Engineering Mathematics I 9789387432109, 9387432106

Table of contents :
Title
Contents
1 Differential Calculus
2 Functions of Several Variables
3 Integral Calculus
4 Multiple Integrals
5 Differential Equations

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Engineering Mathematics I

About the Author

T Veerarajan retired as 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 I

T Veerarajan Former Dean, Department of Mathematics Velammal College of Engineering and Technology Viraganoor, Madurai Tamil Nadu

McGraw Hill Education (India) Private Limited Chennai McGraw Hill Education Offices Chennai 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 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Engineering Mathematics I Copyright © 2018 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 listings (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. 1 2 3 4 5 6 7 8 9

D102739

22 21 20 19 18

Printed and bound in India. ISBN (13): 978-93-87432-10-9 ISBN (10): 93-87432-10-6 Managing Director: Kaushik Bellani Director—Science & Engineering Portfolio: Vibha Mahajan Senior Portfolio Manager—Science & Engineering: Hemant K Jha Associate Portfolio Manager, Science & Engineering: Mohammad Salman Khurshid Production Head: Satinder S Baveja Assistant Manager—Production: Anuj K Shriwastava General Manager—Production: Rajender P Ghansela Manager—Production: 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. Typeset at Text-o-Graphics, B-1/56, Aravali Apartment, Sector-34, Noida 201 301, and printed at Cover Printer:

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Contents Preface Unit 1. Differential CalCUlUs

ix 1.1–1.61

1.1 Functions 1.1 1.2 Limit of a Function 1.2 1.3 Continuity of a Function 1.5 1.4 Differentiability of a Function 1.5 Worked Example 1(a) 1.7 Exercise 1(a) 1.15 1.5 Derivatives 1.18 1.6 Rules of Differentiation 1.23 1.7 Derivatives of Hyperbolic Function 1.26 1.8 Methods of Differentiation 1.29 Worked Example 1(b) 1.30 Exercise 1(b) 1.44 1.9 Maxima and Minima of Functions of One Variable 1.46 Worked Example 1(c) 1.48 Exercise 1(c) 1.57 Answers 1.58 Unit 2. fUnCtions of several variables 2.1 Introduction 2.1 2.2 Total Differentiation 2.1 Worked Example 2(a) 2.4 Exercise 2(a) 2.20 2.3 Jacobians 2.27 2.4 Differentiation Under the Integral Sign 2.31 Worked Example 2(b) 2.33 Exercise 2(b) 2.48 2.5 Maxima and Minima of Functions of Two Variables 2.50 Worked Example 2(c) 2.52 Exercise 2(c) 2.65 Answers 2.67

2.1–2.69

vi

Contents

Unit 3. integral CalCUlUs

3.1–3.53

3.1 Introduction 3.1 3.2 Constant of Integration 3.1 3.3 Techniques of Integration 3.4 Worked Example 3(a) 3.6 Exercise 3(a) 3.13 3.4 Integration of Rational (Algebraic) Functions 3.14 Worked Example 3(b) 3.15 Exercise 3(b) 3.25 3.5 Integration of Irrational Functions 3.26 Worked Example 3(c) 3.27 Exercise 3(c) 3.35 3.6 Integration by Parts 3.36 Worked Example 3(d) 3.39 Exercise 3(d) 3.48 Answers 3.50 Unit 4.

MUltiple integrals

4.1 Introduction 4.1 4.2 Evaluation of Double and Triple Integrals 4.1 4.3 Region of Integration 4.2 Worked Example 4(a) 4.3 Exercise 4(a) 4.15 4.4 Change of Order of Integration in a Double Integral 4.17 4.5 Plane Area as Double Integral 4.18 Worked Example 4(b) 4.21 Exercise 4(b) 4.38 4.6 Line Integral 4.41 4.7 Surface Integral 4.43 4.8 Volume Integral 4.44 Worked Example 4(c) 4.45 Exercise 4(c) 4.57 4.9 Gamma and Beta Functions 4.59 Worked Example 4(d) 4.63 Exercise 4(d) 4.77 Answers 4.79

4.1–4.83

Contents Unit 5.

Differential eqUations

vii 5.1–5.88

5.1 Equations of the First Order and Higher Degree 5.1 Worked Example 5(a) 5.3 Exercise 5(a) 5.12 5.2 Linear Differential Equations of Second and Higher Order with Constant Coefficients 5.14 5.3 Complementary Function 5.14 Worked Example 5(b) 5.19 Exercise 5(b) 5.32 5.4 Euler’s Homogeneous Linear Differential Equations 5.33 5.5 Simultaneous Differential Equations with Constant Coefficients 5.35 Worked Example 5(c) 5.35 Exercise 5(c) 5.47 5.6 Linear Equations of Second Order with Variable Coefficients 5.48 Worked Example 5(d) 5.53 Exercise 5(d) 5.67 5.7 Method of Variation of Parameters 5.71 Worked Example 5(e) 5.73 Exercise 5(e) 5.82 Answers 5.82

Preface

I am deeply gratified with the enthusiastic response shown to all my books on Engineering Mathematics (for first year courses) by students and teachers throughout the country. This book has been designed to meet the requirements of the students of first-year undergraduate course on Engineering Mathematics. The contents have been covered in adequate depth for Semester I of various universities/deemed universities across the country. It offers a balanced coverage of both theory and problems. Lucid writing style supported with step-by-step solutions to all problems enhances understanding of the concepts. The book has ample number of solved and unsolved problems of different types to help students and teachers learning and teaching the subject. I hope that this book will be received by both the faculty and the students as enthusiastically as my other books on Engineering Mathematics. 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) invites suggestions and comments from you, all of which can be sent to [email protected] (kindly mention the title and author name in the subject line). Piracy-related issues may also be reported.

UNIT

1

Differential Calculus 1.1

FUNcTIoNs

Students are familiar, to some extent, with the concept of relation. A relation can be thought of as a relationship of elements of a set to the elements of another set. In other words, when A and B are sets, a subset R of the cartesian product A × B is called a relation from A to B viz., If R is a relation from A to B, R is a set of ordered pairs (a, b) where a Œ A and b Œ b.

1.1.1  Definitions A relation from set X to another set Y is called a function, if for every x Œ X, there is a unique y Œ Y such that (x, y) Œ f. In other words, a function from X to Y is an assignment of exactly one element of Y to every element of X. If y is the unique element of Y assigned by the functions f to the element x of X, we write f(x) = y and say that y is a function of x. If f is a function from X to Y, we may also represent it as f f : X → Y or X  → Y.

Note

(i) Sometimes the terms transformations, mapping or correspondence are also used in the place of function. (ii) If y = f(x), then x is called an argument or preimage and y is called the image of x under f or the value of the function f at x. (iii) X is the domain of f denoted by Df and Y is called the codomain of f (iv) The set of the images of all elements of X is called the range of X denoted Rf ncY Rf ≤ y. (v) x is called the independent variable and y the dependent variable, if y = f(x).

1.1.2  Representation of Functions (1) A function can be represented or expressed by means of a mathematical rule of formula, such as y = x3[≡ f(x)] or (2) it can be represented pictorially by means of two closed circles or two closed ellipses or any two closed curves. This representation is possible only if Df consists of finite number of elements. The elements of Df will be represented by points inside the first closed curve and those of range or co-domain of f by

1.2

Engineering Mathematics I points inside the second closed curve. The points in Df and the corresponding Rf will be connected by directed arrows as explained below: Let Df = (1, 2, 3, 4) and f(1) = b, f(2) = d, f(3) = a and f(4) = b The pictorial representation is shown as follows: 1 2 3 4

a b c d

Fig. 1.1

Here Df = {1, 2, 3, 4}, Rf = {a, b, d} and co-domain of f = {a, b, c, d} (3) A continuous function can be represented by means of a (curve) graph. For example, y = x2[= f(x)] is a continuous function of x; The values of x2 = [f(x)] for different values of x ΠR lie on a parabola as in the figure given below: f (x)

Note

(i) The curve y = x2 drawn here is a one-piece without any break. (ii) Discontinues functions can also be represented graphically but with a break in the neighbourhood of the point of discontinuity.

O

x

Fig. 1.2

 x + 1, in(−1, 0)  For example, f ( x) =  is a discontinues function, with the point     x − 1, in(0, 1) x = 0 as a point of discontinuity. The graph of y = f(x) consists of two line segments with a break near the origin as shown in the figure below:

y

=

x

+

1

y

x

y=

x–

1

–1

Fig. 1.3

Note 1.2

Detailed discussion of continuous functions will be done later.

LImIT oF a FUNcTIoN

x2 − 4 . The value of f(x) can not be found x−2 out at x = 2 by direct substitution though the values of f(x) can be found out for all other values of x, however close they may be to 2. Let us consider the function y = f ( x) =

Differential Calculus

1.3

As may be verified, f(x) assume the values 3.9, 3.99, 3.999, 3.9999, etc. as x take the values 1.9, 1.99, 1.999, 1.9999, etc. Similarly f(x) assume the value 4.1, 4.01, 4.001, 4.0001, etc. as x takes the value 2.1, 2.01, 2.001, 2.0001, etc. 0 But if we put x = 2 in the definition of f(x), we get y = , which is meaningless 0 and usually referred to as an indeterminate form. The definition of limit of a function f(x) does not require that the function f(x) be defined at x = 2. From the above example, we note that to make the difference between f(x) and 4 as small as possible, we have to make the difference between x and 2 correspondingly  x 2 − 4  small. This fact is symbolically put as lim   = 4 x→2  x − 2 

Note

( x − 2)( x + 2) x−2 =x+2

For all value of x ≠ 2, f ( x) =

lim [ f ( x)] = 4

\

xÆ2

The formal definition of the limit of a function is given below: lim [ f ( x)] = l , if and only if for any arbitrarily small positive number Œ, there xÆa

exists another small positive number d, such that f ( x) - l < Œ, whenever 0 < x - a < d . When x Æ a through values greater than a, we say that x approaches a from the right (or from above) and write lim { f ( x)} = l. This is called the right

Note

x→a+

hand limit. Similarly when x Æ a through values less than a (or from below), we { f ( x)} = l. obtain the left hand limit and write xlim → a− If lim{ f ( x)} exists and equals l, it implies that both the left hand and right hand x→a

limits exist and are equal, each equal to l.

1.2.1  S   ome Fundamental Theorems on Limits (stated without  proof) If lim[ f ( x)] = l and lim[ g ( x)] = m, then x→a

(i)

x→a

lim [ f ( x) ± g ( x)] = l ± m

xÆa

Note

This theorem can be extended to the algebraic sum of a finite number of functions. viz., lim[c1 f1 ( x) + c2 f 2 ( x) +  ck f k ( x)] = c1 lim[ f1 ( x)] + c2 lim[ f 2 ( x)] x→a

x→a

x→a

+  ck lim[ f x ( x)] x→a

Engineering Mathematics I

1.4 (ii)

lim[ f ( x) ⋅ g ( x)] = lm

(iii)

 f ( x)  l  = , provided m π 0. lim  x → a  g ( x)    m

x→a

1.2.2  Some Standard Limits  xn − an    n −1  lim    = na , for all rational values of n and a π 0. x→a  x − a      Proof: Let x – a = h. since x Æ a, h Æ 0 1.

 xn − an    ( a + h) n − a n     = lim  lim     x→a  x − a  h→0  h       n     a n 1 + h  − a n      a  = lim   h→0  h 

(1)

Since x can approach a from the left or right, h = x − a → 0 and hence

h

sin θ > cos θ (considering the reciprocate) θ

 sin θ  Taking limits as θ → 0, lim(1) ≥ lim   ≥ lim(cos θ ) θ→0 θ→0   θ  θ → 0  sin θ   sin θ  1 ≥ lim   = 1.  ≥ 1 or lim  θ→0  θ  θ→0  θ  

viz.,

Cor:

 tan θ   sin θ  1   tan θ  lim   = 1×1 = 1  = lim    = 1, since lim  θ→0  θ  θ→0  θ   θ → 0  θ  cos θ   x

3.

 1 lim 1 +  = e, which is the Naperian logarithmic base defined as x→∞  x 1+

1 1 1 + + +  ∞ = 2.71828 1! 2 ! 3!

Proof: x

1 Expanding ÊÁ1 + ˆ˜ by Binomial theorem, we have Ë x¯

Engineering Mathematics I

1.6 x

  1 + 1  = 1 + x ⋅ 1 + x( x − 1) ⋅ 1 + x( x − 1)( x − 2) ⋅ 1 +   x  x 2! 3! x2 x3 1 1 1  1  2 = 1 + 1 + 1 −  + 1 − 1 −  +  x  3!  x  x 2! x

 1 1 1 1 lim 1 +  = 2 + + +  + +  x→∞  x 2 ! 3! n! =e

Note

Obviously, e > 2. 1 1 1 + + + 2 2⋅2 2⋅2⋅2

Also

e 3 3 3

and

x 2 Left

lim{ f ( x)} = lim (1 − h) = 1 x →1

Right

x →1− h h→0

lim{ f ( x)} = lim {2 − (1 + h)} = 1. Also f(1) = 2 – 1 = 1 x →1

x →1+ h h→0

\ f(x) is continuous at x = 1 Left

lim{ f ( x)} = lim {2 − (2 − h)} = 0

Right

lim{ f ( x)} = lim {−2 + 3 (2 + h) − (2 + h) 2 } = 0

x→2

x→2

x → 2− h h→0

x→ 2+h h→0

Also, f(2) = 2 – 2 = 0 \ f(x) is continuous at x = 2 also. Example 1.17 A function f(x) is defined as follows: Discuss the continuity of the function at x = 0 and at x = 1.   −x 2 , if x ≤ 0   f ( x) =  5 x − 4, if 0 < x ≤ 1   2   4 x − 3x, if 1 < x ≤ 2 Left

lim{ f ( x)} = lim {−(−h) 2 } ≡ 0 x→0

Right

x → 0− h h→0

lim{ f ( x)} = lim {5h − 4} = − 4 x→0

x→0+h h→0

\ f(x) is not continuous at x = 0, since left lim π right lim. Left

lim{ f ( x)} = lim {5(1 − h) − 4} = 1 x →1

Right

x →1− h h→0

lim{ f ( x)} = lim {4(1 + h) 2 − 3(1 + h)} = 1; Also f (1) = 1 x →1

x →1+ h h→0

Since left lim = right lim = f(1), f(x) is continuous at x = 1.

Engineering Mathematics I

1.14

Example 1.18 Find the values of a, b, c for which the function defined below is continuous at the origin:  sin(a + 1) x + sin x   , for x < 0   x   f ( x) =  c , for x = 0   1 1    ( x + bx 2 ) 2 − x 2  , for x > 0  3   2  bx  Left

 sin(a + 1)(−h) + sin(−h)   sin(a + 1)h sinh   = lim   lim  + h → 0    −h h h  

x → 0− h h→0

= a + 1 +1 = a + 2 Also, f(0) = c Right

 h + bh 2 − h   { h + bh 2 − h }{ h + bh 2 + h     lim  = lim  h→0   x→0+h 2 bh h      bh h{ h + bh + h h→0  lim h→0

bh 2 2

bh { 1 + bh + 1}

=

1 2

Since the function f(x) is continuous at the origin,

a+2=c= \

a=-

1 2

3 1 and c = ; b is arbitrary, but π 0 2 2

1 Example 1.19 Examine the continuity and differentiability of f ( x) = x 2 sin ( x π 0) x and f(0) = 0 at the origin. Left

   1   1  lim  (−h) 2 sin −  = lim  h 2 sin   x → 0− h   h   h  h → 0   h→0 1 = 0, since sin   is bounded h

Right

 1 lim  h 2 sin   = 0, Aslo f (0) = 0 x→0+h  h  h→0

\ f(x) is continuous at x = 0 Now

 2  1  h sin   − 0   f ( 0 + h ) − f ( 0)    h  = lim  lim   h→0  h → 0    h h   

Differential Calculus

 lim  h→0  

1.15

  1  = lim  h sin   = 0 h→0   h      1   h 2 sin   − 0     −h  f ( 0 − h ) − f ( 0)   = lim   h → 0    −h −h    1  = lim  h sin   = 0 h→0   h  

\ f(x) is also differentiable at the origin. Example 1.20 Examine the continuity and differentiability of f(x) defined below at x = 2. 1 + x, for x ≤ 2  f ( x) =     5 − x, for x > 2 Left

lim {1 + 2 − h} = 3

x → 2− h h→0

Right

lim {5 − (2 + h)} = 3. Also f(2) = 3

x→ 2+h h→0

\ f(x) is continuous at the point x = 2.  f ( 2 + h ) − f ( 2)  Now  lim  h→0  h   = lim h→0

and

5 − ( 2 + h) − 3 = −1 h

 f ( 2 − h ) − f ( 2)   lim  h→0   −h   (1 + 2 − h − 3)   =1 = lim  h→0   −h 

Since the two limits are not equal, f(x) is not differentiable at the point x = 2. EXERCISE 1(a)

Part A (Short Answer Questions) 1. If f(x) = a cos4x + b sin2x + c, show that f (p + x) – f(p – x) = f(x). 2. If x is real, find the domain of the function of f(x) = x2 + x – 12, for which it is negative.

Engineering Mathematics I

1.16

3. If x is real, find the domain of the function f(x) = (x2 – 3x – 18) for which it is positive. Evaluate the following limits:     1 + x − 1− x    4. lim   x→0  x    

  5 x − 4 − x     5. lim  x →1  x −1  

  x     6. lim   x→0   + 2 − 2 x    

 x− a    7. lim  x→a  3 x − 3 a  

 a n + 1 + b n + 1  , 0 < a < b 8. lim  n→∞ a n − b n 

1 + 2 + 3 +  n  9. lim   n→∞  n2

1 − cos mx     10. lim    x → 0  1 − cos nx     

1 − cos x   11. lim    x → 0  x sin x    

12. lim x→0

{

cosec x − cot x x

 sin θ − sin α    13. θlim →α   θ − α 

}

È sin( x + h) - sin( x - h) ˘ 14. lim Í ˙˚ hÆ0 Î h

 cosec x − sec x  15. lim   π  cot x − tan x  x→   4

2 x − tan 2 x) 16. lim(sec π

17. lim[(1 + ax) x ]

b x→

x→0

4 1

18. lim{x x −1 } x →1

19. lim(1 + cos x) 2 sec x x→

π 2

20. lim{(1 + cos x) tan x } x→

π 2

Part B x −1 x−3 21. Find the domain of the function f ( x) = for which f(x) is − 4x + 5 4x − 3 negative, given that x is real. 3x − 2 22. If x is real, find the domain of the function f ( x) = − 2( x ≠ 1) for x −1 which f(x) is negative. 23. If x is real, find the range of the function f ( x) = 24. If x is real, prove that f ( x) =

x2 − 2x + 4 . x2 + 2x + 4

x 1 lies between − and 1. 11 x2 − 5x + 9

Differential Calculus

1.17

25. If x is real, prove that the range of the function f ( x) = 26. If y = f ( x) =

x 2 − 3x + 4  1  is  , 7 . x 2 + 3 x + 4  7 

ax + b , prove that x = f(y). cx − a

12 + 22 +  + n 2  . 27. Evaluate lim   x→∞  n(n 2 + 1)    (12 + 32 + 52 +  (2n − 1) 2  . 28. Evaluate lim   x→∞  n3   (1 + 2 + 3 +  n)(13 + 23 +  + n3 )  . 29. Evaluate lim   n→∞  n3 (12 + 22 +  + n 2   1.3 + 2.4 +  + n(n + 2)  . 30. Evaluate lim  n→∞   12 + 22 +  + n 2   13 + 23 +  + n3 31. Evaluate lim  . n → ∞ 1 ⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ 4 +  n( n + 1)( n + 2)   2x − 1   a x − 1   ⋅  32. Evaluate lim  Hint: Use lim    x→0  x→0 x     1 + x − 1

Ê 1 - cos 2 x + tan 2 x ˆ 33. Evaluate lim Á ˜¯ . xÆ0Ë x sin x È e x + log(1 + x) - (1 - x)-2 ˘ 34. Evaluate lim Í ˙. xÆ0 Î x2 ˚ Ï 8x - 2 x ¸ 35. Evaluate lim Ì ˝. xÆ0 Ó x ˛

{

{Hint: Use expansions}

}

px . x Æ1 2 37. Show that the function f(x) defined below is continuous at x = 0 and at x = 1: Also draw the graph of y = f(x). 36. Evaluate lim (1 - x) tan

Ï - x, when x £ 0 Ô f ( x) = Ì x, when 0 < x < 1 ÔÓ 2 - x, when x ≥ 1 3 38. Test the continuity of the function f(x) defined below at x = . 2   2x − 3 3  , for x ≠   2x − 3 2  f ( x) =   3   for x = 0,   2 

Engineering Mathematics I

1.18

39. Examine the continuity of the function f(x) defined below at x = 0.    sin 1 , for x ≠ 0 f ( x) =   x     0 , for x = 0 40. Find the values of the a and b, if the function f(x) defined below is continuous at x = 3 and at x = 5.  1, if x ≤ 3    f ( x) = ax + b, if 3 < x < 5    if x ≥ 5   7,  x − 1, when x > 1    when x = 1 41. If f ( x) =  0,     1 − x, when x < 1 shown that f(x) is continuous at x = 1, but is not differentiable there. 42. Show that f(x) = | x | is continuous at x = 0, but is not differentiable there. 43. Discuss the continuity and differentiability of the function f(x) defined as 2 x − 3, for 0 ≤ x ≤ 2  f ( x) =  at the point x = 2.  2    x − 3, for x > 2 44. Discuss the continuity and differentiability of the function f(x) defined below at x = 1 and at x = 2. f ( x) = x, for x < 1; = 2 − x, for 1 ≤ x ≤ 2 and = − 2 + 3 x − x 2 for x > 2 45. Discuss the continuity and differentiability of the function f(x) = | x – 1 | + | x – 2 | at the point x = 1 and x = 2.

1.5

DerIvaTIves

Let y be a continuous function of x. Let Dx be a small increment in the value of x and  ∆y  let the corresponding increment in the value of y be Dy. The limit of the ratio    ∆x  as Dx Æ 0, if it exists is called the differential coefficient or derivative of y with re  spect to x and denoted by dy . viz , lim  ∆y  = dy or Dy ∆ x → 0  ∆x  dx dx If y is assumed as f(x), then

{

 ∆y  dy f ( x + ∆x) − f ( x) = lim   = lim dx ∆x → 0  ∆x  ∆x → 0 ∆x

denoted f¢(x) or Df(x). The process of finding

dy is called differentiation. dx

} and is

Differential Calculus

Note

(1)

1.19

dy is a composite symbol and should be understood neither as dx d lim(∆y ) nor as dy ∏ dx. must be interpreted as derivative w.r.t. x. lim(∆x) dx

Ê Dy ˆ lim Á ˜ is found out by the methods discussed in the Dx Æ 0 Ë Dx ¯ previous section.

(2) The

1.5.1  Derivatives of Elementary/Standard Functions We shall follow the 4-step working procedure given below to find the derivatives of dy some elementary functions from first principles, viz., using the definition of or dx f¢(x). (i) In the given function y = f(x), replace x by x +Dx. At the same time y becomes y + Dy. viz., y + Dy = f(x + Dx) (ii) Obtain Dy = f(x + Dx) – f(x) and simplify as far as possible (iii) Obatin

Dy . Dx

dy Ê Dy ˆ (iii) Find lim Á ˜ , which is the required derivative . Ë ¯ Dx Æ 0 Dx dx d n ( x ) = nx n - 1 , where n is rational number dx Let y = xn. Let Dx and Dy be the small increments in x and y respectively. Then y + Dy = (x + Dx)n 1.

\

Dy ( x + Dx)n - x n = Dx Dx dy = dx

lim

Dx Æ 0 x + Dx Æ x

È ( x + Dx)n - x n ˘ Í ˙ , by definition Î ( x + Dx) - x ˚

Ê xn - an ˆ n -1 = nx n - 1, on using lim Á ˜ = na xÆa Ë x - a ¯ viz.,

d n ( x ) = nx n - 1 dx

d (sin x ) = cos x dx Let y = sin x. Let Dx and Dy be the small increments in x and y respectively. Then y + Dy = sin(x + Dx) 2.

Engineering Mathematics I

1.20 \

\

Dy = sin(x + Dx) – sin x Dx ˆ Dx Ê = 2 cos Á x + ˜ sin Ë 2¯ 2   ∆x    sin     ∆y   2  dy ∆x    = lim   = lim   ⋅ cos  x +      ∆x dx ∆x → 0  ∆x  ∆x → 0  2        2    

sin q ˆ Ê = cos x ◊ 1Á∵ lim = 1˜ Ë q Æ0 q ¯ = cos x viz.,

d (sin x) = cos x dx

d (cos x ) = − sin x dx Let y = cos x Then Dy = cos(x + Dx) – cos x  ∆x  ∆x = − 2 sin  x +  ⋅ sin  2  2 3.

\

   ∆x   sin    ∆y   dy ∆x    2  = lim   = lim − sin  x +  ⋅    dx ∆x → 0  ∆x  ∆x → 0  2    ∆x         2   = − sin x ×1 = − sin x

viz.,

d (cos x) = − sin x dx

d (tan x ) = sec 2 x dx Let y = tan x Then Dy = tan(x + Dx) – tan x 4.

sin( x + ∆x) sin x − cos( x + ∆x) cos x sin( x + ∆x) cos x − cos( x + ∆x) sin x = cos( x + ∆x) cos x sin ∆x = cos( x + ∆x) cos x =

\

 sin ∆x  dy 1 = lim   ⋅ ∆ x → 0  ∆x  cos( x + ∆x) cos x dx

Differential Calculus

=1¥ viz., 5.

1 2

cos x

= sec2 x

d (tan x) = sec2 x. dx

d (cot x ) = − cosec 2 x dx

Let Then

y = cot x Dy = cot(x + Dx) – cot x sin x cos( x + ∆x) − cos x sin( x + ∆x) sin x sin( x + ∆x) sin(−∆x) = sin x sin( x +∆x) =

\

  ∆y    sin ∆x   dy 1 = lim   = lim   ⋅ −   ∆ x → 0 ∆ x → 0   ∆x  dx    ∆x  sin x ⋅ sin( x + ∆x)  = 1×

viz.,

1 x − cosec 2 x sin 2 x

d (cot x) = − cosec 2 x dx

d (sec x ) = − sec x tan x dx Let y = sec x Then Dy = sec(x + Dx) – sec x cos x − cos( x + ∆x) = cos ⋅ cos( x + ∆x)  ∆x   ∆x  2 sin  x +   sin  2   2  = cos x ⋅ cos( x + ∆x) 6.

\

   ∆x     sin    sin  x + ∆x    ∆y     2   dy 2   = lim   = lim  ⋅  ∆ x → 0 ∆ x → 0    cos x ⋅ cos( x + ∆x)  ∆x  dx ∆x     2   =

viz.,

sin x ⋅ 1 = sec x tan x cos 2 x

d (sec x) = sec x tan x dx

1.21

1.22

Engineering Mathematics I

d (cosec x ) = − cosec x cot x dx Let y = cosec x Then Dy = cosec(x + Dx) – cosec x sin x − sin( x + ∆x) = sin( x + ∆x) ⋅ sin x   ∆x  ∆x  2 cos  x +  ⋅ sin −   2   2  = sin( x + ∆x) sin x 7.

\

viz.,

   ∆x    ∆x   cos  x +   − sin  ∆y     dy  2 2  = lim   = lim        ∆ x → 0 ∆ x → 0   ∆x  sin( x + ∆x) sin x   ∆x  dx     2   cos x = (−1) sin 2 x = − cosec x cot x d (cosec x) = − cosec x cot x. dx

d x (e ) = e x dx Let y = ex Then Dy = ex + Dx – ex  e x + ∆x − e x   ∆y  dy \  = lim   = lim   dx ∆x → 0  ∆x  ∆x → 0  ∆x 8.

 e∆x − 1  = e x ⋅ lim  ∆x → 0  ∆x     e h − 1   = 1 = e x ×1, since lim  h →1   h  

d x (e ) = e x dx d 1 9. (log e x ) = dx x Let y = logex Then Dy = loge(x + Dx) – (logex)  x + ∆x  = log e   x   ∆x  = log e 1 +   x  viz.,

Differential Calculus   ∆y   1 dy ∆x  = lim   = lim  log e 1 +  ∆ x → 0 ∆ x → 0     dx ∆x x    ∆x

\

x

1 Dx ˆ Dx Ê = lim log e Á1 + ˜ Ë Dx Æ 0 x x¯ x ¸ ÏÔ Ô x 1 D D x Ê ˆ = lim Ìlog e Á1 + ˜ ˝ Ë x Ê Dx ˆ Æ 0 ÔÓ x ¯ Ô˛ Á ˜ Ë x¯

1 ˘ È 1 = log e (e) Í∵ lim (1 + h) h = e˙ ÍÎ h Æ 0 x ˚˙ 1 1 = ◊1 = x x

viz.,

1.6

1 d (loge x) = . dx x

rULes oF DIFFereNTIaTIoN

d (c ) = 0,  where c is a constant dx Let y=c Then Dy = c – c = 0  0   ∆y  dy \ = lim   = lim   = 0 ∆ x → 0 ∆ x → 0  ∆x    dx ∆x 1.

viz.,

2.

d du (c ⋅ u) = c ,  where c is a constant and u is a function of x dx dx

Let Then \

d (any constant) = 0 dx

y = c ◊u Dy = c(u + Du) – cu = cDu  ∆u   ∆y  dy du = lim   = lim c  = c ⋅ ∆ x → 0 ∆ x → 0     dx ∆x ∆x dx

d du dv (u ± v ) = ,  where u and v are functions of x ± dx dx dx Let y=u±v Then Dy = {(u + Du) ± (v + Dv)} – {u ± v} = Du ± Dv 3.

\

ÏÊ Du ˆ Ê Dv ˆ ¸ dy Ê Dy ˆ = lim Á ˜ = lim ÌÁ ˜ ± Á ˜ ˝ dx Dx Æ 0 Ë Dx ¯ Dx Æ 0 ÓË Dx ¯ Ë Dx ¯ ˛

1.23

Engineering Mathematics I

1.24 =

du dv ± dx dx

This result can be extended as follows: If y = au ± bv ± cw ± ..., where a, b, c,... are constants and u, v, w, ... are function of dy du dv dw x, then = ± ± ± dx dx dx dx 4.   Product  rule,  viz., 

d dv du ( uv ) = u +v ,   where  u  are  v  are  different  dx dx dx

function of x Let Then \

y = uv Dy = uDv + vDu – Du . Dv  ∆v   ∆y   ∆u  dy ∆u = lim   = lim u   + lim v + lim   ∆v dx ∆x → 0  ∆x  ∆x → 0  ∆x  ∆x → 0 ∆x ∆x → 0  ∆x  dv du du +v + ×0 dx dx dx dv du =u +v dx dx =u

d dv du (uv) = u + v dx dx dx

viz., Extension

d d (uvw) = {u ⋅ (vw)} dx dx d du (vw) + vw dx dx dw dv du =u v +w + vw dx dx dx dv dw du = uv + vw + wu dx dx dx =u

{

}

 du  dv   v −u    u   d   d d x x 5. Quotient rule, viz.,    =   2     dx v   v    Let Then

y= ∆y =

u v u + ∆u u − v + ∆v v

(or)

Differential Calculus =

v ∆u − u ∆v v(v + ∆v)

 ∆y  dy = lim   = lim dx ∆x → 0  ∆x  ∆x → 0

\

1.25

 ∆v   ∆u    v lim   − u ∆lim ∆x → 0  ∆x  x → 0  ∆x   lim v(v + ∆v) ∆x → 0

du dv du dv v −u −u d d d dx x x x = = v ( v + 0) v2 v

viz.,

du dv −u v d  u  x d dx  = dx  v  v2

6. Derivative of a Function of Function When y is a function of u, where u itself is a function of x, then In this situation, ∆y =

dy dy du = ⋅ . dx du dx

∆y ⋅ ∆u ∆u

 ∆y ∆u  dy du  ∆y  dy × = lim   = lim  ⋅ . = ∆ x → 0 ∆ x → 0  ∆u ∆x  du dx   dx ∆x

\

Extension If y is a function of u, u a function of u, v a function of w and w a function of x, then dy dy du dv dw = ⋅ ⋅ ⋅ . dx du dv dw dx dy 1 = d x  d x     dy  If y is a function of x, say f(x), x can also be considered as a function of y and denoted as f –1(y) ∆y ∆x Then ⋅ =1 ∆x ∆y 7.

 ∆y  ∆x  lim    = 1  ∆x  ∆y 

\

∆x → 0 ∆y → 0

viz.,

 ∆x   ∆y  lim   ⋅ lim   = 1  ∆x  ∆y → 0  ∆y 

∆x → 0

viz.,

1 dy dx dy ⋅ = 1 or = dx dy dx  dx     dy 

Engineering Mathematics I

1.26

1.7 1.

DerIvaTIves oF hyperboLIc FUNcTIoN d (sinh x) = cosh x dx Let

y = sinh x =

\

e x − e− x [by definition] 2

{

}

dy 1 d x d = (e ) − (e− x ) dx 2 dx dx

 x   1 −x  d  =  e − (e )   (−1)      2 dx    x −x e +e = 2 = cosh x 2.

Similarly we can prove that

3.

d (tanh x) = sech 2 x dx Let

\

y = tanh x =

dy = dx

cosh x ⋅

d (cosh x) = sinh x dx

sinh x cosh x

d d (sinh x) − sinh x (cosh x) dx dx [by quotient rule] cosh 2 x

cosh 2 x − sinh 2 x cosh 2 x 1 = = sech 2 x 2 cosh x =

4.

Similarly, we can prove that

5.

d (sech x) = − sech x tanh x dx Let \

y = sech x =

d (coth x) = − cosech 2 x dx

1 cosh x

dy d 1 =− ⋅ (cosh x) dx cosh 2 x dx

Differential Calculus sinh x cosh 2 x = − sech x ⋅ tanh x =−

6. Similarly, we can prove that

d (cosech x) = − cosech x coth x dx

1.7.1  D   erivative of Inverse Circular and Inverse Hyperbolic  Functions 1.

d 1 (sin −1 x) = dx 1 − x2 Let

y = sin −1 x; Then x = sin y

\

dx = cos y dy

\

dy 1 1 1 = = = 2 dx cos y 1 − sin y 1 − x2

2. Similarly, we can prove that 3.

d 1 (tan −1 x) = dx 1 + x2 Let

y = tan −1 x; Then x = tan y

\

dx = sec 2 y dy

\

dy 1 1 1 = = = dx sec 2 y 1 + tan 2 y 1 + x 2

4. Similarly, we can prove that 5.

d 1 (cos−1 x) = − dx 1 − x2

d 1 (cot −1 x) = − dx 1 + x2

d 1 (sec−1 x) = dx x x2 − 1 Let

y = sec−1 x, Then x = sec y

\

dx = sec y tan y dy

\

dy 1 1 = = 2 dx sec y sec y − 1 x x 2 − 1

1.27

Engineering Mathematics I

1.28 6. 7.

1 Similarly, we can prove that d (cosec−1 x) = − dx x x2 − 1 d 1 (cosh −1 x) = 2 dx x −1 Let y = cosh −1 x Then x = cosh y \

dx = sinh y dy

\

dy 1 1 = = (∵ cosh 2 y − sinh 2 y = 1) 2 dx sinh y cosh y − 1 1

=

2

x −1 8.

Similarly, we can prove that

d (sinh −1 x) = dx

1 2

x +1

d 1 (coth −1 x) = − 2 dx x −1

9. Let

y = coth −1 x. Then x = coth y

\

dx = − cosech 2 y dy

\

dy 1 −1 =− = (∵ coth 2 y − 1 = cosech 2 y ) dx cosech 2 y −1 + coth 2 y =−

1 x −1 2

d 1 (tanh −1 x) = dx 1 − x2

10.

Similarly, we can prove that

11.

d 1 (cosech −1 x) = − dx x 1 + x2 Let y = cosech −1 . Then x = cosech y \

dx = − cosech y ⋅ coth y dy

\

dy 1 =− dx x 1 + x2

Differential Calculus 12. Similarly we can prove that

1.8

1.29

d 1 (sech −1 x) = − dx x 1 − x2

meThoDs oF DIFFereNTIaTIoN

Though the rules of differentiation stated earlier can be applied to find the derivatives of many functions some functions can be differentiated using the methods indicated below:

1. Logarithmic Differentiation When the given function y is of the form uv, where u and v are functions of x, we use this method, in which we take logarithms on both sides of y = uv and then differentiate w.r.t. x. This method will also simplify the work on differentiation of a function consisting of a number of products and quotients.

2. Differentiation of Implicit Functions If x and y are implicitly related as f(x, y) = 0, viz., if y cannot be expressed explicity as a function of x, the differentiation of f(x, y) = 0 w.r.t. x is done, noting that y is a dy dy function of x whose derivation is as a mixed . On simplification, we will get dx dx function of x and y.

3. Differentiation by Trigonometric Substitution Some apparently complicated functions of x, in particular, certain inverse trigonometric functions can be simplified by using appropriate trigonometric substitutions for x and then differentiation is performed.

4. Differentiation from Parametric Equations When x and y are both expressed as functions of a parameter, say ‘t’, it is not recovery dy to get y as a function of x, as can be given as a function of t. dx In this case,

dy dy dt dy = ⋅ or dt dx dt dx

dx dt

5. Differentiation of One Function w.r.t. another Function If it is required to find the derivative of u(x) w.r.t. v(x), we treat that u and v are parametrically expressed in terms of the parameter x. du du dv Then = dv dx dx

Engineering Mathematics I

1.30

WorkeD exampLes 1(b)

Example 1.1 Find (i)

1 + 3 x,

dy from first principles, when y = dx x2 + 8 (ii) (iv) e x , and (v) tan −1 (sin x) , (iii) sin3x, 2x + 3

(i) y = 1 + 3 x . Let Dx and Dy be the increments in x and y respectively. Then

∆y = 1 + 3( x + ∆y ) − 1 + 3 x

\

 1 + 3 x + 3∆x − 1 + 3 x  dy  = lim   dx ∆x → 0  ∆x  1 1   1  (1 + 3 x + 3∆x) 2 − (1 + 3 x) 2  3 −1 3  = (1 + 3 x) 2 = = 3 × lim  ∆x → 0  (1 + 3 x + 3∆x ) − (1 + 3 x )  2 1 + 3x   2

(ii) y =

x2 + 8 2x + 3

Let Dx and Dy be the increments in x and y respectively. Then

\

∆y =

( x + ∆x) 2 + 8 x 2 + 8 − 2( x + ∆x) + 3 2 x + 3

=

(2 x + 3)( x 2 + 2 x∆x + ∆x 2 + 8) − (2 x + 2∆x + 3)( x 2 + 8) (2 x + 3)(2 x + 2∆x + 3)

=

(2 x + 3)(2 x∆x + ∆x 2 ) − 2∆x( x 2 + 8) (2 x + 3)(2 x + 2∆x + 3)

∆y (2 x + 3)(2 x + ∆x) − 2( x 2 + 8) = ∆x (2 x + 3)(2 x + 2∆x + 3)  ∆y  (2 x + 3) ⋅ 2 x − 2( x 2 + 8) dy = lim   = dx ∆x → 0  ∆x  (2 x + 3) 2 =

4x2 + 6x − 2x2 − 6 (2 x + 3) 2

=

2 x 2 + 6 x − 16 (2 x + 3) 2

(iii) y = sin3x Let Dx and Dy be the increments in x and y respectively. Then Dy = sin3(x + Dx) – sin3x

Differential Calculus

1.31

= {sin(x + Dx) – sinx} {sin2(x + Dx) + sinx . sin(x + Dx) + sin2x}     ∆x  ∆x  2 2  = 2 cos  x +  {sin ( x + ∆x) + sin x ⋅ sin( x + ∆x) + sin x}  sin    2 2      \

\

Ï Ê Dx ˆ ¸ sin Á ˜ Ô Ë 2 ¯Ô Dy ÔÔ Ê Dx ˆ 2 2 = Ìcos Á x + ˜ ◊ ˝{sin ( x + Dx) + sin x sin( x + Dx) + sin x} Ë ¯ D x 2 Dx Ô Ê ˆ Ô ÁË ˜¯ ÔÓ Ô˛ 2 dy Ê Dy ˆ = lim Á ˜ = {cos x ◊1}{sin 2 x + sin 2 x + sin 2 x} dx Dx Æ 0 Ë Dx ¯ = 3 sin 2 x cos x.

(iv) y = e

x

Let Dx and Dy be the increments in x and y respectively Then

∆y = e

x + ∆x

−e

x 1

 ∆x 2 = e 1 +  −e  x  x

x

 ∆x  x = e x 1 +  − e , expanding by Binomial theorem and omitting  2 x  higher power of Dx ∆x

= e x {e 2 \

x

− 1}

  e∆x / 2 x − 1  ∆y  dy = lim   = e x ⋅ lim   × 2 x ∆x → 0   ∆x dx ∆x → 0  ∆x       2 x  = e x ×1× 2 x

  e mh −1      = 1 ∵ lim h→0    mh   

(v) y = tan–1(x) Let Dx and Dy be the increments in x and y respectively. Then y + Dy = tan–1{(x + Dx)} \ x = tan y and (x + Dx) = tan (y + Dy) \ Dx = tan (y + Dy) – tan y \

   ∆y  dy ∆y  = lim   = lim    ∆ x → 1 ∆ y → 0  ∆x   dx   tan( y + ∆y ) − tan y  {∵ when ∆x → 0, ∆y also → 0}  cos y ⋅ cos( y + ∆y ) ⋅ ∆y  = lim    ∆y → 0   sin ∆y  

Engineering Mathematics I

1.32

 ∆y  2 = cos 2 y ⋅ lim   = cos y ⋅ 1 ∆y → 0  sin y   =

1 1 1 = = 2 2 sec y 1 + tan y 1 + x 2

Example 1.2 Find

dy , when y = sin m x + cos n x . dx

y = sin m x + cos n x \

dy d 1 = × (sin m x + cos n x) m n dx 2 sin x + cos x dx 1

=

m

n

{m sin m − 1 x cos x + n cos n − 1 x ⋅ (− sin x)}

2 sin x + cos x =

sin x cos x(m sin m − 2 − n cos n − 2 x

Example 1.3 Find

2 sin m x + cos n x  a + b cos x  dy , where y = log e  .  a − b cos x  dx

y = log(a + b cos x) − log(a − b cos x) \

dy b sin x −b sin x = − dx a + b cos x a − b cos x  a − b cos x + a + b cos x     = − b sin x    2 2 2   − a b cos x     −2ab sin x = 2 a − b 2 cos 2 x

2   Example 1.4 Find dy , when y = log  x − 1 + x .  x + 1 + x 2  dx  

y = log( x − 1 + x 2 ) − log( x + 1 + x 2 )    1 1 1    1 +  ⋅ (2 x) ⋅ 2 x 1 − − 2 2  2     2 1+ x 2 1+ x      x + 1 + x  2 2      1 + x + x   1 1  1− x − x  = −     2  x + 1 + x 2  x − 1 − x2  1 + x 2       1+ x   1 1 2 =− − =− 2 2 1+ x 1+ x 1 + x2

dy 1 = dx x − 1 + x 2

Differential Calculus

Example 1.5 Find

1.33

dy cos x + sin x , when y = log . dx cos x - sin x

1 y = [log(cos x + sin x) − log(cos x − sin x )] 2 dy 1  − sin x + cos x (− sin x − cos x)   =  − cos x − sin x  dx 2  cos x + sin x =

1  (cos x − sin x) 2 − (cos x + sin x) 2   2  cos 2 x − sin 2 x 

1 1 2(cos 2 x + sin 2 x) = = × 2 2 2 2 cos x − sin x cos x − sin 2 x Example 1.6 Find

dy , when y = ( x + x 2 + a 2 ) n + (−x + x 2 + a 2 )−n . dx

 dy = n( x + x 2 + a 2 ) n − 1 1 +  dx 

  x + a  x

2

 − n(−x + x 2 + a 2 )−n− 1 −1 +   1

=

2

x +a

2

⋅ n( x + x 2 + a 2 ) n +

  x + a  x

2

2

1 2

x +a

2

⋅ n(−x + x 2 + a 2 )−n

ny

=

2

x + a2

Example 1.7 Find

x 2  dy a2 , when y =  x + a 2 + log{x + x 2 + a 2 } .   dx 2 2

dy x x = ⋅ + dx 2 x 2 + a 2 =

2

1  2 x 2 + a 2 + 2  x 2 + a 2 

x2 + a2 a 2  1 +  2 2  x + x 2 + a 2   a2  x 2 + a 2 

 x  ⋅ 1 + 2   x + a 2  

= x2 + a2    b + a + b − a tan x  dy 1 2 . Example 1.8 Find , when y = log   2 2  dx b −a  b + a − b − a tan x   2 

Engineering Mathematics I

1.34 Putting

b + a = c and b − a = d , we get    x x   log c + d tan − log c − d tan  , where c and d    2 2  are constants   d x d x   sec 2 sec 2 1  2 dy 2 2 2  = +   dx cd  c + d tan x c − d tan x   2 2   y=

=

1 cd

1 x c sec 2 ×  2c 2  2 2 2 c − d tan sec 2

=

x 2

(b + a ) − (b − a ) tan 2 sec 2 =

=

 a 1 + tan 2 

x 2

x   2  a+b x 1 + tan 2 2

=

x   2 

1 a + b cos x

dy Ê a + b cos x ˆ , when y = cos -1 Á . dx Ë b + a cos x ˜¯

dy =dx

=− =−

=

x 2

 x  2  + b 1 − tan   2

1  1 − tan 2 

Example 1.9 Find

x   2 

1 2

Ê a + b cos ˆ 1- Á Ë b + a cos x ˜¯ Ï (b + a cos x)(-b sin x) - (a + b cos x) ◊ (- a sin x) ¸ ◊Ì ˝ (b + a cos x)2 Ó ˛ (b + a cos x) (b + a cos x) 2 − (a + b cos x) 2 (a 2 − b 2 ) sin x (b 2 − a 2 ) sin 2 x (b + a cos x)

b2 − a 2 b + a cos x

⋅

(a 2 − b 2 ) sin x (b + a cos x) 2

Differential Calculus Example 1.10 Find dy = dx

 a cos x + b sin x  dy , when y = tan −1  .  b cos x − a sin x  dx

1 2  a cos x + b sin x  1 +    b cos x − a sin x 

{b cos x − a sin x)(−a sin x + b cos x) − (a cos x + b sin x)(−b sin x − a cos x)} (b cos x − a sin x) 2 =

1 (a 2 + b 2 ) (b cos x − a sin x) 2 + (a cos x + b sin x) 2

=

a 2 + b2 =1 a 2 + b2

y dy 2a − y - 2ay - y 2 , show that = . dx y 2a

Example 1.11 If x = 2a sin -1 Differentiating w.r.t. y, we have

dx = 2a ◊ dy

= =

1 1-

2a 2a 2a − y

y 2a

2ay − y =

\

2

d y 1 ( 2a - 2 y ) dy 2a 2 2ay - y 2 1

⋅

a

◊

2a ⋅ 2 y −

y 2ay − y 2

dy 2ay − y 2 = dx y =

y 2a − y y

=

2a − y y

−

(a − y ) 2ay − y 2

a− y 2ay − y 2

1.35

Engineering Mathematics I

1.36 Example 1.12 Find

dy , when y = sinh x . dx

dy d 1 = (sinh x ) dx 2 sinh x dx 1

=

cosh x ,

2 sinh x

d x dx

cosh x

=

4 x sinh x Example 1.13 Find

1 + x  d tan   1 − x  1 + x  dx 2  1 + tan   1 − x  1

dy = dx

=

1 + x  dy , when y = sinh −1 tan  . 1 − x  dx

1 + x  d 1 + x  1 sec 2  ⋅   1 − x  dx 1 − x  1 + x  sec   1 − x 

 (1 − x) − (1 + x)(−1)  1 + x  = sec    1 − x    (1 − x) 2   =

1 + x  2 sec   1 − x  (1 − x) 2

Example 1.14 Find

x 3 x2 + 4 dy , when y = . dx x2 + 3

Taking logarithms, 1 1 log y = log x + log( x 2 + 4) − log( x 2 + 3) 3 2 Differentiating w.r.t x, 1 dy 1 2x x = + − 2 2 y dx x 3( x + 4) x + 3 \

2  dy x 3 x + 4  x  1 2x  = + −   2 dx x2 + 3    x 3( x + 4) x + 3

Example 1.15 Find Let

dy , when y = x sin x + (sin x) x . dx

y = u + v, say.

Differential Calculus dy du dv = + dx dx dx u = xsin x Taking logarithms, log u = sin x log x Then

\

(1)

1 du sin x = + log x ⋅ cos x u dx x

{

}

(2)

dv = (sin x) x ( x cot x + log sin x) dx

(3)

du 1 = x sin x sin x + log x ⋅ cos x dx x x v = (sin x) Taking logarithms, log v = x log sin x \

\ \

1 dv 1 = x⋅ cos x + log sin x v dx sin x

Using (2) and (3) in (1), we get

Example 1.16 Find Let Then

dy . dx

1 1 1+ dy , when y = (1 + x) x + x x . dx

y = u + v, say

dy du dv = + dx dx dx u = (1 +

(1)

1 x) x

Taking logarithms, log u = \

1.37

1 log (1 + x) x

1 du 1 1 1 = ◊ - 2 log(1 + x) u dx x 1 + x x 1

\

du 1 Ï 1 ¸ = (1 + x) x Ì - 2 log(1 + x) ˝ dx Ó x(1 + x) x ˛ 1+

v=x

1 x

Taking logarithms, log v = ÊÁ1 + Ë

1ˆ ˜ log x x¯

(2)

Engineering Mathematics I

1.38 \ \

1 dv  1 1 1 = 1 +  ⋅ − 2 log x   v dx x x x 1 1+ dv x +1 1 =x x − 2 log x dx x2 x

{

Using (2) and (3) in (1), we get

}

(3)

dy . dx

dy Example 1.17 Find , when x and y are connected by the relation xy + yx = c, where d x c is a constant. x y + y x = c, viz., u + v = c du dv + =0 dx dx u = xy log u = y log x

\ \

(1)

1 du y dy = + log x ⋅ u dx x dx

\

y du dy  = x y  + log x   dx dx  x x v=y log v = x log y

\ \

(2)

1 dv x dy = + log y v dx y dx

\

 x dy  dv = y x  + log y   y dx dx 

\

(3)

Using (2) and (3) in (1), we get  x dy  y dy  + log y  = 0 x y  + log x ⋅  + y x   x  dx  y dx  viz.,

( x y log x + xy x − 1 )

\

Example 1.18 Find

dy = − ( yx y − 1 + y x log y ) dx dy ( yx y − 1 + y x log y ) =− dx ( xy x − 1 + x y log x)

dy , when (sin x)cos y = (cos x)sin y . dx

Taking logarithms on both sides of the given relation, we get cos y ⋅ log sin x = sin y ⋅ log cos x

(1)

Differential Calculus

1.39

Differentiating both sides of (1) w.r.t. x, cos y ⋅ cot x − log sin x(− sin y ) viz.,

dy dy = − sin y tan x + log cos x (cos y ) dx dx

(sin y log sin x − cos y log cos x)

dy = − (sin y tan x + cos y cot x ) dx

dy sin y tan x + cos y cot x = dx cos y log cos x − sin y log sin x

\

Example 1.19 If x m y n = ( x + y ) m + n , prove that

dy y = . dx x

Taking logarithms on both sides of the given relation, we get m log x + n log y = (m + n) log( x + y ) Differentiating both sides of (1) w.r.t. x, m n dy 1 + = ( m + n) ⋅ x y dx x+ viz.,

viz., \

m + n  −  x + y

(1)

  1 + dy  y  dx 

n  dy m m + n  = − y  dx x x+ y

(my − nx) dy my − nx ⋅ = y ( x + y ) dx x( x + y ) dy y = dx x

Example 1.20 If 1 − x 2 + 1− y 2 = a ( x − y ), prove that

1 − y2 dy = . dx 1 − x2

Differentiating both sides of the give equation w.r.t. x, we get  −x y dy dy  − ⋅ = a 1 −  2 2  d x dx  1− x 1− y

(1)

Eliminating ‘a’ between the given equation and (1), we get dy x y dy − − 1− 2 2 dx 1− x 1− y dx = 2 2 x y − 1− x + 1− y viz.,

viz.,

 y ( x − y )  dy x( x − y )  1 − x2 + 1 − y 2 −  = 1 − x2 + 1 − y 2 + 2  dx  1 − y  1 − x2  2 2 2 2  (1 − x 2 )(1 − y 2 ) + 1 − y 2 − xy + y 2    dy = 1 − x + (1 − x )(1 − y ) + x − xy     dx 1 − x2 1 − y2

Engineering Mathematics I

1.40

viz.,

\

(1 − x 2 )(1 − y 2 ) + (1 − xy ) dy (1 − x 2 )(1 − y 2 ) + (1 − xy ) ⋅ = dx 1 − y2 1 − x2 dy 1 − y2 = dx 1 − x2

Example 1.21 Find

2   2 x  dy −1 1 − x  . , when y = sin −1   + cos  2 2 1 + x  dx 1 + x 

x = tan q 2    2 tan θ  −1 1 − tan θ   y = sin −1   + cos  2  2 1 + tan θ  1 + tan θ 

Put Then

= sin −1 (sin 2θ ) + cos−1 (cos 2θ ) = 4θ or 4 tan −1 x \

dy 4 = dx 1 + x 2

Example 1.22 Find

dy , when y = cos 2 tan −1 dx

Put

x = cos q

Then

1 − cos θ  y = cos 2 tan −1   1 + cos θ 

1 − x   . 1 + x 

θ 2 2 θ 2 cos 2 2 sin 2

2

= cos tan

−1

 θ = cos 2 tan −1  tan   2 = cos 2 \

dy 1 = . dx 2

Example 1.23 Find Put

1 θ 1 = (1 + cos θ ) = (1 + x) 2 2 2

 x + a  dy  , where a is a constant. , when y = tan −1   1 − ax  dx

x = tan q and

a = tan a

Differential Calculus  tan θ + tan α  −1 y = tan −1   = tan tan(θ + α) 1 − tan θ tan α 

\

= θ + α = tan −1 x + tan −1 a \

dy 1 1 1 = +0= , dx 1 + x 2 x 2 x (1 + x)

Example 1.24 Find

 1 + x 2 + 1 − x 2  dy  . , when y = tan −1   1 + x 2 − 1 − x 2  dx  

Put

x2 = cos q

Then

 1 + cos θ + 1 − cos θ   y = tan −1   1 + cos θ − 1 − cos θ     2 cos θ + 2 sin θ   2 2  = tan −1   θ θ   2 cos − 2 sin  2 2   θ    1 + tan 2    −1  = tan     θ   tan 1 −    2    π θ     tan + tan     −1  4 2 = tan     π θ   1 − tan ⋅ tan    4 2   π θ  = tan −1 tan  +   4 2 π 1 = + cos−1 x 2 4 2

\

dy 1 x −1 = ⋅ ⋅ 2x = − 4 dx 2 1 − x 1 − x4

Example 1.25 Find

dy 3at 3at 2 , when x = and y = . dx 1+ t 1 + t3 x= =

3at 1 + t3

∴

3a (1 − 2t 3 ) (1 + t 3 ) 2

dx 3a{1 + t 3 ) − t ⋅ 3t 2 } = dt (1 + t 3 ) 2

1.41

Engineering Mathematics I

1.42

y= \

dy 3a{(1 + t 3 )2t − t 2 ⋅ 3t 2 } = dt (1 + t 3 ) 2 =

\

3at 2 1 + t3

3a (2t − t 4 ) (1 + t 3 ) 2

dy dy dx 2t − t 4 = ÷ = dx dt dt 1 − 2t 2

Example 1.26 Find

dy t , when x = a (cos t + log tan ) and y = a sin t. dx 2

 t x = a cos t + log tan   2 \

 dx t 1  1  = a − sin t + ⋅ sec 2 ⋅   t  dt 2 2   tan   2     1    =a − sin t +  t t     2 sin cos  2 2      1  a cos 2 t = a − sin t + =  sin t sin t 

\

y = a sin t dy = a cos t dt

\

dy dy dx a cos t = ÷ = × sin t = tan t dx dt dt a cos 2 t

Example 1.27 Find

dy , when y = log(sec θ + tan θ ) and x = sec θ. dx y = log(sec θ + tan θ )

\

dy 1 = (sec θ tan θ + sec 2 θ ) dθ sec θ + tan θ = sec θ x = sec q

\

dx = sec θ tan θ dθ

Differential Calculus \

dy dy dx sec θ = ÷ = = cot θ. dx dθ dθ sec θ tan θ

 3 x − x 3   2   w .r .t . sin −1  Example 1.28 Differentiate tan −1  . 2  1 + x 2   1 − 3 x  Let

 3 x − x 3   2 x   and v = sin −1  u = tan −1   2  1 + x 2   1 − 3 x 

Put

x = tan θ

Then

u = tan −1 (tan 3θ ) and v = sin −1 (sin 2θ )

viz.,

u = 3θ and v = 2θ du du dv 3 = ÷ = dv dθ dθ 2

Example 1.29 Find

−1 −1 du , where u = esin x and v = ecos x . dv

u = esin \

−1

−1 du 1 = esin x ⋅ dx 1 − x2 −1

v = ecos \

x

x

−1 dv 1 = ecos x ×− dx 1 − x2 −1 −1 du du dv = ÷ = − e(sin x − cos x ) dv dx dx

Example 1.30 Differentiate tan −1

Let Put Then

1 + x2 − 1 w.r.t. tan–1x. x

 1 + x 2 − 1    and v = tan −1 x u = tan −1    x x = tan q  sec θ − 1 −1 1 − cos θ   u = tan −1   = tan   sin θ   tan θ     2 sin 2 θ   θ θ   2  = tan −1  tan  = = tan    θ θ 2 2  2 sin cos  2 2 −1

1.43

Engineering Mathematics I

1.44

v = tan −1 (tan θ ) = θ du du dv 1 = ÷ = . dv dθ dθ 2

EXERCISE 1(b)

Part A (Short Answer Questions)

x x 2 x3 dy + + + , show that = 1! 2! 3! dx dy Using first principles, find when dx 1 2. y = 3. y = x 1. If y = 1 +

4. y = log x

y.

x+2 x−2

5. y = x2 sin x

6. y = sin–1 x2 Find

dy using the rules and methods of differentiation, when dx

1 − sin x  7. y = log   1 + sin x 

2 8. y = tan 1 + x + x

9. y = cos3x sin2x 11. y = tan–1(sinh x) 13. x3 + y3 = 3axy

10. y = sin–1(cos x) 12. y = (tan x)cot x 14. y = axy

15. y = sin −1 (2 x 1 − x 2 )

 x  16. y = tan −1    1 − x 2   

17. x = cos3t and y = sin3t 18. x = a(q – sin q) and y = a(1 – cos q) 19. Differentiate u = a sec x w.r.t. v = b tan x 20. Differentiate u = sin 2 x w .r .t . v = cos 2 x Part B Differentiate the following function w.r.t. x: 21.

a 2 + b 2 − 2ab cos x

22.

1 + sin 2 x − 1 − sin 2 x

Differential Calculus

23.

x sin x + cos x x sin x − cos x

24.

1.45 1 − x + x2 1 + x + x2

ax 3 25. e cos bx

 1 + x − 1   26. log e   1 + x + 1 

2 2     a + x + x  27. log   2   + − a x x      

2 2 28. sec log a − x

2

29.

x sin −1 x 1 − x2

2 2 2 2 30. x x − a − a log( x + x − a

cosh x + cos x sinh x + sin x

−1 31. tan (cos x )

32.

1 − x   33. sinh  1 + x 

 ( x + a )( x 2 + b)  2 34.     x3 + c

1

−1

1 x 35. x + x x

36. (sin x) tan x + (tan x)sin x

37. Find

dy , when x and y are connected by (x + y)x = xx + y. dx

38. Find

dy , if sin y = x sin(a + y). dx

39. Find

dy , if (sin x)cos y + (cos y )sin x = 1. dx

40. Find

dy , if (x – c) (y – c) = 1 + c2. dx

41. Find

dy 1+ x , if y = sin 2 cot −1 . dx 1− x

42. Find

dy , y = cos–1(4x3 – 3x). dx

43. Find

dy cos x − sin x , if y = . dx cos x + sin x

44. Find

dy , if y = cos−1[ x 2 − x 3 + x − x 3 ]. dx

Engineering Mathematics I

1.46 45. Find

dy , if x = cos t + t sin t and y = sin t − t cos t. dx

46. Find

dy c , if x = c log tan θ and y = (tan θ + cot θ ). dx 2

47. Find

dy , if x = 3 sin t − sin 3 t and y = 3 cos t − cos3 t. dx

 2 x  −1  2 x  . 48. Differentiate tan −1   w.r.t. sin  1 + x 2  1 − x 2   3 x − x 2   2 x   w.r.t. tan −1  49. Differentiate tan −1  . 2  1 − x 2   1 − 3 x    1 + x2 − 1 − x2     −1 2 50. Differentiate tan −1    w.r.t. cos ( x ). 2 2   1 + x + 1 − x      

1.9

maxIma aND mINIma oF FUNcTIoNs oF oNe varIabLe

A function f(x) is said to have a maximum at the point x = a, if f(a) ≥ f(a + h) for all positive and negative values of h sufficiently near zero. Similarly f(x) is said to have a minimum at the point x = b, if f(b) £ f(b + h) for values of h close to zero. The figure given below represents the graph of f(x), viz., y = f(x) is the equation of the curve shown. The function has maximum at A and C, while it has minimum at B and D. In other words if the continuous function increases algebraically upto a y A C certain value and then decreases, that value is called a maximum value of the function. Similarly if the continuous function D B decreases algebraically upto a certain value and then increases, that value is called a x O x=a x=b minimum value of the function.

Note

Fig. 1.5

(1) There are values of f(x) which are greater then the maximum and values which are less than the minimum. Hence f(a) is a relative maximum value of f(x). This means that f(a) is algebraically greater than f(a – h) and f(a + h) where h is a small positive quantity. Similarly, a relative minimum value of f(x) occurs at x = b. (2) The points A, B, C, D... at which the tangents to the curve y = f(x) are parallel to the x-axis are called stationary points or turning points of the curve y = f(x). The maximum or minimum values of f(x) [at x = a and x = b] are called extreme values of f(x).

Differential Calculus

1.47

Theorem If f(x) is differentiable at x = a and has a maximum or minimum there, then f¢(a) = 0 By definition f (a) ≥ f(a + h), if x = a is a maximum point. By definition f ′(a ) = lim h→0

{ f (a + hh) − f (a) }, since f¢(x) exist at x = a.

This value is zero, for f ( a + h) − f ( a ) f ( a + h) − f ( a ) ≤ 0, if h > 0 and ≥ 0, if h is < 0 h h viz.,

 f ( a + h) − f ( a )  ≤0 lim+  h → 0   h

(1)

and

 f ( a + h) − f ( a )  ≥0 lim−  h → 0   h

(2)

(1) and (2) will be true, if and only if f¢(a) = 0. Similarly if f(b) is a minimum value of f(x), f¢(b) = 0.

Theorem f(a) is an extreme value of f(x) if and only if f¢(x) change sign as x passes through the point x = a. Form the figure, it is clear that f(x) is an increasing function before reaching A and after passing through A it is a decreasing function. dy Thus [slope of the curve y = f(x)] changes sign from positive to negative as x dx passes through A d2 y is negative at A. dx 2 Hence a function y = f(x) is said to have a maximum (minimum) at x = a, if dy (i) = 0 at x = a dx

viz.,

(ii)

d2 y < 0 at x = a for a maximum dx 2 d2 y > 0 at x = a for a minimum dx 2

Working Rule (1) Find f ¢(x) and solve the equation f¢(x) = 0. Let the roots be a, b, c,... (2) Find f ¢¢(x) and find f¢ ¢(a), f ¢¢(b), etc. (3) If f ¢¢(a) < 0, there is a maximum at the point x = a for f(x). If f ¢¢(a) > 0, there is a minimum point for f(x) at x = a.

