Engineering Mathematics-II 9789351343257, 9351343251

Table of contents :
Title
Contents
Engineering Mathematics II
1 Vector Calculus
2 Ordinary Differential Equations
3 Laplace Transforms
4 Analytic Functions
5 Complex Integration
Appendix A
Appendix B
Appendix C

Citation preview

Engineering Mathematics-II

About the Author

T Veerarajan is the retired Dean, Department of Mathematics, Velammal College of Engineering and Technology, Viraganoor, Madurai, Tamil Nadu. A Gold Medalist from Madras University, he has had a brilliant academic career all through. He has 50 years of teaching experience at undergraduate and postgraduate levels in various established Engineering Colleges in Tamil Nadu including Anna University, Chennai.

Engineering Mathematics-II

T Veerarajan Dean, Department of Mathematics, (Retd.) Velammal College of Engineering & Technology Viraganoor, Madurai, Tamil Nadu

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110 016 Engineering Mathematics-II Copyright © 2014, by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listing (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. ISBN (13 digits): 978-9-35-134325-7 ISBN (10 digits): 9-35-134325-1 Vice President and Managing Director: Ajay Shukla Head—Higher Education Publishing and Marketing: Vibha Mahajan Sr Publishing Manager—SEM & Tech. Ed: Shalini Jha Editorial Executive—Acquisitions: Vamsi Deepak Sankar Executive—Editorial Services: Sohini Mukherjee Manager—Production Systems: Satinder S Baveja Production Executive: Anuj K Shriwastava Marketing Manager—Higher Education: Vijay Sarathi Sr Product Specialist—SEM & Tech Ed: Tina Jajoriya Sr Graphic Designer (Cover): Meenu Raghav General Manager—Production: Rajender P Ghansela Production Manager—Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

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Contents Preface

xi

ENGINEERING MATHEMATICS − II 1. Vector Calculus 1.1 Introduction 1.1 1.2 Vector Differential Operator ∇ 1.1 Worked Example 1(a) 1.4 Exercise 1(a) 1.9 1.3 The Divergence of a Vector 1.10 Worked Example 1(b) 1.16 Exercise 1(b) 1.24 1.4 Line Integral of Vector Point Functions Worked Example 1(c) 1.28 Exercise 1(c) 1.35 1.5 Integral Theorems 1.37 Worked Example 1(d) 1.40 Exercise 1(d) 1.54 Answers 1.56

1.1

1.25

2. Ordinary Differential Equations 2.1 Equations of the First Order and Higher Degree 2.1 Worked Example 2(a) 2.3 Exercise 2(a) 2.12 2.2 Linear Differential Equations of Second and Higher Order with Constant Coefficients 2.14 2.3 Complementary Function 2.14 Worked Example 2(b) 2.19 Exercise 2(b) 2.32 2.4 Euler’s Homogeneous Linear Differential Equations 2.33 2.5 Simultaneous Differential Equations with Constant Coefficients 2.35 Worked Example 2(c) 2.35 Exercise 2(c) 2.47 2.6 Linear Equations of Second Order with Variable Coefficients 2.48 Worked Example 2(d) 2.53 Exercise 2(d) 2.67

2.1

Contents

vi

2.7 Method of Variation of Parameters 2.71 Worked Example 2(e) 2.73 Exercise 2(e) 2.82 Answers 2.82

3. Laplace Transforms

3.1

3.1 Introduction 3.1 3.2 Linearity Property of Laplace and Inverse Laplace Transforms 3.3 Laplace Transforms of Some Elementary Functions 3.3 3.4 Laplace Transforms of Some Special Functions 3.7 3.5 Properties of Laplace Transforms 3.10 Worked Example 3(a) 3.12 Exercise 3(a) 3.33 3.6 Laplace Transform of Periodic Functions 3.36 3.7 Derivatives and Integrals of Transforms 3.37 Worked Example 3(b) 3.40 Exercise 3(b) 3.60 3.8 Laplace Transforms of Derivatives and Integrals 3.63 3.9 Initial and Final Value Theorems 3.66 3.10 The Convolution 3.68 Worked Example 3(c) 3.70 Exercise 3(c) 3.92 3.11 Solutions of Differential and Integral Equations 3.96 Worked Example 3(d) 3.96 Exercise 3(d) 3.108 Answers 3.110

4. Analytic Functions 4.1 Introduction 4.1 4.2 The Complex Variable 4.2 Worked Example 4(a) 4.10 Exercise 4(a) 4.22 4.3 Properties of Analytic Functions 4.24 Worked Example 4(b) 4.30 Exercise 4(b) 4.41 4.4 Conformal Mapping 4.43 4.5 Some Simple Transformation 4.46 4.6 Some Standard Transformations 4.49 Worked Example 4(c) 4.57 Exercise 4(c) 4.74 4.7 Bilinear and Schwarz-Christoffel Transformations 4.8 Schwarz-Christoffel Transformations 4.80 Worked Example 4(d) 4.82 Exercise 4(d) 4.91 Answers 4.93

3.2

4.1

4.77

Contents

5. Complex Integration

vii

5.1

5.1 Introduction 4.1 5.2 Cauchy’s Integral Theorem or Cauchy’s Fundamental Theorem 5.2 Worked Example 5(a) 5.7 Exercise 5(a) 5.17 5.3 Series Expansions of Functions of Complex Variable-Power Series 5.20 5.4 Classification of Singularities 5.25 Worked Example 5(b) 5.31 Exercise 5(b) 5.43 5.5 Contour Integration—Evaluation of Real Integrals 5.47 Worked Example 5(c) 5.50 Exercise 5(c) 5.67 Answers 5.68 Appendix A: Solved Model Question Paper I Appendix B: Solved Model Question Paper II Appendix C: Solved Question Paper May/June 2013

A.1–A.8 B.1–B.6 C.1–C.10

Preface

I am deeply gratified by the enthusiastic response shown to all the earlier editions of my book on Engineering Mathematics by the students and teachers throughout the country. This book has been revised to meet the requirements of the first-year undergraduate course on Engineering Mathematics offered to the students of engineering. The contents have been covered in adequate depth for the semester 2 Syllabi of various universities/ deemed universities across the country. The book offers a balanced coverage of both theory and problems. Lucid writing style supported by step-by-step solutions to all problems enhances understanding of the concepts. It has an excellent pedagogy with 612 unsolved problems, 289 solved problems. For the benefit of the students, the solutions to 2012–13 of Anna University is given in appendix along with model question papers. Additionally, the book is accompanied by a website which provides useful supplements including recent solved question papers. I hope that the book will be received by both the faculty and the students as enthusiastically as the earlier editions of the book and my other books. Critical evaluation and suggestions for the improvement of the book will be highly appreciated and acknowledged. T VEERARAJAN

Publisher’s Note: McGraw Hill Education (India) Private Limited looks forward to receiving from teachers and students their valuable views, comments and suggestions for improvements, all of which may be sent to [email protected], mentioning the title and author’s name.

Engineering Mathematics-II 1. Vector Calculus 2. Ordinary Differential Equations 3. Laplace Transforms 4. Analytic Functions 5. Complex Integration

Chapter

1

Vector Calculus

1.1

INTRODUCTION

In Vector Algebra we mostly deal with constant vectors, viz. vectors which are constant

each value of scalar variable t a vector function of the scalar variable t and is denoted as F t.

F, then F is called F (t ) .

r t, i.e. r

r (t ) .

rectangular cartesian co-ordinates (x, y, z), then r is a function of the scalar variables x, y, z, i.e. r r (x, y, z ). A a scalar point function or a vector point function,

that region is called a

.

or a

1.2 VECTOR DIFFERENTIAL OPERATOR

, according

Engineering Mathematics II

1.2

i

j

x

k

y

z

, vector) with i

x

x

,

,

y

i , j,k

z

, it should be noted that i and

. When writing

x

.

1.2.1

Gradient of a Scalar Point Function

Let (x, y, z) i

Note

1.

j

x

k i y z x and shortly denoted as

should not be written as

j

y .

k

z

.

2. When combines with , neither . and . 3. If is a constant, = 0. 4. (c1 1 ± c2 2) = c1 1 ± c2 where c1 and c2 are constants and 1, 2 2

5. 6.

(

)=

1

2

1

1

2

2

+

1

1 2

2

i.e. d

2

, if

2

0.

(u)

u.

Directional Derivative of a Scalar Point Function (x, y, z)

Let P and Q origin O be r (= OP) and r = r. Let and + Then

1

2

7. If v = f (u), then

1.2.2

.

2

d dr

lim r 0

r

r(

OQ)

y, so that PQ

r and PQ P and Q

is called the directional derivative of in the direction OP.

gives the rate of change of dr direction of r .

Chapter 1: Vector Calculus ,

x

,

y

,

1.3

are the directional derivatives of

z

at P(x, y, z) in the

.

1.2.3

Gradient as a Directional Derivative

Let (x, y, z (x, y, z) = c values of c, a family of surfaces, called the level surfaces of the function . , namely = c1 and = c2 r ) . [Refer to Fig. 1.1] P ( r ) and Q ( r R Let the normal at P to the level surface = c1 Q meet = c2 at R. = c2

surfaces = PR = n. If QPR , is almost a right angled triangle,

PQR, which

P

= c1

Fig. 1.1

n = r cos (1) If n is the unit vector along PR, i.e. in the direction of outward drawn normal at P to the surface = c1, then (1) can be written as where r PQ . n n r, or

n dr

d =

d dn dn

= Also

(2)

dn

d =

d n dr dn x

dx

x

[by using (2)]

y

i

y

dy

j

z

z

(3)

dz of Engg. Maths I] k (dx i

dr

dy j

dz k ) (4)

From (3) and (4), dr Since

PQ

d n.dr dn r (or d r ) is arbitrary, d n dn

From (1),

d dn

d dr (cos )

(5)

1.4

Engineering Mathematics II

i.e.

d dr

cos

i.e.

d dr

d dn

d dn cos

i.e.

1 is

d , that is the directional derivative dn

in the direction of n. : is a vector whose magnitude is the greatest directional derivative of and whose direction is that of the outward drawn normal to the level surface = c. WORKED EXAMPLE 1(a) Example 1.1 Find the directional derivative of = x 2yz + 4xz2 PQ, where Q x 2 yz x

4 xz 2

i

y

j

z

4z2 ) i

(2 xyz ( ) (1, The magnitude of (

P (1, 2,

k

x2 z j

(x 2 y

8 xz ) k

8i j 10k )P is the greatest directional derivative of

2 , 1)

at P.

at (1, 2 , 1)

64

1

100

165 units. PQ Directional derivative of

OQ

OP

2i

j

k

in the direction of PQ

along PQ . PQ PQ (8 i

j

10k ) (2 i 4

1

j

k)

1

27

units. 6 Example 1.2 Find the unit normal to the surface x3 (1, 1, 1).

Note

Unit normal to a surface

=c

+ z3

n in the

Chapter 1: Vector Calculus x3

xyz + z3

= x3

xyz + z3

1.5 (x, y, z) = c.

is a vector acting in the direction of the outward drawn normal to the surface = c. (3 x 2

Now (

)(1,1,1)

2i

yz ) i j

n |n | 1 (2 i 3

n

2k

j

(3 z 2

xzj

xy ) k

n (= a vector in the direction of the normal)

2k )

Example 1.3 Find the directional derivative of the function = xy2 + yz2 x log z y2 y2

x log z

(x, y, z) = c. y2 and c

(x, y, z) = x log z

The direction of the normal to this surface is the same as that of x Now (log z ) i 2 y j k z

(

(

)(

)( 2 ,

Directional derivative of

4j

1,2 ,1)

1,1)

k

b (say)

xy 2

yz 3

y2i

(2 xy

i

3j

.

z3 ) j

3 yz 2 k

3k

in the direction of b b |b | (i

3j

3k ) (

4j

k)

16 1 15 17

units.

Example 1.4 Find the angle between the normals to the surface xy = z2 , , 2) and (1, 9, . Angle between the two normal lines can be found out as the angle between the vectors acting along the normal lines. xy = z2 with (x, y, z) = c, we get = xy z2 and c = 0. yi

xj

2 zk

Engineering Mathematics II

1.6 (

)( (

2i

2 , 2 ,2 )

)(1,9 ,

2j

9i

3)

j

4k

6k

n1 (say)

n2 (say)

n1 and n2 If n1 n2 n1 n2

cos

44 24

118

177

11

1

cos

11

177

Example 1.5 Find the angle between the surfaces x2

Identifying x2 we have

2

2

= x2

2

)(6 ,4 ,3) 12 i

Identifying we have

6 ,4 ,3

If

2

and c = 11. 2 zk

2y j 8j

6k

n1

18 with

= c´, and c´ = 18

(y z ) i

(z x) j (y x)k

i

9j

2k

is the angle between the surfaces at (6, 4, 3), then n1 n2 cos n1 n2 48

24

244 86

61 86

cos

24

1

5246 xz2

Example 1.6 Find the e 4x = Identifying 2xz2 we have

,

, 2). 4x = 7 with = c, = 2xz2 3 4x and c = 7.

3

(2 z 2 ( (

18

= 11 with = c, 2 xi

(

y2 z2 = 11 and

)

)(1,

1,2 )

7i

3 y 4) i 3j

3x j

4 xzk

8k

is a vector in the direction of the normal to the surface

= c.

3

Chapter 1: Vector Calculus

1.7

D.R.’s of the normal to the surface ( = c)

7(x

y + 1) + 8(z 7x y

i.e.

Example 1.7 Find the constants a and b, so that the surfaces 5x2 ax2y + bz3 =

Identifying 5x2

2yz

(10 x 9) i

1

(

1 )(1, 1,2 )

Identifying ax2y + bz3 = 4 with we have

2

i

4j

2z j

2 yk

n1 (say)

2k

= c´.

( (

2)

2axyi

2 )(1, 1,2 )

2ai

Since the surfaces cut orthogonally, n1

ax 2 j aj

3bz 2 k 12bk

n2 (say)

n2 .

n1 n2 0 a + 24b = 0 a + 4b = 0

i.e.

(1) ax2y + bz3 = 4 (2)

a + 8b = 4 Solving (1) and (2), we get a = 4 and b = 1. Example 1.8 If r and

9x = 0 and

. = c, 1

9x = 0 with

we have

2yz

x, y, z), a is a constant vector

= x2 + y2 + z2,

(r a ) a and (ii) r grad r

xi

a

a1 i

a2 j

a3 k

r a

a1 x

a2 y

a3 z

a1 i

a2 j

a3 k

Let

grad r a grad r grad Example 1.9 If r (r n )

yj

2 .

zk

a

x2 y2 + z2 2 xi 2 yj 2 zk 2(x 2

y2

z2 ) 2

. x, y, z)

nr n 2 r .

-

Engineering Mathematics II

1.8 r

xi

r2

| r 2 | x2

(r n )

x nr n

From (l),

2r

r x

Similarly,

zk y2

(r n ) i

z2 (r n ) j

y

r i x

1

(1) z

r j y

(r n )k

r k z

(2)

2x r x r y

i.e.

yj

x r y r

and

r z

x i r

y j r

z r

(3)

Using (3) in (2), we have (r n )

nr n

1

nr n 2 ( xi nr

n 2

z k r zk )

yj

r.

Example 1.10 Find the function , if grad ( y2

2 xyz 3 ) i

(3

2 xy

x2 z3 ) j

(6 z 3

3 x 2 yz 2 )k .

( y2

2 xyz 3 ) i

(3

2 xy

x2 z3 ) j

(6 z 3

3x 2 yz 2 )k

x

x y z

y2 3 6z3

i

y

j

z

(2)

k

2 xyz 3 2 xy

(1)

(3)

x2 z3

(4)

3 x 2 yz 2

(5) x (i.e. treating y and z as

constants), = xy2

x2yz3 + a function not containing x

(6)

Chapter 1: Vector Calculus

1.9

Note x we add an arbitrary function of the other variables y and z, i.e. an arbitrary function x. y. x2yz3 + a function not containing y z,

y + xy2 3 4 z 2

x 2 yz 3

(7)

a function not containing z

(8)

. The general form of

is obtained as

follows: be included in the value of . . The last terms indicate that there is a term of a constant. 3 4 3y z xy 2 x 2 yz 3 2

x, y, z, i.e. c.

EXERCISE 1(a) Part A (Short Answer Questions) grad and give its geometrical meaning. 2. If r

x, y, z

3. If r

x, y, z

4. If r

x, y, z

5. Find grad 7. Find the directional derivative of the x

1 r. r ( | r |2 ) 2r . 1 f (r ) f (r )r . r (r )

= 3x2y y3z2. = x3y2z = xy + yz + zx

= x2y2z4 T = xy + yz + zx

x, y, z

Part B 11. If = xy + yz + zx and F grad at the

x 2 yi

T(x, y, z

y 2 zj

z 2 xk

F · grad

2

+ y2 z.

and F

Engineering Mathematics II

1.10

12. Find the directional derivative of = 2xy + z2 direction of i 2 j 2k . 13. Find the directional derivative of = xy2 + yz3 P direction of PQ where Q 14. Find a unit normal to the surface x2y + 2xz 15. Find the directional derivative of the scalar function = xyz in the direction of the outer normal to the surface z = xy 16. Find the angle between the normals to the surface xy3z2 17. Find the angle between the normals to the surface x2 = yz and (2, 4, 1). 18. Find the angle between the surfaces z = x2 + y2 x2 + y2 + z2 = 9 at the 19. Find the angle between the surfaces xy2z = 3x + z2 and 3x2

y2 + 2z = 1 at the

x log z y2 2

xy+z xz2 + x2y = z 22. Find the values of and , if the surfaces x2

= ( + 2)x and 4x2y + z3 = 4

23. Find the values of a and b, so that the surfaces ax3 by2z = (a + 3)x2 and 4x2y z3 (2 xy z 2 ) i (x 2 2 yz ) j 2 (y 2 zx) k . 25. If

2 xyz 3 i

x2 z3 j

3 x 2 yz 2 k

(x, y, z), given that

1.3 THE DIVERGENCE OF A VECTOR If F (x , y , z )

x, y, z) in F , denoted as div

F

F i

F , when F

F1 i

F

F x

F2 j

i

j

x

i Formula for

F

x

k

y F y

j

z k

F F z

F3 k

j

y

k

z

(F1 i

F2 j

F3 k )

Chapter 1: Vector Calculus F1 x

Note and hence

1.3.1

F3 z

F2 y

F

Since F

, F1, F2

and F3

.

Physical meaning of

(i) If V

1.11

Vx i

Vy j

F

Vz k x, y, z), then .

(ii)

V

V

V

V

V

In general, if F

F F , it is called the divergence of F .

1.3.2

Solenoidal Vector

If F is a vector such that F 0 be a solenoidal vector in that region.

1.3.3

Curl of a Vector

If F (x, y, z) is a differentiable vector x, y, z) in the curl of F or the rotation of F , denoted as curl F or rot F F

F i

F x

i

Note

j

x

k

y j

F y

z k

F F z

F is also

Formula for functions):

F, when F

F

i

F1 i

x F1 (i x

F3 k (where F1, F2 and F3 are scalar

F2 j

(F1 i i)

F2 j F2 (i x

F3 k ) j)

F3 (i x

k)

Engineering Mathematics II

1.12

F3 j x

F2 k x F3 y

1.3.4

F2 i z

i

j

k

x F1

y F2

z F3

F3 j x

F1 z

F2 x

F1 k y

Physical Meaning of Curl F

If F

x, y, z) of a rigid body that rotates , then curl F 2 .

1.3.5

Irrotational Vector

If F is a vector such that F 0 be an irrotational vector in that region.

1.3.6

Scalar Potential of an Irrotational Vector

If F function. Let

F

F1 i

Since F

F

0

i.e.

i.e.

F3 y

F2 i z F3 y

F1

F2 j

i

j

k

x F1

y F2

z F3

x

, F2

0

F3 j x

F1 z F2 ; z

F3 k

F2 x

F3 ; x

F1 z

y

and

F2 x

F1 k y F1 y

F3

0

(1)

z

Chapter 1: Vector Calculus F

i

j

x

If F is irrotational and F

1.3.7

k

y

1.13

z

, then

F.

Expansion Formulae Involving Operations by ’s are given below: The

students. 1. If u and v

(u

2. If u and v

(u

3. If (

( F)

F) i

x

u

v.

( F) F x

x

F x

i

F i

F

x

F

F

F

4. If ( F)

(

F)

F

5. If u and v

u v (u v )

i

u v x u v u x

i

u v x

i i [ interchanged]

v)

v.

F

i

Proof:

u

F ( F)

Proof:

v)

u v x

v x v u x

i i

v u x

v curl u

u curl v.

Engineering Mathematics II

1.14

u v x u u

i v

v u x

i v

6. If u and v (u v ) (

Note

v )u

(v

v and v

In this formula,

vx

v Thus v

)u

u

x u x

vx

(

u )v

)v

vy vy

vz

y u y

vz

z u z

, if v

u v

)u

v means div v , but

are not the same,

7. If u and v (v

(u

vx i

v curl u

. (

) = 2

2

2

x2

y2

z2

2

2

2

2 2

x

2

Note

y

2

F

F x2

2

z2

is called the Laplacian of .

2

2

x

i

i x

y

x

y

y z

j

y j

2

(

) = 0.

k

z

k z z 2

z y

2

i

z x

2

x z

2

j

x y

2

y x

= 0.

Note

,

= 0 is called the

F. z2 )=

curl (grad )

2

2

F y2

grad

) =

F resulting in

9. If Proof

u curl v

is called the Laplacian operator and

can also

2 2

vz k

)v .

(u

8. If where

vy j

This result means that (grad ) is always an irrotational vector.

k

Chapter 1: Vector Calculus 10. If F Proof:

1.15

(curl F ) Let F

F1 i

F2 j

i

j

k

x F1

y F2

z F3

curl F

F3 y div curl F

x

(

F) 0 .

F3 k

F2 i z F3 y

F2 z

2

y

2

F3 x y

F3 j x

F1 z

F3 x

F1 z 2

2

F2 x z

F2 x

F3 y x

F1 y z

F1 k y F2 x

z 2

F1 y 2

F2 z x

F1 z y

0.

Note

This result means that (curl F ) is always a solenoidal vector.

11. If F curl (curl F ) Proof: Let Then

F

F1 i

F2 j

F3 y

curl F

(

F)

(

2

F)

F.

F3 k

F2 i z

F3 j x

F1 z

F1 k y

F2 x

curl (curl F ) i

j x

F3 y

y F2 z

2

F2 y x 2

F1 x2

F1 y

2

F3 z x F2 x y

F1 z

z

2

2

z F3 x

F1 z

F2 x

y

k

F1 y2

2

F3 x z

F2 x

F1 y

F3 i x 2

F1 i z2

2

F1 x2

2

F1 y2

2

F1 i z2

Engineering Mathematics II

1.16 F1 x

x

i

2

F

x

F

x

F3 z

F2 y

j 2

F

2

2

2

x2

y2

z2

F1 i

F1 i F

y

k

2

F

z

F1 i

F2 j

F3 k

F.

12. If F grad (div F )

Note

(

F)

(

2

F)

F.

Rewriting the formula (11), this result is obtained. WORKED EXAMPLE 1(b)

Example 1.1 When (1, 2, 3).

= x3 + y3 + z3

xyz

.

,

and

= x3 + y3 + z3 3x yz i

x 3(x 2

yz ) i

3(y 2

3(x 2 yz )] [3 x 6(x y z )

y

i

2

yz ) 3(y

( 3x 3x) i

Note

[3(z 2

xy )]

z zx) 3(z

2

( 3 y 3 y )j

xy ) ( 3z

3z )k

= 0, for any , ( ( . (x 2

(

F ) and

F,

(

)(1, 2 ,3)

15 i

)(1, 2, 3) = 36

Example 1.2 If F F ),

z

k

y 2

xy )k

[3(y 2 zx)]

j x

3(x

3(z 2

zx)j

y2

2 xz ) i (

and (xz F)

3j ( xy

×

21k )(1, 2, 3) = 0. yz )j

(z 2

x 2 )k ,

, 1, 1).

F,

Chapter 1: Vector Calculus

(

F

(x 2

F

(xz (x 2 y 2 2 xz ) x y (2 x 2 z ) ( x z ) 2 z x 5z

F)

x i

y2

2 xz ) i

(x 5 z ) i

(xz

y

2

y

2

2 xz

xz

y xy

yz

z

(y

z )k

(x

y) i

(2 x 2 x) j

(x

y) i

(y

(

F)

(

F ) 0 , for any F ,

x2 )

z

(0)

y

F)

z

x i

(

F )(1,1,1)

2i

(

F )](1,1,1)

k z y

z

F )](1,1,1)

i

2k ; [

(

a1 x

F )] 1,1,1

0;

yj

x, y, (a r ) a , (ii) div (a r ) = 0 and

zk

a3 k , where a1, a2, a3 are constants. a r

5k ;

k

Example 1.3 If is a constant vector and r z (iii) curl (a r ) 2a . xi

z)

k

6; [ (

i

(y

y 0

y

F )(1,1,1)

r

x2

j

x

(

2

z )k

[ (x y )] x 1 0 1 0

(

a2 j

(z 2

k

i

a1 i

z

(x 5 z )k

z

j x

x

Let a

yz )

xy

x 2 )k

5k

F

[

(z 2

yz ) j

(x 5 z ) j

i

Note

xy

1.17

a2 y

a3 z

Engineering Mathematics II

1.18 grad (a r )

x a1 i

(a1 x

a2 y

a2 j

a3 k

a3 z ) i

a a r

i

j

k

a1 x

a2 y

a3 z

(a2 z a3 y ) i div (a r )

x

(a3 x a1 z ) j

(a2 z a3 y )

y

(a1 y a2 x)k

(a3 x a1 z )

z

(a1 y a2 x)

0 i curl (a r )

j

k

x y z a2 z a3 y a3 x a1 z a1 y a2 x (a1

a1 ) i

2(a1 i

(

a2 j

a2 ) j

a2

(a3

a3 )k

a3 k )

2a

(2x 2

Example 1.4 Show that u

x

(2 x 2

(3 x 3 y 3 xy ) j

(4 y 2 z 2

{ (4 y 2 z 2

2 x 3 z )}

2 x 3 z )k is

xyz 2 u is solenoidal.

not solenoidal, but v u

8 xy 2 z ) i

8 xy 2 z )

y

= (4x + 8y2z) + (3x3 = x3 + x

(3 x 3 y 3 xy ) x

z

y2z + 2x3)

x, y, z) u is not solenoidal. v xyz 2 u v

(2 x 3 yz 2

8x2 y3 z3 ) i

(6 x 2 yz 2

16 xy 3 z 3 ) (6 x 4 yz 2

=

(3 x 4 y 2 z 2

3x 2 y 2 z 2 ) j

(4 xy 3 z 4

6 x 2 yz 2 ) (16 xy 3 z 3

2 x 4 yz 3 )k

6 x 4 yz 2 )

x, y, z)

v is solenoidal. Example 1.5 Show that F (y 2 z 2 3 yz 2z )k is both solenoidal and irrotational.

2 x) i

(3 xz

2 xy ) j

(3 xy 2 xz

Chapter 1: Vector Calculus F

x

(y 2

=

z2 x

3 yz

2 x) +

y

(3 xz

1.19

2 xy ) +

(3 xy

z

2 xz

2 z)

x+2

=

x, y, z)

F is a solenoidal vector. i F

j x

(y

2

z

(3 x

2

k

y

3 yz

3 x) i

2 x) (3 xz

(3 y

=

2z

z 2 xy ) (3 xy

2z

3 y) j

2 xz

(3 z

2z)

2y

2y

(2 x 2 z

y

3z ) k

x, y, z)

:. F is an irrotational vector. (y 2

Example 1.6 Show that F

2 xz 2 ) i

i F

(2 xy j

x (y 2

y

1) i

(4 xz

=

2 z ) k is

k

2 xz 2 ) (2 xy

( 1

z) j

z z ) (2 x 2 z

4xxz ) j

(2 y

y

2z)

2 y) k

x, y, z)

F is irrotational. F be . F x x

i

y2

y

j

z

k

2 xz 2

x; = xy2 + x2z2 y

2 xy

x

(1)

z

y; = xy2 z

2 x2 z

yz

y y

2z

(2)

Engineering Mathematics II

1.20

z; = x2z2 yz + z2 + From (1), (2), (3), we get = xy2 + x2z2 Example 1.7 (3 x 2

cz ) j

z

(3)

yz + z2 + c.

Find the values of the constants a, b, c, so that F (3 xz 2

(axy

b z3 ) i

y ) k may be irrotational. For these values of a, b, c, F.

F is irrotational. F i

0

j

k

i.e. x

y

(axy bz 3 ) (3 x 2 i.e.

(3 z 2

( 1 c) i c

,

0

z cz ) (3xz 2

3bz 2 ) j

y)

(6 x ax) k

3z b) = 0, x a = 6, b= 1, c = 1. 2

0

a) = 0

Using these values of a, b, c, F

(6 xy

z3 ) i

Let

(3 x 2

z) j

(3 xz 2

y) k

F. F

x

x 6 xy

z3 ,

i

y y

j 3x 2

z

k

z,

3 xz 2

z

= 3x2y + xz3 = 3x2y = xz3

x

yz

y

yz +

z

y

(1) (2) (3)

From(l), (2) and (3), we get = 3x2y + xz3 Example 1.8 tional and (ii)

If

and is solenoidal. = ( )+ = =0 is irrotational

yz + c is irrota-

Chapter 1: Vector Calculus

1.21

u v v curl u u curl v )= curl ( curl ( = = 0. ) is solenoidal.

(

)

Example 1.9 If r = | r | , where r that

2

x, y, z),

(rn ) = n(n + 1) rn and hence deduce that

(r n ) nr n 2

Now

2

r.

(r n )

( r n) (nr n

2

(r n

2

n n (n

r) rn

) r

2) r n

[since

4

r r

2

(

r)

3r n (xi

r x 3]

2

1 r

(x)

2

zk )

yj y

(y )

z

(z )

(rn) = n [(n r 4 r2 + 3r 2] = n (n+1) r 2

n= 2

i.e.

1 r

( 1) (0) r

3

0

1 r

Example 1.10 If u and v v u) = u 2 v v 2 u. v u) = (u v (v u) = ( u v + u 2v v u + v 2 u) (u v

(u v

=u

2

v

2

v

u.

Example 1.11 If u and v such that uF Given

F

v

F curl F uF F

v 1 u

v

0.

Engineering Mathematics II

1.22

1 u

F

v

1 ( u 1 u 1 u

F curl F

Now

1 u

v)

v

v , since 1 u

v

1 (0) , u

v = 0.

v

[

= 0. Example 1.12 If r = r , where r

x, y, z) with f (r)

2

(ii)

f (r )

f (r )

2 f (r ). r

f (r ) r and r

r2 = x2 + y2 + z2 2r r x

r x

2x

x . Similarly, r

r y

f (r )

Now

r z

y and r i

x

f (r ) r i x

f (r )

x i r

f (r )

2

f (r )

z . r

f (r ) r r f (r ) f (r ) r r f (r ) r r f (r ) r2

r

f (r ) r

f (r )

r

(r ) r

3

f (r ) r

r

3

Chapter 1: Vector Calculus r f (r ) r

2

f (r ) 1 r r r

1.23

3 f (r ) r r i x

(r ) r f (r ) 2

f (r ) 1 2 (r ) r

x i r

1 r r

3 f (r ) r

r 2 f (r ) f (r ) r

Example 1.13 Find f (r) if the vector f (r) r is both solenoidal and irrotational. f (r) r is solenoidal f (r ) r i.e. i.e. i.e. i.e.

f (r ) r

0 f (r )

r

0

f (r ) r r 3 f (r ) 0 r rf (r ) + 3f (r) = 0 f (r ) f (r )

3 r

0

Integrating both sides w.r.t. r, log f (r) + 3 log r = log c i.e. log r3 f (r) = log c c f (r ) r3 f (r) r is also irrotational f (r ) r i.e.

0 f (r ) r

f (r ) r r r

i.e.

(1)

f (r )

0 0

r

r

0 i

j

k

x x

y y

z z

0

f (r ) (0) 0 0 r

i.e. This is true for all values of f (r)

From (1) and (2), we get that f (r) r is both solenoidal and irrotational if f (r) Example 1.14

i.e,

If

= 0, only when 2 = 0. is solenoidal, only when 2 = 0

(2) c

r3 is both solenoidal and

(1)

Engineering Mathematics II

1.24

i.e. is irrotational always From (1) and (2), is both solenoidal and irrotational, when

(2) 2

= 0, i.e. when

is a solution of

Example 1.15 If F is solenoidal, Since F is solenoidal, F 0 F

F

(

= curl curl ( = ( 2

[ { [ {

=

4

2

(

F. (1)

2

F) 2

F

4

(2)

F

F [by (1)]

2

F)

2

F)

F}

2

F )}

( 4

2

F )] , by using (2)

F] ,

by interchanging the and

2

F {by using (1)}

EXERCISE 1(b)

Part A (Short Answer Questions) 2.

F F

7. If r x, y, z) r and curl r . 8. If F 3 xyz 2 i 2 xy 3 j x 2 yz k , F 9. If F (x 2 yz ) i (y 2 2 zx) j (z 2 3 xy ) k F 10. If F (x y 1) i j (x y ) k , show that F F. 11. If F zi x j yk F 0. 12. Show that F (x 2 y ) i (y 3 z ) j (x 2 z )k is solenoidal. 13. Show that F (sin y z ) i (x cos y z ) j (x y )k is irrotational. 14. Find the value of , so that F y 4 z 2 i 4 x3 z 2 j 5 x 2 y 2 k may be solenoidal. 15. Find the value of , if F 2x 5 y i x y j 3 x z k is solenoidal.

Chapter 1: Vector Calculus

1.25

3 2 2 16. Find the value of a, if F (a x y z ) i (a 2)x j (1 a)xz k is irrotational. 17. Find the values of a, b, c, so that the vector F (x y a z ) i (bx 2 y z ) j ( x cy 2 z )k may be irrotational. 18. If and v (u v ) is solenoidal. 19. If 1 and 2 ( 1 2 ( 2 1). 20. If

Part B 21. If u = x2yz and v =

3z2,

( u

v) and

( u

v) at the

22. Find the directional drivative of ( of the normal to the surface xy2 z = 3x + z2 , where = x2 y2 z2 F , ( F ), F, ( F ) and 23. If F 3 x 2 i 5 xy 2 j x y z 3 k , ( F ) at the (1, 2, 3) 24. If is a constant vector and r is the vector of (x, y, z) w.r.t. the ori[(a r ) r ] a r . 25. Prove that F 3 yzi tational. Find also 26. Show that F

(z 2

2 zx j

4 xy k is not irrotational, but (x2 y z3) F is irro-

2 x 3 y) i

(3 x

2y

z) j

(y

2 z x) k is irrotation-

27. Find the constants a, b, c, so that F (x 2 y az ) i (bx 3 y (4 x cy 2 z ) k may be irrotational. For these values of a, b, a, b, c, if F

axyz 3 i

z) j bx 2 z 3 j

cx 2 yz 2 k is irrotational. For these values of a, b, c, 29. If r

(x, y, z) w.r.t. the origin,

1 2 1 2 r and (ii) r r r3 r r 30. Find the value of n, if rn r is both solenoidal and irrotational, when r xi y j zk . (i)

1.4 LINE INTEGRAL OF VECTOR POINT FUNCTIONS C be a curve in that region (Fig. 1.2). P and Q on C be r and r acts at P

Q

F

Let F (x, y, z)

C

P r+

r .Then PQ

r If F with PQ ,

r Fig. 1.2

Engineering Mathematics II

1.26

then F

r

F ( r ) cos

.

In the limit, F dr F dr cos Note Physically F dr dr . Now the integral

F through

F dr

the line integral of F along the curve C.

C

Since

F cos dr , it is also called the line integral of the tangential

F dr C

C

F along C. B

Note

(1)

F dr

C but also on the terminal

A (C )

A and B. B

(2) Physically

F dr

done by the force F in

A (C )

A to B along the curve C. B

F dr

(3) If the value of

curve C, but only

A

A and B, F is called a Conservative vector. F from A to B A and B, the force F is called a Conservative force. C is a closed curve, the line integral is denoted as

F dr . C

(5) When F

F1 i

F dr C

(F1 i

F2 j F2 j

F3 k , F3 k ) (dx i

dy j

dz k )

C

( (F1dx

F2 dy

r

xi

F3 dz )

yj

zk ) -

C

dr , where

(6) C

line integrals.

1.4.1

Condition for F to be Conservative

If F is an irrotational vector, it is conservative.

F C

dr are also

Chapter 1: Vector Calculus Since F

Proof:

1.27 . i.e., F

B

B

F dr A

dr A B

x

A

i

j

y

k

z

(dx i

dy j

dz k )

B

x

A

dx

y

dy

z

dz

B

d A

[ ]BA (B)

(A)

F is conservative. If F F dr 0 .

Note C

[

(B) , as A and B coincide]

(A)

1.4.2

Surface Integral of Vector Point Function

Let S be a two sided surface, one side of which is Let F

F

S. Let dS

S

x, y, z). Let n be the unit vector normal to the surface S at (x, y, z) direction) Let be the angle between F and n . F F n F cos

nˆ ds (x, y, z)

S

Fig. 1.3

dS over the surface S is called the surface integral of F over S and denoted as F cos dS or F n dS . S

S

If

is a vector whose magnitude is dS and whose direction is that of n , then n dS . F n dS can also be written as

S

F dS . S

Engineering Mathematics II

1.28

Note

(1) If S is a closed surface, the outer surface is usually chosen as the (2)

F

dS and

dS , where

S

S

surface integrals. (3) When evaluating

F n dS S

To evaluate a surface integral in the scalar form, we convert it into a double integral and then evaluate. Hence the surface integral F dS is also denoted as S

F dS . S

WORKED EXAMPLE 1(c)

dr , where C is the curve x = t, y = t2, z =

Example 1.1 Evaluate

t) and

C

= x2 y (1 + z) from t = 0 to t = 1. r

xi

yj

dr

dx i

dy j

x 2 y (1

Hence the given line integral I

zk dz k

z ) (dx i

dy j + dz k )

C

x 2 y (1

i

z ) dx

C 1

i

z ) dy

1

t 4 ( 2 t ) dt

j

1

t6 6

7 i 30

8 j 21

If F

t6 j 4 6

0

xyi

z ) dz

C

t 4 (2

1

t ) 2t dt

k

0

t5 i 2 5

x 2 y (1

k

C

0

Example 1.2

x 2 y (1

j

t 4 (2

t ) ( dt )

0

t7 2 7

1

t5 2 5

k 0

t6 6

1

0

7 k. 30 zj

F

x 2 k , evaluate

dr , where C is the curve

C

x = t2, y = 2t, z = t3 from (0, 0, 0) to (1, 2, 1).

F

dr

i xy dx

j k z x2 dy dz

(z dz

x 2 dy ) i

(xy dz

x 2 dx) j

(xy dy

zd x ) k

Chapter 1: Vector Calculus

1.29

The given line integral x 2 dy ) i

[ (zdz

x 2 dx) j

(xy dz

(xy dy

z dx ) k ]

C 1

[ (t 3 3t 2

t 4 2)) dt i

(2t 3 3t 2

t 4 2t ) dt j

(2t 3 2

t 3 2 t ) dt k ]

0

[

t 1

i (3t 5

1

2t 4 ) dt

j

0

t = 1]

(6t 5

1

2t 5 ) dt

k

0

(4t 3

2t 4 ) dt

0

1

t5 t6 t6 2 j 4 i 3 5 0 6 6 9 2 7 i j k 10 3 5

1

k t4 0

t5 2 5

Example 1.3

1

0

F

(x 2

y2

x) i

xy y2 = x. W=

F

F dr C

[(x 2

y2

x) i

[(x 2

y2

x ) dx

(2 xy

y ) j ] (dx i

dy j

C

(2 xy

y ) dy ]

C

Case (i) C is the line y = x. [(x 2

W1 y (dy

y2

x) dx

(2 xy

y ) dy ]

x dx )

1

2 x 2 ) dx

( 0

2 3 Case (ii) C is the curve y2 = x. [(x 2

W2

y2

x) dx

(2 xy

y ) dy ]

2

x (dx

y 2 y dy )

1

(2 y 5

2 y3

y ) dy

0

2 3 Comment force. In fact, F is a conservative force, as F is irrotational.

dz k )

(2xy y ) j y = x, (ii)

Engineering Mathematics II

1.30

F 2 . 3

x2 = y F

Example 1.4 r

cos ti C,

zi

xj

sin t j

yk , when it moves

tk

from t = 0 to t = 2 .

curve as x = cos t, y = sin t, z = t. F

F dr C

(zi

xj

yk ) (dx i

dy j

dz k )

C

(z dx

x dy

y dz )

C 2

t ( sin t )

cos 2 t

sin t dt

0

t cos t (2 3

Example 1.5

sin t 1)

1 t 2

cos t 0

( 1)

F dr , where F

Evaluate

2

sin 2t 2

(sin y ) i

x (1

cos y ) j

zk and

C

C is the circle x2 + y2 = a2 in the xy Given integral

(sin y ) i

x(1

cos y ) j

zk

(dx i

dy j

dz k )

C

sin y dx x

2

y

2

z

a

cos y ) dy

z dz

0

sin y dx x2

x(1

2

y2

x(1

cos y ) dy

a2

Since C y = a sin circle C

C, namely x = a cos , as the variable of integration. To move around the varies from 0 to 2 .

Now, given integral

(sin y dx

x cos y dy )

d(x sin y )

x dy

x dy

2

d a cos 0

sin (a sin )

a 2 cos d

Chapter 1: Vector Calculus a2 2

a cos sin (a sin ) a

1.31

sin 2 2

2

0

2

Example 1.6

F

z3 )i

(2 xy

x2 j

3 xz 2 k ,

guess that the given force F is conservative. Let us verify whether F is conservative, i.e. irrotational. i F

j x

2 xy

k

y z

3

x

z

2

3 xz 2

(3 z 2

( 0 0) i 0

3z 2 ) j

(2 x

2 x) k

F is irrotational and hence conservative. F Since F is irrotational, let F . It is easily found that = x2 y + z3 x + c. (3 ,1,4 )

done by

F

F dr (1, 2 ,1) (3,1, 4)

dr (1, 2, 1) ( 3, 1, 4)

d (1, 2, 1) (3, 1, 4 ) 2, 1)

(x, y, z ) (1,

Example 1.7 j

2 x y zk

x2 y

z2 x

(201 202 .

c)

c

( 3, 1, 4) (1, 2, 1)

( 1 c) F

y (3x 2 y

z2 )i

x(2 x 2 y

z2 )

C. C and the terminal

Engineering Mathematics II

1.32 Since C x0, y0, z0

C is not given. Hence we guess that the given force F is i

F

j x

2

3y x

k

y

2

yz

( 2 xz 0

2

3

2x y

2 xz ) i

(

Since F is irrotational, let F F

z x

z 2 xyz

2 yz

2 yz ) j

2

(6 x 2 y

z2

6x2 y

z 2 )k

.

F dr C

dr C (x0 , y0 , z0 )

d (x0 , y0 , z0 )

(x0 , y0 , z0 )

(x0 , y0 , z0 )

0 12 x 2 yi

A dS , where A

Evaluate

Example 1.8

3 yz j

2 zk and S is

S

x+y+z= Given integral I =

A n dS, where n is the unit normal to the surface S given by z

= c, i.e. x + y + z = 1 x

y

i

j

n

C

1 3

z

(i

k j

B

O

k) A

Fig. 1.4

(12 x 2 yi

I

3 yzj

2 zk )

1

S

1 3

(12 x 2 y S

3 yz

2 z ) dS

3

(i

j

k ) dS

y x

Chapter 1: Vector Calculus

xoy -

S on the dA, where is the angle between the surface S and the

Then dS cos

xoy -

1.33

, i.e. the angle between n and k . dA

dS

n k

dx d y 1 3

n k 1

I=

cos

(12 x 2 y

3

3 yz

2z)

OAB

[ 12 x 2 y

3 y (1

x

dx dy 1 3 S

y)

OAB]

2(1

x

y ) dx dy

OAB

Note

x and y only, z

function of x and y 1 1

y

(12 x 2 y

I 0

S. 3y2

3 xy

5y

2x

2) dx dy

0

1

4 y (1

3y (1 2

y )3

0

3 y 2 (1 y )

y)2

5 y (1

y)

(1

y)2

2(1

y ) dy

49 120 Example 1.9

F dS , where F

Evaluate

yzi

zxj

xyk and S

S

x2 + y2 + z2 Given integral I

F n dS , where n is the unit normal to the surface S given by S

= c i.e. x2 + y2 + z2 = 1. = x2 + y2 + z2 2 xi n

2 yj

2 xi

2 zk

2 yj

4(x

2

y

2(xi

2 zk 2

z2 ) zk )

yj

[

x, y, z) lies on S]

4 1 xi I= S

yj

zk .

