 + x4 5

12·3n−1x, n ≥ 1u(x) = x2 + x4




5. Differentiating both sides givesu′(x) = ex + 2− 3

4 +∫ 1

0tu(t) dt, u(0) = 1

u0(x) = 1u1(x) = ex + 13

8 x

u2(x) = ex + 2732x

u3(x) = ex + 133128x

...u(x) = ex

6. Differentiating both sides givesu′(x) = −e−x + 2 + 3

2 +∫ 0

−1 tu(t) dt, u(0) = 1u0(x) = 1u1(x) = e−x + 5

4x

u2(x) = e−x + 98x

u3(x) = e−x + 1716x

...u(x) = e−x

7. Differentiating both sides givesu′(x) = 1− 1

6x+ 2∫ 1

0tu(t) dt, u(0) = 1

u0(x) = 1u1(x) = 1 + x+ 5

12x2

u2(x) = 1 + x+ 4148x

2

u3(x) = 1 + x+ 185192x

2

...u(x) = 1 + x+ x2

8. Differentiating both sides givesu′(x) = ex − 2x+ 2x

∫ 1

0tu(t) dt, u(0) = 1

u0(x) = 1u1(x) = ex − 1

2x2

u2(x) = ex − 18x

2

u3(x) = ex − 132x

2

...un(x) = ex − 1

2·4n−1x2, n ≥ 1 u(x) = ex



2.4. The Direct Computation Method 25

2.4 The Direct Computation Method

Exercises 2.4

1. We setu(x) = xex + (α− 1)xwhere

α =

∫ 1

0

u(t) dt

Accordingly, we find

α =

∫ 1

0

(tet + (α− 1)t

)dt

This givesα = 1Substituting for α in the first equation we findu(x) = xex

2. We set

u(x) = x2 − 25

12x+ 1 + αx

where

α =

∫ 1

0

tu(t) dt


α =

∫ 1

0

t

(t2 − 25

12t+ 1 + αt

)dt

This gives

α =1

12Substituting for α in the first equation we findu(x) = x2 − 2x+ 1

3. We setu(x) = x sinx+ (α− 1)xwhereα =

∫ π2

0u(t) dt


α =

∫ π/2

0

(t sin t+ (α− 1)t) dt

This givesα = 1Substituting for α in the first equation we findu(x) = x sinx

4. We set




u(x) = e2x − 1

4(e2 + 1)x+ αx

where

α =

∫ 1

0

tu(t) dt


α =

∫ 1

0

t

(e2t − 1

4(e2 + 1)t+ αt

)dt

This gives

α =1

4(e2 + 1)

Substituting for α in the first equation we findu(x) = e2x

5. We setu(x) = sec2x+ (α− π

4 )where

α =

∫ π4

0

u(t) dt


α =

∫ π4

0

(sec2t+ (α− π

4

)dt

This givesα = 1 + π

4

Substituting for α in the first equation we findu(x) = 1 + sec2x

6. We set

u(x) = sin 2x+ (α− 1

2)x

where

α =

∫ π4

0

u(t) dt


α =

∫ π4

0

(sin 2t+ (α− 1

2)t

)dt

This gives

α =1

2Substituting for α in the first equation we findu(x) = sin 2x

7. We set

u(x) = x2 − 1

3x− 1

4+ α(x+ 2)

where




α =

∫ 1

0

u(t) dt


α =

∫ 1

0

(t2 − 1

3t− 1

4+ α(t+ 2)

)dt

This gives

α =1

18Substituting for α in the first equation we find

u(x) = x2 − 5

18x− 5

36

8. We setu(x) = sinx+ cosx+ (α− π

2)x

where

α =

∫ π2

0

tu(t) dt


α =

∫ π2

0

t(

sin t+ cos t+ (α− π

2)t)dt

This gives

α =π

2Substituting for α in the first equation we findu(x) = sinx+ cosx

9. We setu(x) = secx tanx+ (1− α)xwhere

α =

∫ π3

0

u(t) dt


α =

∫ π3

0

(sec t tan t+ (1− α)t) dt

This givesα = 1Substituting for α in the first equation we findu(x) = secx tanx

10. We set

u(x) = x2 − 1

6x− 1

24+

1

2α(1 + x)− 1

2β

where

α =

∫ 1

0

u(t) dt




β =

∫ 1

0

tu(t) dt


α =

∫ 1

0

(t2 − 1

6t− 1

24+

1

2α(1 + t)− 1

2β

)dt

β =

∫ 1

0

t

(t2 − 1

6t− 1

24+

1

2α(1 + t)− 1

2β

)dt

This gives

α =1

3

β =1

4Substituting for α and β in the first equation we findu(x) = x2

11. We set

u(x) = sinx+1

4(α− 1)x

where

α =

∫ π2

0

tu(t) dt


α =

∫ π2

0

t

(sin t+

1

4(α− 1)t

)dt

This givesα = 1Substituting for α in the first equation we findu(x) = sinx

12. We setu(x) = 1 + α lnx+ βwhere

α =

∫ 1

0+u(t) dt

β =

∫ 1

0+ln t u(t) dt


α =

∫ 1

0+(1 + α ln t+ β) dt

β =

∫ 1

0+ln t (1 + α ln t+ β) dt

This gives

α =1

2



2.5. Successive Approximations Method 29

β = 0Substituting for α and β in the first equation we find

u(x) = 1 +1

2lnx

13. u(x) = x3

14. u(x) = 1 +π

4sec2x

2.5 Successive Approximations Method

Exercises 2.5

1. We selectu0(x) = 0Substituting in the original equation we find

u1(x) =11

12x

Substituting u1(x) in the original equation we obtain

u2(x) =11

12x+

1

4

∫ 1

0

11

12xt2dt

so that

u2(x) =122 − 1

122x

Proceeding as before we find

u3(x) =123 − 1

123x

...

un(x) =12n − 1

12nx

Hence, u(x) = limn→∞ un(x) = x


u1(x) =6

7x3


u2(x) =6

7x3 +

5

7

∫ 1

0

6

7x3t4dt

so that

u2(x) =72 − 1

72x3





u3(x) =73 − 1

73x

...

un(x) =7n − 1

7nx3

Hence, u(x) = limn→∞ un(x) = x3


u1(x) =13

3x

u1(x) in the original equation we obtain

u2(x) =13

3x− 1

4

∫ 1

0

13

3xt2dt

so that

u2(x) =143

3× 12x


u3(x) =1729

3× 122x

...

un(x) = 4x+(−1)n+1

3× 12n−1x, n ≥ 1

Hence, u(x) = limn→∞ un(x) = 4x

4. We selectu0(x) = 0Substituting in the original equation we findu1(x) = 1Substituting u1(x) in the original equation we obtain

u2(x) = 1 +

∫ 1

0

xdt

so thatu2(x) = 1 + xProceeding as before we find

u3(x) = 1 +3

2x

u4(x) = 1 +7

4x

...

un(x) = 1 +2n − 2

2n−1x, n ≥ 1

Hence, u(x) = limn→∞ un(x) = 1 + 2x




5. We selectu0(x) = 0Substituting in the original equation we findu1(x) = sinxSubstituting u1(x) in the original equation we obtain

u2(x) = sinx+

∫ π2

0

sinx sin t cos tdt

so that

u2(x) =3

2sinx


u3(x) =7

4sinx

...

un(x) =2n − 1

2n−1sinx, n ≥ 1

Hence, u(x) = limn→∞ un(x) = 2 sinx


u1(x) = −1

2+ sec2x


u2(x) = −1

2+ sec2x+

1

2

∫ π4

0

(−1

2+ sec2t)dt

so thatu2(x) = − π

16+ sec2x


u3(x) = − π2

128+ sec2x

...

un(x) = − πn−1

2× 8n−1+ sec2x, n ≥ 1

Hence, u(x) = limn→∞ un(x) = sec2x


u1(x) = −1

4+ secx tanx

u1(x) in the original equation we obtain

u2(x) = −1

4+ secx tanx+

1

4

∫ π3

0

(−1

4+ sec t tan t)dt




so thatu2(x) = − π

4× 12+ secx tanx


u3(x) = − π2

4× 122+ secx tanx

...

un(x) = − πn−1

4× 12n−1+ secx tanx

Hence, u(x) = limn→∞ un(x) = secx tanx

8. We selectu0(x) = 0Substituting in the original equation we findu1(x) = coshx+ (1− e−1)xSubstituting u1(x) in the original equation we obtain

u2(x) = coshx+ (1− e−1)x−∫ 1

0

xt(cosh t+ (1− e−1)t)dt

so that

u2(x) = coshx− 1

3(1− e−1)x


u3(x) = coshx+1

9(1− e−1)x

...

un(x) = coshx+(−1)n+1

3n−1(1− e−1)x, n ≥ 1

Hence, u(x) = limn→∞ un(x) = coshx

9. We selectu0(x) = 0Substituting in the original equation we findu1(x) = ex − (sinh 1)xSubstituting u1(x) in the original equation we obtain

u2(x) = ex − (sinh 1)x+1

2

∫ 1

−1x(et − (sinh 1)t)dt

so thatu2(x) = ex

Proceeding as before we findu3(x) = ex

...un(x) = ex

Hence, u(x) = limn→∞ un(x) = ex



2.6. Successive Substitutions Method 33

10. We selectu0(x) = 0Substituting in the original equation we findu1(x) = 1

4x+ sinxSubstituting u1(x) in the original equation we obtain

u2(x) = 14x+ sinx− x

∫ π2

0

(1

4t+ sin t)dt

so that

u2(x) = sinx− π2

128x


u3(x) = sinx+π4

128× 32x

...

un(x) = sinx+ (−1)n+1 π2n−2

2n+1 × (4)2n−2x, n ≥ 1

Hence, u(x) = limn→∞ un(x) = sinx

2.6 Successive Substitutions Method

Exercises 2.6

1. Using the successive substitutions method, and noting that f(x) =116 x, λ = 1

4 , K(x, t) = xt, we find

u(x) =11

6x+

1

4

∫ 1

0

11

6xt2dt+

1

16

∫ 1

0

∫ 1

0

11

6xt1

2t2 dt1dt+ · · ·so that

u(x) =11

6x+

11

72x+

11

864x+ · · ·

=11

6x[1 +

1

12+

1

144+ · · ·] = 2x

by finding the sum of the infinite geometric series.

2. Using the successive substitutions method, and noting thatf(x) = 1, λ = − 1

4 , K(x, t) = cosx, we find

u(x) = 1− 1

4

∫ π2

0

cosxdt+1

16

∫ π2

0

∫ π2

0

cosx cos t dt1dt+ · · ·so thatu(x) = 1− π

8cosx+

π

32cosx+ · · ·

= 1− π

8cosx[1− 1

4+

1

16+ · · ·]

= 1− π

10cosx




by finding the sum of the infinite geometric series.

3. Using the successive substitutions method, and noting thatf(x) = 7

12x+ 1, λ = 12 , K(x, t) = xt, we find

u(x) =7

12x+ 1 +

1

2

∫ 1

0

xt(7

12t+ 1)dt+

1

4

∫ 1

0

∫ 1

0

xt2t21(7

12t1 + 1) dt1dt+ · · ·

so that

u(x) = 1 +7

12x+

25

72x

(1 +

1

6+

1

36+ · · ·

)u(x) = 1 +

7

12x+

5

12x = 1 + x

4. Using the successive substitutions method, and noting thatf(x) = cosx, λ = 1

2 , K(x, t) = sinx, we find

u(x) = cosx+1

2

∫ π2

0

sinx cos tdt+1

4

∫ π2

0

∫ π2

0

sinx sin t cos t1 dt1dt+ · · ·so that

u(x) = cosx+1

2sinx+

1

4sinx+

1

8sinx+ · · ·

= cosx+1

2sinx[1 +

1

2+

1

4+ · · ·] = cosx+ sinx


8x2, λ = 1

2 , K(x, t) = x2t, we find

u(x) =7

8x2 +

1

2

∫ 1

0

x2t(7

8t2)dt+

1

4

∫ 1

0

∫ 1

0

7

8x2t3t31 dt1dt+ · · ·

so that

u(x) =7

8x2 +

7

64x2 +

7

512x2 + · · ·

u(x) =7

8x2(

1 +1

8+

1

64+ · · ·

)= x2


10x3, λ = 1

2 , K(x, t) = x3t, we find

u(x) =9

10x3 +

1

2

∫ 1

0

9

10x3t4dt+

1

4

∫ 1

0

∫ 1

0

9

10x3t4t41 dt1dt+ · · ·

so that

u(x) =9

10x3 +

9

100x3 +

9

1000x3 + · · ·

u(x) =9

10x3(

1 +1

10+

1

100+ · · ·

)= x3

7. Using the successive substitutions method, and noting thatf(x) = sinx, λ = 1

2 , K(x, t) = cosx, we find

u(x) = sinx+1

2

∫ π2

0

cosx sin tdt+1

4

∫ π2

0

∫ π2

0

cosx cos t sin t1 dt1dt+ · · ·



2.8. Homogeneous Fredholm Equation 35

so that

u(x) = sinx+1

2cosx+

1

4cosx+

1

8cosx+ · · ·

= sinx+1

2cosx[1 +

1

2+

1

4+ · · ·] = sinx+ cosx

8. Using the successive substitutions method, and noting thatf(x) = 1, λ = 1

2 , K(x, t) = sinx, we find

u(x) = 1 +1

2

∫ π2

0

sinxdt+1

4

∫ π2

0

∫ π2

0

sinx sin t dt1dt+ · · ·so thatu(x) = 1 +

π

4sinx+

π

8sinx+

π

16sinx+ · · ·

= 1 +π

4sinx[1 +

1

2+

1

4+ · · ·]

= 1 + π2 sinx


2 , K(x, t) = sec2x, we find

u(x) = 1 +1

2

∫ π4

0

sec2xdt+1

4

∫ π4

0

∫ π4

0

sec2x sec2t dt1dt+ · · ·so thatu(x) = 1 +

π

8sec2x+

π

16sec2x+ · · ·

= 1 +π

8sec2x+ [1 +

1

2+

1

4+ · · ·]

= 1 +π

4sec2x


5 , K(x, t) = secx tanx, we find

u(x) = 1 +1

5

∫ π3

0

secx tanxdt+1

25

∫ π3

0

∫ π3

0

secx tanx sec t tan t dt1dt+ · · ·so thatu(x) = 1 +

π

15secx tanx+

π

75secx tanx+ · · ·

u(x) = 1 +π

15secx tanx

(1 +

1

5+

1

25+ · · ·

)u(x) = 1 +

π

12secx tanx

2.8 Homogeneous Fredholm Equations

Exercises 2.8

1. Using the direct computation method we find




u(x) = 2αλwhere

α =

∫ 1

0

tu(t)dt

Substituting for u(t) from the above equation we find

α =

∫ 1

0

2αλt dt

Integrating and solving for λ where α is a constant we obtainλ = 1Hence, u(x) = 2α

2. Using the direct computation method we findu(x) = 4αλxwhere

α =

∫ 1

0

u(t)dt


α =

∫ 1

0

4αλt dt

Integrating and solving for λ where α is a constant we obtain

λ =1

2Hence, u(x) = 2αx

3. Using the direct computation method we findu(x) = αλxwhere

α =

∫ 1

0

et u(t)dt


α =

∫ 1

0

αλtet dt

Integrating and solving for λ where α is a constant we obtainλ = 1Hence, u(x) = αx

4. Using the direct computation method we findu(x) = αλ cosxwhere

α =

∫ π2

0

sin t u(t)dt


α =

∫ π2

0

αλ sin t cos tdt



2.8. Homogeneous Fredholm Equation 37

Integrating and solving for λ where α is a constant we obtainλ = 2u(x) = 2α cosx

5. Expanding sin(x+ t) yields

u(x) =2

πλ

∫ π

0

[sinx cos t+ cosx sin t]u(t)dt

Using the direct computation method we find

u(x) =2

πλ[α sinx+ β cosx]

where

α =

∫ π

0

cos tu(t)dt

β =

∫ π

0

sin tu(t)dt


α =2

πλ

∫ π

0

cos t[α sin t+ β cos t]dt

β =2

πλ

∫ π

0

sin t[α sin t+ β cos t]dt

Integrating and solving for λ where α and β are constants we obtainα = λββ = λαHence, we findλ = ±1β = ±αAccordingly, u(x) = ± 2