1.48

Engineering Mathematics I

(4) If f ¢¢(a) = 0, this rule fails for testing whether f(x) is maximum or minimum at x = a. WORKED EXAMPLES 1(c)

Example 1.1 Find the maximum and minimum values of f(x) = 2x3 – 9x2 – 24x – 20. f¢(x) = 6x2 – 18x – 24 = 6(x2 – 3x – 4) = 0, when (x – 4)(x + 1) = 0 \ The turning point of f(x) are x = 4 and x = –1 f¢¢(x) = 6(2x – 3) f¢¢(4) = 6 × 5 > 0 \ f(x) is minimum at x = 4 f¢¢(–1) = 6 × 5 < 0 \ f(x) is maximum at x = –1 \ Maximum of f(x) (at x = –1) = –2 – 9 + 24 – 20 = –7. and minimum value of f(x) (at x = 4) = 128 – 144 –96 – 20 = –132. Example 1.2 Find the maximum and minimum values of y = 3 sin2x + 4 cos2x. dy = 6 sin x cos x − 8 cos x sin x dx = − 2 sin x cos x or − sin 2 x d2 y = − 2 cos 2 x dx 2 The turning points of y are given by sin x = 0 and cos x = 0 viz.,

x = 0 and x =

π 2

 d 2 y   d 2 y     = − < 2 0 and  dx 2   dx 2  π = 2 > 0 x=0 x= 2

π 2 Maximum value of y = 4 and minimum value of y = 3 \ y is maximum at x = 0 and minimum at x =

1 Example 1.3 Find the maximum and minimum values of y = kx 2 log e   ,  x  where x > 0. y = –kx2 log x   dy 1 = k  x 2 ⋅ + 2 x log x    dx x = kx(1 + 2 log x)  2  d2 y = k  x ⋅ + 1 + 2 log x  2  dx  x = k (3 + 2 log x)

Differential Calculus

1.49

dy 1 = 0, when x = 0 and x = (∵ x > 0) dx e  d 2 y   is undefined and  dx 2  x=0 \ y is maximum at x =

 d 2 y    dx 2  x=

1

= k {3 − 2 log e } = − k < 0

2

1 e

Maximum value of y = k ⋅

1 1 log . e e

Example 1.4 Find the maximum and minimum values of y, given by x2y = x3 – 3x2 + 4. x3 − 3x 2 + 4 y= x2 \

dy x 2 (3 x 2 − 6 x) − ( x 3 − 3 x 2 + 4) ⋅ 2 x = dx x4 =

\

8 x4 − 8x or 1 − 3 4 x x

dy = 0, at x 3 = 8 or x = 2. dx d 2 y 24 = 6 > 0, at x = 2. dx 2 x

\ y is minimum at x = 2 and the minimum value of y =

8 − 12 + 4 =0 4

Example 1.5 Find the minimum value of (x2 + y2), given that x and y are connected by the relation ax + by = 0, where a, b, c are constants. c − ax , b using (1) in the given function, whose minimum value is required, we get 1 f ( x) = x 2 + 2 (c - ax)2 b From the given constant y =

2 (c − ax)(−a ) b2  2a 2  2ac f ′( x) = 0, where x 2 + 2  = 2  b  b

f ′( x ) = 2 x +

(1)

(2)

Engineering Mathematics I

1.50 viz., when

x=

ac a + b2 2

f ′′( x) = 2 +

2a 2 > 0, for values of x. b2

\ f(x) is minimum, when x = of f ( x) =

ac bc get from (1), minimum value or y = 2 2 a +b a + b2 2

c2 . a 2 + b2

Example 1.6 A rectangular sheet of metal has four equal square portions removed at the corners and the sides are then turned up so as to form an open rectangular box. Show that, when the volume of the box made is maximum, the depth will be 1 [(a + b) - a 2 - ab + b 2 ], where a and b the sides of the original rectangle. sin b Let x be the side of each square metal removed. x x Then the dimensions of the rectangular box made are a – 2x, b – 2x and x x x \ Volume of the box made, V = x(a – 2x)(b – 2x), where x is the depth of the box Fig. 1.6 viz., V = 4x3 – 2(a + b)x2 + abx (1) We have to find the value of x, for which V is maximum. dV (1) = 12 x 2 − 4(a + b) x + ab dx d 2V = 24 x − 4(a + b) dx 2

(2)

dV = 0, when 12x2 – 4(a + b)x + ab = 0 dx viz., when

x= =

when

x=

4(a + b) ± 16(a + b) 2 − 48ab 24 1 2 2  (a + b) ± a − ab + b  6 d 2V 1 2 2  (a + b) + a − ab + b  = 2 > 0 6 dx

1 2 2  (a + b) − a − ab + b  is the admissible value of the depth for 6 which v is maximum. Hence x =

Differential Calculus

1.51

x , where ( x + k )2 k, the valve resistance is a constant and x is a variable impedance. Show that the output is a maximum when x = k. P, the power output in given by Example 1.7 The power output of a radio valve is proportional to

P=

λx , where l is the constant of proportion. ( x + k )2

 ( x + k ) 2 − x ⋅ 2( x + k )  dP  =λ 4   dx x + k ( )   λ(k − x) ( x + k )3  ( x + k )3 (−1)(k + x) 2 (k − x)  d2 P  =λ   dx ( x + k )3    3(k − x)   = λ −1 −  (k + x)3  =

dP d2 P = 0, when x = k and 2 = − λ < 0 at x = k . dx dx \ P is maximum, when x = k. Example 1.8 A man in a rowboat at P, 5 kms form the nearest point A situated on a straight shore wishes to reach a point B, 6 kms from A along the shore, in the shortest time. Where should be land on AB, if the can row 2 kms per hour and walk 4 kms per hours? L A x B Let L be the point of landing such that AL = x Rowing distance =

x 2 + 25

5 kms

Walking distance = 6 – x The time taken by the man to go from P to B via L is given by 1 2 6− x T= x + 25 + 2 4 x dT 1 = =− dx 2 x 2 + 25 4  2  x2  x + 25 −  2  2 x + 25  d T 1 =  = 2  dx 2 x 2 + 25 

P

Fig. 1.7

25 2 3

( x 2 + 25) 2

Engineering Mathematics I

1.52

dT = 0, where dx viz., viz.,

x 2

=

x + 25

1 2

4x2 = x2 + 25 5 x= 3

5 d 2T 5 \ T is minimum, where x= . > 0, where x = 2 3 dx 3 \ The man should land at L, where AL =

5 kms. to reach B in the shortest times 3

Example 1.9 The horse-power developed by an air-craft travelling horizontally with AW 2 velocity v metres per second is H = + Bv3 , where A, B, Ware constants. Find v for what value of v, the horse-power is minimum? H=

AW 2 + Bv 3 v

AW 2 dH = − 2 + 3Bv 2 dv v 2 d H 2 AW 2 = + 6 Bv v3 dv 2 1

 AW 2 4 dH AW 2  or v =  = 0, if v 4 =  3B  dv 3B  d 2 H   1  dv 2   AW 2 4 > 0 v =   3 B 

1

 AW 2 4  \ H is minimum when v =   3B  Example 1.10 What is the maximum volume of a right circular cone if the sum of the length of its height and base radius is a constant k? Let ‘x’ be the base radius and ‘h’ the height of the right circular cone. Then its volume V =

π 3 x y 3

(1)

Given that x + y = k

(2)

π Using (2) in (1), we have V = x 2 (k − x) 3

(3)

Differential Calculus

1.53

Differentiating (3) w.r.t. x; dV π = (2kx − 3 x 2 ) dx 3 2 dV π = (2k − 6 x) 3 dx 2 dV = 0, where x(2k − 3 x) = 0 dx

2k 3 x = 0 is meaningless.

viz., where x = 0 or

Ê d 2V ˆ 2p ÁË 2 ˜¯ 2 k = 3 (k - 2k ) < 0 dx x = 3

\ V is maximum, when x =

2k . 3

Example 1.11 An open cylindrical vessel is to be made with a given quantity of metal sheet. Find the ratio of the height to the diameter of the base, if the vessel is to have maximum capacity. Let r be the base radius and h the height of the cylindrical vessel, assumed to be open at one end. Given Quantity of metal used = c viz., 2prh + pr2 = c (or) 2rh + r2 = k (1) The volume of the vessel V is given by V = pr2h (2) Using (1) in (2) we have V = π r 2 viz., \

V=

(k − r 2 ) 2r

π (kr − r 3 ) 2

dV π = (k − 3r 2 ) dr 2 d 2V = − 3π r dr 2 dV = 0, when k – 3r2 = 0 dr

viz., when

r=

For this value of r,

k 3

d 2V < 0. dr 2

Engineering Mathematics I

1.54

\ V is maximum, when r =

k 3

k k − r2 = When r = , from (2), h = 3 2r

k 3 or k 2 3

k−

k 3

Thus when V is maximum, h:r=1:1 \ h : d = 1 : 2, where d is the diameter of the base. Example 1.12 A normal window consists of a rectangle surmounted by a semicircle. Given the perimeter of the window to be k, find its height and breadth, if the quantity of light admitted is to be maximum. Let the length and breadth of the rectangular part of the window be l and 2b Then the perimeter of the window is given by

2l + 2b + p b = k

(1)

If the light admitted is to be maximum, the area of the window A = 2lb + pb2 is to be maximum. Using (1) in the value of A, π viz., A = (k − 2b − πb)b + b 2 2 viz.,

 π  π dA  = k − 2b − b ⋅1 − 2 +  b  db  2  2 = k − 4b − πb d2 A = −4 − π < 0 db 2

When b =

k π+4

k , from (1), 2l = k − (π + 2)b π+4 = k − (π + 2), =

b

Fig. 1.8

 π  A = k − 2b − b b  2 

\ A is maximum, when b =

l

k π+4

k 2k or l = π+4 π+4

Differential Calculus \ A is maximum, when l = b =

1.55

k π+4

viz., when the greatest height and breadth of the window are equal. Example 1.13 Find the dimensions of the rectangle of maximum area which can be inscribed in a circle of radius R. When the rectangle inscribed in a circle is to have maximum area, it should be symmetrically situated inside the circle and the four y corners should be on the circle. Choosing the centre of the circle as the origin, its equation become (x, y) x2 + y2 = R2 (1) If 2x, 2y are the length and width of the rectangle, x equation (1) holds good. O We have to find the maximum of A = 4 xy = 4 x k 2 − x 2   dA x2  = 4  k 2 − x 2 − 2 2 dx  k − x  =

Fig. 1.9

4(k 2 − 2 x 2 ) k 2 − x2

 k 2 − x 2 (−4 x) + ( x 2 − 2 x 2 )  d2 A ( x)  = 4  ⋅  2 2 2 dx k −x  k 2 − x 2   −4 x ( k 2 − x 2 ) + x ( k 2 − 2 x 2 )   = 4 3   2 2 2   (k − x )   2 2  −3k x + 2 x   = 4 3    (k 2 − x 2 ) 2    dA k = 0, when x = . dx 2 −3k 2 + 2k 2  d2 A k    < 0 = ⋅ 4  2 3  2 dx 2   k 2 2            2 \ A is maximum, when the length and breadth of the rectangle are equal, each equal to k 2. When

x=

k

,

Engineering Mathematics I

1.56

Example 1.14 Prove that the height of the cylinder of maximum value that can be 2a inscribed in a sphere of radius a is . 3 Let r and 2h be the base radius and height of the cylinder inscribed in the sphere. By symmetry, (r, h) lies on x2 + y2 = a2 y \ r2 + h2 = a2 (1) (r, h) The volume of the cylinder to be maximise is given by V = p r2h = π(a 2 − h 2 )h, from (1) x

O

dV = π (a 2 − 3h 2 ) dh d 2V = 6π h < 0, for all h dh 2

x2 + y2 = a2

Fig. 1.10

dV a = 0, when h = dh 3 2 dV a < 0 at h = dh 2 3 \ V is maximum, when the height of the cylinder 2h =

2a 3

.

Example 1.15 Of all the right circular cones of given slant length l, find the dimensions and volume of the cone of maximum volume. Let r, h, l be the radius, height and slant length of the cone. Given r2 + h2 = constant (–l2) p If V is the volume of the cone, then V = r 2 h 3 p i.e., V = h(l 2 - h 2 ) 3

dV p 2 = (l - 3h 2 ) dh 3 d 2V = - 2p h < 0, for all h dh 2 dV l2 l l ; V is maximum, when h = = 0, when h 2 = or h = dh 3 3 3 When h =

l 3

, from (1), r 2 =

2l 2 2 or r = l 3 3

(1)

Differential Calculus

1.57

π 2 r h 3 π 2 l = ⋅ l2 ⋅ 3 3 3 2π 3 = l 9 3

Maximum volume of the cone =

EXERCISE 1(c) Part A (Short Answer Questions) 1. Find the maximum and minimum values of y = (x – 2)2 (x – 3). x

Ê 1ˆ 2. Find the maximum value of Á ˜ . Ë x¯ 3. Find the maximum and minimum values of (cos x + cos 2x) in (0, 2p). 4. Find the maximum and minimum values of x +

4 . x+2

5. Find the minimum values of x + y, if xy = 1. Part B 6. A battery whose internal resistance is r ohms E.M.F is E sends a current

RE 2 through an internal resistance R. The power is given by W = . Given ( R + r )2 E and r, find R so that W may be maximum. 7. The bending moment M at a distance x from one end of a beam of length 1 1 l uniformly loaded is given by M = wlx − wx 2 , where w = load per 2 2 unit length . Show that the maximum bending moment is at the centre of the beam. 8. The cost of fuel per hour for running a train with uniform spend is proportional to be square of the speed in kms per hour and the cost is Rs. 48/= per hour when the speed is 16 kms per hour, what is the most economical speed if the other fixed charge are Rs. 300 per hour? Take the distance to be covered is a constant. 9. Of all rectangles of given area, show that the square has the least perimeter. 10. A rod AB of given length k slides between two perpendicular lines 0x and 0y. Find when the area of the triangle is maximum.

Engineering Mathematics I

1.58

11. An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. If the experience of lining it with lead is to be least find the ratio of the depth to the side of the base. 12. Given the sum of the area of curved surface and two circular ends of a right circular cylinder is a constant, find the height of the cylinder so that the volume may be a maximum. 13. A wire of length ‘a’ is cut into two parts which are bent respectively in the form of a square and a circle. Show that the least value of the sum of the areas a2 so formed is . 4(π + 4) 14. Find the maximum volume of a right circular cylinder inscribed in a given sphere whose radius is R. 15. Find the attitude and radius of the cone of maximum volume that can be inscribed in a sphere of radius R. 16. a cylinder is inscribed in a cone of height ‘h’ and semi vertical and ‘a’. Prove 4 that the volume of the greatest cylinder thus obtained is π h3 tan 2 α. 27 17. Show that a conical tent of given capacity will require the least amount of canvas when the height 2 time the radius of the base. 18. If the sum of the length of the hypotenuse and another side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between those sides is 60. 19. A sector is cut out of a circular piece of canvas and the bounding radii of the remaining part are drawn together to form a conical tent. What should be angle of the sector cut out, if the tent has maximum volume? 20. A cone is circumscribed to a sphere of radius r. Show that, when the volume of the cone is a minimum, its altitude is 4r.

aNsWers

Exercise 1(a) (2) (–4, 3)

(3) (–•, –3), (6, •)

(6) 2 2

(7)

m2 n2 (16) 0 (10)

3 16 a 2 1 (11) 2 (17) eab

(4) 2

(5) 2

1 2

(9)

(12) cos a

(15)

(8)

(18) e

1 2 1 2 2

(19) e

Differential Calculus  5 (21) − ,  4

(20) e (27)

1 3

(28)

(31) 1

3   4 

1  (23)  , 3 3 

(22) (0, 1)

4 3

(29)

(32) 2 log 2

3 8

(30) 1

(33) 3

2 π (39) not continuous at z = 0 (35) log 4

1.59

(34) –3

(38) not continuous at x =

(36)

3 2

(40) a = 3, b = –8

(43) Continuous but not differentiable (44) Continuous at both point; not differentiable at x = 1, but differentiable x = 2 (45) Continuous, but not differentiable at both the points.

Exercise 1(b) 1

(2) y ′ =

2x

3 2

(3) y ′ = −

4 ( x − 2) e

2 (5) y ′ = x cos x + 2 x sin x

(7) –2 sec x

(8) sec 2 1 + x + x 2

(9) sin x cos2x(2 cos2x – 3 sin2x)

(4) y ′ =

1 2 x log x 2x

(6) y ′ =

1 − x4

1 2 1 + x + x2

⋅ (2 x + 1)

(10) –1

(11) sech x 2

cot x 2 (12) (tan x) ⋅{cosec x(1 − log tan x)} (13)

(14)

y 2 log a 1 − xy log a

(17) –tan t (21)

(23) −

(18) cot

(15) θ 2

ab sin x 2

a + b 2 − 2ab cos x 2( x + sin x + cos x) ( x sin x − cos x) 2

(19)

2 1− x

2

a sin x b

(16)

1 1 − x2

3/ 2 (20) −cot (2x)

(22)

1 − sin 2 x + 1 + sin 2 x

(24)

x2 − 1 3 1 (1 + x + x 2 ) 2 (1 − x + x 2 ) 2

2

ax 3 2 2 (25) xe {2a cos bx − 3bx sin bx }

ay − x y 2 − ax

(26)

1 x 1+ x

(27)

2 2

a + x2

Engineering Mathematics I

1.60 (28) −

(29)

x sec log a 2 − x 2 tan log a 2 x 2 y 2 a −x 2

x 1 − x 2 + sin −1 x 3

2

(1 − x ) (31)

(33)

2 2 (30) 2 x − a

2

− sin x 2 x (1 + cos

2

(32) −

x)

2(1 + cosh x cos x) (sinh x + sin x) 2

− 2 (1 + x) (1 + x 2 ) 1

2   ( x + a )( x 2 + b)    (34)    3   x +c    

 1 x 3 x 2   + 2 − 3  2( x + a ) x + b x + c   

1

x x (1 − log x) (35) x (1 + x) + x2 x

tan x 2 sin x (36) (sin x) {1 + sec x + log sin x) + (tan x) {sin x + cos x log tan x}

sin 2 (a + y ) (38) sin a

x+ y x + log x − − log( x + y ) (37) x x+ y (sin x)cos y cos y cot x + (cos y )sin x cos x log cos y (39) (sin x)cos y sin y log sin x + (cos y )sin x sin x tan y (41) −

(44) −

1 2

3

(42) − 1 1− x

(47) –tan3 t

2

+

1− x

2

1 2 x 1− x

(48) 1

(43) −

(1 + y 2 ) 1 + x2

1 1 + x2

(45) tan t (49)

(40) −

3 2

(46) –cot 2q (50)

1 2

Exercise 1(c) (1) Max =

2 and min = 0 27

1

(2) e e Max = –6 and min = 2

2 (3) Max values 2, 0, 2 and min value − and − 2 3 (8) 40 km/hr.

(10) When OA = OB =

k 2 2

(6) R = r (11) 1 : 2

Differential Calculus

(13)

2k , where k is the total 5.A 3π

(14)

4 2 2 R and radius = R 3 3 (19) 66° approximately (15) altitude =

1.61 4 3 πR 3 3

Unit

2

Functions of Several Variables 2.1

intRODUCtiOn

The students have studied in the lower classes the concept of partial differentiation of a function of more than one variable. They were also exposed to homogeneous functions of several variables and Euler’s theorem associated with such functions. In this chapter, we discuss some of the applications of the concept of partial differentiation, which are frequently required in engineering problems.

2.2

tOtAL DiFFEREntiAtiOn

In partial differentiation of a function of two or more variables, it is assumed that only one of the independent variables varies at a time. In total differentiation, all the independent variables concerned are assumed to vary and so to take increments simultaneously. Let z = f (x, y), where x and y are continuous functions of another variable t. Let ∆t be a small increment in t. Let the corresponding increments in x, y, z be ∆x, ∆y and ∆z respectively. Then

∴

∆z = f ( x + ∆ x, y + ∆y ) − f ( x, y ) = { f ( x + ∆ x, y + ∆y ) − f ( x, y + ∆y )} + { f ( x, y + ∆y ) − f ( x, y )} ∆z  f ( x + ∆ x, y + ∆y ) − f ( x, y + ∆y )  ∆x = ⋅ ∆t  ∆x  ∆t  f ( x, y + ∆y ) − f ( x, y )  ∆y + ⋅ ∆y   ∆t

(1)

We note that ∆x and ∆y → 0 as ∆t → 0 and hence ∆z → 0 as ∆t → 0 dz ∂ f dx ∂ f dy Taking limits on both sides of (1) as ∆t → 0, we have = ⋅ + ⋅ dt ∂ x dt ∂ y dt ( x, y and z are functions of t only and f is a function of x and y). ∴

2.2

Engineering Mathematics I

i.e.,

dz ∂z dx ∂z dy = + [since f ( x, y ) ≡ z ( x, y )]. dt ∂x dt ∂y dt

dz dt

(2)

dx dy   and  is called the total differential coefficient of z.  and also dt dt 

This name is given to distinguish it from the partial differential coefficients ∂z . ∂z and Thus to differentiate z, which is directly a function of x and y, (where ∂ y ∂x x and y are functions of t) with respect to t, we need not express z as a function of t by substituting for x and y. We can differentiate z with respect to t via x and y using the result (2). Corollary 1: In the differential form, result (2) can be written as dz =

∂z ∂z dx + dy ∂x ∂y

(3)

dz is called the total differential of z. Corollary 2: If z is directly a function of two variables u and u, which are in turn functions of two other variables x and y, clearly z is a function of x and y ultimately. ∂z Hence the total differentiation of z is meaningless. We can find only ∂z and by ∂ y ∂x using the following results which can be derived as result (2) given above. ∂z ∂z ∂u ∂z ∂ υ = ⋅ + ⋅ ∂x ∂u ∂x ∂ υ ∂x

(4)

∂z ∂z ∂u ∂z ∂ υ = ⋅ + ⋅ ∂y ∂u ∂y ∂υ ∂y

(5)

We note that the partial differentiation of z is performed via the intermediate ∂z ∂z variables u and u, which are functions of x and y. Hence and are called ∂x ∂y partial derivatives of a function of two functions.

Note

Results (2), (3), (4) and (5) can be extended to a function z of several intermediate variables.

2.2.1

Small Errors and Approximations

 ∆y  dy ∆y dy  dy  approximately or ∆y    ∆x Since lim   = , = ∆ x → 0  ∆x   dx  dx ∆x dx

(1)

If we assume that dx and dy are approximately equal to ∆x and ∆y respectively, result (1) can be derived from the differential relation.

Functions of Several Variables

2.3

 dy  dy =   dx  dx 

(2)

Though (2) is an exact relation, it can be made use of to get the approximate relation (1), by replacing dx and dy by ∆x and ∆y respectively. Let y = f (x). If we assume that the value of x is obtained by measurement, it is likely that there is a small error ∆x in the measured value of x. This error in the value of x will contribute a small error ∆y in the calculated value of y, as x and y are functionally related. The small increments ∆x and ∆y can be assumed to represent the small errors ∆x and ∆y. Thus the relation between the errors ∆x and ∆y can be taken as ∆y  f ′ ( x) ∆x This concept can be extended to a function of several variables. If u = u(x, y, z) or f (x, y, z) and if the value of u is calculated on the measured values of x, y, z, the likely errors ∆x, ∆y, ∆z will result in an error ∆u in the calculated value of u, given by ∆u 

∂u ∂u ∂u ∆x + ∆y + ∆z , ∂x ∂y ∂z

which can be assumed as the approximate version of the total differential relation du =

∂u ∂u ∂u dx + dy + dz ∂x ∂y ∂z

The error ∆x in x is called the absolute error in x, while ∆x is called the x relative or proportional error in x and 100 ∆x is called the percentage error in x. x

Note

2.2.2

Differentiation of implicit Functions

When x and y are connected by means of a relation of the form f (x, y) = 0, x and y are said to be implicitly related or y is said to be an implicit function of x. When x and y are implicitly related, it may not be possible in many cases to express y as a dy single valued function of x explicitly. However can be found out in such cases as dx a mixed function of x and y using partial derivatives as explained below: Since i.e.,

f ( x, y ) = 0, df = 0

∂f ∂f dx + dy = 0, by definition of total differential. Dividing by dx, we have ∂x ∂y ∂f ∂f dy + ⋅ =0 ∂x ∂y dx

Engineering Mathematics I

2.4

 ∂f    dy ∂x =− dx  ∂f   ∂y 

∴

(1)

2 2 2 If we denote ∂f , ∂f , ∂ f , ∂ f and ∂ f by the letters p, q, r, s, t respectively, ∂x ∂y ∂x 2 ∂x ∂y ∂y 2 then dy p (2) =− dx q

d2 y in terms of p, q, r, s, t as given dx 2 below. Noting that p and q are functions of x and y and differentiating both sides of (2) with respect to x totally, we have We can express the second order derivative

dq   dp q −p d2 y  dx dx  = − 2 2  dx q      ∂q ∂q dy   ∂p ∂p dy  p  + ⋅ − q  + ⋅  ∂x ∂y dx   ∂xx ∂y dx  =  q2   − p    − p  p s + t    − q r + s  q   q   ,   =  q2

since

∂p ∂ 2 f ∂p ∂q ∂2 f ∂q ∂ 2 f = = r = = = s = = t. ; ; ∂x ∂y ∂x ∂x ∂y ∂y ∂x 2 ∂y 2 p(qs − pt ) − q (qr − ps ) = q3 =−

( p 2 t − 2 pqs + q 2 r ) q3

WORKED EXAMPLE 2(a)

Example 2.1 1 (i) If u = xy + yz + zx, where x = et, y = e-t and z = , find du t dt du 3 -1 3 (ii) If u = sin (x - y), where x = 3t and y = 4t , show that = dt 1 − t2

Functions of Several Variables

2.5

(i) u = xy + yz + zx ∴

du ∂u dx ∂u dy ∂u dz = ⋅ + ⋅ + ⋅ dt ∂x dt ∂y dt ∂z dt  1 = ( y + z )et + ( z + x) (− e− t ) + ( x + y )  − 2   t  1 1  1  =  e− t +  et −  + et  e− t − et + e− t . 2   t  t t

(

)

1 1 1 1 = 1+ et − e− t − 1− 2 et − 2 e− t t t t t 2 2 sinh t − 2 cosh t. t t

= (ii) u = sin-1 (x - y) ∴

du dx 1 = + 2 dt dt 1 − ( x − y) 1

=

1 − ( x − y)

2

 dy  −  1 − ( x − y )  dt  1

2

(3 − 12t 2 )

(1)

Now 1 − ( x − y ) 2 = 1 − (3t − 4t 3 ) 2 = 1 − 9t 2 + 24t 4 − 16t 6 = (1 − t 2 ) (1 − 8t 2 + 16t 4 ) = (1 − t 2 ) (1 − 4t 2 ) 2

(2)

Using (2) in (1), we get du 1 = × 3(1 − 4t 2 ) 2 2 dt (1 − 4t ) 1 − t 3 = ⋅ 1− t2

 x y z ∂u ∂u ∂u Example 2.2 If u = f  , ,  , prove that x + y + z = 0.  y z x ∂x ∂y ∂z Let

r=

x y z , s = and t = y z x

(1)

Engineering Mathematics I

2.6

∴ u = f (r, s, t), where r, s, t are functions of x, y, z as assumed in (1) ∂u ∂u ∂r ∂u ∂s ∂u ∂t = ⋅ + ⋅ + ⋅ ∂x ∂r ∂x ∂s ∂x ∂t ∂x 1 ∂u ∂u z ∂u = ⋅ + ⋅0 − 2 ⋅ y ∂r ∂s x ∂t

∴

∂u ∂u ∂r ∂u ∂s ∂u ∂t = ⋅ + ⋅ + ⋅ ∂y ∂r ∂y ∂s ∂y ∂t ∂y x ∂u 1 ∂u =− 2⋅ + ⋅ y ∂r z ∂s ∂ u ∂ u ∂ r ∂ u ∂ s ∂u ∂t = ⋅ + ⋅ + ⋅ ∂z ∂ r ∂ z ∂ s ∂ z ∂t ∂z y ∂u 1 ∂ u =− 2⋅ + ⋅ z ∂s x ∂ t From (2), (3) and (4), we have x

(2)

(3)

(4)

∂u ∂u ∂u  x ∂u z ∂u  +y +z = − ∂x ∂y ∂z  y ∂r x ∂t   x ∂u y ∂u   y ∂u z ∂u  +− + + − +   y ∂r z ∂s   z ∂s x ∂t  = 0.

Example 2.3 If z be a function of x and y, where x = eu + e-u and y = e-u - eu, prove that ∂z ∂z ∂z ∂z − = x −y ∂u ∂υ ∂x ∂y ∂z ∂z ∂ x ∂ z ∂ y = ⋅ + ⋅ ∂u ∂x ∂ u ∂ y ∂ u ∂z − u ∂z = eu −e ∂y ∂x ∂z ∂z ∂x ∂z ∂y = ⋅ + ⋅ ∂υ ∂x ∂υ ∂y ∂υ ∂z υ ∂z = − e −υ −e ⋅ ∂x ∂y From (1) and (2), we have ∂z ∂z ∂z ∂z − = (eu + e −υ ) − (e − u − eυ ) ∂u ∂υ ∂x ∂y ∂z ∂z = x −y ∂x ∂y

(1)

(2)

Functions of Several Variables

2.7

Example 2.4 If u = f (x, y), where x = r cos q and y = r sin q, prove that 2

Ê ∂u ˆ Ê ∂u ˆ ÁË ∂x ˜¯ + Á ∂y ˜ Ë ¯ 2

1 Ê ∂u ˆ =Á ˜ + 2 Ë ∂r ¯ r

2

2

Ê ∂u ˆ ÁË ∂q ˜¯ .

∂u ∂u ∂x ∂u ∂y = ◊ + ◊ ∂r ∂x ∂r ∂y ∂r ∂u ∂u = cos q ◊ + sin q ∂x ∂y

(1)

∂u ∂u ∂x ∂u ∂ y = ◊ + ◊ ∂q ∂x ∂q ∂ y ∂q ∂u ∂u = - r sin q ◊ + r cos q ◊ ∂x ∂y 1 ∂u ∂u ∂u = - sin q + cos q ◊ r ∂q ∂x ∂y

i.e.,

(2)

Squaring both sides of (1) and (2) and adding, we get 2

2

2

1 Ê ∂u ˆ Ê ∂u ˆ Ê ∂u ˆ Ê ∂u ˆ ÁË ∂r ˜¯ + 2 ÁË ∂q ˜¯ = ÁË ∂x ˜¯ + Á ∂y ˜ Ë ¯ r Example 2.5 Find the equivalent of

2

∂ 2u ∂ 2u in polar co-ordinates. + ∂x 2 ∂y 2

u = u(x, y), where x = r cos q and y = r sin q ∴ u can also be considered as u(r, q), where r= Now we proceed to find

x2 + y 2

 y and θ = tan−1    x 

∂u ∂u via r and q. and ∂x ∂y ∂ u ∂ u ∂ r ∂ u ∂q = ◊ + ◊ ∂ x ∂ r ∂ x ∂q ∂ x ∂u 1 x + = 2 2 ∂r y2 x +y 1+ 2 x ∂u sin q ∂u = cos q ∂r r ∂q

Ê y ˆ ∂u ◊Á - 2 ˜ ◊ Ë x ¯ ∂q

(1)

Engineering Mathematics I

2.8 From (1), we can infer that

∂ ∂ sin q ∂ ∫ cos q ∂x ∂r r ∂q 2

∂ u

Now

∂x

2

(2)

∂ Ê ∂u ˆ ∂x ÁË ∂x ˜¯

=

∂ sin q ∂ ˆ Ê ∂u sin q ∂u ˆ Ê cos q = Á cos q Á ˜ Ë r ∂q ¯ Ë r ∂q ˜¯ ∂r ∂r 2

= cos 2 q ◊

∂ u 2

- sin q cos q

∂ Ê 1 ∂u ˆ ∂r ÁË r ∂q ˜¯

∂r ∂u ˆ sin q ∂ Ê ∂u ˆ sin q ∂ Ê ◊ cos q ˜ + 2 ◊ Á sin q ˜ r ∂q ÁË ∂r ¯ r ∂q Ë ∂q ¯

∂u ∂u Ê ˆ ÁË∵ r and q are independent and ∂r and ∂q are functions of r and q ˜¯ . = cos 2 q -

Ê 1 ∂2u 1 ∂u ˆ q q sin cos Á r ∂r ∂q - 2 ∂q ˜ 2 r ∂r Ë ¯ ∂2u

∂u ˆ sin q Ê sin q Ê ∂2u ∂2u ∂u ˆ - sin q cos q + 2 Á sin q + cos q Á ˜ 2 ∂r ¯ q ˜¯ r Ë ∂q ∂ r ∂ ∂q r Ë

(3)

∂ u ∂ u ∂ r ∂ u ∂q = ◊ + ◊ ∂y ∂r ∂y ∂q ∂y

Now

∂u 1 Ê 1 ˆ ∂u + ◊ ˜ 2 Á x + y ∂ r 1 + y Ë x ¯ ∂q x2 ∂u cos q ∂u + = siin q ∂r r ∂q

=

y

2

2

(4)

From (4) we infer that ∂ ∂ cos q ∂ ∫ sin q + ∂y ∂r r ∂q ∴

∂2u ∂y

2

=

(5)

∂ Ê ∂u ˆ ∂y ÁË ∂y ˜¯

∂ cos q ∂ ˆ Ê ∂u cos q ∂u ˆ Ê sin q = Á sin q + + Ë r ∂q ˜¯ ÁË r ∂q ˜¯ ∂r ∂r = sin 2 q +

Ê 1 ∂2u 1 ∂u ˆ + sin q cos q Á r ∂r ∂q - 2 ∂q ˜ 2 r ∂r Ë ¯ ∂2u

cos q r

Ê ∂2u ∂u ˆ cos q Ê ∂2u ∂u ˆ sin q + cos q cos q + - sin q Á ˜ Á 2 2 ∂q ∂r ∂r ¯ ∂q ˜¯ r ∂q Ë Ë

(6)

Functions of Several Variables

2.9

Adding (3) and (6), we have ∂2u ∂x 2

+

∂2u ∂y 2

=

∂2u ∂r 2

+

1 ∂u 1 ∂ 2 u + r ∂r r 2 ∂q 2

Example 2.6 Given the transformations u = ex cos y and u = ex sin y and that f is a function of u and u and also of x and y, prove that ∂2 f ∂x 2

+

∂2 f ∂y 2

Ê ∂2 f ∂2 f ˆ = (u 2 + u 2 ) Á 2 + 2 ˜ ∂u ¯ Ë ∂u

∂f ∂f ∂u ∂f ∂u ◊ = ◊ + ∂x ∂u ∂x ∂u ∂x = e x cos y ◊

∂f ∂f + e x sin y ◊ ∂u ∂u

∂f ∂f +u ∂u ∂u

(1)

∂ ∂ ∂ ∫u +u ∂x ∂u ∂u

(2)

=u

∂2 f

∂ ˆ Ê ∂f ∂f ˆ Ê ∂ =Áu +u ˜¯ ÁË u ∂u + u ∂u ˜¯ Ë ∂ u ∂ u ∂x 2

Ê ∂ 2 f ∂f ˆ ∂2 f ∂2 f + uu = u Á u ◊ 2 + ˜ + uu ∂u ∂u ∂u ∂u ∂u ¯ Ë ∂u Ê ∂ 2 f ∂f ˆ + u Áu 2 + ∂u ˜¯ Ë ∂u

(3)

∂ f ∂ f ∂ u ∂ f ∂u = ◊ + ◊ ∂ y ∂ u ∂ y ∂u ∂y ∂f ∂f = - e x sin y ◊ + e x cos y ∂u ∂u ∂f ∂f =-u +u ∂u ∂u ∴

(4)

∂ ∂ ∂ ∫ -u +u ∂y ∂u ∂u ∂2 f ∂y

2

(5)

∂ ∂ ˆÊ ∂f ∂f ˆ Ê = Á -u + u ˜ Á -u +u ˜ Ë ∂u ∂u ¯ Ë ∂u ∂u ¯ = u2

Ê ∂2 f Ê ∂2 f ∂f ˆ ∂f ˆ u u + + u u Á ˜ Á ˜ 2 ∂u Ë ∂u ∂u ∂u ¯ Ë ∂u ∂u ∂u ¯

∂2 f

+ u2

∂2 f ∂u 2

(6)

Engineering Mathematics I

2.10 Adding (3) and (6), we get ∂2 f ∂x 2

+

∂2 f ∂y 2

Ê ∂2 f ∂2 f ˆ = (u 2 + u 2 ) Á 2 + ˜ ∂u 2 ¯ Ë ∂u

Example 2.7 If z = f (u,u), where u = cosh x cos y and u = sinh x sin y, prove that ∂2 z ∂x 2

+

Ê ∂2 z ∂2 z ˆ 2 2 = x + y sinh sin Á 2 + 2˜ ∂y 2 ∂u ¯ Ë ∂u ∂2 z

(

)

∂z etc. ∂x = sinh x ◊ cos y ◊ zu + cosh x sin y ◊ zu

z x = zu ◊ u x + zu ◊ u x , where z x ∫

Since z is a function of u and u, zu and zu are also functions of u and u. Hence to differentiate zu and zu with respect to x or y, we have to do it via u and u. ∴

z xx = cos y ÈÎcosh x ◊ zu + sinh x { zuu ◊ sinh x cos y + zuu cosh x sin y}˘˚ + sin y ÈÎsinh x ◊ zu + cosh x { zuu ◊ sinh x cos y + zuu ◊ cosh x sin y}˘˚

i.e.,

z xx = cosh x cos y ⋅ zu + sinh x ⋅ sin y zυ + sinh 2 x cos 2 y ⋅ zuu + 2 sinh x cosh x sin y cos y ⋅ zuυ + cosh 2 x sin 2 y ⋅ zυυ

(1)

z y = − zu ⋅ cosh x sin y + zυ sinh x cos y z yy = − cosh x [cos y ⋅ zu + sin y {zuu ⋅( − cosh x sin y ) + zuυ ⋅ sinh x ⋅ cos y } + sinh x [− sinh y ⋅ zυ + cos y {− zυu ⋅ cosh x sin y + zυυ ⋅ sinh x cos y }] i.e.,

z yy = − cosh x cos y ⋅ zu − sinh x ⋅ sin y ⋅ zυ + cosh 2 x sin 2 y ⋅ zuu − 2 sinh x cosh x sin y cos y ⋅ zuυ + sinh 2 x cos 2 y ⋅ zυυ

Adding (1) and (2), we get z xx + z yy = (sinh 2 x cos 2 y + cosh 2 x sin 2 y ) ( zuu + zυυ ) = {sinh 2 x (1− sin 2 y ) + (1+ sinh 2 x)sin 2 y }( zuu + zυυ ) = (sinh 2 x + sin 2 y ) ( zuu + zυυ )

(2)

Functions of Several Variables

2.11

Example 2.8 Find dy , when (i) x3 + y3 = 3ax2y and (ii) xy + yx = c. dx (i) f (x, y) = x3 + y3 -3ax2y p=

∂f = 3 x 2 − 6 axy ∂x

q=

∂f = 3 y 2 − 3 ax 2 ∂y

dy p 3( x 2 − 2 axy ) x(2 ay − x) =− =− = 2 dx q 3( y 2 − ax 2 ) y − ax 2 (ii) f (x, y) = xy + yx - c p=

∂f = yx y −1 + y x log y ∂x

q=

∂f = x y log x + xy x −1 ∂y

dy p yx y −1 + y x log x . = − = − x −1 dx q xy + x y log x Example 2.9 If ax2 + 2hxy + by2 = 1, show that f (x, y) = ax2 + 2hxy + by2 - 1

d2 y h 2 − ab . = dx 2 ( hx + by )3

p=

∂f ∂f = 2 ( ax + hy ) ; q = = 2 ( hx + by ) ∂x ∂y

r=

∂2 f ∂2 f ∂2 f = 2 a ; s = = 2 h ; t = = 2b ∂x ∂y ∂x 2 ∂y 2

( ax + hy ) dy p =− =− dx q hx + by

(

2 2 d 2 y − p t − 2 pqs + q r = dx 2 q3

) (Refer to differentiation of implicit functions)

= =

{

− 8b ( ax + hy ) − 16 h ( ax + hy ) ( hx + by ) + 8a ( hx + by ) 2

8 ( hx + by )

2

}

3

1 [2 h {ahx 2 + (ab + h 2 ) xy + bhy 2 } (hx + by )3

− {a 2 bx 2 + 2abhxy + h 2 by 2 } − {ah 2 x 2 + 2abhxy + ab 2 y 2 }]

Engineering Mathematics I

2.12 =

1 [a (h 2 − ab)x 2 + 2h(h 2 − ab ) xy + b(h 2 − ab) y 2 ] (hx + by )3

=

(h 2 − ab) (h 2 − ab ) ⋅1 h 2 − ab . 2 2 ( ax hxy by ) + 2 + = = (hx + byy )3 (hx + by )3 (hx + by )3

 y du if (i) u = sin (x2 + y2), where a2x2 + b2y2 = c2 (i), u = tan-1    x dx 2 2 2 where x + y = a , by treating u as function of x and y only. Example 2.10 Find

(i) u = sin (x2 + y2) du ∂u ∂u dy = + ⋅ dx ∂x ∂y dx

∴

(

(

)

= 2 x cos x 2 + y 2 + 2 y cos x 2 + y 2

) ddyx

(1)

Now a2x2 + b2y2 = c2 Differentiating with respect to x, 2a 2 x + 2b 2 y or

dy =0 dx dy a2 x =− 2 dx b y

(2)

Using (2) in (1), we get  −a2 x  du = 2 x cos x 2 + y 2 + 2 y cos x 2 + y 2 ×  2  dx  b y 

(

)

(

) (b

= 2 x cos x 2 + y 2

(

2

)

)

− a 2 / b2

 y u = tan −1    x d u ∂ u ∂ u dy = + ⋅ d x ∂ x ∂ y dx

(ii)

1  y 1  1  dy − 2+  ⋅ 2  y  x  y 2  x  dx 1+ 2 1+ 2 x x y x dy =− 2 + ⋅ x + y 2 x 2 + y 2 dx

=

∴

x2 + y 2 = a2 dy 2x + 2 y =0 dx

(3)

Functions of Several Variables

2.13

dy x =− dx y

or

(4)

Using (4) in (3), we get  x du y x =− 2 + − dx x + y 2 x 2 + y 2  y  1 =− . y ∂x ∂x ∂y ∂y , , ⋅ Example 2.11 If u = x2 - y2 and u = xy, find the values of and ∂u ∂υ ∂u ∂υ x and y cannot be easily expressed as single valued functions or u and u. Given

x2 - y2 = u

(1)

xy = u

(2)

and

Nothing that x and y are functions of u and u and differentiating both sides of (1) and (2) partially with respect to u, we have 2x

∂x ∂y − 2y =1 ∂u ∂u

(3)

∂x ∂y +x =0 ∂u ∂u

(4)

y Solving (3) and (4), we get

∂x x ∂y y = =− and ∂u 2 x 2 + y 2 ∂u 2 x2 + y 2

(

)

(

)

Differentiating both sides of (1) and (2) partially with respect to u, we have 2x y

∂x ∂y − 2y =0 ∂υ ∂υ

∂x ∂y +x =1 ∂υ ∂υ

Solving (5) and (6), we get ∂x y ∂y x = 2 = 2 and . 2 ∂υ x + y ∂υ x + y 2

(5) (6)

Engineering Mathematics I

2.14

Example 2.12 If x2 + y2 + z2 - 2xyz = 1, show that

dx 1− x

2

+

dy 1− y

2

+

dz 1− z 2

φ ≡ x 2 + y 2 + z 2 − 2 xyz − 1 = 0

Let

(1)

dφ = 0 ∂φ ∂φ ∂φ dx + dy + dz = 0 ∂x ∂y ∂z

∴ i.e.,

= 0.

(2)

2 ( x − yz ) dx + 2 ( y − zx) dy + 2 ( z − xy ) dz = 0

i.e.,

( x − yz )2 = x 2 − 2 xyz + y 2 z 2

Now

= 1 − y 2 − z 2 + y 2 z 2 , from (1)

(

)(

= 1− y 2 1− z 2

)

∴

x − yz =

(1− y ) (1− z )

Similarly,

y − zx =

(1− z ) (1− x )

and

z − xy =

(1− x ) (1− y )

2

2

2

2

2

2

Using these values in (2), we have

(1− y ) (1− z ) dx + (1− z ) (1− x ) dy + (1− x ) (1− y ) dz = 0 2

Dividing by

2

2

2

2

2

(1− x ) (1− y ) (1− z ), we get 2

2

2

dx 1− x

2

+

dy 1− y

2

+

dz 1− z 2

= 0.

W1 where W1 W1 − W2 and W2 are the weights of the body in air and in water respectively. Show that if there is an error of 1% in each weighing, s is not affected. But if there is an error of 1% in Example 2.13 The specific gravity s of a body is given by s =

W1 and 2% in W2, show that the percentage error in s is s= ∴

W1 W1 − W2

W2 . W1 − W2

log s = log W1 − log (W1 − W2 )

Functions of Several Variables

2.15

Taking differentials on both sides, 1 1 1 ds = dW1 − (dW1 − dW2 ) s W1 W1 − W2 ∴ The relation among the errors is nearly 1 1 1 ∆s = ∆W1 − (∆W1 − ∆W2 ) s W1 W1 − W2 100 ∆s 100 ∆W1 1 = − (100 ∆W1 − 100 ∆W2 ) s W1 W1 − W2

or

(1)

(2)

(i) Given that 100 ∆W1 = 1 and 100 ∆W2 = 1 W1 W2 Using these values in (2), we have 100 ∆s 1 = 1− (W1 − W2 ) = 0 s W1 − W2 ∴ s is not affected, viz., there is no error in s. (ii) Given that

100 ∆W1 100 ∆W2 = 1 and = 2. Using these values in (2), we have W1 W2 100 ∆s 1 = 1− (W1 − 2W2 ) s W1 − W2 =

i.e., % error in s =

W2 W1 − W2

W2 . W1 − W2

Example 2.14 The work that must be done to propel a ship of displacement D for a distance s in time t is proportional to s2 D3/2 ÷ t2. Find approximately the percentage increase of work necessary when the distance is increased by 1%, the time is diminished by 1% and the displacement of the ship is diminished by 3%. Given that W = ks2 D3/2/t2, where k is the constant of proportionality. 3 ∴ log W = log k + 2 log s + log D − 2 log t. 2 Taking differentials on both sides, dW ds 3 dD dt =2 + −2 W s 2 D t

Engineering Mathematics I

2.16

∴ The relation among the percentage errors is approximately, 100 ∆W 100 ∆s 3 100 ∆D 100 ∆t = 2× + ⋅ −2× W s 2 D t

(1)

100 ∆s 100 ∆t 100 ∆D = 1, = − 1 and = − 3. s t D

Given that

Using these values in (1), we have 100 ∆W 3 = 2 × 1 + × ( −3) − 2 × ( −1) W 2 = − 0.5 i.e., percentage decrease of work = 0.5. Example 2.15

The period T of a simple pendulum with small oscillations is

given by T = 2π

l . If T is computed using l = 6 cm and g = 980 cm/sec2, find g

approximately the error in T, if the values are l = 5.9 cm and g = 981 cm/sec2. Find also the percentage error. T = 2π

l g

1 1 log T = log 2 + log π + log l − log g 2 2

∴

Taking differentials on both sides, 1 1 1 dT = dl − dg T 2l 2g

∴

 1 l 1 dg   dl − 2g  g  2l  1 l  =π  dl − dg  g g   l g

dT = 2π

0 ⋅1 5⋅9   =π − × ( −1)   5.9 × 981 981 981  i.e.,

Error in T = 0.0044 sec.

(1)

Functions of Several Variables % error in T =

2.17

100 dT T

 dl d g  = 50 −  , by (1) g   l  0 ⋅ 1 ( −1)  = 50 −   5⋅ 9 981  = 0.8984 Example 2.16 The base diameter and altitude of a right circular cone are measured as 4 cm and 6 cm respectively. The possible error in each measurement is 0.1 cm. Find approximately the maximum possible error in the value computed for the volume and lateral surface. 2

D 1 Volume of the right circular cone is given by V = π ⋅   h 3  2

dV =

∴

=

π ( D 2 ⋅ dh + 2 Dh ⋅ dD) 12 π {16 × 0.1 + 2 × 4 × 6 × 0.1} 12

i.e., Error in V = 1.6755 cm3. Lateral surface area of the right circular cone is given by S =π =

∴

dS =

=

D l 2

π D D 2 + 4h 2 4

 π 1 D⋅ (2 D dD + 8h dh + D 2 + 4h 2 dD  4  2 D 2 + 4h 2    4 π {4 × 0 ⋅1 + 24 × 0 ⋅1} + 16 + 144 × 0 ⋅1  4  16 + 144 

= 1.6889 cm 2 .

Engineering Mathematics I

2.18

Example 2.17 The side c of a triangle ABC is calculated by using the measured values of its sides a, b and the angle C. Show that the error in the side c is given by ∆c = cos B ⋅∆a + cos A ⋅∆b + a sin B ⋅∆C. The side c is given by the formula c 2 = a 2 + b 2 − 2ab cos C

(1)

Taking the differentials on both sides of (1), 2c ∆c = 2a ∆a + 2b ∆b − 2 {b cos C ⋅∆a + a cos C ⋅∆b − ab sin C ⋅∆C } , nearly i.e.,

∆c =

( a − b cos C ) ∆a + (b − a cos C ) ∆b + ab sin C ⋅∆C

(2)

c

Now b cos C + c cos B = a ∴

∴ Also ∴ ∴

a − b cos C = cos B c a cos C + c cos A = b

(3)

b − a cos C = cos A c

(4)

b c = sin B sin C b sin C = c sin B ab sin C = a sin B c

(5)

Using (3), (4) and (5) in (2), we get ∆c = cos B ⋅∆a + cos A ⋅∆b + a sin B ⋅∆C Example 2.18 The angles of a triangle ABC are calculated from the sides a, b, c. If small changes δa, δb, δc are made in the measurement of the sides, show that

δA=

a (δ a − δ b cos C − δ c cos B) 2∆

and δB and δC are given by similar expressions, where ∆ is the area of the triangle. Verify that δA + δB + δC = 0. In triangle ABC, cos A =

b2 + c2 − a 2 2bc

(1)

Functions of Several Variables

2.19

Taking differentials on both sides of (1),

(

)

−2 sin A ⋅δ A = bc {2b δ b + 2cδ c − 2aδ a} − b 2 + c 2 − a 2 (bδ c + cδ b)  ÷ b 2 c 2

(b c − c 2

= = =

3

)

(

)

+ a 2 c δ b + bc 2 − b3 + a 2 b δ c − 2abc δ a 2 2

bc

(

)

(

)

c a 2 + b 2 − c 2 δ b + b c 2 + a 2 − b 2 δ c − 2abcδ a 2 2

bc

c ( 2 ab cos C ) δ b + b ( 2 ca cos B ) δ c − 2abc δ a b2c2

,

by formulas similar to (1) 2a (cos C ⋅δ b + cos B ⋅δ c − δ a) bc a δA= (δ a − cos C ⋅δ b − cos B ⋅δ c) bc sin A =

∴

a 1 (δ a − cos C ⋅δ b − cos B ⋅δ c) , since ∆ = bc sin A 2∆ 2

(2)

δB=

b (δ b − cos A ⋅δ c − cos C ⋅δ a) 2∆

(3)

δC =

c (δ c − cos B ⋅δ a − cos A ⋅δ b) 2∆

(4)

= Similarly,

Adding (2), (3) and (4), we get 2∆ (δ A + δ B + δ C ) = ( a − b cos C − c cos B ) δ a + (b − a cos C − c cos A) δ b + ( c − a cos B − b cos A) δ c = ( a − a ) δ a + (b − b ) δ b + ( c − c ) δ c

(∵ b cos C + c cos B = a etc.) =0 ∴

δ A + δ B + δ C = 0.

Example 2.19 The area of a triangle ABC is calculated from the lengths of the sides a, b, c. If a is diminished and b is increased by the same small amount k, prove that the consequent change in the area is given by

Engineering Mathematics I

2.20

δ∆ 2(a − b)k = 2 ∆ c − ( a − b) 2 The area of triangle ABC is given by

∴

∆=

s ( s − a )( s − b)( s − c) , where 2 s = a + b + c

log ∆ =

1 {log s + log( s − a ) + log( s − b) + log( s − c)} 2

Taking differentials on both sides, we get

δ∆ 1  δ s δ s − δ a δ s − δ b δ s − δ c  =  + + +  ∆ 2 s s−a s−b s−c  Since

(1)

2 s = a + b + c, 2δ s = δ a + δ b + δ c i.e.; 2ds = -k + k + 0 = 0, by the given data. ds = 0

∴

(2)

Using (2) in (1), we have

δ∆ 1  k k  =  − ∆ 2  s − a s − b  =

k 2 2  −   (∵ 2 s = a + b + c ) 2 b + c − a c + a − b 

=

 (c + a − b) − (b + c − a )  k ×2  2 [c − (a − b)][c + (a − b)] 

=

2k ( a − b) c − ( a − b) 2 2

EXERCiSE 2(a) Part A (Short Answer Questions) 1. 2. 3. 4. 5.

What is meant by total differential? Why it is called so? If u = sin(xy2), express the total differential of u in terms of those of x and y. If u = xy.yx, express du in terms of dx and dy. If u = xy log xy, express du in terms of dx and dy. If u = axy, express du in terms of dx and dy.

Functions of Several Variables

2.21

6. Find du , if u = x3 y2 + x2 y3, where x = at2, y = 2at. dt 7. Find

du , if u = e xy , where x = dt

8. Find

du , if u = log (x + y + z), where x = e-t, y = sin t, z = cos t. dt

9. Find

dy , using partial differentiation, if x3 + 3x2y + 6xy2 + y3 = 1. dx

a 2 − t 2 , y = sin 3 t.

10. If x sin (x - y) - (x + y) = 0, use partial differentiation to prove that dy y + x 2 cos ( x − y ) . = dx x + x 2 cos ( x − y ) 11. Find

dy , when u = sin (x2 + y2), where x2 + 4y2 = 9. dx

dy , if u = x2y, where x2 + xy + y2 =1. dx Define absolute, relative and percentage errors. Using differentials, find the approximate value of 15 . Using differentials, find the approximate value of 2x4 + 7x3 - 8x2 + 3x + 1 when x = 0.999. What error in the common logarithm of a number will be produced by an error of 1% in the number? The radius of a sphere is found to be 10 cm with a possible error of 0.02 cm. Find the relative errors in computing the volume and surface area. Find the percentage error in the area of an ellipse, when an error of 1% is made in measuring the lengths of its axes. Find the approximate error in the surface of a rectangular parallelopiped of sides a, b, c if an error of k is made in measuring each side. If the measured volume of a right circular cylinder is 2% too large and the measured length is 1% too small, find the percentage error in the calculated radius.

12. Find 13. 14. 15. 16. 17. 18. 19. 20.

Part B 21. If u = f (x - y, y - z, z - x), prove that ∂u + ∂u + ∂u = 0. ∂x ∂y ∂z 22. If f is a function of u, u, w, where u = that ∂f

∂f

∑ u ∂u = ∑ x ∂x .

yz , u = zx , and w = xy show

Engineering Mathematics I

2.22

23. If f = f  y − x , z − x  , show that x 2 ∂f + y 2 ∂f + z 2 ∂f = 0.  xy ∂x ∂y ∂z zx  24. If u = f (x2 + 2yz, y2 + 2zx), prove that

(y

2

− zx

) ∂∂ux + ( x

2

− yz

) ∂∂uy + ( z

2

− xy

) ∂∂uz = 0.

25. If f (cx – az, cy – bz) = 0, where z is a function of x and y, prove that a

∂z ∂z +b = c. ∂x ∂y

26. If z = f (u, u), where u = x + y and u = x – y, show that 2

∂z ∂z ∂z . = + ∂u ∂x ∂y

27. If z = f (x, y), where x = u2 + u2, y = 2uu, prove that u

∂z ∂z ∂z −υ = 2 ( x2 − y 2 ) . ∂u ∂υ ∂x

28. If z = f (x, y), where x = u + u, y = uu, prove that u

∂z ∂z ∂z ∂z +υ = x + 2y . ∂u ∂υ ∂x ∂y

29. If x = r cos q, y = r sin q, prove that the equation to ∂u + 1 tan  π − θ  ∂u = 0. 4  ∂θ ∂r r

∂u ∂u + = 0 is equivalent ∂x ∂y

30. If z = f(u, u), where u = x2 – 2xy - y2 and u = y, show that the equation ∂z ∂z ∂z ( x + y ) + ( x − y ) = 0 is equivalent to ∂υ = 0.. ∂x ∂y 31. If z = f (u, u), where u = x2 - y2 and u = 2xy, prove that 2 2 2   ∂z   ∂z   ∂z   2 2  ∂z    +   = 4( x + y )   +    . ∂x  ∂y  ∂υ   ∂u 2

32. If z = f (u, u), where u = x2 - y2 and u = 2xy, show that  ∂2 z ∂2 z  ∂2 z ∂2 z + 2 = 4( x 2 + y 2 )  2 + 2  . 2 ∂x ∂y  ∂u ∂υ 

Functions of Several Variables

2.23

33. If z = f (x, y) where x = X cos a - Y sin a and y = X sin a + Y cos a, show that ∂2 z ∂2 z ∂2 z ∂2 z + = + . ∂x 2 ∂y 2 ∂X 2 ∂Y 2 34. If z = f (u, u), where u = lx + my and u = ly - mx, show that  ∂2 z ∂2 z  ∂2 z ∂2 z + 2 = (l 2 + m 2 )  2 + 2  . 2 ∂x ∂y  ∂u ∂υ  35. By changing the independent variables x and t to u and u by means of the trans2 formations u = x - at and u = x + at, show that a

2 ∂2 y ∂2 y 2 ∂ y − = 4 a . ∂u ∂υ ∂x 2 ∂t 2

36. By using the transformations u = x + y and u = x - y, change the independent ∂2 z ∂2 z − = 0 to u and u. variables x and y in the equation ∂x 2 ∂y 2 37. Transform the equation

∂2 z ∂2 z ∂2 z +2 + = 0 by changing the independent 2 ∂x ∂y ∂y 2 ∂x

variables using u = x - y and u = x + y. 38. Transform the equation x 2

∂2 z ∂2 z ∂2 z + 2 xy + y 2 2 = 0, by changing the 2 ∂x ∂y ∂x ∂y

independent variables using u = x and υ =

39. Transform the equation

y2 . x

∂2 z ∂2 z ∂2 z − + = 0, by changing the indepen5 6 ∂x ∂y ∂x 2 ∂y 2

dent variables using u = 2x + y and u = 3x + y. 40. Transform the equation

∂ 2u ∂ 2u + = 0, by changing the independent variables ∂x 2 ∂y 2

using z = x + iy and z* = x – iy, where i = 41. Use partial differentiation to find y

− 1.

dy , when (i) xy = yx; (ii) xmyn = dx y

(x + y)m + n; (iii) (cos x) = (sin y)x; (iv) (sec x) = (cot y)x; (v) xy = ex - y. 42. Use partial differentiation to find

d2 y , when x3 + y3 -3axy = 0. dx 2

Engineering Mathematics I

2.24

43. Use partial differentiation to find

d2 y , when x4 + y4 = 4a2xy. dx 2

2 2 2 2 44. Use partial differentiation to prove that d y = abc + 2 fgh − af − bg − ch , 2 3

dx

( hx + by + f )

when ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. 45. Use partial differentiation to prove that = x2.

d2 y b 2 − ac = , when ay2 + 2by + c dx 2 (ay + b)3

46. If x2 – y2 + u2+ 2u2 = 1 and x2 + y2 – u2 – u2 = 2, prove that ∂u = 3 x and ∂x u 2x . ∂υ =− ∂x υ 47. The deflection at the centre of a rod of length l and diameter d supported at its ends and loaded at the centre with a weight w is proportional to wl3/d4. What is the percentage increase in the deflection, if the percentage increases in w, l and d are 3, 2 and 1 respectively. 48 The torsional rigidity of a length of wire is obtained from the formula 8π Il . If l is decreased by 2%, t is increased by 1.5% and r is increased t 2r 4 by 2%, show that the value of N is decreased by 13% approximately. N =

49. The Current C measured by a tangent galvanometer is given by the relation C = k tan q, where q is the angle of deflection. Show that the relative error in C due to a given error in q is minimum when q = 45°. 50. The range R of a projectile projected with velocity u at an elevation q is given υ2 by R = sin 2q. Find the percentage error in R due to errors of 1% in u and g π 1 in q, when θ = . % 6 2 g λ 2π T + , where g and l are con51. The velocity u of a wave is given by υ 2 = 2π ρλ stants and r and T are variables. Prove that, if r is increased by 1% and T is 3π T . decreased by 2%, then the percentage decrease in u is approximately λρυ 2 1 1 1 = − . If equal f υ u errors k are made in the determination of u and u, show that the percentage

52. The focal length of a mirror is given by the formula error in f is 100k  1 + 1  .   u υ

Functions of Several Variables

2.25

53. A closed rectangular box of dimensions a, b, c has the edges slightly altered in length by amounts ∆a, ∆b and ∆c respectively, so that both its volume and ∆a ∆b ∆c = = . surface area remain unaltered. Show that 2 a (b − c) b 2 (c − a ) c 2 (a − b) [Hint: Solve the equations dV = 0 and dS = 0 for ∆a, ∆b, ∆c using the method of cross-multiplication] 54. If a triangle ABC is slightly disturbed so as to remain inscribed in the same circle, prove that ∆a ∆b ∆c + + = 0. cos A cos B cos C 1 55. The area of a triangle ABC is calculated using the formula ∆ = bc sin A. 2 Show that the relative error in ∆ is given by

δ ∆ δb δc = + + cot Aδ A. ∆ b c If an error of 5′ is made in the measurement of A which is taken as 60°, find the percentage error in ∆. 56. Prove that the error in the area Δ of a triangle ABC due to a small error in the measurement of c is given by

δ∆ =

∆1 1 1 1  + + − δ c.  4  s s − a s − b s − c 

57. The area of a triangle ABC is determined from the side a and the two angles B and C. If there are small errors in the values of B and C, show that the resulting error in the calculated value of the area ∆ will be

1 2 (c ∆B + b 2 ∆C ). 2

 1 a 2 sin B sin C    Hint: ∆ = 2 sin( B + C )  

2.2.3

taylor’s Series Expansion of a Function of two Variables

Students are familiar with Taylor’s series of a function of one variable viz. f (x + h) = f ( x) +

h h2 f ′ ( x) + f ′′ ( x) +  ∞, which is an infinite series of powers of h. This idea 1! 2!

can be extended to expand f (x + h, y + k) in an infinite series of powers of h and k.

Engineering Mathematics I

2.26

Statement If f (x, y) and all its partial derivatives are finite and continuous at all points (x, y), then 2

∂ 1 ∂ 1 ∂ ∂ f ( x + h, y + k ) = f ( x , y ) +  h + k  f +  h + k  f ∂y  1!  ∂x 2 !  ∂x ∂y  3

+

∂ 1 ∂ h + k  f + ... ∞ ∂y  3!  ∂x

Proof: If we assume y to be a constant, f (x + h, y + k) can be treated as a function of x only. Then f ( x + h, y + k ) = f ( x, y + k ) +

h ∂f ( x, y + k ) h 2 ∂ 2 f ( x, y + k ) ... + + 1! ∂x 2! ∂x 2

(1)

Now treating x as a constant, f ( x, y + k ) = f ( x, y ) +

2 k ∂f ( x, y ) k 2 ∂ f ( x, y ) ... + + 1! ∂y 2 ! ∂y 2

(2)

Using (2) in (1), we have  k ∂f ( x, y ) k 2 ∂ 2 f ( x, y ) ... f ( x + h, y + k ) =  f ( x , y ) + + +  1! ∂y 2 ! ∂y 2  

= f ( x, y ) +

+

k ∂f ( x, y ) k 2 ∂ 2 f ( x, y ) ... h ∂  f x y ( , ) + + +   1! ∂x  1! ∂y 2 ! ∂y 2 

+

h2 ∂ 2  k ∂f ( x, y ) k 2 ∂ 2 f ( x, y ) ... ... f ( x, y ) + + + + ∞ 2 2 ! ∂x  1! ∂y 2 ! ∂y 2 

 ∂f  1  1  ∂f ∂2 f ∂2 f ∂2 f h + k  +  h 2 2 + 2hk + k 2 2 + ...∞   ∂y  2 !  1!  ∂x ∂x ∂y ∂x ∂y 

∂  ∂  1 ∂ 1  ∂ = f ( x, y ) +  h + k  f +  h + k  ∂y  ∂y  1!  ∂x 2 !  ∂x

2

f + ...∞

(3)

Interchanging x and h and also y and k in (3) and then putting h = k = 0, we have f ( x, y ) = f (0, 0) +

1  ∂f (0, 0) ∂f (0, 0)  1  2 ∂ 2 f (0, 0) +y  + x x 1!  ∂x ∂y  2 !  ∂x 2 + 2 xy

∂ 2 f (0, 0) ∂ 2 f (0, 0)  ... + y2 + ∂x ∂y ∂y 2 

(4)

Functions of Several Variables

2.27

Series in (4) is the Maclarin’s series of the function f (x, y) in powers of x and y. Another form of Taylor’s series of f (x, y) f ( x, y ) = f ( a + x − a , b + y − b ) = f (a + h), (b + k ), say = f ( a, b) + +

1  2 ∂ 2 f ( a, b) ∂ 2 f (a, b) 2 ∂ 2 f (a, b)  ..., + 2 +k h kh   + by (3) 2!  ∂x ∂y ∂x 2 ∂y 2 

= f ( a, b) + +

1  ∂f ( a , b ) ∂f ( a , b )  +k  h 1!  ∂x ∂y 

1 ∂f ( a , b ) ∂f ( a , b )  ( x − a) + ( y − b) ∂y  1!  ∂x

2 2 1 ∂ 2 f ( a, b) 2 ∂ f ( a, b) 2 ∂ f ( a, b)  ... ( ) + y − b ( x − a ) + 2 ( x − a )( y − b ) +  2!  ∂x ∂y ∂y 2  ∂x 2

(5)

(5) is called the Taylor’s series of f (x, y) at or near the point (a, b). Thus the Taylor’s series of f (x, y) at or near the point (0, 0) is Maclaurins series of f (x, y).