(yzi

zxj

xyk ) (xi

yj

zk ) dS

Engineering Mathematics II

1.34 3 xyz dS

z

S

3 xyz

dx dy , n k

R

where R is the region in the xy by the circle x2 + y2 =

B

O

y

A R

dx dy 3 xyz z

1

1

I=

x

y2

Fig. 1.5

3 xy dx dy 0

0

x2 3y 2 0

1

1

3 2

y2

dy 0

1

y 2 ) dy

y (1 0

3 y2 2 2

y4 4

1

0

3 8

Example 1.10

F dS if , F

Evaluate

yzi

2 y2 j

xz 2 k and S is the sur-

S

face of the cylinder x2 + y2 = 9 z = 0 and z = 2. F n dS , where n is the unit normal to the surface S given by Given integral I S

= c, i.e. x2 + y2 = 9

= x2 + y2 2 xi n

2 yj

2 xi 4(x 2(xi

2 yj 2

y2 ) yj )

4 9 1 (xi 3

yj )

[

x, y, z) lies on S]

Chapter 1: Vector Calculus I=

(yzi

2 y2 j

xz 2 k )

1 3

(xyz

2 y 3 ) dS

(xyz

2 y3 )

s

1 3

1 (xi 3

z

yj ) dS

C

B

s

R

1.35

dx dz n j

O A

where R xoz S on the xoz

x Fig. 1.6

1 I= 3

(xyz R

dx dz 2y ) y 3 3

2 y 2 ) dx dz

(xz R 2

3

[xz 0 2

0

x 2 )] dx dz

2(9

0

9 z 18 3 2 9 dz 2 2

9 z2 2 2

36 z

81 0

EXERCISE 1(c) Part A (Short Answer Questions) 2. 3. If F F dS is evaluated. S

6. Evaluate (2, 2).

r dr , where C is the line y = x in the xy C

F xi the curve 2y = x2 from (0, 0) to (2, 2). 8. Prove that the force F

(2 x

yz ) i

(xz

2 yj 3)j

xyk is conservative.

y

Engineering Mathematics II

1.36 9. Evaluate

(yzi

zxj

xyk ) dS , where S is the region bounded by x = 0,

S

x = a, y = 0, y = b and lying in the xoy a closed curve. Part B 11. Evaluate

dr , where

= 2xyz2 and C is the curve given by x = t2, y = 2t,

C

z = t3 from t = 0 to t = 1. 12. Evaluate F dr along the curve x = cos t, y = 2 sin t, z = cos t from t = 0 C

to t

2

, given that F

13. Evaluate

2 xi

yj

zk .

F dr along the curve x = cos , y = sin , z = 2 cos from = 0 C

to

2

, given that F

2 yi

zj

xk

F dr along the curve x = t2, y = 2t, z = t3 from t = 0 to t = 1,

14. Evaluate C

given that F

xyi

zj

x2 k . F

y2 j ,

3 xyi

along the curve y = 2x2 in the xy F

(y

3z ) i

(2 z

x)j

(3 x

x = a cos t, y = a sin t, z

2 y )k , 2at

a, 0, 0) and (0, a, a). 17.

F

(2 y

3) i

xzj

(yz

x)k when it

F dr , where C is the circle x2 + y2 = 4 in the xy

18. Evaluate C

F

(2 x

19.

y

z)i

(x

y

z 2 )j

(3 x F

2y

(x

2

4 z )k . 2

y )i

(x 2

z 2 )j

2

yk , when it

2

x +y =

(3 xz 2

2)k

F

(e x z

F

(y 2 cos x

2 xy ) i

(1 x 2 )j

z3 )i

(e x

(2 y sin x 2

z )k ,

4)j

, 1, 2 along any

Chapter 1: Vector Calculus y2 i C.

F F dS , where F

23. Evaluate

2(xy

x2 j

xyi

1.37

(x

z) j

2 yk , when it

z )k and S

S

x + 2y + z F dS , where F

24. Evaluate

yi

xj

4k and S

zi

xj

3 y 2 zk and S is the surface of the

S 2

x2 + y2 + z = a2 25. Evaluate

F dS , where F S

cylinder x2 + y2 = z = 5.

1.5

z = 0 and

INTEGRAL THEOREMS

3. Gauss Divergence theorem double integral.

s theorem. Gauss Divergence theorem

1.5.1

Green’s Theorem in a Plane

If C (P dx

Q dy )

C

R

R in the xy R, then Q P d x dy x y

P(x, y), Q (x, y)

where C

1.5.2

Stoke’s Theorem C and if F is a S, then

If S F dr C

curl F dS

,

S

where C S.

Engineering Mathematics II

1.38

1.5.3

Gauss Divergence Theorem

If S is a closed surface

V and if F is a vector V, then F dS

(div F ) dv .

S

1.5.4

V

Deduction of Green’s Theorem from Stoke’s Theorem F dr C

curl F dS

(1)

S

S curve C.

R in the xy F

F

Then curl

P(x, y ) i i

j

x P

y Q

Q x

Q(x, y ) j k z 0

P k y

[ P and Q are functions of x and y]

Inserting all these in (1), we get (Pi

Q j ) (dxi

dyj

Q x

dzk )

C

R

(P dx

i.e.

Q dy )

C

(Qx

P k dS y Py ) k n dS

(2)

R

Now n is the unit vector in the outward drawn normal direction to the surface R. R, that lies in the xy z z n , we get n k . Using this in (2), we get (P dx

Q dy )

C

(Qx

Py ) dx dy

R

[ dS =

xy

= dx dy]

Note

If the surface S

1.5.5 F

Scalar form of Stoke’s Theorem Pi

Qj

Rk

P, Q, R are functions of x, y, z.

Chapter 1: Vector Calculus Then

F dr F F dS

(Ry

P dx

Qz ) i

Q dy

(Pz

1.39 (1)

R dz

Rx ) j

(Qx

Py ) k

curl F n dS (Ry

Qz )(n i )dS

(Ry

Qz ) dy dz

(Pz (Pz

Rx )(n j )dS Rx ) dz dx

(Qx

(Qx

Py )(n k )dS

Py ) dx dy

(2)

[ dS cos = dx dy i.e. dS (n k ) = dx dy] s theorem, it reduces to the scalar form (P dx

Q dy

R dz )

[(Ry

C

Qz ) dy dz

(Pz

Rx ) dz dx

(Qx

Py ) dx dy ]

S

(3)

Note

P and Q as functions of x and y only and R = 0 in (3), we get

1.5.6

Scalar form of Gauss Divergence Theorem

F x, y, z.

Pi

Qj

Rk in Divergence theorem, where P, Q, R are functions of

Then div

P x

F F dS

Q y

R z

(1)

F n dS P (n i ) dS P dy dz

Q (n j ) dS

Q dz dx

R (n k ) dS

R dx dyy

(2)

Inserting (1) and (2) in Divergence theorem, we get (Pdy dz

Qdz dx

Rdx dy )

(Px

S

Qy

Rz ) dx dy dz

(3)

V

which is the scalar form of Divergence theorem.

Note 1.5.7

(3) is also called Green’s theorem in space.

Green’s Identities

In Then div (

F = zd} , where =

(

and

) 2

Divergence theorem becomes dS S

( V

2

) dV

(1)

Engineering Mathematics II

1.40 Interchanging

and ,

## }dz $ dS = ### (}d z + dz $ d})d V 2

(l)

S

V

(

) dS

(2)

(2) gives, S

Note

2

(

2

(3)

) dV

V

(1) or (2) is called Green’

. and (3) is called Green’

. WORKED EXAMPLE 1(d) (x 2

Example 1.1

y 2 ) dx +

C

2xy dy], where C is the boundary of the rectangle in the xoy lines x = 0, x = a, y = 0 and y = b. (OR) F rectangular region in the xoy F dr

(x y ) i 2 xy j in the x = 0, x = a, y = 0 and y = b. 2

2

curl F dS

C

S

i curl F

Now

x x

2

y

(2 y

2

j

k

y 2 xy

z 0

2 y) k

We have to verify that [(x 2

y2 )i

2 xy j ] (dxi

dy j

dzk )

4 yk n dS

C

S

[(x 2

i.e.

y 2 ) dx

2 xy dy ]

C

4 y dx dy [Fig. 1.7]

(1)

R

n y

y=b

D

B

x=0

x=a

O

y=0 Fig. 1.7

A

x

k

Chapter 1: Vector Calculus

1.41 [(x 2

L.S. of (1) OA y 0 dy 0

AB x a dx 0

b

x2d x

the boundary C consists of 4 lines] 0

(x 2

2ay d y

0

2 x y dy ]

DO x 0 dx 0

BD y b dy 0

[ a

y 2 ) dx

0

b2 ) d x

0

a

Note x or dy a

x3 3

a

x3 3

a (y 2 ) b0 0

b2 x 0

2 ab 2 b

a

R.S. of (1)

4 y dx d y 0 b

0

4 y (x)0a dy

2a (y 2 )

b 0

0

2 a b2 . [(3x 2

Example 1.2 Verify Green’s

8y 2 ) d x

(4y

6 xy )

C

dy ] , where C

x = 0, y = 0 and x + y = 1.

Green’s theorem is

Q dy ) =

C

For the given integral, [3x 2

(P dx

8y 2 ] d x

(4y

(Qx

Py ) dx dy

R

6 x y ) dy ] =

C

10 y d x dy [Fig. 1.8] R

y B x+y=1

x=0

O

y=0

A

x

Fig. 1.8

[(3x 2

L.S. of (1) OA y 0 dy 0

AB x y 1 x 1 y dx dy

BO x 0 dx 0

8y 2 )d x (4y

6 x y ) dy ]

(1)

Engineering Mathematics II

1.42 1

1

0

2

3x d x

[{3(1 y )

0 1

2

2

8y }( dy ) + {4y 6 y (1 y )}dy ]+

0 1 2

1 2

3x d x

(11y + 4y

0

4 y dy 1

3) dy

0

4 y dy

1

0

11 3

2 3

2

5 3 1 1 y

R.S. of (1)

10 y dx dy 0 1

0

10 y (1 y) dy 0

5 y2

10

y3 3

1

0

5 3

Since L.S. of (1) = R.S. of (1),

. F dr , where F

Example 1.3

(sin x

y)

C

i

cos xj and C is the boundary of the triangle whose vertices are (0, 0),

and 2

2

,0

,1 .

Note

F dr

Evaluating C

integral. F dr

curl F dS , where S

C

S

surface bounded by C. [Fig. 1.9] S

R in the xoy

bounded by C. F dr C

curl F k dx dy [

For the xoy lane, n

R

i curl F

j

x (sin x y ) (sin x

1) k

y cos x

k z 0

k and dS

d x dy ]

Chapter 1: Vector Calculus he given line integral =

1.43 y

(1 + sin x) d x d y

B

R

y=

1

2

0

y 2

2

x

(1 + sin x) d x dy

x= O

1

x

cos x

2

y 2

y dy cos 2

y2 4

2

0 1

0

2

2 y 2

y 2

4

,1

2

2 x

y=0

A

2

dy

,0

Fig. 1.9

sin

y 2

1

0

. [(2x

Example 1.4

y ) dx

C

(x + y ) dy ] , where C is the boundary of the circle x2 + y2 = a2 in the xoy (P d x

Q dy ) =

(Qx

C

[(2x

y) d x

y

Py ) d x dy

R

(x + y ) dy ]

[1 ( 1)] d x dy

C

x O

R

2

d x dy [Fig. 1.10] R

2

Fig. 1.10

area of the region R

2 a2 Example 1.5

F

(2 x y

x 2 ) i (x 2 y 2 ) j and C y = x and x2 = y. 2

F dr

curl F n d S

C

S

i Now curl F

j x

2 xy (

2x 4 xk

k

y x2 2 x)k

x2

y2

z 0

Engineering Mathematics II

1.44

[(2 x y x 2 ) d x (x 2

y 2 ) dy ]

4 x k k dx dy

C

R

4 x d x dy [Fig. 1.11]

(1)

R

B(1, I)

D

y = x2

A O x = y2

Fig. 1.11

L.S. of (1)

[(2 x y OAB y x2 dy 2 x d x

x 2 ) dx

(x 2

y 2 ) dy ]

BDO x y2 d x 2 y dy

1

0

(2 x

5

2

(3 y 4

x )dx

0

2 y5

y 2 ) dy

1

y = x2 and

[ the coordinates of B x = y2] 3 5 1

y

0 y

2

R.S. of (1)

4x dx d y 1

(x 2 )

2

y y2

dy

0 1

(y

2

y 4 ) dy

0

3 5 S Example 1.6 1 (x dy y d x) 2 C

C is given by y2 = 4a x and its x2

y2

a2

b2

1.

Chapter 1: Vector Calculus (P dx

By Green’s theorem,

Q dy )

(Qx

C

P

Py )dx dy

R

y and Q 2

x , we get 2

1 2

y dx)

(x dy

1.45

C

R

1 2

1 dx dy 2

dx dy R

=Area of the region R enclosed by C. 1 2

rectum

(x dy

y dx) [Fig. 1.12]

C

y (a, 2a)

L

1 2

(x dy LOL

y dx )

x=a

L SL

1 2

(x dy

2a

1 2

2a

2a

y dx )

L'

x a dx 0

(a

a)

Fig. 1.12 2a

y2 dy 4a

a dy 2a

2a

y2 dyy 4a

0

(x dy

y dx )

y2 x 4a y dx dy 2a

x

S

O

a dy 0

8 2 a . 3 y

1 2 1 2 1 2

=

(x dy

y dx)

=

2 =0 x =2

=

C

O

(x dy x a cos y b sin

y dx)

[Fig. 1.13] =

2

ab (cos 2 0

.

sin 2 ) d

Fig. 1.13

3 2

Engineering Mathematics II

1.46 Example 1.7 (curl F )

) = 0 and (ii) div

0. curl F dS

F dr

S

F

grad

S2

S1

C

S3

, we have

curl (grad ) dS

grad

S

dr C

C

d

Fig. 1.14

C

=0 S C. [Fig. 1.14] curl (grad ) dS 0, [for any S and hence for any dS ] curl (grad ) =0 (ii) Gauss divergence theorem is F dS , where S is a closed surface enclosing a volume V.

(div F ) dv V

S

F by curl F , we have div (curl F ) dv

curl F dS

V

(1)

S

. S1 and S2 closed curve C as given in Fig. 1.15. Now (1) becomes div (curl F ) dv curl F dS V

. However .S. of (1), as S is a closed surface. , we divide S S1 and S2 by , each of which is bounded by the same

S1

S1

curl F dS

C

S2

F dr

F dr

C

S2

C

Fig. 1.15

Note

If C of the outer normal to S1 S2. div (curl F ) dv V

F dr C

= 0. Since V is arbitrary, div (curl F )

0.

F dr C

Chapter 1: Vector Calculus Example 1.8 Evaluate

(sin z dx

cos x dy

1.47

sin y dz )

C

where C

x (P dx

Q dy

y R dz )

C

dy dz

(Pz Rx ) dz dx P = sin z, Q

(sin z dx

cos x dy

[(Ry

Qz )

S

(Qx Py ) dx dy ] x, R = sin y, we get sin y dz )

C

(cos y dy dz

cos z dz dx

sin x dx dy )

S

sin x dx dy

[

S

S is the rectangle in the z and hence dz = 0]

1

sin x dx dy 0

0

(cos x)0

=2 (x dy dz

Example 1.9 Evaluate

2 y dz dx

3 z dx dy ) where S is the closed

S

x2 + y2 + z2 = a2. Scalar from of divergence theorem is (P dy dz

Q dz dx

R dx dy )

(Px

S

Qy

Rz ) dV

V

P = x, Q = 2y, R = 3z, the given surface integral =

6 dv

6V

V

6

4 3 a 3

= 8 a3. F

Example 1.10

xyi

2 yz j

0, y = 2 and z = 3 above the xoy F dr C

Here

curl F

curl F dS S

i

j

k

x xy

y 2 yz

z xz

zxk where S is the x = 0, x = 1, y =

Engineering Mathematics II

1.48 2 yi

zj

(xy dx

xk

2 yz dy

zx dz ) =

(2 yi

C

zj

(1)

xk ) dS

S

z C

A'

B'

O' ˆn = j

nˆ

j O

y

B

A C'

x Fig. 1.16

S x = 0, x = 1, y = 0, y = 2 and z = 3 and is bounded by the rectangle OAC B lying on the xoy L.S. of (1)

=

(xy dx

2 yz dy

zx dz )

OAC B

xy dx [Fig. 1.17]

( the boundary C

z = 0)

y

OAC B

(xy dx) OA y 0 dy 0

AC x 1 dx 0

C'

B

BO x 0 dx 0

CB y 2 dy 0

0

2 x dx

O

Fig. 1.17

=

. R.S. of (1) =

## + ## + ## + ## + ## (2yir + zjr - xkr ) $ nt ds

`ntx==-0ij

Note

x

A

1

1 ` xnt = = ij

y=0 ct m n = - rj

y=2 ct rm n=j

z=3 ct rm n=k

nt y nt

y Similarly nt for y Using the relevant value of nt R. S. of (1)

j.

j and so on. , we have

2 y dS x 0

y

2 y dS x 1

z dS y 0

z dS y 2

x dS z 3

Chapter 1: Vector Calculus 3

2

3

2

1

2 y dy dz 0

3

1

2 y dy dz

0

0

3

z dz dx

0

0

[ dz etc.]

1.49

0

0

1

0

0

2

x2 dy 2 0

x dx dy

1

0

0

x dx dy

0

S on x = 0 and x 2

2

z dz dx ,

y

( the other integrals cancel themselves)

1

0

1. . F

Example 1.11 z

y 2 zi z 2 xj x 2 yk where S is the x = ±a, y = ±a and z = ±a, in which

a is cut. F dr

curl F dS

C

curl F

Here

i

j

k

x y2 z

y z2 x

z x2 y

(x 2 (y 2 z dx

z 2 x dy

x 2 y dz )

2 zx) i [(x 2

C

S

(y 2

(z 2

2 xy )j (y 2

2 zx) i

2 xy )j

2 yz )k (z 2

(1)

2 yz )k ] dS

S

F

E

G

H

z 0 D

x

y C

A

B Fig. 1.18

S

z

1.18) L.S. of (1)

(y 2 z dx (z

z 2 x dy

a (Fig.

x 2 y dz )

a)

( ABCD

ay 2 dx

a 2 x dy )

[ dz = 0, as z

a]

Engineering Mathematics II

1.50

y=a dy = 0

C

B

x=a dx = 0

x a dx = 0

D

A

y a dy = 0 Fig. 1.19

(

L.S. of (1) AB

BC

CD

a

ay 2 dx

a 2 x dy ) Fig.1.19

a

a

DA

a 3

3

a dy

3

a dx

a

a 3 dx

a dy

a

a

a

4

4a . R.S. of (1) x n

a i

x n

a i

y n

[(x 2 z n

2 zx) i

x 2 ) dS

a

x

(y 2 y

a j

(y 2

2 xy ) j

(x 2

2 zx) dS

a

(z 2 z

a 2 ) dy dz

2az (

2ax

(a 2

2 yz ) k ] ndS

a 2 ) dz dx

a

a a

(a 2

a

a

a

(a 2 a

2az ) dy dz

4ax dz dx a

2ay ) dx dy

a

= 0 + 0 + 4a4 = 4a4. .

a

2 yz ) dS

2ay ) dx dy

a

(2 xy

a

(a 2

4a z dy dz a

(z 2

y

2 xy ) dS

a

(

a

y n

a k

(2 zx x

a j

2ax) dz dx

y 2 ) dS

Chapter 1: Vector Calculus

1.51

F yi 2 yz j y 2 k , where S is the x2 + y2 + z2 = a2 and C is the circular boundary on the xoy-

Example 1.12

j

k

y 2 yz

z y2

i F

x y k.

F dr

F dS

C

Here

S

(

ydx

2 yz dy

y 2 dz )

k dS

C

(1)

S

x2 + y2 = a2

C is the circle in the xoyx = a cos and y = a sin . L.S. of (1) =

( C lies on z = 0)

y dx x2

y2

a2

2

a 2 sin 2

d

0

a2 2

sin 2 2

2

0

2

.

=

k n dS , where

R.S. of (1) = S

n

, where 2 (xi

yj

4 (x

2

xi

yj a

y

= x2

y2

zk ) 2

zk

z2 ) [

S

z dS a

R

z dx dy , where R a n k .

R.S. of (1)

z2

x, y, z) lies on

= a2]

S on the xoy-

dx dy , where R is the region enclosed by x2 + y2 = a2. =

R 2

.

Engineering Mathematics II

1.52

x2 + y2 + z2 = 1, evaluate

Example 1.13 If S is (xi

2 yj

3 zk ) dS .

S

By Divergence theorem, F dS

(div F ) dv

S

V

F dS S

x

V

6

(x)

y

(2 y )

z

(3 z ) dV

dV V

= 6V, where V is the volume enclosed by S. 4 6 3 =8 . 2 y 2 j z 2 k , where Example 1.14 Verify Gauss divergence theorem for F x i S is x = 0, x=a, y=0, y=b, z = 0 and z = c. Divergence theorem is (1) F dS (div F ) dV S

V

S

. L.S. of (1) x n

0

x n

i

a i

y n

2

z n

0 k

2

y j

z k ) n dS x 2 dS

0

x

a

z 2 dS z

b j

c k

x 2 dS x

y n

j 2

(x i z n

0

0

y 2 dS y

y 2 dS

0

y

b

z 2 dS (on using the relevant values z

c

of n ) c

b

a2

a

b2

dy dz 0

c

0

dz dx 0

0

= abc (a + b + c) (2 x

R.S. of (1)

2y

2 z ) dx dy dz

V c

b

a

0

0

0

(2 x

2y

2 z ) dx dy dz

b

a

0

0

c2

dx dy

Chapter 1: Vector Calculus c

1.53

b

(x 2 0

2 z x)0a dy d z

2y x

0

c

(a 2 y

ay 2

2 a z y )b0 d z

(a 2 b

ab 2

2ab z ) dz

0 c

0

a 2 bc

ab 2 c

abc (a

b

abc 2

c)

. x2 i

Example 1.15 Verify divergence theorem for F

zj

yz k over the cube

formed by x = ±1, Divergence theorem is

y = ±1, z = ±1.

F dS

(1)

(div F ) dV

S

V

L.S. of (1) = x n

1 i

x 1 n i

(x 2 i z n

y n

1 j

zj

y 1 n j

x

1

z

1

yz k ) n dS

x 2 dS x

1

z

1

yz dS

yz dS

R.S. of (1) =

(2 x

y ) d x dy dz

v 1

1

(2 x 1

1

y ) dx dy d z

1

1

( 2 y dy d z ) 1

1

0 .

z dS

z dS y

0

1

1 k

1 k

x 2 dS

1

z n

1

y

1

(using the relevant values of n)

Engineering Mathematics II

1.54

EXERCISE 1(d) Part A (Short Answer Questions) 1. State Green’s theorem in (or) the connection between a line integral and a double integral. s theorem (or) the connection between a line integral and a surface integral. 3. State Gauss divergence theorem (or) the connection between a surface integral and a volume integral. s theorem. s theorem. 6. Give the scalar form of divergence theorem. 7. Derive Green’s identities from divergence theorem. s = 0. ( F) 0 . 10. Evaluate (yz dx zx dy xy dz ) where C is the circle given by x2 + y2 + C

z2 = 1 and z = 0. 11. Evaluate

(x dy d z

y dz dx

z d x dy )

S

x2+ y2 + z2 = a2. 12. If C

r

xi

yj

r dr

zk

0.

C

13. If C 0.

dr C

14. If S is a closed surface and r

xi

yj

zk , S

1 r

dS .

(r 2 ) dS .

15. If S is a closed surface enclosing a volume V, evaluate S

16. Evaluate

[(x

2 y )d x

(3 x

y )dy ] , where C is the boundary of a unit

C

(x dy

17. Evaluate

y d x) , where C is the circle x2 + y2 = a2.

C

18. If A

curl F

A dS

0 , where S is any closed surface.

S

19. If S is any closed surface enclosing a volume V and if A A dS

(a + b + c) V .

S

20. If r

xi r dS .

S

yj

zk and S

a xi

by j

c zk ,

Chapter 1: Vector Calculus

1.55

Part B (x 2 dx

xy dy ) , where C

C

is the boundary of the

x = 0, y = 0, x = a, y = a. F

2

x i

xyj

xy-

x = 0, y = 0, x = 2, y = 2. x 2 (1

y ) dx

(x 3

y 3 ) dy

C

where C

x = ± 1 and y = ± 1. s

F dr , when F

(xy

x2 ) i

C

x 2 y j and C is the boundary of the triangle in the xoy y = 0, and y = x.

x = 1, (2 x 2

y 2 ) dx

C

(x 2 y 2 ) dy , where C is the boundary of the region in the xoyenclosed by the x x2 + y2 = 1. s theorem for F (xy y 2 ) i x 2 j in the region in the xoy2 y = x and y = x . y2 = 4ax and x2 = 4ay. xoy bounded by y3 = x2 and y = x. s theorem for F

(y

z

2) i (yz 4) j xzk , where S is x = 0, x = 2, y = 0, y = 2 and z = 2.

(x 2 y 2 ) i 2 xyj xyzk over the surface x = 0, x = a, y = 0, y = b and z = c.

30.

s theorem for F

31.

s theorem for F (2 x y ) i yz 2 j y 2 zk where S is the x2 + y2 + z2 = 1 and C is the circular boundary in the

xoy32. Verify Gauss divergence theorem for

(x 2

F

yz ) i

(y 2

zx) j

(z 2 xy ) k by x = 0, x = 1, y = 0, y = 2, z = 0 and z = 3. F

33. Verify divergence theorem for (i) F

(2 x

z)i

x 2 yj

4 xzi

y2 j

yzk

and (ii)

xz 2 k , when S is the closed surface of the cube

formed by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. (yz 2 i

34. Use divergence theorem to evaluate S

S is the closed surface bounded by the xoy x2 + y2 + z2 = a2

.

zx 2 j

2 z 2 k ) dS , where

Engineering Mathematics II

1.56

2 y2 j

z 2 k ) dS , where S is

the closed surface bounded by the cylinder x2 + y2 = and z = 3.

z=0

35. Use divergence theorem to evaluate

(4 xi s

ANSWERS

Exercise 1(a) (5) (8)

12 i 1 19

(i

(11) 25; (14)

9j 3j

56 i

1 ( 3

-1 (17) cos

(6)

2j

47 k

1 3

1 (2 i 3 11 (13) 3 (10)

k)

-1 (16) cos

11 8

(18) cos -1

3 22

j

27

(15)

13

(i

14 3

(12)

2k )

(7) 5

14

(9)

3k ) 30 j

i

-1 (19) cos

3 21

1

-1 (20) cos

(23) a

16k

(21) 2x

30 7 ;b 3

64 9

(24) x2y

y

z+1=0

xz2 + y2z + c

2j

45 2299 3 7 6 5 ; 2

(22) x 2 yz 3

(25)

20

Exercise 1(b) (7) 3; 0.

(8) 4

(9) 2( i a=4

(17)

1, b= l, c

(23) 80; 80 i

37 j

3k 36k ; 27 i

(22)

54 j

20k ; 0; 74 i

124 21

27 j

(25) x3y2z4 (26) x2 + y2 + 3xy + yz + z2x (27) a = 4, b = 2,

1;

x2 2

3y2 2

z2

2 xy

yz

4 zx

(28) x2yz3

Exercise 1(c) (6) 3

(7) 6

(9)

a 2b2 2

j

k)

k)

1

Chapter 1: Vector Calculus

(10) 0

(11)

2

(13)

4

i

7 6

(15) (18) 8 (21) 4

5 2

(24)

8 i 11

1 j 2

4 j 5

k

1.57 (12) (14)

i 51 70

(16) 2a2

(17) 8

(19) 2

(20)

(22) 0

(23) 27 4

19 2

(25) 90

Exercise 1(d) (10) 0 (16) 5

3

(11) 4 (17) 2

(23)

8 3

(24)

(28)

1 10

(34)

2

1 12 a4

(15) 6V (20) 4 (27) 16 a 2 3

1 j 2

3 k 2

Chapter

2

Ordinary Differential Equations 2.1

EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREE

nth dy dx

n

dy f1 x, y dx fn

If we denote

dy dx

1

n 1

x, y

f2 dy dx

dy x, y dx

f n x, y

n 2

0.

p

pn + f1 (x, y) pn–1 + f2 (x, y) pn–2 + …+ fn–1 (x, y) p + fn(x, y) = 0

(1)

p. y. x.

2.1.1 Type 1—Equations solvable for p n linear factors.

2.2

Engineering Mathematics II

(p – F1) (p – F2) . . . (p – Fn) F1, F2, … , Fn are functions of x and y. Each of these n

p = F1, p = F2 , … , p = Fn, where

n c) = 0,…,

1

(x, y, c) = 0,

2

(x, y,

(x, y, c) = n 1

(x, y, c)

2

(x, y, c) …

n

(x, y, c) = 0.

2.1.2 Type 2—Equations solvable for y y x and p. y = f(x, p)

(1)

x, p

x, p ,

dp dx

(2)

x and p. (x, p, c) = 0 ... (3), where c

p

If p p.

2.1.3 Type 3—Equations solvable for x x y and p. x = f (y, p)

(1)

y, 1 p

dp dy

(2)

(y, p, c) = 0

(3)

y , p,

y and p.

where c p If p p.

Chapter 2: Ordinary Differential Equations

Note

2.3 y = f(x, p) and x

= f(y, p).

y = px + f(p)

y = px+ f(p)

(1)

w. r. t. x, p dp dx

0

p

x

f

p f

(2) or

dp dx

p

x

(3)

0

p=c

(4) y = cx + f(c). p c

p

p the singular solution of the equation (1).

Note x

df dp

p

y = px + f(p) and c

0

y = cx + f(c) and x

df dc

0 y = cx + f(c)

WORKED EXAMPLE 2(a)

Example 2.1 Solve the equation

dy dx

2

p2 – 8p + The equation is (p – 3) (p – 5) = 0 dy 3 or dx

8

dy dx

15

0. p.

dy dx

5

2.4

Engineering Mathematics II y = 3x + c and y = 5x + c.

y – 3x – c = 0 and y – 5x – c = (y – 3x – c) (y – 5x – c) = 0. Example 2.2 Solve the equation p (p + y) = x (x + y). p2 + yp – (x2 + xy) = 0. y

y2

p, p

4 x2

xy

2 y

y

2x

2

2 dy dx dy dx

i.e. or

x

(1) x

y

(2)

x2 c . 2 2 2y – x2 – c = 0 dy y x, (2), dx ex y

i.e.

i.e.

(3) y.

y ex = – x ex dx + c = – x ex + ex + c –x y+x–1–ce =0

(4)

(2y – x2 – c) (y + x – 1 – ce–x) = 0. Example 2.3 Solve the equation p2 – 2py tan x – y2. p2 – 2py tan x – y2 = 0. p, p

2 y tan x

4 y 2 tan 2 x 2

2 y tan x

4 y2 1

y tan x

i.e.

i.e.

4 y2

tan 2 x

(1)

2 sec x

dy dy = y (tan x + sec x) or = y (tan x – sec x). dx dx dy dy = (tan x + sec x) dx (1) and = (tan x – sec x) dx (2) y y y x x + tan x) + y = c sec x (sec x + tan x)

c

Chapter 2: Ordinary Differential Equations

2.5

c (1 sin x) cos 2 x c 1 sin x y (1 – sin x) – c = 0

i.e. y=

x

i.e.

(3)

x + tan x) +

c

c sec x sec x tan x

y

c 1 sin x y (l + sin x) – c = 0

i.e.

(4) y (1 – sin x) – c y (1 +

sin x) – c] = 0. Example 2.4 Solve the equation xp2 – 2py + x = 0. p, p

2y

4 y2 2x

y x

y x

dy dx

i.e.

4 x2 2

1

y = vx, dv v x v v2 dx dv dx 2 x 1 v

i.e.

v2

v

(1)

1 (2) x

1

c

Solutions are y

y2

x2

x2 i.e.

y

y2 y

x2 y2

y

y2

0 and

y

c and cx 2 x2

cx 2

y

y2

x2

c

y2

x2

x2

c

c 0.

Example 2.5 Solve the equation p3 – (x2 + xy + y2) p2 + (x3y + xy3 + x2y2) p – x3y3 = 0.

0

2.6

where

Engineering Mathematics II

2

,

p3 – ( + + ) p2 + ( 2 and .

)p–

+

0,

)=0

( )( )( (p – x2) (p – xy) (p – y2) = 0.

i.e.

dy dx y

i.e.

+

3y

x3 3

x2 ,

c ;l 3

x3

dy dx

x2 2

y

2

ce x

c y

ce 2

/2

1 y

x

c.

0; x

1 y

c

1 y

c

0.

x2

x3

3y

y2

lo c

0; y

c

dy dx

xy and

x

0.

Example 2.6 Solve the equation p2x – 2py – x – 0. p2 x x 2p

y

1 x px 2 p y.

(1)

x

p

i.e.

2p p3

i.e. i.e. i.e.

p

x

dp 2 p dx

p

dp x dx

1 p 2

x

dp dx

xp 2

dp dx

x

dp dx

dp dx

p2

p

x

dp dx

p2 p p p2

1

x

p

x

dp dx 1

0 (

p2

1

0)

Chapter 2: Ordinary Differential Equations dp p

dx x

2.7

lo c

p = cx

i.e.

(2) c2x2 = 2cy + 1.

Example 2.7 Solve the equation 16x2 + 2p2y – p3x = 0. p, nor for x.

y

equation is rewritten as p 3 x 16 x 2 2 p2

y x, p

2

p3

3 p2 x

2p

dp dx

(1) p3 x

32 x

16 x 2 2 p

dp dx

32 px 2

dp dx

p4

i.e.

2 p5

p5

3 p4 x

i.e.

p5

32 p 2 x

i.e.

p 2 p3

dp dx

2 p4 x

dp dx

32 px 2

dp dx

px p 3

32 x

dp dx

dp dx

0

p4 x

32 x

32 p 2 x

p2

i.e. or

px

dp dx

0 (2)

p3 + 32x=0

(3)

p = cx

(4)

(2) is differential equation in p p p dp p

(2)

dx x

p 16x2 + 2c2x2y – c3x4 = 0 i.e. c3x2 – 2c2y equation, 16x2 + 2p2y + 32x2 = 0 i.e. p2y = – 24x2 i.e. p6y3 = – (24)3 x6 i.e. 1024x2y3 + (24)3 x6 = 0 i.e. 16y3 + 9x4

2.8

Engineering Mathematics II

Example 2.8 Solve the equation y = (1 + p) x + p2. y dp dp 1 p x 2p dx dx dp x 2p 1 0 dx dx x 2p dp

p i.e. i.e.

x, (1)

(2)

This is a linear equation in x. xe p

2 pe p dp p

c

p

= – 2 (pe – e ) + c x =2 – 2p + ce–p p

i.e.

(3)

x = 2 – 2p + ce–p and y = (1 + p) (2 – 2p + ce – p) + p2 x = 2 – 2p + ce–p and y = 2–p2 + c (1 + p) e–p.

i.e.

Example 2.9 Solve the equation y = x + p2 – 2p. y and for x. We shall solve the y = x + p2 – 2p . . . x,

Method I:

i.e.

p

1

2 p

1

y.

2p dp dx

2 p

dp dx 1

(1) 0

p–1=0 or

2

dp dx

(2) (3)

1

p

y = x – 1. 2p = x + c

p

y

(4)

x

x

c

2

x

4

i.e. 4 (y + c) = (x + c)2, Method II:

c

x x = y + 2p – p2

(1)'

y, dx dy

1

2

dp dy

2p

dp dy

Chapter 2: Ordinary Differential Equations

i.e.

1 p

1

i.e.

(1

p) 2 p

2 (1

p)

dp dy

2.9

d d 0.

1

p=1 dp 2p dy

or

(5)

1

0

(6)

p p2 = y + c

(7)

x –y + y + c = 2p (x + c)2 = 4 ( y + c) Example 2.10 Solve the equation p2x + py – y4 = 0.

Note

p y x. y4

x

py

(1)

p2

y, p2 4 y3

dx dy

p

y

p 4 y3

1 p

i.e.

p2

i.e.

y 2 y3

p

y

dp dy

4 py 3 p

p2 dp dy

or

dp dy

2 y4

2 py

dp dy

2 y4 2 p 2 y3

py p

dp dy 0

dp 2p 0 dy 2y3 – p = 0

(2) (3)

2dy y

p i.e.

py 2 p

p3

y

p

y4

p4

i.e.

(2) is dp

dp dy

y

c p = cy2

(4)

2.10

Engineering Mathematics II p c2xy4 + cy3 – y4 = 0 2

i.e. c xy + c = y p 4xy6 + y4 = 0 i.e. 4xy2 Example 2.11 Solve the equation p3 – 2 x yp + 4y2 = 0. p nor for y. p2 y

2x

y, p2 y2

2 p i.e. i.e. i.e.

2p y

4 y dp p 2 dy

2 p3 p2 y p3

2 y2

2 y2 2y

2 p dp y dy 2 p

dp dy

p

0

dp dy

p

2dp p

dy y

lo y

c

(1)

4 y dp p 2 dy

0

1 p3 py 2

or

p

4 p

dp dy

2y

i.e. 2 l

p2 y2

4y p

2 y2

0

0

(2)

p3 – 2y2 = 0

(3)

lo c

i.e.

p2 = cy

i.e.

p cpy –2xyp + 4y2 = 0 p (c –2x) = – 4y cy (c –2x)2 = 16y2

i.e. c (c – 2x)2 = 16y, y2 – 2 x yp + 4y2 = 0 i.e.

i.e.

2x3y2 = 27y3 2x3 = 27y,

xp = 3y

(4)

Chapter 2: Ordinary Differential Equations

2.11

Example 2.12 Solve the equaion p3x – p2y –1= 0. x. y y

1 p2

px

y

1 c2

cx

(1)

c, x

0

2 c3

(2)

2 x

(3)

c c3 c2y = c3x – 1 c6y3 = –27 4y3 = –27x2

i.e.

Example 2.13 Solve the equation y = 2px + yp2. x y2 = Y dy dY 2y = dx dx i.e.

2yp = P, say. y, y2 = 2ypx + y2p2

(1)

Px

P2 , 4

(2)

Y

cx

c2 4

(3)

cx

c2 . 4 c 2

(4)

Y General solution of (2) is

2 General solution of (1) is y

c, 0 c

x

Y = – x2

2.12

Engineering Mathematics II

i.e. x2 + y2 Example 2.14 Solve the equation (px – y) (py + x) = 2p. Put X = x2 and Y = y2 dX = 2 x dx and dY = 2y dy y dY y dy i.e. P x p (say) dX x dx x P y

p

or

x2 P i.e.

PX

i.e.

Y

Y PX

y2

2P P 1 2P P 1

x P 1 y

2

x P y

2c c 1 Example 2.15 Solve the equation e4x (p – 1) + e2y p2 = 0 Note I eax and eby ky Y = e where k a and b X = e2x and Y= e2y dY 2e 2 y dy dX 2e 2 x dx dY X i.e. p P, where P = . Y dX 2 General solution is y

cx 2

X2

X P 1 Y

Y

X2 2 P Y2

X = ekx and

0

i.e. XP – Y + P2 = 0 i.e. Y = PX + P2 e2y = ce2x + c2.

EXERCISE 2(a) Part A (Short Answer Questions) p – f1 (x, y)] p – f2 (x, y)]=0. y = f (x, p), x = f (y, p), solution.

y. x.

Chapter 2: Ordinary Differential Equations

2.13

y = px + f (p)? 7. 8. 9. 10. 11.

Solve the equation p2 – 5p + 6 = 0. Solve the equation p2 – (x + y) p + x y = 0. Solve the equation p2 – (ex + e–y) p + ex–y = 0. Solve the equation x yp2 – (x + y) p + 1 =0. Rewrite the equation (y – px) (p – 1) = p

12. Rewrite the equation p

px – y)

13. Rewrite the equation p = sin (y – px) y

px

1 . p

y = px – p2. Part B 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

yp2 + (x – y) p – x – 0 2p2 – (x + 2y2) p + xy2 = 0. xyp2 + (3x2 – 2y2) p – 6xy = 0 xyp2 –(x2 + y2) p + xy = 0. x y 1 p p y x p2 + 2py cot x = y2. x2p2 – 2xyp + (2y2 – x2) = 0. y = – px + x4p2. 4y = x2 + p2. y = 2px + pn. y = 2px – p2. y = x + 2 tan–1 p y = (1+ p) x + ep. y = 3x + p p3 – 4xyp + 8y2 = 0. x = y + p2. y = 2px + 4yp2 y = 2px + y2p3. y2 y = xyp + p2.

y and (ii) x.

y = (x – 1) p + tan–1 p. y p = px.

37. 38. 39. 40.

x2 (y – px) = yp2 X = x2, Y = y2] 2 y = 2px + p y X = 2x, Y = y2] 2 2 (y + px) = px Y = xy] (p – 1) e3x + p3e2y = 0 X = ex, Y = ey]

2.14

Engineering Mathematics II

2.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS 2.2.1

Introduction

nth order with constant

a0

dn y dx n

a1

dn 1 y dx n 1

an

1

dy dx

an y

X,

(1)

where a0 ( 0), a1, a2, . . . , an are constants and X is a function of x. D

d , D2 dx

d2 , dx 2

(a0Dn + a1 Dn–1 + ...+ an–l D + an) y = X or f (D) y = X, where f (D quantity. When X

Dn

dn , dx n (2)

D,

f (D) y = 0

(3)

General Solution of equation (2) is y = u + v, where y = u (3), that contains n y=v u is called the complementary function (C.F.) and v is called the particular integral (P.I.) of the solution of Equation (2).

2.3

COMPLEMENTARY FUNCTION f (D) y = 0 or (a0Dn + a1Dn–1 + . . .+ an) y = 0 auxiliary equation (A.E.) f (m) = 0

or a0mn + a1mn–1 + . . . + an = 0 D m nth

equation in m.

(3) (4)

Chapter 2: Ordinary Differential Equations

Case (i) The roots of the A.E. are real and distinct. m1, m2, …, mn. mx Then the solution of Equation (3) is y c1e 1 c1, c2, …, cn c1e m1 x

u

2.15

cn e mn x , where

c2 e m2 x

cn e mn x

c2 e m2 x

Case (ii) m1, m1, m3, m4, …, mn Then the solution of Equation (3) is y

c1 x

c2 e m1 x

c3 e m3 x

cn e mn x

If three roots of the A.E. are equal, i.e. if ml = m2 = m3 (say), then the solution is c1 x 2

y

c2 x

c3 e

m1 x

c4 e m4 x

cn e mn x

r y

(c1 x r

1

c2 x r

2

cr

1

x

cr ) e m1 x

cr 1e mr

1x

cn e mn x

Case (iii) m1 = + and m2 – . Then the solution of Equation (3) is y = e ax (c1 cos bx + c2 sin bx) + c3 e m3x + g + cn e mnx . Case (iv) i.e. m l = m3 = + and m2 = m4 = – . Then the solution of Equation (3) is y = e ax 7c1 x + c2h cos bx + ^c3 x + c4h sin bx A + c5 e m5x + g + cn e mn x .