πα(sinx± cosx)

6. Expanding cos(x− t) yields

u(x) =2

πλ

∫ π

0

[cosx cos t+ sinx sin t]u(t)dt


u(x) =2

πλ[α cosx+ β sinx]

where

α =

∫ π

0

cos tu(t)dt

β = λ

∫ π

0

sin tu(t)dt


α =2

πλ

∫ π

0

cos t[α cos t+ β sin t]dt

β =2

πλ

∫ π

0

sin t[α cos t+ β sin t]dt




Integrating and solving for λ where α and β are constants we obtainα = λαβ = λβHence, we findλ1 = λ2 = 1

Accordingly, u(x) =2

π(α sinx+ β cosx)

7. Using the direct computation method we findu(x) = αλ secxwhere

α =

∫ π3

0

tan tu(t)dt


α =

∫ π3

0

αλ sec t tan t dt

Integrating and solving for λ where α is a constant we obtainλ = 1Hence we find u(x) = α secx

8. Using the direct computation method we findu(x) = αλsec2xwhere

α =

∫ π4

0

u(t)dt


α =∫ π

4

0αλsec2t dt

Integrating and solving for λ where α is a constant we obtainλ = 1Hence we find u(x) = αsec2x

9. Using the direct computation method we findu(x) = αλsin−1xwhere

α =

∫ 1

0

u(t)dt


α =

∫ 1

0

αλsin−1t dt

Integrating and solving for λ where α is a constant we obtain

λ =2

π − 2

Hence we find u(x) =2

π − 2αsin−1x



2.9. Fredholm Integral Equation of the First Kind 39

10. Using the direct computation method we find

u(x) = αλ(3− 3

2x)

where

α =

∫ 1

0

t u(t)dt


α =

∫ 1

0

αλ(3t− 3

2t2) dt

Integrating and solving for λ where α is a constant we obtainλ = 1

Hence we find u(x) = α(3− 3

2x)

2.9 Fredholm Integral Equation of the First

Kind

Exercises 2.9

1. 13x =

∫ 1

0xtu(t)dt

Using the method of regularization, we obtain

uε(x) = 13εx−

xε

∫ 12

0tuε(t) dt


uε(x) = ( 13ε −

αε )x

whereα =

∫ 1

0tuε(t) dt

This in turn givesα = 1

3+9ε

u(x) = limε→0 uε(x) = x

2. 34x =

∫ 1

0xt2u(t)dt

Using the method of regularization, we obtainuε(x) = 3

4εx−xε

∫ 1

0t2uε(t) dt

Using the direct computation method we finduε(x) = ( 3

4ε −αε )x

whereα =

∫ 1

0t2uε(t) dt


4+16ε

u(x) = limε→0 uε(x) = 3x




3. 12e−x =

∫ 12

0et−xu(t)dt

Using the method of regularization, we obtain

uε(x) = 12εx−

xε

∫ 12

0et−xuε(t) dt

Using the direct computation method we finduε(x) = ( 1

2ε −αε )e−x

whereα =

∫ 12

0exp−tuε(t) dt


2+4ε

u(x) = limε→0 uε(x) = e−x

4. 25x

2 =∫ 1

−1 x2t2u(t)dt

Using the method of regularization, we obtainuε(x) = 2

5εx2 − x2

ε

∫−1 x

2t2uε(t) dtProceeding as before we findα = 4

10+25ε

u(x) = limε→0 uε(x) = x2

5. π2 cosx =∫ π0

cos(x− t)u(t)dtUsing the method of regularization, we obtainuε(x) = π

2ε cosx− απε cosx− β

ε sinxwhereα =

∫ π0

cos(t)uε(t) dtβ =

∫ π0

sin(t)uε(t) dtThis in turn gives

α =π2

2π + 4ε, β = 0

u(x) = limε→0 uε(x) = cosx

6. − π2 cosx =

∫ π0

sin(x− t)u(t)dtUsing the method of regularization, we obtainuε(x) = − π

2ε cosx− απε sinx+ β

ε cosxwhereα =

∫ π0

cos(t)uε(t) dt

β =∫ π0

sin(t)uε(t) dtThis in turn gives

α = − π2ε

π2 + 4ε2, β =

π3

2π2 + 8ε2u(x) = limε→0 uε(x) = sinx



Chapter 3

Volterra IntegralEquations

3.2 Adomian Decomposition Method

Exercises 3.2

1. Using the Adomian decomposition method, we set

u0(x) = 4x+ 2x2

Hence, we find

u1(x) = −∫ x

0

(4t+ 2t2)dt

u1(x) = −2x2 − 2

3x3

and

u2(x) =2

3x3 +

1

6x4

and so on. Substitute the components obtained in the decomposition

u(x) = u0(x) + u1(x) + u2(x) + · · ·

Accordingly, we obtain

41



42 Chapter 3. Volterra Integral Equations

u(x) = 4x+ 2x2 − 2x2 − 2

3x3 +

2

3x3 + · · ·

u(x) = 4x


u0(x) = 1 + x− x2

Hence, we find

u1(x) =

∫ x

0

(1 + t− t2)dt

u1(x) = x+1

2x2 − 1

3x3

and

u2(x) =1

2x2 +

1

6x3 − 1

12x4

u3(x) =1

6x3 +

1

24x4 − 1

60x5


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 1 + 2x


u0(x) = 1

Hence, we find

u1(x) = −∫ x

0

dt

u1(x) = −x

and




u2(x) =1

2!x2

u3(x) = − 1

3!x3


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 1− x+1

2!x2 − 1

3!x3 + · · ·

u(x) = e−x


u0(x) = x

Hence, we find

u1(x) =

∫ x

0

t(x− t)dt

u1(x) =1

3!x3

and

u2(x) =1

5!x5


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = x+1

3!x3 +

1

5!x5 + · · ·

u(x) = sinhx





u0(x) = 3x

Hence, we find

u1(x) = −9

∫ x

0

3t(x− t)dt

u1(x) = − 1

3!(3x)3

and

u2(x) =1

5!(3x)5


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 3x− 1

3!(3x)3 +

1

5!(3x)5 + · · ·

u(x) = sin 3x


u0(x) = 1

Hence, we find

u1(x) = −4

∫ x

0

(x− t)dt

u1(x) = − 1

2!(2x)2

and

u2(x) =1

4!(2x)4


u(x) = u0(x) + u1(x) + u2(x) + · · ·





u(x) = 1− 1

2!(2x)2 +

1

4!(2x)4 + · · ·

u(x) = cos 2x


u0(x) = 1 + x

Hence, we find

u1(x) = −∫ x

0

(1 + t)(x− t)dt

u1(x) = − 1

2!x2 − 1

3!x3

and

u2(x) =1

4!x4 +

1

5!x5


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) =

(1− 1

2!x2 +

1

4!x4 + · · ·

)+

(x− 1

3!x3 +

1

5!x5 + · · ·

)u(x) = cosx+ sinx


u0(x) = 1− x

Hence, we find

u1(x) = −∫ x

0

(1− t)(x− t)dt

u1(x) = − 1

2!x2 +

1

3!x3




and

u2(x) =1

4!x4 − 1

5!x5


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) =

(1− 1

2!x2 +

1

4!x4 + · · ·

)−(x− 1

3!x3 +

1

5!x5 + · · ·

)u(x) = cosx− sinx


u0(x) = 1 + x

Hence, we find

u1(x) =

∫ x

0

(1 + t)(x− t)dt

u1(x) =1

2!x2 +

1

3!x3

and

u2(x) =1

4!x4 +

1

5!x5


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 1 + x+1

2!x2 +

1

3!x3 + · · ·

u(x) = ex





u0(x) = 1− x

Hence, we find

u1(x) =

∫ x

0

(1− t)(x− t)dt

u1(x) =1

2!x2 − 1

3!x3

and

u2(x) =1

4!x4 − 1

5!x5


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 1− x+1

2!x2 − 1

3!x3 + · · ·

u(x) = e−x


u0(x) = 2

Hence, we find

u1(x) =

∫ x

0

2(x− t)dt

u1(x) = 2

(1

2!x2)

and

u2(x) = 2

(1

4!x4)


u(x) = u0(x) + u1(x) + u2(x) + · · ·





u(x) = 2

(1 +

1

2!x2 +

1

4!x4 + · · ·

)u(x) = 2 coshx


u0(x) = 1 + x

Hence, we find

u1(x) =

∫ x

0

(1 + t)dt

u1(x) = x+1

2!x2

and

u2(x) =1

2!x2 +

1

3!x3

u3(x) =1

3!x3 +

1

4!x4


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 1 + 2

(x+

1

2!x2 +

1

3!x3 + · · ·

)u(x) = 2ex − 1


u0(x) = 1− 1

2x2

Hence, we find

u1(x) = −∫ x

0

(1− 1

2t2)(x− t)dt




u1(x) = − 1

2!x2 +

1

4!x4

and

u2(x) = +1

4!x4 +

1

6!x6


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 1 + 2

(− 1

2!x2 +

1

4!x4 + · · ·

)u(x) = 2 cosx− 1


u0(x) = 1 +1

2x2

Hence, we find

u1(x) =

∫ x

0

(1 +1

2t2)(x− t)dt

u1(x) =1

2!x2 +

1

4!x4

and

u2(x) =1

4!x4 +

1

6!x6


u(x) = u0(x) + u1(x) + u2(x) + · · ·


u(x) = 1 + 2

(1

2!x2 +

1

4!x4 + · · ·

)u(x) = 2 coshx− 1




15. Using the modified decomposition method we set

u0(x) = cosx

Hence, we find

u1(x) = (1− esin x)x+ x

∫ x

0

esin t cos tdt

Using the substitution v = sin t, we obtain

u1(x) = 0

and consequently

uk(x) = 0 for k ≥ 1


u(x) = cosx


u0(x) = sec2x

Hence, we find

u1(x) = (1− etan x)x+ x

∫ x

0

etan tsec2tdt

Using the substitution v = tan t, we obtain

u1(x) = 0

and consequently

uk(x) = 0 for k ≥ 1


u(x) = sec2x





u0(x) = coshx

Hence, we find

u1(x) =x

2(1− esinh x) +

x

2

∫ x

0

esinh t cosh tdt

Using the substitution v = sinh t, we obtain

u1(x) = 0

and consequently

uk(x) = 0 for k ≥ 1


u(x) = coshx


u0(x) = sinhx

Hence, we find

u1(x) =1

10(e− ecosh x) +

1

10

∫ x

0

ecosh t sinh tdt

Using the substitution v = cosh t, we obtain

u1(x) = 0

and consequently

uk(x) = 0 for k ≥ 1


u(x) = sinhx


u0(x) = x3




Hence, we find

u1(x) = −x5 + 5

∫ x

0

t4dt

Using the substitution v = cosh t, we obtain

u1(x) = 0

and consequently

uk(x) = 0 for k ≥ 1


u(x) = x3


u0(x) = secx tanx

Hence, we find

u1(x) = (e− esec x) +

∫ x

0

esec t sec t tan tdt

Using the substitution v = sec t, we obtain

u1(x) = 0

and consequently

uk(x) = 0 for k ≥ 1


u(x) = secx tanx

21. Using the noise terms phenomenon we set

u0(x) = 8x− 4x3

Hence, we find




u1(x) = 4x3 − x5

By cancelling 4x3, we find

u(x) = 8x


u0(x) = 8x2 − 2x5

Hence, we find

u1(x) = 2x5 − 27x

8

By cancelling 2x5, we find

u(x) = 8x2


u0(x) = sec2 x− tanx

Hence, we find

u1(x) = tanx+ ln(cosx)

By cancelling tanx, we find

u(x) = sec2 x


u0(x) = 1 + x+ x2 − 14x

4 − 15x

5 − 16x

6

Hence, we find

u1(x) = 14x

4 + 15x

5 + 16x

6 + other terms

By cancelling the noise terms, we find

u(x) = 1 + x+ x2





u0(x) = x sinx+ x cosx− sinx

Hence, we find

u1(x) = sinx− x cosx + other terms


u(x) = x sinx


u0(x) = cosh2 x− 14 sinh(2x)− 1

2x

Hence, we find

u1(x) = 14 sinh(2x) + 1

2x + other terms


u(x) = cosh2 x


Exercises 3.3

1. Differentiating both sides with respect to x givesu′(x) = u(x)

Using the correction functional givesu0(x) = 1,

u1(x) = 1−∫ x0

(u′

0(t)− u0(t))dt

= 1 + x,

u2(x) = 1 + x−∫ x0

(u′

1(t)− u1(t))dt

= 1 + x+ 12!x

2

u3(x) = 1 + x+ 12!x

2 + 13!x

3,and so on. The solution in a series form is given by




u(x) = 1 + x+ 12!x

2 + 13!x

3 + · · ·

that converges to the exact solution

u(x) = ex.

2. Differentiating both sides with respect to x givesu′(x) = 1 +

∫ x0u(t) dt


u1(x) = 0−∫ x0

(u′

0(t)− 1−∫ t0u0(r) dr

)dt

= x,

u2(x) = x−∫ x0

(u′

1(t)− 1−∫ t0u1(r) dr

)dt

= x+ 13!x

3

u3(x) = x+ 13!x

3 + 15!x

5,

and so on. The solution in a series form is given byu(x) = x+ 1

3!x3 + 1

5!x5 + · · ·

that converges to the exact solutionu(x) = sinhx.

3. Differentiating both sides with respect to x givesu′(x) = 3− 9

∫ x0u(t) dt


u1(x) = 0−∫ x0

(u′

0(t)− 3 + 9∫ t0u0(r) dr

)dt

= 3x,

u2(x) = 3x−∫ x0

(u′

1(t)− 3 + 9∫ t0u1(r) dr

)dt

= 3x− 13! (3x)3

u3(x) = (3x)x− 13! (3x)3 + 1

5! (3x)5,and so on. The solution in a series form is given byu(x) = (3x)x− 1

3! (3x)3 + 15! (3x)5 + · · ·

that converges to the exact solutionu(x) = sin(3x).

4. Differentiating both sides with respect to x gives

u′(x) = −4∫ x0u(t) dt

Using the correction functional gives




u0(x) = 1,

u1(x) = 1−∫ x0

(u′

0(t) + 4∫ t0u0(r) dr

)dt

= 1− 2x2,

u2(x) = 1− 2x2 +∫ x0

(u′

1(t) + 4∫ t0u1(r) dr

)dt

= 1− 12! (2x)2 + 1

4! (2x)4

and so on. The solution in a series form is given by

u(x) = 1− 12! (2x)2 + 1

4! (2x)4 + · · ·that converges to the exact solutionu(x) = cos(2x).

5. Using the correction functional givesu0(x) = 1,u1(x) = 1 + x− 1

2!x2

u2(x) = 1 + x− 12!x

2 − 13!x

3 + 14!x

4


u(x) = (x− 13!x

3 + 15!x

5 + · · ·) + (1− 12! (2x)2 + 1

4! (2x)4 + · · ·)that converges to the exact solutionu(x) = sinx+ cosx.

6. Using the correction functional givesu0(x) = 1,u1(x) = 1− x− 1

2!x2

u2(x) = 1− x− 12!x

2 + 13!x

3 + 14!x

4


u(x) = (1− 12! (2x)2 + 1

4! (2x)4 + · · ·)− (x− 13!x

3 + 15!x

5 + · · ·)that converges to the exact solutionu(x) = cosx− sinx.

7. Differentiating both sides with respect to x gives

u′(x) = −x+ cosx+ sinx+∫ x0

(x− t)u(t) dt

Using the correction functional gives

un+1(x) = un(x)−∫ x0

(u′

n(t) + t− cost− sin t−∫ r0

(t− r)un(r) dr)dt

This givesu0(x) = 0,u1(x) = xu2(x) = x− 1

3!x3 + 1

5!x5




3.4. The Series Solution Method 57

u(x) = (x− 13!x

3 + 15!x

5 + · · ·)that converges to the exact solutionu(x) = sinx.