2.3

JACOBiAnS

If u and u are functions of two independent variables x and y, then the determinant ∂u ∂u ∂x ∂y ∂υ ∂υ ∂x ∂y is called the Jacobian or functional determinant of u, u with respect to x and y and is written as  u,υ  ∂(u , υ ) . or J  ∂ ( x, y )  x, y  Similarly the Jacobian of u, u, w with respect to x, y, z is defined as ∂u ∂x ∂υ ∂(u , υ , w) = ∂x ∂ ( x, y , z ) ∂w ∂x

∂u ∂y ∂υ ∂y ∂w ∂y

∂u ∂z ∂υ ∂z ∂w ∂z

2.28

Engineering Mathematics I

Note 1. To define the Jacobian of n dependent variables, each of these must be a function of n independent variables. 2. The concept of Jacobians is used when we change the variables in multiple integrals. (See property 4 given below)

2.3.1

Properties of Jacobians

1. If u and u are functions of x and y, then ∂(u , υ ) × ∂( x, y ) = 1. ∂( x, y ) ∂(u , υ ) Proof: Let u = f(x, y) and u = g(x, y). When we solve for x and y, let x = f (u, u) and y = ψ(u, u). Then ∂u ∂x ∂u ∂y ⋅ + ⋅ ∂x ∂u ∂y ∂u ∂u ∂x ∂ u ∂ y ⋅ + ⋅ ∂x ∂ υ ∂ y ∂ υ ∂ υ ∂x ∂υ ∂y ⋅ + ⋅ ∂x ∂u ∂y ∂u ∂u ∂x ∂ u ∂ y ⋅ + ⋅ ∂x ∂υ ∂ y ∂ υ

Now

∂u ∂u ∂u = ∂υ ∂υ = ∂u ∂υ = ∂υ =

 =1   = 0   = 0   = 1 

∂u ∂x ∂(u , υ ) ∂( x, y ) × = ∂υ ∂( x, y ) ∂(u , υ ) ∂x

∂u ∂x ∂y ∂u × ∂υ ∂y ∂y ∂u

∂x ∂υ ∂y ∂υ

∂u ∂x = ∂υ ∂x

∂u ∂x ∂y ∂u × ∂x ∂υ ∂υ ∂y

∂y by interchanging the rows ∂u , and columns of the ∂y second determinant. ∂υ

=

=

 ∂u ∂x ∂u ∂y   ∂ u ∂ x ∂ u ∂ y   ∂x ⋅ ∂u + ∂y ⋅ ∂u   ∂x ⋅ ∂υ + ∂y ⋅ ∂υ   ∂υ ∂x ∂υ ∂y   ∂υ ∂x ∂υ ∂y   ∂x ⋅ ∂u + ∂y ⋅ ∂u   ∂x ⋅ ∂υ + ∂y ⋅ ∂υ  1 0 [by (1)] 0 1

=1

(1)

Functions of Several Variables

2.29

2. If u and u functions of r and s, where r and s are functions of x and y, then ∂ (u , υ ) ∂ (u , υ ) ∂ ( r , s ) = × ∂ ( x, y ) ∂ ( r , s ) ∂ ( x, y ) Proof: ∂u ∂ (u , υ ) ∂ ( r , s ) ∂r × = ∂υ ∂ ( r , s ) ∂ ( x, y ) ∂r

∂r ∂r ∂u ∂ x ∂y ∂s × ∂υ ∂s ∂s ∂s ∂x ∂y

∂u ∂r = ∂υ ∂r

=

 ∂u ∂r ∂ u ∂ s   ∂ u ∂ r ∂ u ∂ s  + + .  . .  . ∂r ∂x ∂s ∂x   ∂r ∂y ∂s ∂y   ∂ υ ∂r ∂ υ ∂ s   ∂ υ ∂ r ∂ υ ∂ s  + .  . . + .  ∂s ∂x   ∂r ∂y ∂s ∂y  ∂r ∂x

∂u ∂x = ∂υ ∂x

Note

∂r ∂s ∂u ∂x ∂x ∂s , by rewriting the second determinant. × ∂r ∂s ∂υ ∂y ∂y ∂s

∂u ∂y ∂(u , υ ) . = ∂υ ∂ ( x, y ) ∂y

The two properties given above hold good for more than two variables too.

3. If u, u, w are functionally dependent functions of three independent variables x, y, z then

∂(u , υ , w) = 0. ∂ ( x, y , z )

Note

The functions u, u, w are said to be functionally dependent, if each can be expressed in terms of the others or equivalently f (u, u, w) = 0. Linear dependence of functions is a particular case of functional dependence. Proof: Since u, u, w are functionally dependent, f (u, u, w) = 0 Differentiating (1) partially with respect to x, y and z, we have

(1)

f u . u x + fυ . υ x + f w . wx = 0

(2)

f u . u y + fυ . υ y + f w . wy = 0

(3)

f u . u z + fυ . υ z + f w . wz = 0

(4)

Engineering Mathematics I

2.30

Equations (2), (3) and (4) are homogeneous equations in the unknowns fu, fu, fw. At least one of fu, fu and fw is not zero, since if all of them are zero, then f (u, u, w) ≡ constant, which is meaningless. Thus the homogeneous equations (2), (3) and (4) possess a non-trivial solution. ∴ Matrix of the coefficients of fu, fu , fw is singular.

i.e.,

ux υ x uy υ y uz υ z

wx wy = 0 wz

∂(u , υ , w) =0 ∂ ( x, y , z )

i.e.,

Note

The converse of this property is also true. viz., if u, u, w are functions of ∂(u , υ , w) = 0 then u, u, w are functionally dependent. i.e., there x, y, z such that ∂ ( x, y , z ) exists a relationship among them. 4. If the transformations x = x(u, u) and y = y(u, u) are made in the double integral J =

∫∫ f ( x, y) dx dy, then f(x, y) = F(u, u) and

dx dy = J du dυ , where

∂ ( x, y ) . ∂(u , υ )

Proof: dx dy = Elemental area of a rectangle with vertices (x, y), (x + dx, y), (x + dx, y + dy) and (x, y + dy) This elemental area can be regarded as equal to the area of the parallelogram with ∂x ∂y   ∂x ∂x ∂y ∂y   vertices (x, y),  x + du , y + du  ,  x + du + dυ , y + du + dυ and  ∂u ∂u   ∂u ∂υ ∂u ∂υ  ∂x ∂y   dυ , y + dυ  , since dx and dy are infinitesimals.  x + ∂υ ∂v  Now the area of this parallelogram is equal to 2× area of the triangle with vertices ∂x ∂y   ∂x ∂y  (x, y),  x + du , y + du  and  x + dυ , y + dυ    ∂u ∂u  ∂υ ∂υ 

∴

x 1 ∂x dx d y = 2 × x + du 2 ∂u ∂x x+ dυ ∂υ

y 1 ∂y y+ du 1 ∂u ∂y y+ dυ 1 ∂υ

Functions of Several Variables

x ∂x = du ∂u ∂x dυ ∂υ ∂x ∂u = ∂x ∂υ dx dy =

i.e.,

y 1 ∂x du ∂y ∂u du 0 = ∂x ∂u dυ ∂y ∂υ dυ 0 ∂υ

2.31

∂y du ∂u ∂y dυ ∂υ

∂y ∂u du dυ ∂y ∂υ

∂ ( x, y ) du dυ . ∂(u , υ )

Since both dx dy and du du are positive, dx dy = J du dυ , where J = Similarly, if we make the transformations

∂ ( x, y ) ∂(u , υ )

x = x(u , υ , w), y = y (u ,υ , w) and z = z (u ,υ , w) in the triple integral J =

2.4

∫∫∫ f ( x, y, z ) dx dy dz,

then dx dy dz = J du dυ dw, where

∂ ( x, y , z ) . ∂(u , υ , w)

DiFFEREntiAtiOn UnDER tHE intEGRAL SiGn

When a function f(x, y) of two variables is integrated with respect to y partially, viz., b

treating x as a parameter, between the limits a and b, then

∫ f ( x , y ) dy

will be a

a

function of x. Let it be denoted by F(x). Now to find F ' (x), if it exists, we need not find F(x) and then differentiate it with respect to x. F ' (x) can be found out without finding F(x), by using Leibnitz’s rules, given below: 1. Leibnitz’s rule for constant limits of integration If f (x, y) and ∂f ( x, y ) are continuous functions of x and y, then ∂x b  d   ∫ f ( x , y ) dy  = dx  a 

a and b are constants independent of x.

b

∫ a

∂f ( x, y ) dy, where ∂x

Engineering Mathematics I

2.32 Proof: b

∫ f ( x, y) dy = F ( x).

Let

a

b

Then

F ( x + ∆x) − F ( x) =

b

f ( x + ∆x, y )dy − ∫ f ( x, y )dy

∫ a

a

b

=

∫ [ f ( x + ∆x, y) −

f ( x, y )]dy

a b

= ∆x ∫ a

∂f ( x + θ ∆ x , y ) dy, 0 < θ < 1, ∂x

df ( x + θ h )   , 0 < θ < 1  by Mean Value theorem, viz., f ( x + h) − f ( x) = h dx  F ( x + ∆ x) − F ( x) = ∆x

∴

b

∫ a

∂f ( x + θ ⋅ ∆ x , y ) dy ∂x

(1)

Taking limits on both sides of (1) as ∆x → 0, b

F′ ( x ) =

∫

a

d   dx 

i.e.,

b

∫

a

 f ( x , y ) dy  = 

∂f ( x , y ) dy ∂x b

∫

a

∂f ( x, y ) dy ∂x

2. Leibnitz’s rule for variable limits of integration b( x)  d  If f(x, y) and ∂f ( x, y ) are continuous functions of x and y, then  ∫ f ( x , y ) dy  dx  a ( x )  ∂x b( x)

=

∂f ( x , y ) db da dy + f {x, b( x)} − f {x, a ( x)} , provided a(x) and b(x) possess ∂x dx dx a( x)

∫

continuous first order derivatives. Proof: Let

∫ f ( x, y)dy = F ( x, y), so that

∂ F ( x, y ) = f ( x, y ) ∂y

(1)

Functions of Several Variables

2.33

b( x)

∫

∴

f ( x, y )dy = F{x, b( x)} − F{x, a ( x)}

a( x)

d   dx 

b( x)

∫

a( x)

 d d f ( x, y )dy  = F{x, b( x)} − F {x, a ( x)} dx dx    d =  F ( x, y )    dx

y = b( x)

  d −  F ( x, y )    dx

 ∂ ∂ dy  =  F ( x, y ) + F ( x, y ) ⋅  ∂y dx   ∂x

y = a( x)

y = b( x)

 ∂ dy  ∂ F ( x, y ) + F ( x, y )  −  x y dx  ∂ ∂ 

y = a( x)

by differentiation of implicit functions

∂  = F ( x, y )  ∂x 

y = b( x)

y = a( x)

dy   +  f ( x, y )  dx  y = b ( x )  dy   −  f ( x, y )  dx  y = a ( x ) 

∂  = f ( x , y ) dy ∫  ∂x  =

∫ ∂f ( x, y) dy   ∂x 

by (1)

y = b( x)

+ f {x, b( x)}b ′ ( x) − f {x, a ( x)}a ′ ( x) y = a( x)

y = b( x)

+ f {x, b( x)}b ′ ( x) − f {x, a ( x)}a ′ ( x) y = a( x)

b( x)

=

∂f ( x , y ) dy + f {x, b( x)}b ′ ( x) − f {x, a ( x)}a ′ ( x) ∂x a( x)

∫

WORKED EXAMPLE 2(b) Example 2.1 degree.

Expand ex cos y in powers of x and y as far as the terms of the third

f ( x, y ) = e x cos y;

∂f ( x, y ) = f x ( x, y ) = e x cos y; ∂x

Engineering Mathematics I

2.34

∂f ( x , y ) = f y ( x, y ) = −e x sin y. ∂y ∂ 2 f ( x, y ) ∂ 2 f ( x, y ) x = f ( x , y ) = e cos y ; = f yx ( x, y ) = −e x sin y xx ∂x ∂y ∂x 2 ∂ 2 f ( x, y ) = f yy ( x, y ) = −e x cos y. ∂y 2

f xxx ( x, y ) = e x cos y; f xxy ( x, y ) = − e x sin y;

Similarly

f xyy ( x, y ) = − e x cos y; f yyy ( x, y ) = e x sin y f (0, 0) = 1; f x (0, 0) = 1; f y (0, 0) = 0;

∴

f xx (0, 0) = 1; f xy (0, 0) = 0; f yy (0, 0) = − 1; f xxx (0, 0) = 1; f xxy (0, 0) = 0; f xyy (0, 0) = − 1; f yyyy (0, 0) = 0 Taylor’s series of f (x, y) in powers of x and y is

f ( x, y ) = f (0, 0) +

1 {xf x (0, 0) + yf y (0, 0)} + 1!

1 2 {x f xx (0, 0) + 2 xyf xy (0, 0) + y 2 f yy (0, 0)} + 2! 1 3 {x f xxx (0, 0) + 3 x 2 yf xxy (0, 0) + 3 xy 2 f xyy (0, 0) + y 3 f yyy (0, 0)} + ⋅⋅⋅ 3!

∴ e x cos y = 1 +

1 1 {x ⋅1 + y ⋅ 0} + {x 2 ⋅1 + 2 xy ⋅ 0 + y 2 (−1)} 1! 2!

1 3 {x ⋅1 + 3 x 2 y ⋅ 0 + 3 xy 2 ⋅ (−1) + y 3 ⋅ 0} + ⋅⋅⋅ 3! x 1 1 =1+ + + ( x 2 − y 2 ) + ( x 3 − 3 xy 2 ) + ⋅⋅ ⋅ 1! 2 ! 3! +

2.4.1  Verification e x cos y = Real part of e x + iy   x + iy ( x + iy ) 2 ( x + iy )3 = R.P. of 1 + + + + ⋅⋅⋅ , 2! 3! 1!   by exponential theorem

Functions of Several Variables = 1+

2.35

x 1 2 1 + ( x − y 2 ) + ( x 3 − 3 y 2 ) + ⋅⋅⋅ 1! 2 ! 3!

Expand ( x + h)( y + k ) in a series of powers of h and k upto the x+h+ y+k second degree terms. Example 2.2

Let

∴

f ( x + h, y + k ) =

( x + h)( y + k ) x+h+ y+k

f ( x, y ) =

xy . x+ y

Taylor’s series of f (x + h, y + k) in powers of h and k is 1  ∂f ∂f  f ( x + h, y + k ) = f ( x , y ) +  h + k  1!  ∂x ∂y  +

Now

2 1  2 ∂2 f ∂2 f 2 ∂ f  + k h + 2 hk + ⋅⋅⋅ 2 !  ∂x 2 ∂x ∂y ∂y 2 

 ( x + y) − x  y2 fx = y  = 2  2  ( x + y)  ( x + y)  ( x + y) − y  x2 fy = x  = 2  2  ( x + y)  ( x + y) f xx = − =

2 y2 ( x + y ) 2 ⋅ 2 y − y 2 ⋅ 2( x + y ) f = ; xy ( x + y )3 ( x + y)4

2{ y ( x + y ) − y 2 } 2 xy = (xx + y )3 ( x + y )3

f yy = −

2x2 . ( x + y )3

Using these values in (1), we have ( x + h)( y + k ) xy hy 2 kx 2 h2 y 2 = + + − x+h+ y+k x + y ( x + y ) 2 ( x + y ) 2 ( x + y )3 +

2hkxy k 2 x2 − + ⋅⋅⋅ 3 ( x + y ) ( x + y )3

(1)

Engineering Mathematics I

2.36

y

Example 2.3 Find the Taylor’s series expansion of x near the point (1, 1) upto the second degree terms. 1 Taylor’s series of f (x, y) near the point (1, 1) is f ( x, y ) = f (1,1) + 1! 1 ( x − 1) f x (1,1) + ( y − 1) f y (1,1) + ( x − 1)2 f xx (1,1) + 2 ( x − 1) ( y − 1) f xy (1,1) 2!

{

}

}

{

+ ( y − 1) f yy (1,1) + ⋅⋅⋅ 2

(1)

f ( x, y ) = x y ; f x ( x, y ) = yx y −1 ; f y ( x, y ) = x y log x; f xx ( x, y ) = y ( y − 1) x y − 2 ; f xy ( x, y ) = x y −1 + yx y −1 log x f yy ( x, y ) = x y ⋅ ( log x) ⋅ 2

f xx (1,1) = 1; f x (1,1) = 1; f y (1,1) = 0; f xx (1,1) = 0; f xy (1,1) = 1; f yy (1,1) = 0 Using these values in (1), we get x y = 1 + ( x − 1) + ( x − 1) ( y − 1) + ⋅⋅⋅

π Find the Taylor’s series expansion of ex sin y near the point  −1,   4 upto the third degree terms. Example 2.4

 π Taylor’s series of f (x, y) near the point  −1,  is  4

π   π   π 1  π  f ( x, y ) = f  −1,  + ( x + 1) f x  −1,  +  y −  f y  −1,        4 4 1!  4 4  +

π  π 1 2  π  ( x + 1) f xx  −1,  + 2 ( x + 1)  y −  f xy  −1,     2!  4 4 4

2 π   π   +  y −  f yy  −1,   + ⋅⋅⋅   4 4  

f ( x, y ) = e x sin y; f x = e x sin y; f y = e x cos y; f xx = e x sin y; f xy = e x cos y; f yy = − e x sin y; f xxx = e x sin y; f xxy = e x cos y; f xyy = − e x sin y; f yyy = − e x coss y. ∴

1 1 1  π  π  π f  −1,  = ; f x  −1,  = ; f y  −1,  = ;    4 e 2 4 e 2 4 e 2 1 1 1  π  π  π ; f xy  −1,  = ; f yy  −1,  = − ; f xx  −1,  =    4 e 2 4 e 2 4 e 2

(1)

Functions of Several Variables

2.37

1 1  π  π  π f xxx  −1,  = ; f xxy  −1,  = ; f xyy  −1,     4 e 2 4 e 2 4 1 1  π ; f yyy  −1,  = − ⋅ =  4 e 2 e 2 Using these values in (1), we get e x sin y =

1  1 π   1 + ( x + 1) +  y −   4  e 2  1!  +

1  π  π 2  ( x + 1) + 2 ( x + 1)  y −  −  y −     2 !  4 4

+

π  π 1  π 3 2  ( x + 1) + 3 ( x + 1)  y −  − 3 ( x + 1)  y −  −  y −    3!  4 4  4

2

   2

3

  + ⋅⋅⋅ 

Example 2.5 Find the Taylor’s series expansion of x2 y2 + 2x2 y + 3xy2 in powers of (x + 2) and (y - 1) upto the third powers. Taylor’s series of f (x, y) in powers of (x + 2) and (y - 1) or near (-2, 1) is f ( x, y ) = f (−2,1) + +

1 {( x + 2) f x (−2,1)+( y −1) f y (−2,1)} 1!

{

1 2 ( x + 2) f xx (−2,1)+ 2 ( x + 2)( y −1) f xy (−2,1) 2!

}

+ ( y −1) f yy (−2,1) +⋅⋅⋅ 2

f ( x, y ) = x 2 y 2 + 2 x 2 y + 3 xy 2

(1) f ( −2,1) = 6

f x = 2 xy 2 + 4 xy + 3 y 2

f x ( −2,1) = − 9

f y = 2 x 2 y + 2 x 2 + 6 xy

f y ( −2,1) = 4

f xx = 2 y 2 + 4 y

f xx ( −2,1) = 6

f xy = 4 xy + 4 x + 6 y

f xy ( −2,1) = − 10

f yy = 2 x 2 + 6 x

f yy ( −2,1) = − 4

f xxx = 0

f xxx ( −2,1) = 0

f xxy = 4 y + 4

f xxy ( −2,1) = 8

f xyy = 4 x + 6

f xyy ( −2,1) = − 2

f yyy = 0

f yyy ( −2,1) = 0

Engineering Mathematics I

2.38

Using these values in (1), we have x 2 y 2 + 2 x 2 y + 3 xy 2 = 6 +

Example 2.6

1 {−9 ( x + 2) + 4 ( y −1)} 1!

{

}

+

1 2 2 6 ( x + 2) − 20 ( x + 2) ( y − 1) − 4 ( y − 1) 2!

+

1 2 2 24 ( x + 2) ( y − 1) − 6 ( x + 2) ( y − 1) + ⋅⋅⋅ 3!

{

}

Using Taylor’s series, verify that

log (1 + x + y ) = ( x + y ) −

1 1 ( x + y )2 + ( x + y )3 −⋅⋅⋅ 2 3

The series given in the R.H.S. is a series of powers of x and y. So let us expand f (x, y) = log (1 + x + y) as a Taylor’s series near (0, 0) or Maclaurin’s series. fx =

1 1 ; fy = 1+ x + y 1+ x + y

f xx = − f xxx =

1 = f xy = f yy (1 + x + y ) 2 2

(1 + x + y )

3

= f xxy = f xyy = f yyy

f ( 0, 0) = 0; f x ( 0, 0) = f y ( 0, 0) = 1; f xx ( 0, 0) = f xy ( 0, 0) = f yy ( 0, 0) = − 1; f xxx ( 0, 0) = f xxy ( 0, 0) = f xyy ( 0, 0) = f yyy ( 0, 0) = 2 . Maclaurin’s series of f (x, y) is given by f ( x, y ) = f ( 0, 0) + +

{

}

1 xf x ( 0, 0) + yf y ( 0, 0) 1!

{

}

1 2 x f xx ( 0, 0) + 2 xyf xy ( 0, 0) + y 2 f yy ( 0, 0) + ⋅⋅⋅ 2!

Using the relevant values in (1), we have log (1 + x + y ) = ( x + y ) + +

{

{

}

1 − x 2 − 2 xy − y 2 + 2

}

1 2 x 3 + 6 x 2 y + 6 xy 2 + 2 y 3 + ⋅⋅⋅ 6

(1)

Functions of Several Variables = ( x + y) −

Example 2.7

2.39

1 1 ( x + y )2 + ( x + y )3 −⋅⋅⋅ 2 3

If x = r cos q, y = r sin q, verify that ∂ ( x, y ) × ∂ ( r , θ ) = 1. ∂ ( r , θ ) ∂ ( x, y ) x = r cos q; y = r sin q ∂x ∂r = ∂y ∂ (r , θ ) ∂r

∂ ( x, y )

∂x cos θ ∂θ = ∂y sin θ ∂θ

−r sin θ r cos θ

= r (cos 2 θ + sin 2 θ ) = r. r 2 = x 2 + y 2 and θ = tan −1

Now

∴

2r

∂r = 2x ∂x

∴

∂r x = ∂x r

Similarly,

∂r y = ∂y r

y x

∂θ = ∂x

1 −y × y 2 x2 1+ 2 x y −y =− 2 = 2 2 x +y r ∂θ x Similaarly = ∂y r 2

∂r ∂r ∂x ∂y ∂ (r, θ) = ∂θ ∂θ ∂ ( x, y ) ∂x ∂y x r = y − 2 r ∴

y x2 + y 2 r 2 1 r = = 3 = x r r3 r r2

∂ ( x, y ) ∂ ( r , θ ) 1 × = r × = 1. ∂ ( r , θ ) ∂ ( x, y ) r

Example 2.8 If we transform from three dimensional cartesian co-ordinates (x, y, z) to spherical polar co-ordinates (r, q, f), show that the Jacobian of x, y, z with respect to r, q, f is r2 sin q.

Engineering Mathematics I

2.40

The transformation equations are x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. ∂x ∂y ∂z = sin θ cos φ, = sin θ sin φ, = cos θ ∂r ∂r ∂r ∂z ∂x ∂y =− r sin θ = r cos θ cos φ, = r cos θ sin φ, ∂θ ∂θ ∂θ ∂x ∂y ∂z = − r sin θ sin φ, = r sin θ cos φ, = 0.. ∂φ ∂φ ∂φ

Now

∂x ∂y ∂z ∂r ∂r ∂r ∂ ( x , y , z ) ∂x ∂y ∂z = ∂ ( r , θ , φ ) ∂θ ∂θ ∂ θ ∂x ∂y ∂z ∂φ ∂φ ∂φ sin θ cos φ

sin θ sin φ

cos θ

= r cos θ cos φ r cos θ sin φ − r sin θ 0 − r sin θ sin φ r sin θ cos φ = r 2 [sin θ cos φ (0 + sin 2 θ cos φ )− sin θ sin φ ×(0 − sin 2 θ sin φ ) + cos θ (sin θ cos θ cos 2 φ + sin θ cos θ sin 2 φ )] = r 2 [sin 3 θ cos 2 φ + sin 3 θ sin 2 φ + sin θ cos 2 θ ] = r 2 (sin 3 θ + sin θ cos 2 θ ) = r 2 sin θ. Example 2.9

If u = 2xy, u = x2 – y2, x = r cos q and y = r sin q, compute ∂ (u , υ ) . ∂ ( r ,θ )

By the property of Jacobians, ∂ ( u , υ ) ∂ ( u , υ ) ∂ ( x, y ) = × ∂ ( r , θ ) ∂ ( x, y ) ∂ ( r , θ ) =

ux u y

υx υ y

×

xr xθ yr yθ

Functions of Several Variables =

2 y 2 x cos θ × 2 x − 2 y sin θ

2.41

− r sin θ r cos θ

=− 4( y 2 + x 2 )×r (cos 2 θ + sin 2 θ ) =− 4r 3 .

Example 2.10 Find the Jacobian of y1, y2, y3 with respect to x1, x2, x3, if y1 =

x2 x3 xx xx , y2 = 3 1 , y3 = 1 2 x1 x2 x3

∂y1 ∂x1

∂y1 ∂x2

∂y1 ∂x3

∂( y1 , y2 , y3 ) ∂y 2 = ∂( x1 , x2 , x3 ) ∂x1

∂y 2 ∂x2

∂y 2 ∂x3

∂y3 ∂x1

∂y3 ∂x2

∂y3 ∂x3

− =

x2 x3 x12 x3 x2

−

x2 x3

x3 x1

x2 x1

x3 x1 x22

x1 x2

x1 x3

−

− x2 x3 1 = 2 2 2 x2 x3 x1 x2 x3 x2 x3 x2 x2 x2 = 12 22 32 x1 x2 x3

−1 1 1 −1 1

x1 x2 x32

x3 x1 − x3 x1

x1 x2 x1 x2

x3 x1

−xx1 x2

1 1

1 −1

−1 1 1 = 0 0 2 =4 0 2 0

Example 2.11 Express ∫ ∫ ∫ xyz (1 − x − y − z ) dx dy dz in terms of u, u, w given that x + y + z = u, y + z = uu and z = uuw.

Engineering Mathematics I

2.42

The given transformations are x+y+z=u

(1)

y + z = uu and

(2)

z = uuw

(3)

Using (3) in (2), we have y = uu(1 – w) Using (2) in (1), we have x = u(1 – u) dx dy dz = ‫׀‬J‫ ׀‬du du dw, where

∂x ∂u ∂ ( x, y , z ) ∂y J = = ∂(u , υ , w) ∂u ∂z ∂u

∂x ∂υ ∂y ∂υ ∂z ∂υ

∂x ∂w ∂y ∂w ∂z ∂w

1−υ 0 −u = υ (1 − w) u (1 − w) −uυ wu uυ υw = (1 − υ ){u 2υ (1 − w) + u 2υ w} + u{uυ 2 (1 − w) + uυ 2 w} = u 2υ (1 − υ ) + u 2υ 2 = u 2υ ∴

dx dy dz = u 2υ du dυ dw

(4)

Using (1), (2), (3) and (4) in the given triple integral I, we have I = ∫ ∫ ∫ u 3υ 2 w(1 − υ )(1 − w)(1 − u ) u 2υ du dυ dw 1

1

1

= ∫ ∫ ∫ u 7 / 2 υ 2 w1 / 2 (1 − u ) 2 (1 − υ ) 2 (1− w) 2 du dυ dw Example 2.12 Examine if the following functions are functionally dependent. If they are, find also the functional relationship. −1 −1 2 2 (i) u = sin x + sin y; υ = x 1 − y + y 1 − x

(ii) u = y + z; υ = x + 2 z 2 ; w = x − 4 yz − 2 y 2 −1 −1 2 2 (i) u = sin x + sin y; υ = x 1 − y + y 1 − x

Functions of Several Variables

2.43

∂u  1  ∂ u  1  ∂ υ xy = ; = = 1− y 2 − ; ; 2 2     ∂x  1 − x  ∂ y  1 − y  ∂ x 1− x2 xy ∂υ =− + 1− x2 ∂y 1− y2

1 Now

∂(u , υ ) = ∂ ( x, y )

=

1− x 1− y 2 −

1 1− y 2

2

xy 1− x

− xy (1 − x )(1 − y ) 2

2

−

2

xy 1− y

+ 1−1+

2

+ 1− x2 xy

(1 − x )(1 − y 2 ) 2

= 0. ∴ u and u are functionally dependent by property (3). Now

sin u = sin (sin −1 x + sin −1 y ) = sin (sin −1 x) cos (sin −1 y ) + cos (sin −1 x) sin (sin −1 y )

{

}

{

}

= x ⋅ cos cos −1 ( 1 − y 2 ) + cos cos −1 ( 1 − x 2 ) ⋅ y = x 1− y 2 + y 1− x2 = υ. ∴ The functional relationship between u and u is u = sin u. (ii) u = y + z; υ = x + 2 z 2 ; w = x − 4 yz − 2 y 2 ∂u ∂υ ∂w = 0; = 1; =1 ∂x ∂x ∂x ∂u ∂υ ∂w = 1; = 0; = − 4 y − 4z ∂y ∂y ∂y ∂u ∂υ ∂w = 1; = 4 z; = − 4y ∂z ∂z ∂z

Now

0 1 1 ∂(u , υ , w) = 1 0 4z ∂ ( x, y , z ) 1 −4 y − 4 z −4 y

Engineering Mathematics I

2.44

= − {−4 y + 4 y + 4 z} + 4 z = 0. ∴ u, u and w are functionally dependent

υ − w = 2 z 2 + 4 yz + 2 y 2

Now

= 2 ( y + z ) = 2u 2 2

∴ The functional relationship among u, u and w is 2u2 = u - w. π dx π Example 2.13 Given that ∫0 a + b cos x = a 2 − b2 ( a > b) , find π

π

dx

∫ ( a + b cos x)

and

2

0

cos x dx

∫ ( a + b cos x)

2

0

π

π

dx

∫ a + b cos x =

(1)

a − b2 2

0

Differentiating both sides of (1) with respect to a, we get π

∂ 



1

∂

π

∫ ∂a  a + b cos x  dx = ∂a ⋅

a − b2 2

0

π

i.e.,

−dx

∫ ( a + b cos x)

, since the limits of integration are constants

(

2

= π × − 1 / 2 a 2 − b2

2

=

)

−3 / 2

2a

0

π

i.e.,

dx

∫ ( a + b cos x)

πa

(a

0

2

− b2

)

3/ 2

Differentiating both sides of (1) with respect to b, we get π

∂ 

1



∂

π

∫ ∂b  a + b cos x  dx = ∂b ⋅

a − b2 2

0

π

i.e.,

1

∫ − ( a + b cos x)

2

(

× cos x dx = π × − 1 / 2 a 2 − b 2

0

π

i.e.,

cos x

∫ ( a + b cos x) 0

2

dx = −

πb

(a

2

− b2

)

3/ 2

.

) (−2b) −3 / 2

Functions of Several Variables

2.45 x

Example 2.14 By differentiating inside the integral, find the value of 1

Hence find the value of

∫

log (1 + x ) 1+ x2

0

log (1 + xy ) 1+ y 2

0

dy.

dx.

x

f ( x) =

Let

∫

∫

log (1 + xy ) 1+ y 2

0

(1)

dy

Differentiating both sides of (1) with respect to x, we have f ′ ( x) =

x d log ( 1+ xy ) dx dx ∫0 1+ y 2

x

=

∫

0

(

)

log 1+ x 2 d( x) ∂  log ( 1+ xy )  ⋅  dx +  2 2 dx ∂x  1+ y 1+ x 

( by Leibnitz's rule ) x

=

y

∫ ( 1+ xy ) ( 1+ y ) dy + 2

0

x

=

1+ x

)

2

 −x 1  y+ x   + 2  2  2  1+ x ( 1+ xy ) 1+ x  1+ y 

∫ ( 0

(

log 1+ x 2

)

(

)

 log 1+ x 2 ,  dy + 1+ x 2 

by resolving the integrand in the first term into partial fractions

(

log 1 + x 2  1 1 1 x 2 −1  + 1 1 = − log + + ⋅ log + + tan xy y y ( )  2 2 1+ x2 1+ x2 1+ x2 0  1+ x

(

(

)

x

)

1 1 x log 1 + x 2 + tan −1 x = ⋅ 2 1+ x2 1+ x2

(2)

Integrating both sides of (2) with respect to x, we have f ( x) = =

(

) (

)

1 x log 1 + x 2 d tan −1 x + ∫ tan −1 x dx + c 2∫ 1+ x2  1  −1 2x 2 −1 dx   tan x log 1 + x − ∫ tan x ⋅ 2 1+ x2  x tan −1 x +∫ dx + c 1+ x2

(

)

)

Engineering Mathematics I

2.46

(

)

1 tan −1 x ⋅ log 1 + x 2 + c 2

=

(2)

Now putting x = 0 in (2), we get c = f(0) = 0, by (1) ∴

x

f ( x) =

∫

log (1 + xy ) 1+ y

0

2

(

1 tan −1 x ⋅ log 1 + x 2 2

dy =

)

(3)

Putting x = 1 in (3), we get 1

∫

log (1 + y ) 1+ y

0

2

1 tan −1 (1) ⋅ log 2 2

dy =

π log 2 8

=

Since y is only a dummy variable, 1

∫

log (1 + x )

0

1+ x2

dx =

Example 2.15 Show that d da

π log 2 8

d da

a2

∫ tan

a2

0

 x ∫0 tan  a  dx = −1

−1

a2

∫ 0

1  x −1 2   dx = 2a tan ( a ) − log a + 1 . a 2

(

)

 a2  d 2 ∂  x tan −1   dx + tan −1   ⋅ a ,  a ∂a  a  da

( )

by Leibnitz's rule

a2

=

1  −x ⋅  2  dx + 2a tan −1 a 2 a  x 0 1+ 2 a

∫

a2

=− ∫ 0

x dx + 2a tan −1 a x + a2 2

(

)

a2 1 = − log x 2 + a 2  + 2a tan −1 a 0 2

Functions of Several Variables

2.47

 a4 + a2  1 = − log  + 2a tan −1 a 2  a 2 

(

)

1 = 2a tan −1 a − log a 2 + 1 2 ∞

Example 2.16 If I = ∫ e

 a − x2 −    x

2

dx, prove that

0

∞

I = ∫e

 a − x2 −    x

dI = − 2 I . Hence find the value of I. da

2

dx

(1) .

0

Differentiating both sides of (1) with respect to a, we have dI = da

∂  − x2 − 2 ∫0 ∂a e x 

a2

∞

∞

= ∫e

− x2 −

a2 x2

0 0

= ∫e

−

a2

− y2

y2

  dx 

 2a  ⋅  − 2  dx  x  2 dy, on putting x =

∞

∞

= − 2∫ e

 a − y2 −    y

a a or y = y x

2

dy

0

i.e.,

dI = − 2I da

∴

dI = − 2 da I

(2)

Solving, we get log I = log c – 2a ∴

I = ce-2a

(3) ∞

When

a = 0, I = ∫ e − x dx = 2

0

Using (4) in (3), we get c =

π 2

π 2

(4)

Engineering Mathematics I

2.48 I =

Hence

π −2 a e . 2

EXERCiSE 2(b)

Part A (Short Answer Question ) 1. Write down the Taylor’s series expansion of f (x + h, y + k) in a series of (i) powers of h and k (ii) power of x and y. 2. Write down the Maclaurin’s series expansion of (i) f (x, y), (ii) f (x + h, y + k). 3. Write down the Taylor’s series expansion of f (x, y) near the point (a, b). 4. Write down the Maclaurin’s series for e x + y. 5. Write down the Maclaurin’s series for sin (x + y). 6. Define Jacobian. 7. State any three properties of Jacobians. 8. State the condition for the functional dependence of three functions u(x, y, z), u(x, y, z) and w(x, y, z). 9. Prove that

∫ ∫ f ( x, y ) dx dy = ∫ ∫ f ( r cos θ , r sin θ ) ⋅ r dr dθ .

10. Show that

∫ ∫ f ( x, y ) dx dy = ∫ ∫ f {u (1 − υ ), uυ } ⋅ u du dυ.

11. If x = u (1 + u) and y = u(1 + u), find the Jacobian of x, y with respect to u, u. 12. State the Leibnitz’s rule for differentiation under integral sign, when both the limits of integration are variables. d 13. Write down the Leibnitz’s formula for dx stant.

b( x)

∫

f ( x, y ) dy, where a is a con-

a

b

d 14. Write down the Leibnitz’s formula for f ( x, y ) dy, where b is a dx a∫( x) constant. 1

15. Evaluate

d log( x 2 + y 2 ) dx, without integrating the given function. dy ∫0

Part B 16. Expand ex sin y in a series of powers of x and y as far as the terms of the third degree. 17. Find the Taylor’s series expansion of ex cos y in the neighbourhood of the  π point 1,  upto the second degree terms.  4

Functions of Several Variables

2.49

18. Find the Maclaurin’s series expansion of ex log (1 + y) upto the terms of the third degree. −1  y  19. Find the Taylor’s series expansion of tan   in powers of (x - 1) and (y - 1)  x upto the second degree terms. 20. Expand x2y + 3y - 2 in powers of (x - 1) and (y + 2) upto the third degree terms. 21. Expand xy2 + 2x - 3y in powers of (x + 2) and (y - 1) upto the third degree terms. 22. Find the Taylor’s series expansion of yx at (1, 1) upto the second degree terms. 23. Find the Taylor’s series expansion of e xy at (1, 1) upto the third degree terms. 24. Using Taylor’s series, verify that ( x + y)2 ( x + y)4 + −⋅⋅⋅ ∞ 2! 4! 25. Using Taylor’s series, verify that cos ( x + y ) = 1 −

1 tan −1 ( x + y ) = ( x + y ) + ( x + y )3 ⋅⋅⋅ ∞ 3 26. If x = u (1 - u), y = uu, verify that ∂( x, y ) ∂(u , υ ) × =1 ∂(u , υ ) ∂( x, y ) 27. (i) if x = u2 – u2 and y = 2uu, find the Jacobian of x and y with respect to u and u. (ii) if u = x2 and u = y2, find ∂(u , υ ) ∂ ( x, y ) 28. If x = a cosh u cos u and y = a sinh u . sin u, show that ∂ ( x, y ) a 2 = (cosh 2u − cos 2υ ). ∂(u , υ ) 2 29. If x = r cos q, y = r sin q, z = z, find ∂( x, y, z ) ∂(r , θ , z ) 30. If F = xu + u – y, G = u2 + uy + w and H = zu – u + uw, compute ∂( F , G , H ) . ∂(u , υ , w) ∂ (u , υ , w) . 31. If u = xyz, u = xy + yz + zx and w = x + y + z, find ∂ ( x, y , z ) 32. Examine the functional dependence of the functions u =

υ=

x+ y x− y

xy . If they are dependent, find the relation between them. ( x − y)2

and

Engineering Mathematics I

2.50

33. Are the functions u =

x+ y and u = tan-1 x + tan-1 y functionally dependent? 1 − xy

If so, find the relation between them. 34. Are the functions f1 = x + y + z, f2 = x2 + y2 + z2 and f3 = xy + yz + zx functionally dependent? If so, find the relation among f1, f2 and f3. x

35. If

−λ ( x − y )

∫ λe

f ( y )dy = λ 2 ⋅ xe−λx prove that f (x) = λe−λx . [Hint: Differentiate

0

both sides with respect to x]. Use the concept of differentiation under integral sign to evaluate the following: dx ( x 2 + a 2 )2

x  dx Use Hint:  2 ∫ x + a2 0 

x m (log x) n dx

1  m  Hint: Use ∫ x dx 0 

x

36.

∫

0

1

37.

∫

0

∞

38.

∫e

− x2

cos 2ax dx

0

∞

39.

∫ 0

1

40.

∫ 0

2.5

e − ax sin x dx and hence x

∞

∫ 0

sin x dx x

xm − 1 dx, m ≥ 0 .⋅ log x

MAXiMA AnD MiniMA OF FUnCtiOnS OF tWO VARiABLES

Students are familiar with the concept of maxima and minima of a function of one variable. Now we shall consider the maxima and minima of a function of two variables. A function f (x, y) is said to have a relative maximum (or simply maximum) at x = a and y = b, if f (a, b) > f (a + h, b + k) for all small values of h and k. A function f(x, y) is said to have a relative maximum (or simply maximum) at x = a and y = b, if f(a, b) < f (a + h, b + k) for all small values of h and k. A maximum or a minimum value of a function is called its extreme value. We give below the working rule to find the extreme values of a function f (x, y): (1) Find ∂f and ∂f . ∂y ∂x

Functions of Several Variables

2.51

(2) Solve the equations ∂f = 0 and ∂f = 0 simultaneously. Let the solutions be ∂x ∂y (a, b); (c, d);… ∂f ∂f = 0 are called stationary = 0 and ∂y ∂x points of the function f (x, y). The values of f (x, y) at the stationary points are called stationary values of f (x, y). ∂2 f ∂2 f ∂2 f , B = , C = (3) For each solution in step (2), find the values of A = ∂x dy ∂x 2 ∂y 2 2 and ∆ = AC - B . (4) (i) If ∆ > 0 and A (or C) < 0 for the solution (a, b) then f (x, y) has a maximum value at (a, b). (ii) If ∆ > 0 and A (or C) > 0 for the solution (a, b) then f (x, y) has a minimum value at (a, b). (iii) If ∆ < 0 for the solution (a, b), then f (x, y) has neither a maximum nor a minimum value at (a, b). In this case, the point (a, b) is called a saddle point of the function f (x, y). (iv) If ∆ = 0 or A = 0, the case is doubtful and further investigations are required to decide the nature of the extreme values of the function f (x, y).

Note

2.5.1

The points like (a, b) at which

Constrained Maxima and Minima

Sometimes we may require to find the extreme values of a function of three (or more) variables say f (x, y, z) which are not independent but are connected by some given relation f (x, y, z) = 0. The extreme values of f (x, y, z) in such a situation are called constrained extreme values. In such situations, we use f (x, y, z) = 0 to eliminate one of the variables, say z from the given function, thus converting the function as a function of only two variables and then find the unconstrained extreme values of the converted function. [Refer to examples (2.8), (2.9), (2.10)]. When this procedure is not practicable, we use Lagrange’s method, which is comparatively simpler.

2.5.2 Let

Lagrange’s Method of Undetermined Multipliers u = f ( x, y , z )

(1)

be the function whose extreme values are required to be found subject to the constraint

φ ( x, y , z ) = 0

(2)

The necessary conditions for the extreme values of u are ∂f = 0, ∂f = 0 and ∂f = 0 ∂x ∂y ∂z

Engineering Mathematics I

2.52

∂f ∂f ∂f dx + dy + dz = 0 ∂x ∂y ∂z

∴

(3)

From (2), we have ∂φ ∂φ ∂φ dx + dy + dz = 0 ∂x ∂y ∂z

(4)

Now (3) + λ × (4), where λ is an unknown multiplier, called Langrange multiplier, gives  ∂f ∂φ  ∂φ  ∂φ   ∂f  ∂f (5)  + λ  dx +  + λ  dy +  + λ  = 0 ∂x ∂x  ∂y ∂y  ∂z ∂z Equation (5) holds good, if ∂f ∂φ +λ =0 ∂x ∂x

(6)

∂f ∂φ +λ =0 ∂y ∂y

(7)

∂f ∂φ +λ =0 ∂z ∂z

(8)

Solving the Equations (2), (6), (7) and (8), we get the values of x, y, z, λ, which give the extreme values of u.

Note (1) The Equations (2), (6), (7) and (8) are simply the necessary conditions for the extremum of the auxiliary function (f + λf), where λ is also treated as a variable. (2) Lagrange’s method does not specify whether the extreme value found out is a maximum value or a minimum value. It is decided from the physical consideration of the problem.

WORKED EXAMPLE 2(c) Example 2.1 Examine f (x, y) = x3 + 3xy2 - 15x2 - 15y2 + 72x for extreme values. f ( x, y ) = x 3 + 3 xy 2 − 15 x 2 − 15 y 2 + 72 x f x = 3 x 2 + 3 y 2 − 30 x + 72 f y = 6 xy − 30 y f xx = 6 x − 30; f xy = 6 y; f y = 6 x − 30

Functions of Several Variables

2.53

The stationary points are given by f x = 0 and f y = 0 i.e.,

3( x 2 + y 2 − 10 x + 24) = 0

(1)

and

6 y ( x − 5) = 0

(2)

From (2), x = 5 or y = 0 When x = 5, from (1), we get y2 - 1 = 0; ∴ y = ± 1 When y = 0, from (1), we get x2 - 10x + 24 = 0 ∴

x = 4, 6.

The stationary points are (5, 1), (5, -1), (4, 0) and (6, 0) At the point (5, ±1), A = fxx = 0; B = fxy = ±6; C = fyy = 0 Though AC - B2 < 0, A = 0 ∴ Nothing can be said about the maxima or minima of f (x, y) at (5, ±1). At the point (4, 0), A = -6, B = 0, C = -6 ∴

AC - B2 = 36 > 0

and A < 0

∴ f (x, y) is maximum at (4, 0) and maximum value of f (x, y) = 112. At point (6, 0), A = 6, B = 0, C = 6 ∴

AC - B2 = 36 > 0 and A > 0.

∴ f (x, y) is minimum at (6, 0) and the minimum value of f (x, y) = 108. Example 2.2 Examine the function f (x, y) = x3y2(12 - x - y) for extreme values. f ( x, y ) = 12x 3 y 2 − x 4 y 2 − x3 y 3 f x = 36 x 2 y 2 − 4 x3 y 2 − 3 x 2 y 3 f y = 24 x 3 y − 2 x 4 y − 3 x 3 y 2 f xx = 72 xy 2 − 12 x 2 y 2 − 6 xy 3 f xy = 72 x 2 y − 8 x 3 y − 9 x 2 y 2 f yy = 24 x 3 − 2 x 4 − 6 x 3 y The stationary points are given by fx = 0; fy = 0 i.e.,

x2 y2 (36 - 4x - 3y) = 0

(1)

and

x3 y (24 - 2x - 3y) = 0

(2)

Solving (1) and (2), the stationary points are (0, 0), (0, 8), (0, 12), (12, 0), (9, 0) and (6, 4). At the first five points, AC - B2 = 0. ∴ Further investigation is required to investigate the extremum at these points. At the point (6, 4), A = -2304, B = -1728, C = -2592 and AC - B2 > 0. Since AC – B2 > 0 and A < 0, f (x, y) has a maximum at the point (6, 4). Maximum value of f (x, y) = 6912.

Engineering Mathematics I

2.54

Example 2.3 Discuss the maxima and minima of the function f (x, y) = x4 + y4 -2x2 + 4xy – 2y2. f ( x, y ) = x 4 + y 4 − 2 x 2 + 4 xy − 2 y 2 . f x = 4( x 3 − x + y ) f y = 4( y 3 + x − y ) f xx = 4(3 x 2 − 1); f xy = 4; f yy = 4(3 y 2 − 1) The possible extreme points are given by f x = 0 and f y = 0 i.e.,

x3 − x + y = 0

(1)

and

y3 + x − y = 0

(2)

Adding (1) and (2), Using (3) in (1): i.e.,

x3 + y 3 = 0 ∴ y = − x

(3)

x3 − 2 x = 0 x ( x 2 − 2) = 0 ∴ x = 0, + 2 , − 2

and the corresponding values of y are 0, − 2 , + 2 . ∴ The possible extreme points of f (x, y) are (0, 0), (+ 2 , − 2 ) and (− 2 , 2 ). At the point (0, 0), A = -4, B = 4 and C = -4 AC - B2 = 0 ∴ The nature of f (x, y) is undecided at (0, 0). At the points (± 2 , ∓ 2 ), A = 20, B = 4, C = 20 AC - B2 > 0 ∴ f (x, y) is minimum at the points (± 2 , ∓ 2 ), and minimum value of f (x, y) = 8. Example 2.4

1 1 Examine the extrema of f ( x, y ) = x 2 + xy + y 2 + + . x y 1 1 f ( x, y ) = x 2 + xy + y 2 + + x y fx = 2x + y −

1 x2

fy = x + 2y −

1 y2

Functions of Several Variables f xx = 2 +

2.55

2 2 ; f xy = 1; f yy = 2 + 3 x3 y

The possible extreme points are given by fx = 0 and fy = 0. i.e.,

2x + y −

1 =0 x2

(1)

and

x + 2y −

1 =0 y2

(2)

(1) – (2) gives

x− y+

1 1 − =0 y 2 x2

i.e.,

x− y+

x2 − y 2 =0 x2 y 2

i.e.,

( x − y )( x 2 y 2 + x + y ) = 0 x= y

∴

(3)

Using (3) in (1), 3x3 – 1 = 0 1

∴

 1 3 x=  = y  3

1 1    1  3  1  3  At the point   ,    , A = 8, B = 1 and C = 8  3   3  

∴

AC - B2 > 0

1 1   4  1  3  1  3  ∴ f (x, y) is minimum at   ,    and minimum value of f ( x, y ) = 3 3 .  3   3  

Example 2.5 Discuss the extrema of the function f (x, y) = x2 - 2xy + y2 + x3 - y3 + x4 at the origin f ( x, y ) = x 2 − 2 xy + y 2 + x 3 − y 3 + x 4 . f x = 2 x − 2 y + 3x 2 + 4 x3 f y = − 2x + 2 y − 3y2 f xx = 2 + 6 x + 12 x 2 f xy = − 2;

f yy = 2 − 6 y

Engineering Mathematics I

2.56

The origin (0, 0) satisfies the equations f x = 0 and f y = 0. ∴ (0, 0) is a stationary point of f (x, y). At the origin, A = 2, B = –2 and C = 2 AC - B2 = 0

∴

Hence further investigation is required to find the nature of the extrema of f (x, y) at the origin. Let us consider the values of f (x, y) at three points close to (0, 0), namely at (h, 0), (0, k) and (h, h) which lie on the x-axis, the y-axis and the line y = x respectively. f ( h, 0 ) = h 2 + h 3 + h 4 > 0 . f (0, k ) = k 2 − k 3 = k 2 (1 − k ) > 0, when 0 < k < 1 f ( h, h ) = h 4 > 0 Thus f (x, y) > f (0, 0) in the neighbourhood of (0, 0). ∴ (0, 0) is a minimum point of f (x, y) and minimum value of f (x, y) = 0. Example 2.6

Find the maximum and minimum values of f ( x, y ) = sin x sin y sin ( x + y ); 0 < x, y < π . f ( x, y ) = sin x sin y sin ( x + y ) f x = cos x sin y sin ( x + y ) + sin x sin y cos ( x + y ) f y = sin x cos y sin ( x + y ) + sin x sin y cos ( x + y )

i.e.,

f x = sin y sin (2 x + y )

and

f y = sin x ⋅ sin ( x + 2 y ) f xx = 2 sin y cos (2 x + y ) f xy = sin y cos (2 x + y ) + cos y ⋅ sin (2 x + y ) = sin (2 x + 2 y ) f yy = 2 sin x cos ( x + 2 y )

For maximum or minimum values of f (x, y), fx = 0 and fy = 0 i.e., sin y sin (2x + y) = 0 and sin x · sin (x + 2y) = 0 i.e.,

1 [cos 2 x − cos (2 x + 2 y )] = 0 2

and

1 [cos 2 y − cos (2 x + 2 y )] = 0 2

i.e.,

cos 2 x − cos (2 x + 2 y ) = 0

(1)

and

cos 2 y − cos (2 x + 2 y ) = 0

(2)

From (1) and (2), cos 2x = cos 2y. Hence x = y Using (3) in (1), cos 2x – cos 4x = 0 i.e., 2 sin x sin 3x = 0 ∴ sin x = 0 or sin 3x = 0 π 2π ∴ x = 0, π and 3x = 0, π, 2π i.e., x = 0, , 3 3

(3)

Functions of Several Variables

2.57

∴ The admissible values of x are 0, π , 2π . 3 3 π π 2π 2π  Thus the maxima and minima of f (x, y) are given by (0, 0)  ,  and  ,  3 3  3 3  At the point (0, 0), A = B = C = 0 ∴ AC - B2 = 0 Thus the extremum of f (x, y) at (0, 0) is undecided. 3 π π  3 At the point  ,  , A = − 3 , B = − and C = − 3 and AC − B 2 = 3 − > 0.  3 3 2 4

π π As AC – B2 > 0 and A < 0, f (x, y) is maximum at  ,  .  3 3 Maximum value of f ( x, y ) =

3 3 3 3 3 ⋅ ⋅ = . 2 2 2 8

3 3  2π 2π  At the point  , , A = 3, B = and C = 3 and AC − B 2 = 3 − > 0.  3 3  2 4 2π 2π  2 As AC – B > 0 and A > 0, f (x, y) is maximum at  . ,  3 3  3 3 Minimum value of f ( x, y ) = − . 8 Example 2.7

Identify the saddle point and the extremum points of f ( x, y ) = x 4 − y 4 − 2 x 2 + 2 y 2 f ( x, y ) = x 4 − y 4 − 2 x 2 + 2 y 2 f x = 4 x 3 − 4 x; f y = 4 y − 4 y 3 f x x = 12 x 2 − 4; f x y = 0;

f yy = 4 − 12 y 2

The stationary points of f (x, y) are given by fx = 0 and fy = 0 i.e.,

4( x 3 − x) = 0 and 4( y − y 3 ) = 0

i.e.,

4 x( x 2 − 1) = 0 and 4 y (1 − y 2 ) = 0

∴ x = 0 or ±1 and y = 0 or ±1. At the points (0, 0), (±1, ±1), AC - B2 < 0 ∴ The points (0, 0), (1, 1), (1, -1), (-1, 1) and (-1, -1) are saddle points of the function f (x, y). At the point (±1, 0), AC - B2 > 0 and A > 0 ∴ f (x, y) attains its minimum at (±1, 0) and the minimum value is -1. At the point (0, ±1), AC - B2 > 0 and A < 0

Engineering Mathematics I

2.58

∴ f (x, y) attains its maximum at (0, ±1) and the maximum value is + 1. Example 2.8 Find the minimum value of x2 + y2 + z2, when x + y + z = 3a. Here we try to find the conditional minimum of x2 + y2 + z2, subject to the condition x + y + z = 3a

(1)

Using (1), we first express the given function as a function of x and y. From (1), z = 3a - x - y. Using this in the given function, we get f ( x, y ) = x 2 + y 2 + (3a − x − y ) 2 f x = 2 x − 2(3a − x − y ) f y = 2 y − 2(3a − x − y ) f xx = 4; f xy = 2; f yy = 4 The possible extreme points are given by fx = 0 and fy = 0. i.e.,

2x + y = 3a

(2)

and

x + 2y = 3a

(3)

Solving (2) and (3), we get the only extreme point as (a, a) At the point (a, a), AC – B2 > 0 and A > 0 ∴ f (x, y) is minimum at (a, a) and the minimum value of f (x, y) = 3a2. Example 2.9 Show that, if the perimeter of a triangle is constant, its area is maximum when it is equilateral. Let the sides of the triangle be a, b, c. Given that

a + b + c = constant

= 2k, say Area of the triangle is given by A= where

s=

s ( s − a ) ( s − b) ( s − c )

(1) (2)

a+b+c 2

Using (1) in (2), A=

k ( k − a ) ( k − b) ( k − c )

A is a function of three variables a, b , c Again using (1) in (3), we get A=

k ( k − a ) ( k − b) ( a + b − k )

(3)

Functions of Several Variables

2.59

2 A is maximum or minimum, when f (a, b) = A = (k - a) (k - b)(a + b - k) is k maximum or minimum.

f a = (k − b){(k − a ) ⋅ 1 + (a + b − k ) ⋅ (−1)} = ( k − b) ( 2k − 2a − b) f b = (k − a ){(k − b) ⋅ 1 + (a + b − k ) ⋅ (−1)} f aa

= (k − a ) (2k − a − 2b) = − 2 (k − b); f ab = − 3k + 2a + 2b;

f ab = − 2 (k − a ) The possible extreme points of f (a, b) are given by fa = 0 and fb = 0 i.e., (k - b) (2k - 2a - b) = 0 and (k - a)(2k - a - 2b) = 0 ∴ b = k or 2a + b = 2k and a = k or a + 2b = 2k Thus the possible extreme points are given by (i) a = k, b = k; (ii) b = k, a + 2b = 2k; (iii) a = k, 2a + b = 2k and (iv) 2a + b = 2k, a + 2b = 2k. (i) gives a = k, b = k and hence c = 0. (ii) gives a = 0, b = k and hence c = k. (iii) gives a = k, b = 0 and hence c = k. All these lead to meaningless results. Solving 2a + b = 2k and a + 2b = 2k, we get a=

2k 2k and b = 3 3

 2k 2k  , At the point  ,  3 3  A = f aa = −

2k k 2k ; B = f ab = − ; C = f bb = − 3 3 3

AC - B2 > 0 and A < 0  2k 2k  ∴ f (a, b) is maximum at  ,  3 3  Hence the area of the triangle is maximum when a = When a =

2k 2k and b = 3 3 .

2k 2k 2k ,b= ; c = 2k − ( a + b) = 3 3 3

2k Thus the area of the triangle is maximum, when a = b = c = , i.e., when the 3 triangle is equilateral.

Engineering Mathematics I

2.60

Example 2.10 In a triangle ABC, find the maximum value of cos A cos B cos C. In triangle ABC, A + B + C = π. Using this condition, we express the given function as a function of A and B Thus cos A cos B cos C = cos A cos B cos {π − ( A + B )} = − cos A cos B cos ( A + B ) Let

f ( A, B ) = − cos A cos B cos ( A + B ) f A = − cos B {− sin A cos ( A + B ) − cos A sin( A + B) } = cos B sin (2 A + B ) f B = − cos A{− sin B cos ( A + B ) − cos B sin( A + B )} = cos A sin ( A + 2 B ) f AA = 2 cos B cos (2 A + B) f AB = cos B cos (2 A + B ) − sin B sin (2 A + B) f BB

= cos (2 A + 2 B) = 2 cos A cos ( A + 2 B)

The possible extreme points are given by f A = 0 and f B = 0 i.e.,

cos B sin (2 A + B) = 0

(1)

and cos A sin (A + 2B) = 0 Thus the possible values of A and B are given by (i) cos B = 0, cos A = 0; (ii) cos B = 0, sin (A + 2B) = 0; (iii) sin (2A + B) = 0, cos A = 0 and (iv) sin (2A + B) = 0, sin (A + 2B) = 0 π π π π i.e., (i) A = , B = ; (ii) B = , A = 0 or π , (iii) A = , B = 0 or π and 2 2 2 2 (iv) 2 A + B = π , A + 2 B = π or

π π ,B= 3 3 The first three sets of values of A and B lead to meaningless results. A=

Hence A =

π π give the extreme point. ,B= 3 3

1 π π At this point  ,  , A = f AA = − 1; B = f AB = − ; f BB = − 1 and AC = B 2 > 0.  3 3 2 Also A < 0

π and the maximum value 3 π π 2π 1 = − cos ⋅ cos⋅ cos = . 3 3 3 8

∴ f (A, B) is maximum at A = B =

Functions of Several Variables

2.61

Example 2.11 Find the maximum value of xm yn zp, when x + y + z = a. Let f = xm yn zp and f = x + y + z – a. Using the Lagrange multiple l, the auxiliary function is g = (f + λf). This stationary points of g = ( f + λf) are given by gx = 0, gy = 0, gz = 0 and gl = 0 i.e.,

mx m −1 y n z p + λ = 0

(1)

nx m y n −1 z p + λ = 0

(2)

px m y n z p−1 + λ = 0

(3)

x+ y+ z−a = 0

(4)

From (1), (2) and (3), we have − λ = mx m −1 y n z p = nx m y n −1 z p = px m y n z p −1 . m n p m+n+ p i.e., = = = x y z x+ y+ z m+n+ p , by (4) = a ∴ Maximum value of f occurs, when x =

am an ap , y= ,z= m+n+ p m+n+ p m+n+ p

Thus maximum value of f =

a m + n + p ⋅ mm ⋅ nn ⋅ p p (m + n + p) m + n + p

Example 2.12 A rectangular box, open at the top, is to have a volume of 32 c.c. Find the dimensions of the box, that requires the least material for its construction. Let, x, y, z be the length, breadth and height of the respectively. The material for the construction of the box is least, when the area of surface of the box is least. Hence we have to minimise S = xy + 2 yz + 2 zx, subject to the condition that the volume of the box, i.e., xyz = 32. Here f = xy + 2yz + 2zx; f = xyz - 32. The auxiliary function is g = f + lf, where l is the Lagrange multiplier. The stationary points of g are given by gx = 0, gy = 0, gz = 0 and gλ = 0 i.e.,

y + 2 z + λ yz = 0

(1)

x + 2 x + λ zx = 0

(2)

2 x + 2 y + λ xy = 0

(3)

xyz − 32 = 0

(4)

Engineering Mathematics I

2.62

From (1), (2) and (3), we have 1 2 + = −λ z y

(5)

1 2 + = −λ z x 2 2 + = −λ y x

(6) (7)

Solving (5), (6) and (7), we get 4 4 2 , y = − and z = − λ λ λ Using these values in (4), we get 32 − 3 − 32 = 0 λ x=−

λ = −1

i.e.,

x = 4, y = 4, z = 2.

∴

Thus the dimensions of the box and 4 cm; 4 cm and 2 cm.

Example 2.13 Find the volume of the greatest rectangular parallelopiped inscribed 2 2 2 in the ellipsoid whose equation is x + y + z = 1. a 2 b2 c2 Let 2x, 2y, 2z be the dimensions of the required rectangular parallelopiped. By symmetry, the centre of the parallelopiped coincides with that of the ellipsoid, namely, the origin and its faces are parallel to the co-ordinate planes. Also one of the vertices of the parallelopiped has co-ordinates (x, y, z), which satisfy the equation of the ellipsoid. 2 2 2 Thus, we have to maximise V = 8xyz, subject to the condition x + y + z = 1 2 2 2 a b c

Here f = 8xyz and φ =

x2 y 2 z 2 + + −1 a 2 b2 c2

The auxiliary function is g = f + λf, where λ is the Lagrange multiplier. The stationary points of g are given by g x = 0, g y = 0, g z = 0 and g λ = 0 i.e.,

8 yz +

2λ x =0 a2

(1)

8 zx +

2λ y =0 b2

(2)

Functions of Several Variables

2.63

2λ z =0 c2

(3)

x2 y 2 z 2 + + =1 a 2 b2 c2

(4)

8 xy +

2 Multiplying (1) by x, 2λ x = − 8 xyz a2 2 2 Similarly 2λ y = 2λ z = − 8 xyz from (2) and (3) b2 c2

Thus

x2 y2 z2 = = = k say a2 b2 c2

Using in (4), 3k = 1 ∴ k = ∴x =

a 3

, y=

b 3

1 3 c

and z =

∴ Maximum volume =

3

8abc 3 3

.

Example 2.14 Find the shortest and the longest distances from the point (1, 2, -1) to the sphere x2 + y2 + z2 = 24. Let (x, y, z) be any point on the sphere. Distance of the point (x, y, z) from (1, 2, -1) is given by d =

( x − 1)2 + ( y − 2)2 + ( z + 1)2 .

We have to find the maximum and minimum values of d or equivalently d 2 = ( x − 1) + ( y − 2) + ( z + 1) , 2

2

2

subject to the constant x2 +y2 +z2 – 24 = 0 Here

f = ( x − 1) + ( y − 2) + ( z + 1) and 2

2

2

φ = x 2 + y 2 + z 2 − 24 The auxiliary function is g = f + lf, where l is the Lagrange multiplier. The stationary points of g are given by gx = 0, gy = 0, gz = 0 and gλ = 0. i.e.,

2 ( x − 1) + 2λ x = 0

(1)

2 ( y − 2) + 2 λ y = 0

(2)

2 ( z + 1) + 2λ z = 0

(3)

x 2 + y 2 + z 2 = 24

(4)

Engineering Mathematics I

2.64

From (1), (2) and (3), we get x=

1 2 1 , y= ,z= 1+ λ 1+ λ 1+ λ

Using these values in (4), we get 6

(1 + λ ) ∴

= 24 i.e., (1 + λ ) = 2

2

λ=−

1 4

1 3 or − ⋅ 2 2

1 , the point on the sphere is (2, 4, -2) 2 3 When λ = − , the point on the sphere is (-2, -4, 2) 2 When λ = −

When the point is (2, 4, -2), d =

(1)2 + ( 2)2 + ( −1)2

When the point is (-2, -4, 2), d = ∴ Shortest and longest distances are

=

( −3)2 + ( −6)2 + 32

6 =3 6

6 and 3 6 respectively.

Example 2.15 Find the point on the curve of intersection of the surfaces z = xy + 5 and x + y + z = 1 which is nearest to the origin. Let (x, y, z) be the required point. It lies on both the given surfaces. ∴ xy – z + 5 = 0 and x + y + z = 1 Distance of the point (x, y, z) from the origin is given by d = We have to minimize d or equivalently d 2 = x2 + y 2 + z 2 ,

x2 + y 2 + z 2 ⋅

subject to the constraints xy – z + 5 = 0 and x + y + z – 1 = 0.