2.3.1

Particular Integral f (D) y = X

(2)

is the function v, where y = v X and .I. i.e. f (D)

1 1 X , where f (D) f (D)

1 X f (D)

f (D).

X. X is equal

Rule I X= eax, where

is a constant. P.I. =

1 e ax = 1 e ax , f]ag ! 0 f]ag f ] Dg

2.16

Engineering Mathematics II

When f( ) = 0, ( )r is a factor of f(D). r f(D) = ( ) (D), where ( 1 Then e ax P.I. = r ^ D − ah z ]Dg 1 1 e ax F = < z^ah ^ D − ahr 1 $ x r e ax $ = z^ah r! 1 e ax x = e ax and D a 1! 2 1 ax = x e ax . 2e 2! ^ D ah Rule 2 X = sin ax or cos ax, where are used. 1 1 sin ax sin ax 2 (D ) ( a2 ) 1 (D ) When

(–

2

) = 0, (D2 +

(D2) = (D2

2

1

cos ax

2

( a2 )

2

1 2

(D )

(–a2)

0. 1

sin ax

0.

(D2)

) is a factor of

(D2), where

)

cos ax, if ( a 2 )

2

( D )( D 2

2

1

)

sin ax.

1 2

( a ) D

2

2

sin ax

1 cos ax 1 ; 1 cos ax E = z ] D 2g }^ a2h D 2 + a2 D2

1 +

sin

2

=–2

x

cos

x = 2 1 D2 +

and

2

cos

.

=

x sin 2

=

x 2

.

Note

f(D) D2

(D2). – 2, D 3 (aD + b).

2

D, D4

4

f(D

Chapter 2: Ordinary Differential Equations Then

2.17

P.I. = ]1 g sin ax f D 1 sin ax =] aD + bg ]aD bg = 2 2- 2 sin ax, a D -b aD – b).

Then

1

P.I. a

2

2

a

2

2

b2

1

, Rule 3

1 cos x f D

b

1 a

2

2

b2

2

aD b sin x,

a cos x b sin x , since D

a sin x

X =xm, where m

d dx

b cos x .

. P.I.

1 xm f D

Rewrite f(D

(D)], f(D).

Thus

P.I.=

1 k

aD 1 1 aD

k

1

D

xm 1

D

xm

(D)]–1

D, 1 k

aD xm. 1 Note f D , since Dm+1 (xm) = 0, Dm+2 (xm) = 0 and so on. Rule 4 X = e . V(x), where V

1

D

1

to Dm

P.I.= 1 f D

Note Rule 5

1 e xV x f D

e

x

Dm or xm.

, cos

1 f D

V x

V(x) Thie rule is referred to as Exponential shift rule. X = x. V(x), where V(x) or cos

.

2.18

Engineering Mathematics II

P.I =

1 xV x f D

x

1 V x f D

d dD

1 f D

V x

or =x

f

1 V x f D

D 2

f D

V x X = xrV(x), where r

Note X = xr cos ax or xr sin ax. 1 1 art of x r ei x ] x r cos x f D f D

when

= R.P. of

1 x r ei f D

R.P. of ei

1 x r sin x I.P. of ei f D Rule 6 X is any other function of x. 1 P.I X f D 1 X, D m1 D m2 D mn

x

1

x

f D

i

1

x

f D

=

D

m1

xr

i

f D into linear factors.

resolvi A1

xr

An X, D mn

A2 D m2

(1) artial fractions..

1

Consider

i.e.

Its solution is u e

D

m

1

D

m

D

m u

mx

u , say

X

D

e

m X

mx

X

or

D du dx

mu

m u X.

X dx u

Chapter 2: Ordinary Differential Equations u

P.I.

A1e m1 x

e

m1 x

1 D

X dx

m

X

e mx

A2 e m2 x

e

e

m2 x

mx

2.19

X dx

X dx

(2)

An e mn x

e m n x X dx

WORKED EXAMPLE 2(b) Example 2.1 Solve the equation (D2 – 4D + 3) y = sin 3x + x2. A.E. is i.e.

m2 – 4m + 3 = 0. (m – 1) (m – 3) = 0; m = 1, 3 C.F. = c1ex + c2e3x. 1 P. I. = 2 (sin 3 x x 2 ) D 4D 3 =

P. I.1 = =

=

P.I.1

1 (sin 3 x) + 2 D 4D 3 P.I.2 (say)

D2

1 4D

3

9

1 4D

3

D2

1 2 2D 1 2D 2 4D2 1 2D 90

sin 3 x

sin 3 x

3 3

9

sin 3 x

3 sin 3 x

1 6 cos 3 x 90

P.I.2

sin 3 x

3 sin 3 x

1 2 cos 3 x siin 3x 30 1 x2 D2 4D 3 1 x2 D 4 D 31 3

1 4D

3

x2

2.20

Engineering Mathematics II 1 1 3

D 4

1 1 3

D 4

1 1 3

4 D 3

1

D

x2

3

D2 4

D 3

1 2 x 3

D

9

2

x2

13 2 2 D x 9

8 x 3

26 9 y = C.F + P.I.1 + P.I.2

i.e.

y

c1 e x

1 2 cos 3 x sin 3 x 30

c2 e3 x

1 3

Example 2.2 Solve (D2 + 4) y = x4 + cos2 x. A.E. is m2 + 4 = 0. The roots are m = ± i 2. C.F. = A cos 2x + B sin 2x. P.I.1

1 D

2

4

1 1 4

D2 4

1 1 4

D2 4

1 4 x 4 P.I.2

x4

D D

2

4 1

x4 D4 4 x 16 3 2

3x 2

1 2

1

cos 2 x

1 4 2

1 os 2 x co 2

1 1 e0.x 2 D2 4 1 1 2 4 1 1 8

x sin 2 x 2 2 x sin 2 x

1 D2

4

cos 2 x

x2

8 x 3

26 9

Chapter 2: Ordinary Differential Equations

y

A cos 2x

1 4 8

B sin 2 x

6x2

2x4

2.21

x sin 2 x

Example 2.3 Solve (D3 + 8)y = x4 + 2x + 1 + cosh 2x. A.E. is m3 + 8 = 0. i.e. (m + 2) (m2 – 2m + 4) = 0. m

2

2, m C.F. P.I.1

4

or 1 i 3 2 e x (B cos 3 x C sin 3 x)

2x

Ae

1

(x 4

2x

1)

(x 4

2x

D3

8

1 1 8

D3 8

1 1 8

D3 4 (x 8

1 4 (x 8 1 4 (x 8 P.I.2

16

D

2x x

8

2x

1) 1)

3x

1) 1)

e2 x

1 3

1

e

2x

2

1 1 e2 x 3 2 D 8

1 D

2 D

1 1 2x e 2 16

1 1 e 12 D 2

1 1 2x e 2 16

1 x e 12 1!

2

4

2x

2x

1

2x

1 (3 e 2 x 4 x e 2 x ) 96 General solution is y = C.F. + P.I.1 + P.I.2 Example 2.4 Solve (D4 – 2D3 + D2)y = x2 + ex. A.E. is m4 – 2m3 + m2 = 0. i.e. m2(m2 – 2m + 1) = 0 The roots are m = 0, 0, 1, 1.

2D

e

D

e

x

x e 1!

x

2.22

Engineering Mathematics II C.F. = c1 x P.I.1 =

c2 1

2

D D 1

2

c3 x

c2 e x .

x2

1 1 D D2

2

x2

1 1 2 D 3D 2 4 D 3 5 D 4 x 2 D2 1 2 3 4 D 5D 2 x 2 2 D D x4 12 P.I.2

2 x3 3 1

3x 2

D2 D 1 1 12

2

ex

2

ex

1 D 1

8 x 10

1 2 x D

x 2 dx

8 x 10

x2 x e . 2

x2 x e 2! General solution is y

c1 x

c2

c3 x

Note

x4 12

c4 e x

2 x3 3

3x 2

x + 10) in P.I.1

c1x +

c2) of the C.F.

y

c1 x

c2

x4 12

c4 e x

c3 x

2 x3 3

3x 2

x2 x e . 2

Example 2.5 Solve (D2 + 1 )2 y = x4 + 2 sin x cos 3x. A.E. is (m2 + l)2 = 0 The roots are m = i, i, –i, –i C.F. = (c1x + c2) cos x + (c3x+ c4) sin x. P.I.1

1 1

D2

1

D2

2

2

1 2D2 x4 P.I.2

1 D

1

x4 3D 4 x 4

24 x 2 2

x4

2

72. 2 sin x cos 3 x.

Chapter 2: Ordinary Differential Equations 1 D

2

1

sin 4 x sin 2 x

2

1 16 1

2.23

1

sin 4 x

2

4 1

2

sin 2 x

1 n4x sin 225

1 sin 2 x. 9 y = C.F. + P.I.1 + P.I.2. 2 –2x 3 Example 2.6 Solve (D + 6D + 9)y = e x A.E. is m2 + 6m + 9 = 0 i.e. (m + 3)2 = 0 The roots are m = –3, –3 C.F. = (c1x+c2) e–3x 1 e 2 x x3 P.I. = 2 D 3 1

2x

e

D

2

= e

2x

1

e

2x

1 2D

D

3 2

2

x3

yE

l shift rule

x3 3D 2

4 D3 x3

e 2 x x 3 6 x 2 18 x 24 General solution is y = C.F. + P.I. Example 2.7 Solve (D3 3D2 + 3D – 1) y = e–xx3. A.E. is m3 3m2 + 3m –1 = 0 i.e. (m l)3 = 0 The roots are m = 1, 1, 1 C.F. = (c1x2 + c2x + c3) ex 1 e x x3 P.I. 3 D 1 1 e x. x3 3 D 2 3

1 x D 1 e x3 8 2 1 x 1 D D2 e 12 2 3 3 4 8 12 2 4 1 x 5 3 3 e 2 3D 3D 2 D x 16 2 1 x e 2 x 3 9 x 2 18 x 15 16 y = C.F. + P.I.

4 5

D3 3 x 8

2.24

Engineering Mathematics II

Example 2.8 Solve (D2 – 4) y = x2 cosh 2x. A.E. is m2 – 4 = 0 i.e. (m + 2) (m – 2) = 0 The roots are m = – 2, 2 C.F. = A e–2x + B e2x 1 P.I. x 2 cosh 2 x 2 D 4 1 D2 1 2x e 2

x2 2 x e 4 2 1 D

2

2x

e

4

1 2x 1 e x2 2 2 D 4D

1 2x 1 e 1 8 D 1 1 D

D 4

1 2 x x3 e 8 3 1 e 8

2x

(the have

s

1

x2 D2 16

D 4 D2 16 x2 4

x3 3

x3 sinh 2x 12

1 e 2

D 4

1 2x 1 e 1 2 4D

1 e 2

x2

2

D 2 1

2x

D 1 e 2

2

2 2x

D3 2 x 64

x2 4

x2

4D

1 1 4D 1 e 8

D 4

1

x2

2x

D3 2 x 64 x 8

x2 4

1 32 x 8

x2 cosh 2x 16

1 2x 1 e and e 256 256 included in the C.F.)

1

2x

2x

1 32 x sinh 2x 32

are o itted, as they

considered to

Then the G.S. is y

Ae

2x

B e2 x

x 8 x 2 sinh 2 x 96

6 x cosh 2 x 3 sinh 2 x

Example 2.9 Solve (D4 – 2D2 + 1)y = (x + 1)e2x. A.E. is m4 – 2m2 + 1 = 0 i.e. (m2 – 1)2 = 0 The roots are m = ±l, ±l C.F. = (c1x + c2) ex + (c3x + c4) e–x.

Chapter 2: Ordinary Differential Equations 1

P.I. D

2

1

2

e2 x x 1

D 1

e2 x D e2 x

2

1 1 9

1 2x 1 e 9

1

e2 x

4D

3

D D

2

x 1

1

x 1

2

2

4

x 1

3 2D D 3

2

2

2.25

x 1

4

1 2x 8 e 1 D x 1 9 3 5 3

1 2x e x 9

y = C.F. + P.I. Example 2.10

Solve (D2 + 2D – 1)y = (x + ex)2.

A.E. is m2 + 2m –1 = 0. i.e. (m + 1)2 = 2 m

1

2 real roots C.F.

Ae

P.I.

2

D

D2

1

2 x

Be

1 x ex 2D 1 1 x2 2D 1

P.I.1 P.I.2

1 D D 2

P.I.2

D

2

x

Ae

2x

Be

2x

2

e2 x

x2

1 2 D 5D 2 x 2 x2

e

2 xe x

P.I.3 say

1

P.I.1

2 x

1

4 x 10 . 1 e2 x 2D 1

1 2x e 7

1 D( D 2) D 2 D 2

2

x2

2.26

Engineering Mathematics II P.I3

1 2D

D2

1

1

2e x D 2e x 2e x

D

e

1

2

2

D D

1 1 2

x

2 D

1 4D

2

ex 1 x

2 xe x

1

1

x 4

1

x

2

2D x

x

2

y = C.F. + P.I.1 + P.I.2 + P.I.3. Example 2.11 Solve (D2 + 5D + 4) y = e–x sin 2x. A.E. is m2 + 5m + 4 = 0. i.e. (m + 1) (m + 4) = 0. The roots are m = 1, 4 C.F.=Ae–x + B e–4x. 1 e x sin 2x P.I. 2 5D 4 D e

1

x

D e

x

e

x

e

x

1

2

1 D2

3D

5 D

sin 2x 1

4

sin 2x

1 sin 2x 3D 4 3D

4

9D2

16

sin 2x

1 x e 6 cos 2x 52 1 x e 3 cos 2x 26 y = C.F. + P.I.

4 sin 2x 2 sin 2x

Solve (D4 – l)y = cos 2x cosh x Example 2.12 A.E. is m4 – l = 0 i.e. (m – l) (m + l) (m2 + l) = 0 The roots are m = 1, 1, ± i C.F. = c1 ex + c2 e–x + c3 cos x + c4 sin x.

Chapter 2: Ordinary Differential Equations

P.I. =

1 D4

1

1 x e 2

cos 2x

1

2

1 1

4D

3

6D2

D

1 D

4

4D

3

6D2

1 16 D

4D

24

1 x 3D e 8 9D2

2 4

6 sin 2x

3 ex e sin 2x 80 2

4D

x

1 e 8

1 3 sin 2 x sinh x 80

1 e 2

cos 2x

4

1

1

x

16

16 D

24

x

3D 9D

2

2

4 x

cos 2x 6 sin 2x

1 ex e cos 2x 2 80

2 cos 2x

x

cos 2 x cosh x

The y = C.F. + P.I. Solve (D2 4D + 13) y= e2x cos 3x Example 2.13 A.E. is m2 4m +13 = 0 i.e. (m 2)2 = 9 The roots are m = 2 ± i3 C.F. = e2x (A cos 3x + B sin 3x) P.I.

D2

1 4D

13

D e2 x

e 2 x cos 3x 1

e2 x 2

1 D2

9

2

4 D

cos 3x

1 x e 2 x sin 3 x 6

4D

1 cos 2x 3D 2

1 e 320

2 cos 2x x

1

cos 2x

cos 2x

1 e 8

cos 2x

1

x

cos 2x

4D

1 x 1 e cos 2x 8 3D 2

1 x e 320

1 e 2

cos 2x

4

x

1 x e 2 16

x

e

1 D

1 x e 2 D4 1 e 2

ex

2.27

cos 3x 2 e2 x

13 x sin 3x 2 3

coss 2x

2.28

Engineering Mathematics II y = C.F. + P.I.

The

Solve (D 2

Example 2.14 A.E. is

D

1) y

x

e

sin 2

m2 + m + 1 = 0 1 2

The roots are m

3 2

i

x 2

C.F.= e

P.I.

2

3 x 2

A cos

1 D

x 2

D 1

1 1 e 2 D2 D 1 1 e 2 1 e 2

1 e 2

x

1

x

D2

D 1

e

x

1

x

(D 1)

1 e 2

x

3 x 2

1 cos x 2

x

e

B sin

2

(D 1) 1

1

x

D

2

D

1

cos x cos x

cos x

1 x 1 x 1 cos x e e D 2 2 1 x e (1 + sin x) 2 General solution is y = C.F. + P.I. Solve (D2 + 2D + 5) y = ex cos 3 x. Example 2.15 A.E.is The roots are m

m2 + 2m + 5 = 0 1

i2 C.F.

P.I.

e

x

1 D

2

2D 1

2

5

(A cos 2 x

B sin 2 x).

e x cos3 x ex

3 cos x 4

1 cos 3 x 4

2D 5 D 3 x 1 e cos x 2 4 2(D 1) + 5 (D 1) 1 x 1 e cos 3 x 2 4 2(D 1) + 5 (D 1) 1 1 x 1 3 x cos x e e 2 2 4 4 D D 4D 8 4D

8

cos 3 x

Chapter 2: Ordinary Differential Equations

2.29

3 x 1 1 x 1 e cos x e cos 3 x 4 4D 7 4 4D 1 3 x ( 4 D 7) 1 x (4 D 1) e cos x e cos 3 x 2 4 4 16 D 49 16 D 2 1 3 x 1 x e (4 sin x 7 cos x) e (12 sin 3 x 260 580 y = C.F. + P.I. 2 Solve (D + 4D + 8) y = 12e–2x sin x sin 2x

cos 3 x)

Example 2.16 A.E. is m2 + 4m + 8 = 0. The roots are m 2 i2

P.I.

D

1 4D

2

= 6e

2x

= 6e

2x

= 6e

2x

(D

2)

2x

2x

4

(A cos 2 x

(cos x

1 4(D

2

1 D2

2 e 5

6e

8

2x

e

C.F.

(cos x

2)

B sin 2 x)

cos 3 x)

8

(cos x

cos 3 x)

cos 3 x)

1 cos x 3

1 cos 3 x 5

(5 cos x

3 cos 3 x)

y = C.F. + P.I. Solve (D3 – 1) y = x sin x.

Example 2.17

A.E. is m3 – 1 = 0 2 i.e. (m – 1) (m + m + 1) = 0 The roots are m

1,

C.F.

P.I.

1 D x

3

1

x

3

1

sin x

1 D

3 2

i

c1 e x

e

x/ 2

c2 cos

3 x 2

c3 sin

3 x 2

x sin x

1 D

1 2

1

sin x

3D 2 (D

3

1)

2

3 (D

1) 2

sin x sin x

1 xV f (D)

x

1 V f (D)

f (D) f (D)

2

V

2.30

Engineering Mathematics II x

(D 1) sin x D2 1

3 D2 2D 3 cos x. 2

x (cos x sin x) 2 y = C.F. + P.I. 1

P.I.

D

3

D

3

1

1 1

of x e ix )

(I

D

3

I.P. of e ix I.P. of e ix

1

x e ix 1 i )3

(D

1

x

1 D

3

3i D

2

3D

e ix 1 1 i

1

I.P. of

e ix 1 1 i

3D x 1 i

I.P. of

e ix x 1 i

1

i)

(11 2

D

1 3 (cos x 2 2 3 cos x 2

1

x

i

( 3

(cos x

i sin x) x

sin x)

sin x)

x (cos x 2

A.E. is m2 + 4 = 0 i2 The roots are m C.F. = A cos 2x + B sin 2x. 1 D

2

4

x

i

x

R.P. of x 2 ei 2 x

3 (1 2

i (sin x 3 (sin x 2

sin x)

Solve the equation (D2 + 4) y = x2 cos 2x

P.I.

D2 )

3iD

3

1 {(cos x 2

I.P. of

i

1

I.P. of

I.P. of

Example 2.18

sin x

x sin x

1

= I.P. of

1

i)

cos x)} x cos x)

3 (1 i ) 2

Chapter 2: Ordinary Differential Equations R.P. of e i 2 x R.P. of e i 2 x

1 i 2) 2

(D D

1

e i2x 1 4 iD

iD 4

R.P. of

e i2x 1 4iD

iD 4

1 (sin 2 x 4

x3 3

1 4

x2

i i 2 x x3 e 4 3

R.P. of

x2

4iD

R.P. of

R.P. of

x2

4

1 2

2.31

D2 16

iD 3 2 x 64

i 2 x 4 i cos 2 x)

x sin 2 x 8

1 2 x 4

x 8

i 32

x3 3

x 8

i 2 x 4

1 8

1 cos 2 x 8

General solution i s y = C.F. + P.I. Solve (D2 4D + 4) y = 8x2 e 2x sin 2x Example 2.19 A.E. is m2 – 4m + 4 = 0 i.e. (m 2)2 = 0 Roots are m = 2, 2. C.F. = (c1 x+ c2) e2x. P.I.

1 (D

2) 2

8 x 2 e 2 x sin 2 x cos 2 x 2

1 x2 D

8 e 2x

8 e 2x

1 D2

x 2 sin 2 x

sin 2 x 4

2x

2

yi

= e 2x

1 ( 4 x 2 cos 2 x) D

= e 2x

e

4 x2

sin 2 x 2

cos 2 x 4

4 x

cos 2 x 2

sin 2 x 4

2x

2 x 2 ) sin 2 x

4 x cos 2 x

(3

y = C.F. + P.I.

Bernouilli's 1 (2 cos 2 x) D

1 (4 x sin 2xx) D 2x

cos 2 x 8

2

sin 2 x

sin 2 x 8

2.32

Engineering Mathematics II

Solve (D2 + a2) y = sec a x Example 2.20 A.E. is m2 + a2 = 0 ia The roots are m C.F. = A cos a x + B sin a x. 1 P.I. = 2 sec a x D a2 D

1 ia D

1 2ia D ia

ia

sec a x

1 2ia sec a x D ia

1 eia x 2ia

e

ia x

1 e 2ia

sec a x d x 1 D

m

X

ia x

e ia x sec a x d x

e mx .

1 1 ia x e ia x 1 i tan a x d x e 2ia 2ia 1 1 ia x i lo ec a x e ia x x e x 2ia 2ia a x e iax e 2i a

ia x

x sin a x a

1

1

lo

ec a x

cos a x lo a2 General solution is y = C.F. + P.I.

ec a x

a

2

eia x

mx

Xe 1

i tan a x d x

i lo a e

dx

ec a x

ia x

2

EXERCISE 2(b)

Part A (Short Answer Questions) 1. Solve the equation (D2 – D + 1)2 y = 0. D – 1)3 y = 2 cosh x. D2 + a2) y = b cos ax + c sin ax. D2 + 4D + 4) y = x e–2x. D – 3)2 y = x e–2x. D + 1)2 y = e–x cos x. D2 – 2D + 5) y = e x sin 2x. D2 + 4D + 5) y = e–2x cos x.

Chapter 2: Ordinary Differential Equations

2.33

D2 – 2D + 6)y = e x (4 sin x + cos x). 1 D a

f (x).

Part B 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35.

(D3 + D2 + D + 1)y = x2 + 2e–x (D2 + 9)y = x2 + cosh x (D2 + 2D + 1)y = x3 + cos 2x (D2 _ 8D + 9)y = 8 sin 5x + x2 (D2 + 3D + 2)y = 2 sin2 x + 2x2 (D4 + D3 + D2)y = 12x2 + 2 cos 2x cos x (D2 _ l)y = 12ex (x + l)2 (D3 – 6D2 + 12D – 8)y = 16x3 e4x (D3 + 2D2 + D)y = x2e2x (D2 – 4) y = x sinh x (D2 + l)2y = 2x2e–x (D2 – 5D + 4)y = (2x + e–x)2 (D2 – 4D + 3)y = 8ex cos 2x (D4 – 1)y = cos x cosh x (D2 – 2D + 5)y = ex (sin x + cos x)2 x ( D 3 1) y e x cos 2 2 (D2 + 4)y = 4 e2x sin3 x (D2 – 4D + 3)y = sin 3x cos 2x (D2 – 2D + 1)y = x ex sin x (D2 + D)y = x cos x (D2 – 4D + 4)y = x sin x (D2 – l)y = x2 cos x (D2 + 4)2 y = cos 2x (D2 + l)y = x2 sin 2x (D2 + 4)y = 4 tan 2x.

EQUATIONS

a0x n

dn y dxn

a1 x n

1

d n -1 y dxn 1

a n 1x

where a0, a1, … , an are constants and X is a function of x linear differential equation.

dy dx

an y

X

(1)

2.34

Engineering Mathematics II

Note

x and the order

x to t x = et or

x

t

dy dx

dy dt dt d x

dy dx

dy dt

x 1 dy x dt (2)

x, 2

i.e.

x

d y dx 2

dy dx

x2

d2y d x2

x x2

i.e. d dx

D and

d dt

d 2y 1 dt 2 x

dy dx

d2y dt 2

d 2y d x2

d 2y dt 2

dy dt

2

(3)

,

and 2

x D

2

2

(

1) x3 D3 = ( x4 D4 = (

) ( – 2) 1) ( – ) ( – 3) and so on. a0

a1

1

n

an y

2

1

n

1

0 , which is a linear differential equation with

s a0 a x

b an

1

n

d ny d xn

a1 a x

ax b

dy dx

b

an y

n 1

d n -1 y dxn 1

X

(2)

ax + b = et. Equation (2) is called

.

Chapter 2: Ordinary Differential Equations

2.5

2.35

SIMULTANEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

If x and y

t f1 (D) x + f2 (D) y = T1

(1) (2)

z1 ]Dgx + z2 ]Dgy = T2 where f1, f2,

1

,

2

d and T1 and T2 are functions dt .

D

,

of t

2

f2(D)

(D)

y. f1 D

2

D

f2 D

1

D x

2

D T1

f 2 D T2

f D x T

or

which is a linear equation in x and t The value of x y. x and y equal to the order of the resultant equation (3).

Note

x,

y and t

WORKED EXAMPLE 2(c)

2 Example 2.1 Solve the equation x

d 2y dy 4x 2y d x2 dx

x 2 D 2 4 xD 2 y x 2 Put x = et or t =

x and denote 1

i.e.

d dt

x2

1 , where D x2

. 2 y

4 2

3

e2 t

2 y e 2t

e

2t

e

2t

1 . x2 d dx

(3)

2.36

Engineering Mathematics II

A.E. is m2 + 3m + 2 = 0 i.e. (m + 1) (m + 2) = 0 The roots are m = 1, 2.

P.I.

(

t

Ae

C.F.

1 1) (

1 2t e 12 1 2t e 12

Be

2) 1 2

(e 2t

e

t

x = e or t

2

2

(

P.I. =

1)

1] y

sin 2t sin t

1) y

1 ( sin 3t 2

sin t )

x) + B

x).

1

1 (sin 3t sin t ) 12 t 1 1 sin 3t ( cos t ) 2 8 2 1 1 sin (3 lo x lo x os 16 4 2

General solution is y = C.F.+ P.I. Example 2.3 Solve (x2 D2 – 2 xD 4) y x = et or t =

x)

,

A.E. is m2 + 1 = 0. The roots are m = ± i C.F. = A cos t + B sin t = A cos

1)

A.E. is i.e. The roots are m = 4, 1.

x)2 d dt

x

i.e.

x

x) ·

d dt

x

i.e.

)

1 l x2

1 2 x 12

y = C.F. + P.I. Example 2.2 Solve (x D + xD + 1) y 2

2t

B x2

2t

e

2t

te

A x

2t

(

32 t 2

4] y

2 2

,

3

4) y

32t 2

m2 3m 4 = 0 (m – 4)(m + l) = 0 C.F.= A e 4t

Be

t

x)

Chapter 2: Ordinary Differential Equations B x

A x4 P.I.

1 3

4

8 1

(

2

32t 2 1

4

t2

3)

2

81

(

4

3)

3 4

81 8 t2

13 16

2

3 t 2

13 8

2

12

8 lo x

3) 2 t 2

(

16 t2

x) 13]

General solution is y = C.F. + P.I. Example 2.4 Solve(x 2 D 2

x = et or t =

xD 1) y

d dt

x

(

2

2

, 2t

1)] y t 2 e

1) i.e.

2

lo x x.

1) y t 2 e

2t

A.E. is m2 2m + 1 = 0. i.e. (m – l)2 = 0 The roots are m = 1, 1. B ) et

C.F. (At P.I.=

1 e ( 1) 2

=e

2t

=e

2t

1 = e 9 1 = e 9

(A l

2t

x

B)

t2

1 t2 ( 3) 2 1 1 9 3 2t

2t

1 t2

2 3 4 t 3

2

t2 2

3

9 2 3

t2

1 x

2.37

2.38

Engineering Mathematics II 1 3 lo x 27 x 2

2

x 2

4

General solution is y = C.F. + P.I. Example 2.5 Solve (x 2D2 xD + 4) y = x 2 x

et or t

x).

d dt

x

4 y e 2 t sin t

1 i.e.

2

A.E. is The roots are m 1 i 3

m2

4 y e 2t sin t

2

2m + 4 = 0

C.F.= et A cos 3 t x A cos

P.I.

4 2

e 2t

e 2t

B sin

2

1 2

2

sin t

2 4

2

4

sin t

1

sin t 2 3 2 3 sin t 4 2 9

1 2t e 2 cos t 3 sin t 13 1 2 x 2 cos l x 3 sin l 13 General solution is y = C.F. + P.I. Example 2.6 Solve 2 x 3 Put 2x + 3 = et or

t

2

d2 y dy 2 2x 3 12 y 6 x. 2 dx dx x + 3)

Then i.e.

x

3l

e 2t sin t 1

e 2t

e 2t

x

3l

1 2

2

B sin 3 t

2x

3

dy dx

dy 2 dt 2 x 3

dy dx

2 y

x

Chapter 2: Ordinary Differential Equations d2 y dx 2

4

12 y

3 et

3

8

12 y

t

3

2

3 y

2x 4

1

2

2

i.e. A.E. is The roots are m = 3,

2

2 2 4

i.e.

3

1 y

3 e

3 t e 4 3=0

m2 2m

3

1

C.F. Ae3t

Be

P.I.

1 2

2

3 4

t

3

A 2x 3

B 2x 3

3 t e 34

1 t e 4

General solution is y = C.F. + P.I. Example 2.7 Solve (x2 D2 + xD+1) y = x = e or t

3

1

3 2x 3 16

t

2.39

3 4

x

x)

d dt

x

1 y

1

t sin t

i.e.

2

A.E.is The roots are m = ± i C.F. = A cos t + B sin t

m2 + 1 = 0

P.I.=

1 2

1

1 y t sin t

of eit t

I

I.P. of e

it

1 i

2

t 1

i 1 I.P. of e 1 2i 2

I.P. of

i it 1 i e 1 t 2 2

I.P. of

1 t2 sin t i cos t 2 2

t sin t 4

1 2 t cos t 4

it

I.P. of it 2

i it t 2 e 2 2

1

t it 2

2.40

Engineering Mathematics II

1 lo x in lo x 4 General solution is y = C.F. + P.I.

1 4

2

x cos l

x

Example 2.8 Solve (x2 D2 + 4xD + 2) y = sin x. x = et

1) + 4 + 2] y = sin (e–t )

( i.e. A.E. is The roots are

d dt

x=t

2

) y = sin (e–t ) m + 3m + 2 = 0 m = 1, 2. C.F. = A e–t + B e–2t. (

2

A x

B x2 1 1

P.I.= 1

1 1

e

sin e t

2

t

2

sin e t

sin e t e t dt e 1 D m

e

t t

e cos u e e t cos e t

em x

X

sin u du e 2t

2t

2t

sin e t e 2 t dt Xe

mx

dx

u sin u du , u cos u

e t cos e t

et

u

sin u e

2t

sin e t

1 sin x x2 General solution is y = C.F. + P.I.

Example 2.9

d dt

dx 2 x 3 y 5t dt dy 3x 2 y 2 e2 t dt

D, (D + 2)x – 3y = 5t –3x + (D + 2) y = 2e2t (D + 2); (D + 2)2x – 3(D + 2)y = 5 + 10t –9x + 3(D + 2)y = 6e2t

(1) (2)

Chapter 2: Ordinary Differential Equations

2.41

and (2) (D2 + 4D – 5) x = 5(1 +2t) + 6e2 t A.E. is m2 + 4m

(3)

5=0

The roots are m = 1, C.F. = A e t + B e–5t.

5

P.I.=

D2

1 5 1 2t 4D 5

1

1

1

1

D D

x

Aet

13 5

Be

1 2t

4D 1 2t 5 8 5

D2

1 e2 t 4D 5

1

4

5

1 2t 2t

6

6 2t e 7

6 2t e 7

6 2t e 7

6 2t e 7

5t

2t

13 5

6 2t e . 7

y, Method 1 x (D2 + 4D

y C et

5) y = 15t + 8e2 t

De

5t

3t

(4)

12 8 2t e 5 7

Note

In the solution of y used in the solution of x.

A and B x and y x and y should contain

C and D

A and B

x and y in Equation (1), Ae t 5 Be

5t

12 2t e 7

3Ce t 3De

i.e.

2 5t

2 Ae t

2 Be

5t

24 2t 36 e 9t 7 5

3(A – C) et – 3(B + D) e–5t = 0 C = A and D = – B.

12 2t e 7 5t

4t

26 5

2.42

Engineering Mathematics II

and

x

A et

Be

y

A et

Be

6 2t e 7

5t

2t

13 5

3t

12 5

8 2t e 7

5t

Method 2 Dy D2x + 2Dx

D;

3Dy = 5

9x + 6y + 3Dy = 6 e2t

6y y x

x

x

1 x 6

x

A et

x

Be

9 x 5 6e 2 t 6e 2 t

9x 6 2t e 7

5t

2t

5

(5)

13 5

(6)

2

(7)

x w.r.t. t; Aet

x

5 Be

12 2 t e 7

5t

t; Aet

x

i.e.

y

1 6

y

Aet

6 Aet

Be

Example 2.10 Solve Dx and

D; D

2); (D2

i.e. 2

A.E. is m 2m + 2 = 0 The roots are m = 1 ± i

6 Be

8 2t e 7

5t

(D

25 Be

5t

3t

5t

24 2 t e 7

48 2t e 7

18t

(8)

72 5

12 5

2) y = cos 2t

2) x + Dy = sin 2t (D 2) y = cos 2t 2) x + Dy = sin 2t D(D 2) y = 2 sin 2t 2)2x + D(D 2)y = 2cos 2t 2sin 2t. D2 4D + 4) x = 2cos 2t 4sin 2t 2D + 2) x = cos 2t 2sin 2t (D Dx (D D2x (D

(1) (2)

(3)

Chapter 2: Ordinary Differential Equations

2.43

C.F. = et (A cos t + B sin t) P.I.

D

2

1 2D

2

cos 2t 2 sin 2t

1 1 . cos 2t 2 sin 2t 2 D 1 1 10

1 D 1 . 2 5

2 sin 2t 4 cos 2t cos 2t 2 sin 2t

x et A cos t B sin t

cos 2t 2 sin 2t 1 cos 2t 2

1 cos 2t 2

(4)

2Dx – 2x + 2y = sin 2t + cos 2t 2y = 2x – 2x + sin 2t + cos 2t

(5)

t; x = et (A cos t + B sin t) + et (–A sin t + B cos t) + sin 2t

(6)

2y = 2A et sin t – 2B et cos t – sin 2t y = et A sin t – B cos t) –

1 sin 2t 2

(7)

Example 2.11 Solve D2x – Dy – 2x = 2t Dx + 4Dy – 3y = 0 (D2

2)x – Dy = 2t

Dx + (4D

3)y = 0

(1) (2)

D – 3); (D2 – 2) (4D – 3) x –D(4D – 3) y = 8 – 6t

(1)

D2x + D(4D – 3) y = 0

(2)

D

and (2) (4D3 – 2D2 – 8D + 6)x = 8 – 6t i.e.

(2D3 – D2 – 4D + 3)x = 4 – 3t

(3)

2.44

Engineering Mathematics II

A.E. is

2m3 – m2 – 4m + 3 = 0

i.e.

(m – l) (2m2 + m – 3) = 0

i.e.

(m – l) (m – l) (2m + 3) = 0 3 2

The roots are m 1, 1,

1

P.I. D 1

2

B et

At

C.F.

1 1 3

4 3t 2D 3

1 2D 1 1 2 D 4 3t 3 3

3 t 2

Ce 2D 3

1 1 3

1

1 D

2

4 3t

4 D 4 3t 3

1 4 3t 4 3 t. x

B et

At

3 t 2

Ce

(4)

t

Dy 4D2x + Dx – 8x – 3y = 8t y

and

1 4x 3

x

8 x 8t 3 Ce 2

x

At

B et

Aet

x

At

B et

2 Aet

Example 2.12 Solve

3 t A B et

1 Ce 6

3 t 2

3 t 2

9 Ce 4

x, x' and x'' in (5), y

(5)

1 3

d2 x 3x 4 y 0 dt 2 d2 y x y 0 if x = y 1 dt 2

1 3 t 2

Chapter 2: Ordinary Differential Equations dx dt

and

dy dt

0 , when t

2.45

0

(D2 – 3)x – 4y = 0

(1)

x + (D2 + l)y = 0

(2)

2

D + 1); (D2 + 1) (D2 – 3)x – 4 (D2 + 1) y = 0

4x + 4(D2 + 1) y = 0 D4 – 2D2 + 1) x = 0 A.E. is (m2 – l)2 = 0 Roots are m = ±1, ±1 x = (At + B) et+ (Ct + D) e–t

(3)

x' = (At + B)et + A et – (Ct + D) e–t + C e–t

(4)

x'' = (At + B) et + 2 A et + (Ct + D) e–t – 2C e–t

(5)

Solution is x w.r.t. t;

y

1 x 4 1 4

3x 2 At

1 At 2

y

B et B et

1 At 2

2 Aet A t e 2

2 Ct 1 Ct 2

1 Ct 2

B et

D e D e

t

D e

t

t

2C e C e 2

t

t

(6)

(7)

x = 1 when t = 0 in (3), B+ D=1 y = 1 when t = 0 in (6), A B 2

C

D 2

1

(8)

2.46

Engineering Mathematics II A–B–C–D=1

i.e

(9)

= 0 when t = 0 in (4), A+B+C–D=0

(10)

= 0 when t = 0 in (7), –B + D = 0 3 1 , B , C 2 2

A

and

x

1 3t t e 2

y

1 2

(11) 3 1 , D 2 2

1 3t e 2

3 t t e 4

1 2

t

3 t e 4

t

Example 2.13 Solve (D2 – 5)x + 3y = sin t –3x + (D2 + 5) y = t D4

y A.E. is m – 16 = 0 The roots are m = ±2, ±i 2.

16)x = 4 sin t

3t

4

C.F.= A e2t + B e–2t + C cos 2t + D sin 2t P.I.= 4.

1 D

4

16

3

sin t

D

4

16

4 3 D4 sin t 1 15 16 16 3 D4 4 1 sin t 16 16 15

t 1

t 1

t

4 3 sin t t 15 16

and

x

Ae 2t

x'

2 Ae 2t

2B e

x ''

4 Ae 2t

4B e

Be

2t

4 3 t sin t 15 16 4 cos t 2 C sin 2t 2 D cos 2t 15 4 sin t 4 C cos 2t 4 D sin 2t 15

C cos 2t 2t

2t

D sin 2t

(1)

Chapter 2: Ordinary Differential Equations y

1 5 x x sin t 3 A 2t B 2t e e 3C cos 2t 3 3

1 sin t 5

3D sin 2t

2.47

5 t 16

EXERCISE 2(c) Part A (Short Answer Questions) xy + y + 1 = 0 into a linear equation with constant 2. Solve the equation x2y 3. Convert the equation 4. Convert the equation x4 5. Solve the equation x2 6. Solve the equation x3 7. Solve for x

= 0. –1 y = x2 as a linear equation with constant

3 3

2 3x2

Part B D

2

1 as a linear equation with constant

(n + 1)y = 0. 0. and

= 1.

d dx

(x2D2 + 2xD – 20)y = (x2 + 1)2. (x4D3 – x3D2 + x2D)y = 1. (x3D3 – x2D2 + 2xD – 2)y = x). (x2D2 + xD – 9)y = sin 3 x). (x2D2 + 9xD + 25)y x)2. 4 4 3 3 2 2 (x D + 6x D + 9x D + 3xD + l)y x)2. (x2D2 – 3xD + 4)y = x x)2. 4 4 3 3 2 2 (x D + 2x D + x D – xD + 1)y = x2 x. 1 2 2 16. x D xD 3 y cos 2 lo x x 17. (x2D2 + 3xD + 5)y = x cos x). x + 2)2 D2 + 3(3x + 2) D – 36] y = 3x2 + 4x + 1 8. 9. 10. 11. 12. 13. 14. 15.

x + 1)2D2 + (x + 1 ) D + 1 ] y =

x + 1) D

20. (D + 4) x + 3y = t. 2x + (D + 5) y = e2t.

d dt

2.48

Engineering Mathematics II

21. (2D + 1) x + (3D + 1) y = et (D + 5)x + (D + 7) y = 2et. 22. Dx + y = sin t x + Dy = cos t 23. 2D2x Dy 4x = 2t. 2Dx + 4Dy – 3y = 0. 24. D2x + y = 3e2t Dx Dy = 3e2t. 25. (D2 + 4) x + y = 0 (D2 + 1) y – 2x= 1 + cos2 t. 26. D2x 2Dy x = et cos t. D2y + 2Dx y = et sin t. 27. (D2 + 4) x + 5y= t2. (D2 + 4) y + 5x = t + 1.

2.6

x = 2 and y = 0 at t = 0.

LINEAR EQUATIONS OF SECOND ORDER WITH VARIABLE COEFFICIENTS

x.

d2 y dx 2

p x

dy dx

q x y

r x ,

d 2 y is unity and p(x), q(x) and r(x) are functions of x. In dx 2 this section,

.

2.6.1 Method of Reduction of Order-Transformation of the Equation by Changing the Dependent Variable

d2 y dx 2

d2 y dx 2

p x

p x

dy dx

dy dx

q x y

r x

(1)

0

(2)

y = u (x)

(3)

q x y

Chapter 2: Ordinary Differential Equations

2.49

y = u (x) . v(x)

(4)

is a solution of equation (1). =

(5)

and

2

pu)

( i.e.

(

pu)

+ qu) v = r

+(

= r, v

i.e.

v

i.e. v

(6)

2

u u

p v

r u

2u u

r1 , where p1

p1v

r . u

p and r1

(7)

w in 7 , it dw dx

p1 w

r1

(8)

w y to w, we have reduced the order of the

Note

r(x) = 0, then equation 2u u

v i.e.

d v

p v 2u u

v

lo v

x,

0 p

2 lo Au

v

Au

2

u 2

pdx e

e

v and hence the solution of equation (1).

A second order differential equation in y canonical or normal form.

lo A

pdx

(9)

2.50

Engineering Mathematics II

2.6.3

Reduction to canonical or normal form (x) (x)y = r(x) (x)v = g(x), where f (x) =

The second order linear differential equation 1 {p x }2 4

q x y

ve

1/ 2

1 p x and g x 2

p x dx

r x . e1/ 2

n

,

p x dx

Proof: y = uv is a solution of the equation y'' + p(x) y = uv

(x) y = r(x)(1) (2)

x, and The values of y, i.e. i.e.

= + + p( pu) + (

2 (

pu

i.e.

u u

i.e.

lo u c

0, we

tu=e

ny

Thus the s transfor

uv

u

and

u ''

u, u and 1 r it u

12

ve

) + quv = r + qu) v = r

+

(5)

0

p 2 pdx 2

c

pdx

pdx

12

equationn 1 into the canonical

When

v

+

and

2u

Assu

2

(3) (4)

s

e

12

pdx

p2 e 4

,u '

12

pdx

p e 2

12

pdx

p e 2

12

pdx

v

1 u u

pu

qu

Chapter 2: Ordinary Differential Equations p2 4

v

p2 2

p 2 1 2 p 4

i.e.

v

q

i.e.

v

f x v

1/ 2

pdx

1/ 2

pdx

q v

re

1 p v 2

re

2.51

g x , where 1 p x 4

f x

q x

g x

r x e

1/ 2

1 p x and 2

2

p x dx

2.6.4 Method of Reduction of Order—Special Types of Equations d2 y

Type 1. Equations of the form 2 dx absent. dy dx

p, w

et

d2 y dx 2

f x,

dy dx

, in which y is explicitly

dp dx

equation.

p

dp dx

f x, p ,

dy dx

x , c1

y=

(x, c1, c2).