8. Using the correction functional givesu0(x) = 1,u1(x) = 1 + 1

2!x2

u2(x) = 1 + 12!x

2 + 14!x

4


u(x) = 1 + 12!x

2 + 14!x

4 + · · ·that converges to the exact solutionu(x) = coshx.

3.4 The Series Solution Method

Exercises 3.4

1. Substituting u(x) by the series

u(x) =

∞∑n=0

anxn

into both sides of the equation leads to

∞∑n=0

anxn = 2x+ 2x2 − x3 +

∫ x

0

( ∞∑n=0

antn

)dt

Evaluating the regular integrals at the right hand side that involve termsof the form tn, n ≥ 0 yields

a0+a1x+a2x2+a3x

3+ · · · = (a0+2)x+(2+ 12a1)x2+( 1

3a2−1)x3+ · · ·Equating the coefficients of like powers of x in both sides we finda0 = 0a1 = 2a2 = 3ak = 0 for k ≥ 3Accordingly, we findu(x) = 2x+ 3x2





u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = 1 + x− 2

3x3 − 1

2x4 + 2

∫ x

0

t

( ∞∑n=0

antn

)dt

Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 yields

a0 + a1x+ a2x2 + a3x

3 + · · · = 1 + x+ a0x2 + ( 2

3a1 −23 )x3

+( 12a2 −

12 )x4 + · · ·

Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 1a2 = 1ak = 0 for k ≥ 3

Accordingly we find

u(x) = 1 + x+ x2


u(x) =

∞∑n=0

anxn

into both sides of the equation and using the Taylor expansion for sinx,lead to

∞∑n=0

anxn = 1 + 2x− 2

3!x3 −

∫ x

0

( ∞∑n=0

antn

)dt

Evaluating the regular integrals at the right hand side that involve termsof the form tn, n ≥ 0 yields

a0 + a1x+ a2x2 + a3x

3 + · · · = 1 + (2− a0)x− 1

2a1x

2

−(1

3a2 +

2

3!)x3 − 1

4a3x

4 · · ·

Equating the coefficients of like powers of x in both sides we find




a0 = 1a1 = 1

a2 = − 1

2!

a3 = − 1

3!

a4 =1

4!


u(x) =

(1− 1

2!x2 +

1

4!x4 − · · ·

)+

(x− 1

3!x3 +

1

5!x5 − · · ·

)and in a closed form

u(x) = cosx+ sinx

4. Substituting u(x) by the series as shown above yields

∞∑n=0

anxn = 1 + x+

1

2!x2 +

1

3!x3 −

∫ x

0

((x− t)

∞∑n=0

antn

)dt

Evaluating the regular integrals at the right hand side as discussed abovegives

a0 + a1x+ a2x2 + a3x

3 + · · · = 1 + x+ (1

2− 1

2a0)x2

+(1

6− 1

6a1)x3 − 1

12a2x

4 + · · ·

Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 1a2 = 0a3 = 0ak = 0 for k ≥ 3


u(x) = 1 + x


∞∑n=0

anxn = −1−

∫ x

0

( ∞∑n=0

antn

)dt





a0 +a1x+a2x2 +a3x

3 + · · · = −1−a0x−1

2a1x

2− 1

3a2x

3− 1

4a3x

4−· · ·

Equating the coefficients of like powers of x in both sides we finda0 = −1a1 = 1

a2 = − 1

2!

a3 =1

3!

a4 = − 1

4!


u(x) = −(

1− x+1

2!x2 − 1

3!x3 + · · ·


u(x) = −e−x


∞∑n=0

anxn = 1− 2

∫ x

0

( ∞∑n=0

antn

)dt


a0 + a1x+ a2x2 + a3x

3 + · · · = 1− 2a0x− a1x2 −2

3a2x

3 − 1

2a3x

4 − · · ·Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = −2a2 = 2

a3 = −4

3...Accordingly, we find

u(x) = 1− (2x) +(2x)2

2!− (2x)3

3!+ · · ·

and in a closed form




u(x) = e−2x


∞∑n=0

anxn = 1 + xex −

∫ x

0

(t

∞∑n=0

antn

)dt


a0+a1x+a2x2+a3x

3+· · · = 1+x+(1− 1

2a0)x2+(

1

2!− 1

3a1)x3−· · ·

Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 1

a2 =1

2!

a3 =1

3!...


u(x) = 1 + x+1

2!x2 +

1

3!x3 + · · ·

and in a closed formu(x) = ex


∞∑n=0

anxn = x+

∫ x

0

((x− t)

∞∑n=0

antn

)dt


a0 + a1x+ a2x2 + a3x

3 + · · · = x+1

2a0x

2 +1

6a1x

3 +1

12a2x

4 + · · ·

Equating the coefficients of like powers of x in both sides we finda0 = 0a1 = 1a2 = 0

a3 =1

3!...





u(x) = sinhx


∞∑n=0

anxn = 1− 1

2x2 −

∫ x

0

((x− t)

∞∑n=0

antn

)dt


a0+a1x+a2x2+a3x

3+· · · = 1−(1

2+

1

2a0)x2− 1

6a1x

3− 1

12a2x

4+· · ·

Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 0a2 = −1a3 = 0

a4 =1

12...


u(x) = 2

(1− 1

2!x2 +

1

4!x4 − · · ·

)− 1


u(x) = 2 cosx− 1


∞∑n=0

anxn = 1− x−

∫ x

0

((x− t)

∞∑n=0

antn

)dt


a0+a1x+a2x2+a3x

3+ · · · = 1−x− 1

2a0x

2− 1

6a1x

3− 1

12a2x

4+ · · ·

Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = −1



3.5. Converting Volterra Equation to IVP 63

a2 = − 1

2!

a3 =1

3!

a4 =1

4!...


u(x) =

(1− 1

2!x2 +

1

4!x4 − · · ·

)−(x− 1

3!x3 +

1

5!x5 − · · ·


u(x) = cosx− sinx

11. We findu(x) = sinhx12. We findu(x) = sinx

3.5 Converting Volterra Equation to IVP

Exercises 3.5

1. Differentiate both sides using Leibniz rule we obtain

u′(x) = −3u(x)

The equivalent initial value problem is then

u′(x) + 3u(x) = 0, u(0) = 1

Solving this equation and using the initial condition, we find

u(x) = e−3x

2. Differentiate both sides twice using Leibniz rule we obtain

u′(x) =

∫ x

0

u(t) dt

u′′(x) = u(x)





u′′(x)− u(x) = 0, u(0) = 1, u

′(0) = 0

The characteristic equation is

r2 − 1 = 0Solving this equation and using the initial conditions, we find

u(x) = coshx


u′(x) = −1−

∫ x

0

u(t) dt

u′′(x) = −u(x)


u′′(x) + u(x) = 0, u(0) = 1, u

′(0) = −1


r2 + 1 = 0Solving this equation and using the initial conditions, we find

u(x) = cosx− sinx

4. Differentiate both sides using Leibniz rule we obtain

u′(x) = 1 + u(x)


u′(x)− u(x) = 1, u(0) = 0

Solving this equation and using the initial conditions, we find

u(x) = ex − 1


u′(x) = 1 +

∫ x

0

u(t) dt



3.5. Converting Volterra Equation to IVP 65

u′′(x) = u(x)


u′′(x)− u(x) = 0, u(0) = 1, u

′(0) = 1


r2 − 1 = 0Solving this equation and using the initial conditions, we find

u(x) = ex

6. Differentiate both sides four times using Leibniz rule we obtain

u′(x) =

1

2

∫ x

0

(x− t)2u(t) dt

u′′(x) =

∫ x

0

(x− t)u(t)dt

u′′′

(x) =

∫ x

0

u(t)dt

uiv(x) = u(x)The equivalent initial value problem is thenuiv(x)− u(x) = 0, u(0) = 1, u

′(0) = u

′′(0) = u

′′′(0) = 0

The characteristic equation isr4 − 1 = 0 with roots given by r = ±1, ±i

Accordingly we findu(x) = A cosx+B sinx+ C coshx+D sinhxUsing the initial conditions yields

u(x) =1

2(cosx+ coshx)

7. Differentiate both sides four times using Leibniz rule we obtain

u′(x) = 1 +

1

2

∫ x

0

(x− t)2u(t) dt

u′′(x) =

∫ x

0

(x− t)u(t)dt

u′′′

(x) =

∫ x

0

u(t)dt




uiv(x) = u(x)The equivalent initial value problem is then

uiv(x)− u(x) = 0, u(0) = u′′(0) = u

′′′(0) = 0, u

′(0) = 1


r4 − 1 = 0 with roots given by r = ±1, ±iAccordingly we find

u(x) = A cosx+B sinx+ C coshx+D sinhxUsing the initial conditions yields

u(x) = 12 (sinx+ sinhx)


u′(x) = 2x+

∫ x

0

u(t) dt

u′′(x) = 2 + u(x)

The equivalent initial value problem is thenu′′(x)− u(x) = 2, u(0) = 0, u

′(0) = 0

The characteristic equation of the homogeneous part isr2 − 1 = 0This gives uc(x) = A coshx+B sinhxThe particular solution is obtained by substituting up(x) = C where weobtain up(x) = −2

Using the initial conditions in u(x) = uc(x) + up(x),we find

u(x) = 2 coshx− 2


u′(x) = 1 +

1

2x2 −

∫ x

0

u(t) dt

u′′(x) = x− u(x)


u′′(x) + u(x) = x, u(0) = 0, u

′(0) = 1




The characteristic equation of the homogeneous part is

r2 + 1 = 0

This gives uc(x) = A cosx+B sinx

The particular solution is obtained by substituting

up(x) = C +Dx , where we obtain up(x) = x

Using the initial conditions in u(x) = uc(x) + up(x), we find

u(x) = x


u′(x) = 1− 2x+

1

2x2 − 1

3x3 −

∫ x

0

u(t) dt

u′′(x) = −2 + x− x2 − u(x)


u′′(x) + u(x) = −2 + x− x2, u(0) = 0, u

′(0) = 1

The characteristic equation of the homogeneous part is

r2 + 1 = 0This gives uc(x) = A cosx+B sinxThe particular solution is obtained by substituting

up(x) = C +Dx+ Ex2, where we obtain up(x) = x− x2Using the initial conditions in u(x) = uc(x) + up(x), we find

u(x) = x− x2

3.6 Successive Approximations Method

Exercises 3.6

1. We select




u0(x) = 0

Substituting in the original equation, we find

u1(x) = 1


u2(x) = 1−∫ x

0

dt

so that

u2(x) = 1− x


u3(x) = 1− x+1

2!x2

u4(x) = 1− x+1

2!x2 − 1

3!x3

...

Accordingly,

u(x) = limn→∞ un(x) = e−x

2. We select

u0(x) = 0


u1(x) = 1


u2(x) = 1− 9

∫ x

0

(x− t)dt

so that




u2(x) = 1− (3x)2

2!


u3(x) = 1− (3x)2

2!+

(3x)4

4!

...

Accordingly,

u(x) = limn→∞ un(x) = cos(3x)

3. We select

u0(x) = 0


u1(x) = 1 + 2x


u2(x) = 1 + 2x+ 4

∫ x

0

(x− t)(1 + 2t)dt

so that

u2(x) = 1 + 2x+(2x)2

2!+

(2x)3

3!

...

Accordingly,

u(x) = limn→∞ un(x) = e2x

4. We select

u0(x) = 0





u1(x) = 1− 1

4x


u2(x) = 1− 1

4x+

1

16

∫ x

0

(x− t)(1− 1

4t)dt

so that

u2(x) = 1− 1

4x+

( 14x)2

2!−

( 14x)3

3!

...

Accordingly,

u(x) = limn→∞ un(x) = e−14x

5. We select

u0(x) = 0


u1(x) = 2


u2(x) = 2− 2

∫ x

0

(x− t)dt

so that

u2(x) = 2(1− 1

2!x2)

u3(x) = 2(1− 1

2!x2 +

1

4!x4)

...

Accordingly,

u(x) = limn→∞ un(x) = 2 cosx




6. We select

u0(x) = 0


u1(x) = 1


u2(x) = 1− 2

∫ x

0

tdt

so that

u2(x) = 1− x2

u3(x) = 1− x2 +1

2!x4

...

Accordingly,

u(x) = limn→∞ un(x) = e−x2

7. We select

u0(x) = x


u1(x) = x+1

3!x3


u2(x) = x+

∫ x

0

(x− t)(t+t3

3!)dt

so that

u2(x) = x+1

3!x3 +

1

5!x5




...

Accordingly,

u(x) = limn→∞ un(x) = sinhx

8. We select

u0(x) = 0


u1(x) = 1


u2(x) = 1−∫ x

0

(x− t)dt

so that

u2(x) = 1− 1

2!x2

u3(x) = 1− 1

2!x2 +

1

4!x4

...

Accordingly,

u(x) = limn→∞ un(x) = cosx

9. We select

u0(x) = 0


u1(x) = 1 + x





u2(x) = 1 + x−∫ x

0

(x− t)(1 + t)dt

so that

u2(x) = 1 + x− 1

2!x2 − 1

3!x3

u3(x) = 1 + x− 1

2!x2 − 1

3!x3 +

1

4!x4 +

1

5!x5

u3(x) =

(1− 1

2!x2 +

1

4!x4 − · · ·

)+

(x− 1

3!x3 +

1

5!x5 − · · ·

)Accordingly,

u(x) = limn→∞ un(x) = cosx+ sinx

10. We select

u0(x) = 0


u1(x) = 1− x


u2(x) = 1− x−∫ x

0

(x− t)(1− t)dt

so that

u2(x) = 1− x− 1

2!x2 +

1

3!x3

u3(x) = 1− x− 1

2!x2 +

1

3!x3 +

1

4!x4 − 1

5!x5

u3(x) =(1− 1

2!x2 + 1

4!x4 − · · ·

)−(x− 1

3!x3 + 1

5!x5 − · · ·

)Accordingly,

u(x) = limn→∞ un(x) = cosx− sinx

11. We select




u0(x) = 0


u1(x) = 2− x


u2(x) = 2− x+

∫ x

0

(2− t)dt

so that

u2(x) = 2 + x− 1

2!x2

u3(x) = 2 + x+1

2!x2 − 1

3!x3

u4(x) = 2 + x+1

2!x2 +

1

3!x3 − 1

4!x4

Accordingly,

u(x) = limn→∞ un(x) = 1 + ex

12. We select

u0(x) = 0


u1(x) = 1− x− 1

2!x2


u2(x) = 1− x− 1

2!x2 +

∫ x

0

(x− t)(1− t− 1

2!t2)dt

so that

u2(x) = 1−(x+ 1

2!x2 + 1

3!x3)

u3(x) = 1−(x+

1

2!x2 +

1

3!x3 +

1

4!x4)