Note

Here we have two constraint conditions. To find the extremum of f (x, y, z) subject to the conditions f1 (x, y, z) = 0 and f2 (x, y, z) = 0, we form the auxiliary function g = f + l1f1 + l2f2, where l1 and l2 are two Lagrange multipliers. The stationary points of g are given by gx = 0, gy = 0, gz = 0, gλ1 = 0 and gλ2 = 0. In this problem, f = x2 + y2 + z2, f1 = xy – z + 5 and f2 = x + y + z – 1. The auxiliary function is g = f + l1f1 + l2f2, where l1, l2 are Lagrange multipliers. The stationary points of g are given by 2 x + λ1 y + λ2 = 0

(1)

2 y + λ1 x + λ2 = 0

(2)

2 z − λ1 + λ2 = 0

(3)

Functions of Several Variables

2.65

xy − z + 5 = 0

(4)

x + y + z −1 = 0

(5)

Eliminating l1, l2 from (1), (2), (3), we have 2x y 1 2y x 1 = 0 2 z −1 1 i.e.,

x ( x + 1) − y ( y − 2) − ( y + zx ) = 0

i.e.,

x2 − y 2 + x − y − z ( x − y) = 0

( x − y ) ( x + y − z + 1) = 0

i.e.,

x = y or x + y − z + 1 = 0 ∴ Using x = y in (4) and (5), we have

and

z = x2 + 5

(6)

z =1 − 2 x

(7)

2

From (6) and (7), x + 2x + 4 = 0, which gives only imaginary values for x. Hence x + y − z +1 = 0

(8)

Solving (5) and (8), we get x + y = 0

(9)

z =1

(10)

xy = − 4

(11)

and Using (10) in (4), we get

Solving (9) and (11), we get x = ± 2 and y = ± 2. ∴ The required points are (2, -2, 1) and (-2, 2 ,1) and the shortest distance is 3.

EXERCiSE 2(c) Part A (Short Answer Questions) 1. Define relative maximum and relative minimum of a function of two variables. 2. State the conditions for the stationary point (a, b) of f (x, y) to be (i) a maximum point (ii) a minimum point and (iii) a saddle point. 3. Define saddle point of a function f (x, y). 4. Write down the conditions to be satisfied by f (x, y, z) and f (x, y, z), when we extremise f (x, y, z) subject to the condition f (x, y, z) = 0. 5. Find the minimum point of f (x, y) = x2 + y2 + 6x + 12. 6. Find the stationary point of f (x, y) = x2 – xy + y2 – 2x + y. 7. Find the stationary point of f (x, y) = 4x2 + 6xy + 9y2 – 8x – 24y + 4.

Engineering Mathematics I

2.66

2 2 2 2 8. Find the possible extreme point of f ( x, y ) = x + y + + . x y 9. Find the nature of the stationary point (1, 1) of the function f (x, y), if fxx = 6xy3, fxy = 9x2 y2 and fyy = 6x3 y. 10. Given fxx = 6x, fxy = 0, fyy = 6y, find the nature of the stationary point (1, 2) of the function f (x, y). Part B Examine the following functions for extreme values: 11. x3 +y3 - 3axy 12. x3 + y3 -12x - 3y + 20 13. x4 + 2x2 y - x2 + 3y2 14. x3 y -3x2 - 2y2 -4y - 3 15. x4 + x2 y + y2 at the origin 16. x3 y2 (a – x –y) 17. x3 y2 (12 – 3x - 4y)  1 1 18. xy + 27  +   x y

π 2 Identify the saddle points and extreme points of the function xy (3x + 2y + 1). Find the minimum value of x2 + y2 + z2, when (i) xyz = a3 and (ii) xy + yz + zx = 3a2. Find the minimum value of x2 + y2 + z2, when ax + by + cz = p. Show that the minimum value of (a3 x2 + b3 y2 + c3 z2), when 1 1 1 1 + + = , is k 2 (a + b + c)3 . x y z k

19. sin x + sin y + sin ( x + y ), 0 ≤ x, y ≤ 20. 21. 22. 23.

24. Split 24 into three parts such that the continued product of the first, square of the second and cube of the third may be minimum. 25. The temperature at any point (x, y, z) in space is given by T = k x y z2, where k is a constant. Find the highest temperature on the surface of the sphere x2 + y2 + z2 = a2. 26. Find the dimensions of a rectangular box, without top, of maximum capacity and surface area 432 square meters. 27. Show that, of all rectangular parallelopipeds of given volume, the cube has the least surface. 28. Show that, of all rectangular parallelopipeds with given surface area, the cube has the greatest volume. 29. Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube. 30. Find the points on the surface z2 = xy + 1 whose distance from the origin is minimum. 31. If the equation 5x2 + 6xy + 5y2 = 8 represents an ellipse with centre at the origin, find the lengths of its major and minor axes.

Functions of Several Variables

32. 33. 34.

35.

2.67

(Hint: The longest distance of a point on the ellipse from its centre gives the length of the semi-major axis. The shortest distance of a point on the ellipse from its centre gives the length of the semi-minor axis). Find the point on the surface z = x2 + y2, that is nearest to the point (3, -6, 4). Find the minimum distance from the point (3, 4, 15) to the cone x2 + y2 = 4z2. Find the points on the ellipse obtained as the curve of intersection of the surfaces x + y = 1 and x2 + 2y2 + z2 = 1, which are nearest to and farthest from the origin. Find the greatest and least values of z, where (x, y, z) lies on the ellipse formed by the intersection of the plane x + y + z = 1 and the ellipsoid 16x2 + 4y2 + z2 = 16.

AnSWERS

Exercise 2(a) (2) (3) (4) (5) (6)

du = cos (xy2) (y2dx + 2xy dy) du = xy - 1 ∙ yx (y + x log y) dx + xy yx - 1 (x + y log x) dy du = y (1 + log xy) dx + x(1 + log xy) dy du = (y log a) axy dx + (x log a) axy dx 8 a5t6 (4t + 7);

(7) e

a2 − t 2

{

sin 3 t 3 a 2 − t 2 sin 2 t cos t − t sin 3 t / a 2 − t 2

}

(8) (cos t - e-t – sin t)/(e-t + sin t + cos t) 2 2 (9) − x + 2 xy + 2 y 2 2 x + 4 xy + y

x(xy + 4y2 – 2x2)/(x + 2y) 4.984 0.006 cm3; 0.004 cm2 4(a + b + c)k ∂2 z =0 (36) ∂u ∂υ (12) (15) (17) (19)

(38)

∂2 z =0 ∂u 2

(40)

∂ 2u =0 ∂z ⋅ ∂z *

(41)

(i)

y ( y − x log y ) x ( x − y log x)

(iii) y tan x + log sin y log cos x − x cot y

(11)

3 x cos ( x 2 + y 2 ) 2

(14) (16) (18) (20)

3.875 0.0043 2 1.5

∂2 z (37) ∂υ 2 = 0 (39)

∂2 z =0 ∂u ∂υ

(ii) y x (iv)

log cot y − y tan x log sec x + x sec y cosec y

Engineering Mathematics I

2.68 (v)

x− y x(1 + log x)

(42) 2a3xy/(ax – y2)3 (47) 5%

(43) 2a2xy (3a4 + x2y2)/(a2x – y3)3 (50) 2.3%

(55) 5 3π 324

Exercise 2(b) (4) 1 + ( x + y ) +

( x + y ) 2 ... + 2

(11) u + u + 1 (16) y + xy +

(5) ( x + y ) −

1 ( x + y )3 + ... 3!

(15) 2 tan −1  1   y 

x 2 y y 3 ... − + 2 6

2  e  π  ( x − 1) 2 π 1 π   − ( x − 1)  y −  −  y −  + ⋅⋅⋅ 1 + ( x − 1) −  y −  +   4 2 4 2 4 2  

(17)

1 1 y2 1 2 y + xy − + x y − xy 2 + y 3 + ⋅⋅⋅ (18) 2 2 2 3

π 1 1 1 1 − ( x − 1) + ( y − 1) + ( x − 1) 2 − ( y − 1) 2 + ⋅⋅⋅ 4 2 2 4 4 (20) - 10 – 4 (x - 1) + 4(y + 2) – 2(x - 1)2 + 2(x - 1) (y + 2) + (x - 1)2 (y + 2) (19)

(21) - 9 + 3 (x + 2) – 7(y - 1) + 2(x + 2) (y - 1) – 2(y - 1)2 + (x + 2) (y - 1)2 (22) 1 + (y - 1) + (x - 1) (y – 1) + … 1 1 (23) e 1 + ( x − 1) + ( y − 1) + ( x − 1) 2 + 2( x − 1) ( y − 1) + ( y − 1)2 + ( x − 1)3  2 6 3 1 3  + (xx − 1) 2 ( y − 2) + ( x − 1) ( y − 2) 2 + ( y − 2)3  2 6 2  (27) (i) 4(u2 + u2) (ii) 4xy (28) r (30) x(yu + 1 - w) + z – 2uu (31) (x - y) (y - z) (z - x) (32) u2 = u + 1 (33) u tan u

Functions of Several Variables

2.69

2 (34) f 1 = f 2 + 2 f 3

(36)

1 x {tan −1 + (ax) /( x 2 + a 2 )} a 2a 3

(37)

(−1) n n ! (m + 1) n +1

(38)

2 1 π e− a 2

−1  1  π (39) tan   ; a 2

(40) log (1 + m)

Exercise 2(c) (5) (-3, 0)

(6) (1, 0)

(8) (1, 1)

(9) Saddle point

 4 (7)  0,   3 (10) Minimum point

(11) Maximum at (a, a) if a < 0 and minimum at (a, a) if a > 0 (12) Minimum at (2, 1) and maximum at (-2, -1)  3 1 (13) Minimum at  ± ,−  4  2 (14) Maximum at (0, -1)

(15) Minimum at (0, 0)

(16) Maximum at  a , a   2 3

(17) maximum at (2, 1)

(18) Minimum at (3, 3)

(19) Maximum at  π , π  and minimum at  − π , − π     3 3 3 3 1  (20) Saddle point are (0, 0),  − 1 , 0 and  0, −  ; maximum at  2  3  p2 a 2 + b2 + c2

 1 1  − , −  9 6

(21) 3a2; 3a2

(22)

4 (25) ka 8

(26) 12, 12 and 6 metres

(30) (0, 0, 1) and (0, 0, -1)

(31) 4, 2

(32) (1, -2, 5)

(33) 5 5

(34)  1 , 2 , 0 ; (1, 0, 0) 3 3 

(35) 8 ; − 8 3 7

(24) 4, 8, 12

UNIT

3

Integral Calculus 3.1

INTrodUcTIoN

Integration can be considered as the reverse process of differentiation. viz, in integration, we are required to find the function f(x) from its derivative which will be d given as g(x), say. In other words, the process of finding f(x) from g(x), given that dx {f(x)} = g(x) is integration. In this situation, we say that f(x) is the integral of g(x) and write symbolically that The symbol

∫ g ( x)dx = f ( x).

∫ is the symbol of integration, g(x) is called the integrand and dx

indicates the variable (x) with respect to which integration is performed. d (sin x) = cos x For example, ∫ cos x dx = sin x, since dx and

3.2 When

∫ 3x

2

d x = x 3 , since

d 3 ( x ) = 3x 2 . dx

coNsTaNT of INTegraTIoN

∫ 3x

2

dx = x 3 , the result

∫ 3x

2

dx = x 3 + c equally holds good, as

d 3 d ( x ) = ( x3 + c), where c is a constant. As c can take any constant value, it is dx dx called the arbitrary constant of integration. d f ( x) = g ( x), ∫ g ( x)dx = [ f ( x) + c]⋅[ f ( x) + c] is called the In general, when dx indefinite integral of g(x) due to indefinite nature, of c. For convenience, we normally omit c when we evaluate an indefinite integral.

3.2.1  Definite Integrals When

∫ g ( x)dx = f ( x) + c,

then [f(b) – f(a)] is called the definite integral of g(x)

between the limits (or end values) a and b and denoted by the symbol

b

∫ g ( x)dx. a is a

b

called the lower limit and b called the upper limit and is denoted by [f(x)]a .

Engineering Mathematics I

3.2 b

Thus

∫ g ( x)dx = [ f ( x)]

b a

= f (b) − f (a )

a

Note

The constant of integration ‘c’ occurring in the indefinite integral

does not find a place in the definite integral, for if b

∫ g ( x)dx = f ( x) + c,

∫ g ( x)dx = [ f ( x) + c]

b a

Then

a

= {f(b) + c} – f(a) + c = f(b) – f(a) b

Thus to evaluate

∫ g ( x)dx

first get f(x) and omit the arbitrary constant c. Then

a

we substitute b and a for the variable x and obtain f(b) and f(a). Finally we get b

[f(b) – f(a)] =

∫ g ( x) dx a π /2

For example,

∫ cos x dx = [sin x]

π /2 0

= sin

0

π – sin 0 = 1 – 0 = 1 2

2

∫ 3x

and

2

dx = [ x 3 ]12 = 23 − 13 = 8 − 1 = 7.

1

3.2.2  Standard Integrals Using the knowledge of derivatives of elementary/standard functions, the following standard integrals are obtained. Students should not try to derive these results from differentiation results, but remember them as formulas of integral calculus. [Constants of integration are omitted in all the formulas] x n +1 (ax + b) n +1 n (n ≠ −1); Extension: ∫ (ax + b) n dx = 1. ∫ x dx = n +1 (n + 1)⋅ a

∫

1 1 dx ≠ − and x2 x

∫

dx =2 x x

An important note on the extension: In the place of the variable x of the integrand of any standard integral, if we have a simple first degree expression (ax + b), we have to replace x by (ax + b) in the corresponding result in the R.H.S. also and divide it by the coefficient of x in (ax + b), namely ‘a’. viz., if

∫ g ( x)dx = f ( x) ... then

1

∫ g (ax + b) dx = a f (ax + b)

(1) (2)

Integral Calculus

3.3

This result which can be considered as extended standard integral is obtained as follows: 1 If we put ax + b = y in (1), then a dx = dy or dx = dy a dy 1 1 Thus (1) becomes ∫ g ( y ) = ∫ g ( y ) dy = f ( y ) by (1). a a a Particular cases of stand and formula(1):   Extension:  

1. (a)

∫

1 1 dx = − x2 x

(b)

∫

 dx = 2 x  Extension: x 

dx

∫ (ax + b) ∫

2

=−

 1  a (ax + b) 

2 ax + b  dx =  a ax + b 

  dx dx 1 = log e x;  Extension: ∫ = log e (ax + b)   x ax + b a     1 e x dx = e x ;  Extension: ∫ a ax + b dx = e ax +b    a

2.

∫

3.

∫

4.

∫ sin x dx = − cos x;  Extension: ∫ sin (ax + b) dx = − a cos(ax + b)



Note 5. 6.



Extensions are omitted for the remaining standard integrals that follow, as they are obvious.

∫ cos x dx = sin x ∫ tan x dx = log sec x x 2 π x sec x dx = log (sec x + tan x) or log tan  +   4 2 

7.

∫ cosec x dx = − log (cosec x + cot x)

8.

∫

9.

∫ cot x dx = log sin x

Note

1

or log tan

Formulas (6), (7), (8) and (9) are not derived from standard differentiation formulas.

10.

∫ sec

11.

∫ cosec

12.

∫ sec x tan x dx = sec x

2

x dx = tan x 2

x dx = − cot x

Engineering Mathematics I

3.4 13. 14. 15.

∫ cosec x cot x dx =− cosec x ∫ sinh x dx = cosh x ∫ cosh x dx = sinh x dx

dx

∫

17.

∫

 x = cosh −1   or log ( x + x 2 − a 2 )  a  x +a

18.

∫

 x = sinh −1   or log ( x + x 2 − a 2 )  a  x2 + a2

19.

∫

20.

∫

 x dx 1 dx = tan −1   and hence ∫ 2 = tan −1 x 2  a  x +a a x +1

21.

∫

 x − a  dx 1  = log  2 x −a 2a  x + a 

22.

∫a

2

23.

∫

a 2 − x 2 dx =

 x x a2 a2 − x2 + sin −1    a  2 2

24.

∫

x 2 − a 2 dx =

x 2

x2 − a2 −

 x a2 cosh −1    a 2

25.

∫

x 2 + a 2 dx =

x 2

x2 + a2 +

 x a2 sinh −1    a  2

1− x

2

= sin −1 x and

∫

x a

16.

a −x 2

2

= sin −1

dx

2

2

dx

dx x x −1 2

= sec−1 x and

∫

dx x x −a 2

2

=

x 1 sec−1 a a

2

2

 a + x  dx 1  = log  2 −x 2a  a − x 

3.3 TechNIqUes of INTegraTIoN Before we proceed to discuss techniques of integration, we give below two basic properties of integration for which no proof is required as it is obvious.

∫ k f ( x) dx − k ∫ f ( x) dx. If k , k , ..., k are constants, then ∫ [k f ( x) ± k f ( x) ± k f ( x) ± k f ( x)] dx = k ∫ f ( x) dx ± k ∫ f ( x) dx ±  k ∫ f ( x) d x.

(i) If k is a constant, (ii)

1

2

n

1 1

n

n

1

1

2

2

2

2

3 3

m

n

3.3.1  Integration by Substitution If the integral is of the form ∫ F{ f ( x)} ⋅ f ′( x) dx, where f(x) is an elementary/ standard function, the integral can be reduced to a simpler integrable form by putting

Integral Calculus y = f(x) so that dy  f ¢(x)dx. The integral gets reduced to the form which can be done by known methods or by using standard formulas.

3.5

∫ F ( y)} dy,

As particular cases of this rule, we mention a few: To evaluate ∑ ∫ F(xn)xn – 1dx, we put xn = y ∑

∫ F(x ) x dx, we put x

∑

∫ F (log x) ◊

∑ ∑ ∑

2

2

=y

dx , we put log x = y x x x x ∫ F(e ) ◊ e dx, we put e = y

∫ F(sin x) cos x dx, we put sin x = y ∫ F(tan x) sec x dx, we put tan x = y 2

∑

∫ F (sin

∑

∫ F (tan

Note

−1

−1

x) x)

dx 1 − x2

, we put sin −1 x = y

dx , we put tan −1 x = y 1 + x2

The following two particular cases of

∫ F{f (x)}f ¢(x) dx are of importance,

as they will be used in integrating some rational and irrational functions. (i)

∫

(ii)

∫

f ′ ( x) dx → f ( x) f ′ ( x) f ( x)

dx →

∫ ∫

dy = log y → log f ( x) y dy y

= 2 y → 2 f ( x)

3.3.2  Integration by Trigonometric Substitution If the integrand contains a 2 − x 2 , x 2 + a 2 or x 2 − a 2 , it can be reduced to a rational or integrable form by making the trigonometic substitution x = a sin q, x = a tan q or x = a sec q respectively In the first case, we may even put x = a cos q In the second case, we may also put x = a cot q or x = sinh y In the third case, we may also put x = a cosec q or x = cosh y.

3.3.3  Compound Trigonometric Substitution ( x − α) , when b > a, it can (β − x ) be retionalised or reduced to the integrable form by making the substitution x = a cos2 q + b sin2 q. If the integrand contains

( x − α ) (β − x) or

Engineering Mathematics I

3.6

Worked example 3(a) x3

∫

Example 3.1 Evaluate

1 + x4

dx.

The integrand is of the form f (1 + x4) ¥ x3 \ Put 1 + x4 = y

\

4x3dx = dy

∫

I =

1 dy 4 = 1 × 2 y = 1 1 + x4 + c 4 2 y

∞

Example 3.2 Evaluate

∫a 0

∞

I=∫ 0

2

dx . e x + b 2 e− x ∞

dx e x dx = a 2 e x + b 2 e− x ∫0 a 2 e 2 x + b 2

Put ex = y \ exdx = dy, since the integrand is of the form f (ex) ¥ ex.

Note

Instead of evaluating the indefinite integral and using the limits in the end, we can express the limits for the new variable y using the substitution used. Thus, when x = 0, y = 1 and when x = •, y = • ∞

\

I=

∫ 1

dy 1 = 2 2 2 2 a y +b a

∞

∫ 1

dy 2 b y 2 +    a 

∞

 1 a  −1  ay  1  π −1 a   = 2 ×  tan   =  − tan  b  1 b ab  2 b  a Example 3.3 Integrate

1 w.r.t x. x(1 + log x) 2

1 Since the integrand is the product of log x or 1 + log x and its derivative , we put x 1 1 + log x = y \ dx = dy x dy 1 1 + c. \ I = ∫ 2 = − + c or − y y 1 + log x π/4

Example 3.4 Evaluate

∫ sin 3x cos x dx. 0

The integrand can be rewritten as the product of f (sin x) and cos x. So we put sin x = y and \ cos x dx = dy

Integral Calculus When x = 0, y = 0 and when x =

3.7

π 1 , y= 4 2

π/4

\

I =

∫ (3 sin x − 4 sin

3

x) cos x dx

0 1/ 2

=

∫

(3 y − 4 y 3 ) dy

0

3  =  y 2 − y 4   2 0

1/ 2

Example 3.5 Evaluate

I = =

3 1 1 1 = ⋅ − = . 2 2 4 2

1 − sin x 

∫ 1 + sin x  dx.  (1 − sin x) 2   dx = ∫ (sec x − tan x) 2 dx cos 2 x 

∫ 

∫ sec

2

x + (sec 2 x − 1) − 2 sec x tan x  dx

= 2 tan x – 2 sec x – x + c π/2

Example 3.6 Evaluate

∫

1 + sin 2 x dx.

0 π /2

I =

∫

cos 2 x + sin 2 x + 2 cos x sin x dx

0 π /2

I =

∫

cos x + sin x) 2 dx

0 π /2

=

∫ (cos x + sin x) dx = (sin x − cos x)

π /2 0

=2

0

∫ cosec x dx. = ∫ cosec x cosec x dx = ∫ (1 + cot x) cosec x dx

Example 3.7 Evaluate I

8

6

2

2

3

2

As the integrand is of the form f (cot x). cosec2 x, we put cot x = y and cosec2 x dx = –dy \ I = ∫ (1 + y 2 )3 (− dy ) = − ∫ (1 + 3 y 2 + 3 y 4 + y 6 ) dy  5 7 3 = −  y + y + 3 y + y  5 7 

Engineering Mathematics I

3.8

 3 1 3 5 7  = −  cot x + cot x + cot x + cos x  + c   5 7 Example 3.8 Evaluate I =

∫

(sin −1 x)3 1 − x2

dx. 1

As the integrand is of the form f (sin −1 x) ⋅ 1

sin −1 x = y and ∴

1 + x2

I = ∫ y 3 dy = Example 3.9 Evaluate

∫

, we put

dx = dy

y4 1 = (sin −1 x) 4 + c 4 4

tan −1 x dx. 1 + x2

As the integrand if of the form f (tan −1 x) ⋅ tan −1 x = y and ∴ I =∫

1 − x2

1 , we put 1 + x2

1 dx = dy 1 + x2

y dy =

3 2 32 2 y = (tan −1 x) 2 + c 3 3

−1

Example 3.10 Evaluate

∫

esec x x x2 − 1

dx. 1

As the integrand is of the form f (sec−1 x) ⋅ sec−1 x = y and ∴

1 x x2 −1

x x2 − 1 dx = dy −1

I = ∫ e y dy = e y = esec Example 3.11 Evaluate

∫

+c

1− x dx. 1+ x

Multiplying the Nr. and Dr. by I =∫

x

1 − x , we get

1− x 1 − x2

dx

, we put

Integral Calculus

3.9

To reduce the integrand to the integrable form, we make the trigonometric substitution x = sin q and so dx = cos dq 1 − sin θ  \ I = ∫   cos θ dθ  cos θ  = θ + cos θ = sin −1 x + 1 − x 2 + c Example 3.12 Evaluate

dx

∫ (1 − x)

1 − x2

.

Due to the occurrence of 1 − x 2 in the integrand, we put x = sin q and so dx = cos dq cos θ d θ I =∫ \ (1 − sin θ ) cos θ 1 + sin θ d θ = ∫ (sec 2 θ + sec θ tan θ )dθ cos 2 θ = tan θ + sec θ =∫

x

=

Example 3.13 Evaluate

1− x

2

1

+

1− x

x2

∫ (4 + x

2

)

5

2

or

1+ x 1− x

2

or

1+ x +c 1− x

dx. 2

Due to the presence of ( x 2 + 4) 2 in the integrand, we put 1

x = 2 tan θ and so dx = 2 sec 2 θ dθ \

I =∫

4 tan 2 θ ⋅ 2sec 2 θ dθ 5 (4sec 2 θ ) 2

8 tan 2 θ dθ 32 ∫ sec3 θ 1 = ∫ sin 2 θ cos θ d θ 4 =

=

1 sin 3 q , on putting sin q = t 4 3

ˆ 1 Ê x = Á ˜ 12 ÁË x 2 + 4) ˜¯ Example 3.14 Evaluate Due to the presence of dx = a sec q tan q dq

∫

3

or

1 x3 ◊ 2 +c 12 ( x + 4)3/2

x2 − a2 dx. x x 2 − a 2 in the integrand, we put x = a sec q and so

Engineering Mathematics I

3.10

Then

a 2 sec 2 θ − a 2 a sec θ tan θ dθ a sec θ

I=∫

= a ∫ tan 2 θ dθ = a ∫ (sec 2 θ − 1)dθ = a (tan θ − θ ) 2 2      x −a −1   x  +c = a − sec    a  a     Example 3.15 Evaluate

a2 − x2

∫

x4

a 2 − x 2 in the integrand, we put x = a sin q and so

Due to the presence of dx = a cos q cos q dq Then

a cos θ ⋅ a cos θ dθ a 4 sin 4 θ

I =∫ 1 a2 1 = 2 a =

dx.

∫ cot ∫t

2

2

θ ⋅ cosec 2 θ dθ

(−dt ), where t = cot θ

1 t3 1 = − 2 cot 3 θ 2 a 3 3a 3 2 2  1  a − x  = − 2    x 3a  =−

1 (a 2 − x 2 ) 2 =− 2 + c. x3 3a 3

Example 3.16 Evaluate

∫

( x − 3)(7 − x) dx.

We make the compound trigonometric substitution x = 3 cos2q + 7 sin2q Then x – 3 = 3 cos2q + 7 sin2q – 3 = 4 sin2q and 7 – x = 7 – (3 cos2q + 7 sin2q) = 4 cos2q dx = (–6 cos q sin q + 14 sin q cos q)dq = 8 sin q cos q dq Then

I =∫

4sin 2 θ ⋅ 4cos 2 θ × 8sin θ cos θ dθ

= 32 ∫ sin 2 θ cos 2 θ dθ = 8∫ (2 sin θ cos θ ) 2 dθ = 8∫ sin 2 2θ dθ

Integral Calculus

3.11

8 (1 − cos 4θ )dθ 2∫  sin 4θ  = 4θ −   4  =

 x − 3   − 2sin 2θ cos 2θ = 4sin −1   2   x − 3   − 4sin θ cos θ (2cos 2 θ − 1) = 4sin −1   2   x − 3     − 4 x − 3 ⋅ 7 − x 2 × (7 − x) − 1 = 4sin −1   2    2 2 4  x − 3     − ( x − 3)(7 − x)  5 − x  + c = 4sin −1   2   2  2

Example 3.17 Evaluate

∫ 1

x −1 dx. 2− x

\ dx = (–2 cos q sin q + 4 sin q cos q) dq = 2 sin q cos q dq x – 1 = cos2 q + 2 sin2 q – 1 = sin2 q 2 – x = 2 – (cos2 q + 2 sin2 q) = cos2 q

Put x = cos q + 2 sin q 2

2

When

x = 1, cos 2 θ + 2 sin 2 θ + 1 viz., sin 2 θ = 0 ∴ θ = 0

When

x = 2, cos 2 θ + 2 sin 2 θ = 2 viz., cos 2 θ = 0 ∴ θ =

Then

I = ∫ tan θ ⋅ 2 sin θ cos θ dθ

π /2

0 π /2

π /2

= ∫ 2 sin 2 θ dθ = ∫ (1 − cos 2θ )dθ 0

0 π 2

  1 π = θ − sin 2θ  = .  0 2 2 β

Example 3.18 Evaluate

∫ α

dx ( x − α)(β − x)

(β > α).

Put x = α cos 2 θ + β sin 2 θ ∴ dx = 2(β − α ) sin θ cos θ dθ x = θ = α cos 2 θ + β sin 2 θ − θ = (β − α )sin 2 θ β − x = β − (α cos 2 θ + β sin 2 θ ) = (β − α ) cos 2 θ

π 2

Engineering Mathematics I

3.12

x = α, θ = 0 and when x = β , θ =

When

π /2

\

I=∫ 0

Example 3.19 Evaluate

∫

π 2

2(β − α )sin θ cos θ π d θ = 2 ⋅ = π. (β − α )sin θ cos θ 2 x−2 dx. 5− x

Put x = 2 cos 2 θ + 5 sin 2 θ ∴ = dx = 6 sin θ cos θ dθ x − 2 = 3 sin 2 θ and 5 − x = 3 cos 2 θ \

I =∫

3 sin 2 θ ⋅ 6 sin θ cos θ dθ = ∫ 6 sin 2 θ dθ 3 cos 2 θ

= 3∫ (1 − cos 2θ )dθ   1 = 3 θ − sin 2θ    2 = 3θ − 3sin θ cos θ = 3 sin −1

x−2 3 x−2 5− x − ⋅ 3 3 3

= 3 sin −1

x−2 − ( x − 2)(5 − x) + c 3

β

Example 3.20 Evaluate

∫ α

x ( x − α)(β − x)

dx.

Putting x = α cos 2 θ + β sin 2 θ and proceeding as in Example (3.18), π /2

we get

I=∫ 0

α cos 2 θ + β sin 2 θ ⋅ 2(β − α )sin θ cos θ dθ (β − α ) sin θ cos θ

π /2 2  1 + cos 2 θ    + β 1 − cos θ  dθ = 2 ∫ α         2 2 0  π π π = α   + β   or (α + β ) 2 2 2

Integral Calculus

3.13

EXERCISE 3(a) Part A (Short Answer Questions) Integrate the following functions w.r.t. x: x2 (1) x +1

x2 − 4x + 3 (2) x−2

(3) cos mx cos nx

1 (4) 1 − cos x

sin 2 x (5) 1 + cos x

(1 + x ) n (6) x

(7)

1 2 + e x + e− x

(10)

sin x + cos x sin x − cos x

(13)

ax + b +

(8) (11)

a −x 2n

x +1 x + 2x + 8 2

2n

x 1 − (log) 2

x2 + 2x

(12)

x3 + 3x 2 + 2

x 27 (14) 14 x −4 x

sinh x

(16)

sinh x + 5 2

1 (18)

2 (9) sin x ⋅ sin 2 x

1 cx + d

xn −1

(15)

1 x(log x) n

cos x

(19)

9 − sin 2 x

(17)

1 + x4 sec 2 x

(20)

16 + tan 2 x

Part B Evaluate the following integrals: (21)

cos 2 x x 1

(24)

(22) 2

∫

(25)

∫x

a 2 − x 2 dx

∫ (2 − x 0

a

∫ 0

x

(27)

∫ 0

1

∫ 0

1

x 1 − x2

x

0

sin −1 x(1 − x 2 )

0

(30)

x

∫e ∞

a

(29)

(23)

dx

0

(26)

1

sin 3 x

dx + e− x −1

e tan x 1 + x2

a

dx

(28)

∫ (a 0

2

dx 3 + x2 ) 2

dx 2

) 4 − x2

a2 − x2 dx a2 + x2

[Hint: Put x2 = a2 cos2q]

Engineering Mathematics I

3.14 1

(31)

∫ (1 + x ) 2

0

5

(33)

∫ 2 β

(35)

∫ α

15

dx 1 − x2

(32)

8

3

x−2 dx 5− x

(34)

dx

∫ ( x − 3) ∫ 2

x +1

[Hint: Put x + 1 = y2]

dx ( x − 2)(3 − x)

dx x ( x − α )(β − α )

3.4  I  nTEgRATIon of RATIonAl (AlgEbRAIC)  fUNcTIoNs

∫ ax

Integrals of the form

2

dx (lx + m) and ∫ 2 dx are two typical + bx + c ax + bx + c

integrals of rational functions. 1 dx dx , we rewrite it as ∫ (1) To evaluate ∫ 2 (viz., the a x2 + b x + c ax + bx + c a a coefficient of x2 is made unity) dx Then ∫ is re-written in any one of the forms b c x2 + x + a a dx dx dx ∫ ( x + p)2 + q 2 ⋅ ∫ ( x + p)2 − q 2 and ∫ q 2 − ( x + p 2 ) . There are extended standard integral formulas and hence easily evaluated. (i) To evaluate

(lx + m) dx, we express 2 + bx + c

∫ ax

d (ax 2 + bx + c) + B, where A and B are constants to be found dx out in individual problems. lx + m = A ⋅

Then I = A∫

d (ax 2 + bx + c) dx dx dx + B ∫ 2 ax 2 + bx + c ax + bx + c

f ′( x ) dx and hence the result f ( x) is log {f(x)} or log (ax2 + bx + c) and the second integral is evaluated as in case (i). (ii) If the denominator of the integrand f(x) in ∫ f ( x)dx can be factorised, The first of these integrals is of the form

∫

f(x) is split into partial fractions by algebraic method and the integration is performed term by term.

Integral Calculus

3.15

Integration of the trigonometric functions of the following form can be reduced to integration of rational functions, or integration by substitution. dx (i) To evaluate integrals of the form ∫ (where a, b, c are a cos 2 x + b sin 2 x + c constants, we multiply the numerator and denominator of the integrand by sec2q and then the integral can be rewritten as ∫ f (tan θ )sec 2 θ dθ , which can be evaluated by earlier method by making the substitution tan q = u. (ii) To evaluate integrals of the form

dx

∫ a cos x + b sin x + c ,

we express cos x

x and sin x in terms of tan . On simplification, the integral takes the form 2 dt ∫ at 2 + bt + c which can evaluated by the earlier method. (iii) To evaluate integrals of the form

 l cos x + m sin x + n 

∫  l ′ cos x + m′ sin x + n′  dx,

first

  d we put integral in the form Nr. =  A Dr + Bx Dr  where A, B, and C  dx  constants to be found out in individual problems [Nr = numerator and Dr. = denomination] Then the integral takes the form   d   (Dr) I = ∫  A + B dx + c  dx, the result of which is immediately obtained as Dr   dx Ax = B log(Dr.) + c ∫ . The third integral is a problem in the case (ii) Dr

Worked example 3(b)

Example 3.1 Evaluate

∫

xdx . x4 + x2 + 1

The integrand is the product of a f(x2) and x. So we make substitution x2 = y and hence 1 dx = dy. 2 1 dy 1 dy I =∫ 2 2 = ∫ Then 2 y + y +1 2      y + 1  + 1 − 1   2   4 

Engineering Mathematics I

3.16

1 dy 2 2 2∫      y + 1  +  3   2   2   1  y +  1 2 −1  2  tan  = ⋅  3  2 3    2   2 x 2 + 1 1  + c tan −1  =  3 3  =

Example 3.2 Evaluate

dx

∫ 1 + 4x − 4x

2

.

1 dx 1 dx = 4 ∫ 1 + x − x 2 4 ∫ 1 − ( x 2 − x) 4 4 1 dx = ∫ 2 4 1  1 −  x −  2  2

I=

   1 dx 1 2 = ∫ = ⋅ log  2 2 4  1   4 2     −  x − 1     2    2   2 − 2 + 2 2 x  1 = log   +c  4 2  2 + 2 − 2 2 x   1

Example 3.3 Evaluate

∫ 0

1 1 + x −  2  2 1 1  − x +  2  2

dx ; 0 < θ < 1. x + 2 x cos θ + 1 2

1

I =∫ 0 1

=∫ 0

dx ( x + cos θ ) 2 + (1 − cos 2 θ )  1  x + cos θ  dx tan −1  =  2 2  sin θ  0 ( x + cos θ ) + sin θ  sin θ

  −1 1 + cos θ  1  −1  tan   − tan (cot θ )    sin θ  sin θ      θ 1  −1  −1  = tan cot  − tan (cot θ )     sin θ  2  =

π θ  π   −1 1  −1  tan tan  −  − tan tan  − θ   2 2 2  sin θ    π θ π 1  θ  = ⋅  − − + θ = .   sin θ   2 2 2  2sin θ

=

1

Integral Calculus Example 3.4 Evaluate

∫

x8 . x6 + 1

1 3 2 Put x = y and so x dx = dy 3 1 2 y dy 1  1  3 =  dy 1 − 2 2 y + 1 3 ∫  y + 1

I =∫

1 = [ y − tan −1 y ] + c 3 1 2 = [ x − tan −1 x 3 ] + c 3 Example 3.5 Evaluate

∫ 2x

2

x−2 dx. + 3x + 1

d (2 x 2 + 3 x + 1) dx = A(4 x + 3) + B 1 3 11 Comparing like terms; A = and + B = − 2 ∴ B = 4 4 4 Let

Then

x − 2 = A⋅

I =∫

1 11 (4 x + 3) − 4 4 dx 2 x 2 + 3x + 1

1 11 dx = log(2 x 2 + 3 x + 1) − ∫ 4 8 x2 + 3 x + 1 4 2 1 11 dx = log(2 x 2 + 3 x + 1) − ∫ 2 2 4 8      x + 3  −  1   4   4     x + 3 − 1  1 11 4  4 4  = log(2 x 2 + 3 x + 1) − × log   3 1  4 8 2  x + 4 + 4  1 11  2 x + 1  = log(2 x 2 + 3 x + 1) − log  +c 4 4  2 x + 2  Example 3.6 Evaluate Let

x +1

∫ 6x + 7 − x

2

dx.

x + 1 = A(6 − 2 x) + B

Comparing like terms; 1 − 2 A = 1 ∴ A = −

1 and 6 A + B = 1 ∴ B = 4 2

3.17

3.18

Engineering Mathematics I

Then

1 − (6 − 2 x) + 4 1 dx 2 =− log(6 x + 7 − x 2 ) + 4 ∫ 6x + 7 − x2 2 7 − ( x 2 − 6 x)

I =∫

1 dx = − log(6 x + 7 − x 2 ) + 4∫ 2 2 4 − ( x − 3) 2  4 + x − 3  1 1 = − log(6 x + 7 − x 2 ) + 4 × log  + c  4 − x + 3  2 2× 4  1 + x  1 1 = − log(6 x + 7 − x 2 ) + log  + c 2 2  7 − x  Example 3.7 Evaluate

∫

( x 2 + x + 1) dx. x2 − x + 1

Integrand is an improper function. So we express it as the sum of an integer and a proper function.   2x Then I = ∫ 1 + 2  dx x − x + 1  =x+∫

1 ⋅ (2 x − 1) + 1 dx x2 − x + 1

= x + log( x 2 − x + 1) + ∫

dx

     x − 1  +  3   2   2   2 x − 1 2 tan −1  = x + log( x 2 − x + 1) +  + c.   3  3 2

Example 3.8 Evaluate

∫ 4x 0

2

I =∫ 0

3

∫

2

x2 + 1 dx. + 12 x + 7

1 × (12 x 2 + 12) 1 12 dx = {log(4 x 3 + 12 x + 7)}02 3 4 x + 12 x + 7 12 =

Example 3.9 Evaluate

2

1 1 [log 63 − log 7] = log e 9. 12 12

x2 + x + 1 dx. ( x − 1)( x − 2) 2

As the denominator is the product of factors, we split the integrand into partial fractions. Let

x2 + x + 1 A B C = + + 2 ( x − 1)( x − 2) x − 1 x − 2 ( x − 2) 2

Integral Calculus \

x 2 + x + 1 = A( x − 2) 2 + B ( x − 1)( x − 2) + C ( x − 1)

Putting x = 1, A = 3 Putting x = 2, C = 7 Equating coefficients of x2 on both sides, we get A + B = 1 Then

3.19

\ B = –2.

 3 2 7   dx I =∫  − +  x − 1 x − 2 ( x − 2) 2  7 = 3log( x − 1) − 2log( x − 2) − +c x−2

Example 3.10 Evaluate

∫

x2 − 1 dx. x + x2 + 1 4

Though the denominator of the integrand is not directly factorisable, it is made factorisable as explained below: x 4 + x 2 + 1 = ( x 4 + 2 x 2 + 1) − x 2 = ( x 2 + 1) 2 − x 2 = ( x 2 + x + 1)( x 2 − x + 1) Then

I=

x2 − 1 ( x − x + 1)( x 2 + x + 1)

Let

I=

Ax + B Cx + D + x2 − x + 1 x2 + x + 1

2

\ x 2 − 1 = ( Ax + B)( x 2 + x + 1) + (cx + D)( x 2 − x + 1) Equating like coefficients, we get A+C=0 A+B–C+D=1 A+B+C–D=0 B + D = –1 1 Solving these equations, we get A = 1, C = –1, B = − = D 2 \

 1   x − 1 x+   2 2  dx I=∫ 2 − 2  x − x + 1 x + x + 1 =

1  2 x − 1 2 x + 1  −  dx  2 ∫  x 2 − x + 1 x 2 + x + 1

=

 x 2 − x + 1 1 + c log  2  x + x + 1 2

Engineering Mathematics I

3.20 Example 3.11 Evaluate

∫

x dx . x3 + 1

The denominator of the integrand is factorisable as (x + 1) (x2 – x + 1) x A Bx + C Let = + x3 + 1 x + 1 x 2 − x + 1 \

x = A( x 2 − x + 1) + ( x + 1)( Bx + 0)

Putting

x = −1, A = −

1 3

Equating like terms, we get A + B = 0 ∴ B =

Then

I=∫

1 and A + C = 0 3

∴C =

1 3

 1  −   3 + 1 x + 1  dx   2  x + 1 3 x − x + 1

1 3 (2 x − 1) + 1 1 2 2 dx = − log( x +1) + ∫ 3 3 x2 − x + 1 1 1 1 dx =− log( x +1) + log( x 2 − x +1) + ∫ 2 2 3 6 2      x − 1  +  3   2   2   2 x − 1 1 1 1 tan −1  = − log( x + 1) + log( x 2 − x + 1) + +C  3  3 6 3 π/2

Example 3.12 Evaluate

cos x dx

∫ (1 + sin x) (2 + sin x). 0

Put sin x = y and so cos x dx = dy π When x = 0 and , y = 0 and 1 resepectively 2 \

1

I =∫ 0

dy (1 + y )(2 + y )

 1 1 = ∫  − 1 + y 2 +  0 1

  dy, y 

on splitting the integrand into partial fractions. \

I = [log(1 + y ) − log(2 + y )]10  4 2 1 = log − log = log    3 3 2

Integral Calculus ∞

Example 3.13 Evaluate

x

∫ (1 + x)(1 + x

2

0

)

3.21

dx.

x A Bx + C = + 2 (1 + x)(1 + x ) 1 + x 1 + x2

Let \

A(1 + x 2 ) + (1 + x)( Bx + c) = x

Putting

x = −1, 2 A = −1 ∴ A = −

Equating like terms,

Then

1 2

1 2 1 and A + C = 0 ∴ C = 2 A+ B=0 ∴ B=

 1    − 1 x 1 +  dx I = ∫  2 + ⋅ 1 + x 2 1 + x 2  0 ∞

∞

 1  1 1 = − log(1 + x) + log(1 + x 2 ) + tan −1 x   2  0 4 2 ∞ 1     2  1 1  (1 + x ) 4   log = + tan −1 ∞ − tan −1 0  1    2 2  2    (1 + x)  0   1      1 4       x 2 + 1    =π = log    1   4  +1      x  0 1

Example 3.14 Evaluate

dx

∫ sin x + sin 2 x .

I =∫ =∫

dx sin x + 2 sin x cos x dx sin x dx =∫ sin x(1 + 2 cos x) (1 − cos 2 x)(1 + 2cos x)

Put cos x = y and so sin x dx = –dy \ Let

I =∫

−dy (1 − y )(1 + y )(1 + 2 y )

1 A B C = + + (1 − y )(1 + y )(1 + 2 y ) 1 − y 1 + y 1 + 2 y

Engineering Mathematics I

3.22

1 4 1 6 2 = − + 3 , on splitting into partial fractions. 1− y 1+ y 1+ 2 y \

 1 4  1  −   6 I=∫ + 2 − 3  dy 1 − y 1 + y 1 + 2 y  1 1 1 4 = log(1 − y + log(1 + y ) − ⋅ log(1 + 2 y ) 6 2 2 3 1 1 2 = log(1 − cos x) + log(1 + cos x) − log(1 + 2cos x) + C. 6 2 3

Example 3.15 Evaluate

∫ sin

2

dx . x + 6 cos 2 x + 3

Multiplying the numerator and denominator by sec2x, we have

I =∫

sec 2 x dx tan x + 6 + 3(1 + tan 2 x) 2

Since the integrand is the product of f(tan q) and sec2q, we make the substitution tan x = y and so sec2x dx = dy dy I =∫ 2 \ y + 6 + 3(1 + y 2 ) =∫

=

∫ 0 π /2

I=∫ 0

Put

2y dy 1 2 = ⋅ tan −1   2 3 4 3  3 y 2 +    2

 1 −1  2 tan  tan x + C 3  6 π/2

Example 3.16 Evaluate

dy 1 = 4 y2 + 9 4 ∫

dx . a cos x + b 2 sin 2 x 2

2

sec 2 x dx a 2 + b 2 tan 2 x

tan x = y and so sec 2 x dx = dy

When

x = 0, y = 0 and when x =

Then

I =∫

∞

0

dy 1 = 2 2 2 2 a +b y b

π , y=∞ 2

∞

∫ a

dy

2

0

  + y 2  b 

Integral Calculus

3.23

∞

1 b  −1  by  ⋅  tan    a  0 b 2 a   1 π π =  − 0 =   ab 2 2ab

=

Example 3.17 Evaluate

dx

∫ 4 cos x + 3sin x + 5.

x x Expressing cos x and sin x in terms of tan and putting tan = t and so 2 2 x 1 2dt sec 2 ⋅ dx = dt or dx = 2 2 2 1+ t

we get

I =∫

2dt 1+ t2 1 − t 2   2t   + 3 +5 4 2 2  1 + t  1 + t 

dt 4(1 − t ) + 6t + 5(1 + t 2 ) dt dt = 2∫ 2 = 2∫ t + 6t + 9 (t + 3) 2 2 =− +c x tan + 3 2 = 2∫

π/2

Example 3.18 Evaluate

∫ 0

2

π dx 2 . 12 cos x + 9 sin x

x 2dt Putting tan = t and so dx = and changing the limits as 0 and 1, we get 2 1+ t2 1

I =∫ 0

2dt 1+ t2 1 − t 2  2t  + 9× 12 2 1+ t2 1 + t  1

= 2∫ 0 1

=

1 6 ∫0

dt 12 − 12t 2 + 18t dt  3  1 − t 2 − t   2 

1

=

1 dt 2 2 ∫ 6 0  5     − t − 3   4   4 

Engineering Mathematics I

3.24

5   + t − 3  1 2  4 4  = × log    6 5  5 − t + 3   4 4 0 1

1  3 1 log − log   15 2 4 1 = log 6. 15 =

Example 3.19 Evaluate

sin x + 18 cos x

∫ 3 sin x + 4 cos x dx.

Let sin x + 18cos x = A(3sin x + 4cos x) + B (3cos x − 4sin x) Equating like terms, we get 3A – 4B = 1 (1) and 4A + 3B = 18 (2) Solving equation (1) and (2), we get A = 3 and B = 2 Then

3 × (Dr) + 2

I=∫

d (Dr.) dx dx

Dr. = 3 x + 2 log(3sin x + 4 cos x) + c π/2

Example 3.20 Evaluate

dx

∫ 1 + cot x . 0

π /2

I=∫ 0

Let

sin x dx sin x + cos x

sin x = A ⋅ f ( x) + B ⋅ f ′( x), where f(x) is the denominator = A(sin x + cos x) + B(cos x − sin x)

Equating like terms, we get \ Then

A − B = 1 and A + B = 0

1 1 and B = − 2 2 π /2  1 1 f ′( x )   dx I=∫  −  2 2 f ( x)  A=

0

π

1 2 1 =  x − log(sin x + cos x)  2  0 2  1 1 π =  − 0 − (log1 − log1)  2 22 =

π . 4

Integral Calculus

3.25

EXERCISE 3(b) Part A (Short Answer Questions) (1)

1 x + a2

(2)

1 x − a2

(3)

1 a − x2

(4)

1 −1  x + 1 tan    2  2

(5)

x +1 x2 + 9

(6)

x x4 + a4

(7)

1 x( x + 1)

(8)

cos x sin x(1 + sin x)

(9)

1 sin x

(10)

1 cos x

2

2

2

Part B Integrate the following functions w.r.t. x: (11)

x2 x + x3 + 1

(12)

ex e + 2e x + 10

(13)

2log x + 3 x[(log x) 2 + 2log x + 5

(14)

3x + 7 2 x 2 − 3x + 5

(15)

(5 − 4sin x) cos x 1 + 2sin x − sin 2 x

(16)

x2 − x + 1 x2 + x + 1

6

1

(17) Evaluate

∫ 0

2x

1

( x − 3) dx x + 2x − 4

(18) Evaluate

2

∫ 0

( x − 3) dx x + 2x − 4 2

(7 x − 4)

(19) Evaluate

∫

dx x( x + 1)( x + 2)

(20) Evaluate

∫ ( x − 1) ( x + 2) dx

(21) Evaluate

∫

dx x +1

(22) Evaluate

∫

3

2

2x2 + 3 dx x3 − 1

Evaluate the following: cos x dx

(23)

∫ (1 + sin x)(1 + sin

(25)

∫ cos

(27)

2

2

x)

dx x + 2sin 2 x + 3 dx

∫ 1 + 3 sin x + 4 cos x

dx + 1)( x 2 + 4)

(24)

∫ (x

(26)

∫ 2 + 7 cos

3 dx

π

(28)

2

dx

2

∫ 5 + 3 cos x 0

x

Engineering Mathematics I

3.26

(29)

3.5

8 cos x + sin x + 6

∫ 3 cos x + 2 sin x + 4 dx

π /2

(30)

dx

∫ 1 + tan x 0

INTegraTIoN of IrraTIoNal fUNcTIoNs

∫

(i) To evaluate

is rewritten as

or

1 a

∫

dx ax + bx + c 2

1 a

, we make the coefficient of x2 as unity, viz., I

1 dx which is put as a b c x2 + x + a a

∫

dx a − ( x + p)2 2

∫

dx ( x + p)2 ± q 2

.

lx + m

dx, we express lx + m = A × d f ( x) + B, dx ax 2 + bx + c 2 where f(x) = ax + bx + c and A, B are constants to be found out in individual problems.

(ii) To evaluate

Then I = ∫

∫

Af ′( x) + B dx = A × 2 f ( x ) + B ∫ f ( x)

dx . The second intef ( x)

gral is evaluated as in case (i). (iii)

∫

ax 2 + bx + c dx :

Making coefficient of x2 as unity, we get I = a∫

( x + p ) 2 ± q 2 dx or a ∫

q 2 − ( x + p ) 2 dx ,

which are extended standard integrals. (iv)

∫ (lx + m)

ax 2 + bx + c dx

To evaluate this, we put lx + m = Af ′( x) + B, when A and B are constants t to be found out in individual problems and f(x) = ax2 + bx + c. Then

I = ∫ { A ⋅ f ′( x) + B} f ( x) dx

2 3 = A ⋅ [ f ( x)] 2 + B ∫ f ( x) dx 3 Then second integral is evaluated as in case (iii).

Integral Calculus (v) Integral of the form dx

∫ (x − k)

ax + bx + c 2

3.27

can be converted as

∫

dx lx + mx + n 2

, by making

1 the substitution x − k = . y (vi) To evaluate made.

dx

∫ ( px + q)

also, the substitution px + q =

lx + m

1 can be y

2 ax + b , we may put ax + b = y and

(vii) If a part of the integral contains

remove the irrational part from the integrand. 1 , we put x = . Then the y + b) cx + d integral takes the form which can be evaluated by substitution.

(viii) If the integral is of the form

∫ (ax

dx

2

2

Worked example 3(c)

Example 3.1 Evaluate

dx

∫

6x − x2 − 5

I=∫ =∫

.

dx −5 − ( x 2 − 6 x ) dx 2 − ( x − 3) 2 2

 x − 3  = sin −1  +c  2  1

Example 3.2 Evaluate

dx

∫

x + 2x + 2 2

0 1

I=∫ 0

.

dx ( x + 1) 2 + 1

= [log{( x + 1) + x 2 + 2 x + 2}]10 = log(2 + 5) − log(1 + 2)  2 + 5   = log   1 + 2 

Engineering Mathematics I

3.28

dx

∫

Example 3.3 Evaluate

2x − 7x + 5 2

1

I=

2

Let

dx 1 = ∫ 7 5 2 x − x+ 2 2

∫

dx    x − 7  − 9  4  16 2

2

 4 x − 7  9 2 cosh −1   , since a =  3  16 2

1

= Example 3.4 Evaluate

.

x

∫

5x − 4 x 2

x = A⋅

dx.

d (5 x 2 − 4 x) + B dx

A ⋅ (10 x − 4) + B Equating like terms, we 10 A = 1 and so A = −4 A + B = 0 and so B =

Then

I =∫ =

=

1 . 10

2 5

1 2 (10 x − 4) + 10 5 5x2 − 4 x

1 2 × 2 5x2 − 4 x + ∫ 10 5 5 1 2 5 x 2 − 4 x+ ∫ 5 5 5

dx x2 −

4 x 5

dx

     x − 2  −  2   5   5   5 x − 2  1 2 5x2 − 4 x + cosh −1  =  + c.  2  5 5 5 Example 3.5 Evaluate

Let

3x − 2

∫

4 x − 8 x + 13 2

2

dx.

d (4 x 2 − 8 x + 13) + B dx = A × (8 x − 8) + B

3x − 2 = A ×

Equating like terms, A =

3 and B – 3 = –2 and so B = 1. 8

2

Integral Calculus

Then

3 (8 x − 8) + 1 8 dx 4 x 2 − 8 x + 13

I=∫ =

=

3.29

3 1 4 x 2 − 8 x + 13 + ∫ 4 2

dx x2 − 2 x +

13 4

dx

3 1 4 x 2 − 8 x + 13 + ∫ 4 2

 3 ( x − 1) 2 +    2 3 1 2( x − 1) 4 x 2 − 8 x + 13 + sinh −1 = +c 4 2 3 5

Example 3.6 Evaluate

2

x−2 dx. 5− x

∫ 2

Multiplying the numerator and denominator of the integrand by 5

( x − 2)

I=∫

( x − 2)(5 − x)

2

Let

5

dx or

∫ 2

x − 2, we get

( x − 2) −10 + 7 x − x 2

x – 2 = A × (7 – 2x) + B

1 3 Equating like terms, A = − and B = 2 2 \

I =∫

1 3 − × (7 − 2 x) + 2 2 −10 + 7 x − x 2

= − −10 + 7 x − x 2 +

3 2∫

dx −10 + 7 x − x 2 5

I = − { ( x − 2)(5 − x)}52 +

3  −1  2 x − 7  sin    3  2 2  3  π  π  =  − −  2  2  2  5

=0+

=

3π 2

3 2 ∫2

dx  3      −  x − 7   2   2  2

2

dx

Engineering Mathematics I

3.30 Example 3.7 Evaluate

∫

I=∫

x(1 − x) dx. x − x 2 dx

= ∫ −( x 2 − x) dx 1  1 = ∫   −  x −  dx  2  2 2

2

   x − 1     2

   1   2 

2

   x − 1    2  x(1 − x) + sin −1  =  1  2 2   2  (2 x − 1) 1 = x − x 2 + sin −1 (2 x − 1) + c 4 8 Example 3.8 Evaluate

∫ ( x + 1)

x 2 − x + 1 dx.

d 2 ( x − x + 1) + B dx = AX (2 x − 1) + B

x +1= A

Let

Equating like terms, we get 2A = 1 and so A = Then

I =∫

{12 f ′( x) + 32} 3

1 { f ( x)} 2 3 = + ∫ 3 2 2 2

1 3 and –A + B = 1 and so B = 2 2

f ( x) dx, where f ( x) = x 2 − x + 1 2      x − 1  +  3  dx  2   2  2

2    3   1  1      x − x −   1 3   3 2  x 2 − x +1 +  2  sinh −1  2  = ( x 2 − x +1) 2 +    3 2 2 2  3    2      1 3 9 2 x −1 = ( x 2 − x +1) + (2 x −1) x 2 − x + 1 + sinh −1  + c  3  3 8 16

Example 3.9 Evaluate

∫ ( x + 1)

I = ∫ ( x + 1)

x+2 dx. x−2 ( x + 2) x2 + 4

dx, on multiplying the numerator and denominator by x + 2.

Integral Calculus Now I = ∫

x 2 + 3x + 2 x2 − 4

3.31

dx. This integral is of form

∫

lx 2 + mx + n ax 2 + bx + c

dx

To evaluate this integral, we proceed as given below d Let x 2 + 3 x + 2 = A( x 2 − 4) + B ⋅ ( x 2 − 4) + C dx 3 and C = 6 2

Equating like terms, we get A = 1, B =

Then

d 2 ( x − 4) dx 3 x 2 − 4 dx + ∫ dx dx + 6 ∫ 2 2 x −4 2 x −4

I =∫

 x 3 x 2 4 x x − 4 − cosh −1   + × 2 x 2 − 4 + 6 cosh −1   2 2 2 2 2   x 2 x x − 4 + 3 x 2 − 4 + 4 cosh −1   + C. =  2 2 =

Example 3.10 Evaluate

Let

x +1=

∫ ( x + 1)

dx x2 + 4x + 2

1 1 1− y and so dx = − 2 dy and x = y y y −

Then

.

=∫

1 y

=−∫ =−∫

1 dy y2

1 − y  1 − y    + 4  + 2  y   y  2

dy (1 − y ) + 4 y (1 − y ) + 2 y 2 2

dy 1+ 2y − y

2

=−∫

dy ( 2) 2 − ( y − 1) 2

 y − 1  −x −1  +c = cos−1   or cos   2( x + 1)   2    Example 3.11 Evaluate

Let

4x + 1 =

∫ (4 x + 1)

dx 1 − x − x2

.

1  1 1 1− y and so dx = − 2 dy and x =  − 1 ÷ 4 or  y 4y 4y y 

Engineering Mathematics I

3.32

− Then

I=∫

1 − y  1 − y  1 1 −  −   4 y   4 y  y

2

−dy

=∫

16 y − 4 y (1 − y ) − (1 − y ) 2 2

dy 19 y − 2 y − 1 1 dy

=−∫ =−

=−

1 dy 4 y2

2

19 1 19

∫ ∫

y2 −

2 1 y− 19 19 dy

2      y − 1  −  20      19   19 

2

      y − 1        19 −1   =− cosh       19 20             19     1

=−

1

=−

1

=−

1

=−

1

Example 3.12 Evaluate

Let

x+2 =

19

19 19

 38 x + 9     cosh −1    + c.   x 19 5(4 1) +     dx

∫ ( x + 2)

x+3

.

1 1 1 and so dx = − 2 dy. Also x + 3 = 1 + y y y 1 dy y2 =−∫ 1 1 +1 y y −

Then

19 y − 1 cosh −1    20    19 × 1 − 1  x + 4 1  cosh −1     20  76 x + 18   cosh −1      20(4 x + 1)  

I =∫

dy y ⋅ y +1

Integral Calculus dy

=−∫

y +y 2

3.33

dy 2 2      y + 1  −  1   2   2 

=−∫

  1 = − cosh −1 (2 y + 1) or − cosh 2 × + 1 x+2    x + 4  = − cosh −1   + c.  x + 2  Example 3.13 Evaluate

∫ 2+

x x

dx.

Let

x = y 2 and so dx = 2 y dy

Then

I =∫

y ⋅ 2 y dy 2+ y

2  4  = 2 ∫  y − 2 +  dy, as the improper function y has y + 2  y+2 been rewritten

 y2  = 2  − 2 y + 4log( y + 2)  2  = x − 4 x + 8log( x + 2) + c Example 3.14 Evaluate

Let

x=

∫ (x

dx 2

+ 1) x 2 − 4

.

1 1 and so dx = − 2 dy y y 1 dy y2 1   + 1 1 − 4  y 2  y 2 −

Then

I =∫

I=∫ Let

Then

− ydy ( y + 1) 1 − 4 y 2 2

1 − 4 y 2 = u 2 and so − 8 ydy = 2u du I =∫

1 u du 4 1 (5 − u 2 ) ⋅ u 4

=∫

du ( 5) 2 − u 2

Engineering Mathematics I

3.34 =

 5 + u  log   2 5  5 − u 

=

2     5 + 1− 4y   log   2   2 5  5 − 1− 4y     

1

1

     5+ 1 log  =   2 5   5−      1 x 5 + = log   2 5  x 5 − 

              x2 − 4   +c 2 x −4   

4 x2 4 1− 2 x 1−

1 3

Example 3.15 Evaluate

∫ (1 + x 0

dx 2

) 1 − x2

.

1 1 and so dx = 2 dy, when x = 0, y = • and when y y 1 x= , y= 3 3

Let

x=

Then

1 dy y2 I=∫  1 1 ∞  1 + 2  1 − 2 y y   −

3

∞

=∫ 3

y dy ( y + 1) y 2 − 1 2

Let y2 – 1 = u2 and so ydy = u du; when y = 3, u = 2 and when y = •, u = • ∞

I=∫ 2

u du (u + 2)u 2

∞

Then

1  −1 u  =   tan 2  2  2 1  π π  π =  −  = 2  2 4  4 2

Integral Calculus EXERCISE 3(c) Part A (Short Answer Questions) 1. Evaluate

2. Evaluate

3. Evaluate

4. Evaluate

dx

∫

a − x2 2

dx

∫

x − a2 2

dx

∫

x + a2 2

dx

∫

x + 2x + 5 2

dx

5. Evaluate

∫

6. Evaluate

∫

7. Evaluate

∫

8. Evaluate

∫ ( x + 4)

9. Evaluate

∫

4 − 2x − x2 dx x + 3x + 1 2

( x + 3)dx 1 − x2 dx

10. Evaluate

x

dx x 1− x

∫

dx 2ax − x 2

Part B 11. Evaluate

∫ 1

12. Evaluate

∫ 0 β

13. Evaluate

∫ α

1 3 dx x(2 − 3 x) xdx 1 + x4 dx ( x − α)(β − x)

(β > α )

3.35

Engineering Mathematics I

3.36

14. Evaluate

∫

15. Evaluate

∫

(2 x + 5) x − 2 x + 10 2

x +1 dx 2x − 3

2

16. Evaluate

∫ 0

dx

2x 3x − x 2 − 2 x2 + 2x + 3

17. Evaluate

∫

18. Evaluate

∫ ( x + 2)

19. Evaluate

∫

20. Evaluate

∫ (2 x + 3)

21. Evaluate

∫ ( x + 3)

x2 + 1

dx

dx

dx x2 + 6x + 7

dx x 7x − 6x −1 2

dx x+5

dx x−5

1 3

22. Evaluate

dx

∫ (1 + x )

1 − x2

2

0

3.6

dx

23. Evaluate

∫ (x

24. Evaluate

∫

25. Evaluate

∫ ( x + 1)

2

− 1) x 2 + 1

7 x − 10 − x 2 dx x 2 + x + 1 dx

INTegraTIoN by parTs

When u and v are function of x, then by the product rule of differentiation, we have d dv du (uv) = u + v dx dx dx viz.,

u

dv d du = (uv) − v dx dx dx

Integration both side w.r.t. x, we have

∫ u dv = uv − ∫ v du

Integral Calculus

∫ u dv

Evaluation of

depends on

∫ v du.

3.37

Thus when we want to evaluate an

integral which is a product, it should be identified as the product of one factor (u) and the differential of another factor v. The choice of u and v should be carefully made so that

∫ v du

is easier then

∫ u dv.

3.6.1  Improper Integrals b

The definite integral I = ∫ f ( x)dx has meaning only when the limits a and b are finite a

and the integrand f(x) is bounded in the interval [a, b]. Now we extend the definition when the range of integration is infinite or when the integrand is unbounded. Definition: If f(t) is bounded and integrable in a ≤ t ≤ x, where a is a constant and z is any number greater than a and if ∞

∫

∫

f (t )dt = F (t ), then

x

f (t )dt = lim ∫ f (t )dt = lim{F ( x) − F (a )} x →∞

a

x →∞

a

is called the improper or infinite integral of f (t), provided the limit exists. If the limits x

exists, the integral is said to converge. On the other hand, if the integral is said to diverge to +• or said not to exist.

∫

f (t )dt → ∞ as x → ∞,

a

b

Similarly lim

x →−∞

∫

f (t )dt = lim {F (b) − F ( x)}, if the limit exists, is denoted by x →−∞

x b

∫

f (t )dt and the infinite integral is said to converge

−∞ ∞

Finally

∫

a

f (t )dt = lim

x →−∞

−∞

∫ x

x

f (t )dt + lim ∫ f (t )dt is defined the infinite integral x→∞

a

in the in the L.S. is said to converge if both the integrals in the R.S. converge, ‘a’ is arbitrary and the value of the integral does not depend on a.