Extension:

dn y dx n

f x,

dn 1 y can dx n 1

2 Type 2. Equation of the form d y2 dx absent.

dy dx

f y,

dn 1 y dx n 1

dy dx

p and treat p as a function of y. Then dp dy

f y, p ,

p and

in which x is explicitly

d2 y dx 2

dp dx

p

dp dy

2.52

Engineering Mathematics II

dy dx

p

x=

y, c1

(y, c1, c2)

Extension: f{y, , y , ... y(n)}

dy = p dx

p as a function of y.

Type 3. Equations f (x, y, y', y'') = 0, which are homogeneous in y, y' and y'' (but not in x) y

e

solved. When y

zdx

e

, zdx

,y

ze

zdx

and y

z2

z e

zdx

. Thus, the order of the

z

Type 4. Equations f (x, y, , y ) = 0 which are exact, viz. which can be expressed as d 0 . x, y, y dx

d dx

x, y, y

x, y, y'

0 is

c1 , which is

a equation.

Note 0

p0(x)D2 + p1(x)D + p2(x)]y = r(x) 1

+ p2 = 0.

p0 D2 + p1 D + p2) y = D(q0 D + q1) y

Conversely, when

q0 D 2 + (

p0 = q ; p1 + q1 and p2 0 p1 + q1, 0 = 0+ 1 1 = 0 + p2 – 1+ p2 = 0 0 – + p = 0, we have 0 1 2 (p0 D2 + p1D + p2)y = (p0 D2 + p1 = p0 + p1

0

+ q1)

]y

1

1

– y– 1 1

)y y 0 0

Chapter 2: Ordinary Differential Equations = (p0

) + (p1

0

1

= D(p0 ) + D(pl y) – D( = D(p0

l

0

0

2.53

y) – (

0

0

)

y)

y)

Thus (p0 D2 + p1 D + p2)y =

0

1

+ p2 = 0.

WORKED EXAMPLE 2(d) 2(x + 1) + (x + 2)y = (x – 2)e2x,

Example 2.1 Solve the equation the order of the equation.

Important Notes p0(x)

(x) y + p2(x). y =

1

(i) If p0(x) + p1(x) + p2(x) = 0, y = ex is a solution of the equation. (ii) If p0(x) – p1(x) + p2(x) = 0, y e–x is a solution of the equation. (iii) If p1(x) + xp2(x) = 0, y = x is a solution of the equation. p0 = x, p1 = – 2(x + 1) and p2 = x + 2. The condition p0 + p1 + p2 = y = ex equation. y = vex = vex

Then y

x

v ex + 2v ex + vex

y, , x

x(e

– 2 = (x – 2)ex

i.e.,

i.e.,

+ 2ex + exv) – 2(x + 1) (ex + exv) + (x + 2) exv = (x – 2)e2x

dp dx

2 p x

1

dv dx

2 x e , where p x

This is a I.F. = e

2 dx x

e

2 lo x

1 x2

Solution of this equation is p x2

1 x2

2 x e dx x3

3c1

2.54

Engineering Mathematics II 1 x e x2

d

dv dx

i.e.

3c1

1 x e x2

3c1

3c1 x 2

ex

v =c1x3 + c2 +ex

y = ex (c1x3 + c2+ cx) Example 2.2 Solve the equation

d2 y dx 2

dy dx

1 cot x

y cot x

sin 2 x,

y cot x = sin2 x

(1)

+ (l – cot x) – y cot x = 0

(2)

+ (1 – cot x) p0=1; p1 = 1 – cot x; p2 = – cot x p0 – p1 + p2 = 0 y = e–x is a solution of the equation

–x

y = ve Then

–x

– ve–x

and

–x

–2

y, (

–x

–2

–x

–x

+ve–x

and y

+ ve–x) + (1 – cot x) (

–x

– ve–x) – ve–x cot x = sin2 x x

(1+ cot x)

i.e.

dp dx

1 cot x p

I.F. = e e

sin2 x

(3)

e x sin 2 x

(4)

1 cot x dx

x lo

in x

e x sin x

Chapter 2: Ordinary Differential Equations

2.55

Solution of equation (4) is pe x sin x

cos x i.e.

e x dx sin x

e x sin 2 x

dv dx

p

c1

c1

e x c1 sin x sin x cos x

(5)

x, v

c1

c1 x e sin x 2

y

1 2

e x sin x dx cos x

A sin x

e x sin 2 x dx

c2

1 ex sin 2 x 2 5

2 cos 2 x

c2

1 sin 2 x 10

2 cos 2 x

Be x .

cos x

d2 y Example 2.3 Solve the equation 1 x 2 dx 2 reduction of order.

2x

dy dx

2y

2,

x2)

2

2y = 2

(1)

(1 – x2)

–2

+ 2y = 0

(2)

(1 2 p0 = 1 – x ; p1 = –2x; p2 = 2 p1 + p2x = 0. y = x is a solution of

y = vx Then y, , i.e. i.e.

i.e. where

=

+

2

satisfy equation (1) (1 – x2)(

2 ) – 2x(

x(1 – x2)v dp dx

2 x

) + 2vx = 2

x2) – 2x2]v = 2 2x p 1 x2

2 x 1 x2

(3)

2.56

Engineering Mathematics II

Equation (3) is a 2 2x dx x 1 x2

I.F. e e

1 x2

2 lo x

x2 1 x2 Solution of equation (3) is 2

px 2 1 x 2

x1 x x2 dv dx

i.e.

2

x 2 1 x 2 dx

c1

c1 c1

1 1 x

2

1 1 x2

2

x 1 x2 1 1 c1 2 1 x2 x

(4)

x v

c1

1

1 1 x lo 2 1 x

c1 x

c2

1 x 1 x

c1

c2 x

dy dx

y

Solution of equation (1) is y

c1

1 2

x lo

d2y Example 2.4 Solve the equation x 2 dx a solution.

2

x

If y = u is a solution of the equation (x) + q(x)y = 0, then y = uv 0, where v

c1u 2 e

0

y

+ p(x)

1 is x

+ q(x)y =

p x dx

y

2 1 y x

1 y x

0

(1)

Chapter 2: Ordinary Differential Equations p x

Since y

y

2 1 and q x x

2.57

1 x

1 is a solution of equation (1), x

1 v is also a solution of (1), where x 2 1 dx x

c1 x 2 e

v

c1 x 2 e

2 lo x x

c1 x 2 x 2 e x v

1 c1e x x

y

c1e x

c2

c2

or

2 Example 2.5 Solve the equation d y dx 2

solution. y

c1e x

c1e x

xy

2 dy x dx

y

c2

1 sin x v x v

c1 c1

1 sin x is a x

y

0

, where x2

sin 2 x 2 x

sin 2 x

e e

2 dx x ,

1 sin x and p x x

since u c1

2 lo x

sin 2 x

v = c1 cot x + c2

or A cot x + B y

:. i.e. Example 2.6 Solve the equation

2 x

xy

A cos x

d2 y dx

2

2x

dy dx

1 sin x A cot x x

B sin x x2 y

0

B

2.58

Engineering Mathematics II 2

y=0

+2

(1)

+ p(x)

(x) y = r(x)

(2)

We have p(x) = 2x; q(x) = x2; r(x) = 0. p x dx

y = uv, where u

2

e v

x2 2

e p2 4

q

,

p v 2

re

p 2d x

(x2 x2 1) v = 0 =v

i.e. i.e.

= f(v, ) p as a function of v, we have v

dp dv

p

(3) p

dp dv (4)

v

p2 = v2 + c 12 dv dx

p

or

1

sinh or

v

v2

v c1

x

c12

(5)

c2

c1 sinh x c2 ex

c1 Ae x

c2

e 2

Be

x

x c2

The required solution of equation (1) is y

Ae x

Example 2.7 Solve the equation 4 x 2

Be d2 y dx

2

x

e

4x

x2 / 2

dy dx

x2

1 y

0

Chapter 2: Ordinary Differential Equations

1 y x

y 1 ,q x

p

1 1 4

1 x2

and r

1 1 4

1 x2

y

2.59

(1)

0

0

y = uv u 1

y

x

1 2

e

pd x

dx x

1 2

e

1 x

v

v

1 1 4

v

i.e.

p2 4

q

p v 2

1

1

1

x2

4x2

2 x2 v

i.e.

v

A cos 1

Solution of equation (1) is y

x

x 2

A cos

d2 y Example 2.8 Solve the equation x dx 2 = 0.

dy dx

B sin x 2

re

v

0

1 v 4

0

p dx 2

(3)

x 2

B sin

x 2

x2e x

y(0) = – 1 and (0)

y dy dx

p

p as a function of x, we have

dx

x

i.e.

d2 y

dp dx

p

dp dx

1 p x

2

dp ; the equation dx

x2e x

xe x

(1)

2.60

Engineering Mathematics II p.

Equation

p

Solution of (1) is

1 dx x

e

I.F. 1 x

1 x

1 xe x dx x

2c1

p = x (ex + 2c1)

i.e.

dy dx

or

xe x

xe x dx

y

2c1 x dx

c2

1) ex + c1x2 + c2.

y = (x

y(0) = 1, c2 = c1 = 0, the required solution is y = (x Example 2.9 Solve the equation

(2)

2c1 x

d2 y

dy dx

dx 2

(0) = 0 in (2), c1 is

1)ex.

2

1

y(0) = 0, (0) = 0

Method 1 +

2

=1

(1) y

dy dx

p

p as a function of x, we have

dp dx

1

p2

x

c1

x

c1

Solution of equation (2) is dp 1 i.e.

1 1 lo 2 1

p2 p p

Given that p = 0, when x = 0 c1 = 0

d2 y dx

2

dp dx

(2)

Chapter 2: Ordinary Differential Equations Thus we have

1 p 1 p

2.61

e2 x e2 x 1 e2 x 1

dy dx

p

ex ex y=

e e

x

(3)

tanh x

x

x + c2

y

c2 = 0

Solution of equation (1) is y

x.

Method 2 The equation 2

y

=1

(1)

x dy dx

p as a function of y, we have

p

p

dp 1 dy

p2

d2 y dx 2

p

dp dy

(2)

Solution of equation (2) is p dp 1 p2

y

c1

1 lo 1 p 2 2

i.e.

y

c1

Given that y = 0 and p = 0, when x = 0 or when y = 0, p = 0 c1 = 0 1 – p2 = e–2y

Thus i.e.

p

dy dx

1 e

dy 1 e

2y

2y

x c2

(3)

2.62

Engineering Mathematics II e y dy

i.e.

x c2

e2 y 1

cosh–1 (ey)= x + c2

i.e.

x = 0, y = 0.

When

c2 = 0

The required solution of equation (1) is ey = cosh x or y = Example 2.10 Solve the equation x

y lo x

y

d2 y dx 2

dy lo dx

(1) y

p as a function of x, we have y

dp dx p = xev, we have

dp dx

p lo x ev

xev

ev 1 x

i.e.

1 dy x dx

y x

y =f (x, y =p

x.

x

dv dx

p x

dp . dx

(2)

dv dx

dv dx

ev lo ev

v 1

(3)

Solution of (3) is dv v 1

lo c

v = 1 + cx

i.e. i.e.

dx x

lo

p x

1 cx

p

dy dx

xe1

cx

(4)

Chapter 2: Ordinary Differential Equations

y

x e1 c

cx

e1 cx c2

2.63

c

The required solution of equation (1) is c2y = (cx – 1)e1+cx d2 y dx 2

Example 2.11 Solve the equation

dy 1 y dx 2

2

y

y

1 y

2

2

0

(1)

0

y = f(y, ), which does not contain x p as a function of y, we have y

p

dp dy

2 p2 1 y

dp p

i.e.

2 1 y

dy

p

dp . dy

(2)

0

0

solution of (2) is p

y dy dx

i.e.

p c1 1 y

dy 1 y

2

1 1 y

i.e.

c1 x c2

c1 x c2

The required solution of equation (1) is y

1

1 . c1 x c2

c1 2

(3)

2.64

Engineering Mathematics II

Example 2.12 Solve the equation xy 2

xyy –

y (1), we have

e

zdx

d2 y dx 2

2

dy dx

x

y

dy dx

0.

– 0 y, , y .

and hence y

(1)

zdx

ze

z2

and y

zdx

z e

in equation

2

e

zdx

x z2

xz 2

z

z

xz

i.e.

0 z=0

(2)

z = c1x y

e

c1 xdx

y = Ae Bx

or d2 y Example 2.13 Solve the equation y 2 dx yy

y

2

y

e

zdx

and hence y

x2 2

c2

2

dy dx . 1 x2 y

2

1 x2

yy y,

dy dx

e

c1

(1)

and y . ze

zdx

z2

and y

z1 e

zdx

in equation (1), we

have 2 zdx

z1

z2

i.e.

dz dx

2z2

i.e.

1 dz z 2 dx

i.e.

du dx

e

z2

z

0

1 x2 z 1 x2 1 z

1 1 x2 1 1 x2

u

0

(2)

2

2

(3)

Chapter 2: Ordinary Differential Equations 1 z

u

where

2.65

Equation 1

I.F. e e

lo

1 x2

dx

1 x2

x

1 x2

x

Solution of equation (3) is x

1 x2 u

x2

i.e.

1 x 2 dx c1

2 x

x 1 x2

x2

x 1 x2

1 d lo 2 dx

x2

1 2 lo

e

x

y2

x 1 x2

c1

1 x2

x

c1

zdx

y e

or

1 x2

x

lo

x2

c1

1 x2

x

z

1 x2

x

lo

2

x 1 x2

x 1 x

A x2

2

c1

d2 y dx 2

x

2

c1

1 x2

x

lo

c2

1 x

x

lo

x 1 x2

Example 2.14 Show that the equation x

1 x2

x

2

dy dx

y

1 2

c3

B

0

hence solve it. xy + (x + 2) p0y + p1

y=0

2

y =0.

(1)

2.66

Engineering Mathematics II p0 = x, p1 = x + 2, p2 = 1

We have

p0 – p1 + p2 = 0 – 1 + 1 = 0

(xy + y ) + (xy + y ) + y = 0 d xy dx

i.e.

d xy dx d xy dx

or

Solution of (2) is x

dy dx

x 1 y

0

y

0

(2)

c1 x

(3)

c1 dy dx

i.e.

xy

dy dx

1 y x

1

1

1

dx

x I.F. e x lo x e xe x

Solution of equation (3) and hence equation (1) is y x ex xye

i.e. Example 2.15 Solve the equation sin x

d2 y dx 2

x

cos x

c1e x dx c1e dy dx

x

c2 2 sin x y

p0 y + p1y + p2y = 0, we have p0 = sin x, p1 = – cos x and p2 = 2 sin x. p 0 – p 1 + p2 = – sin x – sin x + 2 sin x = 0 d p0 y dx

p1 y

p0 y

cos x

c2

cos x

Chapter 2: Ordinary Differential Equations d dx

i.e.

sin x y

cos x y

cos x y

cos x

x) y y

i.e.

2 lo

x)y = 1 + c1 cosec x

cot x c1 2

cot x

y

i.e.

(1)

1 sin 2 x

in x

y sin 2 x

Solution of (1) is

x) y = sin x + c1

2 cot x dx

I.F. e e

2.67

sin x cos x

c1

cosec3 x dx

cosec x cot x c1 sin 2 x l 2

c2

l

tan

tan x 2

x 2

c2 c2 sin 2 x

cos x

EXERCISE 2(d) Part A (Short Answer Questions) 1. If y = u(x) and y = u(x) v(x) are solutions of the equation y v(x).

q(x) y = r(x 4. When the equation y g(x

p(x) y y

p(x) y

q(x) y = r(x

v

q(x) y

y

p(x) y

v

f(x) v =

1 2 p x dx , what are the values of

f(x) and g(x)? y y

f(x, y

f(y, y f(x, y, y y

y, y y 7. Write down the condition for the equation p0(x) y 8. If the equation p0(x) y

p1(x) y

p2(x) y = r(x

p1(x) y

p2(x) y = r(x)

2.68

Engineering Mathematics II y with y =ex as a solution (ii) y = e–x as a solution? y with y = x as a solution?

Part B

11.

x 1

12. x

13.

d2 y dx 2

d2 y dx 2

d2 y d x2

x 1

16. x

18.

d2 y d x2

d2 y d x2

x2

d2 y d x2

2x

4x d y 2x 1 d x

19. x 2

d2 y d x2

x2

20. x 2

d2 y d x2

3x

21.

d2 y d x2 solution.

2x dy dx

x2

2x x

2

x 1y

ex .

x x

ex .

e x sin x .

0.

2 y

y 1

dy dx

2x 1

5y

dy dx

dy dx

x

x

1 cot x y

2x x 1

d2 y d x2

17. 1

dy dx

cot x

dy dx

3

dy dx

2x 1

2 14. x 2 d y d x2

15.

2 x

dy dx

x 1 y

2y

6(1

4 y 2x 1

0

dy dx

2 y

3y

x

0

x2 )

x3e x .

x 2 (2 x 1)

2 dy dx x

2x2 2x 2 y x3 x 2

0

y = x2 is a

Chapter 2: Ordinary Differential Equations d2 y d x2

x

2 23. d y d x2

y

22.

24.

d2 y d x2

dy dx

4y

28. cos 2 x

2 29. sin x

y = cos x

y = sin 2x

4 tan 2 x

x2

26. x (1 3 x 2 )

1)

y = x3 is a solution.

0

sec x

2 25. x 2 d y d x2

2 27. (x

9y

x tan x

3 y 4

0

y

dy dx

6 xy

0

y

d2 y d x2

2

d2 y d x2

6y 1

d2 y d x2

2y

d2 y d x2

2.69

1 x

cos x is a solution.

1 is a solution. x

y = x – x3

y = tan x is a solution.

2y

y = cot x is a solution.

30. Find the values of a and b if y = x x2

d2 y d x2

2x 1

x

dy dx

ax

dy dx

y cos x

x 3 . For

b y

these values of a and b 31. Solve the equation x sin x

cos x

d2 y d x2

that y = xm is a solution of the equation.

32. x 2

d2 y d x2

x

dy dx

x2

1 y 4

0.

33. x 2

d2 y d x2

x

dy dx

x2

1 y 4

0.

x cos x

0

2.70

Engineering Mathematics II

2 34. x

d2 y d x2

35. 4 x 2

36.

2 x2

d2 y d x2

d2 y d x2

4x

dy dx

d2 y d x2

x

x2 )

38. (1

d2 y x ) d x2

d2 y 39. d x 2 40. y

dy x dx

d2 y d x2

dy dx

43.

d2 y d x2

2

dy dx

d2 y 45. y 2 dx

dy dx

46. y

d2 y d x2

2

0.

y(0) = – 1 and y 2

dy dx

y(0) = 0 and y 1

1 2

1 0.

. 2

dy dx

2

dy dx

d2 y 44. y 2 dx

2)y

4 x 2 sin x .

1)y

sin 2 x

dy dx

dy dx

2x

3

d y d2 y 41. y d x d x2 d2 y y 42. d x 2

(16x 2

dy dx

2

(x 2

3

dy dx

tan x

37. (1

dy dx

x

3

0. d2 y d x2

2

.

dy . dx 2

a2

2

6 xy 2 . 2

dy dx

3

2 y2 . 2

0.

y(0) = – 1 and y

4

.

Chapter 2: Ordinary Differential Equations

47. x 2 .

d2 y d x2

x

dy dx

y

0.

48. 1 x 2

d2 y d x2

3x

dy dx

y

1.

49. x 2

d2 y d x2

3x

dy dx

y

1.

50. x

2.7

1

d2 y d x2

x

2

dy dx

2.71

e x.

y

METHOD OF VARIATION OF PARAMETERS

f(D)y = x, f(D)y =

Solution of the equation

dy dx

Py

Q

(1)

where P and Q are functions of x.

dy dx i.e.

Py dy y

(2)

0

Pdx

lo y

P x

lo c

ce

Pdx

solution of Eq. (2) is y

ce

Pdx

c as a function of x required solution of (1). x, we have

(3)

2.72

Engineering Mathematics II dy dx

cPe

Pdx

dc e dx

P x

c e

Pdx

cPe

Pdx

(4)

Pdx

Qe

c

Pdx

Q e

y

ye

P

Q

dc dx

i.e.

dc e dx

P

A

(5)

Pdx

Pdx

e

.dx

Qe

dx

A

Pdx

Pdx

Q e

.dx

A

(6)

Note

Solution of the equation

d2 y dx 2

P

dy dx

Qy

(1)

R

where P, Q and R are functions of x or constants.

d2 y dx 2

y

P

dy dx

Au

Qy

0

(2)

Bv

(3)

where A and B

y = u(x) and y = v(x) are A and B as functions of x x, we have dy dx

Au

Bv

Au

Bv

(4)

Chapter 2: Ordinary Differential Equations

2.73

We choose A and B such that Au+Bv=0

dy dx

(5)

Au

(6)

Bv

x, we have d2 y dx 2

Au

Bv

Au

(7)

Bv

Since (3) is a solution of Eq. (1), (3), (6) and (7) satisfy (1). (Au + i.e.

+

+

A(

) + P(

) + Q(Au + Bv) = R

)+B(

)

(8)

Since y = u is a solution of Equation (2) u + Pu + Qu = 0 v +

+ Qv = 0

(9) A values of A and B as functions of x.

Notes

the values of A and B

d2 y dx 2

differential equation is unity. A and B

x.

WORKED EXAMPLE 2(e)

Example 2.1 Solve the equation x 2

1

dy dx

4 xy

1 x2

1

2.74

Engineering Mathematics II

x2

1

dy y

i.e.

x2 +

y+2 c

y

i.e.

dy 4 xy xy 4x dx x2 1

x

2

2

1

(1)

0 0

c is the solution of (1)

(2)

c as a function of x 2

x2

dy dx

x2

x, we have

x

2

1 c' x

2

4cx x 2 1

(x2 + l)2 c

i.e.

c 2 x2

1 c' 2

1

1 2x

(3)

4

1

4cx

3

x

2

1

1 2

x2

1

cx (x2 + l) + 4cx (x2 + 1) = (x2 + 1)2

i.e.

1

c=x+k

(4)

x2 + 1)2 y = x + k, where k Example 2.2 Solve the equation

dy dx

2

sec 2 y

x sin 2y = x3 cos2 y

y

dy dx

2 x tan y

x3

(1)

x 3 , which is linear.

(2)

y=z dz dx

2x z

Chapter 2: Ordinary Differential Equations dz dx

2 xz

dz z

or

0

2 xdx

2.75 (3)

0

c – x2

z i.e. the solution of Eq. (3) is

z

ce

x2

(4)

c as a function of x

x, dz dx

– 2cxe–x2 +

–x2

x2

2cxe

ce

x2

(5)

+ 2cxe–x2 = x3 = x3 ex2

i.e.

c = x3 ex2 dx + k 1 2

t e t dt

1 t te 2 1 2 x 2

z

1 2 x 2

tan y

et 1 ex

ke

1

1 2 x 2

1

x2 = t

k,

k 2

k

(6)

x2

ke

x2

where k is Example 2.3 Solve the equation d2 y dx 2

d2 y dx 2 y

y

x cos x,

x cos x

(1)

2.76

Engineering Mathematics II

d2 y dx 2

y

(2)

0

The solution of Eq. (2) is y = A cos x + B sin x,

(3)

where A and B A and B as functions of x, sin x +

– and

A

x+

and B

cos x = x cos x

(4)

sin x = 0

(5)

A

B

1 x sin 2 x and 2

x sin x cos x or x cos 2 x or

1 x 1 cos 2 x 2

(6)

(7)

x A

B

and

y

where c2

1 has 8

assu

x2 4

sin 2 x 4

(8)

c1

1 x sin 2 x 2 2

cos 2 x 4

c2

c1

x cos 2 x 4

1 sin 2 x 8

cos x

c2

x2 4

x sin 2 x 4

1 cos 2 x 8

y

c1 cos x

c2 sin x

1 sin x 8

y

c1 cos x

c3 sin x

x2 sin x 4

i.e. or

x cos 2 x 2

1 2

d as c3 .

(9)

sin x

x2 sin x 4 x cos x 4

x cos x 4

Chapter 2: Ordinary Differential Equations

Example 2.4 Solve the equation

d2 y dx 2

a2 y

2.77

tan ax

d2 y dx 2

a2 y

tan ax

(1)

d2 y dx 2

a2 y

0

(2)

The solution of Eq. (2) is y = A cos ax + B sin ax If we treat A and B as functions of x, –

and B

sin ax + A

and

cos ax = tan ax

ax +

B

(4)

sin ax =0

(5)

1 sin 2 ax a cos ax

A

and

(3)

(6)

1 sin ax a

(7)

x, A

1 sin ax a2

and

B

1 cos ax a2

i.e.

y

c1

c2 i.e.

y

sec ax

l

c1

(8)

c2

1 sin ax a2 1 cos ax a2

c1 cos ax

tan ax

(9)

l

sec ax

tan ax

cos ax

sin ax

c2 sin ax

1 cos ax l a2

sec ax

tan ax

2.78

Engineering Mathematics II

Example 2.5 Solve the equation (2D2 – D – 3) y = 25e–x,

D2y D2

1 D 2

3 y 2

25 e 2

D2

1 D 2

3 y 2

0

3 2

0

m 1

0

1 m 2

m2 3 2

m

or

(1)

x

(2)

3 and 1. 2 Therefore the solution of Equation (2) is m

3

Ae 2

y A and B as functions of x,

and

3

A e2

A

y

2e

c1

Be

(3)

x

and

3 x 3 A e2 2

A

x

5e 5 x 2

2e

Be

x

5 x 2

Be

x

x

and B

3

e2

x

(4)

x

(5) 0

5

c1 and B

5 x 2

25 e 2

5x

c2

c2

5x e

x

Chapter 2: Ordinary Differential Equations 3

c1 e 2

y

i.e.

x

c2 e

d2 y dx 2

Example 2.6 Solve the equation x 2

x

4x

x

2e dy dx

6y

5 xe

2.79

x

x

sin l

. d2 y dx 2 d2 y dx 2

4 dy x dx

d2 y dx 2

6 y x2

4 dy x dx

d2 y dx 2

x2

4x

x=t

i.e.

1 sin l x2

6 y x2 6y

d dt

,

4

6 y

0

2

5

6 y

0

(1)

0 or

dy dx

1

x

0

(2)

(3)

m2 –5m + 6 = 0 m = 2, 3 Therefore the solution of Eq. (2) is

or

y

Ae 2t

Be3t

y

Ax 2

Bx 3

A and B as functions of x,

and

A

and

2A x

3B x 2

A x2

B x3

1 sin l x3

(4)

x and

1 sin (lo x x2

(5)

0

(6)

B

1 sin l x4

x.

2.80

Engineering Mathematics II 1 sin l x3

A A

i.e.

e

2t

x dx

sin t dt

1 sin l x4

c1 and B =

c1 and B

e

3t

sin t dt

x dx

c2

c2 ,

x = t or x = et i.e.

A

i.e.

c1

e

2t

5

2 sin t

cos t and

3 sin t

cos t

B = c2

e 3t 10

A

c1

1 2 sin l 5x2

B

c2

1 3 sin l 10 x 3

x

cos l

x

cos l

x and x

A and B in (4), the required solution of Eq. (1) is y

1 2 sin lo x 5x2

c1

c2

i.e.

y

c1 x 2

c2 x 3

i.e.

y

c1 x 2

c2 x 3

2 5

os lo x

1 3 sin lo x 10 x 3 3 sin l 10

1 sin l 10

Example 2.7 Solve the equation x 2

os lo x

1 5

x

x

cos l

d2 y d x2

x

x2

1 cos l 10

x3

x

x .

dy dx

y

x log x

d2 y d x2 d2 y d x2

1 dy x dx

1 y x2

1 log x x

(1)

Chapter 2: Ordinary Differential Equations d2 y d x2

1 dy 1 y x d x x2 d2 y dy x2 x y 2 dx dx

x = et or t

2.81

0 or (2)

0

d dt

x

1y 0

1 i.e.

2

1

y

(3)

0

Therefore the solution of Eq. (3) is y

At

B et or y

A and B as functions of x, A 1 and

i.e.

lo x

(5)

x

(6)

1 x

1 lo x dx and B x

A

t dt and B

2

(4)

Bx

B x

1 lo x and B x

1 lo x 2

1 x

B

A

A

x

and

A x lo x

A

Ax l

2

x

1 x

x

2

t 2 dt on

x 1 3

c1 and B

dx

x

t

3

c2

A and B in (4), the required solution of Eq. (1) is

i.e.

y

1 lo x 2

y

c1 x lo x

2

c1 x

c2 x

1 x 6

1 lo x 3

x

x

3

3

c2 x

2.82

Engineering Mathematics II EXERCISE 2(e)

1.

dy dx

e x sec x.

y tan x

dy dx

1

x y

3.

d2 y d2 x

y

x sin x

4.

d2 y d2 x

a2 y

4.

d2 y dx 2

2

dy dx

2y

e x tan x

6.

d2 y dx 2

6

dy dx

9y

1 3x e x2

2. x

x

e

sec ax

2 7. x 2 d y dx 2

4x

dy dx

2y

x2

d2 y dx 2

2x

dy dx

4y

32 l

8. x 2

1 x2 2

x .

ANSWERS Exercise 2(a) (7)

y

2x C y

3x C

0

( ) (y – ex – C) (ey – x – C) = 0 (11) y

px

p p 1

;y

Cx

C C 1

(8)

y

x2 2

(10)

y

lo x C

C lo y

x C

0

y2 2

x C

0

(12) y = px – ep; y = Cx – eC

(13) y = px + sin–1 p; y = Cx + sin–1 C

(14) y2 = 4x

(15) x2 = 4y

(16) (y – x – C) (x2 + y2 – C) = 0

(17)

y

x2 4

C x

1 y

C

0

(18) (y – Cx2) (3x2 – y2 – C) = 0

Chapter 2: Ordinary Differential Equations (19) (y – Cx) (x2 – y2 – C) = 0 y (1 + cos x) – C

(23) y

y x

1

(22) sin

C x

(20) (xy – C) (x2 – y2 – C) = 0

y (1 – cos x) – C] = 0 1

x C sin

l

1 4x2

C2; y

y x

x C

l

0 y = x2 + p2 and lo

p

x

p

x x

n n 1 2p 3 C

pn

x

p

2

p

x

C.

p2 3

2C . p

an 1 p 1

ep and y 2

p

x

x

C p2

1

p 1

lo

Ce

p

C and y p2

p

3x

0

p

p

(29) y

2.83

C 1

p e

p

1 1 2

p ep

3 1 Ce3 x

lo

(30) 64y = C(C – 4x)2; 4x3 = 27y (31) of p y=C– p – 1) + 2p + p2] and x= C p – l) + 2p] (32) y2 = 2Cx + 4C2 (33) y2 = 2Cx + C3 y = Cx + C2 (35) y

x1 x

tan

x

1

1 x

(37) y2 =Cx2+ C2 (39) xy= Cx – C2

(36) y

x

(38) y2 = 2Cx + C2 (40) ey = Cex + C3

Exercise 2(b) x

(1) y

e 2 C1 x

(2)

x3 x e 6

(4)

x3 e 6

1 e 8 2x

(6) – e–x cos x

x

C2 cos

3 x 2

C3 x

C4 sin

(3)

3 x. 2

x b sin a x 2a

(5) (x – 2) e2x (7)

1 x xe cos 2 x 4

c cos a x .

2.84

Engineering Mathematics II

1 x e 2 x sin x 2 (9) ex (4 sin x + cos x) (11) y = C1 e–x + C2 cos x + C3 sin x + x2 – 2x + xe–x (8)

(12) y (13) y

C1 cos 3 x Ax

e 4 x Ae

(15) y

C1e

(16) y

x

B e

(14) y

x

C1 x

x2 9

C2 sin 3 x x3

7x

Be

C2 e

2x

C2

e

6x2

2 81

18 x

1 cosh x 10 1 4 sin 2 x 3 cos 2 x 25 1 2 16 x x 2 sin 5 x 9 9

24

1 5 cos 5 x 29

7x

1 cos 2 x 3 sin 2 x 20 x 2

3 x 2

C3 cos

C4 sin

x2

3x

3 x 2

x4

110 81

4 4 x3

1 8 cos 3 x 3 sin 3 x sin x 657 (17) y = C1ex + C2e–x + xex + (3 + 3x + 2x2) (18) y = (C1 x2 + C2 x + C3) e2x + (2x3 – 9x2 + 18x – 15) e4x (19) y

C1

C2 x

C3 e

e2 x 2 x 18

x

7 x 3

11 6

2 x sinh x cosh x 3 9 (21) y = (C1x + C2) cos x + (C3x + C4) sin x + (x2 + 4x + 4) e–x (20) y

C1e 2 x

C2 e

2x

(22) y

C1e x

C2 e 4 x

1 e 18

(23) y

C1e x

C2 e 3 x

e x co 2 x

(24) y

C1e x

C2 e

(25) y

e x C1 cos 2 x

(26) y

C1e

x

x

7 10

x

C4 sin x

C2 sin 2 x

1 x e 4

e 2 C2 cos

3 x 2

5 x 2

x2

21 8

in 2 x

C3 cos x

x x

2 e 5

2x

C3 sin

1 sin x cosh x 5 1 x xe cos 2 x 4

3 x 2

1 xe 6

x

1 e 26

x

2 siin x

3 cos x (27) y

C1 cos 2x

C2 sin 2x

e2 x

3 7 sin x 4 cos x 65 1 12 cos 3 x sin 3 x 145

Chapter 2: Ordinary Differential Equations (28) y

C1e x

(29) y

B ex

Ax A

(31) y

Ax

B e2 x

(32) y

Ae x

Be

(33) y

C1 x

(34) y

A cos x

(35) y

A cos 2 x

(3)

C2 cos 2 x

B

(6) y

A x

x

(7) x

A cos t

(8) y

Ax 4

(9) y

A

(12) y (13) y

1 x4

(14) y (15) y (16) y

B sin t

2

B x3

x

2

3 sin l 40

B

in 2 lo x

y

4

A lo x

x

x .

x

1 25

lo x

os lo x

C

x

D sin l

B

x

B x

3

B

x x

1 2 sin 2 l 20 x

2

2

4 lo x

C lo x x

e

7 os 2 lo x

1 sin 3 l 72

B sin 3 l

B

1 20

2

x

4

2

2

16 25

x

x

78 625

lo x

3

A lo x Ax 3

1 100

Cx 2

x

x A lo x 3

3 l 2

C sin

1 x 7 1 9x

C lo x

B x

A lo x

x

1 4 x lo x 9

A cos 3 l

lo x

2 x cos 2 x

(4)

3 l 2

x2 B

3

x sin 2 x 64

3t

B cos

B x5

Ax

C4 sin 2 x

(2) y

e

Bx n 1 .

(11) y

2 sin x

x sin x.

x

Ax n

A lo x

C3 x

2 11 cos x 125

4 cos x

x 2 cos x

sin x 2

1 24 x cos 2 x 9 x 2 26 sin 2 x . 27 B sin 2 x cos 2 x l sec 2 x tan 2 x .

(5) y

(10) y

cos x

B sin x

1 y

4

cos x

1 1 2

2 cos x .

2 cos x

x 3 sin x 25

x

Exercise 2(c) (1) y A lo x 2

e x x sin x x sin x 2

x

(30) y

Be

1 sin x 20

1 10 cos 5 x 11 sin 5 x 884

C2 e 3 x

2.85

6

D

x

cos 2 l

x

x2

x

4

2

2

t

2.86

Engineering Mathematics II

(17) y

1 A cos 2 l x

(18) y

A 3x

2

(19) y = A (20) x

x

2

B 3x

1 3x 108 x

2

2

x + 1) + B 2t

Ae

Be

y

2 Ae 3

(21) x

5 Ae 3

2t

2

2 lo 3 x

7 os lo x 2

x

1 x + 1)

5 31 1 2t t e 14 196 12 1 9 1 2t Be 7t t e 7 98 6

7t

4B t e 3

2t

x 4 in lo x 65

B sin 2 lo x

1 t e 3

y Ae 2t Bet (22) x = 2 cosh t; y = sin t – 2 sinh t (23) x

A Bt et

y (25) x y (26) x

A B cos t B cos t

t 2

C sin t

C sin t

A cos 2 t

C e 3

At At

Aet Aet

y

1 3

9 2t e 10

6 2t e 10

B sin 2 t

C cos 3 t

B cos t

Ct

B sin t

Be

t

Be

1 t e 4 sin t 3 cos t 25 1 t D cos t e 3 sin t 4 cos t 25

Ct

C cos 3t t

5

A x 1 ex

4 2 5 37 t t . 9 9 81 5 2 4 44 D sin 3t t t . 9 9 81

D sin 3t

C cos 3t

(12) y = ex (c1

x x e 4

x + c2 + x).

1 1 cos 2t 4 D sin 3 t

D sin 3 t

D sin t

Exercise 2(d) (11) y

3t 2

1 2 A cos 2 t 2 B sin 2 t C cos 3 t

y

(27) x

3t 2

2 A 6 B 2 Bt et

y (24) x

Ce

Be x .

Chapter 2: Ordinary Differential Equations (13) y

Ae

x

1 x e cos x 2

Be x

2 sin x cos x

(14) y = (Ax3 + B)e–x (15) y = Axe–2x + Be–x (16) y = e–x (c1 x + c2) c1 3 1 x x lo 2 4 1 x 2x (18) y = c1e + c2x (17) y

c1

3 2 x 2

c2 x

(19) y = c1 xex + c2x + x (x – 1)ex (20) y = c1 x3 + c2x + x3 x + x2 (21) y = c1 xex + c2x2 (22) y = c1 x3 + c2x–3 (23) y= c1 cos x + c2 sin x + x sin x + cos x (24) y= c1 cos 2x + c2 sin 2x – cos 2x (25) y

1 x

cos x A B x tan x lo

os x

1 x

2 (26) y c1 1 x

c2

(27) y c1 x x3

c2 4 3 x 6 x 2

3 x 3 lo

(28) y = c1 tan x – c2 (1 + x tan x) (29) y = c1 (1 – x cot x) + c2 cot x (30) y

Ae 2 x

x2 2

B x

(31) y = c1x – c2 cos x (32) y (33) y

1 x

A cosh x

B sinh x

1

A cos x B sin x x (34) y = xex (Ax + B) 1 1 (35) y sin x A cos 2 x B sin 2 x 3 x (36) y = 2 sin x – sin x cos x – x – 1 (37) y

1 sin 2

(38) y

1 x c1

1

x 1

x x + tan 2x)

2

4

sin

1

x

1 lo 1 c1 x c12

c2

1 x 1 x

1. 6

2.87

2.88

Engineering Mathematics II

(39) x= c1 sin (y – c2) (40) y + c1

y = x + c2

(41) y = c2 ec1 x + c1 1 (42) y = c2 ec1 x + c1 (43) y = a cos x – (a + 1) 3

(44) y = c1 e( x c2 x ) (45) y2 = c1 cosh (2x + c2) (46) y3 = Ax + B B (47) y Ax x (48) y

c1 sin

1

x/ 1

2 1 x2 (49) y 1 x (50) xyex = c1 ex + c2 – x.

x2

c2 / 1

x2

c1 sinh -1 x

c2

1

Exercise 2(e) x 2 2. xy ex = x + c 1. y cos x

5. 6. 7. 8.

c

1 1 2 x sin x x cos x 4 4 1 1 cos ax l cos ax y c1 cos ax c2 sin ax x sin ax. a a2 y = ex (c1 cos x + c2 sin x) – ex cos x x + tan x) y = (c1x + c2) e3x – e3x x. c1 c2 1 2 1 lo x y x x x 2 12 x2 1 2 8 lo x 12 y c1 x 4 c2 x 13. x

3. y 4.

sin 2x 4

c1 cos x

c2 sin x

Chapter

3

Laplace Transforms 3.1

INTRODUCTION

The Laplace transform is a powerful mathematical technique useful to the engineers and scientists, as it enables them to solve linear differential equations with given initial conditions by using algebraic methods. The Laplace transform technique can also be used to solve systems of differential equations, partial transform, we shall discuss below the properties of Laplace transforms and derive the transforms of some functions which usually occur in the solution of linear differential equations.

If f (t) is a function of t

t

e

st

f (t ) dt

Laplace

transform of f(t), provided the integral exists. Clearly the integral is a function of the parameters s. This function of s is denoted as f (s) or F(s) or (s). Unless we have to deal with the Laplace transforms of more than one function, we shall denote the Laplace transform of f (t) as s). Sometimes the letter ‘p’ is used in the place of s. The Laplace transform of f (t) is also denoted as L{f (t)}, where L is called the Laplace transform operator. Thus

L f (t )

(s)

e

st

f (t ) dt.

The operation of multiplying f (t) by e and integrating the product with respect to t Laplace transformation. The function f (t) is called the Laplace inverse transform of s) and is denoted by L { (s)}. Thus

f (t) = L {

s)}, when L {f (t)} =

s)

Engineering Mathematics II

3.2

Note s

-

form. 2. Laplace transforms of all functions do not exist. For example, L (tan t) and 2

L (e t ) for the existence of Laplace transform of a function f (t): Conditions for the existence of Laplace transform If the function f (t for t t (ii) of the exponential order, then L{ f (t)} exists.

Note f (t a t b (i) f (t) is continuous at every point inside each of the sub-intervals and (ii) f (t t approaches the end points of each sub-interval from the interior of the sub-interval. f (t) is said to be of the exponential order, if | f (t M e , for all t some constants M and or equivalently, if lim{e t f (t )} a finite quantity. t

Most of the functions that represent physical quantities and that we encounter in differential equations satisfy the conditions stated above and hence may be assumed to have Laplace transforms.

3.2

LINEARITY PROPERTY OF LAPLACE AND INVERSE LAPLACE TRANSFORMS

L{k f (t) ± k2 f2(t)} = k L{f (t)} ± k2 L{ f2(t)}, where k and k2 are constants. Proof: L{k f (t ) k2 f 2 (t )}

{k f (t ) k2 f 2 (t )}e k

f (t ) e

st

dt

k2

st

dt f 2 (t ) e

k L{f (t )} k2 L{ f 2 (t )}. Thus L is a linear operator. L{k f (t)} = kL{f(t)}, where k is a constant.

st

dt

3.3

Chapter 3: Laplace Transforms If we take L{f (t)} = (s) and L{ f2(t)} = 2(s), the above property can be written in the following form. L{k f (t) ± k2 f2(t)} = k

(s) ± k2 2(s).

L {k

(s) ± k2

(s)} = k · f (t) ± k2· f2(t) = k ·L { (s) ± k2·L { 2(s)} Thus L is also a linear operator L {k

s)} = k L { (s)}, where k is a constant.

2

Note L{f (t) . f2(t

L{f (t)} . L{f2(t)} and

L { (s) .

L ( (s)} × L { 2(s)}

(s

2

2. Generalising the linearity properties, n

n

we get

(i)

L

kr f r (t )

kr L{ f r (t )} r

r

n

n

(ii)

L

kr

r (s)

kr L { r ( s )} r

r

a linear combination of elementary functions whose transforms are known. expressed as a linear combination of elementary functions whose inverse transforms are known.

3.3

LAPLACE TRANSFORMS OF SOME ELEMENTARY FUNCTIONS L{k}

k s

k is a constant,

L{k}

ke

k

e

k ( s k. s

st

dt

st

s )[ e

st

as t

, if s

]

Engineering Mathematics II

3.4 In particular, L L 2. L{e

at

=

L( ) =

s

.

s }

s

a L{e

, where a is a constant, at

}

e

st

e

at

e

(s

a )t

dt

(s

s Inverting, we get L 3. L{e at }

s

Changing a L

s

a

a)

a e

a

(

e (s (s

a )t

a)

) , if (s + a

, if s

at

dt

a.