Accordingly,

u(x) = limn→∞ un(x) = 1− sinhx

3.7 Successive Substitutions Method

Exercises 3.7

1. Substituting λ = 1, f(x) = x, and K(x, t) = 1 into the formula yields

u(x) = x+

∫ x

0

tdt+

∫ x

0

∫ t

0

t1dt1dt+ · · ·

Accordingly we obtain u(x) in a series form

u(x) = x+1

2!x2 +

1

3!x3 + · · ·

and in a closed form, u(x) is given by

u(x) = ex − 1

2. Substituting λ = 1, f(x) = 12!x

2, and K(x, t) = 1 into the formula yields

u(x) =1

2!x2 +

∫ x

0

1

2!t2dt+

∫ x

0

∫ t

0

1

2!t21dt1dt+ · · ·


u(x) = 12!x

2 +1

3!x3 +

1

4!x4 + · · ·


u(x) = ex − x− 1

3. Substituting λ = −1, f(x) =1

3!x3, and K(x, t) = (x−t) into the formula

yields

u(x) =1

3!x3−

∫ x

0

(x−t) 1

3!t3dt+

∫ x

0

∫ t

0

(x−t)(t−t1)1

3!t31dt1dt+· · ·


u(x) =1

3!x3 − 1

5!x5 +

1

7!x7 + · · ·





u(x) = x− sinx

4. Substituting λ = 1, f(x) = 13!x

3, and K(x, t) = (x− t) into the formulayields

u(x) =1

3!x3+

∫ x

0

(x−t) 1

3!t3dt+

∫ x

0

∫ t

0

(x−t)(t−t1)1

3!t31dt1dt+· · ·


u(x) =1

3!x3 +

1

5!x5 +

1

7!x7 + · · ·


u(x) = −x+ sinhx

5. Substituting λ = −1 , f(x) =1

2!x2, and K(x, t) = (x − t) into the

formula yields

u(x) =1

2!x2−

∫ x

0

(x−t) 1

2!t2dt+

∫ x

0

∫ t

0

(x−t)(t−t1)1

2!t21dt1dt+· · ·


u(x) =1

2!x2 − 1

4!x4 +

1

6!x6 + · · ·


u(x) = −1− cosx

6. Substituting λ = −1, f(x) = 1− 1

2!x2, and K(x, t) = 1 into the formula

yields

u(x) = 1− 1

2!x2 −

∫ x

0

(1− 1

2!t2)dt+

∫ x

0

∫ t

0

(1− 1

2!t21)dt1dt+ · · ·

Accordingly we obtain

u(x) = 1− x+ (1

2!x2 − 1

2!x2) + (

1

3!x3 − 1

3!x3) + · · ·

Hence, u(x) is given by

u(x) = 1− x




7. Substituting λ = 2, f(x) = 1, and K(x, t) = 1 into the formula yields

u(x) = 1 + 2

∫ x

0

dt+

∫ x

0

∫ t

0

dt1dt+ · · ·


u(x) = 1 + 2x+(2x)2

2!+

(2x)3

3!+ · · ·


u(x) = e2x

8. Substituting λ = 1, f(x) = 3−2x, andK(x, t) = 1 into the formula yields

u(x) = 3− 2x+

∫ x

0

(3− 2t) dt+

∫ x

0

∫ t

0

(3− 2t)(3− 2t1)dt1dt+ · · ·


u(x) = 3 + x+1

2!x2 +

1

3!x3 + · · ·


u(x) = 2 + ex

9. Substituting λ = −1, f(x) = 2 + 12!x

2, and K(x, t) = (x − t) into theformula yields

u(x) = 2 +1

2!x2 −

∫ x

0

(x− t) (2 +1

2!t2)dt

+

∫ x

0

∫ t

0

(x− t)(t− t1)(2 +1

2!t21)dt1dt+ · · ·


u(x) = 2− 1

2!x2 +

1

4!x4 + · · ·


u(x) = 1 + cosx

10. Substituting λ = −1, f(x) = 1 − x +1

2!x2, and K(x, t) = (x − t) into

the formula yields




u(x) = 1− x+1

2!x2 −

∫ x

0

(x− t) (1− t+1

2!t2)dt

+

∫ x

0

∫ t

0

(x− t)(t− t1)(1− t1 +1

2!t21)dt1dt+ · · ·


u(x) = 1−(x− 1

3!x3 +

1

5!x5 + · · ·

)and in a closed form, u(x) is given by

u(x) = 1− sinx

11. Substituting λ = 1, f(x) = 2 − 1

2!x2, and K(x, t) = (x − t) into the

formula yields

u(x) = 2− 1

2!x2 −

∫ x

0

(x− t) (2− 1

2!t2)dt

+

∫ x

0

∫ t

0

(x− t)(t− t1)(2− 1

2!t21)dt1dt+ · · ·


u(x) = 1 +

(1 +

1

2!x2 +

1

4!x4 + · · ·

)and in a closed form, u(x) is given byu(x) = 1 + coshx

12. Substituting λ = 1, f(x) =1

2!x2, and K(x, t) = (x− t) into the formula

yields

u(x) = 12!x

2+

∫ x

0

(x−t) (1

2!t2)dt+

∫ x

0

∫ t

0

(x−t)(t−t1)(1

2!t21)dt1dt+ · · ·


u(x) =

(1

2!x2 +

1

4!x4 +

1

6!x6 + · · ·

)and in a closed form, u(x) is given byu(x) = −1 + coshx



3.9. Volterra Equations of the First Kind 79

3.9 Volterra Equations of the First Kind

Exercises 3.9

1. Noting that K(x, t) = 5 + 3x− 3t, hence K(x, x) = 5 6= 0

Differentiating both sides of the equation with respect to x yields

10x+ 3x2 = 5u(x) +

∫ x

0

3u(t)dt

or equivalently

u(x) = 2x+3

5x2 − 3

5

∫ x

0

u(t)dt

Using the modified decomposition method, we set

u0(x) = 2x

which gives

u1(x) =3

5x2 − 3

5

∫ x

0

2t dt

Consequently, we find

uk(x) = 0 for k ≥ 1

Accordingly, the exact solution is

u(x) = 2x

2. Noting that K(x, t) = et−x, hence K(x, x) = 1 6= 0


e−x − xe−x = u(x)−∫ x

0

et−xu(t)dt

or equivalently

u(x) = e−x − xe−x +

∫ x

0

et−xu(t)dt





u0(x) = e−x

which gives

u1(x) = −xe−x +

∫ x

0

e−x dt


uk(x) = 0 for k ≥ 1


u(x) = e−x

3. Noting that K(x, t) = 1 + x− t, hence K(x, x) = 1 6= 0


2ex − 1 = u(x) +

∫ x

0

u(t)dt

or equivalently

u(x) = 2ex − 1−∫ x

0

u(t)dt


u0(x) = ex

which gives

u1(x) = ex − 1−∫ x

0

et dt


uk(x) = 0 for k ≥ 1





u(x) = ex

4. Noting that K(x, t) = 2− x+ t, hence K(x, x) = 2 6= 0


2 sinhx− coshx+ 1 = 2u(x)−∫ x

0

u(t)dt

or equivalently

u(x) = sinhx− 1

2coshx+

1

2+

1

2

∫ x

0

u(t)dt


u0(x) = sinhx

which gives

u1(x) =1

2coshx+

1

2+

1

2

∫ x

0

sinh tdt


uk(x) = 0 for k ≥ 1


u(x) = sinhx

5. Noting that K(x, t) = 4 + 3x− 3t, hence K(x, x) = 4 6= 0


4 cosx+ 3 sinx = 4u(x) + 3

∫ x

0

u(t)dt

or equivalently

u(x) = cosx+3

4sinx− 3

4

∫ x

0

u(t)dt





u0(x) = cosx

which gives

u1(x) =3

4sinx− 3

4

∫ x

0

cos tdt


uk(x) = 0 for k ≥ 1


u(x) = cosx

6. Noting that K(x, t) = 1 + x− t, hence K(x, x) = 1 6= 0, x < π2


sec2x+ tanx = u(x) +

∫ x

0

u(t)dt

or equivalently

u(x) = sec2x+ tanx−∫ x

0

u(t)dt


u0(x) = sec2x

which gives

u1(x) = tanx−∫ x

0

sec2tdt


uk(x) = 0 for k ≥ 1





u(x) = sec2x



Chapter 4

FredholmIntegro-DifferentialEquations

4.3 The Direct Computation Method

Exercises 4.3

1. u′(x) = 1

6 + 536x−

∫ 1

0xtu(t)dt, u(0) = 1

6

We first set

α =

∫ 1

0

tu(t)dt

so that the given equation can be rewritten as

u′(x) =

1

6+ (

5

36− α)x

Integrating both sides from 0 to x and using the given condition gives

u(x) =1

6(1 + x) + (

5

36− α)

x2

2

Substituting this expression of u(x) into the first equation for α yields

85



86 Chapter 4. Fredholm Integro-Differential Equations

α =

∫ 1

0

t

(1

6(1 + t) + (

5

36− α)

t2

2

)dt

so that

α =5

36

This gives

u(x) =1

6(1 + x)

2. u′(x) = 1

21x−∫ 1

0xtu(t)dt, u(0) = 1

6

We first set

α =

∫ 1

0

tu(t)dt


u′(x) =

1

21x− αx

Integrating both sides from 0 to x and using the given condition give

u(x) =1

6+ (

1

42− α

2)x2


α =

∫ 1

0

t

(1

6+ (

1

42− α

2)t2)dt

so that

α =5

63

This gives

u(x) =1

6− 1

63x2

3. u′′(x) = − sinx+ x−

∫ π/20

xtu(t)dt, u(0) = 0, u′(0) = 1




We first set

α =

∫ π/2

0

tu(t)dt


u′′(x) = − sinx+ (1− α)x

Integrating both sides from 0 to x twice and using the given conditionsgives

u′(x) = cosx+

1− α2!

x2

and

u(x) = sinx+1− α

3!x3


α =

∫ π/2

0

t

(sin t+

1− α3!

t3)dt

so that

α = 1

This gives

u(x) = sinx

4. u′′(x) = 9

4 −13x+

∫ 1

0(x− t)u(t)dt, u(0) = u

′(0) = 0

We first set

α =

∫ 1

0

u(t)dt

and

β =

∫ 1

0

tu(t)dt





u′′(x) = 9

4 −x3 + αx− β

Integrating both sides from 0 to x twice and using the given conditionsgives

u′(x) = 9

4x−16x

2 + α2 x

2 − βx

and

u(x) = (9

8− β

2)x2 + (

α

6− 1

18)x3


α =1

3, β =

1

4

This gives

u(x) = x2

5. u′(x) = 2sec2x tanx− x+

∫ π/40

xu(t)dt, u(0) = 1

We first set

α =

∫ π/4

0

u(t)dt


u′(x) = 2sec2x tanx+ (α− 1)x


u(x) = sec2x+ α−12 x2


α = 1




This gives

u(x) = sec2x

6. Integrating both sides from 0 to x and using the given condition gives

u(x) = 1− (5 + β)x+ (α2 − 3)x2

where

α =∫ 1

−1 u(t) dt, β =∫ 1

−1 tu(t) dt

Substituting u(x) into α and β and by solving the resulting equationswe find α = 0, β = −2 hence we find

u(x) = 1− 3x− 3x2


u(x) = sinx+ (α2 −12 )x2 + (1− β)x

where

α =∫ π

2

0u(t) dt, β =

∫ π2

0tu(t) dt

Proceeding as before we find α = 1, β = 1 hence we find

u(x) = sinx


u(x) = sinx+ cosx(α2 − 1x2 + (π2 − β)xwhereα =

∫ π2

0u(t) dt, β =

∫ π2

0tu(t) dt

Proceeding as before we find α = 2, β = π2 hence we find

u(x) = sinx+ cosx




4.4 The Adomian Decomposition Method

Exercises 4.4

1. u′(x) = sinhx+ 1

8 (1− e−1)x− 1

8x

∫ 1

0

tu(t)dt, u(0) = 1


u(x) = coshx+1

16(1− e−1)x2 − 1

16x2∫ 1

0

tu(t)dt

Following the Adomian decomposition method, we set

u0(x) = coshx+1

16(1− e−1)x2

u1(x) = − 116x

2

∫ 1

0

t

(cosh t+

1

16(1− e−1)t2

)dt

so that

u1(x) = − 1

16(1− e−1)x2 − 1

16× 64(1− e−1)x2

Cancelling the noise term of u0 that appears in u1, and justifying thatthe remaining non-canceled term in u0 justifies the equation, we thereforefind

u(x) = coshx

2. u′(x) = 1− 1

3x+ x

∫ 1

0

tu(t)dt, u(0) = 0


u(x) = x− 1

6x2 +

1

2x2∫ 1

0

tu(t)dt


u0(x) = x− 1

6x2

u1(x) = 12x

2

∫ 1

0

t

(t− 1

6t2)dt



4.4. The Adomian Decomposition Method 91

so that

u1(x) =7

48x2

and continuing in the same manner we find

u2(x) =7

48× 8x2

u(x) = u0(x) + u1(x) + u2(x) + · · ·

u(x) = x− 16x

2 + 748x

2(1 + 1

8 + · · ·)

we therefore find

u(x) = x

3. u′(x) = xex + ex − x+ x

∫ 1

0

u(t)dt, u(0) = 0


u(x) = xex − 1

2x2 +

1

2x2∫ 1

0

u(t)dt


u0(x) = xex − 1

2x2

u1(x) = 512x

2

and continuing in the same manner we find

u2(x) =5

72x2

u(x) = u0(x) + u1(x) + u2(x) + · · ·

u(x) = xex − 12x

2 + 512x

2(1 + 1

6 + 136 + · · ·

)we therefore find

u(x) = xex

4. u′(x) = x cosx+ sinx− x+ x

∫ π2

0

u(t)dt, u(0) = 0





u(x) = x sinx− 1

2x2 +

1

2x2∫ π

2

0

u(t)dt


u0(x) = x sinx− 1

2x2

u1(x) =1

2x2

Cancelling the noise terms between u0(x) and u1(x) and justifying thatthe non-canceled term of u0(x) satisfies the integral equation gives

u(x) = x sinx

5. u′′(x) = − sinx+ x− x

∫ π2

0

tu(t)dt, u(0) = 0, u′(0) = 1

Integrating both sides from 0 to x twice and using the given conditionsgive

u(x) = sinx+1

3!x3 − 1

3!x3∫ π

2

0

tu(t)dt


u0(x) = sinx+1

3!x3

u1(x) = − 1

3!x3 − π5

160× (3!)2x3


u(x) = sinx

6. u′′′

(x) = 6 + x− x∫ 1

0

u′′(t)dt u(0) = −1, u

′(0) = 1, u

′′(0) = −2

Integrating both sides from 0 to x three times and using the given con-ditions give

u(x) = −1 + x− x2 + x3 + 14!x

4 − 1

4!x4∫ 1

0

u′′(t)dt





u0(x) = −1 + x− x2 + x3 +1

4!x4

u1(x) = − 1

4!x4 − 1

4!× 3!x4


u(x) = x3 − x2 + x− 1

7. u′′′

(x) = − cosx+ x+

∫ π2

0

xu′′(t)dt, u(0) = u

′′(0) = 0, u

′(0) = 1

Integrating both sides from 0 to x three times and using the given con-ditions give

u(x) = sinx+ 14!x

4 +1

4!x4∫ π

2

0

u′′(t)dt


u0(x) = sinx+1

4!x4

u1(x) = − 1

4!x4 +

π3

4!× 48x4


u(x) = sinx

8. Integrating both sides from 0 to x and using the given condition we find

u(x) = sinx+ cosx− x2 + π2x+

∫ π2

0(x

2

2 − xt)u(t)dt


u0(x) = sinx+ cosx

u1(x) = 0

u(x) = sinx+ cosx





u(x) = cosx− x2

2 + π2x− x+

∫ π2

0(x

2

2 − xt)u(t)dt


u0(x) = cosx

u1(x) = 0

u(x) = cosx


u(x) = sinx− cosx+ 2xπ2x+∫ π

2

0(x

2

2 − xt)u(t)dt


u0(x) = sinx− cosx

u1(x) = 0

u(x) = sinx− cosx


Exercises 4.5

1. The correction functional for this equation is given by

un+1(x) = un(x)−∫ x0

(u′

n(t)− sin t− t cos t+ 1−∫ π

2

0un(r) dr

)dt

Using u0(x) = 0 into the correction functional gives the following suc-cessive approximationsu0(x) = 0

u1(x) = u0(x)−∫ x0

(u′

0(t)− sin t− t cos t+ 1−∫ π

2

0u0(r) dr

)dt

= x sinx− xu2(x) = u1(x)−

∫ x0

(u′

1(t)− sin t− t cos t+ 1−∫ π

2

0u1(r) dr

)dt

= (x sinx− x) + (x− π2

8 x)




u3(x) = (x sinx− x) + (x− π2

8 x) + (π2

8 x− · · ·)u(x) = x sinx


un+1(x) = un(x)−∫ x0

(u′

n(t)− cos t+ t sin t− 2−∫ π0un(r) dr

)dt

Using u0(x) = 0 into the correction functional gives the following suc-cessive approximationsu0(x) = 0u1(x) = x cosx+ 2xu2(x) = (x cosx+ 2x) + (π2x− 2x)

u3(x) = (x cosx+ 2x) + (π2x− 2x) + (π4

2 x− π2x) + · · ·

u(x) = x cosx


un+1(x) = un(x)−∫ x0

(u′

n(t)− 2 sec2 x tanx+ π2

32 −∫ π

4

0un(r) dr

)dt


u1(x) = sec2 x+ π4x−

π2

32x

u2(x) = sec2 x+ (x+ π4x)− (π4x+ π2

32x)u3(x) = sec2 x+ x+ · · ·u(x) = x+ sec2 x


un+1(x) = un(x)−∫ x0

(u′

n(t)− 3 + 12t−∫ 1

0run(r) dr

)dt

Using u0(x) = 1 into the correction functional gives the following suc-cessive approximationsu0(x) = 1u1(x) = 1− 6x2 + 7