3.6.2  Integral with Unbounded Integrands b

If f(t) is unbounded at t → a in a ≤ t ≤ b, then

∫

b

f (t )dt = lim

a

∈→ 0

∫

a +∈

provided the limit exists. Similarly if f(t) becomes unbounded as to Æ b in a £ t £ b, then b −∈

b

∫ a

f (t )dt = lim

∈→ 0

∫ a

f (t )dt (∈> 0), provided the limit exists.

f (t )dt (∈> 0),

Engineering Mathematics I

3.38

If both and a and b are points of discontinuity, then b

b−∈

c

∫

∫

f (t )dt = lim

∈→ 0

a

∫

f (t )dt + lim

∈→ 0

a +∈

f (t )dt , a < c < b. If f (t) isunboundedataninterior

c + ∈2 c −∈1

b

point c such that a < c < b, then

∫

f (t )dt = lim

∈1 → 0

a

∫

b

f (t )at + lim

∈2 → 0

a

∫

f (t )dt,

c + ∈2

where Œ1 and Œ2 are two arbitrary positive quantities tending to zero independently. Although the two integrals in the R.S. may not exist independently, their sum may exist when Œ1 = Œ2 = Œ. The value of the sum is called the Cauchy Principal value and written as b b  c −∈  P ∫ f (t )dt = lim  ∫ f (t )dt + ∫ f (t )dt  ∈→ 0  a  a c +∈ If f(t) has a finite number of points of infinite discontinuity in (a, b), say, c1, c2, ..., cn where a ≤ c1 ≤ c2  ≤ cn ≤ b then b

∫

c1

c2

b

a

c1

cn

f (t )dt = ∫ f (t )dt + ∫ f (t )dt +  + ∫ f (t )

a

3.6.3  Comparison Tests for Improper Integrals b

Let

∫

f (t )dt be an improper integral. If there exists a g(t) such that | f(t) | £ g(t) in

a b

a ≤ t ≤ b and ∫ g (t )dt converge, then a

b

∫

f (t )dt also converges.

a b

If there exists function g(t) such that f (t ) ≥ g (t ) in a ≤ t ≤ b and ∫ g (t ) dt b

diverges, then

∫

a

f (t )dt also diverges.

a

Limit Form of Comparison Tests Let f(x) > 0 and g(x) > 0 for all x £ a.  f ( x)   = k , where k ≠ 0, then both the integrals If lim  x → +∞  g ( x )    converge or diverge together.

∞

∫

∞

f ( x)dx and ∫ g ( x)dx

a

a

∞

It k = 0, we may conclude only that the convergence of ∞

∫ a

∫ g ( x)dx a

f ( x)dx.

implies that of

Integral Calculus

3.39

Worked example 3(d)

Example 3.1 Evaluate Let Then

∫xe

3 x2

dx.

1 x 2 = t and so x dx = dt 2 1 I = ∫ tet dt 2 1 td(et ) [note : u = t ; dv = et dt. ∴ v = et ] 2∫ 1 = [tet − ∫ et dt ] 2 2 1 2 x2 = [ x e − ex ] + c 2 =

Example 3.2 Evaluate

∫xe

3 2x

dx.

 e2 x  I = ∫ x 3 d    2  2x   ∵ u = x 3 and dv = e 2 x dx and so v = e 2 x dx = e  ∫  2 

1 =  x 3e 2 x − ∫ e 2 x ⋅ 3 x 2 dx  2 2 x e  1 3 = x 3e 2 x − ∫ x 2 d    2  2 2 1 3 = x 3e 2 x −  x 2 e 2 x − ∫ e 2 x ⋅ 2 xdx   2 4  e2 x  1 3 3 = x 3 e 2 x − x 2 e 2 x + ∫ xd    2  2 4 2 1 3 2x 3 2 2x 3  2x x e − x e +  xe − ∫ e 2 x dx   2 4 4 1 3 2x 3 2 2x 3 2x 3 2x = x e − x e + xe − e + c 2 4 4 8 =

Example 3.3 Evaluate

∫x

2

sin 2 x dx.

 cos 2 x  I = ∫ x 2 d −   2   cos 2 x  ∵ u = x 2 and dv = sin 2 x dx and so v = ∫ sin 2dx = −   2 

Engineering Mathematics I

3.40

x2 cos 2 x cos 2 x − ∫ − ⋅ 2 x dx 2 2  sin 2 x  x2 = − cos 2 x + ∫ xd    2  2 =−

x2 1 cos 2 x +  x sin 2 x − ∫ sin 2 x dx   2 2 2 x 1 1 = − cos 2 x + x sin 2 x + cos 2 x + c 2 2 4 =−

Example 3.4 Evaluate

∫ x tan

2

x dx.

I = ∫ x(sec 2 x − 1)dx = ∫ xd(tan x) − = x tan x − ∫ tan xdx − = x tan x − log sec x −

x2 2

x2 +c 2

2

Example 3.5 Evaluate

∫x

n

log x dx.

1

Let

2 2  x n + 1   I = ∫ x n log x dx = ∫ log x ⋅ d   n + 1 1 1 2  xn + 1  xn + 1 1 log x − ∫ =  ⋅ dx  n +1 x  n + 1 1 1 2

 xn + 1    xn + 1    log x −  =    2    ( n 1) + n +1  1   1 n +1 n +1 2 2 1 log 2 − = + 2 n +1 (n + 1) (n + 1) 2 2

Example 3.6 Evaluate Let

log x

∫ ( x + 1)

2

2

dx.

 1  I = ∫ log x ⋅ d −   1 + x  1 1 1 −∫ − ⋅ dx x +1 x +1 x 1 log x 1  =− +  −  dx x + 1 ∫  x x + 1 = − log x ⋅

x2 2

Integral Calculus =−

Example 3.7 Evaluate

3.41

 x  log x + log  + c  x + 1 x +1

∫ (log x)

2

dx.

Note

The integrand is not a product of two factors. So we assume that u = (log x)2 and dx = dv so that v = x. 1 I = ∫ (log x) 2 d( x) = x(log x) 2 − ∫ x ⋅ 2 log x ⋅ dx Then x = x(log x) 2 − 2 ∫ (log x) d ( x)

{

1 = x(log x) 2 − 2 x log x − ∫ x ⋅ dx x = x(log x) 2 − 2 x log x + 2 x + c Example 3.8 Evaluate Let

∫ x sin

−1

}

x dx.

 x2  I = ∫ sin −1 x ⋅ d   2 =

x 2 −1 x2 1 ⋅ sin x − ∫ dx 2 2 1 − x2

=

x 2 −1 1 sin x − ∫ sin 2 θ dθ , on putting x = sin q 2 2

x2 2 x2 = 2 x2 = 2 x2 = 2 =

Example 3.9 Evaluate

∫

I=∫

1 1 − cos 2θ   dθ   2 ∫  2   1 1 sin −1 x − θ − sin 2θ  4 2 sin −1 x −

1 sin −1 x − (θ − sin θ cos θ ) 4 1 sin −1 x − (sin −1 x − x 1 − x 2 ) + c 4 tan −1 x dx. x2  1 tan −1 x ⋅ d −   x

1 1 1 = − tan −1 x + ∫ ⋅ dx x x 1 + x2

Engineering Mathematics I

3.42

1 x  1 = − tan −1 x + ∫  −  dx  x 1 + x 2  x 1 1 = − tan −1 x + log x − log(1 + x 2 ) + c x 2 Example 3.10 Evaluate

∫ sin

−1

x ⋅ dx.

I = ∫ sin −1 x ⋅ d ( x) = x sin −1 x − ∫ x ⋅

= x sin

−1

x−∫

1 1 − x2

dx

1 − (−2 x) 2 dx 1 − x2

1 = x sin −1 x + × 2 1 − x 2 + c 2 = x sin −1 x + 1 − x 2 + c Example 3.11 Evaluate

Let

x + sin x

∫ 1 + cos x dx.

 x x  2 sin cos   x 2 2  dx I=∫ +   2 x 2 x  2 cos  2 cos 2 2  x x x = ∫  sec 2 + tan  dx 2 2 2   x x = ∫  x ⋅ d  tan  + ∫ tan  dx   2 2  x x x = x tan − ∫ tan dx + ∫ tan dx 2 2 2 x = x tan + c 2

Example 3.12 Evaluate (Note

∫

x 2 − a 2 dx.

This integral and

∫

a 2 − x 2 dx and ∫

in the list of standard integrals.) Let

I =∫

x 2 − a 2 d( x)

x 2 + a 2 dx have been included

Integral Calculus = x x2 − a2 − ∫ x ⋅ = x x2 − a2 −

3.43

2x 2 x2 − a2

x2 − a2 + a2 x2 − a

dx

dx

 = x x 2 − a 2 − ∫  x 2 − a 2 + 

  dx x 2 − a 2  a2

 x = x x 2 − a 2 I − a 2 cosh −1   a \ \

 x 2 I = x x 2 − a 2 − a 2 cosh −1   a I=

 x x 2 a2 x − a 2 − cosh −1   + c a 2 2

Example 3.13 Evaluate I1 = ∫ e ax cos bx dx and I 2 = ∫ e ax sin bx dx. Let

 e ax  I1 = ∫ cos bx ⋅ d    a  1 e ax = e ax cos bx − ∫ (−b sin bx)dx a a =

Let

1 ax b e cos bx + I 2 a a

(1)

 e ax  I 2 = ∫ sin bx d    a  1 e ax = e ax sin bx − ∫ × b cos bx dx a a =

1 ax b e sin bx − I1 a a

(2)

Using (2) in (1), we get 1 b 1 b  I1 = e ax cos bx +  e ax sin bx − I1  a a  a a  \ \

2  1 + b  I = 1 e ax cos bx + b e ax sin bx 1  a 2  a a2

I1 =

e ax (a cos bx + b sin bx) a + b2 2

(3)

Engineering Mathematics I

3.44

Similarly, using (1) in (2) and simplifying, we can get I2 =

e ax (a sin bx + b cos bx) a 2 + b2

(4)

(Note Results (3) and (4) can be treated as formulas and remembered as such, as these will be used frequently in later situations) ∞

Example 3.14 Evaluate

∫e

−3 x

sin 4 x dx.

0

Using the formula (4) derived in Example (3.13) above, we have ∞

∫ 0

∞

  e−3 x e−3 x sin 4 x dx =  (−3 sin 4 x − 4 cos 4 x 2 2  (−3) + 4   0 =0−

Example 3.15 Evaluate Let

1 4 (−4) = 25 25

∫ cosh 3x cos 4 x dx.

 e3 x + e−3 x   cos 4 x dx I = ∫    2 1 1 e3 x cos 4 x dx + ∫ e−3 x cos 4 x dx ∫ 2 2 1 e3 x = (3 cos 4 x + 4 sin 4 x) 2 25 1 e−3 x (−3 cos 4 x + 4 sin 4 x) + c + ⋅ 2 25 =

 e3 x − e−3 x  4  e3 x + e−3 x  3  +  cos 4 x  sin 4 x    25   25 2 2 3 4 = cos 4 x sinh 3x + sin 4 x cosh 3 x + c 25 25 =

Example 3.16 Evaluate Let

I=

∫e ∫e

x

x

(sin x + cos x) dx.

sin x dx +

=

∫ sin x ⋅ d (e

=

∫e

x

x

∫e

x

cos x dx

) + ∫ e x cos x dx

sin x − ∫ e x cos x dx + ∫ e x cos x dx

= ex sin x + c

Integral Calculus Example 3.17 Evaluate

Let

I= =

∫e

x

∫e

x

3.45

1   + log x) dx.  x 

1 ⋅ dx + ∫ e x log x dx x

∫ e d (log x) + ∫ e x

x

log x dx

x x x = e log x − ∫ log x d (e ) + ∫ e log x dx x x x = e log x − ∫ e log x dx + ∫ e log x dx = ex log x + c

Example 3.18 Evaluate

∫

e x ( x 2 + 3 x + 3) dx. ( x + 2) 2

   x +1   I = ∫ ex  dx 1 − 2   ( x + 2)     x +1 = ex − ∫ ex ⋅ dx ( x + 2) 2  1  x x  = e + ∫ e ( x + 1) d  x + 2     e x ( x + 1)  1 x x = e +  x + 2 − ∫ x + 2 d {e ( x + 1)}    e x ( x + 1)  1 x − {e x + ( x + 1) e x } = e +   x+2  ( x + 2)   e x ( x + 1)   x + 1  x x  e +c + − e x  =  e =  x+2   x + 2    Example 3.19 Evaluate

Let

∫e

x

 1 + sin x    d x. 1 + cos x  1

I=

∫

ex ⋅

=

∫

 x x e x d  tan  + ∫ e x tan dx  2 2

x 2 cos 2 2

dx + ∫ e x tan

x dx 2

x x x − ∫ tan ⋅ e x dx + ∫ e x tan dx 2 2 2 x x = e tan + c 2 x = e tan

Engineering Mathematics I

3.46

∞

Example 3.20 Examine the convergence of

∫ log x dx. 1

The integrand (log x) is unbounded as x Æ •, viz. at x = b, where b Æ • ∞

b

∫ log x dx

\

log x dx = blim →∞ ∫

1

1

( x log x − x)1b = blim →∞ = lim [b log b - b + 1] Æ • as b Æ • b Æ•

•

\

∫ log x dx diverges to + • 1 ∞

Example 3.21 p π 1. ∞

∫ a

Test the convergence of the integral

∫ a

1 dx, where a > 0 and xp

b

1 1 dx = lim ∫ p dx b →∞ xp x a  x− p + 1    = blim →∞  − p + 1  a b

 − b−( p − 1)  a1 − p  + = blim  →∞   p −1  p −1 Now

 − b−( p − 1)   → 0, if p > 1 lim   b →∞   p −1  ∞

\

(1)

∫ a

1 a1− p dx = by (1), if p > 1 p −1 xp

viz., I converges if p > 1 and diverges to +•, if p < 1 •

If p = 1,

∫ a

1 dx =  lim (log b) − log a  → •, as b Æ •  b→ •  x

\ The integral diverges. 1

Example 3.22

Examine the convergence of

∫ 0

The integrand has an infinity at the lower limit  1 dx   = lim (2 x )1 \ I = lim  ∫ ∈  ∈→ 0 ∈→ 0 x  0 +∈  [2 − 2 ∈ ] = 2 = lim ∈→ 0

dx x

.

Integral Calculus

3.47

\ The given integral converges to 2. 4

Example 3.23 Find the Cauchy’s principal value of the integral

dx

∫ ( x − 3)

3

x = 3 is

2

the interior point of discontinuity for the integrand or has an infinity at x = 3 4

∫ 2

dx = ∈lim 1 →0 ( x − 3)3

3 −∈1

∫ 2

4

dx dx + lim ( x − 3)3 ∈2 → 0 3 ∫ ( x − 3)3 +∈ 2

3 −∈1

  1 −  = ∈lim 2  → 0 1  2 ( x − 3)  2

  1  + lim − 2 ∈2 → 0   2 ( x − 3)  3 − ∈2 4

 1    + − 1  + lim − 1 + 1  = ∈lim 2 2 ∈2 → 0  1 →0   2 2 ∈1   2 2 ∈2  Both the limits do not exist and hence the integral diverges. But if we put Œ1 = Œ2 = Œ, then 1 1 1 1   = 0, which is the Cauchy’s principal − + I = lim  − 2 ∈→ 0  2 2 2 ∈2  2∈  value of the integral. 3

Example 3.24 Test the convergence of

∫ 0

3

Let

∫ 0

dx ( x − 1)

2 3

( x − 1)

2 3

and find its value.

the denoted by f (x) 3

Choose

dx

g(x) =

∫

dx 2

x3 2    f ( x)   x3    = lim   Now xlim 2  →∞  g ( x )  x →∞     ( x − 1) 3    0

2  −   1  3  1 −   = xlim →∞  x   

=0 3

3

∫

g ( x)dx =

0

0

1

0

x3

3

3

∫

∫

f ( x) dx =

∫ 0

2

dx converges. Hence by comparison test, 1 2

( x − 1) 3

also converges.

Engineering Mathematics I

3.48

1  Required value = 3( x − 1) 3   0 3

= 3 ¥ 21/3 – 3(–1)1/3 = 3{1 + 21/3} ∞

Example 3.25 Test the convergence of I =

x

∫ 1+ x

2

dx.

−∞ ∞

0

x x dx + ∫ dx I= ∫ 2 1 + x 1 + x2 −∞ 0 0

= lim

a → −∞

∫ a

b

x x dx + lim ∫ dx 2 →∞ l 1+ x 1 + x2 0

1  1     = lim  log (1 + x 2 ) + lim  log (1 + x 2 ) a → −∞  2 b→ ∞  2  0  a     0

b

 1  1  = lim − log (1 + a 2 ) + lim  log (1 + b 2 ) a→ ∞  b → ∞     2  2 =0

EXERCISE 3(d) Part A (Short Answer Questions) Evaluate the following integrals: 1.

∫ xe

4.

∫

7.

∫ tan

10.

ax

dx

log x dx x2 −1

x dx

2.

∫ x sin mx dx

3

∫ x cosec

5.

∫ log x dx

6.

∫ x cos

8.

∫ sinh

9.

∫ x sec

−1

x dx

2

2

x dx

x dx

−1

x dx

∫ x log x dx

11. Define improper integral. 12. Explain Cauchy’s principal value of an improper integral. 13. When do you say that an improper integral converges or diverges? 14. State the comparison test used in testing convergence of an improper integral. 15. State the limit form of comparison test used for testing convergence of an improper integral.

Integral Calculus

3.49

Part B Evaluate the following integrals: 16.

∫ e ( x − 2)(2 x + 3)dx

19.

∫x

x

∫e

17.

2x

( x + 1) 2 dx

18.

∫x

21.

∫ x cosec

2

sin 2 2 x dx

π/2 2

3

cos x dx

∫ x sin x cos x dx

20.

0

1

22.

x dx

23.

∫x

x 2 + a 2 dx

26.

∫e

28.

∫ e ( x + 1) log x dx

∫ sin

−1

24. ∫

tan −1 x dx

2

∫

27.

∫ e (sec x(1 + tan x)dx x

1

29.

∫ 0

31. Test the convergence of

e x ( x 2 + 1) dx ( x + 1) 2

∫

30.

1 2

sin(3 x + 1)dx

x

2

xe x dx (1 + x) 2

2x + 3

dx

∫ ( x − 1)

.

2

1

a

32. Test the convergence of

dx

∫

a − x2 2

0

.

1

33. Test the convergence of

1 dx. x −1

∫

4

34. Evaluate Cauchy’s principal value of

dx

∫ ( x − 2)

3

.

1

∞

∫e

−x

35. Discuss the convergence of

x 2 dx, using comparison test.

1 ∞

36. Test the convergence of

∫ 0

1

37. Test the convergence

∫ 0

dx . x 1

38. Test the convergence of

∫ 0 ∞

39. Test the convergence of

sin x dx. x

∫ 1

1 + x2 2 x − x2 e− x dx. x

dx.

x dx

a 2 − x 2 dx

0

25.

2

Engineering Mathematics I

3.50

∞

40. Test the convergence of

sin x dx. x2

∫ 1

aNsWers

Exercise 3(a) x − x + log( x + 1) 2 1  sin(m + n) x sin(m − n) x   + (3)  2 m+n m − n  (1)

(5) x – sin x 1 (8) − (n − 1)(log x)n − 1

(2)

x2 − 2 x − log( x − 2) 2

(4) − cot x − cosec x

2(1 + x ) n + 1 (6) n +1 sin 3 x 3 2 3 (12) x + 3x 2 + 2 3 (9)

(11)

1 log( x 2 + 2 x + 8) 2

(13)

2 2 3 (ax + b) 2 + cx + d 3a c

1 (7) − 1 + e x (10) log(sin x − cos x)

 xn  1 14 2 1 x + log( x14 − 4) (15) sin −1  n  a  14 7 a 1 (16) log{cosh x + cosh 2 x + 4} (17) log{x 2 + 1 + x 4 } 2 (14)

−1 (18) sin (log x)

−1  sin x   (19) sin   3 

−1  tan x   (20) sinh   4 

(21)

1 x + sin 2 x 2

1 π (22) −3 cos x − cos 3 x (23) tan −1 2 − 3 4 2 (25) e − 1 π

(28) (31)

1 a2 2 π 2 2

(34) π

1 3 (26) 3 a  1  1 (29) 4 log   3 (32)

 5 1 log   2  3 π

(35)

αβ

(24)

2π

(27) 1 1 2 π (30) a  −   4 2 (33)

3 π 2

Integral Calculus

3.51

Exercise 3(b) (1)

1 −1 x tan a a

(2)

 x − a  1 log    x + a  2a

(4)

1 x + 2x + 5

(5)

1 1 x log( x 2 + 9) + tan −1 2 3 3

(6)

 x2  1 tan −1  2  2 a  2a

 x  (7) log  x + 1  

2

 a + x  1 log    a − x  2a

 sin x  (8) log 1 + sin x   

  1 + tan x   2  (10) log   1 − tan x   2

x (9) log tan 2

(11)

(3)

 2 x 3 + 1  tan1   3 3 3  2

(12)

1 −1  e x + 1  tan   3  3

1 −1  y + 1 2  where y = log x (13) log( y + 2 y + 5) + tan   2  2 (14)

 4 x − 3  3 37 log(2 x 2 − 3 x + 5) + × tan −1   31  4 2 31

2 (15) 2 log(1 + 2 y − y ) +

 2 + y − 1  , where y = sin x log   2 − y + 1 2 2

2 (16) x − log( x + x + 1) +

1

 2 x + 1 2 tan −1   3  3

(17)

 3 − 5  1 2 1  log   − log   3 + 5  2  4 5

(19)

1 1 log x − log( x + 1) + log( x + 2) 2 2

 x − 1  1  − (20) 2 log  + x + 2 x 1   (21)

 2 x − 1 1 1 1 log( x + 1) − log( x 2 − x + 1) + tan −1   3  3 6 3

(22)

 2 x + 1 5 1 log( x − 1) + log( x 2 + x + 1) − 3 tan −1   3  3 6

(23)

1 1 1 log(1 + sin x) + log(1 + sin 2 x) + tan−1 (sin x ) 2 4 2

Engineering Mathematics I

3.52

(24)

 x 1 −1 1 tan x − tan −1    2 3 6

(26)

  2 1 tan −1  tan x   3 2

(28)

π 4

(25)

 5  tan −1  tan x  2  2 5 1

    x  2 2 + 3  tan − 1  1  2   log    (27)    x 2 6  2 2 − 3  tan − 1     2    

   2 tan x  π   2 (29) 2 x + log(3 cos x + 2 sin x + 4) − 2 3 tan   (30)  4 3  −1

Exercise 3(c) −1 x (2) cosh a

−1 x (3) sinh a

−1  x + 1  (4) sinh  2 

−1  x + 1  (5) sin   5 

−1  2 x + 3   (6) cosh  5 

2 −1 (7) − 1 − x + 3 sin x

−1 (8) tan

−1 (1) sin

x a

−1  x − a   (10) sin  a 

(11)

x 2

1 sin −1 (3 x − 1) 3

1 + 1 − x    (9) − log  1 − 1 − x  1 (12) 2 log(1 + 2 )

2 −1  x − 1  (14) 2 x − 2 x + 10 + 7 sinh   3 

(13) π (15)

 4 x − 1 1 5 cosh −1  2x2 − x − 3 +   5  2 4 2

(17)

x 2 5 x + 1 + 2 x 2 + 1 + sinh −1 x 2 2

(16) 3π

  −1    −( x + 1)  −1 1 + 3 x    (18) cos  (19) − sin     4 x  2 x + 2 ( )      1  1 1 1 1   − log + + + (20)  2x + 3 (2 x + 3) 2 7(2 x + 3)  14 14  (21) −

(23)

 16  1 sin −1  − 1  x + 3  8

(22)

 x 2 − x2 + 1    1   log   2   2   x 2 + x + 1   

π 4 2

Integral Calculus

3.53

 2 x − 7  9 −1  2 x − 7  2   7 x − 10 − x + sin  (24)   3  4  8 (25)

 2 x + 1 3 1 2 1 3 ( x + x + 1) 2 + (2 x + 1) x 2 + x + 1 + sinh −1   3  3 3 16

Exercise 3(d) 1 ax 1 ax (1) a xe − a 2 e

(2) −

(3) −x cot x + log sin x

1 1  (4) − x log x + x 

(5) x log x − x

x2 x 1 (6) 4 + 4 sin 2 x − 8 cos 2 x

x cos mx sin mx + m m2

1 −1 2 −1 2 (7) x tan x − log(1 + x ) (8) x sinh x − x + 1 2 x2 1 2 −1 (9) 2 sec x − 2 x − 1

x2 x2 x log − (10) 2 4

x 2 (16) e (2 x − 5 x − 1)

(17)

e2 x (2 x 2 + 2 x + 1) 4

(18)

(20)

π 8

x 1 (21) − cot 3 x + log sin 3 x 3 9

1 3 1 2 1 1 x − x sin 4 x − x cos 4 x + sin 4 x 6 8 16 64   1 2 sin 3 x cos 3x  sin 3x    (19) 4  x (3 sin x + 8 − 2 x −3 cos x − 9  + 2 −3 sin x − 27   

π (22) 2 − 1

(23)

x3 x2 1 tan −1 x − + log(1 + x 2 ) 3 6 6

x 2 a 2 −1  x  2 (24) 2 a − x + 2 sin  a  (26)

(25)

1 2x + 3 e [2 sin(3 x + 1) − 3 cos(3 x + 1)] 13

x (27) e sec x 1 2 (30) e 3

x (28) e ( x log x −1)

(32)

(33) convergent to 0

(34) (36) (38) (40)

π convergent to 2 3 8 convergent convergent convergent

(31) divergent

(35) convergent (37) divergent (39) convergent

 x x 2 a2 x + a 2 + sinh −1   a 2 2

e (29) 2 − 1

Unit

4

Multiple Integrals 4.1

intRODUCtiOn

When a function f(x) is integrated with respect to x between the limits a and b, we get b

the definite integral

∫

f (x)d x .

a

If the integrand is a function f (x,y) and if it is integrated with respect to x and y repeatedly between the limits x0 and x1 (for x) and between the limits y0 and y1 (for y), y1 x1

we get a double integral that is denoted by the symbol ∫

∫

f (x, y )dx d y .

y0 x0

Extending the concept of double integral one step further, we get the triple integral z1

y1 x1

∫∫∫

f (x, y, z ) dx dy dz

z0 y0 x0

4.2

EVALUAtiOn OF DOUBLE AnD tRiPLE intEGRALS y1 x1

To evaluate

∫∫

f (x, y ) dx dy, we first integrate f (x, y) with respect to x partially,

y0 x0

i.e. treating y as a constant temporarily, between x0 and x1. The resulting function got after the inner integration and substitution of limits will be a function of y. Then we integrate this function of y with respect to y between the limits y0 and y1 as usual. The order in which the integrations are performed in the double integral is illustrated in Fig. 4.1.

Engineering Mathematics I

4.2

y1

x1

y0

x0

f (x, y) dx

dy

Fig. 4.1

Note

Since the resulting function got after evaluating the inner integral is to be a function of y, the limits x0 and x1 may be either constants or functions of y. The order in which the integrations are performed in a triple integral is illustrated in Fig. 4.2.

z1

y1

x1 f (x, y, z) dx

z0

dy

dz

x0

y0

Fig. 4.2

When we first perform the innermost integration with respect to x, we treat y and z as constants temporarily. The limits x0 and x1 may be constants or functions of y and z, so that the resulting function got after the innermost integration may be a function of y and z. Then we perform the middle integration with respect to y, treating z as a constant temporarily. The limits y0 and y1 may be constants or functions of z, so that the resulting function got after the middle integration may be a function of z only. Finally we perform the outermost integration with respect to z between the constant limits z0 and z1. y1

Note

Sometimes

∫∫ y0

z1

x1

x0

y1 x1

∫∫∫

f (x, y, z )dx dy dz is also denoted as

z0 y0 x0

y1

x1

y0

x0

∫ dy ∫

f ( x, y )dx dy is also denoted as z1

y1

x1

z0

y0

x0

∫ dz ∫ dy ∫

f ( x, y )dx and

f (x, y, z ) dx . If these

notations are used to denote the double and triple integrals, the integrations are performed from right to left in order.

4.3

REGiOn OF intEGRAtiOn d φ2 ( y)

Consider the double integral

∫ ∫ c φ1 ( y )

to φ 2(y) and y varies from c to d. i.e. φ 1(y) ≤ x ≤ φ 2(y) and c ≤ y ≤ d.

f (x, y ) dx dy. As stated above x varies from φ 1(y)

Multiple Integrals

4.3

These inequalities determine a region in the xy-plane, whose boundaries are the curves x = φ 1(y), x = φ 2(y) and the lines y = c, y = d and which is shown in Fig. 4.3. This region ABCD is known as the region of integration of the above double integral. y

D

x=

C

y=d

x=

1 (y)

A

2 (y)

B

y=c

x

o Fig. 4.3 b φ 2 (x )

Similarly, for the double integral

∫ ∫

f (x, y )dy dx, the region of integration

a φ 1 (x )

ABCD, whose boundaries are the curves y = φ 1(x), y = φ 2(x) and the lines x = a, x = b, is shown in Fig. 4.4. y y=

2 (x)

x=a

x=b

y=

o

1 (x)

x

Fig. 4.4 z2 ψ2 (z) φ2 (y, z)

f (x, y, z) dx dy dz, the inequalities φ 1(y, z) ≤ x

For the triple integral

z1 ψ1 (z) φ1 (y, z)

≤ φ 2(y, z); ψ1(z) ≤ y ≤ ψ2(z); z1 ≤ z ≤ z2 hold good. These inequalities determine a domain in space whose boundaries are the surfaces x = φ 1(y, z), x = φ 2(y, z), y = ψ1(z), y = ψ2(z), z = z1 and z = z2. This domain is called the domain of integration of the above triple integral. WORKED EXAMPLE 4(a) 1

1

∫∫

(x 2 + y 2 ) dx dy = ∫

0

0

2

Example 4.1 Verify that

1

2  1  L.S. = ∫  ∫ (x 2 + y 2 ) dx  dy  1  0

2

∫ (x 1

2

+ y 2 ) dy dx .

Engineering Mathematics I

4.4 2

=∫ 1

Note

x =1

 x3  + y2  3 

 x dy   x=0

y is treated a constant during inner integration with respect to x. 2  y y3 1  8 = ∫  + y 2 dy =  +  =  3    3 3  1 3 1 2

1  2  R.S. = ∫  ∫ (x 2 + y 2 ) d y  d x  0  1 y=2

1  y3  = ∫  x2 y +  dx  3  y =1 0 

Note

x is treated a constant during inner integration with respect to y. 1  x3 7   7 8 = ∫  x 2 + dx =  + x =  3  3 3  0 3 0 1

Thus the two double integrals are equal.

Note

From the above problem we note the following fact: If the limits of integration in a double integral are constants, then the order of integration is immaterial, provided the relevant limits are taken for the concerned variable and the integrand is continuous in the region of integration. This result holds good for a triple integral also. 2π π

Example 4.2 Evaluate

a

∫∫∫r 0

0

2π

4

sin φ d r dφ dθ .

0

π

a

The given integral = ∫ dθ ∫ dφ ∫ r 4 sin φ d r 0

0

2π

π

= ∫ dθ ∫ 0

0 5 2π

=

a 5

0 a

 r    sin φ dφ  5  5

0

π

∫ dθ ∫ sin φ dφ 0

0

5 2π

=

a 5

∫ ( −cos φ ) 0

2 = a5 5

2π

∫ dθ

4 = πa 5 5

0

π 0

dθ

Multiple Integrals 1 + y2

1

∫ ∫

Example 4.3 Evaluate

0

1

The given integral = ∫ 0

1

=∫ 0

    

0

1 + y2

∫ 0

dx dy . 1 + x2 + y 2

  1 d x  dy (1 + y 2 ) + x 2   x = 1 + y2

  x 1 −1   tan   2 2  1 + y 1 + y  x = 0 1

π 4

∫

=

π 4

 2   log y + 1 + y   0

=

π log (1 + 2 ) 4

=

dy

dy 1 + y2

0

(

1

Example 4.4 Evaluate

4.5

)

1

x

∫ ∫ xy (x + y) dx dy. 0

x

Since the limits for the inner integral are functions of x, the variable of inner integration should be y. Effecting this change, the given integral I becomes 1  x  I = ∫  ∫ xy (x + y ) dy  dx  0  x

y= x

1  y2 y3  = ∫  x 2 + x  dx 3  y = x  2 0

1   x3 1   x4 x 4  = ∫  + x 5 / 2  −  +  dx   2 3 3  2 0 

 x4 2 7 / 2 x 5  x −  =  + 21 6 0 8 1

=

1 2 1 3 + − = 8 21 6 56

Engineering Mathematics I

4.6

1− z

1

∫ ∫ ∫

Example 4.5 Evaluate

0

0

1 1− z

The given integral = ∫

∫

0

0

= = = =

1− y − z

xyz dx dy dz .

0 1− y − z

 x2  yz    2  0

dy dz

1 1− z

1 2

∫∫

yz (1 − y − z ) 2 dy dz

0

0

1 1− z

1 2

∫∫ 0

0 y =1− z

3 4 2   z (1 − z ) 2 y − 2 z (1 − z ) y + z y   3 4  y = 0 2 

1

1 2

∫ 0

1

1

1 2

yz {(1 − z ) 2 − 2 (1 − z )y + y 2 } dy dz

∫  2 z (1 − z ) 0

4

−

dz

 2 1 z (1 − z ) 4 + z (1 − z ) 4  dz  3 4

1 1 1 2 1 =  − +  ∫ z (1 − z ) 4 dz 2  2 3 4  0

=

1

1 24

∫ {1 − (1 − z )} (1 − z )

4

dz

0

 (1 − z )5 (1 − z )6    +  −5 6  0  1  1 1  1 =  − = 24  5 6  720

=

1

1 24

log 2 x x + y

∫∫ ∫

Example 4.6 Evaluate

0

0

e x + y + z dx dy dz .

0

Since the upper limit for the innermost integration is a function of x, y, the corresponding variable of integration should be z. Since the upper limit for the middle integration is a function of x, the corresponding variable of integration should be y. The variable of integration for the outermost integration is then x. Effecting these changes, the given triple integral I becomes, log 2 x x + y

I=

∫∫ ∫ 0 log 2

=

∫ 0 log 2

=

∫ 0

0

e x + y + z dz d y d x

0 x

dx ∫ dy e x + y (e z ) zz == 0x + y 0 x

dx ∫ (e 2 x + 2 y − e x + y ) dy 0

Multiple Integrals log 2

=

∫ 0 log 2

=

4.7 y=x

  e2 y dx e 2 x ⋅ − e x ⋅ e y    2 y=0 1

∫  2 e 0

4x

 3 − e 2 x + e x  dx  2

1  3 =  e 4 x − e 2 x + e x   8 0 4

log 2

= Example 4.7 Evaluate

5 8

∫ ∫ xy dx dy , where R is the region bounded by the line R

x + 2y = 2, lying in the first quadrant. We draw a rough sketch of the boundaries of R and identify R. The boundaries of R are the lines x = 0, y = 0 and the segment of the line x y + = 1 lying in the first quadrant. 2 1 Now R is the region as shown in Fig. 4.5. y

B

C y x + =1 2 1 Q (x1, y)

(x0, y) P R O

A

x

Fig. 4.5

Since the limits of the variables of integration are not given in the problem and to be fixed by us, we can choose the order of integration arbitrarily. Let us integrate with respect to x first and then with respect to y. Then the integral I becomes I = ∫  ∫ xy dx  dy   R

When we perform the inner integration with respect to x, we have to treat y as a constant temporarily and find the limits for x. Geometrically, treating y = constant is equivalent to drawing a line parallel to the x-axis arbitrarily lying within the region of integration R as shown in the figure. Finding the limits for x (while y is a constant) is equivalent to finding the variation of the x co-ordinate of any point on the line PQ. We assume that the y co-ordinates of all points on PQ are y each (since y is constant on PQ) and P ≡ (x0, y) and Q ≡ (x1, y).

Engineering Mathematics I

4.8

Thus x varies from x0 to x1. Wherever the line PQ has been drawn, the left end P lies on the y-axis and hence x0 = 0 and the right end Q lies on the line x + 2y = 2, and hence x1 + 2y = 2 i.e. x1 = 2 − 2y. Thus the limits for the variable x of inner integration are 0 and 2 − 2y. When we go to the outer integration, we have to find the limits for y. Geometrically we have to find the variation of the line PQ, so that the region R is fully covered. To sweep the entire area of the region R, PQ has to start from the position OA where y = 0, move parallel to itself and go up to the position BC where y = 1. Thus the limits for y are 0 and 1. 1 2− 2y

∴

I=∫

∫

0

0

1

=∫ 0 1

=∫ 0

xy dx dy 2− 2y

 x2  y    2 

dy

0

y ( 2 − 2 y ) 2 dy 2 1

= 2 ∫ y (1 − y ) 2 dy 0

 y2 y3 y4  = 2  − 2 +  3 4 0 2 1

=

1 6

4.3.1 Aliter Let us integrate with respect to y first and then with respect to x. Then I = ∫  ∫ xy dy  dx   R

As explained above, to find the limits for y, we draw a line parallel to the y-axis (x = constant) in the region of integration and note the variation of y on this line y B Q (x, y1)

y x + =1 2 1

C

R o

P

(x, y0) Fig. 4.6

A

x

Multiple Integrals

4.9

P(x, y0) lies on the x-axis. ∴ y0 = 0 1 ∴ y1 = (2 − x) 2

Q(x, y1) lies on the line x + 2y = 2.

1 (2 − x) . 2 To cover the region of integration OAB, the line PQ has to vary from OB (x = 0) to AC (x = 2) ∴ The limits for x are 0 and 2. i.e., the limits for y are 0 and

1 ( 2− x ) 2

2

I=∫

∴

∫

0

xy dy dx

0 1

 y 2 2 x    2 0

2

=∫ 0

( 2− x )

dx

2

=

1 x (2 − x) 2 dx 8 ∫0

x3 x4  1 x2 = 4 − 4 +  8 2 3 4 0 1 = 6

2

Example 4.8 Evaluate

∫∫ R

e −y dx dy , by choosing the order of integration suitably, y

given that R is the region bounded by the lines x = 0, x = y and y = ∞. Q (x, ∞)

y

y=∞ (0, y) A x=0

B (y, y) R

y=x

P (x, x) x

o Fig. 4.7

Let

e −y dx d y y

I = ∫∫ R

Suppose we wish to integrate with respect to y first. ∞ ∞

Then

I=∫

∫

0

x

e −y dy d x y

Engineering Mathematics I

4.10

We note that the choice of order of integration is wrong, as the inner integration cannot be performed. Hence we try to integrate with respect to x first. Then

∞

y

I=∫

∫

0

0

∞

=∫ 0

e− y dx dy y

e− y (x)0y dy y

∞

= ∫ e− y dy 0

= (e− y )∞ = 1 0

Note

From this example, we note that the choice of order of integration sometimes depends on the function to be integrated. Example 4.9 Evaluate

∫∫ xy dx dy , where R is the region bounded by the parabola R

y2 = x and the lines y = 0 and x + y = 2, lying in the first quadrant. R is the region OABCDE shown in Fig. 4.8. y

y2 = x

D Q1

C x+y=2 Q2 F

E

P1

O

A

P2

B

x

Fig. 4.8

Suppose we wish to integrate with respect to y first. Then we will draw an arbitrary line parallel to y-axis (x = constant). We note that such a line does not intersect the region of integration in the same fashion throughout. If the line is drawn in the region OADE, the upper end of the line will lie on the parabola y2 = x; on the other hand, if it is drawn in the region ABCD, the upper end of the line will lie on the line x + y = 2. Hence in order to cover the entire region R, it should be divided into two, namely, OADE and ABCD and the line P1 Q1 should move from the y-axis to AD and the line P2 Q2 should move from AD to BF. Accordingly, the given integral I is given by 1

2− x

x

2

I =∫

∫

xy dy dx + ∫

∫

0

0

1

0

xy dy dx

Multiple Integrals [

4.11

∴

the co-ordinates of D are (1, 1) and so the equation of AD is x = 1] 1 5 3 I= + = 6 24 8 Note This approach results in splitting the double integral into two and evaluating two double integrals. On the other hand, had we integrated with respect to x first, the problem would have been solved in a simpler way as indicated below. [Refer to Fig. 4.9] y

y2 = x D E

(y2,

C

y) P

Q (2 – y, y) x+y=2

O

x

B

Fig. 4.9 1 2− y

I=∫ 0

∫

xy dx dy

y2

1

=

1 y {(2 − y ) 2 − y 4 }dy 2 ∫0

=

1 (4 y − 4 y 2 + y 3 − y 5 ) dy 2 ∫0

1

3 8 Note From this example, we note that the choice of order of integration is to be made by considering the region of integration so as to simplify the evaluation. =

Example 4.10 Evaluate

∫∫∫ (x + y + z ) dx dy dz,

where V is the volume of the

V

rectangular parallelopiped bounded by x = 0, x = a, y = 0, y = b, z = 0 and z = c. z C

A' Q

B'

O' O

B P

x

A

C'

Fig. 4.10

y

Engineering Mathematics I

4.12

The region of integration is the volume of the parallelopiped shown in Fig. 4.10, in which OA = a, OB = b, OC = c. Since the limits of the variables of integration are not given, we can choose the order of integration arbitrarily. Let us take the given integral I as I = ∫∫∫ (x + y + z ) dz dy dx V

The innermost integration is to be done with respect to z, treating x and y as constants. Geometrically, x = constant and y = constant jointly represent a line parallel to the z-axis. Hence we draw an arbitrary line PQ in the region of integration and we note the variation of z on this line so as to cover the entire volume. In this problem, z varies from 0 to c. since P ≡ (x, y, 0) and Q ≡ (x, y, c) Having performed the innermost integration with respect to z between the limits 0 and c, we get a double integral. As P take all positions inside the rectangle OAC′B in the xy-plane, the line PQ covers the entire volume of the parallelopiped. Hence, the double integral got after the innermost integration is to be evaluated over the plane region OAC′B. The limits for the double integral can be easily seen to be 0 and b (for y) and 0 and a (for x). a

∴

I=∫

b

∫ ∫ (x + y + z ) dz dy dx

0 a

=∫ 0 a

=∫ 0

c

0

0

   z2  ∫ (x + y)z + 2  dy dx z = 0 0  b

c

   c2    ( + ) c x y +   dy dx ∫   2   0  b

a    y2  c2  =∫ cx +  y + c   dx     2 2   0 0  b

a  bc 2 b 2 c  = ∫ bcx +  dx +  2 2  0

 x 2 bc  = bc + (b + c)x  2  0 2

a

= Example 4.11 Evaluate

abc (a + b + c) 2

∫ ∫ ∫ dx dy dz, V

where V is the finite region of space

(terra-hedron) formed by the planes x = 0, y = 0, z = 0 and 2x + 3y + 4z = 12.

Multiple Integrals

4.13

z C

(x, y, z1)

Q O

y

B P (x, y, 0)

A x

Fig. 4.11

Let I = the given integral.

∫ ∫ ∫ dz dy dx

Let I =

V

The limits for z, the variable of the innermost integral, are 0 and z1, where (x, y, z1) lies on the plane 2x + 3y + 4z = 12. [Refer to Fig. 4.11] 1 z1 = (12 − 2 x − 3 y ) 4

∴

After performing the innermost integration, the resulting double integral is evaluated over the orthogonal projection of the plane ABC on the xy-plane, i.e. over the triangular region OAB in the xy-plane as shown in Fig. 4.12. In the double integral, the limits for y are 0 1 y and (12 − 2 x) and those for x are 0 and 6. 3 B E

6

∴

I = ∫ dx 0

∫ dy ∫ 0

6

=

1 1 (12 − 2 x ) (12 − 2 x − 3 y ) 3 4

1 dx 4 ∫0

dz

0

O

1 (12 − 2 x ) 3

∫

(12 − 2 x − 3 y ) dy

0 y=

1

6  1 3 y 2  3 = ∫ dx  (12 − 2 x)y −  4 0 2  y =0 6

=

1 (12 − 2 x) 2 dx 24 ∫0

 (6 − x ) 3   1  =     6 3 −    0 6

= 12

x y + =1 6 4

(12 − 2 x )

D

A Fig. 4.12

x

Engineering Mathematics I

4.14 Example 4.12 Evaluate

dz dy dx

∫∫∫

, where V is the region of space 1− x 2 − y 2 − z 2 bounded by the co-ordinate planes and the sphere x2 + y2 + z2 = 1 and contained in the positive octant. V

z C (x, y, z1) Q O

y

B

P (x, y, 0) A x Fig. 4.13

Note

In two dimensions, the x and y-axes divide the entire xy-plane into 4 quadrants. The quadrant containing the positive x and the positive y-axes is called the positive quadrant. Similarly in three dimensions the xy, yz and zx-planes divide the entire space into 8 parts, called octants. The octant containing the positive x, y and z-axes is called the positive octant. The region of space V given in this problem is shown in Fig. 4.13. dz dy dx Let I = ∫∫∫ 1− x 2 − y 2 − z 2 V To find the limits for z, we draw a line PQ parallel to the z-axis cutting the volume of integration. The limits for z and 0 and z1, where (x, y, z1) lies on the sphere x2 + y2 + z2 = 1 z1 = 1− x 2 − y 2

(

∴

∴

the point Q lies in the positive octant)

After performing the innermost integration, the resulting double integral is evaluated over the orthogonal projection of the spherical surface on the xy-plane, i.e. over the circular region lying in the positive quadrant as shown in Fig. 4.14. y

B

E x2 + y2 = 1

O

D Fig. 4.14

A

x

Multiple Integrals

4.15

2 In the double integral, the limits for y are 0 and 1− x and those for x are 0 and 1. ∴ 1− x 2 1− x 2 − y 2 1 dz I = ∫ dx ∫ dy ∫ 2 (1− x − y 2 ) − z 2 0 0 0 1− x 2

1

= ∫ dx

∫

0

0 1

π = ∫ dx 2 0

z = 1− x 2 − y 2

  z   dy sin − 1 2 2   1− x − y z = 0 

1− x 2

∫

dy

0

1

=

π 1− x 2 dx 2 ∫0

=

π  x 1 2 −1   1− x + sin x 0 2 2 2 1

π π π2 = × = 2 4 8 EXERCiSE 4(a)

Part A (Short Answer Questions) 2

1. Evaluate 2. Evaluate

1

∫ ∫ 4 xy dx dy. 0

0

b

a

∫∫ 1

1

dx dy . xy

π /2 π /2

3. Evaluate 4. Evaluate

∫ ∫ sin (θ + φ) dθ dφ. 0

0

1

x

∫ ∫ dx dy . 0 0 π sin θ

5. Evaluate 6. Evaluate 7. Evaluate

∫ ∫ r dr dθ. 0

0

1

2

3

∫ ∫ ∫ xyz dx dy dz. 0

0

1

z y+z

0

∫∫ ∫ 0

0

0

dz dy dx.

Engineering Mathematics I

4.16

Sketch roughly the region of integration for the following double integrals: b

8.

a

∫∫

f ( x, y ) dx dy.

−b−a 1

9. 10.

x

∫∫

f ( x, y ) dx dy.

0

0

a

a2 − x2

∫ ∫ 0

f ( x, y ) dx dy.

0

a (b − y ) b b

11.

∫ ∫ 0

f ( x, y ) dx dy.

0

Find the limits of integration in the double integral

∫∫

f ( x, y ) dx dy , where R is in the

R

first quadrant and bounded by 12. x = 0, y = 0, x + y = 1 x2 y 2 13. x = 0, y = 0, 2 + 2 = 1 a b 14. x = 0, x = y, y = 1 15. x = 1, y = 0, y2 = 4x

Part B 4

16. Evaluate

y dx d y and also sketch the region of integration roughly. x2 + y 2 y2 / 4

∫∫ 0

a

17. Evaluate

y

a2 − x2

∫ ∫ 1

18. Evaluate

19. Evaluate

y dx dy and also sketch the region of integration roughly.

a−x

0 1

∫∫

y dx dy and also sketch the region of integration roughly. x2 + y 2

0

x

a

a2 − x2

∫ ∫ 0

a 2 − x 2 − y 2 dx dy.

0

1 1− x 1− x − y

20. Evaluate

∫∫ ∫ 0

21. Evaluate

0

xyz dx dy dz.

0

log 2 x x + log y x+ y+z

∫ ∫∫ 0

0 0

e

dz d y d x .

Multiple Integrals

22. Evaluate y = x. 23. Evaluate

−

∫∫ x e

x2 y

4.17

dx dy , over the region bounded by x = 0, x = ∞, y = 0 and

∫∫ xy dx dy ,

over the region in the positive quadrant bounded

by the line 2x + 3y = 6. 24. Evaluate

∫∫ x dx dy ,

over the region in the positive quadrant bounded

2

by the circle x − 2ax + y2 = 0. 25. Evaluate

∫∫ ( x + y) dx dy, over the region in the positive quadrant bounded

by the ellipse 26. Evaluate

x2 y 2 + =1 . a 2 b2

∫∫ ( x

2

+ y 2 ) dx dy , over the area bounded by the parabola y2 = 4x

and its latus rectum. 27. Evaluate

∫∫ x

2

dx dy , where R is the region bounded by the hyperbola xy = 4,

R

y = 0, x = 1 and x = 2. 28. Evaluate ∫∫∫ ( xy + yz + zx) dx dy dz, where V is the region of space bounV

ded by x = 0, x = 1, y = 0, y = 2, z = 0 and z = 3. dx dy dz 29. Evaluate ∫∫∫ , where V is the region of space bounded by ( x + y + z +1)3 V x = 0, y = 0, z = 0 and x + y + z = 1. 30. Evaluate

∫∫∫ xyz dx dy dz, where V is the region of space bounded by the V

co-ordinate planes and the sphere x2 + y2 + z2 = 1 and contained in the positive octant.

4.4

CHAnGE OF ORDER OF intEGRAtiOn in A DOUBLE intEGRAL

In worked example (1) of the previous section, we have observed that if the limits of integration in a double integral are constants, then the order of integration can be changed, provided the relevant limits are taken for the concerned variables. But when the limits for inner integration are functions of a variable, the change in the order of integration will result in changes in the limits of integration. d g2 ( y )

i.e. the double integral

∫ ∫

c g1 ( y )

b h2 ( x )

f ( x, y ) dx dy will take the form

∫ ∫

f ( x , y ) d y dx ,

a h1 ( x )

when the order of integration is changed. This process of converting a given double integral into its equivalent double integral by changing the order of integration is often called change of order of integration. To effect the change of order of integration, the region of integration is identified first, a rough sketch of the region is drawn and then the new limits are fixed, as illustrated in the following worked examples.

Engineering Mathematics I

4.18

4.5

PLAnE AREA AS DOUBLE intEGRAL

Plane area enclosed by one or more curves can be expressed as a double integral both in Cartesian coordinates and in polar coordinates. The formulas for plane areas in both the systems are derived below: (i) Cartesian System y

Let R be the plane region, the area of which is required. Let us divide the area into a large number of elemental areas like PQRS (shaded) by drawing lines parallel to the y-axis at intervals of ∆x and lines parallel to the x-axis at intervals of ∆y (Fig. 4.15). Area of the elemental rectangle PQRS = ∆x. ∆y. Required area A of the region R is the sum of elemental areas like PQRS. viz.,

S

R

P

Q

A

o

x Fig. 4.15

A = lim (ΣΣ ∆x ∆y ) ∆x → 0 ∆y → 0

= ∫∫ dx dy R

(ii) Polar System R

We divide the area A of the given region R into a large number of elemental curvilinear rectangular areas like PQRS (shaded) by drawing radial lines and concentric circular arcs, where P and R have polar coordinates (r, θ) and (r + ∆r, θ + ∆θ) (Fig. 4.16) Area of the element PQRS = r ∆r ∆θ ∴ ( PS = r ∆θ and PQ = ∆r)

S

Q P

O

x Fig. 4.16

∴ Required area A = lim (ΣΣ r ∆r ∆θ ) ∆r → 0 ∆θ → 0

= ∫∫ r dr dθ. R

4.5.1

Change of Variables

(i) From Cartesian Coordinates to Plane Polar Coordinates If the transformations x = x(u, v) and y = y (u, v) are made in the double integral ∂ ( x, y ) ∫∫ f ( x, y) dx dy , then f ( x, y) ≡ g (u, v) and dx dy = |J| du dv, where J = ∂ (u, v) . [Refer to properties of Jacobians in the Unit 2, “Functions of Several Variables”].

Multiple Integrals

4.19

When we transform from cartesian system to plane polar system, x = r cos θ and y = r sin θ ∂x J = ∂r ∂y ∂r

In this case,

∂x ∂θ = cos θ − r sin θ ∂y sin θ r cos θ ∂θ

= r (cos 2 θ + sin 2 θ ) = r

∫∫

Hence

f ( x, y ) dx dy = ∫

R

∫ g ( r , θ ) r dr dθ R

In particular, Area A of the plane region R is given by A = ∫∫ dx dy = ∫∫ r dr dθ R

R

(ii) From Three Dimensional Cartesians to Cylindrical Coordinates y

Let us first define cylindrical coordinates of a point in space and derive the relations P (x, y, z) between cartesian and cylindrical coordinates (Fig. 4.17). Let P be the point (x, y, z) in Cartesian O y coordinate system. Let PM be drawn ┴ r z θ to the xoy-plane and MN parallel to Oy. Let r NOM =θ and OM = r. The triplet (r, θ, z) are M N called the cylindrical coordinates of P. x Clearly, ON = x = r cos θ ; NM = y = r sin Fig. 4.17 θ and MP = z. Thus the transformations from three dimensional cartesians to cylindrical coordinates are x = r cos θ, y = r sin θ, z = z. In this case, xr ∂ ( x, y , z ) J= = yr ∂ (r , θ, z ) zr

xθ yθ zθ

xz cos θ − r sin θ 0 y z = sin θ r cos θ 0 zz 0 0 1

=r Hence dx dy dz = r dr dθ dz and

∫∫∫

f ( x, y, z ) dx dy dz = ∫∫∫ g (r , θ , z ) r dr dθ dz

V

V

Engineering Mathematics I

4.20

In particular, the volume of a region of space V is given by

∫∫∫ dx dy dz = ∫∫∫ V

Note

Whenever

∫∫∫

r dr dθ dz

V

f (x, y, z) dx dy dz is to be evaluated throughout the

volume of a right circular cylinder, it will be advantageous to evaluate the corresponding triple integral in cylindrical coordinates. (iii) From Three Dimensional Cartesians to Spherical Polar Coordinates

r

Let us first define spherical polar coordinates of a point in space and derive the relations between Cartesian and z spherical polar coordinates (Fig. 4.18). Let P be the point whose Cartesian P (x, y, z) coordinates are (x, y, z). Let PM be θ drawn ┴ r to the xOy-plane. Let MN be parallel to y-axis. Let OP = r, the y O angle made by OP with the positive z-axis = θ and the angle made by OM with x-axis = φ . M The triplet (r, θ, φ) are called the N x spherical polar coordinates of P. Since OMP = 90° , MP = z = r cos Fig. 4.18 θ, OM = r sin θ, ON = x = r sin θ cos φ and NM = y = r sin θ sin φ . Thus the transformations from three dimensional cartesians to spherical polar coordinates are x = r sin θ cos φ , y = r sin θ sin φ , z = r cos θ In this case, ∂ ( x, y , z ) 2 J= = r sin θ ∂ (r , θ, φ ) [Refer to example (2.8) of Worked example set 2(b) in Unit 2 “Functions of Several Variables.”] Hence dx dy dz = r2 sinθ dr dθ d φ and

∫∫∫

f (x, y, z ) dx dy dz =

V

∫∫∫ g (r , θ, φ) r

2

V

sin θ dr dθ d φ . In particular, the volume of a region of space V is given by

∫∫∫ dx dy dz = ∫∫∫ r V

Note

Whenever

∫∫∫

2

sin θ dr dθ dφ .

V

f (x, y, z ) dx dy dz is to be evaluated throughout the

volume of a sphere, hemisphere or octant of a sphere, it will be advantageous to use spherical polar coordinates.)

Multiple Integrals

4.21

WORKED EXAMPLE 4(b) a a

Example 4.1 Change the order of integration in evaluate it.

∫∫ 0 y

x x + y2 2

dx d y

and then

The region of integration R is defined by y ≤ x ≤ a and 0 ≤ y ≤ a. i.e. it is bounded by the lines x = y, x = a, y = 0 and y = a. The rough sketch of the boundaries and the region R is given in Fig. 4.19. y y=a After changing the order of integration, the given integral I becomes x x=y I = ∫∫ dy dx 2 2 x=a + x y R Q (x, x) R The limits of inner integration are found by treating x as a constant, i.e. by drawing a line parallel to the y-axis in the region of integration O (x, 0) P y = 0 as explained in the previous section. a x

Thus

I = ∫∫ 0 0 a

x x +y 2

2

0

Fig. 4.19

dy dx

{ (

= ∫ x log y + y 2 + x 2

x

)}

y=x

dx y=0

a

= ∫ x [log (x + x 2 ) − log x] dx 0

 x2  a2 log (1 + 2 ) = log (1 + 2)) ⋅   =  2  2 0 a

1 1

x dx dy and then evaluate it. 2 2 x y + 0 x Note Since the limits of inner integration are x and 1, the corresponding variable of integration should be y. So we rewrite the given integral I in the corrected form first. Example 4.2 Change the order of integration in ∫ ∫

1 1

I = ∫∫ 0 x

x dy d x x2 + y 2

The region of integration R is bounded by the lines x = 0, x =1, y = x and y = 1 and is given in Fig. 4.20. The limits for the inner integration (after changing the order of integration) with respect to x are fixed as usual, by drawing a line parallel to x-axis (y = constant) 1 y

∴

I = ∫∫ 0 0

x dx d y x2 + y 2

Engineering Mathematics I

4.22 y

y=1

(0, y) P x=0

x=1 Q (y, y)

R y=x x

O Fig. 4.20 1

=∫ 0

x= y

1   log (x 2 + y 2 ) dy  2  x=0 1

=

1 log 2 ∫0

=

1 log 2. 2

 2 y 2     y 2  dy

a (b − y ) b b

Example 4.3 Change the order of integration in ∫ 0

∫

xy dx dy

and then

0

evaluate it. a The region of integration R is bounded by the lines x = 0, x = (b − y ) or b x y + = 1 , y = 0 and y = b and is shown in Fig. 4.21. a b y y=b

b (a – x) a

x,

Q y x + =1 a b

x=0 R O

P

x

y=0 (x, 0) Fig. 4.21

After changing the order of integration, the integral becomes I = ∫∫ xy dy dx . R The limits are fixed as usual. b (a − x ) a a

I=∫ 0

∫ 0

xy dy dx

Multiple Integrals a

=∫ 0

=

4.23

b (a − x )

 y 2 a x    2 

dx

0

b2 2a 2

a

∫ x (a − x)

2

dx

0

 2 x2 x3 x 4  a − 2a +   2 3 4  0  a 2 b 2  1 2 1  =  − +  2  2 3 4 

a

b2 = 2 2a

=

a 2b2 24 a

Example 4.4 Change the order of integration in

∫

b 2 2 a −x a

∫

0

integrate it.

x 2 dy dx and then

0

The region of integration R is bounded by the lines x = 0, x = a, y = 0 and the 2 2 2 x2 y 2 b 2 a − x 2 i.e. the curve y = a − x , i.e. the ellipse 2 + 2 = 1 a a b b2 a2 and is shown in Fig. 4.22.

curve y =

x2 + y2 = 1 a2 b2 (0, y) P x=0

a b

Q R

O

x=a x

y=0 Fig. 4.22

After changing the order of integration, the integral becomes I=∫

∫x R

The limits are fixed as usual. b

I=∫ 0

a 2 b − y2 b

∫

x 2 dx dy

0 a

 x 3 b = ∫    3  b

0

0

b2 − y 2

dy

2

dx dy

b2 – y2, y

Engineering Mathematics I

4.24 =

=

b

a3 3b3 a3 3b3

∫

3

(b 2 − y 2 ) 2 dy

0 π/2

∫b

4

cos 4 θ dθ

(on putting y = b sin θ)

0

a 3b 3 1 π × × × 3 4 2 2 π = a 3b 16 =

a

Example 4.5 Change the order of integration in

a2 − y2

∫ ∫ 0

y dx dy and then

a− y

evaluate it. The region of integration R is bounded by the line x = a − y, the curve x = a 2 − y 2 , the lines y = 0 and y = a. i.e. the line x + y = a, the circle x2 + y2 = a2 and the lines y = 0, y = a. R is shown in Fig. 4.23. y x, a2 – x2

Q

R x2 + y2 = a2 (x, a – x) P

x

O

y x + =1 a a Fig. 4.23

After changing the order of integration, the integral I becomes, I=∫

∫

y dy dx

R a2 − x2

a

=∫

∫

y dy dx

a− x

0

 y2  = ∫    2  a

1 2

dx

a− x

0

=

a2 − x2

a

∫ (2ax − 2 x 0

2

) dx

Multiple Integrals

4.25

 x 2 x3  = a −   2 3 0

a

=

a3 ⋅ 6

. 4

Example 4.6 Change the order of integration in

2 x

∫ ∫ dy dx 0

and then evaluate it.

x2 4

x2 i.e. the parabola 4 x2 = 4y, the curve y = 2 x i.e. the parabola y2 = 4x and the lines x = 0, x = 4. R is The region of integration R is bounded by the curve y =

shown in Fig. 4.24. y

A

x2 = 4y

y2 =

4x y2 ,y P 4

Q (2

y, y)

O

Fig. 4.24

The points of intersection of the two parabolas are obtained by solving the equations x2 = 4y and y2 = 4x. 2  2 Solving them, we get  x  = 4 x  4  i.e x (x3 − 64) = 0 ∴ x = 0, x = 4 and y = 0, y = 4 i.e. the points of intersection are O(0, 0) and A(4, 4). After changing the order of integration, the given integral I = ∫∫ d x d y R 4 2 y

=∫ 0

∫

dx d y

y2 4

4  y2  = ∫ 2 y −  d y 4   0

Engineering Mathematics I

4.26

 4 3 y3  =  y 2 −   3 12 0

4

32 16 − 3 3 16 = 3 =

2 2 a a+ a + y

Example 4.7 Change the order of integration in

∫

∫

xy dx dy and then

0 a − a2 − y2

evaluate it. The region of integration R is bounded by the curve x = a ∓ a 2 − y 2 , i.e. the circle (x − a)2 + y2 = a2 and the lines y = 0 and y = a. The region R is shown in Fig. 4.25. y x, 2ax – x2 (x – a)2 + y2 = a2 Q

R

O (x, 0) P

C (a, 0) Fig. 4.25

After changing the order of integration, the integral I becomes I = ∫∫ xy dy dx R 2a

2 ax − x 2

=∫

∫

0

0

2a

 y2  x    2 

=∫ 0

xy d y d x 2 ax − x 2

dx

0

2a

=

1 (2ax 2 − x 3 )dx 2 ∫0

1  x3 x 4  = 2a −  2  3 4 0

2a

=

2 4 a ⋅ 3

x

Multiple Integrals

4.27 1 2− y

Example 4.8 Change the order of integration in

∫∫ 0

xy dx dy

and then

y

evaluate it. The region of integration R is bounded by the lines x = y, x + y = 2, y = 0 and y = 1. It is shown in Fig. 4.26. After changing the order of integration, y B (1, 1) the integral I becomes I = ∫∫ xy dy dx

x+y=2

x=y

R

To fix the limits for y in the inner x integration, we have to draw a line parallel O C A to y-axis (since x = constant). The line drawn Fig. 4.26 parallel to the y-axis does not intersect the region R in the same fashion. If the line segment is drawn in the region OCB, its upper end lies on the line y = x; on the other hand, if it is drawn in the region BCA, its upper end lies on the line x + y = 2. In such situations, we divide the region into two sub-regions and fix the limits for each sub-region as illustrated below: I = ∫∫ xy dy dx + ∫∫ xy dy dx ∆ OCB 1

=∫ 0 1

=∫ 0 1

=∫ 0

∆BCA

2 2− x

x

∫

xy dy dx + ∫

∫

0

1

0

2  y2  x   dx + ∫  2  x

1

0

xy dy dx 2− x

 y2  x    2 

dx

0

2

x3 x dx + ∫ (2 − x) 2 dx 2 2 1

 x4  1 x 4  4  =   + 2 x 2 − x 3 +  8 0 2  3 4 1 1

2

1 5 = + 8 24 =

1 3

Example 4.9 Change the order of integration in

a 2a − x

∫ ∫ 0

evaluate it. 2

xy dy dx and then

x2 a

The region of integration R is bounded by the curve y = x , i.e. the parabola x2 = ay, a the line y = 2a − x, i.e. x + y = 2a and the lines x = 0 and x = a. It is shown in Fig. 4.27.

4.28

Engineering Mathematics I y D

y = 2a C

(0, y) P2

Q2 (2a – y, y)

E

B (a, a) Q1

(0, y) P1

ay, y

A x

O Fig. 4.27

After changing the order of integration, the integral I becomes I = ∫∫ xy dx dy R

When we draw a line parallel to x-axis for fixing the limits for the inner integration with respect to x, it does not intersect the region of integration in the same fashion. Hence the region R is divided into two sub-regions OABE and EBCD and then the limits are fixed as given below: I = ∫∫ xy dx dy + ∫∫ xy dx dy OABE a

=∫ 0

EBCD 2a 2a − y

ay

∫

xy dx dy + ∫

∫

0

a

0

xy dx dy

Note

The co-ordinates of the point B are obtained by solving the equations x + y = 2a and x2 = ay. B ≡ (a, a) and the equation of EB is y = a. a  x2  I = ∫ y    2  0 0

ay

2a − y

2a  x2  dy + ∫ y    2  a 0

dy

=

a 2a  1  2 2  y y + y a − y y d (2 ) d a ∫  2  ∫0  a

=

2a  a  1   y 3   2 2 4a 3 y 4   3 4   + y y a a 2 − +      = a . 2   3 0  3 4 a  8  

Example 4.10 Change the order of integration in each of the double integrals 1

2

2 2 dx dy dx dy and hence express their sum as one double integral and 2 2 ∫ ∫ x y + x2 + y 2 0 1 1 y and evaluate it. The region of integration R1 for the first double integral I1 is bounded by the lines x = 1, x = 2, y = 0 and y = 1.