.

, where a is a constant, if s a

s > a.

a in (2), this result follows. The corresponding inverse result is

s

4. L(t n ) =

e at

a (n sn

)

, if s

L(t n ) =

n

e

e

sn

st

x

t n dt

x s

e

n

x

dx , on putting st = x s

x n dx

3.5

Chapter 3: Laplace Transforms (n sn

) , if s

n+

>

In particular, if n is a positive integer, (n

n!

)

n! , if s sn

L(t n )

n!

Inverting, we get L L

s

n

n is a positive integer.

s n!

t n or

n

tn

L

Changing n to n If n

L

s

n

In particular, L(t ) 5. L(sin at )

(n)

(n

)!

t n , if n is a positive integer.

tn L

or

s2

sn

t.

s2

a s2

a2

L(sin at )

e

st

e

st

s2

sin at dt

s sin at

(

a2

s s2

e

a2

st

a cos at )

sin at

a s2

a2

a s2 [ e

sin at and e

Inverting this result we get L

a2 cos at tend to zero at t a s

2

a2

sin at .

s

e

st

cos at

Engineering Mathematics II

3.6 s

6. L(cos at )

s2

a2

L(cos at )

st

e

cos at dt

st

e s2

s cos at

(

a2 s

s

2

a

s s

2

st

e

2

a sin at ) a

cos at

s

2

a2

e

st

sin at

, as per the results stated above.

a2

s

Inverting the above result we get L

s

2

cos at .

a2

L(cos at + i sin at) = L(eiat) s ia s s2 s

i.e., L(cos at ) + iL (sin at )

s

2

a

ia a2 a

i

2

, by result (3).

s

2

a2 s

Equating the real parts, we get L(cos at )

s2

Equating the imaginary parts, we get L(sin at ) 7. L(sinh at ) =

, by linearity property.

a2

. a

s

2

a2

a s2

L(sinh at )

a2 L

2

2

(e at

L (e at )

2 s

a

a s

2

a2

at

e

)

L (e

s

at

a

, if s > |a|

) , by linearity property. , if s > a and

3.7

Chapter 3: Laplace Transforms

L(sinh at

iL(sin i at) [

sin

i sinh ]

ia , by result (5) s i a

i

2

2

2

a s

2

a

2

a

Inverting the above result we get L 8. L(cosh at )

s2

a2

sinh at.

s s

2

a

L(cosh at )

2

L

2

2

ea t

L ea t

2 s

2

a2

at

at

L e

a

s s

e

s

, by linearity property,

, if s > |a|.

a

, if s > |a|.

L(cosh at) = L(cos iat) [ cos i = cosh s

2

s

2

]

s , by result (6) i2a2 s a2

Inverting the above result we get L

3.4

s s

2

a2

cosh at.

LAPLACE TRANSFORMS OF SOME SPECIAL FUNCTIONS

The function f (t )

, when t

a

, when t

a, where a

function and is denoted by ua (t) or u (t

a)

, is called Heavyside’s unit step

Engineering Mathematics II

3.8

when t

In particular, u (t ) Now

, when t

L{ua (t )}

st

e

ua (t ) dt

a

e s t ua (t ) dt

e s t ua (t ) dt a

st

e

dt ,

ua (t)

a

e

st

e

S

as

, assuming that s > .

S a

In particular, L{u (t )}

, which is the same as L

S

Inverting the above result, we get L

e

as

s

ua (t ) .

lim f (t ) , where f (t h

h , when a h 2 , otherwise

f (t )

t

a

h 2 is called Unit Impulse

Function or Dirace Delta Function and is denoted by Now L{ a (t )}

L lim f (t )

,where f(t

h

lim L{ f (t )} h

e

lim h

a

e

h

a

h

st

h

h 2

e h

f (t ) dt

h 2

lim

lim

st

dt

st

a

h 2

s

a

h 2

(t) or (t

a

a).

3.9

Chapter 3: Laplace Transforms h 2

s a

lim

e

e

lim

es h / 2

e sh

h

2 sinh as

e

lim

sh 2

s cosh

f (t )

u a

h 2

(t )

h

u

h 2

a

, when t

(t )

as

e

as

u

h 2

a

when

t

a

h and hence u a 2

when

a

h 2

t

L{f (t )}

a

h

a

h 2

s

lim cosh

sh 2

h

(t ) ,since

h 2

(t )

h s 2

a

e

h 2

(t ) = ,

(t )

u

h 2

a

L u a h s 2

a

e

s

s

2 sinh e

h

(t )

h 2

L u

h

lim

h and u a 2

a

as

sh

sh 2

s h/2

sh

h

e

h 2

sh

h

as

s a e

h 2

(t )

sh 2 , by L’ Hospital’s rule.

e

as

Engineering Mathematics II

3.10

2 sinh L

a

(t )

lim L f (t )

as

e

h

lim h

sh 2

sh

e as Inverting the above result, we get L {e s}

3.5

(t). When a

(t).

a

PROPERTIES OF LAPLACE TRANSFORMS

If L{ f(t)} =

(s), then L{ f (at )}=

s t and L f a a

a

= a (as ) .

Proof (s )

L{ f (t )}=

and L{ f (at )}=

e

e

s

a

a

x a

st

st

f (t ) dt

f (at ) dt [ f(at) is a function of t]

f (x)

d x , putting x = at and making necessary changes. a

e

(s / a ) x

f (x) dx

e

(s / a ) t

f (t ) dt , changing the dummy variable x as t.

s in (2).

Thus

e

L{ f (at )}

Changing a to

a

(2)

s in the integral a (s), that in (2) is equal to (s/a).

(s a)

in (3) or proceeding as in the proof given above, we have a

L{ f ( t a )}

a (as ) .

(3)

Chapter 3: Laplace Transforms

If L{ f (t)} =

L{e f (t)} =

(s), then

at

L{e f (t)} =

and

3.11

(s + a) a).

(s

Proof L{ f (t )} and

L{ e

at

f (t )}

e

st

e

st

f (t ) dt [e

at

(s )

f (t )]dt [ e

e (s

(s

a )t

f (t) is a function of t] (2)

f (t ) dt

a)

Changing a a in the above result, we get L{e at f (t )}= (s a) .

Note lowing way: L{e

at

f (t )}= (s a ) =[ (s )]s

s +a

L{ f (t )}s ‘s

s +a

s + a’ means that s is replaced by (s + a).

which is e , we ignore e , a function of s and change s into (s + a) in it. Similarly, L{eat f (t)} = L{f (t)s s a 2. The above property can be stated in terms of the inverse Laplace operator as follows: If L { (s )}= f (t ) , then L { (s a )}= e at f (t ) From this form of the property, we get the following working rule: L { (s

a )}= e

at

L { (s )}

s + a), inverse transform of the corresponding function of s and multiply it with e . Similarly, L { (s

a )}= e at L { (s )} .

Engineering Mathematics II

3.12

3. The above property is called so, as it concerns shifting on the s-axis by a (or

a), i.e., replacing s by s + a (or s

).

The second shifting property, that follows, concerns shifting on the t-axis by a i.e., replacing t by

If

L{ f (t)} =

(s), then L{ f (t

a)ua(t)} = e

(s),

where a is a positive constant and ua(t) is the unit step function. Proof L{ f (t

a )ua (t )}=

st

e

f (t

a )ua (t )dt

a

e

st

f (t

a )ua (t ) dt

e

st

f (t

a )ua (t ) dt

a

e

st

e

s (x a )

e

as

e

e

as

(s )

f (t

a ) dt ,

ua(t).

a

st

f (x) dx , putting

and effecting consequent changes

f (t ) dt , changing the dummy variable x as t.

Note L{f (t a) ua(t)} = e as L{f (t)} 2. The above property can be stated in terms of the inverse Laplace operator as follows: If L { (s)} = f (t), then L {e as (s)} = f (t a) ua(t). From this form of the property, we get the following working rule. L {e

as

(s)} = L { (s)}t

t

one of which is e , we ignore e , factor as a function of t, replace t by (t

a

· ua(t). a) in it and multiply by ua(t).

WORKED EXAMPLE 3(a) Example 3.1 Find the Laplace transforms of the following functions:

3.13

Chapter 3: Laplace Transforms

(i)

(ii)

2

t

f (t )

for t for

,

, e k t,

f (t )

t

t

for

,

for t

sin t ,

for

a a t

f t

(iii)

for t

,

f (t )

t, 5,

L{f (t )}

e

st

f (t ) dt

L{f (t )}

e

st

odt

(iv)

(i)

for t

e

2

t

2

4

2

f (t) = (

2

e

st

s3

s

fo 2

dt

e st s2

t

2 e s3

s

t

st

e

st

, u (t )

Thus

2

t

s 2 e s3

t

t

for

t

for t >

u (t)

L{f (t)} = e · L(t2), by the second shifting property. 2 e s3

s

a

(ii)

L{f (t )}

e

st

e kt dt

e

st

dt

a

e

s k t

s k

a

s k

{

e

a s k

}

Engineering Mathematics II

3.14

Consider e kt { Thus

e k t, ,

ua (t )}

f (t) = e k t

kt

= e kt

k(

ak

st

e

st

s2

e

L{e kt } , by the second shifting property

as

a s k

}.

sin t dt

s sin t

2

s2

· ua(t)}

s k

{

e

L{f (t )}

a)

a s k

e

s k

· ua (t)

· L{e k (

e ak e

s k

(iii)

ua (t) a) + ka

L{f (t)} = L(e k t)

s k

for t a for t a

s

e

2

cos t

4

(iv)

L{f (t )}

st

te

dt

st

5e

dt.

4

e

t

t{ Thus

f (t)

e

u4 (t )} ( (

4s

e

4

e st s2

s s

s

st

s2

4s

s2

e

u4 (t )

e

5

e

st

s

4s

s2

s

e

4 4s

4s

s2

t, 5,

5) u4(t) 4) u4(t) + u4(t)

for t 4 for t 4

3.15

Chapter 3: Laplace Transforms 4s

L (f (t )} L(t ) e s2

s2

L(t ) 4s

e

s

L{u4 (t )} e

4s

.

Example 3.2 Find the Laplace transforms of the following functions: (i) (i)

t t

cos t . t

, (ii) sin t , (iii) L

t

L(t

t

) 2 L(t )

s 2

(ii)

2 s s

s

s

t

3!

3

....

5!

3!

5!

t 5 2 ....

L (t 3 2 ) 52

32

52

32

3! s

5!

L(t 5 2 ) .... 72

....

5! s 7 2

3 3! 2 2

2

s

5 3 5! 2 2 2

2

2 s3 2 2 s3 2 (iii)

! 4s

t

cos t t t

2

t

2!

t

2!

2! 4 s

s

e

t

n n

5

t

32 s

n

.

t3 2

L(t )

s

and

3!

t

L sin t

s3 2

s

t sin t

32

2

4!

4!

4

.....

t 3 2 .....

......

s2

....

Engineering Mathematics II

3.16

L

cos t t

/

L(t

)

2!

L t

32 s

2! s

s

L t3 2

4!

52

32

.....

4! s 5 2

.....

4! 2 2 s 2

2! 2s

.....

2

s s

! 4s

.....

2! 4s

s

e

Example 3.3 Find the Laplace transforms of the following functions: (i) (t3 + 3e2t t)e te t)3; 2t 3 (iii) e cosh 2t; (iv) cosh at cos at; 3 t (v) sinh sin t 2 2 (i)

L{(t3 + 3e2t 5 sin 3t)e t} = L(t3 + 3e2t

t)s

3! s4

3

6

(ii)

s

2

)2

(s

L(e 2t cosh 3 2t )

8 8 =

L{e

s

s2

s

L(l + te t)3 = L(l + 3t e t + 3t2 e

s

9

s

3 4

= L(l) + 3L(t)s

iii

3

5

s 2

s

s

2t

L{e 4t

8 s 4

(e6t

+ t3 e t)

+ 3L(t2)s

s

(s

6 2)3

L e

3e 2t

3 3e 3 s

t

2s

e 2t

2t

e

4

e

2t

2 2t

8t

}

3 s

+ L(t3)s

6 (s 3) 4

3e

4t

s+2

s 8

e

6t

)}

3

s+3

3.17

Chapter 3: Laplace Transforms

(iv)

L cosh at cos at

L cos at

2

s 2 s 2 s

ea t

L

L cos at

s a

a

s a

2

s

s a 2a 2 2as

2

a) s 2

(s

cos at

2

s

2

a

at

e

s

2a 2

2

s

s

a

a a

2

a2

s a 2a 2 2as

2

2as

s2

2a

a) s 2

(s 2 2

2a 2

2as

4a 2 s 2

s3 . s 4 4a 4

(v)

3 t L(sinh sin t) 2 2 L

2

et 2

L sin

t 2

e 2

sin

3 t 2

3 t 2

s

s

2

3 2 2

2 s 3 4 s2

s

3 2 s4

s s2

3 t 2

L sin

s

s

2

3 2 34

s

s2

s

2

34

Example 3.4 Find the Laplace transforms of the following functions: (i) eat cos (bt + c); (ii) e 2t cos2 3t; (iii) et sin3 2t; (iv) e t sin 2t cos 3t; (v) e3t sin 2t sin t. (i) L{eatcos (bt + c)}= L{cos (bt + c)}s s a = L{cos c cos bt c sin bt}s s a (s a ) cos c b sin c (s a ) 2 b 2 (s a ) 2 b 2

Engineering Mathematics II

3.18 (ii)

L{e 2t cos2 3t}= L(cos2 3t)s L

s + 2

cos 6t

2

s

s 2 s (iii)

2

s

L

36

sin 6t

4

3 2 4 s2 4

(iv)

2

s

3 sin 2t 4

3 2 s2

2

2

2

L{et sin3 2t} = L(sin3 2t)s

s

s

s

36

s

6 4 s2 s2

2s 5

2 s 37

L{e t sin 2t cos 3t} = L(sin 2t cos 3t)s L

sin 5t sin t

2

s

s

s

s

5 2 s2 2 s (v)

s2

25

s

5 2 s 26

2

s2

L{e3t sin 2t sin t} = L(sin 2t sin t)s L

2

s

2s

2

s

cos t cos 3t s

s

s 3

s

s2

s2

9

s 3 2 2 s s

s

s 3

s 3 s s 8 2

Example 3.5 Find the Laplace transforms of the following functions: 2 (i) (t · u (t); (ii) sin t · u (t); (iii) e · u2(t) 2 (i) L{(t u (t)} = e s · L{t2}, by the second shifting property 2 e s3 (ii)

s

L{sint · u (t)} = L{sin (t = {sin (t

) · u (t)} ) · u (t)}

3.19

Chapter 3: Laplace Transforms s

e

e s2 (iii)

L(sin t) s

L{e t u2(t)} = L{e 6

e

t

6

2s

e

u2(t)} 3t

L {e

2s

e

}=

6

.

s 3

Example 3.6 Find the Laplace transforms of the following functions: (i) t sin at; (ii) t cos at; (iii) te t sin 3t (i)

L (t sin at) = L{t I.P. of e iat} = I.P. of L{t e iat} = I.P. of L(t)s

s

ia

I.P. of s2

I.P. of

I.P. of

2

s ia

a2

s2

a

s ia

i

2 2

s2

a2

2 2

2 as s2

a2

2

2 as s (ii)

2

L{te iat}

L(t cos at

(iii) L(t e

t

2

a2

s2

a2

s2

a2

2

t

sin 3t) = I.P. of L {t e

e i3t} i3)t

= I.P. of L {t · e = I.P. of L(t)s

}

(s +

I.P. of s

4 i3 s

2

4

i3)

2

6 s s

4

4 2

2

9

i6 s

9

I.P. of s

I.P. of

4

2

2

9

s

4 i3

s

4

4

2

2 2

9

Engineering Mathematics II

3.20 Example 3.7

L(sin t) and L(cos t), L(cos 3t). (ii) Given that L t sin t

L t cos

L sin

2s

and L t cost

2

s2

s2 s2

2

L(t sin at) and

t . a

(i) L sin t

s2

L sin

t 2

L f

2

2

s

t a

a L f (t )

s

as

, by the charge of sccale property

2 s2 s

L cos t

s

2

s/3

L cos3t

2

3 s/

L f at

a

L f t s

s

2

9

2s s

a

, by the charge of scale property

.

(ii) Given that L t sin t

L (t sin at )

s a

s

2

2

L(at sin at ) 2s/ a

a a

2

s / a2

2 as s2

a2

2

2

, by change of scale property

t and 2

3.21

Chapter 3: Laplace Transforms s2

Given that L t cos t

2

s2

L t cos

t a

a L

t t cos a a

a4 s2

a2 2

a2 s2

, by change of scale property

2

2

as =

2

as

a a

s2

a2

or

2

s

.

2

a2

Example 3.8 Find the inverse Laplace transforms of the following functions: (i)

(v)

5/ 2

s

2

s

4

s2

4

e

e s ; (ii)

2s

2s 3

; (iii)

3

se

3a 4 s

s

s 3

5

; (iv)

s2

a2

e

bs

;

4s

e

(i) From the second shifting property, we have L {e as (s)} = L { (s)t t a ua(t) L

e

s

s

2

L

5/ 2

Now L s

2

s t

e

5/ 2

[ L { (s + 2) = e e

2t

2 L 2t

u (t )

5/ 2 t

t

s5 / 2 L { (s

t3/ 2

L

5/ 2 e

2t

t2 e

3

e

4

s

s

2

5/ 2

3

n

n n

n

/

3

4

L

t3/ 2

3 2 2

tn

sn

t

2t

/

3/ 2

e

t

u (t )

(2)

Engineering Mathematics II

3.22

when t

, or

4 3 (ii) L

Now L

t

e 2s (2 s 3)3

L

(2 s 3)3

L

8 8

e2

s (s 3)5

Now L

3

2!

3

(t 2 )

(iv) L

s

24

(3a 4s ) e s2 a2

2) 2 u2 (t )

(t

s

u (t ) t

t

e

3 s

5

s4 3!!

t3

e3t (4t 3

24

e

(2)

(s 3) + 3 (s 3)5

L

e 3t

(s 3)5

t

e2 t2

(s 3)5

e tL

s

t2

s

L

e 3t L

L

t 2

s3

t

e2

se s (s 3)5

(iii) L

u2 (t ) t

t

e2 L

e 2s (2 s 3)3

, when t

(s 3 2 )3

3

8

)

(2 s 3)3

3

L

(t

)3/2 e

(t

bs

)

{ (t

L

s5 3

4!

t4

3t 4 )

)

(2)

) 4 } u (t )

(t

3a 4 s s2 a2

ub (t ) t

t b

3.23

Chapter 3: Laplace Transforms 3a 4 s s2 a2

Now L

a

3L

s2

= 3 sin at (3a 4s ) e s2 a2

L

s

s 4 s2 4

L

s 4 s2 4 s

s

2

a2 (2)

b) 4 cos a (t

L

L

s2

at

[3 sin a (t

( s 4) e s2 4

(v) L

Now

bs

s

4L

a2

4

b)]ub (t )

u4 (t ) t

t 4

2L

2

2 s

4

= cosh 2t + 2 sinh 2t ( s 4) e s2 4

L

4s

{cosh 2(t

4) 2 sinh 2(t

(2)

4)} u4 (t )

Example 3.9 Find the inverse Laplace transforms of the following functions: (i)

(v)

e s ; ( s 2) ( s 3) (s s2

(i) L

(ii)

e 2s e ; (iii) s (s 2 ) s (s 2

s

2

4)

;

(iv)

e s ( s 2) ( s 3)

(s

2) (s

L

( s 2) ( s 3)

s

3

;

3)

, we resolve

u (t ) t

t

(s

2) (s

fractions and then use the linearity property of L operator. A B Let ( s 2) ( s 3) s 2 s 3 A (s + 3) + B (s

By the usual procedure, we get A

L

s

3s

)e s . 2s 5

L

Then

e 2

s 2 s 3

L

s 2

5

, B

s 3

5

3)

into partial

Engineering Mathematics II

3.24

L

5

(e 2 t

5

L

e s ( s 2) ( s 3)

L

2

Now L

s 2 (s 2

)

s 2 (s 2

)

e

(t

e

5

e 2s s (s 2 )

(ii) L

s 2

)

u (u

)

L

e 2s s 2 (s 2 ) e s (s 2

(iii) L

Let Then

{(t

s 3

(2)

(t

e

)

u

u (t )

u2 (t )

)

t 2

t

s2

u

L

s2

L

)

s 2 (s 2

=t

L

3t

5

s2

s2

t

(2)

2) sin (t 2} u2 (t )

s

s (s 2

L

4)

s (s 2

A s

4)

u (t )

4)

t

t

Bs C s2 4

A(s2 + 4) + s (Bs + c) A

L

s (s 2

4)

4

,B

4 s

L

4

(

4 4

and C s

s2

cos t )

4

(2)

3.25

Chapter 3: Laplace Transforms s

e s (s 2

L

4

4)

Now

s

L

2

3

s

3

s2

t

3

3

e s2

L

L

e

t

e

2t

s2

(s

s2

u3 (t )

s

3

2) 2

32

t

t 3

, by the shifting property

32 3

L

s

2

32 (2)

sin 3t

3s

s

3

3 (s s2

(v) L

e

2 ( t 3)

)e s 2s 5

s

2

s

2s 5 t

s

2

2s 5

(s

L

e

sin 3(t 3) u3 (t )

L

s

Now L

L

s

e

L

)} u (t )

3s

e

L

(iv)

cos (t

{

(s

)

s

2

t

) 2

s

L

u (t ) t

22

2

, by the shifting property

= e t cos 2t

L

(s s2

)e s 2s 5

e

(t

)

(2)

cos 2(t

) u (t ) .

Example 3.10 Find the inverse Laplace transforms of the following functions: (i)

(iv)

s2 s 2 2 s 2 5s 2 ; (ii) ; (iii) (s s ( s 3) ( s 2) ( s 2) 4

2

s (s

2

) (s

2

)

; (v)

s )2 (s 2

s (s

2

) (s

2

) (s 2

9)

.

)

;

Engineering Mathematics II

3.26

s2 s 2 , we resolve the given function of s into partial s ( s 3) ( s 2)

L

fractions and then use the linearity property of L operator. Let

s2 s 2 s ( s 3) ( s 2)

A s

B

C

s 3

s 2

A

L

3

4

, B

s2 s 2 s ( s 3) ( s 2) 4 3

(ii) To resolve

and C

2 . 5

/ s

/ 5 s 3

L

3t

e

2/5 s 2

2 2t e . 5

2 s2 5 s 2 into partial fractions, we put s ( s 2) 4

x, so that

s = x + 2. 2 s2 5 s 2 ( s 2) 4

Then

2) 2

2 (x

x2

2 ( s 2) 2 2s

2

(s

5s 2)

2 4

2L

2e e e

2t

2t

2) 2

3x 2 x4 3 2 x3 x 4

x2

L

5(x x4

( s 2)3

( s 2)

2t

L

2t 2t

s

2

L

2

t2

t

2t

e

2

2!

( s 2) 4

3!

2

3

t3

t

3

(s

L

2)

s

3

L

3

e

2t

(s

L

s

2)

4

4

3.27

Chapter 3: Laplace Transforms

6

e 2t ( t

s

(iii) Let

(s

s

s2

B(s2

A

B

(s

e 2 t e 2

u (u

s2 s2 ) (u

Now let

D /

s2 2

2

Thus

2

s2 L

s2

sin t.

) (u

A u

)

B

s2 s2

9 s s2

t

8

s2

s2

s2

9

s2

s2

s2

9

8 /

and C

72

/ 2 s2 9

s2 sin 3t .

9

into partial fractions as shown in (iv).

(u

A u

u

, B

sin t

s2

s2

9 / s2

L

s2

C

u

A

(v) To resolve

=s

into partial fractions and then replace u by s2.

u (u

L

2

/ )2

(s

L

t

D

is a function of s2, we put s2 = u, we resolve

s2 )

t

Cs s2

)2

Cs + D) (s

L

)2 s2

B (s

; C

2

s

L

t3 )

A

)2 s 2

A(s

(iv) Since

39t 2

) (u

B u

) (u

C u

9

, say.

9)

Engineering Mathematics II

3.28 A

24

s2

s2

2

2

, B

and C

s2

/ 4 s2

9

s2

s s

s

s

2

s

/ 5 4

/ s2 9

s s

2

24 s

9

s

L

.

2

s

2

24

9

s

2

s s

L

s

s

2

L

2

s

4

2

9

s s

2

4 L

24

cos t

cos 2t

s s

2

cos 3t

Example 3.11 Find the inverse Laplace transforms of the following functions: (i) (iii) (v)

(i)

s 49 s

2

28s

s3

a3

s4

s s2

L

;

;

4s 2

2

s s

(iv)

s4

2

4

s

s 49 s 2

2s3

(ii)

28s

49

2 e 7

2 7

s

2 L 7

s

2 t 7

2

2 7

s

s2 5 7

s

2 L 7

L

2 7

2

s L s2

2

3 7 3 7

2

3 7 3 7 3 7

2

4 s 7

s s

;

5 7 49

2

;

9

3.29

Chapter 3: Laplace Transforms 2 t 7

2 e 7

2s3 4s 2 s s 2 (s 2 s 2)

(ii) Let As(s2

s + 2) + B(s2

A

L

3 cos t 7

2s3

4

4s 2

s2 s2

, B

2

4

2

t 2

4

A s

L

L

t 2

s

2

9 s 4

2

t/

e L

t 2

(iii)

Let

s3

a3

3

3

A s

a

s

7 2

L

2

5 7 4

s 7 2

2

9 7 t cos 4 2

as

s2

2

s

2

5 7 7 t sin 4 2

a2

Bs C s as a 2 a)(Bs + C

3a 2

,B

3a 2

9 s 4 4 2 s s 2

35 8

2

9 4

s

4

2

a

A(s2 + as + a2) + (s

A

et / 2

a) s 2

(s

L

2

s

4

Cs D s2 s 2

9 and D 4

,C

s

4

B s2

s + 2) + (Cs + D)s2 = 2s3 + 4s2

s s

3 sin t 7

and C

2 . 3a

7 2 7 2

2

Engineering Mathematics II

3.30

L

s3

a3

L

3a 2

3a

s

e at

2

L

a

3a

s

L

2

e

3a 2

3a 2

3a (iv)

s4

4

e at

2

s4

4s 2

s2

2s

As

Let s

4

4

s

2

3 at 2

4s 2

s2

2 s2

2s Cs

s

2

2

2

3a 2

2

cos

4

B 2s

(As + B) (s2

at 2

e

3 a 2

L s

2

3a 2

s

at 2

e

2a 2

a 2

s

at

2 3aa a2

s 3a 2 2 s as

3 sin

2

2

(2s) 2

2 D

2s

2

s + 2) + (Cs + D)(s2 + 2s

By the usual procedure, we get A

L

s

4

4

B

8

L

8 8

s

L e

t

C s

2

4 2s

(s (s

) )2

L

s s2

2

D

8

L

8

s

L

et L

2

s 2s

(s (s s s2

3 at 2

4 2

) )2

3.31

Chapter 3: Laplace Transforms t

e

8

et

s s2

et

sint

2

e

cost

cos t sinh t )

s s4

sin t )

t

2

(sin t cosh t

4 s4

t

e

4

(v)

sin t ) et (cos t

(cos t

s

s2

s2

s

2

2

s2

s s2 Let

s

s s2

4

As s

s2

s

2

B

Cs D s2 s

s

(As + B) (s2 s By the usual procedure, we get

L

s

4

s s2

Cs + D) (s2 + s

B

A

2

2

L

s

2 s2

D

, C

s

L

2

2

s

3 2

2

et / L

2

s 2 2

3

2 3

sin

et / 2

e

t/2

L

s2

2

2

e

t/

s s

2

2

t sin t 2

.

s

L

2

L

3 t nh . t sin 2 2

2

.

s s

3 2

2

3 t 2

sin

L

3 2

2

s2

Example 3.12 (i) If L

2

s

2

3 2

2

s

2

a2

2

Engineering Mathematics II

3.32 s2

(ii) Given that L

(iii) Given that L

4

s2

4

s2

4

L

t cosh 2t

2

(sin 2t

2

s2 s2

2t cos 2t )

2

L

s2

9

2

.

(i) By change of scale property, L{ f (at )}=

s a

a

L { (s/a)} = a L { (s)}t s

L

Given

s

s/a

L

s2

s s

2

a

a

at sin at 2

a2 t sin at 2

s

a L

L

a

2

2

s a2 i.e.,

t sin t 2

2

2

at

2 2

t sin at . 2a

2 2

(ii) By change of scale property, L{ f (t/a)} = a (as) L { (as )}= L ( (s )}t a Given

s2

L

s L

i.e.,

4

2

4 4

2

(2s ) 2 4 [(2 s ) 2 4]2 L

s2 s

L

2

2

s2 s2

2

(2)

t /a

t cosh 2t t cosh t , by 2 2 t cosh t 4

t cosh t .

(2)

Chapter 3: Laplace Transforms

(iii) Given L

s2

L 2s 3

i.e.

L

L

2

2

(sin 2t

2

4

3 L 2

4

s2

9

s2

9

2

2

s2

3.33

2t cos 2t )

4

, by

2 t

(2)

3 t 2

3 (sin 3t 3t cos 3t ) 32

54

(sin 3t

3t cos 3t )

EXERCISE 3(a) Part A

2. State the conditions for the existence of Laplace transform of a function. 3. Give two examples for a function for which Laplace transform does not exist. 4. State the change of scale property in Laplace transformation. 6. State the second shifting property in Laplace transformation. 7. Find the Laplace transform of unit step function. 8. Find the Laplace transforms of unit impulse function. 9. Find L{ f (t)}, if f (t )

e 2t , for t , for t

L{ f (t)}, if f (t )

, for t t , for t , for t

L{ f (t)}, if f (t )

sin 2t , fo , for t

L{ f (t)}, if f (t )

cos t , fo t , for t

t

2

Engineering Mathematics II

3.34

, L{ f (t)}, if f (t )= cos t

L{ f (t)}, if f (t )= t and L

L

2 , for t 3

sin t , fo t,

t

for

2 3 2 3

t

for t

t

L{(t u3(t)} and L{u (t) sin 2 L{t u2 (t)}

(t

Find the Laplace transforms of the following functions: at + b)3 t+ ) 2 3 3t 2t 22. sin 2t cos t 23. cos 3t cos 2t 24. sinh3 t 25. cosh2 2t 2

26. (t 28.

e

et cosh 2t

2

sinh 2t

2t

27.

e

cos 3t

29.

sin t sinh t

e3(t + 2) t2 cosh t Find the Laplace inverse transforms of the following functions: 32. e /s2 (a e s e s)/s 34.

e 2s s 3

35. s

e

36.

se s s2 9

2

s Find f (t) if L{ f (t)} is given by the following functions: 37. 39.

(s

45.

38.

s2

2s s3

3

s 2s 3 2s s2

43.

)

3/2

4

3 4

s (s + a ) s2

s 3 s

s 42. 44.

2

5

s 6 s2 9 s2

2 s +5

2 sin 3t 3

3.35

Chapter 3: Laplace Transforms Part B Find the Laplace transforms of the following functions: at at )/ t 46. e3t(2t + 3)3 47. e ( t 3 48. e sinh t 49. sin at cosh at 3 cos 2t et sin t)2 et 52. 54. 56. 58.

e kt sin ( t + ) et cost cos 2t cos 3t t cos 2t t e2t sin 3t

at sinh at

53. e t sin 3t cos t 55. e t sin 2t sin 3t sin 4t 57. te t cos t 4s

59. Given that L(t sin 2t ) s L(t cos 3t )

2

s2 s

2

4

2

L(t sin t)

2

L(t cos 2t).

9 9

Find the Laplace inverse transforms of the following function: s2 4 s 2 3s 5 62. 3 2 s 3s 2 s s s2 s 2 s2 63.

3s 5 64.

3

s

s s

2

s s4

65.

67.

69.

s2

2s s

77.

s

2

s

2

3

68.

s

2

s

s

2s 9 s 6 s 34 ls

s 2

s

4

6s

26

s

3

4a s

s

as

2

4

4

s

4

72.

74.

2

64 s2 s2

76.

2

2

s 2

s s

75.

2

s

2

s

73.

66.

8s 2

m bs c

2

s3 s s

4

4s s4

4 3

3 s2

9

Engineering Mathematics II

3.36

s

78. If L s

79. If L

2

2

s2

9

2

2

s

2

9

t sin 3t 6

2

a2

2

L

(sin 3t 3t cos 3t )

s 2

s

L s

54

L

3.6

t sinh t

2

s2

.

s

L s

2

4

2

LAPLACE TRANSFORM OF PERIODIC FUNCTIONS

f (t) is said to be a periodic function, if there exists a constant P that f (t + P) = f (t), for all values of t. Now f (t + 2P) = f (t + P + P) = f (t + P) = f (t), for all t. In general, f (t + nP) = f (t), for all t, when n is an integer (positive or negative). P is called the period of the function. Unlike other functions whose Laplace transforms are expressed in terms of an t function f (t) with period P can be expressed in terms of the integral of e st f (t) over P), as established in the following theorem.

If f (t) is a piecewise continuous periodic function with period P, then P

L f (t )

e

st

e

Ps

f (t ) dt. .

Proof: L f (t )

e

st

f (t ) dt

e

st

f (t ) dt

P

e

st

f (t ) dt

P

t = x + P, e

st

f (t ) dt

e

s x

P

f (x

dt dx and the limits for x

.

P ) dx

P

e

sP

e

sx

f x dx

[ f(x + P) = f(x)]

3.37

Chapter 3: Laplace Transforms

e

sP

e

sP

e

st

f (t ) dt , on changing the dummy variable x to t.

(2)

L f (t )

P

L f (t )

e

st

f (t ) dt

e

st

f (t ) dt

e

sP

L f (t )

P

e

(

Ps

) L f (t )

P

L f (t )

3.7

Ps

e

st

e

f (t ) dt

DERIVATIVES AND INTEGRALS OF TRANSFORMS

The following two theorems, in which we differentiate and integrate the transform function

s

L{f (t)} with respect to s

L{t f (t)} and L

t

f (t )

s L{tn f (t)} and L

If

tn

f t

, where n is a positive integer.

L{f (t)} =

s

then L{t f (t

' (s).

Proof : L{f (t)} =

Given: i.e.

s ...

e

st

f (t ) dt

(s ) s,

d ds

e

st

f (t ) dt

d (s ) ds

...

with respect to t and differentiation with respect to s d e ds

st

f (t ) dt

(s )

(2)

Engineering Mathematics II

3.38

i.e.

te

st

f (t ) dt

(s )

i.e.

e st [t f (t )]dt

(s )

i.e.

L{t f (t )}

(s )

n times with respect to s, we get L{t n f (t )}

(

)n

dn ds n

(s )

or

(

)n

(n )

(s ) .

Note d (s ) ds d L{ f (t )} ds

L{t f (t )}

we ignore ‘t then we differentiate this function of s with respect to s Extending the above rule, L{t 2 f (t )}

)2

(

t’, s;

d2 L{ f (t )} and in general ds 2

dn L{ f (t )} . ds n 2. The above theorem can be stated in terms of the inverse Laplace operator as follows: L{t n f (t )}

If

)n

(

L { (s)} = f (t),

then L { (s)} = t f (t). From this form of the theorem, we get the following working rule: L { (s )}

t

L { (s )}

This rule is applied when the inverse transform of the derivative of the given function can be found out easily. In particular, the inverse transforms of functions of s that contain logarithmic functions and inverse tangent and cotangent functions can be found by the application of this rule.

3.39

Chapter 3: Laplace Transforms

If L{ f (t )}

(s ) , then L

t

(s ) ds , provided lim t

f (t ) s

t

f (t ) exists.

Proof: L{ f (t)} =

Given:

i.e.

e

st

(s)

f (t ) dt

(s ) s between the limits s and

st

e

f (t ) dt ds

s

st

ds f (t ) dt

s

e

i.e.

t

(s ) ds s

st s

f (t ) dt

t i.e.

(2)

(s ) ds s

e

s s st

e

(

(s ) ds s

(s ) ds ,

) f (t ) dt s

assuming that s > i.e. i.e.

e

st

L

L

f (t ) dt t

s

f (t ) t

s

t2

f (t )

(s ) ds (s ) ds

L

t t

f (t ) (s ) d s ds

s

s

(s )ds ds s

s

Generalising this result, we get L

t

n

(s ) (ds ) n

f (t ) s

s

, we have

s

Engineering Mathematics II

3.40

Note

L

t

f (t )

L{ f (t )}d s s

t s and

we ignore

t integrate this function of s with respect to s between the limits s and Extending the above rule. We get; L

L

t2 tn

f (t )

,

.

L{ f (t )}ds ds and in general s

s

L{ f (t )}(ds ) n .

f (t ) s

s

s

2. The above theorem can be stated in terms of the inverse Laplace operator as follows: L { (s )}

If then

L

(s ) ds s

f (t ) ,

t

f (t ) .

From this form of the theorem, we get the following working rule: L { (s )}

t L

(s ) d s s

This rule is applied when the inverse transform of the integral of the given function with respect to s between the limits s and can be found out easily. In particular, the inverse transforms of proper rational functions whose numerators s and denominators are squares of second degree expressions in s can be found by applying this rule WORKED EXAMPLE 3(b)

Example 3.1 Find the Laplace transform of the “saw-tooth wave” function f (t) f (t) = kt, t< The graph of f (t f (t) is P,

f (t )

k t P

t < P.

3.41

Chapter 3: Laplace Transforms f (t) k

t 2

3

Fig. 3.1

By the formula for the Laplace transform of a periodic function f (t) with period P, P

L{f (t )}

st

e

Ps

e

f (t ) dt

For the given function, L{f (t )}

k e

k s2

e st s2

s s

e s

e s s2

s s

e

k e

dt

st

e

t

s

k e

st

kt e

s

e

s

s2 s

e

s2

s

s

ke

e s)

s(

Example 3.2 Find the Laplace transform of the “square wave” function f (t) f (t) = k in

t

k in a and

a t

2a

f (t +2a) = f (t) for all t.

f (t + 2a) = f (t) means that f (t) is periodic with period 2a. The graph of the function is shown in Fig. 3.2. For a periodic function f (t) with period P, P

L{f (t )} =

e

Ps

e

st

f (t ) dt

Engineering Mathematics II

3.42 f (t) k

t O

a

2a

3a

4a

–k

Fig. 3.2

For the given function; a

L{f (t )}

e k e s(

=

k s

dt

k ) e st dt

( a

e 2a as

k e

2 as

e

st

a

e

s )

k s

2a st

ke

2 as

as

e

e as ) 2 )( +e

as

e ) e as )

st

2a

s

a

as

as

e

as

e

2 as

]

)

as / 2

k (e s (e as / 2

e e

as / 2 as / 2

) )

k as tanh s 2 Example 3.3 Find the Laplace transform of “triangular wave function f (t) whose graph is given below in Fig. 3.3. y = [= f (t)] A

a

B t O

a

2a

3a

4a

5a

Fig. 3.3

From the graph it is obvious that f (t) is periodic with period 2a. f (t) t 2a, and AB.

OA

3.43

Chapter 3: Laplace Transforms OA Equation of OA is y = t

t

a

AB passes through the point B(2a, Equation of AB is y

=

t

y = 2a

or

2a)

t in a

t

2a.

f(t) [= y] can be taken as f(t) = t,

t

a

t, in a

= 2a

t

2a

f(t + 2a) = f(a).

and

2a

Now L{f (t )} =

e

2a s

e

2a s

st

e

f (t ) dt

a

e

e

dt

( 2a

st

t )e

dt

a

2a s

e s (

t

a e s

as

e e

2a s

e ) e as )

tanh

a

e st s2

s

as

( s2 (

st

e

2a s

2

s2

2a st

te

( a

as

as

e s2

2a s

s (

2

s (

)

e

a s/2

e s 2 ea s / 2

2

t)

2a s

s

e

a e s

2

e st s2

st

s as

as

e s2

as 2

e ) e )( e as

e e

as

)

a s/2 a s/2

as 2

Example 3.4 Find the Laplace transform of the f (t) whose graph is given in Fig. 3.4. f (t)

a

2

/

2 /

3 / Fig. 3.4

4 /

t

2a

a

Engineering Mathematics II

3.44

From the graph, it is obvious that f(t) is a periodic function with period 2 graph of f (t)

2

f (t) is given by f (t) = a sin t = in t 2

and

f t

Now

L { f (t )}

. The

, a and

t /

f (t ) . 2

e

a e 2 a e 2 (s

e

2 s

2

(s

f (t ) dt

s

e st sin t dt

s

2

e

2

s

e 2

)

st 2

a )( e

a 2

st

2 s

( s sin t

)

cos t

s

e

s

e

a 2 s

(s

2

2

)

e

s

Example 3.5 f (t), f (t) = |sin We note that

f (t +

) = |sin = |sin

t|, t (t + / )| t|

= f (t) f (t) is periodic with period f (t) is always positive. The graph of f (t) is the sine curve as shown in Fig. 3.5.

/

2 /

3 / Fig. 3.5

t

3.45

Chapter 3: Laplace Transforms

Now

L { f (t )}

(s

s

e

st

e

| sin t | dt

s

e

st

e

sin t dt [ sin t

e

st

e

s

2

s 2

2

e

)(

s2

2

s2

2

e e

2

( s sin t

s

e

)

s 2

coth

t

]

cos t )

s

s

2

2

e e

s 2

e e

s 2

in

,

s 2

s 2

Example 3.6 Find the Laplace transforms of the following functions: (i) t cosh3 t; (ii) t cos 2t cos t; (iii) t sin3 t; (iv) (t sin at)2. 3 (i) L{t cosh t} L t

8

et

e

t

3

2

L {t (e3t

3et

3e

t

e

3t

)

d L (e3t 3et 3e t e 3t ) 8 ds d 3 3 8 ds s 3 s s s 3 3 8 (s 3) 2

(s

3 )2

(s

Note got the same result. (ii) L(t cos 2t cos t )

L

2

t (cos 3t 2

cos t )

d L (cos 3t ds d s 2 ds s 2

cos t ) s s2

)2

(s 3) 2

s s

Engineering Mathematics II

3.46

s 2 9 2s 2 2 (s 2 9) 2 s2 9 2 (s 2 9) 2 (iii)

L(t sin 3 t )

2s (s

2

3 s 2 (s 2

(

d2 2 ds 2 d 2 ds

)2

) )2

2s (s 9) 2

2

2

(s 2

9) 2

d2 {L( cos at )} ds 2 s 2 s s 4a 2

)2

s

2

s 2 4a 2 (s 2 4a 2 ) 2 4a 2 ) 2 2 s (s 2 4a 2 ) 2 (ss 2 (s 2 4a 2 ) 4

2 2 s3

(s 2

2 2 s3

2s ( a 2 s 2 ) (s 2 4a 2 )3

s

)2

cos at 2

L {(t sin at ) 2 } L t 2

2

s2

3 sin t sin 3t 4 4 d {L (3 sin t sin 3t )} 4 ds d 3 3 2 2 4 ds s s 9

L

3 4

(iv)

s2 (s 2

s2 (s 2

3

s ( 2a 2 s 2 ) (s 2 4a 2 )3

Example 3.7 Use Laplace transforms to evaluate the following; (i)

(i)

te

2t

sin 3t dt ;

e

st

(t sin 3t ) dt

(ii)

L(t sin 3t )

te

3t

cos 2t dt

4a 2 ) 2 s

Chapter 3: Laplace Transforms d L (sin 3t ) ds d 3 2 ds s 9

L (t sin 3t )

Now

6s (s 2

st

te

sin 3t t

3.47

6s (s

(2)

9) 2

2

9) 2

,s

(3)

Putting s = 2 in (3), we get 2t

te

(ii)

st

e

Now

.

sin 3t dt

(t cos 2 t ) dt

L(t cos 2t ) d L(cos 2t ) ds

L(t cos 2t )

d s d s s2 4

te

st

cos 2t t

s2 4 ( s 2 4) 2

(2)

s2 4 ,s ( s 2 4) 2

(3)

Putting s = 3 in (3), we get te

3t

cos 2t dt

5

.