2x

u2(x) = 1− 6x2 + 196 x

u3(x) = 1− 6x2 + 5518x

u(x) = 1 + 3x− 6x2





un+1(x) = un(x)−∫ x0

(u′

n(t)− et + 1−∫ 1

0run(r) dr

)dt


u1(x) = ex − 12x

u2(x) = ex − 16x

u3(x) = ex − 118x

un(x) = ex − 12×3n−1x

u(x) = ex


un+1(x) = un(x)−∫ x0

(u′

n(t)− 1912 − 2t+ 3t2 −

∫ 1

0un(r) dr

)dt


u1(x) = 1 + x2 − x3 − 712x

u2(x) = 1 + x2 − x3 − 1924x

u3(x) = 1 + x2 − x3 − 4348x

u(x) = 1− x+ x2 − x3

4.6 Converting to Fredholm Integral

Equations

Exercises 4.6

1. u′(x) = −x sinx+ cosx+ (1− π

2 )x+

∫ π2

0

xu(t)dt, u(0) = 0


u(x) = x cosx+ (1− π2 )x

2

2 + x2

2

∫ π2

0

u(t)dt

or equivalently



4.6. Converting to Fredholm Integral Equations 97

u(x) = x cosx+ (1− π

2)x2

2+ α

x2

2

where

α =

∫ π2

0

u(t)dt

Substituting for u(x) from the above equation yields

α = −(1− π2 )

so that

u(x) = x cosx

2. u′′(x) = −ex + 1

2x+

∫ 1

0

xtu(t)dt, u(0) = 0, u′(0) = −1


u(x) = −ex + 1 + x3

12 + x3

6

∫ 1

0

tu(t)dt

or equivalently

u(x) = −ex + 1 + x3

12 + αx3

6

where

α =

∫ 1

0

tu(t)dt


α = − 12

so that

u(x) = 1− ex

3. u′′(x) = − sinx+ cosx+ (2− π

2)x−

∫ π2

0

xtu(t)dt,

u(0) = −1, u′(0) = 1





u(x) = sinx− cosx+ (2− π

2)x3

6− x3

6

∫ π2

0

tu(t)dt,

or equivalently

u(x) = sinx− cosx+ (2− π

2)x3

6− x3

6α,

where

α =

∫ π2

0

tu(t)dt,


α = (2− π

2)

so that

u(x) = sinx− cosx

4. u′(x) =

7

6− 11x+

∫ 1

0

(x− t)u(t)dt, u(0) = 0


u(x) = 76x−

112 x

2 + 12x

2

∫ 1

0

u(t)dt− x∫ 1

0

tu(t)dt,

or equivalently

u(x) = 76x−

112 x

2 + α 12x

2 − xβ,

where

α =

∫ 1

0

u(t)dt,

and

β =

∫ 1

0

tu(t)dt,



4.6. Converting to Fredholm Integral Equations 99


α = −1

β = − 56

so that

u(x) = 2x− 6x2

5. u′(x) = 1

4x+ cos 2x−∫ π

4

0

xu(t)dt, u(0) = 0


u(x) = 18x

2 + 12 sin 2x− 1

2x2

∫ π4

0

u(t)dt,

or equivalently

u(x) = 18x

2 + 12 sin 2x− 1

2αx2

where

α =

∫ π4

0

u(t)dt


α = 14

so that

u(x) = 12 sinx



Chapter 5

VolterraIntegro-DifferentialEquations

5.3 The Series Solution Method

Exercises 5.3

1. u′(x) = 1− 2x sinx+

∫ x

0

u(t)dt, u(0) = 0

Using the series method, noting that a0 = 0 by using the given condition,we findu(x) = a1x+ a2x

2 + a3x3 + a4x

4 + · · ·and henceu′(x) = a1 + 2a2x+ 3a3x

2 + 4a4x3 + · · ·

Substituting into both sides of the equation, integrating the right hand sideand equating the coefficients of like terms of x in both sides givea0 = 0, a1 = 1, a2 = 0, a3 = − 1

2! , a4 = 0, a5 = 14!

Accordingly, u(x) in a series form is given byu(x) = x(1− 1

2!x2 + 1

4!x4 − · · ·)

and in a closed formu(x) = x cosx

2. u′(x) = −1 + 1

2x2 − xex −

∫ x

0

tu(t)dt, u(0) = 0

Using the series method, noting that a0 = 0 by using the given condition,

101



102 Chapter 5. Volterra Integro-Differential Equations

we findu(x) = a1x+ a2x

2 + a3x3 + a4x

4 + · · ·and henceu′(x) = a1 + 2a2x+ 3a3x

2 + 4a4x3 + · · ·

Substituting into both sides of the equation, integrating the right hand sideand equating the coefficients of like terms of x in both sides givea0 = 0, a1 = −1, a2 = − 1

2! , a3 = − 13! , a4 = − 1

4!

Accordingly, u(x) in a series form is given byu(x) = −(x+ 1

2!x2 + 1

3!x3 + 1

4!x4 + · · ·)

and in a closed formu(x) = 1− ex

3. u′′(x) = 1− x(cosx+ sinx)−

∫ x

0

tu(t)dt, u(0) = −1, u′(0) = 1

Using the series method, noting that a0 = −1, a1 = 1 by using the givenconditions, we findu(x) = −1 + x+ a2x

2 + a3x3 + a4x

4 + · · ·and henceu′(x) = 1 + 2a2x+ 3a3x

2 + 4a4x3 + · · ·

u′′(x) = 2a2 + 6a3x+ 12a4x

2 + 20a5x3 + · · ·

Substituting into both sides of the equation, using the Taylor expansions ofsinx and cosx, integrating the right hand side and equating the coefficientsof like terms of x in both sides givea0 = −1, a1 = 1, a2 = 1

2! , a3 = − 13! , a4 = − 1

4!

Accordingly, u(x) in a series form is given byu(x) = −(1− 1

2!x2 + 1

4!x4 − · · ·) + (x− 1

3!x3 + 1

5!x5 + · · ·)

and in a closed formu(x) = − cosx+ sinx

4. u′′(x) = −8− 1

3 (x3 − x4) +

∫ x

0

(x− t)u(t)dt, u(0) = 0, u′(0) = 2

Using the series method, noting that a0 = 0, a1 = 2 by using the givenconditions, we findu(x) = 2x+ a2x

2 + a3x3 + a4x

4 + · · ·and henceu′(x) = 2 + 2a2x+ 3a3x

2 + 4a4x3 + · · ·

u′′(x) = 2a2 + 6a3x+ 12a4x

2 + 20a5x3 + · · ·

Substituting into both sides of the equation, integrating the right hand sideand equating the coefficients of like terms of x in both sides givea0 = 0, a1 = 2, a2 = −4, a3 = a4 = · · · = 0Accordingly, u(x) is given byu(x) = 2x− 4x2




5. u′′(x) = 1

2x2 − x coshx−

∫ x

0

tu(t)dt, u(0) = 1, u′(0) = −1

Using the series method, noting that a0 = 1, a1 = −1 by using the givenconditions, we findu(x) = 1− x+ a2x

2 + a3x3 + a4x

4 + · · ·and henceu′(x) = −1 + 2a2x+ 3a3x

2 + 4a4x3 + · · ·

u′′(x) = 2a2 + 6a3x+ 12a4x

2 + 20a5x3 + · · ·

Substituting into both sides of the equation, using the Taylor expansions ofcoshx, integrating the right hand side and equating the coefficients of liketerms of x in both sides givea0 = 1, a1 = −1, a2 = 0, a3 = − 1

3! , a4 = 0, a5 = − 15!

Accordingly, u(x) in a series form is given byu(x) = 1− (x+ 1

3!x3 + 1

5!x5 + · · ·)

and in a closed formu(x) = 1− sinhx

5.4 The Adomian Decomposition Method

Exercises 5.4

1. u′′(x) = 1 + x− 1

3!x3 +

∫ x

0

(x− t)u(t)dt, u(0) = 1, u′(0) = 2

Applying L−1 to both sides and using the initial conditions we findu(x) = 1 + 2x+ 1

2!x2 + 1

3!x3 − 1

5!x5 + L−1

(∫ x0

(x− t)u(t)dt)

whereL−1(.) =

∫ x0

∫ x0

(.)dtdtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtain

u0(x) = 1 + 2x+1

2!x2 +

1

3!x3 − 1

5!x5

u1(x) = L−1(∫ x

0

(x− t)(1 + 2t+1

2!t2 +

1

3!t3 − 1

5!t5)dt

)so thatu1(x) = 1

4!x4 + 1

60x5 + 1

6!x6 + · · ·

Accordingly u(x) in a series form is given by

u(x) = x+

(1 + x+

1

2!x2 +

1

3!x3 +

1

4!x4 +

1

5!x5 + · · ·

)and in a closed formu(x) = x+ ex




2. u′′(x) = −1− 1

2!x2 +

∫ x

0

(x− t)u(t)dt, u(0) = 2, u′(0) = 0

Applying L−1 to both sides and using the initial conditions we findu(x) = 2− 1

2!x2 − 1

4!x4 + L−1

(∫ x0

(x− t)u(t)dt)

whereL−1(.) =

∫ x0

∫ x0

(.)dtdtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 2− 1

2!x2 − 1

4!x4

u1(x) = L−1(∫ x

0

(x− t)(2− 1

2!t2 − 1

4!t4)dt

)so thatu1(x) = 1

12x4 − 1

6!x6 − 1

8!x8


u(x) = 1 +

(1− 1

2!x2 +

1

4!x4 − 1

6!x6 + · · ·

)and in a closed formu(x) = 1 + cosx

3. u′(x) = 2 +

∫ x

0

u(t)dt, u(0) = 2

Applying L−1 to both sides and using the initial conditions we findu(x) = 2 + 2x+ L−1

(∫ x0u(t)dt

)whereL−1(.) =

∫ x0

(.)dtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 2 + 2xu1(x) = x2 + 1

3x3


u(x) = 2

(1 + x+

1

2!x2 +

1

3!x3 + · · ·

)and in a closed formu(x) = 2ex

4. u′(x) = 1−

∫ x

0

u(t)dt, u(0) = 1

Applying L−1 to both sides and using the initial conditions we findu(x) = 1 + x− L−1

(∫ x0u(t)dt

)where




L−1(.) =∫ x0

(.)dtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 1 + xu1(x) = − 1

2!x2 − 1

3!x3


u(x) =

(1− 1

2!x2 +

1

4!x4 + · · ·

)+

(x− 1

3!x3 +

1

5!x5 + · · ·

)and in a closed formu(x) = cosx+ sinx

5. u(iv)(x) = −x+ 12!x

2 −∫ x

0

(x− t)u(t)dt,

u(0) = 1, u′(0) = −1, u

′′(0) = 0, u

′′′(0) = 1

Applying L−1 to both sides and using the initial conditions we findu(x) = 1− x+ 1

3!x3 − 1

5!x5 + 1

6!x6 − L−1

(∫ x0

(x− t)u(t)dt)

where

L−1(.) =

∫ x

0

∫ x

0

∫ x

0

∫ x

0

(.)dtdtdtdt

Using the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 1− x+ 1

3!x3 − 1

5!x5 + 1

6!x6

u1(x) = L−1(∫ x

0

(x− t)(1− t+1

3!t3 − 1

5!t5 +

1

6!t6)dt

)so thatu1(x) = − 1

6!x6 + 1

7!x7 + · · ·


u(x) = 1−(x− 1

3!x3 +

1

5!x5 − 1

7!x7 + · · ·

)and in a closed formu(x) = 1− sinx


Exercises 5.5


un+1(x) = un(x)−∫ x0

(u′

n(t) + t− 12 t

2 −∫ t0un(r) dr

)dt.

We find the following successive approximations:u0(x) = 0,




u1(x) = u0(x)−∫ x0

(u′

0(t)− 2et + 1−∫ t0u0(r) dr

)dt

= −2− x+ 2ex,

u2(x) = u1(x)−∫ x0

(u′

1(t)− 2et + 1−∫ t0u1(r) dr

)dt,

= x+ x2 + 12x

3 + 13!x

4 + 14!x

5 + · · ·,u3(x) = u2(x)−

∫ x0

(u′

2(t)− 2et + 1−∫ t0u2(r) dr

)dt

= x+ x2 + 12x

3 + 13!x

4 + 14!x

5 + 15!x

6 · · ·.This gives the exact solutionu(x) = xex.


un+1(x) = un(x)−∫ x0

(u′

n(t)− 2 cos t+ 12 t

2 −∫ t0un(r) dr

)dt.

We find the following successive approximations:u0(x) = 0,

u1(x) = u0(x)−∫ x0

(u′

0(t)− 2 cos t+ 12 t

2 −∫ t0u0(r) dr

)dt

= 2 sinx− 13!x

3,

u2(x) = u1(x)−∫ x0

(u′

1(t)− 2 cos t+ 12 t

2 −∫ t0u1(r) dr

)dt

= 2x− 13!x

3 + 15!x

5 + · · ·.

This gives the exact solutionu(x) = x+ sinx.

3. Proceeding as before we find the following successive approximations:u0(x) = −1,

u1(x) = −1 + x+ 12x

2 − 13!x

3 − 14x

4,

u2(x) = −1 + x+ 12x

2 − 14x

4 + · · ·

This gives the exact solutionu(x) = x− cosx.

4. Proceeding as before we find the following successive approximations:u0(x) = 1,

u1(x) = 1 + x+ 12!x

2 − 13!x

3,

u2(x) = 1 + x+ 12!x

2 + 14!x

4 + 16!x

6 + · · ·.

This gives the exact solutionu(x) = x+ coshx.

5. Proceeding as before we find the following successive approximations:



5.6. Converting to Volterra Equations 107

u0(x) = 1 + x,

u1(x) = 1 + 2x+ 12!x

2 + 13!x

3 + · · ·,

u2(x) = 1 + 2x+ 12!x

2 + 13!x

3 + 14!x

4 + 15!x

5 + · · ·.

This gives the exact solutionu(x) = x+ ex.

6. Proceeding as before we find the following successive approximations:u0(x) = 1− x,

u1(x) = 1− x− 12!x

2 + 13!x

3 + 14!x

4 + · · ·,

u2(x) = 1− x− 12!x

2 + 13!x

3 + 14!x

4 − 15!x

5 + · · ·.

This gives the exact solutionu(x) = cosx− sinx.

5.6 Converting to Volterra Equations

Exercises 5.6

1. u′′(x) = 1 +

∫ x

0

(x− t)u(t)dt, u(0) = 1, u′(0) = 0

Integrating both sides twice from 0 to x, using the given conditions andconverting the resulting multiple integrals to a single integral we obtain

u(x) = 1 + 12!x

2 + 13!

∫ x0

(x− t)3u(t)dtUsing the Adomian decomposition method, (or any other method), wherewe set

u0(x) = 1 +1

2!x2

u1(x) =1

4!x4 +

1

6!x6


u(x) = 1 +1

2!x2 +

1

4!x4 +

1

6!x6 + · · ·


u(x) = coshx




2. u′(x) = 1−

∫ x

0

u(t)dt, u(0) = 0

Integrating both sides from 0 to x, using the given condition and con-verting the resulting multiple integral to a single integral we obtain

u(x) = x−∫ x0

(x− t)u(t)dtUsing the series solution method, (or any other method), where we set

u(x) = a0 + a1x+ a2x2 + · · ·

in both sides of the resulting integral equation we find

a0 = 0, a1 = 1, a2 = 0, a3 = − 13! , a4 = 0, a5 = 1

5! , · · ·

Accordingly, u(x) in a series form is given by

u(x) = x− 1

3!x3 +

1

5!x5 + · · ·


u(x) = sinx

3. u′′(x) = x+

∫ x

0

(x− t)u(t)dt, u(0) = 0, u′(0) = 1


u(x) = x+ 13!x

3 + 13!