∫∫

Multiple Integrals

4.29

The region of integration R2 for the second double integral I2 is bounded by the lines x = y, x = 2, y = 1 and y = 2. R1 and R2 are shown in Fig. 4.28. y

E (2, 2)

(x, x) Q2

(x, 1) Q1

(1, 1) D

A P1 (x, 0)

O

C (2, 1)

P2 (x, 1)

B

x

Fig. 4.28

After changing the order of integration,

and

2

1

I1 = ∫

∫

1

0

2

x

I2 = ∫

∫

1

1

dy dx x2+y2 dy dx x2+y2

Adding the integrals I1 and I2, we get 2 x 1 dy dy  I = ∫ dx  ∫ 2 +   0 x + y 2 ∫1 x 2 + y 2  1 2

x

= ∫ dx ∫ 1

0

dy x +y2 2

2 1 = ∫  tan −1  x 1 2

=∫ 1

y=x

y  dx  x  y = 0

π dx π = log 2 . 4 x 4

Example 4.11 Find the area bounded by the parabolas y2 = 4 − x and y2 = x by double integration. The region, the area of which is required is bounded by the parabolas (y − 0)2 = − (x − 4) and y2 = x and is shown in Fig. 4.29. Required area = ∫

∫ dx dy

OC AB

Engineering Mathematics I

4.30 =2∫

∫ dx dy , by symmetry

o AB

y

2 2 4− y

=2∫ 0

∫

dx dy

y2

(y2,

2

= 2 ∫ (4 − y − y ) dy 2

O

x

C

2

  4 = 2 4 2 − 2    3 =

(4 – y2, y) A (4, 0)

2

0

 2  = 2 4 y − y 3   3 0

y)

(2, 2) B

(2,– 2) y2

=4–x

y2 = x

Fig. 4.29

16 2 square units 3

Example 4.12 Find the area between the circle x2 + y2 = a2 and the line x + y = a lying in the first quadrant, by double integration. The plane region, the area of which is required, is shown in Fig. 4.30. y

Required area = ∫∫ dx dy

C

ABC

=∫

∫

a

(

B ( a2– y2, y)

dx dy O

a− y

0

=∫

(a – y, y)

a2 − y2

a

A

x

)

a2 − y2 − a + y dy

0

y =   2 =

a2 y y2  a2 − y2 + sin −1 − ay +  2 a 2 0

Fig. 4.30

a

a2 π a2 a2 ⋅ − a2 + = (π − 2) 2 2 2 4

Example 4.13 Find the area enclosed by the lemniscate r2 = a2 cos 2θ, by double integration. As the equation r2 = a2 cos 2θ remains unaltered on changing θ to − θ, the curve is symmetrical about the initial line. The points of intersection of the curve with the initial line θ = 0 are given by r2 = 2 a or r = ± a.

Multiple Integrals

4.31

Since r2 = a2 cos 2α = a2 cos 2 ( π − α), the curve is symmetrical about the line π θ= . 2 On putting r = 0, we get cos 2θ = 0. Hence θ = ± π , ± 3π . Hence there is a loop 4 4 π π 3π of the curve between θ = − and θ = and another loop between θ = − and 4 4 4 3π . θ= 4 Based on the observations given above the lemniscate is drawn in Fig. 4.31.

D C

O

P

A

B

X

Fig. 4.31

Required area = 4 × area OABC (by symmetry) = 4 ∫∫ r d r dθ OA B

When we perform the inner integration with respect to r, we have to treat θ as a constant temporarily and find the limits for r. Geometrically, treating θ = constant means drawing a line OP arbitrarily through the pole lying within the region of integration as shown in the figure. Finding the limits for r (while θ is a constant) is equivalent to finding the variation of the r coordinate of any point on the line OP. Assuming that the θ coordinates of all points on OP are θ each (since θ is constant on OP), we take O ≡ (0, θ) and P ≡ (r1, θ); viz., r varies from 0 to r1. Now wherever OP be drawn, the point P(r1, θ) lies on the lemniscate. Hence r12 = a2 cos 2θ or r1 = a cos 2θ (since r coordinate of any point is + ve) Thus the limits for inner integration are 0 and a cos 2θ. When we perform the outer integration, we have to find the limits for θ. Geometrically, we have to find the variation of the line OP so that it sweeps the area of the region, namely OABC. To cover this area, the line OP has to start from

Engineering Mathematics I

4.32

the position OA (θ = 0) and move in the anticlockwise direction and go up to  π π OD θ =  . Thus the limits for θ are 0 and .  4 4 π 4 a cos 2θ

∴ Required area = 4 ∫

∫

0

0

π 4

=4∫ 0

r d r dθ

 r2    2  0

a cos 2θ

dθ

π 4

= 2a

2

∫ cos 2θ dθ 0 π

= a 2 (sin 2θ )04 = a 2 Example 4.14 Find the area that lies inside the cardioid r = a (1 + cos θ) and outside the circle r = a, by double integration. The cardioid r = a (1 + cos θ) is symmetrical about the initial line. The point of intersection of the line θ = 0 with the cardioid is given by r = 2a, viz., the point (2a, 0). Putting r = 0 in the equation, we get cos θ = − 1 and θ = ± π. Hence the cardioid lies between the lines θ = − π and θ = π. The point of intersection of the line  π π θ = is a,  . C  2  2 G Noting the above properties, the B cardioid is drawn as shown in Fig. 4.32. All the points on the curve r = a have x A F O the same r coordinate a, viz., they are at the same distance a from the pole. D Hence the equation r = a represents a H circle with centre at the pole and radius E equal to a. y Noting the above points, the circle r = a is drawn as shown in Fig. 4.32. Fig. 4.32 The area that lies outside the circle r = a and inside the cardioid is shaded in the figure. Both the curves are symmetric about the initial line. Hence the required area = 2×AFGCB

Multiple Integrals π 2

= 2∫

4.33

r2

∫ r d r dθ , where (r , θ) lies on the circle r = a and (r , θ) 1

0

2

r1

lies on the cardioid r = a (1 + cos θ) π 2

a (1+ cos θ )

= 2∫

∫

0

a

π 2

a (1+ cos θ )

r dr dθ

 r2  = 2∫   2 a 0 

dθ

π 2

=a

2

∫ [(1 + cos θ )

2

−1] dθ

0 π 2

 1 + cos 2θ  = a 2 ∫ 2cosθ +  dθ   2 0 π

 2 θ 1 = a  2 sin θ + + sin 2θ   0 2 4  2

 π  a2 (π + 8) = a 2 2 +  =  4 4 a

Example 4.15 Express

a

∫∫ 0

y

x 2 dx dy

( x2 + y 2 )

3/ 2

in polar coordinates and then evaluate it.

The region of integration is bounded by the lines x = y, x = a, y = 0 and y = a, π whose equations in polar system are θ = , r = a sec θ, θ = 0 and r = a cosec 4 θ respectively. The region is shown in Fig. 4.33.

θ

=

π

r = a cosec θ B

O

4

θ=0

r = a sec θ

A

x

Fig. 4.33

Putting x = r cos θ, y = r sin θ and dx dy = r dr dθ in the given double integral I, we get

Engineering Mathematics I

4.34

r 3 cos 2 θ dr d θ r3

I = ∫∫ OAB

π / 4 a sec θ

=

∫ ∫ 0

cos 2 θ dr dθ

0

π/4

=

∫ cos

2

θ ⋅ [r ]0a sec θ dθ

0 π/4

= a ∫ cos θd θ = a [sin θ ]0π /4 =

a 2

0

a

Example 4.16 Transform the double integral

a2 − x2

∫∫ 0

ax − x

dx dy a − x2 − y 2 2

2

in polar

coordinates and then evaluate it. 2 2 2 The region of integration is bounded by the curves y = ax − x , y = a − x and the lines x = 0 and x = a.

y = ax − x 2 is the curve x2 + y2 − ax = 0      x − a  + (y − 0) 2 =  a    2   2 2

i.e.,

2

a  a i.e. the circle with centre at  ,0 and radius  2  2

y = a 2 − x 2 is the curve x2 + y2 = a2 i.e. the circle with centre at the origin and radius a. The polar equations of the boundaries of the region of integration are r2 − ar cos θ = 0 or r = a cos θ, r = a, r = a sec θ and θ = π . The region of integration is 2 shown in Fig. 4.34. Putting x = r cos θ, y = r sin θ and dx dy = r dr dθ in the given double integral I, we get π/2

I=∫ 0

a

∫ a cos θ

r dr dθ a2 − r 2

 1  2 2 =∫ dθ , on putting a2 − r2 = t − × 2 a − r    2  a cos θ  0  π/2

a

Multiple Integrals

4.35

r=a θ=π 2

r = a sec θ

r = a cos θ a 2

O

a

x

Fig. 4.34 π/2

=

∫

π

a sin θ d θ = −a [cos θ ]02 = a

0

Example 4.17 By transforming into cylindrical coordinates, evaluate the integral

∫∫∫ (x

2

+ y 2 + z 2 ) dx dy dz taken over the region of space defined by x2 + y2 ≤ 1

and 0 ≤ z ≤ 1. z

The region of space is the region enclosed by the cylinder x2 + y2 = 1 whose base radius is 1 and axis is the z-axis and the planes z = 0 and z = 1. The equation of the cylinder in cylindrical coordinates is r = 1. The region of space is shown in Fig. 4.35. Putting x = r cos θ, y = r sin θ, z = z and dx dy dz = r dr dθ dz in the given triple integral I, we get I = ∫∫∫ (r 2 + z 2 ) r dr dθ dz ,

z=1

z=0 y x

V

where V is the volume of the region of space. 1 2π 1

=∫ 0

∫ ∫ (r 0

2

+ z 2 ) r dr dθ dz

0

2 r4  + z 2 r  dθ dz ∫  4 2 0 0 1

1 2π

=∫ 0

1 2π

=∫ 0

1

1

∫  4 + 2 z 0

1

2

  dθ dz 

1 1  = 2π ∫  + z 2  dz 4 2  0

Fig. 4.35

Engineering Mathematics I

4.36

 z z3  = 2π  +   4 6  0 5 = π 6 1

Note

The intersection of z = constant c and the cylinder x2 + y2 = 1 is a circle with centre at (0, 0, c) and radius 1. The limits for r and θ have been fixed to cover the area of this circle and then the variation of z has been used so as to cover the entire volume.] Example 4.18 Find the volume of the portion of the cylinder x2 + y2 = 1 intercepted between the plane z = 0 and the paraboloid x2 + y2 = 4 − z. z

y

x Fig. 4.36

Using cylindrical coordinates, the required volume V is given by V = ∫∫∫ r dr dθ dz , taken throughout the region of space. Since the variation of z is not between constant limits, we first integrate with respect to z and then with respect to r and θ. Changing to cylindrical coordinates, the boundaries of the region of space are r = 1, z = 0 and z = 4 − r2. 2 2π 1 4 − r

∴

V =∫ 0

∫ ∫ 0

dz r dr dθ

0

2π 1

=∫ 0

∫ r (4 − r

2

) dr dθ

0

2π 2π  7 7 r4  = ∫  2r 2 −  dθ = ∫ dθ = π   4 2 4 0 0  0 1

a

Example 4.19 Evaluate

∫ ∫ 0

polar coordinates.

a2 − x2

0

a2 − x2 − y2

∫ 0

xyz dz dy dx , by transforming to spherical

Multiple Integrals

4.37

2 2 2 The boundaries of the region of integration are z = 0, z = a − x − y or x2 + y2 +

I = ∫∫∫ r 3 sin 2 θ cos θ sin φ cos φ. r 2 sin θ d r d θ dφ

θ

r

z2 = a2, y = 0, y = a 2 − x 2 or x2 + y2 = a2, x = 0 and x = a. From the boundaries, we note that the region of integration is the volume of the positive octant of the sphere x2 + y2 + z2 = a2. By putting x = r sin θ cos φ , y = r sin θ sin φ , z = r Z cos θ and dx dy dz = r2 sin θ dr dθ dφ, the given triple integral I becomes P

V

O

where V is the volume of the positive octant of the sphere r = a, which is shown in Fig. 4.37. To cover the volume V, r has to vary from 0 to π a, θ has to vary from 0 to and φ has to 2 π vary from 0 to . 2

π 2

Thus

I=∫

π 2

0

M x Fig. 4.37

a

∫∫r

0

y

5

sin 3 θ cos θ sin φ cos φ d r d θ dφ

0

π 2

π 2

a

0

0

0

= ∫ sin φ cos φ dφ . ∫ sin 3 θ cos θ d θ . ∫ r 5 d r

π

π

∴

[

the limits are constants]

 sin 2 φ  2  sin 4 θ  2  r 6   ⋅  ⋅  =  2  0  4  0  6  0 =

a

1 6 a ⋅ 48

Example 4.20 Evaluate

∫∫∫

1− x 2 − y 2 − z 2 dx dy dz , taken throughout the

volume of the sphere x2 + y2 + z2 = 1, by transforming to spherical polar coordinates. Changing to spherical polar coordinates, the given triple integral I becomes I = ∫∫∫ 1− r 2 r 2 sin θ d r d θ dφ V

Engineering Mathematics I

4.38

z

θ

r

P (r, θ, )

O

y

M x Fig. 4.38

To cover the entire volume V of the sphere, r has to vary from 0 to 1, θ has to vary from 0 to π and φ has to vary from 0 to 2π. 2π

Thus

I=∫ 0

π

1

∫∫ 0

1− r 2 r 2 dr ⋅ sin θ d θ ⋅ dφ

0

2π

π

π 2

= ∫ dφ ∫ sin θ d θ ∫ sin 2 t cos 2 t d t , by putting 0

0

0

r = sin t in the innermost integral 1 π 3 1 π = 2 π×( − cos θ )π0 × ⋅ − ⋅ ⋅  2 2 4 2 2 π 1 1 = 4π× × = π 2 4 4 4 EXERCiSE 4(b) Part A (Short Answer Questions) a

1. Change the order of integration in

∫∫ 0 1

2. Change the order of integration in

3. Change the order of integration in

0

y

a

a

∫∫

f (x, y ) dx dy . f (x, y ) dy dx.

x y

∫∫ 0

f (x, y ) dy dx.

0 1

∫∫ 0 1

4. Change the order of integration in

x

f (x, y ) dx dy.

0

1 1− y

5. Change the order of integration in

∫ ∫ 0

0

f (x, y ) dx dy .

Multiple Integrals

4.39

a a− x

6. Change the order of integration in

7. Change the order of integration in

∫∫ 0

1

1− x 2

∫ ∫ 0

a

8. Change the order of integration in

f (x, y ) dy dx.

0

f ( x , y ) d y dx .

0 a2 − y2

∫ ∫ 0

f ( x , y ) dx dy .

0

1 2 x

9. Change the order of integration in

∫∫ 0

f ( x, y ) dy dx.

0

∞ 1/ y

10. Change the order of integration in

∫∫ 0

f ( x , y ) dx dy .

0

Part B Change the order of integration in the following integrals and then evaluate them: a

11.

13.

a

x dx d y x2 + y2

∫∫ 0

y

∞

x

0

xe

dy d x

14.

0

∫∫e 0

2x+ y

dy d x

16.

0

∫∫ 0

3

19.

( x + y ) dy dx

2

x 4a 4− x

∫ ∫ 0

21.

( x + y ) dy dx

3

20.

xy dx dy

0 y

∫∫

y2

y dx d y x2 + y 2

a2 − y2

∫ ∫

y 2 dx dy

0

2 4/ x

dx dy

22.

∫ ∫ xy dy dx 1

0

2 − x2

∫ ∫ 0

∫ ∫

−a

0 5 2 x 9

23.

4− y2

x

∫∫ 1

x

2

a

1 5

3

0

0

+ y 2 ) dy d x

e− y dy dx y

∫∫

1

18.

2

x

0

2a a

17.

∫ ∫ (x 0

1 1− x

15.

2

∞ ∞

x2 − y

∫∫

2

12.

dy dx 1

x

24. Change the order of integration in each of the double integrals 2 2− x

and

∫∫ 1

0

evaluate it.

x

∫ ∫ xy dy dx 0

0

xy dy dx and hence express their sum as one double integral and

Engineering Mathematics I

4.40

0

25. Change the order of integration in each of the double integrals 1

0

∫ ∫ (x

2

+ y2 )

−1 − x

1

∫ ∫ (x

dy dx and

1

2

2

+ y ) dy dx and hence express their sum as one double

x

integral and evaluate it. Find the area specified in the following problems (26−35), using double integration: 26. The area bounded by the parabola y = x2 and the straight line 2x − y + 3 = 0. 27. The area included between the parabolas y2 = 4a (x + a) and y2 = 4a(a − x). 28. The area bounded by the two parabolas y2 = 4ax and x2 = 4by. 29. The area common to the parabola y2 = x and the circle x2 + y2 = 2. x3 and its asymptote. 2− x The area of the cardioid r = a (1 + cos θ). The area common to the two circles r = a and r = 2a cos θ. The area common to the cardioids r = a (1 + cos θ) and r = a (1 − cos θ). The area that lies inside the circle r = 3a cos θ and outside the cardioid r = a (1 + cos θ). The area that lies outside the circle r = a cos θ and inside the circle r = 2a cos θ. Change the following integrals (36 − 40), into polar coordinates and then evaluate them:

30. The area bounded by the curve y 2 = 31. 32. 33. 34. 35.

a2 − x2

a

36.

∫ ∫ 0 x

∫∫ 0

a

dx dy

37.

0

∫∫

−∞ −∞

2a

x3 d x d y x +y 2

39.

2

a

xdx d y

∫ ∫ (x 0

∞ ∞

40.

)

0

a

38.

(

− x2 + y2

e

2

y

2 a x − x2

∫ ∫ 0

+ y2 )

0

xdx d y x2 + y 2

dx d y

(a

2

+ x2 + y 2 )

3/2

Evaluate the following integrals (41− 45) after transforming into cylindrical coordinates: 41.

∫∫∫ (x + y + z ) dx dy dz , where V is the region of space inside the cylinder V

x2 + y2 = a2, that is bounded by the planes z = 0 and z = h. 42.

∫∫∫ (x

2

+ y 2 ) dx dy dz , taken throughout the volume of the cylinder x2 + y2 = 1

that is bounded by the planes z = 0 and z = 4. 43.

∫∫∫ dx dy dz ,

taken throughout the volume of the cylinder x2 + y2 = 4

bounded by the planes z = 0 and y + z = 3.

Multiple Integrals

∫∫∫ dx dy dz ,

44.

4.41

taken throughout the volume of the cylinder x2 + y2 = 4

bounded by the plane z = 0 and the surface z = x2 + y2 + 2.

∫∫∫ dx dy dz ,

45.

taken throughout the volume bounded by the spherical

surface x + y + z2 = 4a2 and the cylindrical surface x2 + y2 − 2ay = 0. Evaluate the following integrals (46-50) after transforming into spherical polar coordinates: d xd yd z 46. ∫∫∫ 2 , taken throughout the volume of the sphere x2 + y2 + x + y2 + z2 z2 = a2. d xd yd z 47. ∫∫∫ , taken throughout the volume contained in the 1− x 2 − y 2 − z 2 positive octant of the sphere x2 + y2 + z2 = 1. 2

2

∫∫∫ z dx dy dz ,

48.

where V is the region of space bounded by the

V

sphere x2 + y2 + z2 = a2 above the xOy-plane. a2 − x2

a

49.

4.6

∞ ∞ ∞

a2 − y2 − z 2

∫ ∫

∫

0

0

0

50. ∫

x dx dy dz

0

∫∫ 0

0

d xd yd z

(a

2

+ x2 + y 2 + z 2 )

5/ 2

LinE intEGRAL

The concept of a line integral is a generalisation of the concept of a definite b

integral

∫

f (x) dx.

a

In the definite integral, we integrate along the x-axis from a to b and the integrand f(x) is defined at each point in (a, b). In a line integral, we shall integrate along a curve C in the plane (or space) and the integrand will be defined at each point of C. The formal definition of a line integral is as follows. Definition  Let C be the segment of a continuous curve joining A(a, b) and B(c, d) (Fig. 4.39). y

B (c, d) Q

(xr, yr)

(xr –1, yr – 1) P

(ξr, ηr)

A (a, b)

x

O Fig. 4.39

Engineering Mathematics I

4.42

Let f (x, y), f1(x, y), f2(x, y) be single-valued and continuous functions of x and y, defined at all points of C. Divide C into n arcs at (xi, yi) [i = 1, 2, . . . (n − 1)] Let x0 = a, xn= c, y0 = b, yn = d.  ) =∆s , Let xr − xr−1 = Δxr, yr − yr−1 = Δyr and the arcual length of PQ (i.e. PQ r where P is (xr−1, yr−1) and Q(xr, yr). Let (ξr , ηr) be any point on C between P and Q. n

lim ∑ f ( ξ r , η r ) ∆s r

Then

n →∞

r =1 n

lim ∑  f1 (ξr , ηr ) ∆ xr + f 2 (ξ r , η r ) ∆ y r 

or

n →∞

r =1

is defined as a line integral along the curve C and denoted respectively as

∫ f (x, y) ds

or

C

4.6.1

∫ [f (x, y) dx + f 1

2

(x, y ) dy ]

C

Evaluation of a Line integral

Using the equation y = φ (x) or x = ψ(y) of the curve C, we express c

dx + f 2 (x , y ) dy ] either in the form it, which is only a definite integral. If the line integral is in the form

d

∫ g (x) dx or in the form ∫ h(y) dy a

∫ [f (x , y) 1

C

and evaluate

b

ds

∫ f (x, y) ds, it is first rewritten as ∫ f (x, y) dx dx = C

C

 dx  ds f (x, y ) dy = ∫ f (x, y ) 1+   dy and then ∫ ∫  dy  dy C C C evaluated after expressing it as a definite integral.  dy  f (x, y ) 1+   dx or as  dx 

2

2

4.6.2

Evaluation when C is a Curve in Space

The definition of the line integral given above can be extended when C is a curve in space. In this case, the line integral will take either the form f 2 (x, y, z ) dy + f 3 (x, y, z ) dz ] or the form

∫ f (x, y, z ) ds.

∫ [f (x, y, z ) dx + 1

C

When C is a curve in

C

space, very often the parametric equations of C will be known in the form x = φ1(t), y = φ 2(t), z = φ 3(t). Using the parametric equations of C, the line integral can be expressed as a definite integral. In the case of ds ∫ f (x , y , z ) dt dt , where C ds = dt

 dx   dy   dz    +   +   .  dt   dt   dt  2

2

2

∫ f (x, y, z ) ds , it is rewritten as C

Multiple Integrals

4.7

4.43

SURFACE intEGRAL

The concept of a surface integral is a generalisation of the concept of a double integral. While a double integral is evaluated over the area of a plane surface, a surface integral is evaluated over the area of a curved surface in general. The formal definition of a surface integral is given below. Definition Let S be a portion of a regular two-sided surface. Let f (z, y, z) be a function defined and continuous at all points on S. Divide S into n sub-regions Δs1, n

ΔS2, . . ., ΔSn. Let P(ξr, ηr, ζr) be any point in ΔSr. Then lim

∑ f (ξ , η , ζ ) ∆S

n→∞ ∆Sr → 0 r =1

r

r

r

r

is called the surface integral of f(x, y, z) over the surface S and denoted as

∫ f (x, y, z ) dS

or

S

4.7.1

∫∫ f (x, y, z ) dS . S

Evaluation of a Surface integral

Let the surface integral be

∫∫ f (x, y, z ) dS

, where S is the portion of the surface

S

whose equation is φ (x, y, z) = c (Fig. 4.40). z

S

O

y R

x Fig. 4.40

Project the surface S orthogonally on xoy-plane (or any one of the co-ordinate planes) so that the projection is a plane region R. The projection of the typical elemental surface ∆S (shaded in the figure) is the typical elemental plane area ∆A (shaded in the figure). We can divide the area of the region R into elemental areas by drawing lines parallel to x and y axes at intervals of ∆y and ∆x respectively. Then ∆A = ∆x · ∆y. Then ∆x · ∆y = ∆S cos θ, where θ is the angle between the surface S and the plane R (xoy-plane), i.e. θ is the angle between the normal to the surface S at the typical point (x, y, z) and the normal to the xoy-plane (z-axis). From Calculus, it is

Engineering Mathematics I

4.44

known that the direction ratios of the normal at the point (x, y, z) to the surface  ∂φ ∂φ ∂φ  φ (x, y, z) = c are  ,   ∂x ∂y , ∂z  . The direction cosines of the z-axis are (0, 0, 1) ∂φ ∂z cos θ = 2 2 2   ∂φ   ∂φ   ∂φ    +   +    ∂x   ∂y   ∂z 

∴

∆S =

Thus

∫∫

∴

φx2 + φy2 + φz2 φz

f (x, y, z ) dS = ∫∫ f ( x, y, z ) ⋅

S

∆x ∆y . φx2 + φy2 + φz2 φz

R

dx dy

Thus the surface integral is converted into a double integral by using the above relation, in which the limits for the double integration on the right side are fixed so as to cover the entire region R and the integrand is converted into a function of x and y, using the equation of S.

Note

Had we projected the curved surface S on the yoz-plane or zox-plane then the conversion formula would have been

∫∫

f (x, y, z )dS = ∫∫ f (x, y, z ) ⋅

S

∫∫

or

R

f (x, y, z )dS = ∫∫ f (x, y, z ) ⋅

S

R

φx2 + φy2 + φz2 φx

φx2 + φy2 + φz2 φy

dy dz

dz dx ,

respectively.

4.8 VOLUME intEGRAL Definition Let V be a region of space, bounded by a closed surface. Let f(x, y, z) be a continuous function defined at all points of V. Divide V into n subregions ΔVr by drawing planes parallel to the yoz, zox and xoy-planes at intervals of Δx, Δy and Δz respectively. Then ΔVr is a rectangular parallelopiped with dimensions Δx, Δy, Δz. Let P(ξr , ηr ,ζr) be any point in ΔVr . n

Then lim

∑ f (ξ , η , ζ ) ∆V

n→∞ ∆Vr → 0 r =1

r

r

r

r

is called the volume integral of f(x, y, z) over the

region V (or throughout the volume V) and denoted as

Multiple Integrals

∫

f (x, y, z ) dv

or

∫∫∫

4.45

f (x, y, z ) dx dy dz

V

V

4.8.1 triple integral versus Volume integral A triple integral discussed earlier is a three times repeated integral in which the limits of integration are given, whereas a volume integral is a triple integral in which the limits of integration will not be explicitly given, but the region of space in which it is to be evaluated will be specified. The limits of integration in a volume integral are fixed so as to cover the entire volume of the region of space V. Though the line integral and surface integral have been defined in the scalar form in this unit, they are also defined in the vector form.

Note

WORKED EXAMPLE 4(c)

Example 4.1

Evaluate

∫ [(3xy

2

+ y 3 ) dx + (x 3 + 3 xy 2 ) dy ] where C is the

C

parabola y2 = 4ax from the point (0, 0) to the point (a, 2a). The given integral I= 2

∫

y = 4 ax

[(3xy 2 + y 3 ) dx + (x 3 + 3 xy 2 ) dy ]

y

A (a, 2a) C

In order to use the fact that the line integral is evaluated along the parabola y2 = 4ax, we x (0, 0) O use this equation and the relation between dx and dy derived from it, namely, 2y dy = 4a dx and convert the body of the integral either to the form f(x) dx or to the form φ (y) dy. Then Fig. 4.41 the resulting definite integral is evaluated between the concerned limits, got from the end points of C. The choice of the form f(x) dx or φ (y) dy for the body of the integral depends on convenience. In this problem, x is expressed as 1 y 2 more easily than expressing y 4a as 2 ax .

Note

From y2 = 4ax, we get y = ±2 ax . Since the arc C lies in the first

quadrant, y is positive and hence y = 2 ax . 2a  1 2 2  1  y 1 2 2  y ⋅ y + y 3 dy +  y 6 + 3⋅ y ⋅ y  dy  Thus I = ∫ 3 ⋅ 3  4a     2 a 4a 64a 0 (As the integration is done with respect to y, the limits for y are the y co-ordinates of the terminal points of the arc C).

Engineering Mathematics I

4.46

5  3 1 I = ∫  y 4 + 2 y 5 + y 6 dy 3  4a  8a 64a 0 2a

1  1 1 =  y 5 + y 7 y6 +  4a 16a 2 448a 3 0

2a

= Example 4.2 Evaluate

86 4 a 7

∫ [(2x − y) dx + (x + y) dy] , where C is the circle x

2

+ y2 = 9.

C

In this problem the line integral is evaluated around a closed curve. In such a situation the line integral is denoted as

∫

[(2x − y ) dx + (x + y ) dy ] , where a small circle is put across the integral symbol.

C

When a line integral is evaluated around a closed curve, it is assumed to be described in the anticlockwise sense, unless specified otherwise. (Fig. 4.42) In the case of a line integral around a closed curve C, any point on C can be assumed to be the initial point, which will also be the terminal point. Further if we take x or y as the variable of integration, the limits of integration will be the same, resulting in the value ‘zero’ of the line integral, which is meaningless. Hence whenever a line integral is evaluated around a closed curve, the parametric equations of the curve are used and hence the body of integral is converted to the form f (t) dt or f (θ) dθ. In this problem, the parametric equations of the circle x2 + y2 = 9 are x = 3 cos θ and y = 3 sin θ. ∴ dx = −3 sin θ d θ and dy = 3 cos θ d θ. y B

θ=π

θ=

A'

π 2

θ=0 x A θ = 2π

B' 3π θ= 2 Fig. 4.42 2π

The given integral =

∫ [(6 cos θ − 3 sin θ ) ( − 3 sin θ d θ ) 0

Multiple Integrals

4.47

+ (3 cos θ + 3 sin θ ) (3 cos θ dθ )] 2π

= 9 ∫ (1− sin θ cos θ ) dθ 0

 sin 2 = 9 θ −  2

θ   0

2π

= 18π Example 4.3 Evaluate

∫ xy ds ,

where C is the arc of the parabola y2 = 4x

C

between the vertex and the positive end of the latus rectum. I = ∫ xy

Given integral

C

ds dx dx

2

Equation of the parabola is y = 4x dy 2 = dx y

Differentiating with respect to x,

 dy  4 ds = 1 +   = 1 + 2  dx  dx y 2

∴

y2 + 4 dx y

I = ∫ xy

∴

C

1

= ∫ x 4 x + 4 dx 0

2

= 2∫

1

(t 2 −1) ⋅ t ⋅ 2t dt, on putting x + 1 = t2

2

= 4 ∫ (t 4 − t 2 ) dt 1

t5 t3  = 4  −   5 3 1 =

Example 4.4 Evaluate

2

8 (1 + 2 ) 15

∫ (y dx − x dy), where C is the boundary of the triangle 2

2

C

whose vertices are (−1, 0), (1, 0) and (0, 1) (Fig. 4.43). C is made up of the lines BC, CA and AB.

Engineering Mathematics I

4.48

Equations of BC, CA and AB are respectively y = 0, x + y = 1 and −x + y = 1. Given integral =

∫

∫

+

BC y=0 dy = 0

∫

+

CA x + y =1 dy =−dx

(y 2 dx − x 2 dy )

AB − x + y =1 dy = dx

y A (0, 1)

B (– 1, 0)

C (1, 0)

O

x

Fig. 4.43 −1

0

= 0 + ∫ [(1− x) 2 + x 2 ]dx + ∫ [(1+ x) 2 − x 2 ]dx 1

0 −1

0

= ∫ (1− 2 x + 2 x 2 ) dx + ∫ (1+ 2 x) dx 0

1

 2 x 3   + (x + x 2 )−0 1 =  x − x 2 +  3  0

1

2 =− 3

∫ [x

Example 4.5 Evaluate

2

y dx + (x − z ) dy + xyz dz ] , where C is the arc of the

C

parabola y = x2 in the plane z = 2 from (0, 0, 2) to ( 1, 1, 2). Given integral =

∫

[x 2 y dx + (x − z ) dy + xyz dz ]

2  y = x     z = 2 

=

∫

y=x

[x 2 y dx + (x − 2) dy ] 2

[∴ dz = 0, when z = 2]

 x5 2 x3  = ∫ [x + (x − 2) 2 x]dx =  + − 2 x 2    5 3

1

1

4

0

0

=−

17 15

Example 4.6 Evaluate

∫ (x dx + xy dy + xyz dz ) ,

where C is the arc of the

C

twisted curve x = t, y = t2, z = t3, 0 ≤ t ≤ 1. The parametric equations of C are x = t, y = t2, z = t3 ∴ dx = dt, dy = 2t dt, dz = 3t2 dt on C. Using these values in the given integral I,

Multiple Integrals

4.49

1

I = ∫ (t + t 3 ⋅ 2t + t 6 ⋅ 3t 2 ) dt 0

t2 t5 t9  =  + 2 + 3   2 5 9 0 1

=

17 30

∫ (x

Example 4.7 Evaluate

2

+ y 2 + z 2 ) ds, where C is the arc of the circular helix

C

x = cos t, y = sin t, z = 3t from (1, 0, 0) to (1, 0, 6 π ) The parametric of equations of C are x = cos t, y sin t, z = 3t. dx dy ∴ =− sin t, = cos t , dt dx

dz =3 on C. dt

 dx   dy   dz    +   +    dt   dt   dt  2

ds = dt

2

2

= sin 2 t + cos 2 t + 9 = 10 2π

I = ∫ (cos 2 t + sin 2 t + 9t 2 )

Given integral

0

Note

ds dt dt

The point (1, 0, 0) corresponds to t = 0 and (1, 0, 6 π ) corresponds to t = 2 π . I = (t + 3t 3 ) × 10 2π

0

= 2 10π (1+12π 2 ) Example 4.8 Evaluate

∫∫ xyz dS ,

where S is the surface of the rectangular

S

parallelopiped formed by x = 0, y = 0, z = 0, x = a, y = b and z = c (Fig. 4.44). z C

z=c

A' O'

B'

y=b y=0

O B A

x

z=0

y

C'

Fig. 4.44

Since S is made up of 6 plane faces, the given surface integral I is expressed as

Engineering Mathematics I

4.50

I = ∫∫ + ∫∫ + ∫∫ + ∫∫ + ∫∫ + ∫∫ (xyz dS ) x=0

x=a

y=0

y =b

z =0

z=c

Since all the faces are planes, the elemental curved surface area dS becomes the elemental plane surface area dA. On the planes x = 0 and x = a, dA = dy dz. On the planes y = 0 and y = b, dA = dz dx. On the planes z = 0 and z = c, dA = dx dy. I = ∫∫ + ∫∫ (xyz dy dz ) + ∫∫ + ∫∫ (xyz dz dx) ∴ x=0

x=a

y=0

y =b

+ ∫∫ + ∫∫ (xyz dx dy ) z =0

z=c

Simplifying the integrands using the equations of the planes over which the surface integrals are evaluated, we get b

a

c

b

I = a∫

c

∫

yz dy dz + b ∫

∫

zx dz dx + c ∫

0

0

0

0

0

a

∫ xy dx dy 0

On the plane face O′A′CB′ (z = c), the limits for x and y are easily found to be 0, a and 0, b. Similarly the limits are found on the faces O′B′AC′ (x = a) and O′C′BA′ (y = b).]

Note

Now

b2 c2 c2 a2 a 2 b2 ⋅ + b⋅ ⋅ + c ⋅ ⋅ 2 2 2 2 2 2 abc (ab + bc + ca ) = 4

I=a

Example 4.9 Evaluate

∫∫ (y + 2 z − 2) dS , where S is the part of the plane 2x + S

3y + 6z = 12, that lies in the positive octant (Fig. 4.45). z C

B

O

y

A x Fig. 4.45

Rewriting the equation of the (plane) surface S in the intercept form, we get

Multiple Integrals

4.51

x y z + + =1 6 4 2 ∴ S is the plane that cuts off intercepts of lengths 6, 4 and 2 on the x, y and z-axes respectively and lies in the positive octant. We note that the projection of the given plane surface S on the xoy-plane is the triangular region OAB shown in the two-dimensional Fig. 4.46. y

B 2x + 3y = 12

4

O

x

A

6 Fig. 4.46

Converting the given surface integral I as a double integral, I = ∫∫ (y + 2 z − 2)

φ 2x + φ 2y + φ 2z φz

∆OAB

dx dy ,

where φ = c is the equation of the given surface S. Here φ = 2x + 3y + 6z ∴ φ x = 2; φ y = 3; φ z = 6. ∴

I = ∫∫ (y + 2 z − 2) ∆OAB

=

4 + 9 + 36 dx dy 6

7 (y + 2 z − 2) dx dy 6 ∆∫∫ OAB

(1)

Now the integrand is expressed as a function of x and y, by using the value of z (as a function of x and y) got from the equation of S, i.e. from the equation 2x + 3y + 6z = 12 1 Thus z = (12 − 2 x − 3 y ) (2) 6 Using (2) in (1), we get I=

7 1 (6 − 2x) dx dy 6 ∆∫∫ 3 OAB

Engineering Mathematics I

4.52

4

7 = ∫ 18 0 =

3 6− y 2

∫

(6 − 2 x) dx dy

0

4 7  9 2 28 9 y − y  dy = 18 ∫0  4 3

Example 4.10 Evaluate

∫∫ z

3

d S , where S is the positive octant of the surface

S

of the sphere x2 + y2 + z2 = a2 (Fig. 4.47) z C

O

y

B

A x Fig. 4.47

The projection of the given surface of the sphere x2 + y2 + z2 = a2 (lying in the positive octant) in the xoy - plane is the quadrant of the circular region OAB, shown in the two-dimensional Fig. 4.48. y x2 + y2 = a2

B (0, y)

a2 – x2, y

O

A

x

Fig. 4.48

Converting the given surface integral I as a double integral. I = ∫∫ z

3

φ 2x + φ 2y + φ 2z

dx dy, φz where φ ≡ x2 + y2 + z2 = a2 is the equation of the given spherical surface. φ x = 2x; φ y = 2y; φ z = 2z. OAB

∴

I = ∫∫ z 3 OAB

4 (x 2 + y 2 + z 2 ) dx dy 2z

= ∫∫ z 2 x 2 + y 2 + z 2 dx dy OAB

Multiple Integrals = a ∫∫ (a 2 − x 2 − y 2 ) dx dy

4.53 ∴ [ (x, y, z) lies on x2 + y2 + z2 = a2]

OAB a2 − y2

a

= a∫

∫

0

0

a

= a∫ 0

(a 2 − y 2 − x 2 ) dx dy

3  2  (a − y 2 ) x − x  3  

x = a2 − y2

dy x=0

a

=

3 2 a ∫ (a 2 − y 2 ) 2 dy 3 0

=

2 5 a 3

π/2

∫ cos

4

θ d θ, on putting x = a sin θ.

0

2 5 3 1 π a ⋅ ⋅ ⋅ 3 4 2 2 π 5 = a ⋅ 8 =

Example 4.11 Evaluate

∫∫

y (z + x) dS , where S is the curved surface of the

S

cylinder x2 + y2 = 16, that lies in the positive octant and that is included between the planes z = 0 and z = 5 (Fig. 4.49). z E

C

D O

B

y

A x z

Fig. 4.49

D

C

We note that the projection of S on the xoy-plane is not a plane (region) surface, but only the arc AB of the circle whose centre is O and radius equal to 4. For converting the given surface integral into a double integral, the projection of S must be a plane region. Hence we project S on the zox-plane (or yoz- O plane). The projection of S in this case is the rectangular region OCDA, which is shown in Fig. 4.50.

5

x 4

A

Fig. 4.50

Engineering Mathematics I

4.54

Converting the given surface integral I as a double integral, φx2 + φy2 + φz2

I = ∫∫ y ( z + x )

φy

OADC

dz dx,

where φ ≡ x2 + y2 = 16 is the equation of the given cylindrical surface. φx = 2x; φ y = 2y; φz = 0. ∴

4(x 2 + y 2 )

I = ∫∫ y (z + x)

2y

OADC

dz d x ∴ [ (x, y, z) lies on x2 + y2 = 16]

= 4 ∫∫ (z + x) dz dx OADC 5

4

=4∫

∫ (z + x) dx dz .

0

0

5

= 4 ∫ (4 z + 8) dz 0

= 8( z 2 + 4z)50 = 360. Example 4.12 Evaluate

∫∫∫ xyz dx dy dz , where V is the region of space inside the V

x y z + + =1. a b c Vide worked Example 4.11 in the section on ‘Double and triple integrals’ for fixing the limits of the volume integral. tetrahedron bounded by the planes x = 0, y = 0, z = 0 and

a

 x  x y b1 −  c1 − −   a   a b 

I= ∫

∫

∫

0

0

0

a

 x b1 −   a 

=∫

∫

0

0

c2 2

a

=

=

=

c2 2 c2 2

xyz dz dy dx

a

∫ 0 a

dy d x

0

bt

∫∫ 0

 x y c1 − −  a b 

 z 2   xy    2 

0

 y x xy t −  dy dx , where t = 1−  b  a 2

 y 2 2t y 3 1 y 4   dx x t 2 − + 2  2 b 3 b 4 0 bt

1

2

1

∫  2 − 3 + 4  b 0

2

xt 4 dx

Multiple Integrals =

=

a

b2 c2 24

∫ 0 a

b2 c2 24

∫ 0

4.55

 x x 1−  dx  a  4

  x   x 4 a 1− 1−  ⋅ 1−  dx   a   a   

  x 5  x 6   1−  1−     a   ab 2 c 2   a   = +  6  24  − 5  a a   0

a

 1 1   −   5 6 

=

a 2 b2 c2 24

=

1 2 2 2 a b c ⋅ 720

Example 4.13 Express the volume of the sphere x2 + y2 + z2 = a2 as a volume integral and hence evaluate it. [Refer to Fig. 4.51] z

y

O

x Fig. 4.51

Required volume = 2 × volume of the hemisphere above the xoy-plane. Vide worked Example 4.12 in the section on ‘Double and Triple Integrals’. a2 − x2

a

Required volume = 2 ∫

∫

−a − a 2 − x2

∫

dz dy dx

0

a2 − x2

a

= 2∫

−a

a2 − x2 − y2

∫

(a 2 − x 2 ) − y 2 dy dx

− a2 − x2

Taking a2 − x2 = b2, when integration with respect to y is performed, a

b

V = 2∫

∫

− a −b

b 2 − y 2 dy dx

Engineering Mathematics I

4.56 a

b

=4∫

∫

b 2 − y 2 dy d x

[ ∵ b 2 − y 2 is an even function of y ]

−a 0

a y 2 b2 b − y 2 + sin −1 = 4 ∫  2 2 −a 

y   dx b 0 b

a

= π ∫ ( a 2 − x 2 ) dx −a

 x3  = 2π a 2 x −   3 0

a

4 = πa 3 3 Example 4.14 Evaluate

∫∫∫ (x + y + z ) dx dy dz , where V is the region of space V

inside the cylinder x2 + y2 = a2 that is bounded by the planes z = 0 and z = h [Refer to Fig. 4.52]. z (x, y, h)

Q

(x, y, 0)

P

y

x Fig. 4.52

Note

The equation x2 + y2 = a2 (in three dimensions) represents the right circular cylinder whose axis is the z-axis and base circle is the one with centre at the origin and radius equal to a. a

I= ∫

a2 − x2

h

∫ ∫ ( x + y + z ) dz dy dx

−a − a 2 − x2 0 a

=∫

−a −

a2 − x2

2  (x + y) h + h  dy dx ∫  2  a2 − x2 

a

= 2h ⋅ ∫

−a

a2 − x2

∫ 0

   x + h dy dx  2  [by using properties of odd and even functions]

Multiple Integrals

4.57

a  h = 2 h ∫  x +  a 2 − x 2 dx  2 −a a

=2h

2

∫

a 2 − x 2 dx

a 2 − x 2 is odd and

[∵ x

a 2 − x 2 is even ]

0

x a2 x = 2 h 2  a 2 − x 2 + sin −1  2 a  0 2

a

π = a2 h2 2

EXERCiSE 4(c) Part A (Short Answer Questions) 1. 2. 3. 4. 5. 6. 7. 8.

Define a line integral. What is the difference between a definite integral and a line integral? Define a surface integral. What is the difference between a double integral and a surface integral? Define a volume integral. What is the difference between a triple integral and a volume integral? Write down the formula that converts a surface integral into a double integral. Evaluate ∫ (x 2 dy + y 2 dx) where C is the path y = x from (0, 0) to (1, 1). C

9. Evaluate

∫

(x 2 + y 2 ) ds, where C is the path y = −x from (0, 0) to (−1, 1).

C

∫ (x dy − y dx) , where C is the circle x

2

10. Evaluate

+ y2 = 1 from (1, 0) to

C

(0, 1) in the counterclockwise sense. 11. Evaluate ∫∫ dS , where S is the surface of the parallelopiped formed by S

x = ± 1, y = ± 2, z = ± 3. [Hint:

∫∫ dS gives the area of the surface S] S

12. Evaluate

∫∫ dS, where S is the surface of the sphere x

2

+ y2 + z2 = a2.

S

13. Evaluate

∫∫ dS, where S is the curved surface of the right circular cylinder S

x2 + y2 = a2, included between z = 0 and z = h. 14. Evaluate ∫∫∫ dV , where V is the region of space bounded by the planes V

x = 0, x = a, y = 0, y = 2b, z = 0 and z = 3c.

Engineering Mathematics I

4.58 [Hint:

∫∫∫ dV gives the volume of the region V] V

15. Evaluate

∫∫∫ dV , where V is the region of space bounded by x

2

+ y2 +

V

z2 = 1. 16. Evaluate

∫∫∫ dV , where V is the region of space bounded by x

2

+ y2 = a2,

V

z = −h, z = h. Part B (1, 3)

17. Evaluate

∫

[x 2 y dx + (x 2 − y 2 ) dy ] along the (i) curve y = 3x2, (ii) line y = 3x.

(0, 0 )

18. Evaluate

∫ [(x

− y 2 + x) dx − (2x y + y ) dy ] from (0, 0) to (1, 1), when C is

2

C

(i) y2 = x, (ii) y = x. (a , 0 )

19. Evaluate

∫

( y 2 dx − x 2 dy ) along the upper half of the circle x2 + y2 = a2.

(- a , 0 )

20. Evaluate

∫

(x dy − y dx) , where C is the ellipse

C

x2 y2 + = 1 and described a2 b2

in the anticlockwise sense. 21. Evaluate

∫ [(x

2

− y 2 ) dx + 2 xy dy ] , where C is the boundary of the rectangle

C

formed by the lines x = 0, x = 2, y = 0, y = 1 and described in the anticlockwise sense. 22. Evaluate

∫ [(3x

2

− 8 y 2 ) dx + ( 4 y − 6 xy ) dy ], where C is the boundary of the

C

region enclosed by y2 = x and x2 = y and described in the anticlockwise sense. 23. Evaluate

∫ (x − y

2

) ds, where C is the arc of the circle x = a cos θ, y = a

C

π sin θ ; 0 ≤ θ ≤ . 2 24. Evaluate

∫ x ds, where C is the arc of the parabola x

2

= 2y from (0, 0 ) to

C

 1  1,  .  2  25. Evaluate

∫ [ xy dx + (x

2

+ z ) dy + (y 2 + x) dz ] from (0, 0, 0) to (1, 1, 1) along

C

the curve C given by y = x2 and z = x3. 26. Evaluate

∫ [ (3x

2

+ 6 y ) dx −14 yz dy + 20 xz 2 dz ] , where C is the segment of

C

the straight line joining (0, 0, 0) and (1, 1, 1).

Multiple Integrals 27. Evaluate

∫ [3x

2

4.59

dx + (2 xy − y ) dy − z dz ] from t = 0 to t = 1 along the curve

C

C given by x = 2t2, y = t, z = 4t3. 28. Evaluate

∫ xy ds

along the arc of the curve given by the equations x = a tan π θ, y = a cot θ, z = 2 a log tan θ from the point θ = to the point 4 π θ= . 3 29. Evaluate ∫ (xy + z 2 )ds , where C is the arc of the helix x = cos t, y = sin t, C

z = t from (1, 0, 0) to (−1, 0, π ). x y z 30. Find the area of that part of the plane + + = 1 that lies in the a b c    Hint: Area of the surface = positive octant. ds  ∫ ∫    S 31. Evaluate ∫∫ z dS , where S is the positive octant of the surface of the sphere S

x2 + y2 + z2 = a2. 32. Evaluate ∫∫ xy dS , where S is the curved surface of the cylinder x2 + y2 =

a2, 0 ≤ z ≤ k, included in the positive octant. 33. Find the volume of the tetrahedron bounded by the planes x = 0, y = 0, z = 0, x y z + + =1. a b c 34. Evaluate ∫∫∫ z dx dy dz , where V is the region of space bounded by the V

sphere x2 + y2 + z2 = a2 above the xoy-plane. 35. Evaluate

∫∫∫ ( x

2

+ y 2 ) dx dy dz , where V is the region of space inside the

V

cylinder x2 + y2 = a2 that is bounded by the planes z = 0 and z = h.

4.9

GAMMA AnD BEtA FUnCtiOnS ∞

Definitions  The definite integral

∫e

−x

x n − 1 dx exists only when n > 0 and when it

0

exists, it is a function of n and called Gamma function and denoted by Γ(n) [read as “Gamma n”]. ∞

Γ(n) = ∫ e− x x n − 1 dx

Thus

0 1

∫x

The definite integral

m −1

(1 − x) n − 1 dx exists only when m > 0 and n > 0 and

0

when it exists, it is a function of m and n and called Beta function and denoted by β (m, n) [read as “Beta m, n”]. 1

Thus

β (m, n) = ∫ x m − 1 (1 − x) n − 1dx: 0

Engineering Mathematics I

4.60

∞

Γ(1) = ∫ e− x dx = ( − e− x )∞ 0 =1.

Note

0 1

β(1, 1) = ∫ dx = 1 . 0

4.9.1

Recurrence Formula for Gamma Function ∞

Γ(n) = ∫ e− x x n − 1 dx 0

∞

−x = − (x n− 1 e− x )∞ x n − 2 dx 0 + ∫ (n − 1) e

[integrating by parts]

0

 xn −1  = (n − 1) Γ(n − 1), since lim  x  = 0 n → ∞  e  This recurrence formula Γ(n) = (n − l) Γ (n − 1) is valid only when n > 1, as Γ(n − 1) exists only when n > 1.

Cor. Γ(n + 1) = n !, where n is a positive integer. Γ(n + l) = n Γ(n) = n (n − l) Γ(n − l) = n (n − l) (n − 2) Γ (n − 2) =..................... = n (n − l) (n − 2) ... 3.2.1 Γ(l) = n ! ( Γ(l) = 1) ∴

1. Γ(n) does not exist (i.e. = ∞), when n is 0 or a negative integer. 2. When n is a negative fraction, Γ(n) is defined by using the recurrence formula. i.e. when n < 0, but not an integer, 1 Γ(n) = Γ(n + 1) n For example, Γ( − 3.5) = 1 Γ( − 2.5) ( − 3.5) 1 1 Γ( −1.5) = ⋅ ( − 3.5) ( − 2.5) 1 = Γ( − .5) (3.5) (2.5) ( −1.5) Γ (0.5) = (3.5) (2.5) (1.5) (0.5)

Note

The value of Γ(0.5) can be obtained from the table of Gamma functions, though its value can be found out mathematically as given below. 1 Value of Γ    2 

Multiple Integrals

4.61

1 − 1 ∞ By definition, Γ   = ∫ e−t t 2 dt  2  0 ∞

= ∫ e− x ⋅ 2

0

1 ⋅ 2 x dx x

(on putting t = x2)

∞

= 2 ∫ e− x dx 2

0 ∞ ∞   1  Γ   = 2 e− x2 dx ⋅ 2 e− y 2 dy ∫ ∫   2  2

0

∴

Now

[

the variable in a definite integral

0

is only a dummy variable] ∞ ∞

=4∫ 0

∫e

−(x 2 + y 2 )

dx d y

(1)

0

∴ [ the product of two definite integrals can be expressed as a double integral, when the limits are constants]. Now the region of the double integral in (1) is given by 0 ≤ x < ∞ and 0 ≤ y < ∞, i.e. the entire first quadrant of the xy-plane. Let us change over to polar co-ordinates through the transformations x = r cos θ and y = r sin θ. Then dx dy = |J| d r d θ = r d r d θ The region of the double integration is now given by 0 ≤ r < ∞ and 0 ≤ θ ≤ π . 2 Then, from (1), we have π /2 ∞    1   −r 2  Γ    = 4 ∫ ∫ e r d r dθ       2  2

0

0

∞

 1 2 = 4 ∫ dθ − e−r   2 0 0 π /2

π /2

= 2 ∫ dθ 0

=π 1 Γ   = π  2 

∴

4.9.2

Symmetry of Beta Function β (m, n) = β (n, m) 1

β (m , n) = ∫ x m − 1 (1 − x) n − 1 dx

By definition,

0 a

Using the property

∫ 0

a

f (x) dx = ∫ f (a − x) dx in (1), 0

(1)

Engineering Mathematics I

4.62

1

β (m, n) = ∫ (1− x) m−1 {1 − (1 − x) n−1 dx 0 1

= ∫ x n −1 (1− x) m −1 dx 0

= β (n, m).

4.9.3 trigonometric Form of Beta Function 1

By definition, β (m, n) = ∫ x m −1 (1− x) n −1 dx 0

Put x = sin2 θ

∴ dx = 2 sin θ cos θ dθ

The limits for θ are 0 and

π . 2

π/2

β (m, n) = ∫ sin 2 m − 2 θ ⋅ cos 2 n − 2 θ ⋅ 2 sin θ cos θ dθ

∴

0 π/2

= 2 ∫ sin 2 m −1 θ ⋅ cos 2 n −1 θ dθ 0 π/2

1 θ ⋅ cos 2 n −1 θ dθ = β (m, n) 2 0 The first argument of the Beta function is obtained by adding 1 to the exponent of sin θ and dividing the sum by 2. The second argument is obtained by adding 1 to the exponent of cos θ and dividing the sum by 2.

∫ sin

Note

2 m −1

π/2

∫ sin

Thus

0

4.9.4

p

1  p + 1 q + 1 θ cos q θ dθ = β  ,  2  2 2 

Relation Between Gamma and Beta Functions β (m, n) =

Γ ( m) Γ ( n) Γ ( m + n) ∞

Consider

∞

Γ ( m) Γ ( n) = ∫ e

−t

t

m −1

dt ⋅ ∫ e− s ⋅ s n −1 ds

0

0

In the first integral, put t = x and in the second, put s = y . 2

2

∞

∴

∞

Γ (m) ⋅ Γ (n) = 2 ∫ e− x x 2 m −1 dx ⋅ 2 ∫ e− y y 2 n −1 dy 2

0

2

0

Multiple Integrals

4.63

∞ ∞

=4∫

∫e

−( x 2 + y 2 )

0

x 2 m−1 ⋅ y 2 n−1 dx dy

0

π/2 ∞

=4∫

∫e

−r 2

0

(r cos θ ) 2 m −1 ⋅ (r sin θ ) 2 n −1 r dr dθ

0

[changing over to polar co-ordinates] ∞

π/2

= 4 ∫ cos 2 m −1 θ sin 2 n −1 θ dθ ⋅ ∫ e−r r 2 m+ 2 n−2 ⋅ r dr 2

0

0 ∞

= β (m, n) ∫ e−r r 2 ( m+ n−1) ⋅ 2r dr 2

0 ∞

= β (m, n) ⋅ ∫ e−u ⋅ u m+ n −1 du

[putting r2 = u]

0

= β (m, n) ⋅ Γ (m, n) β (m, n) =

∴

Γ ( m) Γ ( n) ⋅ Γ ( m + n)

  1  Γ    1 1    2  1  Putting m = n = in the above relation, β  ,  =  2 2  Γ (1) 2

Cor.

2

2

  1  1 1    Γ    = β  ,    2 2     2  

∴

π/2

= 2 ∫ sin 0 θ ⋅ cos 0 θ dθ 0

=π 1 Γ   = π  2 

∴

WORKED EXAMPLE 4(d) ∞

Example 4.1 Prove that

∫e

− ax

x n −1 dx =

0 1

Hence find the value of

0

∞

In

∫e

− ax

0

∫

  1  x q −1  log     x 

Γ (n) , where a and n are positive. an

p −1

x n −1 dx , put ax = t, so that dx =

dx. dt a

Engineering Mathematics I

4.64 ∞

∴

∫ 0

n −1

∞ t  e−ax x n −1 dx = ∫ e−t    a  0

=

⋅

dt a

∞

1 an

∫e

−t

t n −1 dt

0

1 = n Γ(n) a 1 1 In I = ∫ x q −1 log    x  0

(1)

p −1

dx ,

1 = ey x i.e. x = e−y Then dx = − e−y dy Also the limits for y are ∞ and 0. put

0

∴

I = ∫ e−(q −1)y ⋅ y p −1 ⋅ ( − e− y ) dy ∞

∞

= ∫ e−qy y p −1 dy 0

=

1 ⋅ Γ(p ) [by (1)] . qp ∞

Example 4.2 Prove that β (m , n) = ∫ 0 1

Hence deduce that β (m , n) = ∫ 0

x m −1 dx . (1+ x) m + n

m −1

x + x n −1 dx . (1+ x) m + n

1

By definition, β (m , n) = ∫ t m −1 (1− t ) n −1 dt

(1)

0

In (1), put t =

1 x . Then dt = dx (1+ x) 2 1+ x

  ∵x = t   1− t 

When t = 0, x = 0; when t = 1, x = ∞ Then (1) becomes, m −1

 x   β (m, n) = ∫  1+ x  ∞

0

∞

=∫ 0 1

=∫ 0

n −1

 1   ⋅ 1+ x 

⋅

1 dx (1+ x) 2

x m −1 dx (1+ x) m + n

(2) ∞

x m −1 x m −1 dx + ∫ d m+n (1+ x) (1+ x) m + n 1

(3)

Multiple Integrals ∞

In

∫ 1

4.65

1 1 x m−1 dx , put x = . Then dx = − 2 dy m+n y y (1 + x)

When x = 1, y = 1; when x = ∞, y = 0 ∞

∫

∴

1

1 0  1 x m−1 y ( m−1) ⋅ − 2  dy dx = ∫ m+n m+n  (1 + x)  y   1  1  1 +  y  1

=∫ 0 1

=∫ 0 1

=∫ 0

y m+n dy (1 + y ) m+ n ⋅ y m+1 y n−1 dy (1 + y ) m+ n x n−1 dx (1 + x) m+ n [changing the dummy variable]

(4)

Using (4) in (3), we have 1

β(m, n) = ∫ 0

x m−1 + x n−1 dx . (1 + x) m+ n

1

∫x

Example 4.3 Evaluate

m

(1− x n ) p dx

in terms of Gamma functions and

0 1

hence find

∫ 0

dx 1− x n

. 1

I = ∫ x m (1− x n ) p dx,

In

0

xn = t; n−1 nx dx = dt 1 dt ∴ dx = ⋅ 1 n 1− n t When x = 0, t = 0; when x = 1, t = 1. 1 m 1 1 −1 I = ∫ t n (1− t ) p ⋅ ⋅ t n dt n 0 put then

=

1 n

1

∫

t

m +1 −1 n

⋅ (1− t ) p dt

0

 1  m +1 = β  , p + 1   n  n

Engineering Mathematics I

4.66

 m + 1 Γ  ⋅ Γ( p + 1) 1  n  =  n  m + 1 Γ + p + 1  n  1

∫ 0

1

dx 1− x n

−

= ∫ x 0 (1− x n )

1 2

(1)

dx

0

1 Here m = 0, n = n, p = − . 2 Using (1); we have 1

∫ 0

1 1 Γ  ⋅ Γ   1  n   2  dx = 1− x n n Γ  1 + 1   n 2  1 Γ    n  π = ⋅  1 1 n Γ  +   n 2 

Example 4.4 Prove that β (n, n) =

(or)

β ( n, n ) =

π Γ( n ) .  1 22 n−1 Γ n +   2

 1 1 ⋅ β n,  22 n−1  2  π/2

β (n, n) = 2 ∫ sin 2 n−1 θ ⋅ cos 2 n−1 θ dθ

[using trigonometric form]

0 π/2

2 n−1

= 2 ∫ (sin θ cos θ )

dθ

0 π/2

=2∫ 0

= =

1 2 2 n− 2 1 2 2 n− 2

2 n−1

 sin 2θ    2 

dθ

π/2

∫

sin 2 n−1 2θ dθ

0 π

∫ sin 0

2 n−1

φ

dφ , putting 2θ = φ 2

Multiple Integrals

4.67

π/2   π 2 n−1   f (sin φ ) dφ = 2 f (sin φ ) dφ  sin φ d φ 2 n− 2 ∫ ∫ ∫   2  0 0 0  1 1  1 = 2 n−2 ⋅ β n,  2  2  2  1 1 = 2 n−1 β n,   2  2 π/2

1

=

1 Γ(n) ⋅ Γ    2  1 = 2 n−1 ⋅  1 2 Γ n +   2 π ⋅ Γ( n )  1 22 n−1 ⋅ Γ n +   2

=

∞

∫xe

n −a 2 x 2

Example 4.5 Show that

dx =

0 ∞

Deduce that

∫e

−a 2 x2

 n + 1 1 . Γ  n +1  2  2a

π . Hence show that 2a

dx =

0 ∞

∞

∫

cos ( x 2 ) dx = ∫ sin ( x 2 ) dx =

0

0

∞

In I = ∫ x n e−a x dx , put ax = t ; then dx = 2 2

0

1 π 2 2

dt 2a t

When x = 0, t = 0; when x = ∞, t = ∞. n ∞  t  −t dt  I = ∫   e ∴  a  2a t 0  =

1 2a n+1

∞

∫

t

n−1 2

⋅ e−t dt

0

 n + 1 1 = n+1 Γ   2  2a In (1), put n = 0. ∞

∫e

−a x

Then

0

In (2), put a =

2 2

1− i 2

1 Γ    2  π dx = = 2a 2a

; then a 2 = −i

(1)

(2)

Engineering Mathematics I

4.68 ∞

∫e

∴

ix2

π

dx =

2 (1− i )

0

=

π 2 2

(1 + i )

Equating the real parts on both sides, ∞

∫ cos (x

2

1 π . 2 2

) dx =

0

Equating the imaginary parts on both sides, ∞

∫ sin (x

2

) dx =

0

1 π . 2 2

Example 4.6 Evaluate (i) (ii)

∫

b a

∫

a

−a

(x − a ) m−1 (b − x) n−1 dx and (a + x) m−1 ⋅ (a − x) n−1 dx in terms of Beta function. b

(i) In I1 = ∫ (x − a ) m−1 (b − x) n−1 dx , a

put x − a = y; then dx = dy When x = a, y = 0; when x = b, y = b − a b− a

∴

I1 =

∫

n−1

y m−1 {(b − a ) − y }

dy

0 b− a

n−1

 y  = (b − a ) n−1 ∫ y m−1 1−   − b a  0 y = t ; then dy = (b − a) dt b−a When y = 0, t = 0; when y = b − a, t = 1.

In (1), put

1

n−1

I1 = (b − a ) m+ n−1 ∫ t m−1 (1 − t )

∴

0

= (b − a ) m+ n−1 β (m, n) a

(ii) In I 2 = ∫ (a + x) m−1 (a − x) n−1 dx , −a

put a + x = y; then dx = dy When x = − a, y = 0; when x = a, y = 2a. 2a

∴

I 2 = ∫ y m−1 (2a − y ) n−1 dy 0

dt

dy

(1)

Multiple Integrals 2a

= (2a ) n−1

∫ 0

4.69

n−1

 y y m−1 1−   2a 

(2)

dy

y = t ; then dy = 2a dt. 2a When y = 0, t = 0; when y = 2a, t = 1. In (2), put

1

I 2 = (2a ) m+ n−1 ⋅ ∫ t m−1 (1− t ) n−1 dt .

∴

0

= (2a )

m + n−1

β (m,n)

∞ − x2

Example 4.7 Prove that

∫

e

x

0 ∞

In I1 = ∫ 0

e− x

∞

dx ×∫ x 2 e − x dx = 4

0

2

x

dx , put x2 = t; then dx =

π 4 2

.

dt dt = 2x 2 t

When x = 0, t = 0; when x = ∞, t = ∞ ∞ −t ∞ 3 − 1 e dt I1 = ∫ 1/ 4 ⋅ = ∫ e−t ⋅ t 4 dt ∴ t 2 t 2 0 0 1 1 = Γ   2  4 ∞

In I 2 = ∫ x 2 e− x dx , put x4 = s; then dx = 4

0

ds ds = 3/ 4 3 4x 4s

When x = 0, s = 0; when x = ∞, s = ∞. ∞

∞

∴

I2 = ∫

s e−s ⋅

0

1 − 1 ds 4 −s = s e ds 4 s 3 / 4 4 ∫0

1  3 = Γ   4  4 ∞

∴

∫ 0

e −x

2

∞

4 1 1  3 dx × ∫ x 2 e − x dx = Γ   Γ   8  4  4 x 0

From Example 4.4; β (n, n) = i.e.