Example 3.8 Find the Laplace transforms of the following functions: (i) te 4t sin 3t; (ii) t cosh t cos t; (iii) te (i) L {te 4t sin 3t} [ L(t sin 3t )] s s 4

2t

sinh 3t (iv) t 2e cos t.

Engineering Mathematics II

3.48 Now

d L(sin 3t ) ds

L(t sin 3t )

3 d d s s2 9 6s ( s 9) 2

(2)

2

L {te

4t

sin 3t}

(s 2

6s 9) 2

s

s

4

6 (s 4) 2 (s 8s 25) 2

Note

The same problem has been solved by using an alternative method in Worked Example (6) in Section 3(a). t (ii) L {t cosh t cos t} L (et e t ) cos t 2 2 L (t cos t )

Now

[ L(t cos t ) s

(2)

(s )2 2 ( s 2 2 s 2) 2 2 (s

Now

t

2

2s 2 s 2) 2

sinh 3t}= L{t sinh 3 t}s

L (t sinh 3t )

]

)2

s2

(iii) L{t e

s

d s ds s 2 s2 (s 2

L (t cosh t cos t )

L (t cos t ) s

s

(s )2 ( s 2 2 s 2) 2 s2 (s

2

2s 2 s 2) 2

s+2

d L (sinh 3t ) ds d 3 ds s 2 9 6s ( s 9) 2 2

(2)

3.49

Chapter 3: Laplace Transforms L {te

2t

L {te

2t

sinh 3t}

{( s

6 ( s 2) 2) 2 9}2

2t

sinh 3t} L te

2

2

L { tet

( e 3t

2 ( s 2) 2

(iv) L{t2 e t cos t} = [L (t2 cos t)]s L (t 2 cos t ) (

)

}

( s 5) 2

s 24 )2 (s )2

2 (s

Now

3t

e

5t

te

6 ( s 2) )2 (s )2

(s

6 ( s 2) )2 (s )2

(s

s+

d2 L (cos t ) ds 2 s 2 s

)2

d2 ds 2

d s2 d s (s 2 )2 2 (s3 (s 2

3s ) )3

(2)

{( s )3 ( s )} ( s 2 2 s 2)3

L {t 2 e t cos t}

Example 3.9 Find the inverse Laplace transforms of the following functions: a ; s

(i) lo

(ii) log

s s

2

(iii) log

s s( s

)

(i) L { ( s )}

L log

a s

(iv) s log

;

t

L {

L log

2

2

a b

2

2

s s

( s )} s a s

t

L

d s a log ds s

;

k (k is a constant)

Engineering Mathematics II

3.50

t t t

L log

s2 s2

a2 b2

t t

d log (s a ) log s ds

L L

s a

(e at

L

)

s t

e at )

(

d [log(s ds

a 2 ) log(s 2

s

L

s

2

a

b 2 )]

2s s b2

2

2

2 (cos bt cos at ) t

L log

s2 s (s

t

)

t t t

L

s log

s s

k

L

t t t

s

2

s

(2 cos t e

L L L L

) log s log(s

)]]

s

L

(

t

d [log (s ds

t

e

s t

cos t )

d [s log(s ds s

) s log (s

log(s

s s

)

s

s

s

t

s s

2

t

t

s s

)] L (k ) log(s

L [log(s L

s

)

k (t )

) log(s

s

k (t )

)] k (t )

3.51

Chapter 3: Laplace Transforms

Note s

[L

s

and L

s

do not exist, as

s

s

s s s

s

s

and

as

s

s

are improper rational

2s , which is a proper s2

rational function.

t

cosh t

2 sinh t t2

(et

t2

e t)

2 cosh t t

k (t )

k (t )

Example 3.10 Find the inverse Laplace transforms of the following functions:

2

(iii) cot (i)

s

L { (s )}

t

(as )}

L {

t t

L

tan

s

L

d cot (as ) ds

L

a a2 s2 a2 s2

L

a b

2 s2

(s )}

a L t t

t

t

s

L

/a ( / a)2

2

t s

d tan ds

a b

/b

L

s

a

2

b t

a b

(iv) tan

;

L {cot

(ii)

s

(ii) tan

(i) cot (as);

L

(s

b a)2

b2

sin

t . a

;

Engineering Mathematics II

3.52

t (iii)

L

e

at

sin bt .

2

cot

s

t

t

L

d cot ds

2

L

4 (s

t t (iv)

L

tan

s2

t

t

L

L

(s 2

4

(s 2 4 s2

2s

2

)

)2

2

)2

(s

d tan ds

L

s s4

(s

e t sin 2t.

4 L t Consider

.

s

s2

4 s4

2 s2

.

4 s

3

s s4

(2)

4

s 2) 2 (2 s ) 2 s 2 s 2) (s 2 2 s 2s

2

2)

, by resolving into partial fractions

4 (s L

s s

4

4

4

2

)2

et sin t

(s

)2

e t sin t

sin t sinh t

Using (3) in (2), we have L

tan

2 s2

2 sin t sinh t t

(3)

3.53

Chapter 3: Laplace Transforms Example 3.11 Find the Laplace transforms of the following functions: (i)

(v)

(i)

at

sinh t e ; (ii) t sin t t L

L

e

; (iii)

t

cosbt ; (iv) t

2 sin 2t sin t ; t

2

.

f (t ) t

s

sinh t t

s

L {f (t )}ds

L (sinh t ) ds ds

s2

s

2 2

2

log log

log

s s

s

s s

2

L

e

log

at

s 2

2

log

s s

log

s s

s

log

ebt t

2

s

s

(ii)

e at

bt

s s

2

L (e

at

e

bt

s

a

s b

log

s s

) ds

s

s

log

log

s a s b a s b s

ds log

s a s b

log

s b s a

s

s

Engineering Mathematics II

3.54

lo

s b s a

= log

(iii)

L

e at cosbt t

L (e at

cos bt ) ds

s

s s

s a

s

2

log ( s a )

s

2

log ( s 2

s a

log

2

s2

s

log

b 2 /s 2 log

b2 ) s

a /s

lo

2 sin 2t sin t t

2

b

log

(IV) L

ds

b2

s a

log

2

s b s a

log

b2

s2 b2 s a

s

s2 b2 ( s a )2

s2 b2 ( s a )2

log

L(2 sin 2t sin t ) ds, by rul

)

s

L( cos t

cos3t ) ds

s

s s

2

2

s

2

log

log

s s s2 s2 9 s2 s2 9

2

ds

s

s

2

log

s2 s2 9

3.55

Chapter 3: Laplace Transforms

s2 /s 2

log

2

2

s2 9 s2

log

s

log

2

2

log

s2 9 s2

2

2 (v)

L

L

f (t ) t2 sin 2 t t2

s 9 s2

log

(2)

L{ f (t )} ds ds s

s

L (sin 2 t ) ds ds s

s

cos t ds ds 2

L s

s

s 2

s

2

s

s

2

s log

s

2

s2

ds ds

4 4

ds,

s2

by putting a s log

4

s2 4 s2

s s

s

b = 2 in (iii) above.

2s s

2

4

2 ds s by integrating by parts.

s s2 4 log 4 s2 s s2 log 2 4 s 4

L

L

Now

L

log

4 s2

s

cot

s s2 log 2 4 s 4 s/4

s s2 4 log 4 s2

cot

s 2

8 4

s

s2

4

ds

, say s

s 2

(3)

Engineering Mathematics II

3.56

log

lim ( L)

s 2 /4 s

4 s2

log lim

s

s 2 /4

4 s2

s

log (e )

s

log

(4)

Using (4) in (3), we have L

sin 2 t t2

s s2 log 2 4 4 s

s 2

cot

.

Example 3.12 Use Laplace transforms to evaluate the following: e t sin 3 t dt ; t

(i)

cos at cos bt dt ; t

(iii)

(i)

e t sin 3 t dt t

L

sin 3 t t

st

e

sin 3 t dt t

s

2

dt .

s

cot

ds

( 3)2

3

s

cot

3

s

cot

dt

t

4t

L(sin 3 t ) ds

3

s

e

s

s

t

2t

s

3

e sin t

e

(iv)

sin 3 t t

L

Now

sin 2 t dt ; t et

(ii)

(2)

3

3

s

3

.

3.57

Chapter 3: Laplace Transforms

(ii)

sin 2 t dt t et

e e

sin 2 t dt t

t

sin 2 t dt t

st

s

2

L

Now

sin 2 t t

L

sin t t

cos t 2t

L

2

s

L(

cos t ) ds

s

s 2

2

2

2

2

2

sin 2 t dt t et

(iii)

cos at

cos bt t

s2

s

s

s

log s log

2

4

s s

2

4

/s

log

2

dt

2

2

s

2

s2

s

log s

2

s

log

s2

2

4

s2

log

4 s

4 s

4 (2)

s

log 5

e L

s

2

log

log

ds

4

st

4

log 5

cos at

cos bt t

cos at

s

cos bt t

dt

s

Engineering Mathematics II

3.58

Now

L

cos at cosbt t

L(cos at cos bt ) ds s

s s2

s

2

2

cos at cos bt dt t e

(iv)

2t

s2 s2

log

b a

dt

e

e

2t

e

e

t

t

2

2

s

log

b2 a2

s2 s2

a2 b2 (2)

. e

st

2t

e

4t

dtt

t 2t

e

s

4t

s

2t

e

s

2

s

4t

) ds

s

s s

2 4

s s

2 4

s log s

4 2

4t

dt

ds

s

a b2

L (e

log

e

2

t

log

2t

a2 b2

b2

4t

s

e

s2

s2 s2 s s2

log

log

L

L

log

4t

e t

Now

s a2

log 2

4

ds

s

log s

s s

2 4 (2)

Chapter 3: Laplace Transforms

3.59

Example 3.13 Find the inverse Laplace transforms of the following functions: (i) (iii) (v)

s (s

2

a )

;

s

(ii)

(s ) ; 2 s 5) 2

(s 2

(s

2 2

(iv)

(s

2

4) 2

;

s2 3 ; (s 2 4 s 5) 2

s 2 s 8) 2

2

(i)

L { (s )} = t L

(s ) ds s

L

s (s

2 2

2

a )

s

=t L

(s

s

a2

t L 2

x

t L 2

a 2 )2

ds

dx , on putting s2 + a2 = x 2 x2

tL s2

2

s2

s2

a2

a2

t sin at. 2a (ii)

L

s (s

2

4)

2

s

t L

(s

s

t L 2

s2

ds

4) 2

2

4

, as in (i) above.

t sinh 2t 4

Note

The inverse transform in this case can also found out by resolving the given function into partial fractions. (iii)

L

(s 2

(s ) 2 s 5) 2

= 4et L

s s

(s

s

4L

2

22 ) 2

) 2 + 22

2

Engineering Mathematics II

3.60

4e t

t sin 2t , as in problem (i) 4

= tet sin 2t. s2 3 (s 4 s 5) 2

(iv) L

2

L

s

L

L

(v)

(s

2

2

(s

e

2t

=e

t

t

e

t

4L

4s 5 )

4L

2

2t

sin t 4e

s (s

2

2 4 s 5) 2

s 2 {(s + 2) 2 +

2

t sin t , as in problem (i) 2

t) sin t.

s 2 s 8) 2 e

(s 2 + 4 s 5) (4 s 8) (s 2 4 s 5) 2

L

L

L

(s 2

s [(s ) 2

2 2

]

s 32 ) 2

,

t sinh 3t , proceeding as in problem (ii) 6

t t e sinh 3t 6

EXERCISE 3(b) Part A

2. Find the Laplace transform of f(t) = t t f (t f(t) 3. State the relation between the Laplace transforms of f(t) and t · f(t). 4. State the relation between the inverse Laplace transforms of (s) and 5. State the relation between the Laplace transforms of f(t) and

t 6. State the relation between the inverse Laplace transform of integral. Find the Laplace transforms of the following functions: 7.

2a

t sin at

8. t cos at

9. sin kt

(s).

f (t ) .

kt cos kt

(s) and its

3.61

Chapter 3: Laplace Transforms t a2

kt + kt cos kt

cos at )

cos kt

kt sin kt . 2 Find the inverse Laplace transforms of the following functions: log

s s

log

s s

s

log a b

log

s s

cos at t

25.

s2 s2

log

s

a tan s s Find the Laplace transforms of the following functions: e t sin at 22. 23. t t

24.

a s

lo

4

et t

sin 2 t t

Part B Find the Laplace transforms of the following periodic functions: f (t )

26.

E, i , in

given that

f t

2 n

f (t ) = E

27.

t

E

t

E n

,

f t t

T 2

= , in T/2 . given that f (t + T) = f (t) 28. f (t) = et t and f (t + 2 ) = f (t) t 29. f (t ) = sin < t < 2 and f (t + 2 ) = f (t) 2 f (t) = |cos ,t /

Hint: f (t )is periodic with period (

)and 0

/

e /2

st

( cos t ) dt

/2

e

st

| cos t | dt

e 0

st

cos t dt

Engineering Mathematics II

3.62

f(t) = t

t t

,

t

,

f(t + 2 ) = f(t)

given that

f(t) = sin t,

32.

= in f(t +2 ) = f(t).

given that

f (t )

33.

f t

given that 34.

2

in

t

sin

t , in

2

t

,

f (t ) .

f(t) = t

t

=2

t, in

t

,

f(t + 2 ) = f(t).

given that

f(t) = t =

35.

t t, in

t

,

f(t + 2 ) = f(t).

given that

Find the Laplace transforms of the following functions: 36. t sinh3 t 37. t cos3 2t 38. t sin 3t sin 5t 39. t sin 5t cos t t cos 2t)2 t2 sin t cos 2t 42. t2 e t sin 3t 43. te3t cos 4t 44. t2 e 3t cosh 2t 45. t sinh 2t sin 3t Find the inverse Laplace transforms of the following functions: a2 s2

46. log

49. s log

52. cot

s s

47. log

a a a

s

tan

a

b

s2

a2

s

b

2

2s

48. log

tan

s

s 2 s2

2

2 3

53. tan (s2)

Find the values of the following integrals, using Laplace transforms: 54.

te

2t

cos 2t dt

55.

t 2 e t sin t dt

56.

e

t

e t

3t

dt

3.63

Chapter 3: Laplace Transforms cost ) e t

57.

t

dt

e

58.

at

cosbt t

e

dt 59.

2t

sint sinht dt t

Find the Laplace transforms of the following functions: e t 63.

t

cosat t

sin 3t sint t

62.

sin2t

2

t

cost t2

64.

Find the Laplace inverse transforms of the following functions: s s 2 s 65. 66. 67. (s 2 )2 (s 2 4 s 5) 2 (s 2 a 2 ) 2 68.

(s (s 2

a)2 a 2 )2

69.

s2 (s 2

s s

s

6 )2

(s

2

4 s

5) 2

3.8 LAPLACE TRANSFORMS OF DERIVATIVES AND INTEGRALS integrals of a function f(t) in terms of the Laplace transform of f(t). These results will be used in solving differential and integral equations using Laplace transforms.

If f(t) is continuous in t f t) the range t f(t) and f t) are of the exponential order, then L{ (t)} = sL{f(t f

Proof: The given conditions ensure the existence of the Laplace transforms of f(t) and (t). L{f (t )} =

st

e

f (t ) dt

e st d [f (t )]

e t

st

li [e f

=sL {f(t

f (t ) st

f (t )]

s e f

st

f (t ) dt ,on integration by parts.

) + s L{f (t )

sL {f(t)} [ f(t) is of the exponential order] f

Engineering Mathematics II

3.64

f (t) by f t) we get L{f (t)}= sL {f (t f = s [sL {f(t f f = s2 L{f (t s f f

Note f(t) and f t) are continuous in

(t) is piecewise f(t), f (t) and (t) are

of the exponential order.

L{ f(n)(t)} = sn L{f(t

sn f

sn

f (n

Note f(t) f(n) (t and f(t), (t),

t , f (n) (t) are of the exponential order.

3. If we take L { f (t)}= L{

n

(t)} = s (s

(s), f

If we further assume that f L{

(t)}= s

s)

(5)

In terms of the inverse Laplace operator, (5) becomes L {s (s)}=

(t)

(6)

From result (6), we get the following working rule: d L {s (s ) = L { (s ) , provided that dt f L { (s)}t which is ‘s’, we ignore ‘s’, we call it f (t), verify that f (t), which is the required inverse transform. 4. In a similar manner, from result (2) we get L {s 2 (s )} = f

d2 dt 2

L { (s )} , provided that

f f(t) = L { (s)}.

If f(t) exponential order, then

Chapter 3: Laplace Transforms

3.65

t

L

f (t ) d t

s

L{ f (t )}

.

Proof: t

g (t ) =

Let

f (t ) dt

g t) = f (t) Under the given conditions, it can be shown that the Laplace transforms of both f(t) and g(t) exist. Now by the previous theorem, L{g t)}=sL{g(t g t

sL

i.e.,

f (t ) dt

f (t ) dt

L{ f (t )}

t

L

t

f (t ) d t

s

L{ f (t )}

t

L

f (t ) dt dt

s2

L{ f (t )} , as explained below.

t

f (t ) dt

Let

g (t )

.

t

L

t

g (t ) dt

s

L{ g (t )}

t

t

i.e., L

f (t ) dt dt

s

L

f (t )dt

s s s2

L{ f (t )} , again by

L{ f (t )}

(l) (2)

Generalising (2), we get t

L

t

t

f (t ) (dt ) n

sn

L{f (t )}

(3)

Engineering Mathematics II

3.66

Note L{ f (t)} =

(s),

t

L

f (t ) dt

s

(s )

(4) L operator, as

t

L

(s )

s

(5)

f (t ) dt

From (5), we get the following rule: t

L

s

(s )

L { (s )} dt

.

which is

, we ignore s s it with respect to t t. 2. In a similar manner, from (2) above, we get t

L

s2

t

L { (s )} dt dt .

(s )

t

L

3.

f (t ) dt a

s

L{ f (t )} +

s

f (t ) dt a

t

g (t ) =

If we let

f (t ) dt and g (t ) = f (t ), a

we get

L{g (t )} = sL{g (t )} g (

i.e.,

L{ f (t )}= sL

t

f (t ) dt

f (t ) dt

a

a

t

or

L

f (t ) dt a

3.9

s

L{ f (t )} +

s

f (t ) dt . a

INITIAL AND FINAL VALUE THEOREMS

We shall now consider two results, which are derived by applying the theorem on Laplace transform of the derivative of a function.

3.67

Chapter 3: Laplace Transforms

lim[ f (t )] and lim[s (s )], where t

(s ) = L { f (t )}

s

.

lim[ f (t )] and lim[s (s )] t

s

If the Laplace transforms of f (t) and (t) exist and L{ f(t)} =

(s), then

lim[f (t )] = lim[s (s )] t

s

Proof: L{f t)} = s s f s s L{f t)} + f

We know that

st

e

lim [s (s ) = lim s

f (t ) dt

e

s

st

lim {e

s

f

f (t ) dt

st

f (t )} dt

)

f( ,

f( ,

assuming that the conditions for the interchange of the operations of integration and taking limit hold. i.e.

lim [s (s )] =

f(

s

= lim [ f (t )]. t

If the Laplace transforms of f(t) and (t) exist and L{f (t)} = (s), then lim [ f (t )] = lim [s (s )] , provided all the singularities of {s (s)] are in the left t

s

half plane Rl(s)< .

Proof: (t)} = s (t)}+f

We know that s (s) e

st

f (t ) dt

s

f f

)

Engineering Mathematics II

3.68

lim [s (s )] = lim

e

s

s

lim{e

st

f (t )dt

st

f (t )}dt

s

f( )

f ( ) , assuming that the conditions for

the interchange of the operations of integration and taking limit hold. i.e.

lim [s (s )]

f (t )dt

s

f( )

[ f (t )]

f( )

lim f (t )]

f ( )+ f ( )

t

Thus

lim [f (t )] = lim [s (s )]

t

s

3.10 THE CONVOLUTION f (s ) and g (s ) in terms of the inverse transforms of f (s ) and g (s ) . The convolution or convolution integral of two function f (t) and g (t), t t

f (u ) g (t

u ) du

It is denoted as f (t) * g (t) or (f * g) (t) t

i.e.

f (t ) * g (t ) =

f (u ) g (t

u ) du

t

f (t

t

u ) g [t

(t

t

(u ) du

(t

u ) du

t

g (u ) f (t g (t )* f (t ).

u ) du

u )] du , on using the result

3.69

Chapter 3: Laplace Transforms Thus the convolution product is commutative.

If f (t) and g (t) are Laplace transformable, L{f (t) * g(t)} = L{f (t)}·L{g (t)}

then

Proof: e st {f (t ) * g (t )} dt

L{f (t )* g (t )}

,

t st

e

f (u ) g (t

u )du dt,

f (u ) g (t

u )du dt

t st

e

u u = t, t

t=

and is shown in the Fig. 3.6.

=

t

u t

u

t

(u, u)

u) t u Fig. 3.6

L{f (t ) * g (t )}

e

st

f (u ) g (t

u ) dt du

(2)

u

In the inner integral in (2), on putting t we get, L{f (t ) * g (t )}

e e

su

s (u v )

f (u )

u = v and making the consequent changes,

f (u ) g (v) dv du e

sv

g (v)dv du

Engineering Mathematics II

3.70

e

su

f (u )du

e

st

f (t )dt

e

st

e

sv

g (v)dv

g (t )dt ,

on changing the dummy variables u and v. = L{f (t)}·L{g (t)}

Note

f (s ) and L{g (t )}

If L{ f (t ) }

g (s ) , the convolution theorem can be put as

(3) L{f (t ) * g (t )} f (s ) g (s ) In terms of the inverse Laplace operator, result (3) can be written in the following way. L {f (s ) g (s )}

f (t ) * g (t ) t

f (u ) g (t

u ) du

(4)

functions of s is equal to the convolution product of the inverses of the individual functions.

Example 3.1 transforms of (i) e at, (iii) cos2 t, (i)

(ii) sin at, (iv) t n (n is a positive integer)

L{f t)} = sL{f(t

f (t) = e at L ae at) = sL (e i.e. (s + a) L(e at

f

Putting

L(e

at

)

s

at

a

L{f t)} = s2 L{f(t

(ii)

sf

f

f (t) = sin at in (2), we get

Putting L

a2 sin at) = s2L (sin at

s

a

3.71

Chapter 3: Laplace Transforms i.e.

(s2 + a2) L (sin at) = a a

L(sin at )

s a2 Putting f(t) = cos2 t

(iii)

2

t sin t) = sL (cos2 t s·L (cos2 t L (sin 2t)

L i.e.

2 s

2

2

s s2

2 4

s2 2 s (s 2 4)

L(cos 2 t ) L{f (n ) (t )}

(iv)

4

s n L{f (t )} s n f ( )

sn

f (n

f ( )

)

( )

Putting f (t) = tn in (3) and noting that f (n) (t) = n! f( )

and

f (n

f ( )

)

( )

, we get

L{n!} = sn L (tn) n

n!L

i.e.

n!

i.e.

L (tn)

s n L(t n )

s

n!

L(t n )

sn

Example 3.2 Find the Laplace transform of t

L

/

)

s

3/ 2

L(t

In the result L{f t)} = sL {f(t t

f (t ) L

2

t

s

.

t

) 2s3 / 2

f

, we get

2s3 / 2

L

(3 / 2) s3 / 2

/ 2)

2

t

(3)

Engineering Mathematics II

3.72

2 s .

L

t

L

t

s

L

t/

t

L( t / ) ds s

s

ds

2s 3/ 2

s

s

s

Example 3.3 Using the (i) L(t cos at) and hence L(sin at

.

at cos at) and L(cos at

at sin at)

(ii) L(t sinh at) and hence L (sinh at + at cosh at) and L(cosh at (i)

L f (t )

s 2 L f (t )

sf

f

)

f t) = cos at

Then

L

2

a t cos at

at sin at and f

a2 t cos at

f (t) =

a sin at.

2

2a sin at} = s L {t cos at f

(s2 + a2) L(t cos at

a L(sin at) 2a 2 s a2 s2 a2 s2 a2 2

L(t cos at ) Now

at sinh at ) .

f(t) = t cos at in

Put

i.e.

2

L(sin at

at cos at )

s2 a2 (s 2 a 2 ) 2 a s2

a2

a (s 2 a 2 ) (s 2 a 2 ) 2

f

3.73

Chapter 3: Laplace Transforms a (s 2 +a 2 ) (s 2 (s 2

a2 )

a 2 )2

2a 3 (s 2 a 2 ) 2 f (t) = t cos at in the result L{f t)} = sL{f(t f at sin at} = sL (t cos at)

Taking L{cos at

s(s 2 a 2 ) (s 2 a 2 ) 2 Put f (t) = t sinh at f t) = sinh at + at cosh at and f f t) = a2t sinh at + 2a cosh at L{a2t sinh at + 2a cosh at} =s2L(t sinh at) [ f f i.e. (s2 a2) L(t sinh at) = 2a L(cosh at) (ii) Then

2as s a2 2

L(t sinh at )

(s 2

2as a 2 )2

L{f t)} = sL {f(t

In the result have L(sinh at

at cosh at )=

f

2as 2 [ (s 2 a 2 ) 2 L(cosh at

Now

f

2

L(cosh at ) =

s s

2

we put f(t) = t sinh at, we

a2

at sinh at ) a L(t sinh at) 2 a 2as 2 2 (s a 2 ) 2

s (s 2 a 2 ) a 2 s (s 2 a 2 ) 2

s3 . (s a 2 ) 2 2

Example 3.4 Find the inverse Laplace transforms of the following functions: (i)

s (s

2)

4

;

(ii)

s2 ; (s 2)3

Engineering Mathematics II

3.74

(iii)

s

(v) (i)

L

s 4s

2

;

5

(iv)

s )(s 2

s2

Now

f (t )

(s

2) 4

f (t )

e

t

e

2t

3)

;

(s) , provided L { (s)} vanishes at t

s

L

s 2)(s

4)

d L dt

s (s)

(s

L

L

(s

2) 4

and then apply

s4

3!

t3

6

2t

t 3e

We note that f s

L

6

(

= t 2e 6 (ii) L

d2 L dt 2

s (s ) f

L

2t 3 e 2t

2t

f(t) = L f (t )

e tL e 2t

f (t ) = We note that f

3t 2 e

L

s3

2!

t2

2

t 2 e 2t

2 2t

2

(2t e + 2te 2t ) f

L

s2 (s

2t

)

(s ) , provided

f

f (t )

2t

(3 2t ).

s2 (s 2)3

Now

By (2),

d te dt 6

2) 4

(s

2)3

d2 t 2 e 2t dt 2 2

{ (s)}.

(s

2)3

(2) and then apply rule (2).

3.75

Chapter 3: Laplace Transforms d 2 [(t t )e 2t ] dt 2(t 2 t )e 2t ( t (2t 2 L

s

s 4s

2

Now

f (t )

and

f L

By (l),

4s

5

)2

(s

= e sin t . s 4s

s2 =e =e

L

s2

L

)e 2t

4t

,

5

f (t )=L

)e 2t

(s

t t

cos t (cos t

s 2) (s

f (t )

Now

d (e 2t sin t ) dt

5

t

e

sin t t) f (t )

3)

L

(s

L

=e

s

2) (s

s

2

L

(s

2) (s

3)

and

3)

3

,

by resolving the function into partial fractions. e t

t

f

and L

(s

s 2) (s

3)

d (e dt

= 3e L

and

(s 2

s ) (s 2

)

t

t

3t

e e

)

t

f (t )

L

(s 2

) (s 2

)

Engineering Mathematics II

3.76

/

f (t )=L

Now

/

s2

s2

4

,

by resolving the function into partial fractions. 3

sin t

6

sin 2t

f

and L

s ) (s 2

(s 2

)

d sin t dt 3

6

(cos t

cos 2t )

3

sin 2t

Note

We have solved the problems in the above example by using the working rule derived from the theorem on Laplace transforms of derivatives. They can be solved by elementary methods, such as partial fraction methods, discussed in Section 3(a) also. Example 3.5 Find the inverse Laplace transforms of the following functions. (i)

(iii)

(v)

(i) Let

s2 ; (s 2 a 2 ) 2 )2

(s (s 2

(ii)

5) 2

2s

;

(iv)

s3 ; a 2 )2 (s 2

s2 (s 2

(s 3) 2 . (s 2 6 s 5) 2 f (t )

L

s (s

2

a 2 )2

t sin at 2a We note that f d L { (s )} , provided f dt f(t) = L { (s

L {s (s )}

Now where L

s2 (s 2 a 2 ) 2

L

s

s (s

2

d t sin at dt 2a

a 2 )2

4) 2

;

Chapter 3: Laplace Transforms

2a

(sin at

f (t )

(ii) Let

3.77 (2)

at cos at ) s2 (s a 2 )2

L

2

2a

(sin at

at cos at ) , by (2)

we note that f s3 a 2 )2 (s

L

NOW

2

L

s

s2 a 2 )2 (s 2

d (sin at dt 2a 2 (iii)

L

)2

(s (s 2

5) 2

2s

e tL

f (t )

)2

(s )2

{(s

2 2

}

s2 (s 2 22 ) 2

4 (iv) Let

at sin at ) .

(2 cos at

L

at cos at ) , by rule (l)

e t (sin 2t

2t cos 2t ) , by (2)

s

L

(s

2

4) 2

t sinh 2t 4 f

We note that

Now

L

s2 (s

2

4)

2

L

s

d L dt

s (s

2

4) 2 s

(s 2

4) 2

d t sinh 2t dt 4 4

(sinh 2t

2t cosh 2t )

(3)

Engineering Mathematics II

3.78 3) 2 6 s 5) 2

(s

(v) L

(s 2

3) 2

(s

L

3) 2

{(s

4}2

s2

e tL

(s

2

4) 2

e3t (sinh 2t 2t cosh 2t ) , by (3). 4 Example 3.6 Find the Laplace transforms of the following functions: t

t

(i)

te

4t

sin 3t dt ;

(ii) e

e

4t

sin 3t dt ;

(iv)

t

t

(iii) t

t

(v) e

4t

t

(vi)

t

e t sin t dt .

t

(i) L

f (t ) dt

s

L{ f (t )}

by the theorem on Laplace transform of integral t

t e 4t sin 3t dt

L

s

L{te

4t

sin 3t}

6(s 4) s (s 8s 25) 2 2

t

(ii) L e

t

4t

t sin 3t dt

L

t sin 3t dt s

s 4

t

Now

L

sin t dt ; t

t

sin t dt ; t

t

e

t sin 3t dt ;

t sin 3t dt

s

s

L(t sin 3t ) , by rule (l) d L(sin 3t ) ds d s ds s 2

9

3.79

Chapter 3: Laplace Transforms

s

(s 2

2s 9) 2

{(s

6 4) 2

9}2

6 (s 2

9) 2

(3)

Using (3) in (2), we get t

Le

4t

t sin3t dt

(s

6 8s 25) 2

2

t

(iii) L t

e

4t

sin 3t dt t

d L ds

4t

e

sin 3t dt

(4)

t

Now

e 4t sin 3t dt

L

s

s

L(e

4t

sin 3t )

[L(sin 3t )]s

s (s s3

s 4

4) 2 9 3 8s 2 25s

(5)

Using (5) in (4), we get t

d ds s 3

e 4t sin 3t dt

Lt

3 8s 2

25s

2

3(3s s 25) (s 3 8s 25s ) 2 3(3s 2 s 2 (s 2 t

(iv)

Now

L

e t sin t dt t

L

e t sin t t

s

L

s 8s

25) 25) 2

e t sin t t

L(e t sin t ) ds s

s

(s

ds )2

Engineering Mathematics II

3.80

{ cot (s

)}s

cot (s

(7)

)

Using (7) in (6), we get t

t

e

L

t

(v)

t

Le

t

L

Now

sin t dt t

cot (s

s

t

sin t dt t

L

sin t dt t

L

s

)

sin t dt t

(8) s

s

sin t t L(sin t ) ds

s

s

ds s

s

s

2

s

cot s

(9)

Using (9) in (8), we get t t

Le

sin t dt t

cot (s

s

t

(vi)

L

)

t

e t sin t dt

t

e t sin t dt ds

L s

t

e t sin t dt

L

Now

L{e t sin t}

s

)2

s (s

t

L

t

e t sin t dt s

s (s 2

ds 2s

2) s

s

2 s

s

2

2s

2

ds ,

on resolving the integrand into partial fractions. s 2

s

s

(s

)2

(s

)2

ds

3.81

Chapter 3: Laplace Transforms

log s

2

s log

log {(s

} cot (s

)

s

2

2

) s

s

log

2

4

2

cot (s

2s 2s s2

)

2

s

2 2

cot (s

).

Example 3.7 Find the inverse Laplace transforms of the following functions: (i)

(iv)

(vii)

s (s

2

2)

3

;

s (s

2

(s

2) (s 2

54 ; s 3 (s 3)

(ii)

2

a )

;

(v)

4s

s

s s2

2

(iii)

;

(vi)

s (s 2

4s 5s

2

s (s

2 )(s

5)

.

)

;

)

Note functions into partial fractions and applying elementary methods. However we shall solve them by applying the following working rule and its extensions. t

L

s

(s )

L ( (s ) dt

t

(i) L

s (s

L

2) 3

2)3

(s

dt

t

e

t

e

2t

L

dt

s3

t

2

t

2 e

2

t 2 dt

2t

2t

2

e

2t

2

4

e

2t

t

8 by Bernoulli’s formula.

2

e

2 8

[

2t

t 2

( t2

t 2

4

t

)ee

4 2t

]

Engineering Mathematics II

3.82

(ii) L

54 s (s 3) 3

t t t

L

54 t

t

dt dt dt ,

s 3

t

e 3 t dt dt dt

54 t

t

54 t

t

e 3t 3

dt dt

t

(e3t t

) dt dt t

e 3t 3

t dt

t

6

(e3t

e 3t 6 3 2e 3 t

3t

) dt t

3t 2 2 9t 2

t 6t 2.

We can avoid the multiple integration by using the following alternative method. L

54 s (s 3) 3

54

L

3

s 3 3 (s 3) 54e3t L

s (s 3)3

,

t

54e

t

L

(s 3)3

dt

t

54e3t

e

3t

L

s3

t

54e3t

2

t2 e

3t

dt

dt

3.83

Chapter 3: Laplace Transforms

3t

27e t

e

2

3t

e [2 e 2e

2t

3

3t

3t

3t

9t

2

(9t

2

3t

e

9

6t

2

e

3t

t

27

2)]

6t 2

t

(iii)

L

s (s 2

L

4s 5

s2

dt ,

4s 5

. t

L

dt

)2

(s

t 2t

e

sint dt

e 2t (2sin t + cos t ) 5

t

(iv)

L

s 2 (s 2

e

[

5

2t

( sin t

t

cos t )]

t

L

a2 )

t

s2

dt dtt ,

a2

t

a t

sin at dt dt cosat a

a

t

dt

t

(

a2

a3

(v)

L

s

2

s s2

t

sin at a

t

a2

cos at ) dt t

(at sin at ).

t

L

s s2

dt dt ,

Engineering Mathematics II

3.84

t

t

(cos t

sin t ) d t d t

t

(sin t cos t )t d t t

(sin t cos t

) dt t )t

( cos t sin t

t cos t sin t

(vi)

t

5s 2 s 2 (s ) (s

L

t

L

)

t

(s

5s 2 dt dt , ) (s )

t

4

L

s

s

dt dt ,

2

by resolving the function into partial fractions. t

t

(et

2t

4e

) dt dt

t

(et

2e

(et

e

2t t

) dt

t

(et (vii)

L

L e

t

e

t

(s

2) (s 2

4s

)

(s

2){(s

2) 2

9}

s (s 2

9)

L

s2

9

t

3

e

2t

sin 3t dt

2t

t )t

) dt et

e

2t

, by the first shifting property.

t

L

e

2t

dt ,

t 2

3.85

Chapter 3: Laplace Transforms

3 9

e

2t

e

2t

cos 3t 3 (

t

cos t )

Example 3.8 Find the inverse Laplace transforms of the following functions: (i)

(iv)

(i)

(s 2

a 2 )2

(s 2

4) 2 L

;

(ii)

;

(v)

(s 2

a 2 )2

s (s 2

a 2 )2

s (s 2

4) 2

(iii) (s 2

;

;

(vi)

2 s 5) 2 ;

(s 2

2 s 3) 2

.

s

L

s (s 2

a 2 )2

t

s

L

(s

a 2 )2

2

d t ,as

t

L

s

(s )

L { (s )} d t t

t sin at d t 2a

2a 2 a3 (ii)

L

s (s 2

a 2 )2

t

(siin at at cos at )

s 2 (s 2 t

t

a 2 )2 s

L

t

sin at a2

s

L t

cos at a

t

(ss

2

a 2 )2

dt dt ,

t sin at d t d t 2a

t

2 a3

(sin at at cos at ) d t , by (2)

(2)

Engineering Mathematics II

3.86

cos at a

2a 3

(s 2

2s

5) 2

L

t

2

)2

(s e

L

(s 2

4) 2

e t (sin 2t (iv)

L

(s

2

4)

2

s (s

2

t

4) 2

(s 2

dt

4) 2

t sinh 2t dt 4

cosh 2t 2

sinh 2t 4

(2t cosh 2t

sinh 2t )

t

4

s( s 2

4) 2 s

L

L

2t cos 2t ), by problem (i)

s

L t

(v)

t

(2 2 cos at at sin at ).

2a 4 L

cos at a2

( 2 cos at at sin at )t

2a 4

(iii)

sin at a t a

(3)

s

L

s2 (s2

t

4) 2

t

L

t

t

t

s (s

2

4) 2

dt dt ,

t sinh 2t dt dt 4

t

(2t cosh 2t

sinh 2t ) dt ,

by (3)

3.87

Chapter 3: Laplace Transforms

2 t

sinh 2t 2

(s 2

L

2 s 3) 2

t

t sinh t cosh 2t )

(

L

cosh 2t 2

cosh 2t ]t

nh 2t [t sin

(vi)

cosh 2t 4

2

)2

(s et L

(s 2

4) 2

,

et (2t cosh 2t

sinh 2t ), by problem (iv).

Example 3.9 a) f(t) = (t + 2)2 e ; (b) f (t )

L

2) 2

s (s

(ii) If L(e cos2 t) = (iii) If L{ f (t )}

lim [ s ( s )] and lim [ s ( s )] . s

s (s

)( s

s

lim [ f (t )] and lim [ f (t )] .

)

t

t

f(t) = (t2 + 4t + 4)e

(i) (a) ( s) s ( s) Now

s

2

L{ f (t )}

(s

4 )

2s

3

(s

4s

(s

)

3

(s

)

2

4 and slim [ s ( s )]

lim [ f (t )]

and lim [ s ( s )]

t

)

s

4s s

lim [ f (t )] t

4 2

4

4

(s

2) 2

s

t

(i) (b)

f (t )

L

s (s

t

te

dt

2) 2

L

dt

Engineering Mathematics II

3.88

t

)

s

lim [ f (t )]

t

lim [ s ( s )]

4

s

L(e t cos2 t) =

(ii)

2t

e

lim [ s ( s )]

t

and

2t

2) 2

(s

lim [ f (t )]

Now

4 te

(

t

2t

e

2

4 s (s)

2t

e

(s) t

f (t) = e cos2 t

i.e.,

lim [ s ( s )] s

lim [e t cos 2 t ]

.

t

By the initial value theorem, lim [ s ( s )]

s

(iii)

L{ f (t )}

s (s)

lim [e t cos 2 t ]

.

t

s (s

(s

)( s

)( s

)

)

By the initial value theorem, lim [ f (t )] t

lim [ s ( s )]

s

lim [ f (t )] lim [ s ( s )]

t

s

2

Example 3.10 Use convolution theorem to evaluate the following t

t

u2 e

(i)

a (t

u)

du ;

t

(i)

(ii)

sin u cos (t

t

u2 e

a (t

u)

du is of the form

f (u ) g (t

u ) du ,

u ) du .

3.89

Chapter 3: Laplace Transforms f(t) = t2 and g(t) = e

where

at

t

u2 e

i.e.

a (t

u)

du

(t 2 ) * (e

at

)

By convolution theorem, t

u2 e

L

a (t u )

du

L(t ) 2 L(e

s3 s t

u2 e

a (t

u)

du

at

)

a 2 s3 ( s a)

L

e

at

e

at

e

at

e

at

e

at

2 s ( s a )3

L t

L

2 dt ( s a )3

t

a3

t 2 e at dt e at t a

e at 2t 2 a

e at 2 3 a

t 2 e at a

2t e at a2

2e at a3

2

{a 2 t 2

2at

sin u cos (t u ) du is of the form

t

sin u cos (t u ) du

i.e.

(sin t ) * (cos t )

By convolution theorem, t

L

at

}

f (u ) g (t

f(t) = sin t and g(t) = cos t

where

2e

2 a3

t

t

(ii)

2

t

sin u cos (t u ) du

L(sin t ) L(cos t )

u ) du ,

Engineering Mathematics II

3.90

s (s 2

)2

t

sin u cos (t u ) du

s

L

(s

2

)2

t sin t 2 Example 3.11 the following functions: (i)

(iv) (i)

;

(s

) (s

(s 2

4 2s

5) 2

s

s 2

L

)

s

(ii)

; (v)

; (iii)

(s 2

a 2 )2

(s 2

s2 s s ) (s 2

L

*L

s

e t *e

s2 a 2 ) (s 2

(s 2

2)

b2 )

;

.

, by convolution theorem

s 2

2t

t u

e

e

2 (t

u)

du

t

ii

L

s (s 2

a 2 )2

e

2t

e

2t

e u du (et

)

t

e

2t

e s

L

L

s2

a2 s2

s2

a2

a2

*L

s s2

a2

, by convolution theorem

sin at * (cos at )

a t

a

sin au cos a (t u ) du t

2a 2a

[sin at (sin at ) u

sin (2au at )] du cos( 2au 2a

at )

t

3.91

Chapter 3: Laplace Transforms

2a 2a (iii) L

(s

2

t sin at

2a

(cos at cos at )

t sin at.

s2 a 2 )( s 2

s

L

2

b )

s

s

L

(s 2

2

a

s

2

b2

s

*L

a2 )

s 2

s2

, by convollution theorem

b2

= ( cos at ) * ( cos bt ) t

cos au cos b(t u ) du t

{cos [(a b) u

2

bt ] cos[(a b) u bt ]} du t

2 a b 2 a b

sin{(a b) u (sin at sin bt )

2 a b

a b

2a sin at 2 a b2 2

a2

(iv)

b2

s

2

2s 5

L

(s

a b a b

sin at

sin{(a b) u bt}

(sin at

a b

sin bt )

a b

sin bt

2b sin bt a b2 2

(a sin at b sin bt ).

4

L

bt

2

2

L

2 )2

s

2

*L

2

2s 5

(s

s

2 )2

2

2s 5

, by convolution theorem.

(e t sin 2t )*(e t sin 2t ) t

e u sin 2u e

(t u )

sin 2(t u ) du

Engineering Mathematics II

3.92

t

2 2 2

s2

L

s

2

s

(s

)

e

t

e

t

[cos (4u 2t ) cos 2t ] dv sin (4u 2t ) (cos 2t ) u 4 4

(sin 2t

s 2

s

L

t

t

sin 2t ) t cos 2t

e t (sin 2t 2t cos 2t ).

4 (v)

e

s

2

s s

2

s 2s

2 s

2

s

*L

2

L

s

2

t

(e cos t )*(cos t ) t

e

u

cos u cos (t u ) du

t

2 2 2 5

e u [cos t cos t

cos (2u t )] du

u t

e

2 5 e t)

cos t (

e t (sin t 3 cos t )

e u { cos (2u t ) 2 sin (2u t )}

e t (2 sin t cos t ) (2 sin t

5

(sin t

t

cos t )

3 cos t ).