∫ x0

(x− t)3u(t)dtUsing the successive approximation method, (or any other method), wherewe set

u0(x) = 0

so that

u1(x) = x+1

3!x3

u2(x) = x+1

3!x3 +

1

5!x5 +

1

7!x7



5.6. Converting to Volterra Equations 109

un(x) =∑k=2nk=1

x2k−1

(2k−1)!

Accordingly u(x) in a closed form

u(x) = sinhx

4. u′(x) = 2− 1

2!x2 +

∫ x

0

u(t)dt, u(0) = 1


u(x) = 1 + 2x− 13!x

3 +∫ x0


u(x) = a0 + a1x+ a2x2 + · · ·


a0 = 1, a1 = 2, a2 = 12! , a3 = 1

3! , a4 = 14! , a5 = 1

5! , · · ·


u(x) = x+(1 + x+ 1

2!x2 + 1

3!x3 + · · ·


u(x) = x+ ex

5. u′(x) = 1−

∫ x

0

u(t)dt, u(0) = 1


u(x) = 1 + x−∫ x0


u(x) = a0 + a1x+ a2x2 + · · ·


a0 = 1, a1 = 1, a2 = − 12! , a3 = − 1

3! , a4 = 14! , a5 = 1

5! , · · ·





u(x) =(1− 1

2!x2 + 1

4!x4 + · · ·

)+(x− 1

3!x3 + 1

5!x5 + · · ·


u(x) = cosx+ sinx

6. u′′(x) = 1 + x+

∫ x

0

(x− t)u(t)dt, u(0) = u′(0) = 1


u(x) = 1 + x+ 12!x

2 + 13!x

3 + 13!

∫ x0

(x− t)3u(t)dtUsing the Adomian decomposition method, (or any other method), wherewe set

u0(x) = 1 + x+1

2!x2 +

1

3!x3

u1(x) =1

4!x4 +

1

5!x5 + · · ·


u(x) = 1 + x+1

2!x2 + +

1

3!x3 +

1

4!x4 +

1

5!x5 + · · ·


u(x) = ex

5.7 Converting to Initial Value Problems

Exercises 5.7

1. u′(x) = ex −

∫ x

0

u(t)dt, u(0) = 1

Differentiating both sides with respect to x and using Leibniz rule weobtain the nonhomogeneous differential equation

u′′(x) + u(x) = ex, u(0) = 1, u

′(0) = 1

Solving the initial value problem we find



5.7. Converting to Initial Value Problems 111

u(x) = A cosx+B sinx+ 12ex

Using the initial conditions we find

u(x) = 12 (cosx+ sinx+ ex)

2. u′(x) = 1−

∫ x

0

u(t)dt, u(0) = 0

Differentiating both sides with respect to x and using Leibniz rule weobtain the homogeneous differential equation

u′′(x) + u(x) = 0, u(0) = 0, u

′(0) = 1


u(x) = A cosx+B sinx


u(x) = sinx

3. u′′(x) = −x− 1

2!x2 +

∫ x

0

(x− t)u(t)dt, u(0) = 1, u′(0) = 1

Differentiating both sides twice with respect to x and using Leibnizrule we obtain the nonhomogeneous differential equation

u(iv)(x)− u(x) = −1, u(0) = u′(0) = 1, u

′′(0) = 0, u

′′′(0) = −1


u(x) = A cosx+B sinx+ C coshx+D sinhx+ 1


u(x) = 1 + sinx

4. u′′(x) = 1− 1

2!x2 +

∫ x

0

(x− t)u(t)dt, u(0) = 2, u′(0) = 0





u(iv)(x)− u(x) = −1, u(0) = 2, u′(0) = 0, u

′′(0) = 1, u

′′′(0) = 0




u(x) = 1 + coshx

5. u′′(x) = − 1

2!x2 − 2

3x3 +

∫ x

0

(x− t)u(t)dt, u(0) = 1, u′(0) = 4


u(iv)(x)− u(x) = −1− 4x, u(0) = 1, u′(0) = 4, u

′′(0) = u

′′′(0) = 0


u(x) = A cosx+B sinx+ C coshx+D sinhx+ 1 + 4x


u(x) = 1 + 4x

6. u′′(x) = −x− 1

8x2 +

∫ x

0

(x− t)u(t)dt, u(0) =1

4, u′(0) = 1


u(iv)(x)− u(x) = − 14 , u(0) = u

′(0) = 1, u

′′(0) = 0, u

′′′(0) = −1




u(x) = 14 + sinx



5.8. The Volterra Integro-Differential Equations of the First Kind 113

7. u′(x) = 1 + sinx+

∫ x

0

u(t)dt, u(0) = −1

Differentiating both sides with respect to x and using Leibniz rule weobtain the nonhomogeneous differential equation

u′′(x)− u(x) = cosx, u(0) = −1, u

′(0) = 1


u(x) = Aex +Be−x − 12 cosx


u(x) = 14ex − 3

4e−x − 1

2 cosx

5.8 The Volterra Integro-Differential

Equations of the First Kind

Exercises 5.8

1. Differentiating both sides leads to

u′ = cosx− sinx− 1−∫ x0u′(t) dt

The correction functional for this equation is given by

un+1(x) = un(x)−∫ x0

(u′

n(t)− cos t+ sin t+ 1 +∫ t0u′

n(r) dr)dt.

We find the following successive approximations:

u0(x) = 1,

u1(x) = cosx+ sinx− x,

u2(x) = 1− 12x

2 + 14!x

4 + · · ·,

u3(x) = 1− 12x

2 + 14!x

4 − 16!x

6 + · · ·,

This gives the exact solution




u(x) = cosx.


u′ = −1−∫ x0u′(t) dt


un+1(x) = un(x)−∫ x0

(u′

n(t) + 1 +∫ t0u′

n(r) dr)dt.


u0(x) = 1,

u1(x) = 1− x,

u2(x) = 1− x+ 12x

2,

u3(x) = 1− x+ 12!x

2 − 13!x

3 + · · ·,


u(x) = e−x.


u′ = sinhx+ coshx−∫ x0u′(t) dt


un+1(x) = un(x)−∫ x0

(u′

n(t) +− sinh t− cosh t+∫ t0u′

n(r) dr)dt.


u0(x) = 0,

u1(x) = coshx+ sinhx− 1,

u2(x) = x,

u3(x) = x+ 13!x

3 + 15!x

5 + · · ·,



5.8. The Volterra Integro-Differential Equations of the First Kind 115


u(x) = sinhx.


u′ = 3ex − 2−∫ x0

(u+ u′(t)) dt


un+1(x) = un(x)−∫ x0

(u′

n(t)− 3et + 2 +∫ t0(un(r) + u

′

n(r)) dr)dt.


u0(x) = 1,

u1(x) = 1 + x+ 12x

2 + · · ·,

u2(x) = 1 + x+ 12x

2 + 13!x

3 + · · ·,

u3(x) = 1 + x+ 12!x

2 + 13!x

3 + 14!x

4 + · · ·,


u(x) = ex.


u′ = cosx+ x+ 12x

2 −∫ x0

(u+ u′(t)) dtThe correction functional for this equation is given by

un+1(x) = un(x)−∫ x0

(u′

n(t)− cos t− t− 12 t

2 +∫ t0(un(r) + u

′

n(r)) dr)dt.


u0(x) = 1,

u1(x) = 1 + x+ · · ·,

u2(x) = 1 + x− 12x

2 + 14!x

4 + · · ·,

u3(x) = 1 + x− 12x

2 + 14!x

4 − 16!x

6 + · · ·,





u(x) = x+ cosx.


u′ = 12 coshx+ 3

2 sinhx− 12 −

12

∫ x0

(u+ u′(t)) dtThe correction functional for this equation is given by

un+1(x) = un(x)−∫ x0

(u′

n(t)− cos t− t− 12 t

2 +∫ t0(un(r) + u

′

n(r)) dr)dt.


u0(x) = 1,

u1(x) = 1 + 12x

2 + 112x

3 + 116x

4 + · · ·,

u2(x) = 1 + 12x

2 + 132x

4 + · · ·,

u3(x) = 1 + 12!x

2 + 14!x

4 16!x

6 + · · ·,


u(x) = coshx.



Chapter 6

Singular IntegralEquations

6.2 Abel’s Problem

Exercises 6.2

1. π(x+ 1) =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = π(x+ 1) into the formula

u(x) =1

π

d

dx

∫ x

0

f(t)√x− t

dt

we find

u(x) =d

dx

∫ x

0

t+ 1√x− t

dt

Using the substitution y = x− t so that dt = −dy we find

u(x) =d

dx

∫ x

0

x− y + 1

y12

dy

Integrating the right hand side yields

u(x) =d

dx

(2(x+ 1)y1/2 − 2

3y3/2

)|y=xy=0

117



118 Chapter 6. Singular Integral Equations

u(x) = 2√x+ 1√

x

This result can also be obtained by integrating the integral involving tby using Appendix B.

2. π2 (x2 − x) =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = π2 (x2 − x) into the formula

u(x) =1

π

d

dx

∫ x

0

f(t)√x− t

dt

we find

u(x) =1

2

d

dx

∫ x

0

t2 − t√x− t

dt


u(x) =1

2

d

dx

∫ x

0

(x− y)2 − (x− y)

y12

dy


u(x) =√x( 4

3x− 1)


3. 1 + x+ x2 =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = 1 + x+ x2 into the formula

u(x) =1

π

d

dx

∫ x

0

f(t)√x− t

dt

we find

u(x) =1

π

d

dx

∫ x

0

1 + t+ t2√x− t

dt


u(x) =1

π

d

dx

∫ x

0

1 + (x− y) + (x− y)2

y12

dy



6.2. Abel’s Problem 119


u(x) = 1π√x

(1 + 2x+ 83x

2)


4. 38πx

2 =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = 38πx

2 into the formula

u(x) =1

π

d

dx

∫ x

0

f(t)√x− t

dt

we find

u(x) =3

8

d

dx

∫ x

0

t2√x− t

dt

Using the substitution t = xsin2θ so that dt = 2x sin θ cos θdθ we find

u(x) = x32


5. 43x

32 =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = 43x

32 into the formula

u(x) =1

π

d

dx

∫ x

0

f(t)√x− t

dt

we find

u(x) =4

3π

d

dx

∫ x

0

t32

√x− t

dt


u(x) = x





6. 815x

52 =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = 815x

52 into the formula

u(x) =1

π

d

dx

∫ x

0

f(t)√x− t

dt

we find

u(x) =8

15π

d

dx

∫ x

0

t52

√x− t

dt


u(x) = 12x

2


7. x3 =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = x3, f(0) = 0, f′(x) = 3x2, α = 1

2 , sin(απ) = 1 intothe formula of the generalized Abel’s integral equation

u(x) =1

π

∫ x

0

f′(t)√x− t

dt

we find

u(x) =1

π

∫ x

0

3t2√x− t

dt

Using the substitution y = x − t so that dt = −dy or using AppendixB we find

u(x) =16

5πx

52

8. x4 =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = x4, f(0) = 0, f′(x) = 4x3, α = 1

2 , sin(απ) = 1 intothe formula of the generalized Abel’s integral equation



6.2. Abel’s Problem 121

u(x) =1

π

∫ x

0

f′(t)√x− t

dt

we find

u(x) =1

π

∫ x

0

4t3√x− t

dt


u(x) =128

35πx

72

9. x+ x3 =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = x+x3, f(0) = 0, f′(x) = 1+3x2, α = 1

2 , sin(απ) =1 into the formula of the generalized Abel’s integral equation

u(x) =1

π

∫ x

0

f′(t)√x− t

dt

we find

u(x) =1

π

∫ x

0

1 + 3t2√x− t

dt


u(x) =2√x

π(1 +

8

5x2)

10. sinx =

∫ x

0

1√x− t

u(t)dt

Substituting f(x) = sinx, f(0) = 0, f′(x) = cosx into the formula of

the generalized equation

u(x) =1

π

∫ x

0

f′(t)√x− t

dt

Using the approximation cosx ≈ 1, we find

u(x) ≈ 2π

√x




6.3 Generalized Abel’s Problem

Exercises 6.3

1. Notice that α = 16 , f(x) = 36

55x116 . Therefore, we set

u(x) = sin(απ)π

ddx

∫ x0

3655 t

116

(x−t)56dt = x.

2. Notice that α = 13 , f(x) = 243

440x113 . Therefore, we set

u(x) = sin(απ)π

ddx

∫ x0

243440 t

113

(x−t)23dt = x3.

3. Notice that α = 34 , f(x) = 24x

14 . Therefore, we set

u(x) = sin(απ)π

ddx

∫ x0

24t14

(x−t)14dt = 6.

4. Notice that α = 23 , f(x) = 3πx

13 + 9x

43 , u(x) = π + 4x.

5. Notice that α = 56 , f(x) = 36

7 x76 , u(x) = x.

6. Notice that α = 23 , f(x) = 27x

73 + 9x

43 , u(x) = 4x+ 14x2.

6.4 The Weakly Singular Volterra Equations

Exercises 6.4

1. u(x) =√x− πx+ 2

∫ x

0

1√x− t

u(t)dt, I = [0, 2]

Using the Adomian decomposition method, we substitute

u(x) =∑∞n=0 un(x)

into both sides of the integral equation, hence we obtain

u0(x) + u1(x) + u2(x) + · · · =√x− πx

+2

∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt

Setting the zeroth component by



6.4. The Weakly Singular Volterra Equations 123

u0(x) =√x− πx

yields the first component

u1(x) = 2∫ x0

1√x−t (√t− πt)dt

so that

u1(x) = πx− 83π

2x32

by using Appendix B.

Canceling the noise term between u0(x) and u1(x) yields the exact so-lution

u(x) =√x

2. u(x) =3

8πx2 + x

32 −

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · = 3

8πx2 + x

32

−∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) = 38πx

2 + x32


u1(x) = −∫ x0

1√x−t (t

3/2 − 3

8πt2)dt

so that




u1(x) = − 38πx

2 − 8120πx

5/2

Canceling the noise term between u0(x) and u1(x) yields the exactsolution

u(x) = x3/2

3. u(x) = 12 −√x+

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · = 12 −√x

+

∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) = 12 −√x


u1(x) =√x− 1

2πx


u(x) = 12

4. u(x) =√x− 1

2πx+

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · =√x− 1

2πx




+

∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) =√x− 1

2πx


u1(x) = 12πx−

23πx

3/2


u(x) =√x

5. u(x) = x5/2 − 516πx

3 +

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · = x5/2 − 516πx

3

+

∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) = x5/2 − 516πx

3


u1(x) = 516πx

3 − 27πx

7/2


u(x) = x5/2




6. u(x) = x3 + 3233x

7/2 −∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · = x3 + 3233x

7/2

−∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) = x3 + 3233x

7/2


u1(x) = − 3233x

7/2 − 35132πx

4


u(x) = x3

7. u(x) = 1 + x− 2√x− 4

3x3/2 +

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · = 1 + x− 2√x− 4

3x3/2

+

∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt





u0(x) = 1 + x− 2√x− 4

3x3/2


u1(x) = 2√x+ 4

3x3/2 − πx− 1

2πx2

Canceling the noise terms between u0(x) and u1(x) yields the exactsolution

u(x) = 1 + x

8. u(x) = 1 + 2√x−

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · = 1 + 2√x

−∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) = 1 + 2√x


u1(x) = −2√x− πx


u(x) = 1

9. u(x) = x2 + 1615x

52 −

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)





u0(x) + u1(x) + u2(x) + · · · = x2 + 1615x

52

−∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) = x2 + 1615x

52


u1(x) = − 1615x

5/2 − 13πx

3


u(x) = x2

10. u(x) = 2π

√x+ 15

16x2 − x− x5/2 +

∫ x

0

1√x− t

u(t)dt, I = [0, 2]


u(x) =∑∞n=0 un(x)


u0(x) + u1(x) + u2(x) + · · · = 2π

√x+ 15

16x2 − x− x5/2

+

∫ x

0

1√x− t

(u0(t) + u1(t) + · · ·) dt


u0(x) = 2π

√x+ 15

16x2 − x− x5/2


u1(x) = x+ x5/2 − 43x

3/2 − 516πx

3





u(x) = 2π

√x+ 15

16x2

11. We use the recurrence relation

u0(x) = 1 + x,

u1(x) = − 43x

34 − 16

21x74 +

∫ x0

1+t

(x−t)14u(t) dt = 0.


u(x) = 1 + x.


u0(x) = x+ x2,

u1(x) = − 94x

43 − 243

140x103 +

∫ x0

t+t2

(x−t)14u(t) dt = 0.


u(x) = x+ x2.


u0(x) = 1 + 3x2.