∴

1 2

2 n−1

Γ(n) Γ(n) 1 Γ(n) ⋅ π = 2 n−1 ⋅  1 Γ(2n) 2 Γ n +   2  π Γ(2n) 1 Γ(n) ⋅ Γ n +  =  22 n−1 2

 ⋅ β n, 

1   2 

(1)

Engineering Mathematics I

4.70 Putting n = 1 , we get 4

1 π ⋅ Γ    2 

1  3 Γ  ⋅ Γ   =  4   4 

−

2 =π 2

1 2

(2)

Using (2) in (1); ∞ − x2

∫

e

x

0 ∞

Example 4.8 Evaluate

∞

dx × ∫ x 2 e− x dx = 4

0

x m−1

∫ (1 + x

=

)

In I = ∫ 0

dx and deduce that

∫ 0

∞

π . Hence show that  mπ  n sin    n  ∞

∞

n p

0

π 2 π . = 8 4 2

dx

∫ 1+ x

4

=

0

π

x m−1 dx. 1+ xn

.

2 2

x m−1 1 dx , put t = (1 + x n ) p 1+ xn

1 1− t ∴ nx n−1 dx = − 2 dt t t When x = 0, t = 1; when x = ∞, t = 0 Then x n =

1

∴

I=∫

t

 m−1  −  n 

0

=

1 n

1

∫

t

⋅ (1− t ) t− p

m p − −1 n

m−1 n

dt

⋅ nt 2 ⋅ t

m

−

(n −1) n

(1− t )

n −1 n

−1

⋅ (1− t ) n dt

0

1  m m β  p − ,   n  n n  m m Γ  p − ⋅ Γ   1  n   n  = Γ(p ) n Putting p = 1 in (1), we get ∞ x m −1 dx 1  m   m  ∫ 1 + x n = n Γ 1− n  Γ  n  0 =

 πm  π = cosec    n  n

(1)

   H int : Use Γ(α ) Γ (1− α ) = π   sin απ  

(2)

Multiple Integrals

4.71

Taking m = 1 and n = 4 in (2), we get ∞ π dx π ∫ 1+ x 4 = 4 ⋅ cosec  4  0

=

π 2 2

∫∫ x

Example 4.9 Find the value of

y n−1 dx dy , over the positive quadrant of

x2 y 2 + = 1 , in terms of Gamma functions. a 2 b2

the ellipse

Put

m−1

y

x y = X and = Y a b

x

O

a b dX and dy = dY . Fig. 4.53 2 X 2 Y The region of double integration in the xy-plane is given by x ≥ 0, y ≥ 0 and Then dx =

x2 y 2 + ≤1 , shown in Fig. 4.53. a 2 b2 ∴ The region of integration in the XY-plane is given by X ≥ 0, Y ≥ 0 and X + Y ≤ 1, shown in Fig. 4.54. The given integral I=∫

∫ (a

m −1

X)

n −1

⋅(b Y )

=

ambn 4 ambn 4

∫∫

X

m −1 2

Y

n −1 2

ab

Y

dX dY

4 X Y

∆O AB

=

B

X+Y=1 O

dX d Y

A

∆O AB 1 1−Y

∫∫ 0

X

m −1 2

Y

n −1 2

Fig. 4.54

dX d Y

0

I=

ambn 4 m

=

a b 2m

n

1

n

−1

∫ Y2 ⋅ 0 1

n

2 ( X m 2 )0 1−Y dY m

−1

m2 ∫ Y 2 ⋅(1−Y ) dY 0

n m  a b β ⋅ , + 1 2 2  2m n m  Γ  ⋅ Γ  +1 a m b n  2   2  = ⋅ m n  2m Γ  + +1 2 2  m

=

n

X

Engineering Mathematics I

4.72

m  m   n  Γ ⋅ Γ  a m b n 2  2   2  = ⋅ m n  2m Γ  + +1 2 2   m   n  Γ  ⋅ Γ    2   2 a mb n ⋅ =  m + n  4 Γ  +1  2  Example 4.10 Find the area of the astroid x 2/3 + y tions. By symmetry of the astroid, required area A =

2/3

= a 2/3, using Gamma func-

y

4 × area of OACB = 4 ∫∫ dx dy

B

OACB

 x Put    a 

23

 y = X and   a

23

C

=Y O

i.e. x = aX 3/2 and y = aY 3/2 3 3 aX 1 2 and dy = aY 1 2 dY 2 2 The region of integration in the xy-plane is

A

x

∴ dx =

Fig. 4.55

 y  x given by x ≥ 0, y ≥ 0 and   +   ≤ 1 , as shown in Fig. 4.55.  a   a  ∴ The region of integration in the XY-plane is given by X ≥ 0, Y ≥ 0 and X + Y ≤ 1 as shown in Fig. 4.56. 23

23

Y

∴

9 A = 4× a 2 ∫ ∫ X 1 2 Y 1 2 dX dY 4 ∆O PQ 1 1−Y

= 9a 2 ⋅ ∫

∫

0

0

Q

O

X

12

Y

12

d X dY 1−Y

1 2  = 9 a 2 ∫ Y 1 2  X 3 2  3 0

Fig. 4.56

dY

0

1

= 6 a 2 ∫ Y 1 2 (1−Y ) 3 2 dY 0

 3 5 = 6 a 2×β  ,   2 2  3  5 Γ  Γ     2 = 6 a 2× 2 Γ (4)

P

X

Multiple Integrals

4.73 ∴

1 1 1 3 1 = a 2 × Γ  × × × Γ   2 2  2 2 2 3 = πa 2 8

[

Γ(4) = 3!]

 1  ∵Γ   = π    2  

12 Example 4.11 Evaluate ∫∫ [xy (1− x − y )] dx dy , over the area enclosed by the lines x = 0, y = 0 and x + y = 1 in the positive quadrant. 1 1− x

Given Intergral I = ∫

∫

0

0

x 1 2 y 1 2 (1− x − y )1 2 dy dx,

y

a

1

= ∫ x 1 2 dx ∫ y 1 2 (a − y )1 2 dy 0

0

where a = 1 − x.

(1)

a

Consider

∫

y m−1 (a − y ) n−1 dy

x

O

0

Fig. 4.57

n −1

a  y = a n −1 ∫ y m −1 1−   a

dy

0

1

= a n −1 ∫ a m −1 z m −1 (1− z ) n −1 a dz 0

  putting y = z    a

= a m+ n−1 ⋅ β (m, n)

Note

This result (2) will be of use in the following worked examples also.  3 Using (2) in (1)  note that m = n =  , 2   1  3 3 I = ∫ x1 2 (1− x) 2 β  , dx  2 2 0

 3 3  3  = β  , ×β  , 3  2 2  2   3  3  3 Γ  ⋅ Γ   Γ  ⋅ Γ (3)  2  2   = × 2 9 Γ (3) Γ    2  3 1 1 π × π ×Γ   2 2 =2 7 5 3  3  × × ×Γ  2 2 2  2  =

2π 105

(2)

Engineering Mathematics I

4.74

Example 4.12 Show that the volume of the region of space bounded by the x y z abc . + + = 1 is 90 a b c

coordinate planes and the surface Required volume is given by

Vol = ∫∫∫ dz dy dx , where V is the region of space given. V

Put

x = X, a

y z = Y, =Z b c

i.e. x = aX2, y = bY2, z = cZ2 ∴ dx = 2aX dX, dy = 2bY dY, dz = 2cZ dZ ∴ Vol = ∫

∫ ∫ 8abc XYZ dZ dY dX , where V′ is the region of space in XYZ-space V'

defined by X ≥ 0, Y ≥ 0, Z ≥ 0, X + Y + Z ≤ 1 [Refer to Fig. 4.58] Z 1

∴

Vol = 8 abc ∫ dX 0

1− X

∫ dY ∫ XYZ dZ 0

0 1− X

1

= 8 abc ∫ X dX 0

∫ 0

1− X −Y

Z2  Y dY    2 0

Y

1− X

1

= 4 abc ∫ X dX 0

1− X −Y

∫

Y ⋅ (1 − X − Y ) 2 dY

X

0

Fig. 4.58 1

= 4 abc ∫ X (1 − X ) 4 ⋅ β (2, 3) dX [by step (2) of Example (4.11)] 0

Γ(2) Γ(3) ⋅ β (2, 5) Γ(5) 1× 2 Γ(2) Γ(5) = 4 abc × × Γ(7 ) 24 abc 1× 24 = × 720 3 abc = 90 = 4 abc ⋅

Example 4.13 Evaluate

∫∫∫

dx dy dz 1 − x2 − y 2 − z 2

, taken over the region of space

in the positive octant bounded by the sphere x2 + y2 + z2 = 1. Put x2 = X, y2 = Y, z2 = Z

Multiple Integrals dX

dx =

∴

2 X

, dy =

dY 2 Y

4.75 , dz =

dZ 2 Z

The region of integration in xyz-space is defined by x ≥ 0, y ≥ 0, z ≥ 0 and x2 + y2 + z2 ≤ 1. ∴ The region of integration V in the XYZ-space is defined by X ≥ 0, Y ≥ 0, Z ≥ 0 and X + Y + Z ≤ 1. 1 1 −1 −1 −1 ∴Given integral I = ∫∫∫ X 2 Y 2 Z 2 (1 − X − Y − Z ) 2 dX ⋅ dY ⋅ dZ 8 V = =

1 8 1 8

1

∫

X

−

1− X

1 2

dX

∫

Y

−

1 2

1− X −Y

dY

0

0 1

∫X

dX

∫

Y

1 − 2

1 2

-

1

(1 − X − Y − Z ) 2 dZ 1

dY (1 − X − Y ) 2

0

0

−

0

1− X

1 − 2

∫

Z

+

1 −1 2

1 ⋅ β  ,  2

1   2 

[by step (2) of Example (4.11)] = =

1

π 8

∫

π 4

1

X

−

1 2

−

1 2

0

∫X

 dX 2Y 

1 1− X 2

   0

  1 ∵ β  ,   2

 1   = π  2

(1 − X )1/2 dX .

0

π  1 3  β ,  4  2 2  1  3 Γ   Γ   π  2   2  = ⋅ Γ(2) 4 =

=

π2 8

Example 4.14 Evaluate

∫∫∫

a 2 b 2 c 2 − b 2 c 2 x 2 − c 2 a 2 y 2 − a 2 b 2 z 2 dx dy dz ,

V

where V is the region defind by x ≥ 0, y ≥ 0, z ≥ 0 and z  y  x Put   = X ,   = Y ,   = Z  c   b   a  2

i.e.

2

2

x2 y2 z2 + + ≤1 . a2 b2 c2

x=a X, y=b Y, z=c Z dx =

a

dX , dy =

b

dY , d z =

c

dZ 2 X 2 Y 2 Z The region V′ of integration in the XYZ-space is defined by X ≥ 0, Y ≥ 0, Z ≥ 0, X + Y + Z ≤ 1. ∴

Engineering Mathematics I

4.76

∴

abc dX dY dZ

Integral = abc ∫∫∫ 1 − X − Y − Z

8 X

V ' 2

=

2

a b c 8

2

2 2 2

=

abc 8

1

∫

X

1 − 2

1− X

∫

dX

0

Y

1 − 2

∫X

1 − 2

1− X

∫

dX

0

Z

dY

∫

Z

−

1 2

1

(1 − X − Y − Z ) 2 dZ

0

0

1

Y

1− X −Y

Y

1 − 2

0

1 dY (1 − X − Y ) ⋅ β  ,  2

3   2 

[by step 2 of Example (4.11)] 1 abc ⋅ β  ,  2 8 2 2 2

=

3  ⋅ X 2  ∫0 1

1 − 2

3 1  dX (1 − X ) 2 ⋅ β  , 2  2 

[by step 2 of Example 4.11] 1 3 1  1 5 abc = ⋅ β  ,  ⋅ β  , 2 ⋅ β  ,   2 2   2   2 2  8 1 5 1  3 1 Γ   ⋅ Γ   Γ   ⋅ Γ(22) Γ   ⋅ Γ   2 2 2  2   2      2  2  2 a b c ⋅ = ⋅ ⋅ 5 Γ(2) Γ(3) 8 Γ    2  2 2 2

=

π 2 a 2 b2 c2 ⋅ 32

Example 4.15 Find the value of

∫∫∫ x

l −1

y m −1 z n −1 (1− x − y − z ) p -1 dx dy dz , taken

over the interior of the tetrahedron bounded by x = 0, y = 0, z = 0 and x + y + z = 1, in terms of Gamma functions. 1− x

1

Given integral = ∫ x l − 1 dx

∫

0

0

1

1− x

= ∫ x l −1 dx 0

∫

1− x − y

y m −1 dy

∫

z n −1 (1− x − y − z ) p - 1 dz ,

0

y m −1 (1− x − y ) n + p -1 ⋅ β (n, p ) dy ,

0

[by step (2) of Example (4.11)] 1

= β (n, p ) ∫ x l −1 (1− x) m + n + p −1 β (m, n + p ) dx , 0

[by step (2) of Example (4.11)] = β (n, p ) ⋅ β (m, n + p ) ⋅ β (l , m + n + p ) Γ(n) Γ(p ) Γ(m) Γ(n + p ) Γ(l ) ⋅ Γ(m + n + p ) = ⋅ ⋅ Γ(n + p ) Γ(m + n + p ) Γ(l + m + n + p ) Γ(l ) Γ(m) Γ(n) Γ(p ) = Γ(l + m + n + p )

Multiple Integrals

4.77

EXERCiSE 4(d)

Part A (Short Answer Questions) ∞

∫xe

4 − x2

1. Prove that

dx =

0 ∞

2. Evaluate

∫ xe

− x3

0

3 π. 8

 5 dx , given that Γ  = 0.902 .  3  π/ 2

3. Find the value of

∫ sin x cos 3

5/2

5

θ cos 7 θ dθ .

x dx .

0 π/ 2

4. Find the value of

∫ sin 0 π/ 2

5. Find the value of

∫

tan θ dθ in terms of Gamma functions.

0

1 1  3 cot θ dθ = Γ Γ  . 2  4   4 

π/ 2

6. Prove that

∫ 0

π/ 2

7. Find the value of

π /2

∫

sin θ dθ × ∫

0

0

π/ 2

8. Prove that

∫ 0

0

1

9. Prove that

π/ 2

cos x dx × ∫

∫ 0

  1  log     x    ∞

10. Find the value of

∫ 0

∞

11. Assuming that

∫ 0

dx cos x

dθ sin θ

⋅

=π .

n −1

dx = Γ(n) .

xn dx ( n >1) . nx

x α−1 π π , prove that Γ (α ) ⋅ Γ(1− α ) = , dx = 1+ x sin απ sin απ

where α is neither zero nor an integer. [Hint: put x = tan2 θ] ∞

12 Find the value of ∫ e−kx dx . 2

−∞

13. Prove that β (m + 1, n) = m . β (m, n + 1) n 14. Prove that β (m + 1, n) β (m, n + 1) = β (m, n) . m n m+n

Engineering Mathematics I

4.78 ∞

x m −1 dx in terms of a Beta function. ( a + bx ) m + n

∫

15. Find the value of

0

 a   H int: put x = t   b 

16. Prove that β(m + 1, n) + β(m, n + 1) = β(m, n). 2

−

∫

17. Find the value of

(8 − x 3 )

1 3

dx in terms of Gamma functions.

0 a

18. Prove that 19. 20. 21. 22.

∫x

m

(a − x) n dx = a m + n + 1 ⋅ β (m + 1, n + 1) .

0

Define Gamma and Beta functions. Derive the recurrence formula for the Gamma function. When n is a positive integer, prove that Γ(n + 1) = n! State the relation between Gamma and Beta functions and use it to find the 1 value of Γ  .  2 

Part B 23. Prove that

∞ ∞

∫∫e

−(ax 2 + by 2 )

0

x 2 m−1 y 2 n−1dx dy =

0

1 m

4a b n

Γ(m) Γ(n) . 1

24. When n is a positive integer and m > −1, prove that

∫x

m

(log x) n dx =

0

(−1) n n! . ( m + 1) n +1 1

25. Prove that

∫ 0

x m−1 (1− x) n −1 β( m, n) . dx = n m +n ( a + bx ) a (a + b) m    H int: Put x = z   a + bx a + b  

  26. Express β n + 1 , n + 1  in terms of Gamma functions in two different ways  2 2   1 π Γ(2 n +1) and hence prove that Γn +  = 2 n .  2  2 Γ( n +1) ∞

27. Prove that

∫

∞ − x2

xe

dx × ∫

0

28. Prove that

0

2

dx =

x

∞

∞

∫

x e− x dx × ∫ x 2 e− x dx = 8

1

∫ 0

0

x 2 dx 1− x

π

4

1

×∫ 0

dx 1− x

4

=

.

2 2 π

4

0

29. Prove that

e− x

16 2

π . 4

.

Multiple Integrals ∞

30. Evaluate (i)

∫ 0

dx , (ii) 1+ x4

31. Find the value of

∫∫

∞

∫ 0

4.79 ∞

x 2 dx x 2 dx and (iii) ∫ (1 + x 4 )3 (1 + x 4 ) 2 0

[Hint: put x2 = tan2 θ] x y dx dy , taken over the area x ≥ 0, y ≥ 0, x + y ≤ 1 in m

n

terms of Gamma functions, if m, n > 0. 32. Find the value, in terms of Gamma functions, of ∫∫∫ x m y n z p dx dy dz taken over the volume of the tetrahedron given by x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z ≤ 1. 2/ 3 2/ 3  y  x 33. Find the area in the first quadrant enclosed by the curve   +   = 1  b   a  and the co-ordinate axes.

∫∫ x

y n−1 (1− x − y ) p −1 dx dy , taken over the area in the first quadrant enclosed by the lines x = 0, y = 0, x + y = 1. x y z 35. The plane + + =1 meets the axes in A, B and C. Find the volume of the a b c tatrahedron OABC. x2 y 2 z 2 36. Find the volume of the ellipsoid 2 + 2 + 2 = 1 . a b c 37. Find the volume of the region of the space bounded by the co-ordinate planes n n n  x  y  z and the surface   +   +   =1 and lying in the first octant. a b c 34. Evaluate

m−1

∫∫∫

xyz (1− x − y − z ) dx dy dz, taken over the tetrahedral volume in the first octant enclosed by the plane x = 0, y = 0, z = 0 and x + y + z = 1.

38. Evaluate

∫∫∫ x

yz dx dy dz , taken throughout the volume in the first x y z octant bounded by x = 0, y = 0, z = 0 and + + =1 . a b c

39. Evaluate

40. Evaluate

2

∫∫∫ xyz dx dy dz, taken over the space defined by x ≥ 0, y ≥ 0, z ≥ 0

2 2 2 and x + y + z ≤1 . 2 2 2 a b c

AnSWERS

Exercise 4(a) (1) 4 1 (4) 2

(2) log a log b π (5) 4

(3) 2 9 (6) 2

Engineering Mathematics I

4.80

y y=b

(7)

1 2

x=a x

(8) x = – a

O y=–b Fig. 4.59

y y y=x

x=1

(9) R y=0

O

(10) R

x=0

x

x

O

y=0

Fig. 4.60

Fig. 4.61

x y + =1 a b

(11) x = 0 O

y=0

x2 + y2 = a2

1 1− x

∫∫

(12)

0

x

1 1− y

f ( x, y ) dy dx (or)

0

∫∫ 0

f ( x , y ) dx dy

0

Fig. 4.62 b 2 a − x2 a

a

(13) ∫

∫

0

(14) (15)

f ( x, y ) dy dx (or)

0 1

0

0

2

1

∫∫ 0

1

f ( x, y ) dx dy (or)

y2 4

πa 3 6

π 3 a 2

(27) 6

(29)

f ( x , y ) d y dx

x

1 2 x

f ( x, y ) dx dy (or)

8 19 (21) log 2 − 3 9 (24)

f ( x , y ) d x dy

0

1

∫∫ 0

(16) 2 log 2 (19)

∫ ∫ 0

y

∫∫

a 2 b − y2 b

b

1 (8 log 2 − 5) 16

∫∫ 0

f ( x , y ) d y dx

0

(17)

a3 6

(20)

1 720

(22) 1 (25)

1 ab (a + b) 3

(28)

33 2

(30)

1 48

(18)

π 4

(23)

3 2

(26)

344 105

Multiple Integrals

4.81

Exercise 4(b) a

(1) (3)

a

1

∫∫ 0

y

a

y

f ( x , y ) dx dy

∫∫ 0

f ( x , y ) dx dy

(2) (4)

(7)

∫∫

1

∫∫

f ( x, y ) dy dx.

x

f ( x , y ) d y dx

(6)

∫∫

f ( x , y ) dx dy

0

0

1

1− y 2

a

a2 − x2

∫ ∫

f ( x , y ) dx dy

(8)

0

f ( x , y ) d y dx

0

∞ 1/ x

f ( x , y ) dx dy

(10)

y2 4

f ( x , y ) d y dx

0

16 (12) 3 1 2 (15) 2 (e−1) 1 log 2 (18) 2

(14) 1 9 3 a 5 π 4 a (20) 8 π (23) 4 (17)

(13) 1 (16) 2 (19)

(21) 3 (24)

32 3 π 1 + (29) 2 3  2π 3  (32) a 2  −  2   3 (26)

3 πa 2 4 a4 log 1+ 2 (38) 4

(

(22) 8 log 2

3 8

(25)

2 3

16 (28) 3 ab

(30) 3 π

(31)

a2 (3π − 8) 2

(

2 π 1− e−a 4 4 3 a (39) 3

(36)

)

241 60

16 2 (27) 3 a

(33)

(35)

π 2 2 a h 2

∫∫ 0

πa 4

(44) 16 π

∫ ∫ 0

1

∫∫ 0

(41)

1

f ( x , y ) d y dx

0

2

(11)

0

0

0

(9)

0

a a− y

1 1− x

(5)

∫∫ 0

0

x

)

(42) 2 π (45)

16 3 a (3π − 4) 9

3 πa 2 2

(34) π a2 πa 4 2π (40) a (37)

(43) 12 π (46) 4 π a

Engineering Mathematics I

4.82 π2 8 π (50) 6a 2 (47)

(48)

π 4 a 4

(49)

π 4 a 16

Exercise 4(c) (8)

2 3

(10)

(11) 88

(12) 4πa2

(14) 6 abc

(15)

(17) −

69 29 ;− 10 4

(20) 2πab πa  2 (23) a 1−   4 13 (26) 3 (29)

π 2 (13) 2πah

(9) 1

2 3 π 3

πa 3 (31) 4 π 4 a (34) 4

4 π 3 2 2 (18) − ; − 3 3

(16) 2πa2h 4 3 a 3 3 (22) 2

(19)

(21) 4

(

)

1 2 2 −1 . 3 13 (27) 6

(24)

(25) (28)

163 70 2a 3 3

1 2 2 a b + b2 c2 + c2 a 2 2 ka 3 abc (32) (33) 6 2 π 4 a h (35) 2

(30)

Exercise 4(d) (2) 0.456 (5)

(12) (22)

1  1   3  Γ Γ  2  4   4 

π k π

8 (3) 77

1 (4) 120 1 Γ (n +1) (10) (log n) n +1

(7) π

(15) (30)

1 β (m, n) a bm n

π

5π 2 2 2 8 2 128 ;

π

;

(17)

1  1   2  Γ  ⋅ Γ   3  3   3 

Multiple Integrals (31) Γ(m + 1) Γ(n + 1) / Γ(m + n + 3) (32) Γ(m + 1) ⋅ Γ(n + 1) Γ( p + 1) / Γ(m + n + p + 4) 3πab (33) (34) Γ(m) Γ(n) Γ( p ) / Γ(m + n + p ) 32 abc 4 π abc (35) (36) 6 3    Γ  1    (37) abc      n 

3

(39) a3b2c2/2520

 3 3n 2 Γ   (38) π2/1920  n  (40) a2b2c2/48

4.83

UNIT

5

Differential Equations 5.1

EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREE

In the lower classes, the students have studied differential equations of the first order and first degree, such as variable separable equations and linear equations. Now we shall study differential equations of the first order and degree greater than or equal to two. The general form of the differential equation of the first order and nth degree is  dy n  n−1  n−2   + f1 ( x, y ) dy  + f 2 ( x, y ) dy  +   dx   dx   dx  dy + f n−1 ( x, y ) + f n ( x, y ) = 0. dx If we denote

dy by p for convenience, the general equation becomes dx pn + f1 (x, y) pn–1 + f2 (x, y) pn–2 + …+ fn–1 (x, y) p + fn(x, y) = 0

(1)

Since (1) is an equation of the first order, its general solution will contain only one arbitrary constant. Solution of (1) will depend on solving one or more equations of the first order and first degree. To solve (1), it is to be identified as an equation of any one of the following types and then solved as per the procedure indicated below: (i) (ii) (iii) (iv)

Equations solvable for p. Equations solvable for y. Equations solvable for x. Clairaut′s equations.

5.1.1 Type 1—Equations solvable for p If equation (1) is of this type, then the L.H.S. of (1) can be resolved into n linear factors. Then (1) becomes

5.2

Engineering Mathematics I

(p – F1) (p – F2) . . . (p – Fn) = 0, from which we get p = F1, p = F2 , … , p = Fn, where F1, F2, … , Fn are functions of x and y. Each of these n equations is of the first order and first degree and can be solved by methods already known. Let the solutions of the above n component equations be f1 (x, y, c) = 0, f2 (x, y, c) = 0,…, fn (x, y, c) = 0. Then the general solution of (1) is got by combining the above solutions and given as f1 (x, y, c) f2 (x, y, c) …fn (x, y, c) = 0.

5.1.2 Type 2—Equations Solvable for y If the given differential equation is of this type, then y can be expressed explicitly as a single valued function of x and p. i.e. the equation of this type can be re-written as y = f(x, p)

(1)

Differentiating (1) with respect to x, we get  dp  p = φ  x, p,   dx 

(2)

Equation (2) is a differential equation of the first order and first degree in the variables x and p. It can be solved by methods already known. Let the solution of (2) be y (x, p, c) = 0 ... (3), where c is an arbitrary constant. If we eliminate p between (1) and (3), the eliminant is the general solution of the given equation. If p cannot be easily eliminated between (1) and (3), they jointly provide the required solution in terms of the parameter p.

5.1.3 Type 3—Equations solvable for x If the given differential equation is of this type, then x can be expressed explicitly as a single valued function of y and p. i.e the equation of this type can be re–written as x = f (y, p)

(1)

Differentiating (1) with respect to y, we get  1 dp  = φ  y, p,  p dy  

(2)

Equation (2) is a differential equation of the first order and first degree in the variables y and p. It can be solved by methods already known. Let the solution of (2) be y (y, p, c) = 0

(3)

where c is an arbitrary constant. If we eliminate p between (1) and (3), the eliminant is the general solution of the given equation. If p cannot be easily eliminated between (1) and (3), they jointly provide the required solution in terms of the parameter p.

Differential Equations

5.3

Note

Some differential equations can be put in both the forms y = f(x, p) and x = f(y, p). Both these forms may lead to the required solution. Sometimes one of the forms will lead to the required solution more easily than the other.

5.1.4  Type 4—Clairaut′s Equations An equation of the form y = px + f(p) is called Clairaut′s equation. Clairaut′s equation is only a particular case of type–2 equation. Hence it can be solved in the same way in which a type–2 equation is solved, as explained below: Let the Clairaut′s equation be y = px+ f(p) (1) Differentiating (1) w. r. t. x, dp dx

(2)

f ′ ( p) + x = 0

(3)

p = p + { x + f ′ ( p )} ∴

dp = 0…(2) or dx

Solving (2), we get p = c (4) Eliminating p between (1) and (4), we get the general solution of (1) as y = cx + f(c). Thus the general solution of a Clairaut′s equation is obtained by replacing p by c in the given equation. Eliminating p between (1) and (3), we also get a solution of (1). This solution does not contain any arbitrary constant. Also it cannot be obtained as a particular case of the general solution. This solution is called the singular solution of the equation (1).

Note

The singular solution of (1) is the eliminant of p between y = px + f(p) and df x+ = 0 . Equivalently, the singular solution of (1) is the eliminant of c between dp df y = cx + f(c) and x + = 0 . Hence, we observe that if the general solution of (1), i.e. dc y = cx + f(c) represents a family of straight lines, then the singular solution represents the envelope of the family of straight lines.

WORKED EXAMPLE 5(a)  dy 2 dy Example 5.1 Solve the equation   − 8 + 15 = 0 .  dx  dx The given equation is p2 – 8p + 15 = 0, which is solvable for p. The equation is (p – 3) (p – 5) = 0 dy dy ∴ = 3 or =5 dx dx

5.4

Engineering Mathematics I

Solving these equations, we get y = 3x + c and y = 5x + c. To get the general solution of the given equation, we rewrite the solutions as y – 3x – c = 0 and y – 5x – c = 0 (in which the R.S. = 0) and combine them as (y – 3x – c) (y – 5x – c) = 0. Example 5.2 Solve the equation p (p + y) = x (x + y). The given equation is p2 + yp – (x2 + xy) = 0. Solving for p, p =

− y ± y 2 + 4 ( x 2 + xy ) 2 − y ± ( y + 2 x)

2

= i.e. or

2 dy =x dx dy =−x− y dx

(1) (2)

x2 c + . 2 2 i.e. 2y – x2 – c = 0 (3) dy + y = − x, which is a linear equation of the first order in y. Rewriting (2), dx Integrating factor = ex Hence the solution of (2) is y ex = ∫ – x ex dx + c = – x ex + ex + c –x i.e. y+x–1–ce =0 (4) Combining (3) and (4), the required general solution is (2y – x2 – c) (y + x – 1 – ce–x) = 0. Example 5.3 Solve the equation p2 – 2py tan x – y2. The given equation is p2 – 2py tan x – y2 = 0. Solving (1), y =

Solving for p, p= =

2 y tan x ± 4 y 2 tan 2 x + 4 y 2 2 2 y tan x ± 4 y 2 (1 + tan 2 x )

(1)

2 = y ( tan x ± sec x ) dy dy = y (tan x + sec x) or = y (tan x – sec x). dx dx dy dy i.e. = (tan x + sec x) dx (1) and = (tan x – sec x) dx (2) y y Integrating (1), we get, log y = log sec x + log (sec x + tan x) + log c i.e. y = c sec x (sec x + tan x) ∴

Differential Equations

5.5

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

i.e.

(3)

Integrating (2), we get, log y = log sec x – log (sec x + tan x) + log c c sec x sec x + tan x c = 1 + sin x

y=

i.e.

i.e.

y (l + sin x) – c = 0 (4) Combining (3) and (4), the required general solution is [y (1 – sin x) – c] [y (1 + sin x) – c] = 0. Example 5.4 Solve the equation xp2 – 2py + x = 0. Solving the given equation for p, we get p=

2 y ± 4 y 2 − 4 x2 2x

 y dy y = ±   − 1  x  dx x 2

i.e.

(1)

(1) is a homogeneous equation. Putting y = vx, (1) becomes dv v+x = v ± v2 − 1 dx dv dx =± i.e. 2 x v −1 Integrating (2), we get,

(

(2)

)

log v + v 2 − 1 = ± log x + log c ∴ Solutions are y + y 2 − x2 x2

= c and

y + y 2 − x2 = c

y + y 2 − x 2 − cx 2 = 0 and y + y 2 − x 2 − c = 0 i.e. ∴ The general solution of the given equation is

{y +

y 2 − x 2 − cx 2

} {y +

}

y 2 − x2 − c = 0 .

Example 5.5 Solve the equation p3 – (x2 + xy + y2) p2 + (x3y + xy3 + x2y2) p – x3y3 = 0.

5.6

Engineering Mathematics I

The given equation is of the form p3 – (α + β + γ) p2 + (αβ + βγ+ γα) p – αβγ = 0, where α = x2, β = xy and γ= y2. ∴ The given equation can be rewritten as (p – α) (p – β) (p – γ) = 0 i.e. (p – x2) (p – xy) (p – y2) = 0. dy dy dy = x 2 , = xy and = y2 dx dx dx Solving these equations, we get 1 x3 c x2 y= + ; log y = + log c; − = x − c. y 3 3 2 ∴

i.e.

3 y − x 3 − c = 0; y − ce x

2

/2

= 0; x +

1 − c = 0. y

∴ The required general solution is x 2     3 2   x + 1 − c = 0. 3 y − x − c y − ce ( )   y   

Example 5.6 Solve the equation p2x – 2py – x – 0. Rewriting the given equation, we have p2 x − x 1  x =  px −  2p 2 p We identify the equation as one solvable for y. Differentiating (1) w.r.t. x, we get y=

(1)

 dp    p − x  1  dp  dx  − p = p + x  2 2 dx  p       

i.e.

dp   p−x   dp  dx  − 2p = p + x dx p2 p 3 = xp 2

i.e. i.e.

x

dp dp − p+x dx dx

dp 2 ( p + 1) − p ( p 2 + 1) = 0 dx

i.e. Solving this equation, we get

x

dp =p dx

(  p 2 + 1 ≠ 0)

Differential Equations

5.7

dp dx = + log c p ∫ x

∫

i.e. p = cx Using (2) in the given equation, the required solution is c2x2 = 2 c y + 1.

(2)

Example 5.7 Solve the equation 16x2 + 2p2y – p3x = 0. The given equation cannot be solved for p, nor for x. As it is solvable for y, the given equation is rewritten as p 3 x − 16 x 2 y= (1) 2 p2 Differentiating (1) w.r.t. x,   dp dp p 2  p 3 + 3 p 2 x − 32 x − ( p 3 x − 16 x 2 ) 2 p   dx dx 2p = 4 p dp dp dp − 32 p 2 x − 2 p 4 x + 32 px 2 dx dx dx

i.e.

2 p5 = p5 + 3 p 4 x

i.e.

p 5 + 32 p 2 x = p 4 x

i.e.

p 2 ( p 3 + 32 x ) − px ( p 3 + 32 x )

dp dp + 32 px 2 dx dx

p 2 − px

i.e.

dp =0 dx

dp =0 dx

p3 + 32x=0

or

(2) (3)

(2) is differential equation in p and (3) is an algebraic equation. If we eliminate p between the given equation and the solution of (2), we will get the general solution of (1). If we eliminate p between (1) and (3), we will get the singular solution of (1). (2) can be rewritten as

dp dx = p x

Solving this, we get p = cx (4) Eliminating p between the given equation and (4) we have 16x2 + 2c2x2y – c3x4 = 0 i.e. c3x2 – 2c2y – 16 = 0, which is the required general solution. Using (3) in the given 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 = 0, which is the singular solution.

5.8

Engineering Mathematics I

Example 5.8 Solve the equation y = (1 + p) x + p2. Treating the given equation as one solvable for y and differentiating w.r.t. x, dp dp p =1+ p + x + 2p dx dx dp i.e. ( x + 2 p) + 1 = 0 dx dx + x =− 2p i.e. dp This is a linear equation in x. Solving (2),

(1)

(2)

xe p = ∫ − 2 pe p dp + c

= – 2 (pe p – e p) + c i.e. x =2 – 2p + ce–p It is not easy to eliminate p between (3) and the given equation. ∴ The parametric equations of the general solution are given as x = 2 – 2p + ce–p and y = (1 + p) (2 – 2p + ce – p) + p2 i.e. x = 2 – 2p + ce–p and y = 2–p2 + c (1 + p) e–p.

(3)

Example 5.9 Solve the equation y = x + p2 – 2p. This equation can be treated as one solvable for y and for x. We shall solve the equation in both ways. Method I: Let us assume y = x + p2 – 2p . . . (1) as solvable for y. Differentiating (1) w.r.t. x, dp (1) p = 1 + ( 2 p − 2) dx i.e. ∴

2 ( p − 1)

dp − ( p − 1) = 0 dx

p–1=0

(2)

dp =1 dx Eliminating p between (1) and (2), we get the singular solution y = x – 1. Solving (3), we have 2p = x + c or

2

(3)

(4)

( x + c)

2

− ( x + c) 4 i.e. 4 (y + c) = (x + c)2, which is the general solution of (1). Method II: Treating (1) as one solvable for x and rewriting, we have x = y + 2p – p2 Differentiating (1)′ w.r.t. y, dx dp dp =1+ 2 − 2p dy dy dy Eliminating p between (1) and (4), y − x =

(1)'

Differential Equations

i.e.

1 dp −1 = 2 (1− p ) p dy

i.e.

  dp   (1− p )  2 p −1  = 0.   d y    

5.9

∴

p=1

(5)

or

dp 2p − 1= 0 dy

(6)

Eliminating p between (1)′ and (5), we get the same singular solution as before. Solving (6), we have p2 = y + c (7) Using (7) in (1)′, we get x –y + y + c = 2p Squaring and again using (7), we get the general solution as (x + c)2 = 4 ( y + c) Example 5.10 Solve the equation p2x + py – y4 = 0. Note Though the equation appears to be solvable for p, the component equations are not easily solvable. Also the given equation is not solvable for y. Hence we treat it as one solvable for x. Rewriting the given equation, y 4 − py x= (1) p2 Differentiating (1) w.r.t. y,  dp  dp p 2 4 y 3 − p − y  − ( y 4 − py ) 2 p dy  dy  dx = 4 dy p  dp  dp p 4 y 3 − p − y  − (2 y 4 − 2 py )  dy  dy  1 = p p3 dp p 2 = 4 py 3 − p 2 − (2 y 4 − py ) dy d p y (2 y 3 − p ) − 2 p (2 y 3 − p ) = 0 dy dp y −2p = 0 dy 2y3 – p = 0

i.e. i.e. i.e. ∴ or

(2) (3)

(2) is dp = 2dy p

y

Integrating, log p = 2 log y + log c i.e.

p = cy2

(4)

5.10

Engineering Mathematics I

Eliminating p between (1) and (4), we get c2xy4 + cy3 – y4 = 0 i.e. c xy + c = y, which is the general solution of the given equation. 2

Eliminating p between (1) and (3), we get 4xy6 + y4 = 0 i.e. 4xy2 + 1 = 0, which is the singular solution of the given equation. Example 5.11 Solve the equation p3 – 2 x yp + 4y2 = 0. The given equation is neither solvable for p nor for y. Rewriting the given equation, 2x =

p2 4 y + y p

(1)

Differentiating (1) w.r.t. y, 2 p 2 2 p dp 4 4 y dp =− 2 + + − p y dy p p 2 dy y i.e. i.e. i.e.

 2 p 4 y  dp  2 p 2  − 2 + − =0  p  dy  p y 2   y 2 ( p3 − 2 y 2 ) ddpy − py1 2 ( p3 − 2 y 2 ) = 0 p2 y

(p

3

∴

  dp − 2 y2 )  2 y − p = 0 dy   dp 2y − p=0 dy

(2)

p3 – 2y2 = 0

(3)

or Solving (2),

∫

2dp dy =∫ + log c p y

i.e. 2 log p = log y + log c i.e. p2 = cy Let us eliminate p between (1) and (4). Using (4) in (1), cpy –2xyp + 4y2 = 0 i.e. p (c –2x) = – 4y Squaring and using (4), cy (c –2x)2 = 16y2 i.e. c (c – 2x)2 = 16y, which is the general solution of the given equation. Using (3) in (1), 2y2 – 2 x yp + 4y2 = 0 i.e. xp = 3y Cubing and using (3), 2x3y2 = 27y3 i.e. 2x3 = 27y, which is the singular solution of the given equation.

(4)

Differential Equations

5.11

Example 5.12 Solve the equaion p3x – p2y –1= 0. The equation may be treated as one solvable for x. However if we treat it as one solvable for y, it becomes a Clairaut′s equation. 1 Rewriting the given equation, we get y = px − 2 , which is a Clairaut′s equation. p ∴ Its general solution is 1 (1) y = cx − 2 c Differentiating (1) partially w.r.t. c, 2 0=x+ 3 (2) c Eliminating c between (1) and (2), we get the singular solution of the given equation. From (2), 2 c3 = − (3) x c2y = c3x – 1 = –2 – 1, using (3) ∴ c6y3 = –27 i.e. 4y3 = –27x2, using (3) This is the singular solution of the given equation. Example 5.13 Solve the equation y = 2px + yp2. The equation may be treated as one solvable for x. However we can convert it as a Clairaut′s equation by means of the transformation y2 = Y dy dY ∴ 2y = dx dx From (l),

i.e. 2yp = P, say. Multiplying through out the given equation by y, it becomes y2 = 2ypx + y2p2 On using the transformation, (1) becomes P2 , Y + Px + 4 which is a Clairaut′s equation ∴ General solution of (2) is c2 Y = cx + 4 2 c 2 ∴ General solution of (1) is y = cx + . 4 Differentiating (3) partially w.r.t. c, c 0=x+ 2 Eliminating c between (3) and (4), we get, Y = – x2

(1) (2)

(3)

(4)

5.12

Engineering Mathematics I

i.e. x2 + y2 = 0, which is the singular solution of (1). Example 5.14 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 Then the given equation becomes or

p=

( x P − y ) xy ( P + 1) = 2 xy P 2

2

2P P +1 2 P , which is a Clairaut′s equation. i.e. Y = PX − P +1 2c ∴ General solution is y 2 = cx 2 − c +1 Example 5.15 Solve the equation e4x (p – 1) + e2y p2 = 0 Note In problems involving eax and eby we make the transformations X = ekx and ky Y = e where k is the H.C.F. of a and b In the given equation, put 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 Then the given equation becomes X2 X  X 2  P − 1 + Y ⋅ 2 P 2 = 0 Y Y  i.e. XP – Y + P2 = 0 i.e. Y = PX + P2, which is a Clairaut′s equation. ∴ The general solution of the given equation is e2y = ce2x + c2. i.e.

PX − Y =

EXERCISE 5(a) Part A (Short Answer Questions) 1. Explain briefly how to find the general solution of the equation [p – f1 (x, y)] [p – f2 (x, y)]=0. 2. Explain briefly how to solve the equation, y = f (x, p), if it is solvable for y. 3. Explain briefly how to solve the equation x = f (y, p), if it is solvable for x. 4. Write down the standard form of a Clairaut′s equation and give its general solution.

Differential Equations

5.13

5. 6. 7. 8. 9. 10. 11.

How will you find the singular solution of the equation y = px + f (p)? What does the singular solution represent geometrically? 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 as a Clairaut′s equation and write down its general solution. 12. Rewrite the equation p = log (px – y) as a Clairaut′s equation and give its general solution. 13. Rewrite the equation p = sin (y – px) as a Clairaut′s equation and give its general solution. 1 14. Find the singular solution of y = px + . p 15. Find the singular solution of y = px – p2. Part B Solve the following equations: 16. yp2 + (x – y) p – x – 0 17. 2p2 – (x + 2y2) p + xy2 = 0 18. xyp2 + (3x2 – 2y2) p – 6xy = 0 19. xyp2 –(x2 + y2) p + xy = 0 x y 1 20. p − = − p y x 21. p2 + 2py cot x = y2 22. x2p2 – 2xyp + (2y2 – x2) = 0 23. y = – px + x4p2 24. 4y = x2 + p2 25. y = 2px + pn 26. y = 2px – p2 27. y = x + 2 tan–1 p 28. y = (1+ p) x + ep 29. y = 3x + log p, by considering the equation as one solvable for (i) y and (ii) x. 30. p3 – 4xyp + 8y2 = 0 31. x = y + p2 32. y = 2px + 4yp2 33. y = 2px + y2p3 34. y2 log y = xyp + p2 35. Find the singular solution of the equation y = (x – 1) p + tan–1 p. 36. Find the singular solution of the equation y + log p = px. By suitable transformations, reduce the following equations into Clairaut′s equations and hence solve them: 37. x2 (y – px) = yp2 [Put X = x2, Y = y2] 38. y = 2px + p2y [Put X = 2x, Y = y2] 39. (y + px)2 = px2 [Put Y = xy] 40. (p – 1) e3x + p3e2y = 0 [Put X = ex, Y = ey]

5.14

Engineering Mathematics I

5.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS 5.2.1

Introduction

Students are already familiar with formation and solutions of some types of linear differential equations of second and higher order with constant coefficients. Before we take up the study of remaining types of such equations, we shall recall the various notions and working rules related to the solutions of such equations. This will be of use to pursue the study of not only the remaining types of linear differential equations with constant coeffcients, but also linear differential equations with variable coefficeints and simultaneous differential equations. The general form of a linear differential equation of the nth order with constant coefficients is a0

dn y d n−1 y dy a + +  + an−1 + an y = X , 1 n n−1 dx dx dx

(1)

where a0 (≠ 0), a1, a2, . . . , an are constants and X is a function of x. If we use the differential operator symbols D ≡

d d2 dn , D 2 ≡ 2 , D n ≡ n , dx dx dx

equation (1) becomes (a0Dn + a1 Dn–1 + ...+ an–l D + an) y = X

(2)

or f (D) y = X, where f (D) is a polynomial in D, which may be treated as an algebraic quantity. When X = 0, (2) becomes f (D) y = 0

(3)

(3) is called the homogeneous equation corresponding to equation (2). General Solution of equation (2) is y = u + v, where y = u is the general solution of (3), that contains n arbitrary constants and y = v is a particular solution of (2), that contains no arbitrary constants. u is called the complementary function (C.F.) and v is called the particular integral (P.I.) of the solution of Equation (2).

5.3

COMPLEMENTARy FUNCTION

To find the C.F. of the solution of Equation (2), we require the general solution of Equation (3). To get the solution of f (D) y = 0 or (a0Dn + a1Dn–1 + . . .+ an) y = 0 (3)¢ we first write down the auxiliary equation (A.E.) f (m) = 0 or a0mn + a1mn–1 + . . . + an = 0 (4) which is obtained by simply replacing D by m in the operator polynomial, and then by equating it to zero. The auxiliary equation is an nth degree algebraic equation in m.

Differential Equations

5.15

The solution of Equation (3)¢ depends on the nature of roots of the A.E. (4) as explained below. Case (i) The roots of the A.E. are real and distinct. Let the roots of the A.E. be m1, m2, …, mn. m x mx m x Then the solution of Equation (3)¢ is y = c1e 1 + c2 e 2 +  cn e n , where c1, c2, …, cn are arbitrary constants. ∴ C.F. of the solution of Equation (2) is given by u = c1e m1 x + c2 e m2 x +  + cn e mn x Case (ii) The A.E. has got real roots, some of which are equal. Let the roots of the A.E. be m1, m1, m3, m4, …, mn (the first two roots are equal). 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 y = (c1 x 2 + c2 x + c3 ) e

m1 x

+ c4 e m4 x +  + cn e mn x ⋅

In general, if r roots of the A.E. are equal, then the solution of (3)¢ becomes. y = (c1 x r −1 + c2 x r −2 +  cr −1 x + cr ) e m1 x + cr +1e mr+1 x +  + cn e mn x ⋅ Case (iii) Two roots of the A.E. are complex. As complex roots occur in conjugate pairs, let m1 = α + iβ and m2= α – iβ. 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) Two pairs of complex roots of the A.E. are equal. i.e. ml = m3 = α + iβ and m2 = m4 = α – iβ. 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 .

5.3.1

Particular Integral

The particular integral (P.I.) of the solution of the equation f (D) y = X (1) is the function v, where y = v is a particular solution of (1) containing no arbitrary constants. The particular integral depends on the value of the R.H.S. function X and 1 1 is defined as P.I. = is the inverse operator of f (D). X , where f (D) f (D)  1    X i.e. f (D)   = X.     f (D)   We give below the rules and working procedures to be adopted when X is equal to some special function: Rule I X= eax, where α is a constant. P.I. =

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

5.16

Engineering Mathematics I

When f(α) = 0, (D – α)r is a factor of f(D). Let f(D) = (D – α)r f (D), where f (α) ≠ 0. 1 Then P.I. = e ax r ^ D - ah z ]Dg 1 1 e ax F = < z^ah ^ D - ahr 1 $ x r e ax $ = z^ah r In particular,

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 α is a constant. In this case, the following rules are used. 1 sin ax 1 sin ax = 2 φ(D ) φ ( – a2 ) 1 cos ax 1 cos ax, if φ ( – a 2 ) ≠ 0. = φ ( D2 ) φ ( – a2 ) When f (– α2) = 0, (D2 + α2) is a factor of f (D2) Let f (D2) = (D2 + α2) y (D2), where y (–a2) ≠ 0. ∴

1 1 sin α x. sin αx = φ (D 2 ) ψ (D 2 ) (D 2 + α 2 ) =

Similarly Now

1 ψ (– α 2 )

1 sin α x D2 + α 2

1 cos ax 1 ; 1 cos ax E = z ] D 2g }^- a2h D 2 + a2 x 1 sin α x = – cos α x 2α D2 + α 2 x = 2 × Intergral of sin α x.

and

1 cos α x = x sin α x D2 + α 2 2α =

Note

x × Intergral cos α x. 2

When finding the P.I., the above rules are to be applied in parts, as f(D) will not be, in general, of the form f(D2). This means that we have to first replace D2 by – α2, D3 by – α2D, D4 by α4, etc. After this has been done, f(D) will take the form (aD + b).

Differential Equations

5.17

1 sin ax f] Dg 1 sin ax =] aD + bg ]aD bg = 2 2- 2 sin ax, a D -b

Then P.I. =

on multiplying the numerator and denominator by (aD – b). P.I. =

Then

(

1

− a α 2 + b2 2

)

(aD − b)sin α x,

using the rule in the denominator. =−

Similarly, Rule 3

(a α 2

1 2

+b

2

)

(aα cos α x − b sin α x), since D ≡

d ⋅ dx

1 1 cos α x = 2 2 (aα sin α x + b cos α x). f ( D) a α + b2

X =xm, where m is

a positive integer. P.I. =

1 xm f ( D)

Rewrite f(D) in terms of a standard binomial expression of the form [1 ± f(D)], by taking out the constant term or the lowest degree term from f(D). 1 Thus P.I.= xm k  aD 1 ± φ ( D ) −1 1  = 1 ± φ ( D ) x m k  aD –1 Now expand [1 ± f(D)] in a series of ascending powers of D, by using binomial −1 1  theorem, so that the simplified expansion of 1 ± φ ( D ) may contain terms up k  aD to Dm and then operate by each term on xm. 1 Note The ultimate expansion of need not be considered beyond the Dm f ( D) term, since Dm+1 (xm) = 0, Dm+2 (xm) = 0 and so on. Rule 4 X = eαx . V(x), where V is of the form sin βx, cos βx or xm. P.I.=

1 1 eα xV ( x ) = eα x ⋅ V ( x) f ( D) f ( D + α)

1 V(x) is evaluated by using the rule (3) or (4). f ( D + α)

Note Rule 5

Thie rule is referred to as Exponential shift rule. X = x. V(x), where V(x) is of the form sin αx or cos αx.

5.18

Engineering Mathematics I

P.I =

1 1 d  1  xV ( x ) = x ⋅ V ( x) +   V ( x) f ( D) f ( D) dD  f ( D ) or

= x⋅

f ′ ( D) 1 V ( x) − V ( x) 2 f ( D) f D ( ) { }

By repeated applications of this rule, we can find the P.I. when X = xrV(x), where r is a positive integer. Note Instead of applying the Rule (5), we may adopt the following alternative procedure to find the P.I., when X = xr cos ax or xr sin ax. 1 1 [Real Part of x r eiα x ] x r cos α x = f ( D) f ( D) = R.P. of

1 x r eiα x f ( D)

= R.P. of eiα x ⋅

1 xr f ( D + iα )

Similarly, 1 1 x r sin α x = I.P. of eiα x ⋅ xr ⋅ f ( D) f ( D + iα ) Rule 6 X is any other function of x. 1 P.I = X f ( D) 1 = X, ( D − m1 )( D − m2 )( D − mn ) resolving f ( D ) into linear factors.   A A A  X , n 1 2 =  (1) + + +   D − m1 D − m2 D − mn  spliting into partial fractions.. 1 X = u , say D−m     1 ∴ X ( D − m)   = ( D − m) u    ( D − m)   du − mu = X . i.e. ( D − m)u = X or dx This is a linear equation of the first order. Consider

∴ Its solution is u e−mx =

∫e

−mx

X dx

The usual arbitrary constant is not added in the R.H.S, as u is a part of the P.I. of the main problem.

Differential Equations 1 X = e mx ⋅ ∫ e−mx X d x D−m Inserting (2) in (1), the required u=

∴

P.I. = A1e m1 x ∫ e−m1 x X ⋅ dx + A2 e m2 x ∫ e−m2 x X dx +  + An e m n x ⋅ ∫ e−m n x X dx

WORKED EXAMPLE 5(b) Example 5.1 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 1 1 x2 = 2 (sin 3 x) + 2 D − 4D + 3 D − 4D + 3 = P.I.1 + P.I.2 (say) P. I.1 =

1 sin 3 x D2 − 4D + 3

=

1 sin 3 x −9 − 4 D + 3

=−

1 sin 3 x 2 (2 D + 3)

1 (2 D − 3) sin 3 x =− ⋅ 2 4D2 − 9 =

1 (2 D − 3) sin 3x 90

=

1 (6 cos 3x − 3 sin 3x) 90

1 (2 cos 3x − siin 3x) 30 1 x2 P.I.2 = 2 D − 4D + 3 1 x2 =  D (4 − D )   3 1−  3   =

5.19

(2)

5.20

Engineering Mathematics I D (4 − D )  1 = 1 −  3  3 

−1

x2

2  D (4 − D ) D 2 (4 − D )  2 1 = 1 + + x 3 3 9    1 4 13  = 1 + D + D 2  x 2  3  3 9

1 8 26  =  x 2 + x +   3 3 9 ∴ The general solution of the given equation is y = C.F + P.I.1 + P.I.2 i.e.

y = c1 e x + c2 e3 x +

1 1 8 26 ( 2 cos 3x − sin 3x ) +  x 2 + x +  30 3 3 9

Example 5.2 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 x4 D +4 2

−1

D 2  1  = 1 + 4  4 

x4

D2 D 4  4 1  x = 1 − + 4  4 16  1 3 =  x 4 − 3 x 2 +  4  2 P.I.2 = =

1 cos 2 x D +4 2

 1  1 1 os 2 x  + co  D + 42 2 2

 1  1 1 = ⋅ 2 e0.x + 2 cos 2 x   2  D + 4 D +4   1 x sin 2 x   + ⋅   4 2 2 

=

1 2

=

1 (1 + x sin 2 x) 8

Differential Equations

5.21

∴ General solution of the given equation is y = A cos 2x + B sin 2 x +

(

1 4 − 6 x 2 + 2 x 4 + x sin 2 x 8

)

Example 5.3 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, m =

2 ± 4 − 16

C.F. = A e −2 x P.I.1 =

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

1 (x 4 + 2 x + 1) D3 + 8 −1

1 D 3   (x 4 + 2 x + 1) = 1 + 8  8  D 3  4 1  (x + 2 x + 1) = 1 − 8  8  1 4 (x + 2 x + 1) − 3 x  8  1 = (x 4 − x + 1) 8 =

P.I.2 =

1  e 2 x + e −2 x    2 D +8  3

=

 1 1 1 −2 x  3  e2 x + e 2 D +8  D + 2) D2 − 2D + 4 (  

=

 1  1 2x 1 1 e −2 x   e + ⋅ 2  16 12 D + 2 

=

1 2

(

 1 2 x 1 x −2 x   16 e + 12 ⋅ 1! e   

1 (3 e 2 x + 4 x e −2 x ) 96 ∴ General solution is y = C.F. + P.I.1 + P.I.2 Example 5.4 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.

)

  1 x e αx = e αx  ∵ 1!   D−α

5.22 ∴

Engineering Mathematics I C.F. = (c1 x + c2 ) + (c3 x + c2 ) e x . P.I.1 =

1 D ( D −1)

2

2

x2 =

1 −2 1− D ) x 2 2 ( D

1 (1 + 2 D + 3D 2 + 4 D3 + 5D 4 ) x 2 D2  1  2 =  2 + + 3 + 4 D + 5 D 2  x 2  D  D 4 3  1 2  2x x ∵ x = x 2 dx = + + 3 x 2 + 8 x + 10 ∫   D 12 3 =

P.I.2 =

1 D ( D −1)

2

2

ex

=

1 1 ex ⋅ 2 1 ( D −1)2

=

x2 x e 2!

∴ General solution is y = (c1 x + c2 ) + (c3 x + c4 ) e x +

x 4 2 x3 x2 + + 3 x 2 + 8 x + 10 + e x . 12 3 2

Note

The two terms (8x + 10) in P.I.1 can be merged with the two terms (c1x + c2) of the C.F. ∴ The general solution may be given as y = (c1 x + c2 ) + (c3 x + c4 ) e x +

x 4 2 x3 x2 + + 3x 2 + e x . 12 3 2

Example 5.5 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 (i.e. two pair’s of equal complex conjugate roots) ∴ C.F. = (c1x + c2) cos x + (c3x+ c4) sin x. P.I.1 =

1

(1 + D 2 )

2

x4

−2

= (1 + D 2 ) x 4 = (1− 2 D 2 + 3D 4 ) x 4 = x 4 − 24 x 2 + 72. 1 2 sin x cos 3 x. P.I.2 = 2 2 ( D +1)

Differential Equations = =

1

(D

2

5.23

(sin 4 x − sin 2 x)

+ 1) 1

2

(−16 + 1)

2

sin 4 x −

1

(−4 + 1)

2

sin 2 x

1 1 n 4 x − sin 2 x. sin 225 9 ∴ General solution of the given equation is y = C.F. + P.I.1 + P.I.2. Example 5.6 Solve (D2 + 6D + 9)y = e–2x x3 =

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 x 3 P.I. = 2 ( D + 3) = e−2 x ⋅

1

( D − 2 + 3)

2

x 3 , by Exponential shift rule

−2

= e−2 x (1 + D ) x 3 = e−2 x (1− 2 D + 3D 2 − 4 D 3 ) x 3 = e−2 x ( x 3 − 6 x 2 + 18 x − 24) ∴ General solution is y = C.F. + P.I. Example 5.7 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 x 3 ⋅ P.I. = 3 ( D −1) 1 x3 = e− x . 3 ( D − 2) −3  D 1 = − e− x ⋅ 1−  x 3  8 2  1 1  D D2 D3  = − e− x ⋅ + 4 ⋅ 5 ⋅  x 3 1⋅ 2 + 2 ⋅ 3 + + 3 ⋅ 4 ⋅ 8 1⋅ 2  2 4 8   1 5  = − e− x 2 + 3D + 3D 2 + D 3  x 3  16 2  1 −x = − e (2 x 3 + 9 x 2 + 18 x + 15) 16 ∴ the general solution is y = C.F. + P.I.

5.24

Engineering Mathematics I

Example 5.8 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. = 2 x 2 cosh 2 x D −4 x2 2 x e + e−2 x D2 − 4 2 1 1 1 1 x 2 + e−2 x ⋅ = e2 x ⋅ x2 2 2 2 2 ( D + 2) − 4 ( D − 2) − 4 1

=

(

)

=

1 2x 1 1 1 e ⋅ 2 x 2 + e−2 x ⋅ 2 x2 2 2 D + 4D D − 4D

=

−1 −1 D  1 −2 x 1  1 2 x 1  D  2 x2 + ⋅ x e e ⋅ 1 + 1 −     4  2 2 4 D  (−4 D)  4 

=

1 2 x 1  D D2 D3  2 1 −2 x e ⋅ 1 − + − x − e 4 16 64  8 8 D  ×

=

=

1  D D2 D3  2 1 + + + x 4 16 64  D 

 x3 x 2 1  x  − + −   3 4 8 32    x3 x2 x 1  1 − e−2 x  + + +  4 8 32  8  3 1 2x e 8

x3 x2 x sinh 2x − cosh 2x + sinh 2x 12 16 32

1 2x 1 −2 x e and − e are omitted, as they may be considered to 256 256 have been included in the C.F.)

(the terms −

Then the G.S. is y = A e−2 x + B e 2 x +

(

x 8 x 2 sinh 2 x − 6 x cosh 2 x + 3 sinh 2 x 96

Example 5.9 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.

)

Differential Equations P.I. =

1

(D

2

= e2 x ⋅

−1)

2

e 2 x ( x + 1) = e 2 x ⋅

1

{D

2

+ 4 D + 3}

2

5.25 1

{( D + 2) −1} 2

2

( x + 1)

( x + 1) −2

  D ( D + 4)  1  = e2 x ⋅  1 +   9 3    

( x + 1)

=

 1 2x   2D  e ⋅ ( D + 4)( x + 1) 1−  9 3    

=

1 2 x  8  e 1− D( x + 1) 9  3 

=

5 1 2 x  e  x −  3 9 

Then the general solution is y = C.F. + P.I. Example 5.10

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. = A e P.I. = =

( −1+ 2 ) x

+Be

( −1 − 2 ) x

(

1 x + ex D + 2D − 1 2

)

(

= e− x A e

2x

+ B e−

2x

)

2

(

1 x 2 + e 2 x + 2 xe x D2 + 2D − 1

)

= P.I.1 + P.I.2 + P.I.3 ( say ) P.I.1 =

1

− {1 − D ( D + 2 )}

{

}

= − 1 + 2 D + 5D 2 x 2

(

)

= − x 2 + 4 x + 10 . P.I.2 =

1 1 e2 x = e2 x 7 D + 2D − 1 2

{

x 2 = − 1 + D( D + 2) + D 2 ( D + 2 )

2

}x

2

5.26

Engineering Mathematics I 1 2 xe x D2 + 2D − 1 1 x = 2e x ⋅ 2 ( D + 1) + 2 ( D + 1) − 1

P.I3 =

= 2e x ⋅

1 x D + 4D + 2 2

−1

  D ( D + 4) 1  = 2e x ⋅  1 +  x  2 2     x = e ⋅ {1 + 2 D} x = e x ⋅ ( x + 2) Then the general solution is y = C.F. + P.I.1 + P.I.2 + P.I.3. Example 5.11 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 D + 5D + 4 = e− x ⋅

1

( D − 1) + 5( D − 1) + 4 2

= e− x ⋅

1 sin 2x D 2 + 3D

= e− x ⋅

1 sin 2x 3D − 4

= e− x ⋅

(3D + 4) 9 D 2 − 16

sin 2x

sin 2x

1 −x e (6 cos 2x + 4 sin 2x ) 52 1 = − e− x (3 cos 2x + 2 sin 2x) 26 Then the general solution is y = C.F. + P.I. =−

Solve (D4 – l)y = cos 2x cosh x. Example 5.12 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.

Differential Equations

P.I. =

 e x + e− x  1   x cos 2   2 D4 − 1

=

1 x 1 1 1 cos 2x + e− x ⋅ cos 2x e ⋅ 4 4 2 2 1 1 1) − 1 + − D − D ( ) (

=

1 x 1 e ⋅ 4 cos 2x 3 2 D + 4D + 6D2 + 4D +

=

1 −x 1 e ⋅ 4 cos 2x 3 2 D − 4D + 6D2 − 4D

1 x 1 1 1 e ⋅ cos 2x + e− x ⋅ coss 2x 2 16 − 16 D − 24 + 4 D 2 16 + 16 D − 24 − 4 D

1 1 1 1 = − ex ⋅ cos 2x + e− x ⋅ cos 2x 8 3D + 2 8 3D − 2

(3D − 2) (3D + 2) 1 1 = − ex ⋅ cos 2x + ⋅ e− x ⋅ cos 2x 2 8 8 9D − 4 9D2 − 4 =

5.27

1 x 1 −x e (−6 sin 2x − 2 cos 2x ) − e (−6 sin 2x + 2 cos 2x ) 320 320

=−

 e x + e− x   e x − e− x  1 3   − sin 2x  cos 2x   2 80 2  80  

1 (3 sin 2 x sinh x + cos 2 x cosh x) 80 ∴ The general solution is y = C.F. + P.I. Solve (D2 - 4D + 13) y= e2x cos 3x. Example 5.13 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) 1 e 2 x ⋅ cos 3x D 2 − 4 D + 13 1 = e2 x ⋅ cos 3x 2 ( D + 2 ) − 4 ( D + 2 ) + 13

P.I. =

= e2 x ⋅ =

1 x sin 3x cos 3x = e 2 x ⋅ ⋅ 2 3 D2 + 9

1 x e 2 x sin 3 x 6

5.28

Engineering Mathematics I

∴ The general solution is y = C.F. + P.I. Example 5.14 A.E. is

Solve (D 2 + D + 1) y = e− x sin 2 m2 + m + 1 = 0

x . 2

1 3 The roots are m =− ± i 2 2 −

C.F.= e

x 2

   A cos 3 x + B sin 3 x   2 2  

1− cos x  e− x   2  D + D +1  1 1 1 =  2 e− x − 2 e− x cos x  2  D + D +1 D + D +1  1 −x 1 −x 1 cos x = e − e ⋅ 2 2 (D −1) 2 + (D −1) +1

P.I. =

1

2

1 1 −x 1 −x cos x e − e ⋅ 2 2 2 D − D +1  1  1 1 = e− x − e− x ⋅  cos x  −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 5.15 =

A.E. is The roots are m = −1 ± i 2

m2 + 2m + 5 = 0 C.F. = e− x (A cos 2 x + B sin 2 x).