Part A

f (t) and f t). Under what conditions does this relation hold good? 2. Express L {s (s)} interms of L { (s)}. State the condition for the validity of your answer. t

3. Express L

t

f (t ) dt dt in terms of L{f(t)}.

3.93

Chapter 3: Laplace Transforms 4. State the relation between L { (s)} and L

(s ) .

s2

5. State the initial value theorem in Laplace transforms. mutative. * g (t) = g (t), when g (t) = t. 9. State convolution theorem in Laplace transforms. of the following functions: 2 e at . cos a t t Find the inverse Laplace transforms of the following functions: s2

s 2)3

(s

s (s a ) 2 + b 2

(s

)3

(s

s ) (s

)

Find the Laplace transforms of the following functions: t

t

sin t dt t

et t

t

t

dt t

t

t e t dt

2 os t dt t

e t sin t d t

22.

t sin t d t

Find the inverse Laplace transforms of the following functions: 23.

25.

s (s s (s 2

24.

a)

26.

a2 )

27. If L f (t )

s 3 ) (s

(s

28. If L { (s )}= ( 2

2e

s 2 (s

)

s (s 2

)

lim ( f (t )} and lim{ f (t )} .

) t

t

e

t

t

lim (s (s )} and lim {s (s )}. s s

tn 29. Show that * * * ......* (n times) = , where * denotes Convolu(n )! tion. t

L f (t )

s2

, evaluate

f (u ) f (t u ) du .

Engineering Mathematics II

3.94

ing functions: (s 33.

) (s

32.

)

34.

s (s 2 + a 2 ) s2

35. 2 / (s

)2

(s

s 2 (s +

.

Part B L (t sin a t) and hence a t sin a t) and L (sin a t + a t cos a t). L (t cosh a t) and L (sinh a t + a t cosh a t) and L (a t cosh a t a t).

L (2 cos at

38. Find L

39. Find L

( s + a) (s + b)

(s

L

) (s

s2

s ( s + a) (s + b)

L

) (s 3)

a2 s2

s2

L

(s

) (s 3)

.

s

L

b2

) (s

s2

a2 s2

b2

s2

and L

s2

a2 s2 s

L s s2

L s2

4

a2

43. Given that L

t sin 2 t 4

2

4

s3

L s2 s s2

s2 s2

2

,

2

42. Given that L

L

b2

2

,

a2

s

4

s2

s3

L

=

3) 2

(s

L

s

2

3

t sinh at 2a

2

s2

4

4

and

2

4

a2

2

and

(sin t cosh t

L

s s (s

cos t sinh t )

a 2a )

2

.

L

s s4

4

,

3.95

Chapter 3: Laplace Transforms

L

s2 s4

s3

and L

4

s4

4

.

Find the Laplace transforms of the following functions: t

t t

45 . e

t e sin t dt

44.

t

t

t t

46. t

e sin t dt t

2t

e

47.

sin 3t dt t

t

sin 3t dt t

2t

48. e

t sin t dt

49.

e

t

2t

sin 3t dt

Find the inverse Laplace transforms of the following functions: s

s s

2

: Consider the function as

s 2 (2s 53. 56. 59.

(s

4s 7 3) (3s

) (s

2

52.

5)

s

54.

2)

57.

s (s 2 + 9) 2 s (s 2

a 2 )2

s (s 2 s

s s2

2

62. f (t )

t

s)

25) 55.

4

(s 2 + 6s (s 2

f (t) = (2t + 3)2 e

6s

s s (s 2

2

58.

(s 2 + 9) 2 (s 2

a 2 )2

4s)2

L

s (s

4)3

63. Use convolution theorem to evaluate t

e

u

sin (t

u ) du

t

64. Evaluate

65. 67.

cos a u cosh a (t

s (s + 4) (s 2 + 9) 2

(s 2 + 4) 2

66. 68.

u ) d u , using convolution theorem.

s2 (s + a 2 ) 2 2

(s +

s 2 + 4)

Engineering Mathematics II

3.96

69.

3.11

s 2 (s

s4

)3

4

SOLUTIONS OF DIFFERENTIAL AND INTEGRAL EQUATIONS

differential (both ordinary and partial) and integral equations. We shall apply this and a few integral and intergo-differential equations. The advantage of this method is that it gives the particular solution directly. This means that there is no need the classical approach.

tion in y (t), simultaneously using the given initial conditions. This gives an algebraic equation in y (s ) = L{y (t )} .

Note

L{y (n ) (t )} = s n y (s )

sn y (

sn

y

) ........ y (n

)

( .

2. We solve the algebraic equation to get y (s ) as a function of s. 3. Finally we take L {y (s )} to get y(t). The various methods we have disL {y (s )} . The procedure is illustrated in the worked examples given below:

Example 3.1 Using Laplace transform, solve the following equation di L, R, E and a are constants. L Ri E e a t ; i dt Taking Laplace transforms of both sides of the given equation, we get, L L{i (t )} + RL{i (t )} = EL {e i.e.,

i.e.,

L{s i (s ) i ( } + Ri (s )

(Ls

R ) i (s )

i (s )

(s

at

E s

E s

a

E a ) (Ls

}

R)

a

, where i (s ) = L{i (t )}

3.97

Chapter 3: Laplace Transforms

E

R aL s

aL R

a

s

R L

Taking inverse Laplace transforms i (t )

E L R aL E (e R aL

s at

L

a e

Rt L

s

R L

)

Example 3.2 Solve + 8y = e 2 t, y Taking Laplace transforms of both sides of the given equation, we get [s 2 y (s ) s y ( )

i.e.,

(s 2

y ( )]

4 s 8) y (s )

y (s )

[s y (s )

s 2

( s

s 2

)

s (s 2) s 2 A s 2 4 s 2 4 s 2 4 s 2

y

y ( )] 8 y (s )

4

4 4

L

e 2t e 2t (

4s 8

Bs C s 4s 8 2

s2

4s 8 s

s

2

4s 8

s

s 2 s2 4s 8 s2 4s 8 7 s 4 2 s2 4s 8 7 (s 2) 6 4 (s 2) 2 4 4

s 2 e 2t

e tL

7 s 6 4 s2 4

7 cos 2t 3 sin 2t 4 cos t

2 sin 2t )

Engineering Mathematics II

3.98

2 Example 3.3 Solve y y y = (t ,y y Taking Laplace transforms of both sides of the given equation, we get,

[s 2 y (s ) i.e.,

(s 2

i.e.,

y (s ) = =

sy ( ) s

s

) y (s )

s (s

4

y ( )]

)

s (s

)

2

2

s (s

)

6 tet

s (s

4e t

t et

3et

t et

)

2

3

s (s

)

2

3

)2

s (s

t

t

t et d t d t

2

)2

s (s 2

2

t

t e t dt

t

t

t et d t d t d t

2

t

(t et

et

t

(t et

)+2

et

t

(t et

)dt t

=

2 (t et

2et 2 (t et

et

3 t et

2t

5

7et

t et

t2

t

t

(t et

2) + 2 t2 2

3 et

2t

2et

t

2) d t

3)

Example 3.4 Solve y y = sin t, y y Taking Laplace transforms of both sides of the equation, we get [s 2 y (s ) y (s ) =

sy ( )

s2

y ( )]+ 4 y (s ) =

4 s2

2

4 s2

y

2

4 2

y (s ) =

2

s2

2

4

s2

2

Inverting, we have, sin 2t

If 2 (s

y=2L

4) 2 (s 2

4) 2

)

2

2

2

t

s

2 )

4et

2 s2

2 2

t

y

2 s3

s

6 (s

y ( )]+ y (s ) = L(t 2

[s y (s )

sin t , if

2

et

)dt dt

3.99

Chapter 3: Laplace Transforms

8

(sin 2t

2t cos 2t )

Example 3.5 Solve the equation and

2y = 3 cos 3t

t, y

Taking Laplace transforms and using the giving initial conditions, we get (s 2

s

2) y (s ) =

3s

33

2

2

s 9 s 6 s 2 3s s2 9

y (s ) =

6s 2 s

2

6

9

3s

9 (s

2) s

As B C s2 9 s 3 s2 9 s 2 t

y = sin 3t

,

) D s

by the usual procedure

s

+ et.

Example 3.6 Solve the equation (D2 + 4D +

y = e t sin t, y

d . dt Taking Laplace transforms and using the given initial conditions, we get Dy=

D

t

(s 2

s

3) y (s ) =

y (s ) =

s2

2s

s2

2

2s

2 s2

s

3

As

B Cs D s 3 2s 2 s 2 2s 3 2s 7 , 85 s 2 2 s 2 s 2 s 3 s2

A, B, C, D by the usual procedure. (s 85 (s y

85

)+9 )

2

e t { 2 cos t

2(s 2) 7 (s 2) 2 9 9 sin t}+ e

2t

2 cos 3t

7 sin 3t 3

Engineering Mathematics II

3.100

Example 3.7 Solve the equation (D2 + 6D + 9) x = 6 t2 e t, x t Taking Laplace transforms and using the initial conditions, use get, (s 2

6s

9) x (s ) = x (s ) = x

(s

3)3

(s

3)5 t

e

2

L

s5

Dx

.

3t

t4 e

9y = cos 2t, y

Example 3.8 Solve the equation

y(

Note

In all the problems discussed above the values of y and y’ at t given. Hence they are called initial conditions. In fact, the differential equation with such initial conditions is called an initial value problem. But in this problem, the value of y at t t = /2 are given. Such conditions are called boundary conditions and the differential equation itself is called a boundary value problem. towards the end using the condition y ( Taking Laplace transforms, we get s 2 y (s )

i.e.

(s 2

sy( )

9) y (s ) = y (s ) =

y ( ) + 9 y (s ) = s s

2

4

s

4 s2

y Given

y

5

s2

9

s 5 s2

A

cos 2t

s2

s2

9

9

s2

A 9

A sin 3t 3

4 cos 3t 5

2

i.e.

5 y

5

A 3

cos 2 t

A

9

s

s 4

4

A , where

s

s s2

s s2

5

4 cos 3 t 5

4 sin 3 t . 5

s2

9

3.101

Chapter 3: Laplace Transforms Example 3.9 Find the general solution of the following equation 2 2 y = f(t). Taking Laplace transforms, we get [s 2 y (s ) i.e.

(s 2

i.e.

sy ( )

y ( )]

y ( )]+ k 2 y (s ) = f (s )

k [s y (s )

ks

k 2 ) y (s ) = sy ( ) +[y ( )

(s

k ) 2 y (s ) = As

ky ( )]+ f (s )

B + f (s )

[As y y (s ) =

As (s k ) 2 A (s

B (s k ) 2

k ) + (Ak (s k ) 2

C s

B)

f (s ) (s k ) 2

C k

C e kt

y

f (s ) (s k ) 2

k )2

(s

(s

C t e kt

f (s ) k )2

L

f (s )

, where C = A and C2 = Ak + B

k )2

(s

y = (C + C2t) ekt + f (t) * t ekt

i.e.

t

y

i.e.

(C +C2 t ) e

kt

u ) ue ku du .

f (t

Example 3.10 Solve the equation (D3 + D) x = 2, x = 3, D x = at t = Taking Laplace transforms and using the initial conditions, we get (s 3

s ) y (s ) = 3s 2

y (s ) =

2 s

s

3s s

2

2 s

2

s s

2

2

s s2

t

y

3 cos t

sin t

3 cos t

sin t

t

sin t dt

t

sin t d t d t

2 t

cos t sin t = cos t sin t

cos t ) + 2 2 (t 2 t.

sin t )

(

cos t ) dt

D2x =

Engineering Mathematics II

3.102

Example 3.11 Solve the equation

d4 y d x4

d3 y d x3

, y

dy dx

2 and

d2 y d x2

d3 y d x3

at x

Note Change in the independent variable from t to x makes no difference in the procedure.) Taking Laplace transforms and using the initial conditions, we get (s 4

s 3 )y (s ) 2 s 3 s2 s2 s

y ( s)

y (s)

i.e.

s A s

s2

s

B s2

C s

.

s

x + ex

y

Example 3.12 Solve the simultaneous differential equation

and

dy dt

2 x cos 2t , x

dy dt

2 x sin 2t

y

Taking Laplace transforms of both sides of the given equation and using the given initial conditions, we get sy ( s ) 2 x ( s )

and

sx ( s ) 2 y ( s )

x (s)

x

2 s

2

4 s

s

2

(2)

4

s s

2

2

4

sin 2t

s

2

4

and y ( s )

cos 2t and y

2 s

2

4

t.

Example 3.13 Solve the simultaneous equations 2x y x = 2t and x y x y = t2 t, x Taking Laplace transforms of both the equations, we get [ sx ( s )

] [ sy ( s )

]

x (s)

2 and s2

y

3.103

Chapter 3: Laplace Transforms [ sx ( s )

[ s y ( s)

]

i.e.

x (s)

]

y ( s)

s3 2 s2

(2 s 3) x ( s ) sy ( s ) ( s 2) x ( s ) ( 2 s

and

x ( s)

s 3 s

s

s

s3

s2

3

(2)

s

s s

s

8

25 8 5s 3

8

78 s 35

9 e 8

x Eliminating y we get 5x x

) y (s)

3

s

s2

s 3 34 s

54 5s 3

7 3 5t e 8

t

(3)

y = t2 + 3t y = 5x 5

9 e 8

t2

t, 9 e 8

y

i.e.

t2

x

t

t 3

t

e5

9 e 8

t

t

7 53 t e 8 on using (3)

49 e 8

3 t 5

t2

3t 4 .

Example 3.14 Solve the simultaneous equations Dx + Dy = t and D2x y = e t; x = 3, Dx y t Taking Laplace transformed of both the equations, we get sx ( s ) 3 sy ( s ) s 2 x ( s ) 3s

2

y ( s)

i.e.

x (s)

y ( s)

and

s 2 x ( s)

y ( s)

s2

and

s s3 s

s 3s 2

(2)

Engineering Mathematics II

3.104

x (s)

(s

) s

s

2 s

2 2

x

i.e.

s

e

t

e

t

2.

s s

2 2 s2

2

x

3s 2

3 2

e

2

3

s s

s

s

3s

3

5

2

2 s

2

3

s s

s

2

s s

2

t

t

3 sin t dt

3 sin t 2

5 cos t 2

cos t

2

s

2

s3 s 2

5 cos t 2

3 sin t 2

s

2

2

3

2

3 sin t 2

t

2

3 s s

2

2

cos t

t2 2

t

t

sin t dt dt d t t2 2

cos t

2

(3)

3 e t sin t cos t 2 2 2 From the given second equation we have y=x e t

(4)

x

2

e

3 sin t 2

t

2

cos t.

Example 3.15 Solve the simultaneous equations D2x Dy = cos t and Dx + D2y t; x Dx y Dy t Taking the Laplace transforms of both equations, we get s 2 x ( s ) sy ( s )

and

sx ( s )

s s

s 2 y (s) 2

x ( s)

s

2

(2)

s2

s s

2

s s s

s2

s

2 2

2 s s2

s

2

2

s s2

s2 s s2

2

2

3.105

Chapter 3: Laplace Transforms 2

y (s)

and

s

2

2

s2 t

t

x cos t

2 sin t d t

cos t

cos t

L

s2

2L

s2

2

dt

t

sin t sin t

t cos t dt

t sin t

y sin t 2

and

2

sin t t cos t

= t cos t. Example 3.16 Show that the solution of the equation L

di dt

t

Ri

i dt

C

E, i

E e at sin t ,if 2 L E te at ,if L E e at sinh kt ,if 2 kL

i

R , 2L

a

where

L, R, E are constants] is given by

R2

2

LC

4 L2

Note

and k2

2

.

The given equation is an integro-differential equation, as the unknown (dependent variable) i occurs within the integral and differential operations.] Taking Laplace transforms of the given equation, we get Lsi ( s )

Ri ( s )

LCs 2

i.e. i ( s)

Cs

RCs EC 2

LCs RC s E L s2 R s L LC

i (s)

E s

i (s)

EC

Engineering Mathematics II

3.106

E L

R 2L

s

E L (s i (t ) =

2

LC

a)2

E e L

at

R2 4 L2

2

,iif

2

sin t

If

2

If

i (s )

E L (s

i (t )

E te L

2

and

2

=

a)2 at

,

i (s )

E L (s

i (t )

E e Lk

a)2 at

k2

sinh kt .

Example 3.17 Solve the simultaneous equations 3 + 2 + 6x t

y

y 3

x dt

3 sin t ,

cos t

x (0) = 2 and y

Taking Laplace transforms of both the equations, we get 3[s x (s ) 2] 2[s y (s ) 3] y (s )

6) x (s )

s y (s )

s y (s ) 3 i.e.

(3s

3 x (s ) (s s x (s ), we have

and

[3(s

i.e.

2) (s

)

x (s ) 3 x (s ) s

] x (s ) x (s )

) y (s )

and s s

s 3 s 2

2 s (s 3) s

6s

2 3 s

2 s 3

2

2

3

2

s

3

2

(2)

Chapter 3: Laplace Transforms 2 sin t 3

x

2e

3t

3.107

.

y (s ) , we have y (s )

3s 2

5s 2

(s

y

) s

e

s

2

s 3

s

2

2 cos t + sin t.

Example 3.18 Solve the integral equation y (t ) =

t

t2 2

u y (t u ) du

Noting that the integral in the given equation is a convolution type integral and taking Laplace transforms, we get y (s ) =

L(t ) L{ y (t )}

s3 s3

s2 s

2

y (s )

s2

y (s )

or y (s )

s3

s

s2

t

y (t )

sin t dt t.

Example 3.19 Solve the integral equation t

y (t )

a sin t 2

y (u ) cos (t u ) du.

Taking Laplace transforms, y (s )

i.e.

y (s )

i.e.

(s ) 2 y (s ) s2

a s2 a s2 a s2

2 L { y (t )} L (cos t ) 2s s2

y (s )

Engineering Mathematics II

3.108

a

y (s )

i.e.

2

s

at e t.

y (t )

Example 3.20 Solve the integro-differential equation t

y (t )

t

y (t u ) cos u d u , y ( )

Taking Laplace transforms, s

s y (s ) 4

s

i.e.

s

2

s

2

y (s )

s2

i.e.

y (s )

y (t )

y (s )

s2

4

s2

4 s2 s5

4 s

5 s3

4

5 2 t 2

s5

24

t4.

Part A Using Laplace transforms, solve the following equations: x + x = 2 sin t, x t 2. x ,x 3. y y = t, y 4. y + y = y

=

t

5.

y

y (t ) d t

e

t

t

6. x

x (u ) d u t 2

2t

Chapter 3: Laplace Transforms

3.109

t

7. x

x (t ) d t

cos t

sin t

t

8. x

x (t ) d t t

9. y

e

2u

y (t u ) d u

t

y

y (u ) sin (t u ) d u t

f (t ) cos t

e

u

f (t u ) d u

t

y (t ) t

sin u y (t u ) d u

Part B Solve the following differential equations, using Laplace transforms: + 3 +2x = 2 (t2 +t x x y y y = 2e t, y y x x x t, x x D2 + D y t, y Dy = 2 at t x x x = e t (cos t t), x x y y y = 8 et sin t, y y x x x = t 2 et, x x y y = t cos 2t, y y x x t, x x( 22. y y t, y ( y 23. (D2 + a2) x = f (t). 24. (D3 D) y = 2 cos t, x = 3, Dx = 2, D2x at t 25. x x x x e3t, x x x 26. (D4 a4) y y y y y Solve the following simultaneous equations, using Laplace transforms: 27. x y = et; y x = sin t, given that x y 28. y = sin t; y x t, given that x = 2 and y = t 29. x x y= t; y x+y= t, given that x = 2 and y = 3 at t Dx + Dy + x y = 2; D2x + Dx Dy = cos t, given that x Dx = 2 and y at t D 2x + y t; D2y + x = 5 cos2t, given that x = Dx = Dy y t

Engineering Mathematics II

3.110

t

32. x

y

2e t

x

e t ; 2x

y

2et (t

y dt

3

3) , given that x

y Solve the following integral equations, using Laplace transforms: t

33. x

3x

2

34. y

4y

5

35. x

x

36. y

y

3

37. x (t )

4t

3

38. y (t )

e

x dt

t, x ( )

y dt

e t, y( )

.

cos t , x ( )

.

.

t

t

x dt t

y dt

3e

2t

sin 3t , y ( )

.

t

x (u ) sin (t

u ) du .

t

t

39.

x(u ) t

u

t

y (u ) cos (t

2

du

u ) du .

t2 .

t

t

y (u ) y (t

u ) du

2 y (t ) + t

2

ANSWERS

t 3. tan t ; e

.

. (

2

9.

e

s2

s e

s

)

s

s 2 s

2

4

s2

e

s 2

e

{

(s

2)

}.

2s

. (

e

2 s

)

2 s

2

.

3.111

Chapter 3: Laplace Transforms s

e

s2

;

s

2s

s2

4 s2

2

2

4 e s

23. 25. 27. 29.

4 s

2 s

2

s s

4

2

36

s

s s

2

2

3 4 s

s

2 s 2) 2

(s

9

2

.

s

t

37. 2

s2

2

6a 3 s4

6a 2b s3

2 s

s2

36

9

s2

s

e s.

3ab 2 s2

.

24.

2 . s

26. 28.

3 2 s

2

b3 . s

. 3

8 s

3

s

2

s

s

2 )3

(s

)2

(s

3 4 s

3

.

s .

s

3

96

a

(

32. (t

e t.

38.

at

).

2

46. s

a s

4t

(2

a

u (t). 3t 2 ) .

t ) e 2t .

42. cosh 3t + 2 sinh 3t. 44.

45. e3t cos t.

s

a) ua (t).

t 3 (2 3 sin 2t. 2

)3

(s

34. e3(t . u2 (t) 36. sin t + sin (t

t 3 e 3t / 2

e

)3

(s

2 cos 2t

47.

22.

4

33. u2 (t u3 (t). 35. cos 3 (t u (t)

43.

;

e

s2

s

2s 4

2e3 2s 3

39.

.

3s

e

)+

s

( cos + s sin )

3s 2

2s

s

+e

s2

s

2 s3 s

2 s/3

48.

2

e t sin 2t . 8

(s

8 s

9 3)

3

2

(s 3 s

3)

2

s

3

3 s

2

s

4

.

.

Engineering Mathematics II

3.112

49.

4a 3 s 4 4a 4 (s

3)

55.

(s

)2

(s

4 (s

2) 2

s2

4

s2

2

4 6 (s

58.

s

s2

4

2

t

3 2) 2

(s

9

.

2s

s

)2

2s 5 s

9

2

s

7

e

e

s

3

2

bt / 2 a

59.

2

2s 2

s2

sin 2t

3

2

. 5 e 2

2t

cosh

sin 3t . 67. cos t 69.

m l

b 2a

3 2a

e

2 cos 2 t t/2

sin

3 cos

t . 2a

bt / 2 a

kt 2a

m l

b 2a

2a kt , sinh 2a k

k2.

ac > m l

.

9 2 t e t. 2

t

ac < bt / 2 a

.

+ t + 2e + t e .

t). t 2a

2

2et

63.

8

2)

2s

2

cos

if b2 e

s (s 2

9 2) 2

(s

25

.

. 6

)2

(s

5 2) 2

(s 57.

(2 cos 5t

if b2

a

2

+ 2e2t + te2t.

66. sin t

a

)2

(s

et + 5e2t. e

a

3 s2 s

2

4

62. 2e

68. e

.

36

53.

2)

s2

s

.

8

) 4

56.

3) 2

(s

4

cos + (s + k ) sin (s k ) 2 2

52.

54.

3 3) 2

(s

4

2 2) s 2 4 s

(s

b t , if b2 2a

ac = .

cos 3 t . 3 t 2

sin

3 t. 2

Chapter 3: Laplace Transforms t

e 72.

73. 74. 75.

4a 3

2a

s

(sin at cosh at

t t cosh . 2 2

3 sin

3 t . 2

cos at sinh at).

77.

t sinh at

79.

2 3 2

2

s e a2

s2

a2

s

.

. 2

7.

(sin t t cos t ) .

9.

s s2

s2

k

2 2

s

k

s2

bt

2 2

at

2 cos 2t t

2

.

a2

s2 s2

a2

at

.

et

t

sin at t

k2

e t e

.

3s 2

t

e t

2

2sinht . t

s3 2

a2 2k 3

2ks 2

e

3 t t cosh . 2 2

sin

t sin 2t .

s2

8.

3 t 2

(sinh 2t cos 2t + cosh 2t sin 2t).

4

4

2.

et / 2 cos

sin t sinh t

2

76. cos 78.

t

e

3

(cos 3t + 2 sin 3t).

.

t cos t

sin t . t (s/a).

2

.

3.113

Engineering Mathematics II

3.114 s

22. log

24.

2

s s2

log

E 26.

(s

e 2

29.

s

2

35. 36.

37.

38.

s

2 s

s

2

s

s

e

s 2

3 s2 2

4

s2

4

2 s2

s

(s

45. 3

4

)

e

cosech

. /(s2 +

2

) (e

s/

2

/

e

(s

3) 2

)2

64

2

64

36

s2

s 48 s 2 4

s2

3 3

s2

6s 7 . 6 s 25) 2

s

s

2

3

2

s

.

42.

44. s

2 s

3) 2

.

2

s 2

2 s

2

s2 s

s

36 36

s

9

(s 2

s

e

(s

2

2

2

2

s2

2

s

9 s2

.

s

2

3

(s 4

s2

s

s . 2

8 (s 3)

4 s

4 2

).

33.

3

39. s

43.

s

e e s).

e

s2

log

s2

s2

tanh

s2

s

e

.

s

E sT tanh . s 4

27.

]/(s

s2

2

4

coth ( s)

2

e

34.

25.

.

2 s/n

e

.

s/E

e

s

a2 s

s

23. log

.

(s 2

(s

s

2

s

)3

(s

3) 2

3

s2 (s 2

2 s

3

.

) 3)

(s

3

.

.

5)3

.

s 2

.

3.115

Chapter 3: Laplace Transforms

46.

2 ( t

47.

2 (e t

49.

2 sinh a t t2

cos a t ) bt

e

t

57. 59.

a

(t )

t

sin 3t.

52.

sin

t

e

e 2t ) .

t . 2 bt

sin at .

56. log 3.

2

b . a s log s

log 2 .

2

58. log

8 2

63.

2t

2a cosh at t

2 (cos t t

48.

2 t t sin sinh t 2 2

53. 55.

cos a t ).

4

log log

s2

a2 s

2

s2 s2

4

62.

.

64. s log

65.

t sin t . 2

66.

67.

t 2t e sin t . 2

68.

69. e

t

t2 2

s s2

e at (a sin b t

s t

b cos b t )

.

s2 s s

s.

cot

2

a

sin at t sin a t. 4t

sinh t..

s

t

4

s2

t sinh at . 2a

t e 2

2

b

4

t) sin t.

8. No; * t

log

.

.

a

s t) e

t

2

a2 t2 2

et

e

t

t

e t.

.

Engineering Mathematics II

3.116

s

23.

s.

cot

s (s

)

(

e

a

s

log

s

at

)..

at

24. e

s

s

s (s 2

2

2s

.

2)

s

+t

36.

37. 38. 39.

2a s a2

s2

a2

s2

a2

2

43.

a2

b2

44. 47. 49.

2

s

at

sin b t

);

s2

s2

)

2s

2

s

2 3

3 s2 log 26

2

a2 at

(a e

s s

cos a t ).

be

bt

).

a2

b2

(cos b t

cos a t );

b sin bt ). t t sin 2 t ); e 3t sin 2 t . 4

(sin t cosh t 45.

.

2

(

a2

9 3t e . 2

4e 2t

sin a t ;

.

2

. in t

;

.

a t cosh a t ); (a t sinh a t 2 2

).

2

2a 3

a b

a

(cosh a t

t + e t.

a2

2 t sin 2 t ); (cos 2 t

(sinh a t

cot

a2

(a sin at

(s

; 2

e 3t ; e t 2 2

sin t sinh t;

s s

a2

s2 e

(sin 2 t

2

33.

.

2a s 2

; 2

2a s 2

e 2t

b2 b

2a

s2

; 2

a2

4 42.

et

2s3

bt

(e

32. t et. t

; 2

s2

a b

).

.

s 2

2

a2 2

3t

s2

2

22.

t. (et e 4 34. e t + t

log

48. 3

6

2 cosh a t );

t e 2a

at

sinnh at.

cos t sinh t ); cos t cosh t. 2 s

2

s cot

2s 2

2

.

46. 2

3 2

3

.

s

cot s

2

.

3s 2

4s

2

s s

2s

2

2

2

2

.

3.117

Chapter 3: Laplace Transforms e t.

t 4 e 9

53. e

57. 59.

t

(sin 3t

54 2

3t

e

65.

2 5

69.

(3 sin 4 t

2t

3 t cos 3 t ).

56.

t cos t ).

58.

64. 66.

2 t cos 2 t ).

4t

t

6)

5 e 2

y

2 e 3

x

e

2t

e

t

2 2t

sin 2 t ). 3 t sin 3 t ). sinh a t ) .

2a 2a

(sin a t

sinh a t ).

(sin a t

a t cos a t ).

t

+ sin 2 t

t.

(sin t cosh t

cos t sinh t ).

2. x = t e t. 5. y t)e t. 8. x = e 2 t.

3. y = e t t 6. x = 2t. 9. y t. y

f (t) = cos t + sin t

t

x

(a t cosh a t

2a 3

4

t2 2

x = t2

cos 2 t

(2 2 cos 3 t

68. 2 e

3.

x=e t t + sin t. 4. y e t. 7. x = cos t. y

t

(

sinh 2 t ).

cos 5 t )

e t 2 (t 2

4

a t sinh a t ).

e t ).

sin t

(cos 2t

4cos 4 t )]. 54.

(2 t cosh 2 t

(cos t

73 . 225

t

t).

(sin t

(sin 2 t

67.

7

5t /3

(2 2 cosh a t

2a 4 e

63.

3t

[4 e

52.

55.

3 e 25

3t /2

e

3t

4 t e 3

cos t

y

t

2 cos t 3 cos 2t

3 sin t 2

25

( e

sin t . sin 2t .

t (sin t 2

cos t )

t

te

t

t

t3 . 6

e 4t )

Engineering Mathematics II

3.118 y = 2 (sin t cosh t t

x

t sinh t).

4

t

2 et

y

t.

22. y

x = 2t

5 sin t 9 4

(t

4 sin 2 t 9

t cos 2 t . 3

) sin 2 t .

t

23. x

A cos a t

B sin a t

24. y = 3 sinh t + cosh t 25. x = 2e 3 t 27. x

t

( e

2

sin a u f (t

26. y

35. x

+e

sin t

2 2

36. y e

2t

cos t

) e t. t

{(

t) e

2t

3 cos 3t

37. x t 39. x (t ) y (t) =

(cos a t

cosh a t ).

t sin t ).

34. y

t 2 e 3t. t.

2

e

t

2

e

cos t}. 7 sin 3t 3

3 sin 2 t . 2 t

2

cos t t cos t );

28. x = 2 cosh t; y = 2 sinh sin t. 29. x t t 2 + e t; y = 2t x = t + sin t; y = t + cos t. x = sin t + cos 2t; y = sin t 32. x = t e t e t; y = e t + e t. 33. x

u ) d u.

sint + 2.

e t.

(et +2sin t

2

y

t2et

a

3 t sin 3t 2

t cos 3t .

38. y (t) = e 2t

8 32 t 3

t

t)2.

2t

(cos t

3 sin t ).

Chapter

4

Analytic Functions 4.1 INTRODUCTION Before we introduce the concept of complex variable and functions of a complex

If x and y are real numbers and i denotes 1, then z x iy is called a Complex x is called the real part of z z R z); y is called the imaginary part of z z Iz The Complex number x iy is called the conjugate of the Complex number z

x

iy

z.

zz

x2

y2

r 2 , where r is z z

modulus of the Complex number z. Also z = z ; R z

z 2

, I z

z z 2i

Y P x, y) r, )

r y O

x

M

X

Fig. 4.1

The complex number z x iy P x, y) with reference to a pair of rectangular coordinate axes OX and OY, which are to each complex number, there is a unique point in the XOY corresponding to each point in the XOY-plane, there is a unique complex number. The XOY-plane, the points in which represent complex numbers, is called the Complex plane or Argand plane or Argand diagram.

Engineering Mathematics II

4.2

If the polar coordinates of the point P modulus of z or z and Since x = r cos

MOP

tan

1

r, y x

), then r amplitude of z

OP

x2

y2

z

and y = r sin ,

z x iy r i sin ) or re i r i sin ) is called the modulus-amplitude form of z and re i is called the polar form of z

4.2

THE COMPLEX VARIABLE z x iy is called a complex variable, when x and y are two independent

The Argand plane in which the variables z the z point z

4.2.1

z is referred to as the

Function of a Complex Variable

If z x iy and w u iv are two complex variables such that there exists one or more values of w, corresponding to each value of z in a certain region R of the z-plane, then w is called a function of z and is written as w f z) or w f x iy When w u iv f z) f x iy u and v are functions of the variables x and y w z2, then u iv x iy)2 u w

Thus

x2 y2) i xy) x2 y2 and v 2xy f z) u x, y) iv x, y)

If z is expressed in the polar form, u and v are functions of r and value of z, there corresponds a unique value of w, then w is called a single-valued function of z 1 are single-valued functions of z z z, there correspond more than one value of w, then w is called a multiple valued function of z w z and w z) are multiple valued functions of z w z is four valued and w z z w

4.2.2

z2 and w

Limit of a Function of a Complex Variable

The single valued function f z) is said to have the limit l i ) as z tends to z0, if f z z0 z0) such that the values of f z) are as close to l as desired for all values of z z0, but different from z0 lim f z )} = l z

z0

Chapter 4: Analytic Functions

4.3

Note

Neighbourhood of z0 is the region of the z-plane consisting of the set of points z for which z z0 , where z z0 as centre and lim { f z } = l ,

z

z0

, such that f z) l 0

z

, whenever

z0

Note

In real variables, x x0 implies that x approaches x0 along the x-axis or a line parallel to the x x complex variables, z z0 implies that z approaches z0 curved) joining the points z and z0 that lie in the z Y

z

z0 X O

Fig. 4.2

Thus, in order that lim { f z } z

z0

f z) should approach the same value l,

when z approaches z0 along all paths joining z and z0

4.2.3

Continuity of f z)

The single valued function f z) is said to be continuous at a point z0, if lim { f z } = f z0

z

z0

This means that if a function f z) is to be continuous at the point z0, the value of f z) at z0 and the limit of f z) as z z0 A function f z) is said to be continuous in a region R of the z-plane, if it is If f z) u x, y) iv x, y) is continuous at z0 x0 iy0, then u x, y) and v x, y) will x0, y0 x, y) is said to be continuous at x0, y0 ), if lim x y

4.2.4

x0 y0

x, y )] =

x0 y0) , in whatever manner x

x0 and y

Derivative of f z)

The single valued function f z) is said to be differentiable at a point z0, if

y0

Engineering Mathematics II

4.4

lim z

f z0

z

f z0

z

0

This limit is called the derivative of f z) at z0 z0 z z or z z z0, f z0 On putting z0

f z

lim

z

f z0

z z0

z0

z f z

4.2.5

z0 On putting

f z

lim z

z

f z

z

0

Analytic Function

A single valued function f z) derivative at z0 A function f z at R

z0, if it possesses a z0 R of the z a regular function or a holomorphic

function A point, at which a function f z) is called a singular point or singularity of f z

4.2.6

Cauchy-Riemann Equations f z

Theorem If the function f z) u u v and , , x y x

Proof f z) u x, y)

u x, y)

R of the z

iv x, y)

R of the z-plane, then u x

v y

iv x, y) f' z

v and y

u y

v x

R z in R L

lim z

f z

0

L takes the same value, when z In particular, L takes the same value when z QSP

z z

f z

QRP and

Chapter 4: Analytic Functions

4.5

Y Q z + z)

S

y P z)

R

x

X O Fig. 4.3

It is evident that when z takes the increment z, x and y take the increments x and y L

x

lim i

y

u x

x, y

y

iv x

L,

L1

u x

lim x

x, y

iv x

u x

x, y

u x, y

x

0

i

L,

L2

u x, y

lim y

v x

x, y x

y

iv x, y

y

u x, y

1 u x, y i

0

u y

y u x, y y

v x, y

iv x, y

y y

v y

L L1 and L2 u u v and , , x y x L

v x, y

i y

lim i

i

L2, corresponding to the path QSP, y

0

y

iv x y

v x

letting x Thus

u xy

x

lim u x

iv x, y

y

x, y

0

x

i

u x, y

L1, corresponding to the path QRP, x

letting y Thus

x, y + y x

0

R v y

R

v x, y

Engineering Mathematics II

4.6 L1

L2

u v i x x u x

v y

u y

and

v u i y y

v x

R

Note Cauchy-Riemann equations

Thus w

f z z) z) ux ivx or vy iuy , where ux , uy , vx, vy f z), then dw dz

Thus

L1 or L2.

z) is also denoted as d w dz u v i x x x w x

dw dz

Also

L,

i i

u iv

u y

y w i y

v y u iv

the above theorem, we have assumed that w f z)

derived conditions do not w f z). In other words, the two conditions w f z). w f z) R

are

Theorem The single valued continuous function w f z) u x, y) iv x, y) a region R of the z-plane, if the four partial derivatives ux, vx, uy and vy have the ux

vy and uy

vx

R.

Chapter 4: Analytic Functions Proof Consider

u

u x

x, y

[u x

y)

x, y

x ux x

u x, y)

y)

u x, y

x, y

1

4.7

y) u x, y y) . y uy x, y y), 2

y)

u x, y)]

where 0 , 2 1 ux is a continuous function of x and y, ux x

1

x, y

y)

0 as x and

1

ux x, y)

y

, where

0 or as z

uy is a continuous function of x and y, uy x, y

2

y)

uy x, y) 0 as z

2

u where

1

and

, where

2

0

x [ux x, y)

y[uy x, y)

]

],

2

0 as z

2

vx and vy, we have v

x [vx x, y)

where w

u i v ux ivx) x

ux 1

1

uy

w

ivy) y

ivx) x uy i ) and vy ux

2

x

z and

y

1

x z

x z

1

i

x )

2

y z

1

1

iux) y

ivx x x 1

i y) y 2

y, where z

2

1 1

y z

and

2

i

2

vx

vx

2

x

z x z

1

2

ux and uy

ux x i y) ux ivx) z

Now

i

ivy) y

ivx) x

w u x ivx z

],

0 as z

and

Now

y [vy x, y)

]

y z

1

x x

2 2

y y

y,

Engineering Mathematics II

4.8 x z

1

y z

2

z

0[

1

and

z

2

dw u x ivx dz w w

f z)

R.

f z)

4.2.7

R.

C.R. Equations in Polar Coordinates

When z is expressed in the polar form rei , where w u iv, are functions of r and u r, ) and v r, ), assuming that w

Theorem If the function w

f z)

u r, )

u and v, u r, )

iv r, ) is

iv r, )

z-plane,

u u v v , , , r r u 1 v r r

v r

and

1 r

u

Proof f z)

u r, ) z in R

z) L

L

lim z

f z

z z

0

lim i

re

u r

iv r, )

R

f z

r,

iv r

r, re

0

u r,

iv r ,

i

L take the same value, in whatever manner z In particular, L z

rei )

rei ) e r, if

z

i

z

r

L,

0, if z

L1,

Chapter 4: Analytic Functions

Thus L1

u r

lim r

e

r, e

0

u r,

i

i

r

e

i

v r, r

rei ) rei i , if r

z

if r L

L2

r,

u v i r r

i

z

Thus

v r

4.9

u r,

lim

u r, re

0

1 e r

i

u

i

Since L

L2,

i

z tending

i

i

v r,

v r, re

i

i

v

L1 and L2

R.

u u v v , , , r r

R.

Since L L1

e

i

L2

u v i r r

1 e r

i

u

i

v

u 1 v r r v r

and

1 u r

R

Note f z Thus, when f z) z) e

i

ur

ivr )

where ur , u , vr , v w f z), then

z) u r, ) or

L, iv r,

1 e r

i

v

iu ,

L1 or L2.