Proceed as before, to get the exact solution

u(x) = 1 + 3x2.


u0(x) = 5− x.

Proceed as before, to get the exact solution

u(x) = 5− x.




6.5 The Weakly Singular Fredholm Equations

Exercises 6.5

1. We first decompose f(x) into two parts as

f0(x) = x2

f1(x) = − 1615x

52 + 2

√x−15 + 8x

√x−1

15 + 16x2√x−115

We then use the modified recurrence relation

u0(x) = x2,

u1(x) = − 1615x

52 + 2

√x−15 + 8x

√x−1

15 + 16x2√x−115 +

∫ 1

0u(t)√x−tu0(t) dt = 0.

This gives the exact solution by

u(x) = x2.


f0(x) = 10

f1(x) = 20(x− 1)12 − 20(x+ 1)

12


u0(x) = 10,

u1(x) = 20(x− 1)12 − 20(x+ 1)

12 +

∫ 1

−1u0(t)

(x−t)12dt = 0.

This gives the exact solution by

u(x) = 10.


f0(x) = 10x

f1(x) = −9x53 + 6(x− 1)

23 + 9x(x− 1)

23




6.5. The Weakly Singular Fredholm Equations 131

u0(x) = 10x,

u1(x) = −9x53 + 6(x− 1)

23 + 9x(x− 1)

23 +

∫ 1

0u(t)

(x−t)13u0(t) dt = 0.

This gives the exact solution byu(x) = 10x.


f0(x) = 3 + 10x− 9x13

f1(x) = − 452 x

43 + 33

2 (x− 1)13 + 45

2 x(x− 1)13


u0(x) = 3 + 10x− 9x13 ,

u1(x) = − 452 x

43 + 33

2 (x− 1)13 + 45

2 x(x− 1)13 +

∫ 1

0u0(t)

(x−t)23dt = 9x

13 + · · ·.

Cancelling the noise term gives the exact solution byu(x) = 3 + 10x.


f0(x) = x+ x2 − 165 x

54 (1 + 8

9x)

f1(x) = 845 (x− 1)

14 (7 + 22x+ 16x2)


u0(x) = x+ x2 − 165 x

54 (1 + 8

9x),

u1(x) = 845 (x−1)

14 (7+22x+16x2)+

∫ 1

0u(t)

(x−t)34dt = 16

5 x54 (1+ 8

9x)+ · · ·.

Cancelling the noise term gives the exact solution byu(x) = x+ x2.



Chapter 7

Nonlinear FredholmIntegral Equations

7.2 Nonlinear Fredholm Integral Equations

7.2.1 The Direct Computation Method

Exercises 7.2.1

1. u(x) = 1 +1

2λ

∫ 1

0

u2(t)dt

This equation can be rewritten as

u(x) = 1 +1

2λα

where

α =∫ 1

0u2(t)dt

Substituting for u(x) from the above equation we obtain

α =

∫ 1

0

(1 +

1

2λα

)2

dt

Integrating the definite integral yields

λ2α2 + (4λ− 4)α+ 4 = 0

133



134 Chapter 7. Nonlinear Fredholm Integral Equations

Solving the quadratic equation for α gives

α =2(1− λ)± 2

√1− 2λ

2λ2

Substituting for α in the above equation for u(x) we find

u(x) =1±√

1− 2λ

λ

Examining the solution u(x) we obtained leads to

λ = 0 is a singular point

λ =1

2is a bifurcation point

2. u(x) = 1− λ∫ 1

0

u2(t)dt


u(x) = 1− λα

where

α =∫ 1

0u2(t)dt


α =

∫ 1

0

(1− λα)2dt


λ2α2 − (1 + 2λ)α+ 1 = 0


α =(1 + 2λ)± 2

√1 + 4λ

2λ2




7.2. Nonlinear Fredholm Integral Equations 135

u(x) = −1±√

1 + 4λ

2λ



λ = −1


3. u(x) = 1 + λ

∫ 1

0

tu2(t)dt


u(x) = 1 + λα

where

α =∫ 1

0tu2(t)dt


α =

∫ 1

0

t (1 + λα)2dt


λ2α2 + (2λ− 2)α+ 1 = 0


α =(1− λ)±

√1− 2λ

λ2


u(x) =1±√

1− 2λ

λ






λ =1


4. u(x) = 1 + λ

∫ 1

0

t2u2(t)dt


u(x) = 1 + λα

where

α =∫ 1

0t2 u2(t)dt


α =

∫ 1

0

t2 (1 + λα)2dt


λ2α2 + (2λ− 3)α+ 1 = 0


α =(3− 2λ)±

√9− 12λ

2λ2


u(x) =3±√

9− 12λ

2λ

Examining the solution u(x) we obtained leads to the conclusion that


λ =3


5. u(x) = 1 + λ

∫ 1

0

t3u2(t)dt





u(x) = 1 + λα

where

α =∫ 1

0t3u2(t)dt


α =

∫ 1

0

t3 (1 + λα)2dt


λ2α2 + (2λ− 4)α+ 1 = 0


α =(2− λ)± 2

√1− λ

λ2


u(x) =2± 2

√1− λ

λ



λ = 1 is a bifurcation point

6. u(x) = 2− 4

3x+

∫ 1

0

xt2 u2(t)dt


u(x) = 2 + (α− 4

3)x

where

α =∫ 1

0t2 u2(t)dt





α =

∫ 1

0

t2(

2 + (α− 4

3)t

)2

dt


α =4

3


u(x) = 2

7. u(x) = sinx− π8 +

1

2

∫ π/2

0

u2(t)dt


u(x) = sinx+ ( 12α−

π8 )

where

α =∫ 1

0u2(t)dt


α =

∫ 1

0

(sin t+ (

1

2α− π

8)

)2

dt


α =π

4


u(x) = sinx

8. u(x) = cosx− π8 +

1

2

∫ π/2

0

u2(t)dt


u(x) = cosx+ ( 12α−

π8 )




where

α =∫ 1

0u2(t)dt


α =

∫ 1

0

(cos t+ (

1

2α− π

8)

)2

dt


α =π

4


u(x) = cosx

9. u(x) = x− 18 +

1

2

∫ 1

0

tu2(t)dt


u(x) = x+ ( 12α−

18 )

where

α =∫ 1

0tu2(t)dt


α =

∫ 1

0

t

(t+ (

1

2α− 1

8)

)2

dt


α =1

4,

67

12


u(x) = x, x+8

3

10. u(x) = x2 − 110 +

1

2

∫ 1

0

u2(t)dt





u(x) = x2 + ( 12α−

110 )

where

α =∫ 1

0u2(t)dt


α =

∫ 1

0

(t2 + (

1

2α− 1

10)

)2

dt


α =1

5,

43

15


u(x) = x2, x2 +4

3

11. u(x) = x− 56 +

∫ 1

0

(u(t) + u2(t)

)dt


u(x) = x+ (α− 56 )

where

α =∫ 1

0

(u(t) + u2(t)

)dt


α =

∫ 1

0

((t+ (α− 5

6) + (t+ (α− 5

6))2)dt


α =5

6,−1

6





u(x) = x, x− 1

12. u(x) = x− 1 + 34

∫ 1

0

(2t+ u2(t)

)dt


u(x) = x+ ( 34α− 1)

where

α =∫ 1

0

(2t+ u2(t)

)dt


α =

∫ 1

0

(2t+ (t+ (

3

4α− 1))2

)dt


α =4

3,

16

9


u(x) = x, x+ 13

7.2.2 The Adomian Decomposition Method

Exercises 7.2.2

1. u(x) = 1 + λ

∫ 1

0

tu2(t)dt, λ ≤ 1

2

Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-

lowing the procedure of the Adomian polynomials discussed before, we set

u0(x) = 1

Accordingly, the first components is given by




u1(x) = λ∫ 1

0tdt =

1

2λ

u2(x) = λ∫ 1

0t(2u0(t)u1(t))dt = 1

2λ2

u3(x) = 58λ

3

Consequently the solution in a series form is

u(x) = 1 +1

2λ+

1

2λ2 +

5

8λ3 + · · ·

2. u(x) = 1 + λ

∫ 1

0

t3u2(t)dt, λ ≤ 1



u0(x) = 1


u1(x) = λ∫ 1

0t3dt =

1

4λ

u2(x) = λ∫ 1

0t3(2u0(t)u1(t))dt = 1

8λ2

Consequently the solution in a series form is

u(x) = 1 +1

4λ+

1

8λ2 + · · ·

3. u(x) = 2 sinx− π8 +

1

8

∫ π/2

0

u2(t)dt


lowing the modified decomposition method discussed before, we set

u0(x) = 2 sinx


u1(x) = −π8 +1

8

∫ π/2

0

4sin2tdt = 0




u2(x) = 0

u3(x) = 0

Consequently the solution is

u(x) = 2 sinx

4. u(x) = 2 cosx− π8 +

1

8

∫ π/2

0

u2(t)dt



u0(x) = 2 cosx

Accordingly, the first component is given by

u1(x) = −π8 +1

8

∫ π/2

0

4cos2tdt = 0

u2(x) = 0

u3(x) = 0


u(x) = 2 cosx

5. u(x) = secx− x+ x∫ π/40

u2(t)dt



u0(x) = secx


u1(x) = −x+ x∫ π/40

sec2tdt = 0

u2(x) = 0




u3(x) = 0


u(x) = secx

6. u(x) = 32x+

3

8

∫ 1

0

xu2(t)dt



u0(x) = 32x


u1(x) = 932x

u2(x) = 27256x

Consequently the solution in a series form is given by

u(x) = 32x+ 9

32x+ 27256x+ · · ·

so that

u(x) ≈ 2x

7. u(x) = x2 − 1

12+

1

2

∫ 1

0

tu2(t)dt



u0(x) = x2


u1(x) = − 1

12+

1

2

∫ 1

0

t5dt = 0

u2(x) = 0




Consequently the solution is given by

u(x) = x2

8. u(x) = x− π

8+

1

2

∫ 1

0

1

1 + u2(t)dt



u0(x) = x


u1(x) = −π8

+1

2

∫ 1

0

1

1 + t2dt = 0

u2(x) = 0

Consequently the solution is given by

u(x) = x

9. u(x) = x− 1 +2

π

∫ 1

−1

1

1 + u2(t)dt



u0(x) = x


u1(x) = −1 +2

π

∫ 1

−1

1

1 + t2dt = 0

Consequently the solution, after justifying that u0(x) justifies the equa-tion, is given by

u(x) = x

10. u(x) = x− 1

4ln 2 +

1

2

∫ 1

0

t

1 + u2(t)dt






u0(x) = x


u1(x) = −1

4ln 2 +

1

2

∫ 1

0

t

1 + t2dt, = 0


u(x) = x

11. u(x) = sinx+ cosx− π + 2

8+

1

4

∫ π/2

0

u2(t)dt



u0(x) = sinx+ cosx


u1(x) = −π + 2

8+

1

4

∫ π/2

0

(1 + sin 2t)dt− 0.


u(x) = sinx+ cosx

12. u(x) = sinhx− 1 +

∫ 1

0

(cosh2t− u2(t)

)dt



u0(x) = sinhx





u1(x) = 0


u(x) = sinhx

13. u(x) = cosx+ 2−∫ 1

0

(1 + sin2t+ u2(t)

)dt



u0(x) = cosx


u1(x) = 0


u(x) = cosx

14. u(x) = secx− x+

∫ 1

0

x(u2(t)− tan2t

)dt



u0(x) = secx


u1(x) = 0


u(x) = secx




7.2.3 The Variational Iteration Method

Exercises 7.2.3

1. u(x) =3

4x+

∫ 1

0

xtu2(t) dt

Differentiating both sides of this equation with respect to x yields

u′(x) = 3

4 +∫ 1

0tu2(t)dt, u(0) = 0.


un+1(x) = un(x)−∫ x0

(u′

n(t)− 34 −

∫ 1

0ru2n(r) dt

)dt

where we used λ = −1 for first-order integro-differential equations.This gives the following successive approximations

u0(x) = 0,u1(x) = 0.75x,u2(x) = 0.8906x,u3(x) = 0.9483x,u4(x) = 0.9748x,

... .This gives the exact solution by u(x) = x

2. u(x) = x2 − 1

6x+

∫ 1

0

xtu2(t) dt

Differentiating both sides of this equation with respect to x yields

u′(x) = 2x− 1

6 +∫ 1

0tu2(t)dt, u(0) = 0.


un+1(x) = un(x)−∫ x0

(u′

n(t)− 2t− 16 −

∫ 1

0ru2n(r) dt

)dt

This gives the following successive approximations



7.3. Nonlinear Fredholm Integral Equations of the First Kind 149

u0(x) = 0,u1(x) = x2 − 0.1667x,u2(x) = x2 − 0.0597x,u3(x) = x2 − 0.0230x,u4(x) = x2 − 0.0090x,

... .

This gives the exact solution by u(x) = x2

3. u(x) = x2 − 1

8x+

∫ 1

0

xtu2(t) dt

Differentiating both sides of this equation with respect to x and pro-ceeding as before, we obtain the following successive approximations

u0(x) = 0,u1(x) = x3 − 0.1250x,u2(x) = x3 − 0.0377x,u3(x) = x3 − 0.0122x,u4(x) = x3 − 0.0040x,

... .

This gives the exact solution by u(x) = x3

4. u(x) = x− 1

5x2 +

∫ 1

0

x2t2u2(t) dt

Differentiating both sides of this equation with respect to x and pro-ceeding as before, we obtain the exact solution by u(x) = x+ x2

5. u(x) = x+34

105x2 +

∫ 1

0

x2t2u2(t) dt

Differentiating both sides of this equation with respect to x and pro-ceeding as before, we obtain the exact solution by u(x) = x.