Engineering Mathematics II

4.10

dw dz

e

dw dz

Also

i

u iv

r i e r i e r i e r

i

u

e

i

w r

iv

i

u iv w

i

f z r, ) in R ur , u , vr, v must exist

w

f z) u r, )

w iv

y

0

in the region R. WORKED EXAMPLE 4 a)

Example 4.1

li z

x2 y x2 y 2

f z ) , when f z

0

f z) x Thus

li y x

0 0

f z)

lim x

This does not mean that li z

0

0

0 x2

lim 0 x

f z) li z

li z

f z

0

0

0

0

0

f z ) can be 0, as per the mathe-

f z)

0

z 0

, whenever 0

Now, using polar coordinates, f z

0

r cos r 2 cos 2

sin sin 2

r | cos 2 || sin | r, since cos f z

0

1 and sin

, when r
c

u

O

c

x

O

Fig. 4.14

The image of the region y v c 2 u v2 c u2

v2)

u2

1 2c

v

2

1 2c

v

u2

v or

u2

1 2c

c

2

1 2c

v [ c

v2 2

c < 0]

2

, whose centre is 0,

y

1 2c

v

O

1 ) 2c

x y>c

u

O ) Fig. 4.15

y

1 v u2

u2

and radius is 1 2c

v2

1 v2

v or u2 v 2)2 22, u2 v 2)2 22

The image of y

u2 v2 u2 v

1 v u 2 v2 2 v or u2 1)2

v

1)2

Chapter 4: Analytic Functions y

4.65

v y = 1/2

u

y = 1/ x

O

Fig. 4.16

Example 4.6 x2 w

1 is the lemniscate R2 z

1

1 under the transformation w

z2 is

cos w

x

1 under the transformation

cos 2 z

the cardioid R

y2

iy

1 1 or z z w 1 1 R R ei

i sin )

The transformation equations are x y

and

1 cos R 1 sin R x2

y2 w-plane in polar co-ordinates

as

1 cos 2 R2

1 sin 2 R2

1

R2

cos 2 z 1 x 1 iy 1 x 1)2 y2 1 or x2 y2 2x 0 2 r 2r cos 0 or r 2 cos The transformation is w z2 R ei r2 ei2 R r2 and 2 Eliminating r and r2 R

2

cos2 ) cos

Engineering Mathematics II

4.66

Example 4.7 w-plane of the region of the z the straight lines x 1, y 1 and x y 1 under the transformation w z2 w z2 u iv x iy)2 x2 y2 i2 xy The transformation equations are u x2 y2 and v 2xy w z2, we get the image of the line 2 x 1 as the parabola v u 1) and the image of the line y 1 as the parabola v2 u The image of the line x y x and y u v u v

and and Eliminating x v

u

1 1

u

x2 2x 2x 2x

x)2 x) x

1

2 1 u2 v or u2 v 2 which represents a parabola in the w v2

1/2), x 1, y 1), v2

u

1 and x y 1 is the region u 1) and u2 v 1/2) v

v2 u–

y

B

y=1

u + 1)

A' u

O

2

= –2

y=

B' x =1

v2

u

x+

C

C'

1)

1

1/

C"

A

v–

O

2)

Fig. 4.17

Note

The region corresponding region is that which contains the images of the points A and C Example 4.8 x y 1 and y

u 1, u

B

i) x 1, v 1 and v 2 under the transformation w z2

Chapter 4: Analytic Functions

4.67 z2,

w images of the lines x 1, x y 1 and y 2 v2 u , v2 u , v2 u 1) and v2 The given rectangular region in the z 0 and /2 images of 0 and /2 the transformation equations of w

u

w-plane, since the 0 and , as one of z2 in the polar form is 2 v C'

y D y=2

C

x=1 A

D'

B'

x

A'

y=1 B x

O

u

O

Fig. 4.18

z2 are u

w v 2xy. The images of u 1, u x2 y2 1, x2 y2

, v 1 and v xy 1/2 and xy

x2

y2 and

1 in the z y

D'

v D v=2 C u =1 u A v=1 B O

C'

A' B' B" u

A"

C" D"

)

Fig. 4.19

O

x

Engineering Mathematics II

4.68

Both the region in the z-plane are valid images of the given rectangular region in the w This is because we are transforming from the w-plane to the z the transformation z w1/2, Example 4.9

ez:

w

y x, y half of the strip 0 y The transformation equations of the mapping w Rei ex iy ex eiy R and log R

, y

ez

ex y y x is or R e ,

y The image of the y 0 is The image of the region 0 y half of the w

x 0 is R 0 and that of the line y

w is ,

x semicircle w

y

0, between y

0 and y

is the

1, v v

y

u

O x O

|w|=1

Fig. 4.20 v

y

|w|=1

y= O x O

Fig. 4.21

u

Chapter 4: Analytic Functions y and R 1 or w w

circle w

y

4.69 y

and x

is mapped onto the exterior of the w

unit circle w

v

y y=

u O x O

Fig. 4.22

Example 4.10 x

2

and y

c in the z-plane under the transformation w y y=c

C

v

B

x = – /2

sin z

x = /2 S2

S1 O

D

y=–c

C' D'

x

B' S2'

O

S1'

A'

A

) Fig. 4.23

The transformation equations of the mapping function w u sin x cosh y and v cos x sinh y The part AS1 of the side AB c y u cosh y and v Now u is a decreasing function of y y segment

1

AS1 of the u-axis in the w

c

y v

0, since

sin z are

du dy

0 and cosh c

x

/2 and

sinh y

0 , when

u S1 B

Engineering Mathematics II

4.70

v 0 and 1 u cosh c, u The image of the line segment BC, u

1

y

[When

/2

x

x

2

/2

v2 sinh 2 c

1, v

/2

x

/2

0

/2 and c 0, cos x 0 and sinh c BC is the upper half of the ellipse

u2 cosh 2 c line segments

c,

) of the

sin x cosh c and

v cos x sinh c, BC is the elliptic arc u2 cosh 2 c

1

v2 sinh 2 c , 2

1 CS2, S2D and DA of the z-plane are the ) and the lower half of the 2

2

segments of the u w Example 4.11 Show that the transformation w cos z maps the segment of the x x /2 into the segment 0 u 1 of the u maps the strip y 0, 0 x /2 into the fourth quadrant of the w The transformation equation of w cos z u iv x iy) cos x cosh y i sin x sinh y u cos x cosh y v sin x sinh y The image of the x y u cos x and v 0 Since we consider the segment of the x x /2, 0 u 1 Thus the image of the segment of y 0, 0 x /2, is the segment of v 0 u x 0, x /2 and y The image of x u cosh y and v u 1 and v The image of x u 0 and v sinh y Since y 0 for the given strip, v 0 Thus the image of x /2, y 0 is u 0, v 0 The image of y 0, 0 x /2 is v 0, 0 u v

0, 0

u

v

0, u

0 w

l; v 0, u 1 and u and u 0, v 0

0, v

0

0,

Chapter 4: Analytic Functions

4.71 v

y

x = /2

x=0

u

O

x

O y=0

v'

Fig. 4.24

Example 4.12 under the transformation w sinh z The transformation equations of w sinh z i sin i x iy) i sin i ix y) sinh x cos y i cosh x sin y u sinh x cos y and v cosh x sin y The image of x u 0 and v sin y u 0 and 1 2 v 1 2

x u

y iv

x

iy)

y

The image of x u 2

u sinh 2

y and v

v os

sin y

2

1, u

2

0, since cos

The image of y u

1 2

sinh x

1

v

and

2

cosh x 1 2

v2 u 2

v

0)

y v2 u2 the v

v-axis, the ellipse and the rectangular y D

y=

v C B' D'

x

x=0

x

O A

O A'

B y=–

Fig. 4.25

C' u B'

Engineering Mathematics II

4.72 Example 4.13

a 2 b2 z

Show that the transformation w z

transforms the

1 a b in the z-plane into an ellipse of semi-axes a, b in the w 2 k2 In the discussion of the transformation w z , we have proved that the z image of the circle z c r c is the ellipse whose equation is circle z

u2

v2 2

k2 c

c

c k2

Putting c

k2 c

c

k2 c

The image of w

z

a2

k2 c

b2

a2

b2

1 a b 2

1

2

and c

1 a b 2

1 a b 2

we get

a and

1 1 a b a b b 2 2 1 1 z a b r a b 2 2

/ z is the ellipse

u2 a2

v2 b2

under the transformation

1, whose semi-axes are a and b

Example 4.14 Prove that the region outside the circle z 1 whole of the w-plane under the transformation w z z

1 maps onto the k2 , we can prove z

Proceeding as in the discussion of the transformation w z that the image of the circle z r 2 u 2 and that the image of the circle z u a

2

1 a

v 2

a

1 a

2

1 1 a2

r

a is the ellipse

1

The semi axes of this ellipse are f a) Now f a

a,

2

a

1/a and g a

a2 1 0 , when a a2

1 and

a and hence when a f g a) are increasing functions of a, when a The region outside the circle r r a, where 1 a

1 a

a a)

1

1/a2

0, for all

Chapter 4: Analytic Functions

4.73

u2

v2

1 a

a

2

a

1 a

1, 1 a

2

, is

the entire w r

1 maps onto the entire w 1 ze 2

Example 4.15 Show that the transformation w maps the upper half of the interior of the circle z w

iv

and

1 rei e 2

1 e re

is real,

1 ze 2

1 ze

are

i

u

1 re 2

1 re

cos

v

1 re 2

1 re

sin

0,

and r

u

1 re 2

1 re

u

1 2

z z-plane consists of The image of

, where

e onto the lower half of the

The transformation equations of the mapping function w

u

e z

e or r e

e

and v 2

u

1 0 2

r e

2 and v

1

/2

/2

r e 0 or u

u

cos 1 u

1 and v

e and v 0, 1 and v r

e

z-plane

is the entire u r When r

e , re

0

1 and v u

The image of r

2 and v

1 1 re

1

e and 0

Engineering Mathematics II

4.74

1 re

re 1 re 2

1 re

0

0 , since sin is positive when 0

sin

v z|

e

z-plane

maps onto the lower half of the w v

y |z| = e

O

u

x

O

Fig. 4.26

EXERCISE 4 c) Part A

w z is w

w

f z)

e z

1 ze 2

w2 mation w w w

z

z z

) z ). under the transfor-

sin z

2

i

i) z

z 2z

a under the following transformations w i) z i w 2z

w i

x transformation w

iz

i.

2 under the transformation w iz y-axis under the

Chapter 4: Analytic Functions y

w

4.75

1 under the transformation w i)z xy 1 under the transformation w z x2 y2 a2 under the transformation

i)z y

w

1 z

w

1/z

c, when c

x

c, when c

0 under the transformation

0 under the transformation

i z in the z-plane under the transformation

y

mx under the transformation w

zn, where n

w

x

1 and y z

1 under the transformation w iz2 w z2 under the transformation w z2

arg z

u and v under the transformation w iz2 x 1 under the transformation w x-axis under the transformation w cos z y-axis under the transformation w cos z x-axis under the transformation w sinh z y-axis under the transformation w sinh z w

a z 2

ez

1 z )

under the transformation w

z

1 z

Part B i) w i

z w

2i az

z i, z 1 i, z 1 i under w iz 2 i z 1 i i, 1 i and

b, where a

2 1 i and b

z|

i

a maps onto the interior of an ellipse ib y in the w-plane under the transformation w x , 0 b a. Is the a w w z-

16

16

1 maps the circle z z

Engineering Mathematics II

4.76 i y

y

0

1, under the transformation w

1 2c

x, y z

Note

w

0 and

0, 1, 1

i

y

a

z-plane z is to

z maps the region of the z-plane a a/2 onto the upper half of the w x , y 0 under the transformation

a/2

x

sin

cos z x

transformation w

cosh z

from 0 to i onto the segment 0 2 strip x 0, 0 y /2 /2 under the transformation w w z a of the u upper w 1 ze 2

e z

y

0

sin z w

w

1

The properties of the transformation w = log z are identical with ez, if we interchange z and w w cos z w sinh z w

y

x and y

z2

and i under the transformation w z2 a 0, show that w z/a) onto the upper half of the w w log z maps onto the horizontal strip v

those of w

0,

1 z y

under the transformation w

x

, where

u

i) the segment of the y-axis

1 of the u w

z sinh z a z 2

x 1 z

1 in the upper z

2

/2

y

where a is a positive constant a u z-plane onto the

is real, maps the interior of the circle z

1

onto the exterior of an ellipse whose major and minor axes are of lengths 2 cosh and 2 sinh

Chapter 4: Analytic Functions Hint: The image of the circle z

4.77

1 is the ellipse

u2 cosh 2

z

v2 sinh 2

1

a, ), where a 1 is a

point outside the ellipse, since u a, )

u

) and v a, )

v , ) .

4.7 BILINEAR AND SCHWARZ CHRISTOFFEL TRANSFORMATIONS 4.7.1

Bilinear Transformation

az b , where a, b, c, d are complex constants such that cz d ad bc 0 is called a bilinear transformation. It is also called Mobius or linear fractional transformation. The transformation w

Now d w dz

ad bc cz d 2

ad

bc

z-plane becomes a critical

The transformation can be rewritten as w

a c

ad bc . c cz d

the transformation takes the form w

ad

bc

0,

a , which has no meaning as a mapping c ad bc ad bc)

is called the determinant The inverse of the transformation w

az b is z cz d

dw b , which is also a cw a

The images of all points in the z z

d c

for the point w

w-plane is a unique point, except a If we assume that the images of the points z c

d and w c

a c

w- and z

To discuss the transformation w

az b cz d

we express it as the combination of simple transformations discussed in the When c

4.78

Engineering Mathematics II a c

w

bc ad c

1 cz d

If we make the substitutions w1 cz d 1 w2 w1 then

w

Aw2 B

where

A

bc ad c

and

B

a c

z-plane onto the w1-plane, from the w1-plane onto the w2-plane and from the w2-plane onto the w

c

w

a z d

b d

d 0

w Az B.

If the image of a point z under a transformation w called a point or an invariant point w f z) w

f z) is itself, then the point is z f z). az b cz d

az b z cz d

As this is a quadratic equation in z,

Note w1, w2, w of z1, z2, z of the z Let the bilinear transformation required be az b cz d

w

w

a b z d d c z 1 d

or

Az B Cz 1

Chapter 4: Analytic Functions

4.79

Since the images of z1, z2 and z are w1, w2 and w

and

w1

Az1 B C z1 1

w2

Az2 B C z2 1

w

Az Cz

B 1 A, B, C

Solving them we get the values of A, B, C

given,

If z1, z2, z , z are four points in the z-plane, then

z1 z2 z

z

z

z

z

z

is called the

cross- ratio of these points. Cross-ratio property of a bilinear transformation The cross-ratio of four points is w1, w2, w , w are the images of z1, z2, z , z transformation, then w1 w2 w

w

z1 z2 z

z

w

w

z

z

w

w

z

z

a bilinear

Proof az b Let the bilinear transformation be w cz d wi w j

Then

and

a zi b c zi d

azj b

ad bc zi

zj

czj

c zi d c z j

d

ad bc

d 2

w1 w2 w

w

w

w

w

w1 w2 w

w

z1 z2 z

z

w

w

z

z

w

w

w

z1 z2 z

c z1 d c z2 d c z ad bc

2

z

z

c z1 d c z2 d c z

z

z

z

d cz z

d

z

d cz

d

Engineering Mathematics II

4.80

Note

To get the bilinear transformation that maps z2, z , z of the z-plane onto w2, w , w , we assume that the image of z under this transformation is w and make use of the invariance of the cross-ratio of the four points z, z2, z , z w w

w

w

z z

z

z

w w

w

w2

z z

z

z2

z

in the form w

4.8

z2, z , z w w2, w , w Now, w, we get the required bilinear transformation

az b cz d

SCHWARZ-CHRISTOFFEL TRANSFORMATION

w-plane onto x-axis) z-plane is called Schwarz-

the x

Christoffel transformation. x1, x2 xn that are points on the x-axis such that x1 xn, are the images of the vertices w1, w2 wn and are n dw A z x1 dz

1

2

1

z x2

1

n

z xn

1

x2 x w-plane

,

where A Proof A w)

v A w)

P w) 2

A2 w2)

1

A1 w1)

y

–

1

u

O

P'

A1'

A2'

z

x1

x2

0

Fig. 4.27

The given transformation is 1

dw

A z x1

1

2

z x2

1

n

z xn

1

dz

A'

A'

x

x

x

Chapter 4: Analytic Functions

am

n

d w)

1

amp A

2

z x1 )

1 am

4.81

1 amp z x2

…+

1 amp z xn

z1z2 ...) z1) z2 ) .... zk) k z)] Let the image of P w) be z) When P moves towards A1, moves towards 1. As long as z) is on the left of x1), z x1, z x2 z xn 1 z x1) z x2) z xn) But once z) has crossed 1 x1), x1 z x2, is a positive real 1 z x1) z x2) z x) amp z xn) A z [ A is a constant and dz z) 0] Thus when z croses 1, z x1) to 0 or undergoes . [

Increase in amp

1

w

1

)

1

P w), moving along PA1, reaches A1, it changes its direction through an angle in the anticlockwise sense and then starts moving along A1A2 P w) reaches A2, it turns through an angle and then starts moving along A A 2 2 Pw z) moves along the x the w-plane in the anticlockwise sense, the interior of the z-plane should lie to the left of the person, when he or she walks along the corresponding path in the z x corresponding area is the upper half of the z w-plane is mapped onto the upper half of the z

Note 1

as w A z x1

1

2

z x1

1

n

z xn

1

dz B

where B z-plane onto the real axis of the w

z and w x1, x2,

xn can be chosen

Engineering Mathematics II

4.82

xn, If we take A

n

xn

, where

is a constant, the transformation can be

1

written as n

dw dz

1

z x1

As xn dw dz

1

2

z x2

n 1

1

z xn

xn z xn

1

1

1

, the transformation reduces to 1

z x1

1

2

z x2

n 1

1

z xn

1

1 n

This means that, if xn

z xn

1

is absent in the

WORKED EXAMPLE 4 d) Example 4.1

w

z i Prove iz 1

z and its image w, form a set of four

z i iz 1

z iz2 z

z i z

i i)

0 0

i and z

z2

or

iz

0

i. Taking z1 z , z2 w

i, the cross-ratio of the four points z1, z2, z and z z z1 , z2 , z , z

z i iz 1

z i i i z2 z i

i i z i iz 1

i z

i

2 z 8i

z i ,z iz 1

i and

Chapter 4: Analytic Functions i i z2

z

i

i i z2

z

i

4.83

a constant, independent of z Example 4.2 the z-plane into the points 0, 1, i of the w Taking z1 1 i, z2 i, z i and w1 z, z1, z2, z ), we have w 0 1 i w i 1 0

z 2 i

w

w

i, w

w, w1, w2, w )

i z i 2 z 1 i i z i 2 z 2 2i i z i 2 z 2 2i

1

2 z 2 2i i

i z

i

2 z 2 2i i z i i z 0, z 1 and z

z, z1, z2, z )

w w1 w2

w

z z1 z2

z

w w

w1

z z

z1

w

z

w

i of

i and using the

2

of the w1 and w i

To avoid the substitution of w then put

1, w

1 2i 1 i

z 1 i

i w i w

Example 4.3 the z into the points w

0, w2

i

z 1 i 2 z 2 i 1 2i

w i w 1

i,

1 w

Engineering Mathematics II

4.84

w w1 w2 w ww Using the given values z1

1

1 w

w1

i, z2

z w 1 11

z

z1

1, w1

0, w2

1 and

z i 2 ,i z 1 1 i

w

z z1 z2

z

w w

w1

z z

z1

z

w

1, z’

z

z z1 z z

1

zz

z1

0 and w1

1 z i, w2

w

0, we get 1 i z i z 1

, where z

1 and w

1 z

i, we get

1 1 1

w i w i

1 i z 1 i

2w 2i

1 i z

w

z

z z

w w1 w2

Using the values z1 0, z2 w i 1 i w i 1 i

z z1 z2

1 i

1 i

Nr Dr Dr Nr

1 i z

1 i z i 1 1 i 1 i z

w=

or

z i iz 1

Example 4.4 If a, b w a w b 1

1

w a

z a

k

z a , where k is a constant, if a z b

c , where c is a constant, if a a and b are a and b b

z a z

b

w b w

a

z b z

a

w a w b

z

b w

a

z

a w

b

b w, a, w , b)

w a w

z a z b

b;

z, a, z , b)

Chapter 4: Analytic Functions w a w b

k

z

a is w

a

cz

d)

c z a

z a , where k z b a, the bilinear transformation can assumed as

z a cz d

w a

z

4.85

1]

0 d 1 c

z

0 a,

d 1 c

a

d

ca 1 z a cz ca 1

w a 1 w a

c z a 1 z a

1 w a

1 z a

c, where c

Example 4.5 Show that the transformation w w

z 1 maps the unit circle in the z 1

z The image of the unit circle w z 1 z 1 x 1) iy x 1)2 y2 2x 2x

1,

z

z

1

x 1) iy x 1)2 y2

2x or x 2x or x

w 0, which is the right half of the z w

z

1

z

1

1 is the left half of the

z Example 4.6 Show that the transformation w circle z 1 onto the lower half of the w onto the interior of the circle w

z i 1 iz

i) the interior of the z-plane

Engineering Mathematics II

4.86

z i 1 iz

w w z

iwz

z

i

iw)

w

i

w i 1 iw

z

z z z w i 1 iw

w

1 u

i v

2

u

2

iw

v)

iu

2

2

1)
4. OR

( z + 1)dz , where C is the circlt 20. (a) Using Cauchy's residue theorem, valuate 2 z + 2 z + 4 |z + 1 + i| = 2. C 2 d (b) Evaluate (0 < a < 1), using contour integration. 1 - 2 a sin + a 2 0

Ú

Ú

SOLUTIONS PART-A 1.

2.

— = —(2 xy + z 2 ) = 2 yi + 2 xj + 2 k ; (— )(1,-1, 3) = - 2 i + 2 j + 6 k — ◊b 14 Dir. der. in the dir. of i + 2 j + 2 k = , where b = i + 2 j + 2 k = 3 |b | units. � 4 By divergence theorem, r ◊ dS = (div r )dv = 3V = 3 ¥ =4 3

ÚÚ S

3.

P.I. =

1 D +a 2

2

ÚÚÚ V

(b cos ax + c sin ax) = b ◊ =

x sin ax x Ê cos ax ˆ + c ◊ ◊Á 2 a 2 Ë a ˜¯

x (b sin ax – c cos ax) 2a

Engineering Mathematics II

B.4 4.

5.

6.

Putting x = et and xD = D¢, the given equation becomes [D¢(D¢ – 1) – D¢ + 1] y = 0; i.e., (D2 – 2D¢ + 1)y = 0. \ Solution is y = (At + B)et, i.e., y = (A log x + B)x. 1 ; By I.V.T., lim [ f (t )] = lim [ s (s)] = 0. s(s + 1)(s + 2) t Æ0 s Æ• 1 By F.V.T., lim [ f (t )] = lim [ s (s)] = t Æ• sÆ0 2 ( s) =

x¢(t) – x = et; x(0) = 0; Taking Laplace transforms, [sL(x) – x(0)] – L(x) = \ (s – 1) L(x) =

1 s -1

1 1 i.e., L(x) = . s -1 (s - 1)2

ÏÔ 1 ¸Ô t -1 Ê 1 ˆ \ x = L-1 Ì = e L Á 2 ˜ i.e., x = tet . 2˝ Ës ¯ ÓÔ (s - 1) ˛Ô 7.

Let u = log (x2 + y2); ux =

2x

; uy =

2y

x +y x + y2 - 2y If v is the conjugate of u, u + iv is analytic. Then vx = –uy = 2 and vy x + y2 2x = ux = 2 ; dv = vx dx + vy dy x + y2 - 2y Ê yˆ \v= dx = 2 tan -1 Á ˜ Ë x¯ x 2 + y2 2

2

2

Ú

8.

9.

The given circle x2 + y2 = 1 is |z| = 1. \ The image of |z| = 1 under the given transformation is |w| = |3 + 4i| |z| i.e., |w| = 5. Ê z + 2ˆ ÁË z ˜¯ dz; The pole z = 0 lies inside C, i.e., |z| = 2 and (z + 2) is analytic in C C Ê z + 2ˆ \ By C.I.F, Á dz = 2 i ( z + 2)z = 0 = 4 i. Ë z ˜¯

Ú

Ú

C

10.

z2 ; The isolated singularities are z = ia and z = –ia, which ( z - ia ) ( z + ia ) are simple poles. 2 2 2 1 ÔÏ z Ô¸ i a = ia (Res)z = ia = lim Ì ˝= z Æ ia Ô z + ia Ô Ó ˛ 2ia 2 f(z) =

2 2 2 1 ÔÏ z Ô¸ i a (Res)z = –ia = lim Ì = - ia ˝= z Æ - ia Ô z - ia Ô 2 Ó ˛ - 2ia

B.5

Appendix B PART-B 11. (a) Worked example in the book (b) Worked example in the book. 12. (a) Worked example in the book. 13. (a) Worked example in the book. (b) Worked example in the book. 14. (a) {(x + 1)2D2 + (x + 1) D + 1}y = 4 cos log (x + 1) Put x + 1 = et or t = log (x + 1). dy dy 1 Then = ◊ or (x + 1) D ∫ and (x + 1)2D2 ∫ ( – 1) dx dt x + 1 \ The given equation becomes { ( – 1) +

+ 1}y = 4 cos t or (

C.F. = A cos t + B sin t; P.I. =

2

+ 1)y = 4 cos t. 4t ◊ 4 cos t = sin t 2 +1

1 2

\ G.S. is y = A cos log (x + 1) + B sin log (x + 1) + 2 log (x + 1). sin log (x + 1) (b) Worked example in the book. 15. (a) (i) Worked example in the book. ÏÔ Ê s2 + a 2 ˆ ¸Ô 1 d È (ii) L-1 Ìlog Á 2 = - L-1 log (s2 + a 2 ) - log (s2 + b2 )˘˚ 2 ˜˝ Î t ds ÓÔ Ë s + b ¯ ˛Ô 1 È -1 = - ÍL t ÍÎ

ÔÏ 2 s Ô¸ - L-1 Ì 2 2˝ ÔÓ s + a Ô˛

ÔÏ 2 s Ô¸˘ 2 ˙ = (cos bt - cos at ) Ì 2 2˝ ÔÓ s + b Ô˛˙˚ t

(b) Worked example in the book. ÈÏÔ ¸Ô 10 -1 16. (a) L-1 ÍÌ ˝=L 2 ÎÍÓÔ (s + 1)(s + 4) ˛Ô by convolution theorem.

ÏÔ 5 2 ¸Ô˘ -1 ◊ 2 Ì ˝˙ = L s + 1 s + 4 ˛Ô˚˙ ÓÔ

Ï 5 ¸ -1 ÏÔ 2 ¸Ô Ì ˝* L Ì 2 ˝, Ó s + 1˛ ÓÔ s + 4 ˛Ô

t

Ú

-t

= (5e ) * (sin 2t ) = 5e - (t - u ) sin 2u du 0

t

È e ˘ = 5e ◊ Í (sin 2u - 2 cos 2u)˙ ÍÎ 1 + 4 ˙˚ 0 -t

u

-t t = e ÈÎe (sin 2t - 2 cos 2t ) + 2 ˘˚ = sin 2t – 2 cos 2t + 2e–t.

(b) Worked example in the book.

Engineering Mathematics II

B.6

u = e–2xy sin (x2 – y2); ux = e–2xy 2x cos (x2 – y2) – 2y e–2xy sin (x2 – y2)

17. (a)

uy = e–2xy(–2y)cos (x2 – y2) – 2xe–2xy sin (x2 – y2) f ¢(z) = ux + i vx = ux – i uy = 2z cos z2 + 2iz sin z2 2

= 2z (cos z2 + i sin z2) or 2 zeiz + iz 2 + ic \ f(z) = sin z2 – i cos z2 = –i (cos z2 + sin z2) = - ie 2

u + iv = - ie + i ( x + iy ) + ic = - ie -2 xy {cos (x2 – y2) + i sin (x2 – y2) + ic \ v = –e–2xy cos (x2 – y2) + c (b) Worked example in the book. 18 (a) Worked example in the book. It is proved, in the book, that u + iv is not analytic. If u + iv is analytic only, u and v are conjugate harmonic of each other. Hence the required result. (b) Worked example in the book. 19 (a) Worked example in the book. z u -2 (b) f ( z ) = ; Putting z + 2 = u or z = u – 2, f(z) = f ( z ) = ; ( z - 1)( z - 2) (u - 3)(u - 4) 2 1 ; f(z) = u-4 u-3 Case (i) -1 -1 1Ê uˆ 1Ê uˆ 1 + 1 f(z) = 2 ÁË 4 ˜¯ 3 ÁË 3 ˜¯ •

\

f (z) =

Case (ii): f(z) = -

n

-1 1 Ê z + 2ˆ + 2 n = 0 ÁË 4 ˜¯ 3

Â

1Ê 12 ÁË

1 \ f(z) = 2

uˆ 4 ˜¯

-1

-

•

•

 n=0

1Ê 1u ÁË

3ˆ u ˜¯

n

1 Ê z + 2ˆ ÁË 4 ˜¯ - z + 2 n=0

Â

n

Ê z + 2ˆ ÁË 3 ˜¯ , valid in |z + 2| < 3 -1

•

n

Ê 3 ˆ ÁË z + 2 ˜¯ , valid in 3 < |z + 2| < 4. n=0

Â

Case (iii): 2 f(z) = u f(z) =

4ˆ Ê ÁË 1 - u ˜¯

2 z+2

•

-1

-

1 u

3ˆ Ê ÁË 1 - u ˜¯ n

-1

Ê 4 ˆ 1 ÁË z + 2 ˜¯ - z + 2 n=0

Â

•

n

Ê 3 ˆ ÁË z + 2 ˜¯ , valid in |z + 2| > 4. n=0

Â

20 (a) Worked example in the book. (b) Worked example in the book. The letter ‘x’ used in the worked example has been changed as ‘a’

Appendix

C

Solved Question Paper B.E./B.Tech. DEGREE EXAMINATIONS, May/June 2013 (Second Semester) (Civil Engineering) MA 2161/MA 22/080030004 – Mathematics – II (Common to all Branches) (Regulation 2008) Times: 3 Hours

Maximum: 100 Marks Answer ALL Questions PART-A (10 ¥ 2 = 20 Marks)

1.

Find the particular integral of (D2 – 2D + 1)y = cosh x.

2.

Solve x 2

3.

d2 y 2

+ 4x

dy + 2 y = 0. dx

dx Find the directional derivative of

= xyz at (1, 1, 1) in the direction of

i + j + k. 4. 5. 6. 7.

If A and B are irrotational, prove that A ¥ B is solenoidal. 1 Find the image of the line x = k under the transformation w = . z 6z - 9 Find the fixed points of mapping w = . z Evaluate

Ú C

3z 2 + 7 z + 1 1 dz, where C is |z| = . z +1 2 1 - e2 z

at z = 0.

8.

Find the residue of

9.

Find the Laplace transform of

10.

z4

t

. et Verify initial value theorem for the function f(t) = ae–bt.

Engineering Mathematics II

C.2

PART-B (5 ¥ 16 = 80 marks) 11. (a) (i) Solve the differential equation variation of parameters. (ii) Solve: (3 x + 2)2

d2 y dx

2

d2 y dx 2

+ 3(3 x + 2)

+2

dy e- x + y = 2 by the method of dx x (8)

dy - 36 y = 3 x 2 + 4 x + 1. dx

(b) (i) Solve the simultaneous differential equations:

dx + 5 x - 2 y = t; dt

dy + 2 x + y = 0. dt (ii) Solve x 2

d2 y dx

2

+ 4x

(8)

(8) dy 1 + 2 y = x2 + 2 . dx x

(8)

OR 12. (a) Verify Stoke’s theorem for the vector field F = (2 x - y) i - yz 2 j - y 2 z k over the upper half surface x2 + y2 + z2 = 1, bounded by its projection on the xy-plane (16) (b) Verify divergence theorem for F = x 2 i + z j + yz k over the cube formed by the planes x = ±1, y = ±1, z = ±1. (16) 13. (a) (i) Prove that the function u = ex(x cos y – y sin y) satisfies Laplace’s equation and find the corresponding analytic function f(z) = u + iv. (8) (ii) Find the Bilinear transformation which maps z = 0, z = 1, z = • into the points w = i, w = 1, w = –i. (8) OR 1 (b) (i) Find the image of |z – 2i| = 2 under the transformation w = . z

(8)

Ê ∂2 ∂2 ˆ (ii) If f(z) is an analytic function of z, prove that Á 2 + 2 ˜ | f ( z )|2 ∂y ¯ Ë ∂x = 4|f ¢(z)|2.

(8) OR

14. (a) (i) Expand the function f ( z ) =

z2 - 1 z 2 + 5z + 6

in Laurent’s series for |z| > 3. (8)

Appendix C

(ii) Evaluate

Ú C

sin z 2 + cos z 2 dz , where C is |z| = 3. ( z - 1) ( z - 2)

C.3

(8)

OR •

(b) (i) Evaluate

Ú (x 0

2

x 2 dx 2

cos 3

Ú 5 - 4 cos

(ii) Evaluate

, a > 0, b > 0.

(8)

using contour integration.

(8)

+ a 2 ) ( x 2 + b2 )

0

15. (a) (i) Find L[t2 e–3t sin 2t].

(8)

(ii) Find the Laplace transform of the square-wave function (or Meander function) of period a defined as (8) a Ï when 0 < t < ÔÔ1, 2 f (t ) = Ì Ô - 1, when a < t < a. ÔÓ 2 OR (b) (i) Using convolution theorem find the inverse Lapalce transform of 4 . (8) (s2 + 2 s + 5)2 (ii) Solve y ¢¢ + 5y ¢ + 6y = 2 given y ¢(0) = 0 and y(0) = 0 using Laplace transform (8)

SOLUTIONS PART-A 1.

P.I. =

= 2.

1 ( D - 1)

2

cosh x =

1 1 ◊ (e x + e - x ) 2 ( D - 1)2

1 Ê x2 x 1 - x ˆ 1 e + e ˜ = (2 x 2 e x + e - x ) 2 ÁË 2 4 ¯ 8

Put x = et or t = log x. Then xD ∫ =

d dt

and x2D2 ∫ ( – 1), where D =

d and dx

Engineering Mathematics II

C.4

The given equation becomes [ ( – 1) + 4 + 2] y = 0 i.e., (

2

+ 2 + 2) y = 0. A.E. is (m + 1)2 = –1

\ m = –1 ±i \ Solution is y = e–t (A cos t + B sin t) or y = 3.

1 (A cos log x + B sin log x) x

— = yz i + zx j + xy k ; (— )(1,1,1) = i + j + k Dir. der. of in the dir. of i + j + k = Component (projection) of (— )(1, 1, 1) in the dir. of i + j + k = (i + j + k ) ◊

4.

(i + j + k ) 3

= 3 units.

—◊ ( A ¥ B) = B ◊ Curl A - A ◊ Curl B = B ◊ O - A ◊ O (∵ A & B are irrorational) = 0 = A ¥ B is solenoidal.

\ 5.

w=

1 1 1 u v u or x + iy = z = = = -i 2 \x = 2 2 z w u + iv u2 + v 2 u +v u + v2 2

1ˆ Ê Ê 1ˆ \ The map of x = k is k(u2 + v2) – u = 0 i.e., Á u - ˜ + (v - 0)2 = Á ˜ Ë Ë 2k ¯ 2k ¯

2

1 Ê 1 ˆ i.e., it is the circle whose centre is Á , 0˜ and radius . Ë 2k ¯ 2k 6.

7.

6z - 9 are given by z2 – 6z + 9 = 0 or (z – 3)2 = 0 z \ The only fixed point is z = 3. The fixed points of w =

Ú C

3z 2 + 72 + 1 3z 2 + 72 + 1 is analytic on and inside C, since dz ◊ f ( z ) = z +1 z +1

1 |z| = — 2 z = –1

0

C.5

Appendix C 1 . 2

z = –1 lies outside C i.e., |z| =

\ By Cauchy’s integral theorem, I = 0

8.

1 - e2 z z4

=

ÏÔ 2 z 4 z 2 8 z 3 16 z 4 1 - Ì1 + + + + + 1 2 3 4 ÓÔ

¸Ô ˝ ˛Ô

z4

ˆ Ô¸ 4 1 8 1 ˆ Ê 16 32 ÔÏÊ 2 1 z + … •˜ ˝ = - ÌÁ ◊ 3 + ◊ 2 + ◊ ˜ + Á + 2 z 3 z¯ Ë 4 5 ¯ Ô˛ ÔÓË 1 z \ Required residue at (z = 0) is coefft of 9.

10.

4 1 in the expn = . 3 z

1 Ê tˆ L Á t ˜ = L (e - t . t ) = [ L (t )]s Æ s + 1 = . Ëe ¯ (s + 1)2 - bt L{f(t)} = L (a e ) =

a = ( s) s+b

L.S. of I.V.T. = lim [ f (t )] = a t Æ0

Ê as ˆ Ê 1 ˆ = a lim Á = a. R.S. of I.V.T. = lim [ s ◊ (s)] = lim Á ˜ s Æ• s Æ• Ë s + b ¯ s Æ• Ë 1 + b /s ˜ ¯ Hence I.V.T. is verified PART-B (5 ¥ 16 = 80 marks) 11. (a) (i) The corresponding homogeneous equation is (D2 + 2D + 1) y = 0 Solution of this equation is y = (Ax + B) e–x Treating A and B as functions of x, A¢ and B¢ are given by 1 -x (2) (1) and A¢(e–x – x e–x) + B¢(–e–x) = 2 e x 1 -x 1 \ A = - + C1 (1) + (2) gives A¢ e–x = 2 e x x A¢ x e–x + B¢ e–x = 0

using A¢ =

1 x

2

in (1), we get B¢ = -

1 \ B = - log x + C2 x

1ˆ Ê -x -x \ Required solution is y = Á C1 - ˜ x e + (C2 - log x ) e Ë x¯

Engineering Mathematics II

C.6

i.e., y = C1 x e–x + (C2 – 1) e–x – log x e–x or y = (C1 x + C3) e–x – log x e–x. (ii) [(3x + 2)2 D2 + 3(3x + 2) D – 36] y = 3x2 + 4x + 1. Put 3x + 2 = et or t = log(3x + 2) \ (3x + 2) D ∫ 3 and (3x + 2)2 d d and = dx dt \ The given equation becomes [9 ( – 1) + 9 –36] y 1 1 1 1 2 2 = (9 x + 12 x + 4) - i.e., (9 2 –36) y = (3 x + 2) 3 3 3 3 D2 = 9 ( – 1) where D ∫

1 2t 2 (e - 1); A.E. is m – 4 = 0; m = ±2 27 C.F. = A e2t + B e–2t = A(3x + 2)2 + B(3x + 2)–2; 2

or (

P.I. =

= \

- 4) y =

1 È 1 e2 t Í 27 Î ( - 2) ( + 2)

˘ 1 È 1 2t 1 ˘ (1)˙ = Í t e + 4˙ - 4 ˚ 27 Î 4 ˚

1 2

1 [(3 x + 2)2 log (3 x + 2) + 1] 108

G.S. is y = C.F + P.I.

(b) (i) (D + 5)x – 2y = t; 2x + (D + 1)y = 0; Eliminating y; (D2 + 6D + 9)x = t + 1 1 t +1 A.E. is (m + 3)2 = 0; m = –3, –3; C.F = (At + B) e–3t; P.I = ( D + 3)2 =

1 9

Dˆ Ê ÁË 1 + 3 ˜¯

-2

(t + 1) =

- 3t \ x = ( At + B) e +

y =

1Ê 2D ˆ 1Ê 1ˆ 1(t + 1) = Á t + ˜ Á ˜ 9Ë 3 ¯ 9Ë 3¯

1 1 t+ . Substituting in the first equation, 9 27

1 Ï dx 4 8˘ ¸ 1È -3t -3t Ì + 5 x - t ˝ = Í{- 3 At + ( A - 3 B)} e + 5( At + B) e - t + ˙ 2 Ó dt 9 27 ˚ ˛ 2Î

A -3t 2 4 e - t+ 2 9 27 (ii) Worked example in the book. i.e., y = ( At + B) e -3t +

12. (a) Curl F =

i ∂ ∂x

j ∂ ∂y

2 x - y - yz 2

k ∂ ∂z - y2 z

= k ; Stoke’s theorem is

C.7

Appendix C

Ú F ◊ dr = ÚÚ curl F ◊ dS …

(1)

S

C

C is the projection on the xy-plane of the upper surface of x2 + y2 + z2 = 1 is the circle x2 + y2 = 1 L.S. of (1) = 2

Ú

2

(2 x - y) dx = 2

x + y =1

Ê 2 = Á - sin Ë R.S. of (1) =

Ú (2 cos

+

1 2

-

R.S. of (1) =

=

—

ÚÚ k ◊ nˆ ds, where nˆ = |— 2( x i + y j + z k ) 4( x 2 + y 2 + z 2 )

|

, where

= x 2 + y2 + z2

= xi + y j + zk

dx dy

ÚÚ z ds = ÚÚ z nˆ ◊ k S

=

2

1 ˆ sin 2 ˜ ¯ 4

S

=

- sin ) ( - sin ) d

0

ÚÚ z R

R

dx dy = z

ÚÚ dx dy = Area of the circle R = R

Since L.S. of (1) = R.S. of (1), stoke’s theorem is verified. (b) Worked example in the book. 13. (a) (i) u = ex (x cos y – y sin y); ux = (ex + x ex) cos y – ex y sin y uxx = (2ex + x ex) cos y – ex ◊ y sin y uy = –x ex sin y –ex (y cos y + sin y); uyy = –x ex cos y – ex (2 cos y – y sin y) \ uxx + uyy = 0; i.e., u satisfies Laplace equation. f ¢(z) = ux – i uy, by C.R. equation = (z + 1) ez – i ¥ 0, by M.T. principle. f (z) =

Ú (z + 1) e dz + c = (z + 1) e – 1 e + c = ze + c z

z

z

z

(ii) Worked example in the book. (b) (i) The image of |z – 2i| = 2 under w =

1 1 or z = is given by z w

Engineering Mathematics II

C.8

1 1 - 2i (u + iv) |(2 v + 1) - 2iu|2 - 2i = 2; i.e., = 2; i.e., = 4; w u + iv |u + iv|2 i.e., (2v + 1)2 + 4u2 = 4(u2 + v2); i.e., 4v + 1 = 0 or v = –1/4. (ii) Worked example in the book. 14. (a) (i) f(z) = 1 -

5z + 7 3 8 3Ê 2ˆ =1+ = 1 + Á1 + ˜ Ë ( z + 2) ( z + 3) z+2 z+3 z z¯

•

= 1+ 3

Â

(-1)n 2 n

n=0

z n +1

•

-8

Â

(-1)n ◊ 3n

-

8Ê 3ˆ 1+ ˜ Á Ë z z¯

-1

(1)

z n +1

n=0

-1

2 3 < 1 and < 1; i.e., |z| > 2 |z| |z|

This Laurent’s expansion (1) is valid in and |z| > 3; i.e., in |z| > 3. (ii)

Ú c

sin z 2 + cos z 2 dz; The singularities of the integrand f(z) are z = 1 ( z - 1) ( z - 2)

and z = 2, which are simple poles of f(z) lying within C, i.e. the circle |z| =3 Ê sin z 2 + cos z 2 ˆ cos = =1 [Res. of f(z)]z = 1 = Á ˜ z-2 -1 Ë ¯ z =1

Ê sin [Res. of f(z)]z = 2 = Á Ë

z ˆ Ê cos 4 ˆ =Á ˜ =1 ˜ ¯z=2 Ë 1 ¯

z + cos z -1 2

By Cauchy’s residue theorem,

2

Ú f (z) dz = 2

i(1 + 1) = 4 i

C

•

(b) (i) Worked example in the book. 2

(ii) I =

Ú (x 0

cos 3 d

Ú 5 - 4 cos

x 2 dx 2

+ a 2 ) ( x 2 + b2 )

=

2 ( a + b)

In the circle |z| = 1, z = ei ;

0

d =

\ I=

Ú

C

dz ; cos iz

=

z2 + 1 z6 + 1 ; cos 3 = 2z 2 z3

Ê z 6 + 1ˆ dz Á 2 z3 ˜ iz Ë ¯ Ê z 2 + 1ˆ 5 - 4Á ˜ Ë 2z ¯

=-

1 ( z 6 + 1) dz ◊ 3 ; 4i z ( z - 2) ( z - 1/2) C

Ú

The singularities of the integrand f(z) which lie inside C are z = 1/2 (a simple pole) and the triple pole at z = 0.

C.9

Appendix C 1 +1 6 65 2 [Res. f(z)]z = 1/2 = = - ; By Laurent’s expansion, 1 3 12 ¥2 23 Ê 3 1ˆ f(z) = Á z + 3 ˜ Ë z ¯

Ê ÁË 1 -

zˆ (1 - 2 z ) 2 ˜¯

ˆ Ê 1ˆÊ z z2 = Á z3 + 3 ˜ Á 1 + + + …˜ (1 + 2 z + 4 z 2 + …) Ë 2 4 z ¯Ë ¯ Ê 3 1ˆ = Áz + 3 ˜ Ë z ¯

5 21 2 Ê ˆ ÁË 1 + 2 z + 4 z + …˜¯

[Res. f(z)]z = 0 = coefft of

1 21 in the expansion = z 4

By Cauchy’s residue the theorem, I =

i Ê 65 21ˆ ¥ 2 i Á+ ˜= Ë 12 4 4 ¯ 12

15. (a) L[t2 e–3t sin 2t]; Consider L(t2 sin 2t) 2 = ( -1)

d2 Ê 2 ˆ d Ê 4 s ˆ -4 (4 - 3s2 ) = = Á ˜ Á ds Ë (s2 + 4)2 ˜¯ ds2 Ë s2 + 4 ¯ (s2 + 4)3

2 ÔÏ 4(3s - 4 Ô¸ \ L{e3t t 2 sin 2t} = Ì 2 3 ˝ ÓÔ (s + 4) ˛Ô s Æ s + 3

=

4{3 (s + 3)2 - 4} {(s + 3)3 + 4}3 L { f (t )} =

(ii)

=

=

1 1 - e - as

1 1 - e - as

4{3 s2 + 18s + 23} (s2 + 6 s + 13)3 a È a /2 ˘ st Í 1◊ e dt - 1◊ e - st dt ˙ Í ˙ a /2 Î0 ˚

Ú

Ú

ÈÊ - st ˆ a /2 Ê - st ˆ a ˘ e Í e ˙ +Á Á ˜ ÍË - s ¯ s ˜¯ ˙ Ë 0 a /2 ˚ Î

1 s(1 - e

=

- as

)

{e - as - e - as /2 - e - as /2 + 1} =

1 s(1 - e

- as

)

1 Ê 1 - e - as /2 ˆ 1 Ê e as /4 - e - as /4 ˆ 1 Ê as ˆ = ◊Á = = tanh Á ˜ Ë 4¯ s Ë 1 + e - as /2 ˜¯ s ÁË e as /4 + e - as /4 ˜¯ s

(1 - e - as /2 )2

Engineering Mathematics II

C.10

(b) Worked example in the book (ii) y¢¢(t) + ty¢(t) + 6y(t) = 2; Taking Laplace transforms. [s2 L(y) – sy(0) – y¢(0)] + 5[sL(y) – y(0)] + 6L(y) = i.e., L(y) = \ y=

2 s(s 5s + 6) 2

=

1 2 - e -2 t + e -3t . 3 3

2 s

2 1/3 1 2/3 = + s(s + 2)(s + 3) s s+2 s+3