7.3 Nonlinear Fredholm Integral Equations

of the First Kind

Exercises 7.3

1. We set the transformation




v(x) = u2(x), u(x) =√v(x)

This gives

vε(x) = 1ε ( 49

60 − α)x

where

α =∫ 1

0tvε(t) dt

Proceeding as in the examples, we find

α = 4960(1+3ε)

This gives

u(x) = 7√20

√x

Also

u(x) = x+ x2


v(x) = u3(x), u(x) = 3√v(x)

This gives

vε(x) = 1ε ( 1

120 − α)x2

where

α =∫ 1

0t2vε(t) dt


α = 1120(1+5ε)

This gives



7.3. Nonlinear Fredholm Integral Equations of the First Kind 151

u(x) = 3

√x2

24

Also

u(x) = x− x3


v(x) = u3(x), u(x) = 3√v(x)

This gives

vε(x) = 1ε (1− α)e2x

where

α =∫ 1

0e−6tvε(t) dt


α = 1−e−4

1−e−4+4ε

This gives

u(x) = 3

√4e2x

1−e−4

Also

u(x) = e2x


v(x) = u4(x), u(x) = 4√v(x)

This gives

vε(x) = 1ε (1− α)ex

where

α =∫ 1

0e−4tvε(t) dt





α = 1−e−3

(1−e−3+3ε)

This gives

u(x) = 4

√3ex

1−e−3

Also

u(x) = ex


v(x) = u2(x), u(x) =√v(x)

This gives

vε(x) = 1ε ( 1

4 − α)x2

where

α =∫ 1

0tvε(t) dt


α = 14(1+4ε)

This gives

u(x) = x

Also

u(x) = lnx


v(x) = u2(x), u(x) =√v(x)



7.4. Weakly-Singular Nonlinear Fredholm Integral Equations 153

This gives

vε(x) = 1ε ( 2

125 − α)x2

where

α =∫ 1

0t2vε(t) dt


α = 2125(1+5ε)

This gives

u(x) =√25 x

Also

u(x) = x lnx

7.4 Weakly-Singular Nonlinear Fredholm

Integral Equations

Exercises 7.4

1. We decompose f(x) into

f0(x) = x

f1(x) = − 1615x

52 + 2

5

√x− 1(1 + 4

3x+ 83x

2)

Consequently, we set the modified recurrence relation as

u0(x) = x

u1(x = f1(x) +∫ 1

0u2(t)√|x−t|

dt = 0

The exact solution is u(x) = x.





f0(x) = x

f1(x) = − 3235x

72 + 2

7

√x− 1(1 + 6x+ 8x2 + 16x3)


u0(x) = x

u1x = f1(x) +∫ 1

0u3(t)√|x−t|

dt = 0



f0(x) = 1− x

f1(x) = −2√x(1− 4

3x+ 815x

2) + 1615

√x− 1(1− 2x+ x2)


u0(x) = 1− x

u1x = f1(x) +∫ 1

0u2(t)√|x−t|

dt = 0



f0(x) = 1 + x

f1(x) = − 32x

23 (1 + 6

5x+ 920x

2) + 140 (x− 1)

23 (123 + 90x+ 27x2)


u0(x) = 1 + x

u1x = f1(x) +∫ 1

0u2(t)

(|x−t|)13dt = 0

The exact solution is u(x) = 1 + x.



7.4. Weakly-Singular Nonlinear Fredholm Integral Equations 155

5. We decompose f(x), then we use

u0(x) = 4√

cosx

u1x = f1(x) +∫ π

2

0u4(t)

(| sin x−sin t|)13dt = 0

The exact solution is u(x) = 4√

cosx.

6. We decompose f(x), then we use

u0(x) =√

sinx

u1x = f1(x) +∫ π

2

0u2(t)

(| cos x−cos t|)14dt = 0

The exact solution is u(x) =√

sinx.



Chapter 8

Nonlinear VolterraIntegral Equations

8.2 Nonlinear Volterra Integral Equations

8.2.1 The Series Solution Method

Exercises 8.2.1

1. u(x) = x2 + 110x

5 − 12

∫ x

0

u2(t)dt

Substituting u(x) by the series

u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = x2 +

1

10x5 − 1

2

∫ x

0

( ∞∑n=0

antn

)2

dt

Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers ofx in both sides yields

a2 = 1, ak = 0, k 6= 2

157



158 Chapter 8. Nonlinear Volterra Integral Equations


u(x) = x2

2. u(x) = x2 + 112x

6 − 12

∫ x

0

tu2(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = x2 +

1

12x6 − 1

2

∫ x

0

t

( ∞∑n=0

antn

)2

dt


a2 = 1, ak = 0, k 6= 2


u(x) = x2

3. u(x) = 1− x2 − 13x

3 +

∫ x

0

u2(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = 1− x2 − 1

3x3 +

∫ x

0

( ∞∑n=0

antn

)2

dt




8.2. Nonlinear Volterra Integral Equations 159

a0 = a1 = 1, ak = 0, k ≥ 2


u(x) = 1 + x

4. u(x) = 1− x+ x2 − 23x

3 − 15x

5 +

∫ x

0

u2(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = 1− x+ x2 − 2

3x3 − 1

5x5 +

∫ x

0

( ∞∑n=0

antn

)2

dt


a0 = a2 = 1, ak = 0, k 6= 0, 2


u(x) = 1 + x2

5. u(x) = x2 + 114x

7 − 12

∫ x

0

u3(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = x2 +

1

14x7 − 1

2

∫ x

0

( ∞∑n=0

antn

)3

dt

Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers of




x in both sides yields

a2 = 1, ak = 0, k 6= 2


u(x) = x2

6. u(x) = 12 + e−x − 1

2e−2x −

∫ x

0

u2(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn =

1

2+ e−x − 1

2e−2x −

∫ x

0

( ∞∑n=0

antn

)2

dt

Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor expansions of the exponentialfunctions and equating the coefficients of like powers of x in both sides yields

ak = (−1)kk! , k = 0, 1, 2, ...


u(x) = e−x

7. u(x) = 1− 32x

2 − x3 − 14x

3 +

∫ x

0

u3(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = 1− 3

2x2 − x3 − 1

4x3 +

∫ x

0

( ∞∑n=0

antn

)3

dt





a0 = a1 = 1, ak = 0, k ≥ 2


u(x) = 1 + x

8. u(x) = sinx− 12x+ 1

4 sin 2x+

∫ x

0

u2(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = sinx− 1

2x+

1

4sin 2x+

∫ x

0

( ∞∑n=0

antn

)2

dt

Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor series expansion of the trigono-metric functions and equating the coefficients of like powers of x in bothsides yields

a0 = 0

a1 = 1

a2 = 0

a3 = − 13!

a2k = 0, a2k+1 = (−1)k 1(2k+1)! , k = 0, 1, 2, ....


u(x) = sinx




9. u(x) = cosx− 12x−

14 sin 2x+

∫ x

0

u2(t)dt


u(x) =

∞∑n=0

anxn


∞∑n=0

anxn = cosx− 1

2x− 1

4sin 2x+

∫ x

0

( ∞∑n=0

antn

)2

dt

Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor series expansion of the trigono-metric functions and equating the coefficients of like powers of x in bothsides yields

a0 = 1

a1 = 0

a2 = − 12!

a3 = 0

a2k = (−1)k 1(2k)! , a2k+1 = 0, k = 0, 1, 2, ....


u(x) = cosx

10. u(x) = ex + 12x(e2x − 1

)−∫ x

0

xu2(t)dt


u(x) =

∞∑n=0

anxn





∞∑n=0

anxn = ex +

1

2x(e2x − 1

)−∫ x

0

x

( ∞∑n=0

antn

)2

dt

Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor expansions of the exponentialfunctions and equating the coefficients of like powers of x in both sides yields

ak = 1k! , k = 0, 1, 2, ...


u(x) = ex

8.2.2 The Adomian Decomposition Method

Exercises 8.2.2

1. u(x) = 3x+1

24x4 − 1

18

∫ x

0

(x− t)u2(t)dt

Using the decomposition u(x) =∑∞n=0 un(x) into both sides and follow-

ing the procedure of the modified decomposition method discussed before,we set

u0(x) = 3x


u1(x) =1

24x4 − 1

18

∫ x

0

(x− t)u20(t)dt

so that

u1(x) = 0

Consequently, the other components vanish, hence

u(x) = 3x

2. u(x) = 2x− 1

2x4 +

1

4

∫ x

0

u3(t)dt






u0(x) = 2x


u1(x) = −1

2x4 +

1

4

∫ x

0

u20(t)dt

so that

u1(x) = 0


u(x) = 2x

3. u(x) = sinx+1

8sin 2x− 1

4x+

1

2

∫ x

0

u2(t)dt



u0(x) = sinx


u1(x) =1

8sin 2x− 1

4x+

1

2

∫ x

0

u20(t)dt

so that

u1(x) = 0


u(x) = sinx

4. u(x) = x2 +1

5x5 −

∫ x

0

u2(t)dt






u0(x) = x2


u1(x) =1

5x5 −

∫ x

0

u20(t)dt

so that

u1(x) = 0


u(x) = x2

5. u(x) = x+

∫ x

0

(x− t)u3(t)dt


lowing the Adomian decomposition method discussed before, we set

u0(x) = x

Accordingly, the first components are given by

u1(x) =1

20x5

u2(x) =1

720x9

Hence

u(x) ≈ x+ 120x

5 + 1720x

9 + · · ·

6. u(x) = 1 +

∫ x

0

(x− t)2u2(t)dt






u0(x) = 1


u1(x) =1

3x3

u2(x) =1

90x6

Hence

u(x) ≈ x+ 13x

3 + 190x

6 + · · ·

7. u(x) = 1 +

∫ x

0

(x− t)2u3(t)dt



u0(x) = 1


u1(x) =1

3x3

u2(x) =1

60x6

Hence

u(x) ≈ x+ 13x

3 + 160x

6 + · · ·

8. u(x) = x+

∫ x

0

(x− t)2u2(t)dt



u0(x) = x


u1(x) =1

6x5




u2(x) =1

756x9

Hence

u(x) ≈ x+ 16x

5 + 1756x

9 + · · ·

9. u(x) = 1 +

∫ x

0

(t+ u2(t)

)dt


lowing the Adomian decomposition method discussed before, we setu0(x) = 1


u1(x) = 2x

u2(x) =5

2x2

Hence

u(x) ≈ 1 + 2x+ 52x

2 + · · ·

10. u(x) = 1 +

∫ x

0

(t2 + u2(t)

)dt



u0(x) = 1


u1(x) = x+ 13x

3

u2(x) = x2 +1

3x3 +

1

6x4

Hence

u(x) ≈ 1 + x+ x2 23x

3 + 16x

4 + · · ·

11. u(x) = secx+ tanx+ x−∫ x

0

(1 + u2(t)

)dt, x <

π

2






u0(x) = secx


u1(x) = 0


u(x) = secx

12. u(x) = tanx− 14 sin 2x− 1

2x+

∫ x

0

1

1 + u2(t)dt, x <

π

2

Using the decomposition

u(x) =∑∞n=0 un(x)

into both sides and following the procedure of the modified decompositionmethod discussed before, we set

u0(x) = tanx


u1(x) = 0


u(x) = tanx

8.2.3 The Variational Iteration Method

Exercises 8.2.3

1. u(x) = x− 120x

5 +∫ x0

(x− t)u3(t) dt

Differentiating both sides of this equation, and using Leibniz rule, wefind




u′(x) = 1− 1

4x4 +

∫ x0u3(t) dt, u(0) = 0

The correction functional is

un+1(x) = un(x)−∫ x0

(u′

n(ξ)− 1 + 14ξ

4 −∫ ξ0u3n(r) dr

)dξ

This will give the following successive approximations

u0(x) = 0,u1(x) = x− 1

20x5,

u2(x) = 2x− 120x

5 + ( 15x

5 − 1480x

9) + 120800x

13 + · · · ,u3(x) = 2x− 1

480x9 + ( 1

480x9 + 1

20800x13)− 1

20800x13 + · · · ,

... .Cancelling the noise terms gives the exact solution by u(x) = x.

2. u(x) = x2 − 156x

8 +∫ x0

(x− t)u3(t) dt

Differentiating both sides of this equation, and using Leibniz rule, wefind

u′(x) = 2x− 1

7x7 +

∫ x0u3(t) dt, u(0) = 0

The correction functional is

un+1(x) = un(x)−∫ x0

(u′

n(ξ)− 2ξ + 17ξ

7 −∫ ξ0u3n(r) dr

)dξ

This will give the following successive approximations

u0(x) = 0,u1(x) = x2 − 1

56x8,

u2(x) = x2 − 310192x

14 + · · · ,u3(x) = x2 + · · · ,

... .

Cancelling the noise terms gives the exact solution by u(x) = x2.

3. u(x) = x+ x2 − 112x

4 − 110x

5 − 130x

6 +∫ x0

(x− t)u2(t) dt

Proceeding as before gives the following successive approximations




u0(x) = 0,u1(x) = x+ x2 − 1

12x4 − 1

10x5 − 1

30x6,

u2(x) = x+ x2 + noise terms,... .

Cancelling the noise terms gives the exact solution by u(x) = x+ x2.

4. u(x) = 1 + x− 12x

2 − 13x

3 − 112x

4 +∫ x0

(x− t)u2(t) dt

Proceeding as before and by cancelling the noise terms, we obtain theexact solution by u(x) = 1 + x.

5. u(x) = 1 + x− 13x

3 − 14x

4 − 130x

5 +∫ x0

(x− t)2u2(t) dt

Proceeding as before and by cancelling the noise terms, we obtain theexact solution by u(x) = 1 + x.

8.3 Nonlinear Volterra Integral Equations of

the First Kind

8.3.1 The Series Solution Method

Exercises 8.3.1

1. Using the Taylor series of the left side, and substituting the series formof u(x) yields

112x

4− 190x

6+ 11260x

8+· · · =∫ x0

(x−t)[a0 + a1t+ a2t

2 + a3t3 + · · ·

]2dt

Integrating the right side, and equating the coefficients of like powersof x

a0 = 0, a1 = ±1, a2 = 0, a3 = ∓ 13! , · · ·

Consequently, the exact solution is given by

u(x) = ± sinx


12x

2 + 13x

3 + 16x

4 + 115x

5 + · · · =∫ x0

(x− t)[a0 + a1t+ a2t

2 + a3t3 + · · ·

]2dt



8.3. Nonlinear Volterra Integral Equations of the First Kind 171


a0 = ±1, a1 = ±1, a2 = ± 12! , a3 = ± 1

3! , · · ·


u(x) = ±ex


x+ 32x

2+ 32x

3+ 98x

4+ 2740x

5+· · · =∫ x0

[a0 + a1t+ a2t

2 + a3t3 + · · ·

]3dt


a0 = 1, a1 = 1, a2 = 12! , a3 = 1

3! , · · ·


u(x) = ±ex

4. Using the Taylor series of the left side, and substituting the series formof u(x) yields12x

2− 13x

3+ 115x

5− 2315x

7+· · · =∫ x0

(x−t)[a0 + a1t+ a2t

2 + a3t3 + · · ·

]2dt


a0 = ±1, a1 = ∓1, a2 = ∓ 12! , a3 = ± 1

3! , · · ·


u(x) = ±(cosx− sinx)




8.3.2 Conversion to a Volterra Equation of the SecondKind

Exercises 8.3.2

1. Differentiating both sides, and using Leibniz rule gives

v(x) = x6 + 17x

7 −∫ x0v(t) dt

This gives

v0(x) = x6, v1(x) = 0

This in turn gives

u(x) = ±x3


v(x) = x8 + 19x

9 −∫ x0v(t) dt

This gives

v0(x) = x8, v1(x) = 0

This in turn gives

u(x) = ±x4


v(x) = x12 + 113x

14 −∫ x0v(t) dt

This gives

v0(x) = x12, v1(x) = 0

This in turn gives

u(x) = ±x4




8.4. Nonlinear Weakly-Singular Volterra Equation 173

v(x) = 32e

2x − 12 −

∫ x0v(t) dt

This gives

v0(x) = e2x, v1(x) = 0

This in turn gives

u(x) = ±ex

5. Proceeding as before gives

v0(x) = x2e2x, v1(x) = 0

This in turn gives

u(x) = ±xex


v0(x) = x2e2x, v1(x) = 0

This in turn gives

u(x) = ±xex


v0(x) = e−2x, v1(x) = 0

This in turn gives

u(x) = ±e−x

8.4 Nonlinear Weakly-Singular Volterra

Equation

Exercises 8.4

1. Decompose f(x) into two parts, then set:




u0(x) = x3, u1(x) = − 7293080x

283 +

∫ x0

A0(t)

(x2−t2)13dt = 0

This gives

u(x) = x3


u0(x) = x4, u1(x) = − 2312048πx

12 +∫ x0

u3(t)√x2−t2 dt = 0

This gives

u(x) = x4


u0(x) = 5√

1 + x4, u1(x) = −π2 (1 + 38x

4) +∫ x0

u5(t)√x2−t2 dt = 0

This gives

u(x) = 5√

1 + x4


u0(x) = 4√

1 + x+ x3, u1(x) = −(π2 + x+ 23x

3) +∫ x0

u4(t)√x2−t2 dt = 0

This gives

u(x) = 4√

1 + x+ x3


u0(x) = 5√

cosx, u1(x) = −2√

sinx+∫ x0

u5(t)√sin x−sin t dt = 0

This gives

u(x) = 5√

cosx


u0(x) = 4√

sinx, u1(x) = 2√

cosx− 1 +∫ x0

u4(t)√cos x−cos t dt = 0

This gives

u(x) = 4√

sinx


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