P.I. =

1 D + 2D + 5 2

e x cos3 x

3  1 e x  cos x + cos 3 x    4 4 D + 2D + 5 3 1 = ex⋅ cos x 2 4 (D + 1) + 2(D + 1) + 5 =

1

2

+ =

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

1 1 1 3 x cos x + e x ⋅ 2 cos 3 x e ⋅ 2 4 4 D + 4D + 8 D + 4D + 8

Differential Equations

5.29

3 x 1 1 1 e cos x + e x ⋅ cos 3 x 4 4D + 7 4 4D −1 ( 4 D − 7) 3 1 (4 D + 1) = ex ⋅ cos x + e x ⋅ 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 − cos 3 x) 260 580 ∴ The general solution is y = C.F. + P.I. Solve (D2 + 4D + 8) y = 12e–2x sin x sin 2x. Example 5.16 =

A.E. is m2 + 4m + 8 = 0. The roots are m = − 2 ± i 2 C.F. = e−2 x (A cos 2 x + B sin 2 x)

∴

1 6e−2 x (cos x − cos 3 x) D + 4D + 8 1 (cos x − cos 3 x) = 6e−2x ⋅ 2 (D − 2) + 4(D − 2) + 8

P.I. =

2

1 (cos x − cos 3 x) D2 + 4   1 1     cos x + cos 3 x    5   3 

= 6e−2 x ⋅ = 6e−2 x =

2 −2 x e (5 cos x + 3 cos 3 x) 5

∴ The general solution is y = C.F. + P.I. Solve (D3 – 1) y = x sin x. Example 5.17 A.E. is m3 – 1 = 0 2 i.e. (m – 1) (m + m + 1) = 0 ∴ The roots are m = 1, −

1 3 ±i 2 2

 3 3 C.F. = c 1 e x + e− x / 2 c2 cos x + c3 sin  2 2

∴ P.I. =

1 D −1 3

 x 

x sin x

  1 1 f ′(D)   = ⋅ V V sin − x ∵ xV x   2 3 3 2 f (D)  f (D) (D − 1) D −1 { f (D)}   1 3 sin x + sin x =−x⋅ D +1 (D + 1) 2

= x⋅

1

sin x −

3D 2

5.30

Engineering Mathematics I

3 (D − 1) sin x + 2 sin x D2 − 1 D + 2D + 1 3 x = (cos x − sin x) − cos x. 2 2 ∴ The general solution is y = C.F. + P.I. Alternative method for finding P.I. = − x⋅

P.I. = =

1 D −1 1 3

D −1 3

= I.P. of

x sin x (Imaginary part of x e ix ) 1 D −1 3

= I.P. of e ix ⋅ = I.P. of e ix ⋅

x e ix 1

(D + i )3 − 1

x

1 D + 3i D − 3D − i − 1 3

2

x

−1   D ( − 3 + 3iD + D 2 )  x 1 −   1 + i  

= I.P. of −

e ix 1+ i

= I.P. of −

e ix  3D   x 1 − 1 + i  1 + i 

= I.P. of −

e ix  3    x − 1 + i  1 + i 

  3 (11 − i )   (cos x + i sin x)   x − (1 − i )    2 2       3 1   = I.P. of − {(cos x + sin x) + i (sin x − cos x)}   x − (1− i )    2 2        3 1  3 =− (cos x + sin x) +  x −  (sin x − cos x)   2 2  2  = I.P. of −

=−

Example 5.18

3 x cos x + (cos x − sin x) 2 2

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

A.E. is m2 + 4 = 0 The roots are m = ± i 2 ∴ C.F. = A cos 2x + B sin 2x. P.I. =

1 D +4 2

R.P. of x 2 ei 2 x

Differential Equations

5.31

1 x2 (D + i 2) 2 + 4 1 = R.P. of e i 2 x ⋅ 2 x2 D + 4iD = R.P. of e i 2 x ⋅

−1

iD  e i 2 x  2 1 −  x 4 4 iD  e i 2 x  iD D 2 iD 3  2  x = R.P. of − − 1 + 4iD  4 16 64 

= R.P. of

= R.P. of −

i i2x e 4

 x3   + i x 2 − x − i    3 4 8 32 

 x 3  1 x  i  2 1  (sin 2 x − i cos 2 x)   −  +  x −   4 8  4  8   3   x 3 x  1 1  2 1  =   −  sin 2 x +  x −  cos 2 x      4 3 8 4 8      = R.P. of

∴ General solution i s y = C.F. + P.I. Solve (D2 - 4D + 4) y = 8x2 e 2x sin 2x. Example 5.19 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 e 2x ⋅



1 D

= e 2 x ⋅  1

8 x 2 e 2 x sin 2 x = 8 e 2 x ⋅

1 D2

x 2 sin 2 x

   2  − cos 2 x    − 2 x  − sin 2 x  + 2 cos 2 x   x      4 8  2    (by applying Bernouilli's formula) ( −4 x 2 cos 2 x) +

 1 1 (4 x sin 2 x) + (2 cos 2 x)  D D

D          = e 2 x −4  x 2  sin 2 x  − 2 x  − cos 2 x  + 2  − sin 2 x      8 4     2      − cos 2 x   − sin 2 x   +4   −   x   + sin 2 x      2 4       = e 2 x (3 − 2 x 2 ) sin 2 x − 4 x cos 2 x    ∴ The general solution is y = C.F. + P.I.

5.32

Engineering Mathematics I

Solve (D2 + a2) y = sec a x. Example 5.20 A.E. is m2 + a2 = 0 The roots are m = ± ia ∴ C.F. = A cos a x + B sin a x. 1 P.I. = 2 sec a x D + a2 =

1

( D − ia) ( D + ia)

sec a x

 1 1     =  2ia − 2ia  sec a x  D − ia D + ia     =

1 ⋅ eia x 2ia

∫e

−ia x

sec a x d x −

1 −ia x e 2ia

∫e

ia x

sec a x d x

  ∵ 1 X = e mx . X e −mx d x   D−m    1 1 ia x = e (1 − i tan a x) d x − e−ia x (1 + i tan a x) d x 2ia 2ia   1 −ia x  1 ia x  i i = e x + log sec a x e  x − log sec a x −       2ia 2ia a a  eia x + e−ia x  1 x  e iax − e−ia x    =   − 2 log sec a x   a   2i 2 a 

∫

∫

=

∫

1 x sin a x − 2 cos a x log sec a x a a

∴ General solution is y = C.F. + P.I.

EXERCISE 5(b)

Part A (Short Answer Questions) 1. 2. 3. 4. 5. 6. 7. 8.

Solve the equation (D2 – D + 1)2 y = 0. Find the particular integral of (D – 1)3 y = 2 cosh x. Find the particular integral of (D2 + a2) y = b cos ax + c sin ax. Find the particular integral of (D2 + 4D + 4) y = x e–2x. Find the particular integral of (D – 3)2 y = x e–2x. Find the particular integral of (D + 1)2 y = e–x cos x. Find the particular integral of (D2 – 2D + 5) y = e x sin 2x. Find the particular integral of (D2 + 4D + 5) y = e–2x cos x.

Differential Equations

5.33

9. Find the particular integral of (D2 – 2D + 6)y = e x (4 sin x + cos x). 10. Find a formula for 1 f (x). D−a Part B Solve the following differential equations. 11. (D3 + D2 + D + 1)y = x2 + 2e–x 12. (D2 + 9)y = x2 + cosh x 13. (D2 + 2D + 1)y = x3 + cos 2x 14. (D2 _ 8D + 9)y = 8 sin 5x + x2 15. (D2 + 3D + 2)y = 2 sin2 x + 2x2 16. (D4 + D3 + D2)y = 12x2 + 2 cos 2x cos x 17. (D2 _ l)y = 12ex (x + l)2 18. (D3 – 6D2 + 12D – 8)y = 16x3 e4x 19. (D3 + 2D2 + D)y = x2e2x 20. (D2 – 4) y = x sinh x 21. (D2 + l)2y = 2x2e–x 22. (D2 – 5D + 4)y = (2x + e–x)2 23. (D2 – 4D + 3)y = 8ex cos 2x 24. (D4 – 1)y = cos x cosh x 25. (D2 – 2D + 5)y = ex (sin x + cos x)2 x 26. ( D 3 + 1) y = e − x cos 2 2 27. (D2 + 4)y = 4 e2x sin3 x 28. (D2 – 4D + 3)y = sin 3x cos 2x 29. (D2 – 2D + 1)y = x ex sin x 30. (D2 + D)y = x cos x 31. (D2 – 4D + 4)y = x sin x 32. (D2 – l)y = x2 cos x 33. (D2 + 4)2 y = cos 2x 34. (D2 + l)y = x2 sin 2x 35. (D2 + 4)y = 4 tan 2x

5.4  E   ULER′S HOMOGENEOUS LINEAR DIFFERENTIAL  EQUATIONS The equation of the form

a0x n

dn y d n -1 y dy + a 1 x n −1 +  + a n −1 x +an y= X n dx dx d x n −1

(1)

where a0, a1, … , an are constants and X is a function of x is called Euler′s homogeneous linear differential equation.

5.34

Engineering Mathematics I

Note

In each term in the L.H.S. of Equation (1), the power of x and the order of the derivative are equal. It is because of this property that the equation is called a homogeneous equation. Equation (1) can be reduced to a linear differential equation with constant coefficients by changing the independent variable from x to t by means of the transformation. x = et or t = log x as explained below: dy dy dt 1 dy = ⋅ = d x dt d x x dt x

∴

dy dy = d x dt

(2)

Differentiating both sides of (2) w.r.t. x, x

x2

i.e.

d 2 y dy d 2 y 1 + = d x 2 d x dt 2 x d2y dy d 2 y + = x dx dt 2 d x2 x2

i.e.

d 2 y d 2 y dy = 2 −  by ( 2 )  dt  d x2 dt

(3)

d d by D and by θ , dx dt (2) gives xD = θ and (3) gives x2 D2 = θ2 – θ = θ (θ – 1) Similarly we can show that x3 D3 = θ (θ– 1) (θ – 2) x4 D4 = θ (θ – 1) (θ –2) (θ – 3) and so on. Denoting

(

)

 If this transformation is made, then Eqn. (1) becomes  a0 θ (θ − 1) θ − n − 1 +   a1 θ (θ − 1)  θ − n − 2 +  an  y = 0 , which is a linear differential equation with  constant coefficients and can be solved by methods discussed in the previous section. The more general form of Euler′s homogeneous equation is

(

)

n -1 d ny y n −1 d + a a x + b + ( ) 1 n −1 n dx dx dy + an −1 (a x + b) + an y = X (2) dx Equation (2) can be reduced to a linear differential equation with constant coefficients by the substitution ax + b = et. Equation (2) is called Legendre′s linear differential equation.

a0 (a x + b)

n

Differential Equations

5.5

5.35

SIMULTANEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

If x and y are two dependent variables and t is the independent variable, then the pair of equations of the form f1 (D) x + f2 (D) y = T1 (1) (2) z1 ]Dgx + z2 ]Dgy = T2 d and T1 and T2 are functions dt of t is called a pair of simultaneous differential equations. It is not possible to solve for the two dependent variables (unknowns), if only one of the above equations is given. If there are more than 2 dependent variables, we should have as many equations as the number of dependent variables. To solve Equations (1) and (2) simultaneously, we proceed as in solving simultaneous algebraic equations. We operate both sides of (1 ) by f2 (D) and both sides of (2) by f2(D) and subtract to eliminate y. Thus we get  f1 ( D ) φ2 ( D ) − f 2 ( D )φ1 ( D ) x = φ2 ( D )T1 − f 2 ( D )T2 or f ( D ) x = T (3)   where f1, f2, f1, f2, are polynomials in the operator D ≡

which is a linear equation in x and t with constant coefficients and can be solved by the methods discussed already. The value of x obtained by the solution of (3) is substituted either in (1) or (2) to get the value of y. The number of arbitrary constants that appear in the values of x and y should be equal to the order of the resultant equation (3). If more arbitrary constants are introduced in the process of solving the equations, the extra ones should be expressed in terms of the other constants. Note We can also eliminate x, get a linear equation in y and t and solve it first.

WORKED EXAMPLE 5(c)

2 Example 5.1 Solve the equation x

(

d 2y dy 1 + 4x + 2 y = x 2 + 2 . x d x2 dx

)

2 2 2 The given equation is x D + 4 xD + 2 y = x +

1 d , where D ≡ 2 dx x

d by θ. dt Then the given equation becomes θ (θ − 1) + 4θ + 2 y = e 2 t + e−2t   Put x = et or t = log x and denote

i.e.

(θ

2

+ 3θ + 2) y = e 2t + e−2t

5.36

Engineering Mathematics I

A.E. is m2 + 3m + 2 = 0 i.e. (m + 1) (m + 2) = 0 ∴ The roots are m = -1, -2. C.F. = A e −t + B e −2t =

∴

A B + 2 x x

1 (e 2t + e−2t ) (θ + 1) (θ + 2) 1 2t 1 e − e −2t = 12 θ+2 1 1 2t 1 2 = e − t e −2 t = x − 2 log x 12 12 x

P.I. =

∴ The general solution is y = C.F. + P.I. Example 5.2 Solve (x2 D2 + xD + 1) y = sin (2 log x) · sin (log x). d by θ , the given equation becomes Putting x = et or t = log x and denoting dt [θ (θ − 1) + θ + 1] y = sin 2t sint (θ 2 + 1) y =

i.e.

1 (sin 3t + sin t ) 2

A.E. is m2 + 1 = 0. The roots are m = ± i ∴ C.F. = A cos t + B sin t = A cos (log x) + B sin (log x). 1 1 (sin 3t + sin t ) θ +1 2  t 1  1 =  − sin 3t + (− cos t ) 2 2    8 1 1 = − sin (3 log x) − log x cos (log x) 16 4

P.I. =

2

∴ General solution is y = C.F.+ P.I. Example 5.3 Solve (x2 D2 – 2 xD - 4) y = 32 (log x)2 Putting x = et or t = log x and denoting

d by θ , the given equation becomes dt

[θ (θ − 1) − 2θ − 4] y = 32 t 2 i.e. A.E. is i.e. ∴ The roots are m = 4, -1. ∴

(θ 2 − 3θ − 4 ) y = 32t 2 m2 - 3m - 4 = 0 (m – 4)(m + l) = 0 C.F.= A e 4t + B e−t

Differential Equations = A x4 + P.I. =

5.37

B x

1 32t 2 θ 2 − 3θ − 4 −1

  2  θ = −8 ⋅  1− (θ − 3) t     4  θ  θ2 =− 8 1+ (θ − 3) + (θ − 3) 2  t 2  4  16    3θ 13  =− 8 1− + θ 2  t 2  4 16   3 13  =− 8 t 2 − t +   2 8 =−[8 (log x) 2 −12 (log x) +13] ∴ General solution is y = C.F. + P.I. 2

 log x  Example 5.4 Solve(x 2 D 2 − xD + 1) y =  ⋅  x.  Putting x = et or t = log x and denoting

d by θ, the given equation becomes dt

[θ (θ −1) − θ +1)] y = t 2 e−2t i.e.

(θ 2 − 2θ +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. ∴

C.F. = (At + B ) et = (A log x + B) ⋅ 1 e−2t ⋅ t 2 (θ −1) 2 1 t2 = e−2t ⋅ (θ − 3) 2

P.I.=

1  θ = e−2t ⋅ 1−  9  3 

−2

t2

 2θ θ2 1 = e −2t 1+ + 3 ⋅  t 2  9 3 9   1 4 = e −2t t 2 + t +   9 3

2   3 

1 x

5.38

Engineering Mathematics I =

{

1 2 3( log x ) + 4 log x + 2 27 x 2

}

∴ General solution is y = C.F. + P.I. Example 5.5 Solve (x 2D2 - xD + 4) y = x 2 sin (log x). Putting x = et or t = log x and denoting d by θ , the given equation becomes dt θ (θ −1)− θ + 4 y = e 2 t sin t  

(θ

i.e.

− 2θ + 4) y = e 2t sin t

m2 - 2m + 4 = 0

A.E. is The roots are m =1 ± i 3 ∴

2

(

) = x { A cos ( 3 log x ) + B sin (

C.F.= et A cos 3 t + B sin 3 t

P.I. =

3 log x

)}

1 e 2t sin t θ − 2θ + 4 2

= e 2t ⋅

1

(θ + 2) − 2 (θ + 2)+ 4 2

sin t

1 sin t θ 2 + 2θ + 4 1 sin t = e 2t ⋅ 2θ + 3

= e 2t ⋅

= e 2t ⋅

2θ − 3 sin t 4θ 2 − 9

1 2t e (2 cos t − 3 sin t ) 13 1 = − x 2 {2 cos (log x )− 3 sin (log x )} 13 ∴ General solution is y = C.F. + P.I. =−

d2 y dy − 2 ( 2 x + 3) −12 y = 6 x. 2 dx dx The given equation is a Legendre′s linear equation. Put 2x + 3 = et or t = log(2x + 3) Example 5.6 Solve ( 2 x + 3)

2

dy dy 2 = ⋅ dx dt 2 x + 3

Then i.e.

(2 x + 3)

dy =2θ y dx

Differential Equations

(2 x + 3)

2

Similarly

5.39

d2 y = 4θ (θ −1) y dx 2

The given equation then becomes i.e.

 4θ (θ −1)− 2×2θ −12 y = 3(e t − 3)   2 (4θ −8θ −12) y = 3(e t − 3)

(θ

i.e. A.E. is The roots are m = 3, - 1 ∴

3 t (e − 3) 4 m2 -2m - 3 = 0

2

− 2θ − 3) y =

C.F.= Ae3t + B e−t = A(2 x + 3) + 3

B 2x +3

1 3 t (e − 3) θ 2 − 2θ − 3 4  3 1 =  − e t +1 4  4 

P.I.=

3 3 (2 x + 3)+ 16 4

=−

∴ General solution is y = C.F. + P.I. Example 5.7 Solve (x2 D2 + xD + 1) y = log x sin (log x). Putting x = et or t = log x and denoting d by θ , the given equation becomes dt

θ (θ −1) + θ +1 y = t sin t  

(θ

i.e. A.E.is The roots are m = ± i ∴

2

+1) y = t sin t

m2 + 1 = 0 C.F. = A cos t + B sin t P.I.=

1 Imaginary part of eit ⋅ t θ +1 2

= I.P. of e ⋅ it

−1

1

1  iθ  t = I.P. of e ⋅ 1−  t 2iθ  2  +1 it

(θ + i )

2

 t 2 it  i 1  iθ  i = I.P. of − e it ⋅ 1+  t = I.P. of − e it  +  2 2  2 2  θ  2 = I.P. of

 t 2 it  1 (sin t − i cos t ) +  2  2 2

t 1 = sin t − t 2 cos t 4 4

5.40

Engineering Mathematics I

1 1 2 log x ⋅ sin (log x )− (log x ) cos (log x ) 4 4 ∴ General solution is y = C.F. + P.I. =

Example 5.8 Solve (x2 D2 + 4xD + 2) y = sin x. Putting x = et or log x = t and denoting d by θ , the given equation becomes dt

[q (q - 1) + 4q + 2] y = sin (e–t ) i.e. A.E. is The roots are ∴

(q 2 + 3q + 2) y = sin (e–t ) m2 + 3m + 2 = 0 m = - 1, - 2. C.F. = A e–t + B e–2t. A B + x x2 1 P.I.= sin (e t ) (θ +1)(θ + 2) =

 1 1   sin (e t ) =  −  θ +1 θ + 2  = e−t

∫ sin (e ) e t

t

dt − e−2t

∫ sin (e )⋅ e t

2t

dt

  ∵ 1 X = e m x X e−m x d x  ∫  D−m    = e−t

∫ sin u du − e

−2 t

⋅ ∫ u sin u du , putting et = u

= − e−t cos u − e−2t (− u cos u + sin u ) =− e−t cos (e t ) + e−t cos (e t )− e−2t sin (e t ) 1 sin x x2 ∴ General solution is y = C.F. + P.I. =−

Example 5.9 Solve the simultaneous equations dx + 2 x − 3 y = 5t dt dy − 3x + 2 y = 2 e2 t dt d by D , the given equations become Denoting dt (D + 2)x – 3y = 5t –3x + (D + 2) y = 2e2t Operating (1) by (D + 2); (D + 2)2x – 3(D + 2)y = 5 + 10t Multiplying (2) by 3; –9x + 3(D + 2)y = 6e2t

(1) (2) (1)′ (2)′

Differential Equations

5.41

Adding (1)′ and (2)′, we get (D2 + 4D – 5) x = 5(1 +2t) + 6e2 t

(3)

A.E. is m2 + 4m - 5 = 0 The roots are m = 1, - 5 ∴ C.F. = A e t + B e–5t. P.I.=

1 1 e2 t 5 (1+ 2t ) + 6 ⋅ 2 D2 + 4D −5 D + 4D −5 −1

 D ( D + 4)     =−1× 1−    5    

6 7

(1+ 2t )+ e2 t

 6 2t  4D    =−1× 1+ (1+ 2t ) + e   5 7      8 6 =−1+ 2t +  + e 2 t  5 7 =− 2t −

∴

13 6 2t + e 5 7

x = A e t + B e−5t − 2t −

13 6 2 t + e . 5 7

To find y, we may follow any one of the following two methods: Method 1 If we eliminate x between Equations (1) and (2), we will get (D2 + 4D - 5) y = 15t + 8e2 t

(4)

Solving (4) in the usual manner, we will get

y = C et + D e −5t − 3t −

12 8 2t + e 5 7

Note

In the solution of y, we should not use the same arbitrary constants A and B used in the solution of x. Though the values of x and y have been found out, they are expressed in terms of four arbitrary constants. The solutions for x and y should contain as many constants (in this problem, it is 2) as the order of the Equation (3) or (4). Hence the values of C and D should be expressed in terms of A and B as explained below: Inserting the values of x and y in Equation (1), 12 2t 12 2t 26     t − 5t t − 5t  Ae − 5 Be + 7 e − 2  +  2 Ae + 2 Be + 7 e − 4t − 5      24 36   −  3Ce t + 3De − 5t + e 2t − 9t −  = 5t 7 5  

i.e.

3(A – C) et – 3(B + D) e–5t = 0

∴

C = A and D = – B.

5.42

Engineering Mathematics I

∴ The required solutions of the given equations are x = A et + Be −5t +

6 2t 13 e − 2t − 7 5

y = A et − B e −5t +

and

8 2t 12 e − 3t − 7 5

Method 2 We eliminate Dy from Equation (1) and (2). Operating (1) by D; D2x + 2Dx - 3Dy = 5 Multiplying (2) by 3; -9x + 6y + 3Dy = 6 e2t

6 y + x′′ + x′ − 9 x = 5 + 6e 2t

Adding;

y=−

∴ x is given by

1  x′′ + x′ − 9 x − 6e 2 t − 5 6

(5)

6 2t 13 e − 2t − 7 5

(6)

12 2 t e −2 7

(7)

24 2 t e 7

(8)

x = A et + B e −5t +

Differentiating x w.r.t. t; x′ = Aet − 5e −5t + Further differentiating w.r.t. t; x′′ = Aet + 25 Be −5t + Using (6), (7) and (8) in (5) and simplifying, y=−

i.e.

1 6

48 2t 72   t −5t  − 6 A e + 6 Be − 7 e + 18t + 5   

y = Aet − Be −5t +

8 2t 12 e − 3t − 7 5

Example 5.10 Solve Dx - (D - 2) y = cos 2t. (D - 2) x + Dy = sin 2t Dx - (D - 2) y = cos 2t (D - 2) x + Dy = sin 2t Operating (1) by D; D2x - D(D - 2) y = -2 sin 2t Operating (2) by (D - 2); (D - 2)2x + D(D - 2)y = 2cos 2t - 2sin 2t. Adding; (2D2 - 4D + 4) x = 2cos 2t - 4sin 2t i.e. (D2 - 2D + 2) x = cos 2t - 2sin 2t 2 A.E. is m - 2m + 2 = 0 The roots are m = 1 ± i

(1) (2)

(3)

Differential Equations C.F. = et (A cos t + B sin t)

∴ P.I. =

1 ( cos 2t − 2 sin 2t ) D − 2D + 2 2

= − = ∴

5.43

1 1 1 ( D − 1) . ( cos 2t − 2 sin 2t ) = − . ( cos 2t − 2 sin 2t ) −5 2 D +1 2

1 1 ( −2 sin 2t − 4 cos 2t − cos 2t + 2 sin 2t ) = − cos 2t 10 2

x = et ( A cos t + B sin t ) −

1 cos 2t 2

(4)

Adding (1) and (2); 2Dx – 2x + 2y = sin 2t + cos 2t ∴

2y = 2x – 2x′ + sin 2t + cos 2t

(5)

Differentiating both sides of (4) w.r.t. t; x′ = et (A cos t + B sin t) + et (–A sin t + B cos t) + sin 2t

(6)

Using (4) and (6) in (5), we get, 2y = 2A et sin t – 2B et cos t – sin 2t 1 sin 2t 2 Now (4) and (7) constitute the solutions of the given equations. y = et(A sin t – B cos t) –

∴

(7)

Example 5.11 Solve D2x – Dy – 2x = 2t. Dx + 4Dy – 3y = 0 Rewriting the given equation, (D2 - 2)x – Dy = 2t

(1)

Dx + (4D - 3)y = 0

(2)

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

(1)′

D2x + D(4D – 3) y = 0

(2)′

Operating (1) by (AD – 3);

Operating (2) by D:

Adding (1)′ and (2)′, we get (4D3 – 2D2 – 8D + 6)x = 8 – 6t i.e.

(2D3 – D2 – 4D + 3)x = 4 – 3t

(3)

5.44

Engineering Mathematics I

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

∴ The roots are m =1, 1, − 3 2 3 − t 2

C.F. = ( At + B ) et + Ce

∴ P.I.=

−1

1

( D −1) (2 D + 3) 2

1  2D  −2 (4 − 3t ) = 1+  (1− D) (4 − 3t ) 3

3 

1  2 D  1 4  = 1− 1+ 2 D )(4 − 3t ) = 1+ D(4 − 3t )  (  3  3  3  3  =

1 (4 − 3t − 4) 3

=− t . ∴

x = ( At + B ) et + Ce

3 − t 2

−t

(4)

Eliminating Dy from (1) and (2), we get 4D2x + Dx – 8x – 3y = 8t ∴

y=

1 (4 x ′′ + x ′ −8 x −8t ) 3

Form (4)

3 −3t x ′ = ( At + B ) et + Aet − Ce 2 −1 2

and

3 − t 9 x ′′ = ( At + B ) et + 2 Aet + Ce 2 4

Using the values of x, x' and x'' in (5), 3 − t 1 1 y = {( 3 − t ) A − B} et − C e 2 − 6 3

Example 5.12 Solve

d2 x − 3 x − 4 y = 0. dt 2 d2 y + x + y = 0 if x = y = 1 dt 2

(5)

Differential Equations

5.45

dx dy = = 0 , when t = 0 dt dt

and The given equations are

(D2 – 3)x – 4y = 0

(1)

x + (D2 + l)y = 0

(2)

2

Operating (1) by (D + 1); (D2 + 1) (D2 – 3)x – 4 (D2 + 1) y = 0 Multiplying (2) by 4; 4x + 4(D2 + 1) y = 0 (D4 – 2D2 + 1) x = 0

Adding, A.E. is (m2 – l)2 = 0 Roots are m = ±1, ±1 ∴ Solution is

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)

Differentiating x w.r.t. t;

Differentiating further

From (1), y=

1 ( x ′′ − 3x) 4

1 = −2 ( At + B ) et + 2 Aet − 2 (Ct + D ) e−t − 2C e−t  4 =−

∴

1 A C 1 ( At + B) et + et − (Ct + D) e−t − e−t 2 2 2 2

y ′ =−

1 1 ( At + B) et + (Ct + D) e−t 2 2

(6)

(7)

Using the condition x = 1 when t = 0 in (3), B+ D=1 Using the condition y = 1 when t = 0 in (6), A− B C + D − =1 2 2

(8)

5.46

Engineering Mathematics I

i.e

A–B–C–D=1

(9)

Using the condition x′ = 0 when t = 0 in (4), A+B+C–D=0

(10)

Using the condition y′ = 0 when t = 0 in (7), –B + D = 0

(11)

Solving equations (8), (9), (10), and (11), we get 3 1 3 1 A= , B = , C =− , D = 2 2 2 2 Using these values in (3) and (6), the required particular solutions are 1+ 3t  t 1− 3t  −t x =  e +  e  2   2  1 3  1 3  y =  − t  et +  + t  e−t  2 4   2 4 

and

Example 5.13 Solve (D2 – 5)x + 3y = sin t. –3x + (D2 + 5) y = t Eliminating y from the given equations, we get (D4 - 16)x = 4 sin t - 3t A.E. is m4 – 16 = 0 The roots are m = ±2, ±i 2. ∴

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

1 3 sin t − 4 t D − 16 D − 16 4

=−

4 3  D4  sin t + 1−  15 16  16 

=−

3  D4  4 1 sin t + + 16  16  15

=−

4 3 sin t + t 15 16

−1

t −1

t

4 3 sin t + t 15 16 4 2t −2 t x ' = 2 Ae − 2 B e − 2 C sin 2t + 2 D cos 2t − cos t 15 4 x '' = 4 Ae 2t + 4 B e−2t − 4 C cos 2t − 4 D sin 2t + sin t 15 x = Ae 2t + B e−2t + C cos 2t + D sin 2t −

∴ and

(1)

Differential Equations

5.47

From the given equation, 1  5 x − x ′′ + sin t  3 A B 1 5 = e 2t + e−2t + 3C cos 2t + 3D sin 2t − sin t + t 16 3 3 5

y=

EXERCISE 5(c) Part A (Short Answer Questions) 1. Transform the equation xy′′ + y′ + 1 = 0 into a linear equation with constant coefficients and hence solve it. 2. Solve the equation x2y′′ – xy′ + y = 0. 3. Convert the equation xy″ – 3y′ + x–1y = x2 as a linear equation with constant coefficients. 4. Convert the equation x4y″ '– x3y″ + x2y′ = 1 as a linear equation with constant coefficients. 5. Solve the equation x2y″ – 2nxy′ + n(n + 1)y = 0. 6. Solve the equation x3y″ ' + 3x2y″ + xy′ + y = 0. 7. Solve for x from the equations x′ – y = t and x + y′ = 1. Part B   Solve the following equations  D ≡ d   d x  (x2D2 + 2xD – 20)y = (x2 + 1)2 (x4D3 – x3D2 + x2D)y = 1 (x3D3 – x2D2 + 2xD – 2)y = cos (2 log x) (x2D2 + xD – 9)y = sin 3 (log x) (x2D2 + 9xD + 25)y = (log x)2 (x4D4 + 6x3D3 + 9x2D2 + 3xD + l)y = (1 + log x)2 (x2D2 – 3xD + 4)y = x (log x)2 (x4D4 + 2x3D3 + x2D2 – xD + 1)y = x2 log x 1 16. x 2 D 2 − xD − 3 y = cos ( 2 log x ) x 17. (x2D2 + 3xD + 5)y = x cos (log x) 18. [(3x + 2)2 D2 + 3(3x + 2) D – 36] y = 3x2 + 4x + 1 8. 9. 10. 11. 12. 13. 14. 15.

(

)

19. [(x + 1)2D2 + (x + 1 ) D + 1 ] y = 4 cos log (x + 1)  d Solve the following simultaneous equations:  D ≡   dt  20. (D + 4) x + 3y = t 2x + (D + 5) y = e2t

5.48

Engineering Mathematics I

21. (2D + 1) x + (3D + 1) y = et (D + 5)x + (D + 7) y = 2et 22. Dx + y = sin t x + Dy = cos t given that x = 2 and y = 0 at t = 0 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

5.6

LINEAR EQUATIONS OF SECOND ORDER WITH VARIABLE COEFFICIENTS

In the previous section we have discussed the solution of Euler′s homogeneous linear differential equations of the second (and higher) order, which are a particular case of linear equations of second order with variable coefficients, that are functions of x. The general form of such a differential equation will be taken as d2 y dy + p ( x) + q ( x) y = r ( x) , 2 d x dx 2

in which the coefficient of d y is unity and p(x), q(x) and r(x) are functions of x. In dx 2 this section, we shall discuss a few methods of solving such equations.

5.6.1 Method of Reduction of Order-Transformation of the Equation by Changing the Dependent Variable Let the given equation be d2 y dy + p ( x) + q ( x) y = r ( x) 2 dx dx

(1)

Let us assume that one solution of the corresponding homogeneous equation, namely, d2 y dy + p ( x) + q ( x) y = 0 2 d x dx is known. Let it be

y = u (x)

(2) (3)

Differential Equations

5.49

y = u (x) . v(x)

We then assume that is a solution of equation (1).

(4)

From (4), we get

y′ = uv′ + u′v

(5)

and

y″ = uv″ + 2u′v′+ u′′v

(6)

Using (4), (5) and (6) in (1), we get uv″ + (2u′ + pu) v′ + (u″ + pu′ + qu) v = r i.e. uv″ + (2u′ + pu) v′ = r, by (2) and (3) i.e.

 u′  r v ′′ + 2 + p v ′ =  u  u

i.e.

v ′′ = p1v ′ = r1 , where p1 =

2u ′ r + p and r1 = . u u

(7)

Putting v ′ = w in (7) , it becomes dw + p1 w = r1 dx

(8)

Now equation (8) is a linear equation of the first order in w, which can be solved. Thus by changing the dependent variable y to w, we have reduced the order of the equation by one.

Note

Had the given equation (1) been homogeneous, viz., r(x) = 0, then equation (7) would have become  2u ′  v ′′ +  + p v ′ = 0  u  i.e.

d (v ′ ) v′

 2u ′  = − + p  u 

Integrating both sides with respect to x, we get log v ′ = −2 log u − ∫ pdx + log A  − = log  Au −2 e ∫ 

∴

− v ′ = Au −2 e ∫

pdx

pdx

   (9)

Solving equation (9), we get v and hence the solution of equation (1).

5.6.2  Definition A second order differential equation in y not containing the term in the first derivative y′ is said to be in the canonical or normal form.

5.50

5.6.3

Engineering Mathematics I

Reduction to Canonical or Normal Form

The second order linear differential equation y″ + p(x) y′ + q(x)y = r(x) can be transformed to the canonical or normal form v″ + f(x)v = g(x), where f (x) = 1 1 q ( x ) − {p ( x )}2 − p ′ ( x ) and g ( x ) = r ( x ) . e1/ 2 4 2 y = ve

−1 / 2 ∫ p ( x) dx

∫ p ( x) dx

, by using the substitution

⋅

Proof: Let us assume that y = uv is a solution of the equation y'' + p(x) y′ + q(x) y = r(x)(1) Differentiating y = uv (2) with respect to x, we get y′ = u′v + uv′ (3) and y′′ = u′′v + 2u′v′ + uv″ (4) The values of y, y′ and y″ given in steps (2), (3), (4) should satisfy equation (1). i.e. u″v + 2u′v′ + uv″ + p(u′v + uv′) + quv = r i.e. uv″ + (2u′ + pu) v′ + (u″ + pu′ + qu) v = r (5) Equation (5) should be in the canonical form, viz., it should not contain the v′ term.

∴

2u ′ + pu = 0

i.e.

u′ p =− u 2

i.e.

log u = −∫ −1 2

Assuming c = 0, we get u = e

∫

pdx

−1 2

Thus the substitution y = uv = ve

pdx +c 2

∫

pdx

transforms equationn (1) into the canonical form −1 2

When

u=e

and

u '' =

∫

pdx

p −1 2 pdx ,u ' = − ⋅ e ∫ 2

p 2 −1 2 ∫ pdx p ′ −1 2 ∫ pdx ⋅e − ⋅e 4 2

Using the values of u, u′ and u″ in the canonical form, viz., in v′′ + v=

1 r it becomes u

1 ( u ′′ + pu ′ + qu ) u

Differential Equations

5.51

 p2 p′ p2  1/ 2 pdx v ′′ +  − − + q v = re ∫   4 2 2  1/ 2 pdx 1 1  v ′′ + q − p 2 − p ′ v = re ∫  4 2 

i.e.

v ′′ + f ( x )⋅ v = g ( x) , where

i.e.

2 1 1 { p ( x)} − 2 p ′ ( x) and 4 1 / 2 p( x)dx g ( x ) = r ( x )⋅ e ∫

f ( x) = q ( x)−

5.6.4 Method of Reduction of Order—Special Types of Equations æ dy ö d2 y = f çç x , ÷÷ , in which y is explicitly absent. 2 çè dx ÷ø dx

Type 1. Equations of the form 2 Putting dy = p, we get d y2 = dp dx dx dx

dp = f ( x , p ) , which is only a first order The equation gets transformed as dx equation. Solving this transformed equation, we get p=

dy = φ ( x , c1 ) dx

Again, solving this equation, we get y = ψ (x, c1, c2). Extension: n−1  d n−1 y  dn y  x,  can be solved by putting d n−y1 = p and = f Equations of the form n n−1   dx  dx  dx

reducing the order successively.

Type 2. Equation of the form We put

æ dy ö d2 y = f çç y , ÷÷ in which x is explicitly absent. 2 çè dx ÷ø dx

d 2 y dp dp dy =p = p and treat p as a function of y. Then 2 = dx dx dy dx

The equation becomes

dp = f ( y , p ) , which is a first order equation. dy

5.52

Engineering Mathematics I

Solving this transformed equation, we get

dy = φ ( y, c1 ) dx Solving this equation further, we get p=

x = ψ (y, c1, c2) Extension: Equations of the form f{y, y′, y′′, ... y(n)} = 0 can be solved by the above technique, viz., by putting

dy = p and treating p as a function of y. dx

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

zdx

solved. When y = e ∫

, the order of the equation can be reduced by one and hence zdx

, y ′ = ze ∫

zdx

and y ′′ = ( z 2 + z ′) e ∫

zdx

. Thus, the order of the

transformed equation in the dependent variable z will be one less than that of the given equation.

Type 4. Equations f (x, y, y′, y′′) = 0 which are exact, viz. which can be expressed as d {φ ( x , y , y ¢ )} = 0 . dx d {φ ( x , y , y ′)} = 0 is φ ( x , y , y ' ) = c1 , which is a dx first order equation, solving which we get the solution of the given second order equation. The first integral of the equation

Note

The equation [p0(x)D2 + p1(x)D + p2(x)]y = r(x) is exact if and only if p′′0 – p′1 + p2 = 0. Let (p0 D2 + p1 D + p2) y = D(q0 D + q1) y = [q0 D2 + (q′0 + q1)D + q′1] y Comparing like terms, we get p0 = q0′; p1 = q′0+ q1 and p2 = q′1 Differentiating both sides of p1 = q′0 + q1, we get p′1 = q′′0 + q′1 = p′′0 + p2 ∴ p′′0– p′1+ p2 = 0 Conversely, when p′′0 – p′1 + p2 = 0, we have (p0 D2 + p1D + p2)y = (p0 D2 + p1D + p′1 – p′′0)y = p0 y′′ + p1 y′+ p′1y – p′′0 y

Differential Equations

5.53

= (p0 y′′ + p′0 y′) + (p1 y′+ p′1 y) – (p′′0 y + p′0 y′) = D(p0 y′) + D(pl y) – D(p′0 y) = D(p0 y′+ pl y – p′0 y) Thus (p0 D2 + p1 D + p2)y = 0 is an exact equation, when p′′0 – p′1 + p2 = 0.

WORKED EXAMPLE 5(d) Example 5.1 Solve the equation xy″ – 2(x + 1)y′ + (x + 2)y = (x – 2)e2x, by finding one solution of the corresponding homogeneous equation by inspection and reducing the order of the equation.

Important Notes To find one solution of the equation p0(x) y″ + p1(x) y′ + p2(x). y = 0, the following hints may be useful: (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. In the given problem, p0 = x, p1 = – 2(x + 1) and p2 = x + 2. The condition p0 + p1 + p2 = 0 is satisfied. ∴ y = ex is a solution of the homogeneous equation corresponding to the given equation. Let y = vex be a solution of the given equation. y′ = vex + v′ex

Then

y′′= v′′ex + 2v′ex + vex Using these values of y, y′, y″ in the given equation, it becomes x(exv″ + 2exv′ + exv) – 2(x + 1) (exv′ + exv) + (x + 2) exv = (x – 2)e2x i.e.,

i.e.,

xv″ – 2v′ = (x – 2)ex  2 dp 2 dv − p = 1−  e x , where p =  x  dx x dx

This is a linear equation of the first order. 2

I.F. = e

∫ − x dx

= e−2 log x =

1 x2

Solution of this equation is 1 p 2 = ∫  2 − 3  e x dx + 3c1 2  x x x 

5.54

Engineering Mathematics I 1  = ∫ d  2 e x  + 3c1  x  =

1 x e + 3c1 x2

dv = 3c1 x 2 + e x dx

i.e. Solving this equation, we get

v =c1x3 + c2 +ex ∴ The solution of the given equation is y = ex (c1x3 + c2+ cx) d2 y dy Example 5.2 Solve the equation + (1− cot x ) − y cot x = sin 2 x , by the 2 dx dx method of reduction of order. The given equation is y″ + (1 – cot x) y′ – y cot x = sin2 x (1) Here p0=1; p1 = 1 – cot x; p2 = – cot x We observe that p0 – p1 + p2 = 0 ∴ y = e–x is a solution of the equation y″ + (l – cot x)y′ – y cot x = 0

(2)

–x

Let y = ve be a solution of equation (1) Then

y′ = v′ e–x – ve–x

and

y″ = v″e–x – 2v′e–x+ve–x

Using these values of y, y′ and y″ in (1), we have (v″ e–x – 2v′ e–x + ve–x) + (1 – cot x) (v′ e–x – ve–x) – ve–x cot x = sin2 x i.e.

v″ – (1+ cot x) v′ = ex sin2 x

(3)

Putting v′ = p in (3), it becomes dp − (1 + cot x ) p = e x sin 2 x dx (4) in a linear equation of the first order − (1+ cot x )dx I.F. = e ∫

= e− x−log sin x =

e− x sin x

(4)

Differential Equations

5.55

∴ Solution of equation (4) is pe− x e− x = ∫ e x sin 2 x ⋅ dx + c1 sin x sin x = − cos x + c1 p=

i.e.

dv = e x (c1 sin x − sin x cos x ) dx

(5)

Integrating (5) with respect to x, 1 e x sin 2 x dx + c2 2∫ c 1 ex = 1 e x (sin x − cos x ) − ⋅ (sin 2 x − 2 cos 2 x ) + c2 2 2 5

v = c1

∫e

x

sin x dx −

∴ The solution of the given equation (1) is y = A(sin x − cos x ) −

1 (sin 2 x − 2 cos 2 x) + Be− x . 10

d2 y dy Example 5.3 Solve the equation (1− x 2 ) 2 − 2 x + 2 y = 2, by the method of dx x d reduction of order. The given equation is (1 - x2)y″ – 2xy′ + 2y = 2 Here p0 = 1 – x2; p1 = –2x; p2 = 2

(1)

We observe that p1 + p2x = 0. ∴ y = x is a solution of (1 – x2)y″ – 2xy′ + 2y = 0

(2)

Let y = vx be a solution of equation (1). Then

y′ = v′x + v; y″ = v″x + 2v′

y, y′,y″ satisfy equation (1) i.e.

(1 – x2)(v″x + 2v′) – 2x(v′x + v) + 2vx = 2

i.e.

x(1 – x2)v′′+ [2(1 - x2) – 2x2]v′ = 2

i.e.

dp  2 2 x  2 + − p=  dx  x 1− x 2  x (1− x 2 )

where

p = v′

(3)

5.56

Engineering Mathematics I

Equation (3) is a linear equation of the first order. 2

I.F. = e =e

2x 

∫  x − 1− x2 dx

(

2 log x + log 1− x 2

(

= x2 1 − x2

)

)

∴ Solution of equation (3) is

(

) ∫ x 1−2 x2 ⋅ x2 (1− x2 ) dx + c1 ( )

px 2 1− x 2 =

= x 2 + c1 c1 1 dv = + 2 2 dx 1− x x 1− x 2 1 1 1  = + c1  2 +  2   1− x 1− x 2  x

i.e.

(

)

(4)

Integrating (4) with respect to x 1 + x  c1 1 − + c2 v = (c1 + 1) log   1− x  x 2 ∴ Solution of equation (1) is  c + 1 1 + x  − c + c2 x y =  1  x log   1− x  1  2  d2y dy 1 Example 5.4 Solve the equation x 2 + (2 − x ) − y = 0 , given that y = is dx dx x a solution. [Refer to the note under the discussion of the method of reduction of order] If y = u is a solution of the equation y″ + p(x)y′ + q(x)y = 0, then y = uv will also be a solution of y″ + p(x) y′ + q(x)y = −

0, where v ′ = c1u−2 e

∫ p( x) dx

The given equation can be rewritten as 2  1 y ′′ +  −1 y ′ − y = 0  x  x

(1)

Differential Equations Here p ( x ) =

Since y =

y=

5.57

2 1 −1 and q ( x ) = − x x

1 is a solution of equation (1), x

1 v is also a solution of (1), where x v′ = c1 x 2 e

2  −  −1 dx x 

∫

= c1 x 2 e −2 log x + x = c1 x 2 ⋅ x −2 e x = c1e x v = c1e x + c2 ∴ ∴ The general solution of equation (1) is y=

(

1 c1e x + c2 x

)

or

xy = c1e x + c2

2 1 Example 5.5 Solve the equation d y + 2 dy + y = 0 given that y = sin x is a x dx 2 x dx

solution. y =

1 sin x ⋅ v is also a solution of the given equation, where x v ′ = c1 = c1

x2 sin 2 x x2 sin 2 x

−

⋅e

2

∫ x dx , since u = 1 sin x and p ( x) = 2 x

−2 log x

⋅e

=

x

c1 sin 2 x

Integrating, we get v = -c1 cot x + c2

or A cot x + B

:. The general solution of the given equation is y = i.e.

1 sin x ( A cot x + B ) x

xy = A cos x + B sin x

d2 y dy Example 5.6 Solve the equation + 2 x + x 2 y = 0 , by reducing it to the 2 dx dx canonical form.

5.58

Engineering Mathematics I

The given equation is y″ + 2xy′ + x2y = 0

(1)

Comparing equation (1) with the standard equation y″ + p(x) y′ + q(x) y = r(x)

(2)

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

Putting y = uv, where u = e

∫

p ( x) dx 2

= e− x

2

2

, in (1), it becomes

 p 2 p ′  v ′′ + q − −  v = re ∫  4 2 

p 2d x

[Refer to the discussion of reduction to normal form] i.e. v″ + (x2 - x2 -1) v = 0 i.e. v″ = v

(3)

which is of the form v″ = f(v, v′) dp Putting v′ = p and treating p as a function of v, we have v ′′ = p dv ∴ The equation becomes p

dp =v dv

(4)

Solving this equation, we get p2 = v2 + c 12 p=

or

dv = v 2 + c12 dx

(5)

v Solving equation (5), we have sinh−1   = x + c2  c1  v = c1 sinh ( x + c2 )

or

 e x + c2 − e −( x + c2 )  = c1   2   = Ae x + Be − x ∴ The required solution of equation (1) is

(

)

y = Ae x + Be− x e− x Example 5.7 Solve the equation 4 x 2 to the normal form.

d2 y dx

2

+ 4x

2

/2

(

)

dy + x 2 −1 y = 0 , by reducing it dx

Differential Equations

5.59

The given equation can be rewritten as 1 1 1 y ′′ + y ′ + 1− 2  y = 0  x 4 x  Here p =

1 ,q = x

(1)

1  1 1− 2  and r = 0 4 x 

Let y = uv be a solution of (1), where −1 2

u=e

Putting y =

1 x

∫ pd x = e−1 2 ∫

dx x

=

1 x

v in (1), it becomes  p 2 p ′  ∫ v ′′ + q − −  v = re  4 2 

i.e.

  1  1 1 1  v ′′ +   1− 2  − 2 + 2  v = 0  2 x    4  x  4x

i.e.

1 v ′′ + v = 0 4

p dx 2

(3)

x x Solving equation (3), we have v = A cos + B sin 2 2 ∴ Solution of equation (1) is y =

1  x x  A cos + B sin  2 2 x

d 2 y dy Example 5.8 Solve the equation x − = x 2 e x , given that y(0) = – 1 and y′(0) 2 dx dx = 0. The given equation does not contain y explicitly. d 2 y dp Putting dy = p and treating p as a function of x, we have ; the equation = dx dx 2 becomes dx

x

i.e.

dp − p = x2e x dx dp 1 − p = xe x dx x

(1)

5.60

Engineering Mathematics I

Equation (1) in a linear equation of first order in p. −

I.F. = e p⋅

Solution of (1) is

1 = x

1

∫ x dx = 1

x

∫

1 ⋅ xe x dx + 2c1 x

i.e.

p = x (ex + 2c1)

or

dy = xe x + 2c1 x dx

(2)

Solving equation (2), we have y=

∫ xe

x

dx +

∫ 2c x d x + c 1

2

i.e., the solution of the given equation is y = (x - 1) ex + c1x2 + c2. Using the condition y(0) = -1, c2 = 0. Using the condition y′(0) = 0 in (2), c1 is arbitrary. Taking c1 = 0, the required solution is y = (x - 1)ex. Example 5.9 Solve the equation

 dy 2   = 1 , given that y(0) = 0, y′(0) = 0 + dx 2  dx 

d2 y

Method 1 The given equation y″ + y′2 = 1

(1)

can be considered as one not containing y explicitly. dy d 2 y dp = p and treating p as a function of x, we have = dx dx dx 2 Then equation (1) becomes Putting

dp = 1− p 2 dx Solution of equation (2) is dp

∫ 1− p i.e.

2

= x + c1

1 + p  1  = x + c1 log  2  1− p 

Given that p = 0, when x = 0 ∴

c1 = 0

(2)

Differential Equations Thus we have

5.61

1+ p = e2 x 1− p p=

∴

=

dy e 2 x −1 = dx e 2 x +1 e x − e− x = tanh x e x + e− x

(3)

Solving (3), we have y = log cosh x + c2 Using the condition y(0) = 0, we get c2 = 0 ∴ Solution of equation (1) is y = log cosh x. Method 2 The equation y″ + y′2 = 1

(1)

can be considered as one not containing x explicitly. dy d2 y dp =p = p and treating p as a function of y, we have 2 dx dy dx Then equation (1) becomes Putting

p

dp =1− p 2 dy

(2)

Solution of equation (2) is p dp

∫ 1− p i.e.

2

= y + c1

1 − log (1− p 2 )= y + c1 2

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 = 1− e−2 y dx

Solving equation (3),

∫

dy 1− e−2 y

= x + c2

(3)

5.62

Engineering Mathematics I

∫

i.e.

e y dy e 2 y −1

= x + c2

cosh–1 (ey)= x + c2

i.e. When

x = 0, y = 0.∴ c2 = 0

∴The required solution of equation (1) is ey = cosh x or y = log cosh x.  1 dy  d 2 y dy = ⋅ log  2  x dx  dx dx

Example 5.10 Solve the equation x

The given equation is y ′′ =

 y′ y′ log    x  x

(1)

It is of the form y″ =f (x, y′), which does not contain y explicitly. Putting y′ = p and treating p as a function of x, we have y ′′ =

dp . dx

Then equation (1) becomes  p dp p = log    x  dx x

(2)

dp dv = ev + xev dx dx Then equation (2) becomes Putting p = xev, we have

 dv  ev 1+ x  = ev log (ev )  dx  i.e.

x

dv = v −1 dx

(3)

∴ Solution of (3) is dv

∫ v −1 = ∫ i.e.

dx + log c x

v = 1 + cx

i.e.

 p log   =1+ cx  x 

∴

p=

dy = xe1+ cx dx

(4)

Differential Equations

5.63

Solving equation (4), we have y=

x ⋅ e1+ cx e1+ cx − 2 + c′ c c

∴ The required solution of equation (1) is c2y = (cx – 1)e1+cx+c′ d2 y 2  dy  +   = 0 The given equation is 2 1− y  dx  dx 2

Example 5.11 Solve the equation

y ′′ +

2 ⋅ y ′2 = 0 1− y

(1)

It is of the form y″ = f(y, y′), which does not contain x explicitly. Putting y′ = p and treating p as a function of y, we have y ′′ = p

dp . dy

Then equation (1) becomes p

dp 2 p2 = 0 + dy 1− y

(2)

dp 2 dy = 0 + p 1− y

i.e. ∴ solution of (2) is

log p – 2 1og (l – y) = log c1 dy 2 = p = c1 (1− y ) dx

i.e. Solving (3), we have

∫ i.e.

dy

(1− y )

2

= c1 x + c2

1 = c1 x + c2 1− y

∴ The required solution of equation (1) is y =1−

1 . c1 x + c2

(3)

5.64

Engineering Mathematics I  dy  d2 y dy − x   − y = 0 . 2   dx  dx dx 2

Example 5.12 Solve the equation xy

The given equation xyy″ – xy′2 – yy′ = 0 is homogeneous of degree 2 in y, y′, y″. Putting y = e (1), we have

∫ zdx

(1)

∫ and hence y ′ = ze  ∫ zdx   e  

2

{x ( z

2

zdx

and y ′′ = ( z 2 + z ′) e ∫

zdx

in equation

+ z ′)− xz 2 − z } = 0

i.e.

xz′ – z = 0

(2)

Solving (2), we get

z = c1x

∴

c1 + c2 c1 xdx y = e∫ =e 2

or

y = Ae Bx

x2

2

dy 2 y d 2 y  dy  dx . Example 5.13 Solve the equation y 2 +   = dx  dx  1+ x 2 The given equation yy ′′ + y ′ 2 = yy ′

1+ x 2

(1)

is homogeneous of degree 2 in y, y′ and y″. Putting y = e ∫ have

zdx

∫ and hence y ′ = ze

 ∫ zdx   e   i.e.

i.e.

i.e.

2

zdx

∫ 2 1 and y ′′ = ( z + z ) e

zdx

in equation (1), we

   z   2 1 2 ( z + z ) + z − = 0 2   + 1 x     dz z + 2z2 − =0 dx 1+ x 2 −

(2)

1 dz 1 1 + ⋅ =2 z 2 dx z 1+ x 2 du 1 + u=2 dx 1+ x 2

(3)

Differential Equations u=

where

5.65

1 z

Equation (3) is a linear equation of the first order. I.F. = e =e

1

∫

1+ x 2

dx

(

log x + 1+ x 2

) = x+

1+ x 2

Solution of equation (3) is

(x+

)

)

(

1 + x 2 u = ∫ 2 x + 1 + x 2 dx + c1

)

(

= x 2 + x 1 + x 2 + log x + 1 + x 2 + c1

i.e.

z=

=

∴

x + 1+ x 2

(

)

x 2 + x 1+ x 2 + log x + 1+ x 2 + c1

{

) }

(

1 d log x 2 + x 1+ x 2 + log x + 1+ x 2 + c1 2 dx

y = e∫

zdx

{

) }

(

1 2 log x 2 + x 1+ x 2 + log x + 1+ x 2 + c1 + c2

=e

{

(

or

2

{

2

(

1 2

) } ⋅c

+ x + c1 = x + x 1+ x + log x + 1+ 2

3

) }

y 2 = A x 2 + x 1+ x 2 + log x + 1+ x 2 + B

Example 5.14 Show that the equation x

d2 y dy + ( x + 2) + y = 0 is exact and 2 dx dx

hence solve it. The given equation is xy″ + (x + 2)y′ + y = 0 Comparing equation (1) with p0y″ + p1y′ + p2y =0.

(1)

5.66 We have Now

Engineering Mathematics I p0 = x, p1 = x + 2, p2 = 1 p0″ – p1′ + p2 = 0 – 1 + 1 = 0

Thus the condition for exactness is satisfied. Now equation (1) can be rewritten as (xy″ + y′) + (xy′ + y′) + y′ = 0 i.e.

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

or

∴ Solution of (2) is x

i.e.

(2)

dy + ( x + 1) y = c1 dx c dy  1 + 1+  y = 1 d x  x x

(3)

Equation (3) is a linear equation of the first order. 

1

∫ 1 +  d x I.F. = e  x  = e x+logx = xe x ∴ Solution of equation (3) and hence equation (1) is y ⋅ x ⋅ e x = ∫ c1e x dx + c2 xye x = c1e x + c2

i.e. Example 5.15 Solve the equation ( sin x )

d2 y dy − ( cos x ) + 2 ( sin x ) y = cos x 2 dx dx

Comparing the given equation with p0 y″ + p1y′ + p2y = 0, we have p0 = sin x, p1 = – cos x and p2 = 2 sin x. Now p″0 – p′1 + p2 = – sin x – sin x + 2 sin x = 0 Hence the given equation is exact. It can be rewritten as d ( p0 y ′ + p1 y − p0′ y ) = cos x dx

Differential Equations

5.67

d ( sin x ) y ′ − ( cos x ) y − ( cos x ) y  = cos x dx 

i.e.

∴ The first solution of the given equation is (sin x) y′ – 2(cos x) y = sin x + c1 i.e.

y′ – 2(cot x)y = 1 + c1 cosec x

(1)

Equation (1) is a linear equation of the first order − 2 cot x dx I.F. = e ∫ 1 = e −2 log sin x = sin 2 x ∴ Solution of (1) is

y = − cot x + c1 ∫ cosec3 x dx + c2 sin 2 x = − cot x +

i.e.

c1 2

x   − cosec x cot x + log tan 2  + c2  

y = − sin x cos x +

c1  2 x sin x log tan − cos  2 2

 x  + c2 sin 2 x 

EXERCISE 5(d) Part A (Short Answer Questions) 1. If y = u(x) and y = u(x) v(x) are solutions of the equation y″ + p(x) y′ + q(x) y = 0, write down the first order equation satisfied by v(x). 2. When is a second order linear differential equation said to be in the canonical form? 3. Write down the transformation which will convert the equation y″ + p(x) y′ + q(x) y = r(x) into the normal form. 4. When the equation y″ + p(x) y′ + q(x) y = r(x) is transformed as v″ + f(x) v = g(x) by the substitution y = v exp  −1 / 2 ∫ p ( x ) dx  , what are the values of   f(x) and g(x)? 5. What is the substitution to be made to convert the equation y″ = f(x, y′) and y″ = f(y, y′) into first order equation? Indicate the difference in the subsequent procedures. 6. Write down the substitution to be made to convert the equation f(x, y, y′, y″) = 0 that is homogeneous in y, y′, y″, into a first order equation. 7. Write down the condition for the equation p0(x) y″ + p1(x) y′ + p2(x) y = r(x) to be exact. 8. If the equation p0(x) y″ + p1(x) y′ + p2(x) y = r(x) is exact, what is its first integral?

5.68

Engineering Mathematics I

9. How will you identify a second order linear differential equation in y with variable coefficients that has (i) y =ex as a solution (ii) y = e–x as a solution? 10. How will you identify a second order linear differential equation in y with variable coefficients that has y = x as a solution? Part B Solve the following equations by the method of reduction of order, after finding one solution of the corresponding homogeneous equation by inspection:

11.

( x + 1)

12. x

d2 y dy − ( 2 x − 1) + ( x − 1)y = e x dx dx 2

d2 y dy − cot x ⋅ − (1 − cot x ) y = e x sin x 2 dx dx

13.

14. x 2

15.

d2 y dy + x ( x − 2 )y = 0 + 2x ( x − 1) dx d x2

( x − 1)

16. x

d2 y dy +x + y =1 2 dx dx

d2 y dy + ( 2 x + 1) + ( x + 1) y = 0 dx d x2

(

2

17. 1 − x 2

18.

d2 y dy − 2 ( x + 3) + ( x + 5 )y = e x dx dx 2

) dd xy − 2 x dd xy + 2 y = 6(1 − x ) 2

2

d2 y 4x d y 4 − + y=0 d x2 2x − 1 d x 2x − 1

19. x 2

d2 y dy + ( x + 2 )y = x3 e x − x2 + 2x dx d x2

20. x 2

d2 y dy − 3x + 3y = x 2 (2 x − 1) 2 dx dx

(

)

Solve the following equations by the method of reduction: 21.

d2 y  x2 − 2x − 2  d y  2x2 − 2x − 2  2 + −   y = 0 , given that y = x is a d x 2  x 2 + x  d x  x3 + x 2  solution.

Differential Equations

5.69

22.

d2 y dy +x − 9 y = 0 , given that y = x3 is a solution. 2 dx dx

23.

d2 y + y = sec x , given that y = cos x is a solution of the corresponding d x2 homogeneous equation.

24.

d2 y + 4 y = 4 tan 2 x , given that y = sin 2x is a solution of the corresponding d x2 homogeneous equation.

2 1 25. x 2 d y +  x 2 − x tan x − 3  y = 0 , given that y = cos x is a solution. x 4 d x2 

26. x (1 + 3 x 2 )

27. (x 2 − 1)

1 d2 y dy is a solution. +2 − 6 xy = 0 , given that y = 2 x dx dx

d2 y − 6 y = 1 , given that y = x – x3 is a solution of the corresponding d x2

homogeneous equation. 28. cos 2 x ⋅

2 29. sin x

d2 y = 2 y , given that y = tan x is a solution. d x2

d2 y = 2 y , given that y = cot x is a solution. d x2

30. Find the values of a and b if y = x is a solution of the homogeneous equation corresponding to the equation x 2

d2 y dy + ( ax + b )y = x 3 . For − 2 x (1 + x ) dx d x2

these values of a and b, solve the equation completely. 31. Solve the equation ( x sin x + cos x )

d2 y dy − x cos x + y cos x = 0 , given 2 dx dx

that y = xm is a solution of the equation. Reduce the following equations to the canonical form and hence solve them: 32. x 2

d2 y dy  2 1 +x − x + y=0 4 d x  d x2

33. x 2

d2 y dy  2 1 +x + x − y=0 4 d x  d x2

5.70

Engineering Mathematics I

2 34. x

d2 y dy + (x 2 + 2 x + 2)y = 0 − 2 x2 + x 2 dx dx

35. 4 x 2

(

)

3 d2 y dy 2 2 4 x 1 4 x y x + + (16 − ) = sin x dx d x2

Reduce the order of the following equations by suitable transformations and hence solve them: 36.

d2 y dy + tan x ⋅ = sin 2 x , given that y(0) = – 1 and y′(0) = 0. dx d x2

37. (1 − x 2 )

π2 d2 y dy y (1) = − x = , given that y(0) = 0 and . 1 4 dx d x2 2

d2 y  d y  38. (1 + x ) +  +1= 0 d x2  d x  2

dy d2 y 39. d x 2 = x  d x   

3

2

40. y

3

d2 y  d y   d y  −  +  =0 d x2  d x   d x  3

d y d2 y  d y   d2 y  41. y ⋅ ⋅ =   + d x d x2  d x   d x2 

2

2

d2 y  d y  dy y 42. d x 2 −  d x  = d x   2

2

 d2 y   d y  2 43.  2  +   = a , given that y(0) = – 1 and y′(0) = 0. dx  dx 2

d2 y  d y  2 44. y 2 −   = 6 xy x d dx   2

d2 y  d y  2 45. y 2 +   = 2y dx dx 2

46. y

dy d2 y +2  =0 d x2 dx

Differential Equations

5.71

Show that the following equations are exact and hence solve them 47. x 2 .

d2 y dy +x −y=0 2 dx dx

48. (1− x 2 )

d2 y dy − 3x − y =1 2 dx dx

49. ( x 2 + 1)

d2 y dy + 3x + y =1 dx d x2

50. x

5.7

d2 y dy + y = e− x + ( x + 2) dx d x2

METHOD OF VARIATION OF PARAMETERS

The method of Variation of Parameters is another method for solving a linear differential equation, either of the first order or of the second order. If the given equation is of the form f(D)y = x, this method can be applied to get the general solution, provided the corresponding homogeneous equation, viz. f(D)y = 0 can be solved by earlier methods. The procedures to solve linear equations of the first and second orders are the following. Solution of the equation

dy + Py = Q dx

(1)

where P and Q are functions of x. The homogeneous equation corresponding to equation (1) is dy + Py = 0 dx i.e.

∴

(2)

dy = −Pdx y − Pdx log y = −∫ Pdx + log c = log ce ∫

∴ solution of Eq. (2) is − Pdx y = ce ∫

(3)

Now we treat the arbitrary constant c as a function of x and assume that (3) is the required solution of (1). Differentiating (3) with respect to x, we have

5.72

Engineering Mathematics I − Pdx dy dc −∫ Pdx = c ⋅ e ∫ ⋅ (−P ) + ⋅e dx dx

(4)

Since (3) is assumed as the solution of (1), (3) and (4) satisfy (1) − Pdx − Pdx dc − Pdx ∴ −cPe ∫ + e ∫ + cPe ∫ = Q dx

− Pdx dc = Qe ∫ dx

i.e.

c = ∫ Q ⋅ e∫

∴

Pdx

.dx + A

(5)

Using (5) in (3), the required general solution of (1) is  − Pdx  ∫ Pdx y = e ∫  ∫ Qe ⋅ dx + A     ye ∫

Pdx

= ∫ Q⋅e

∫ Pdx

.dx + A

(6)

Note

Solution (6) should not be treated as a formula and hence should not be directly used in problems. The procedure used in obtaining (6) alone should be used in solving problems. Solution of the equation

d2 y dy +P + Qy = R dx dx 2

(1)

where P, Q and R are functions of x or constants. The homogeneous equation corresponding to equation (1) is d2 y dy +P + Qy = 0 2 dx dx

(2)

Let the general solution of equation (2) be y = Au + Bv

(3)

where A and B are arbitrary constants (parameters) and y = u(x) and y = v(x) are independent particular solutions of Eq. (2). Now we treat A and B as functions of x and assume (3) to be the general solution of (1). Differentiating (3) with respect to x, we have dy = ( Au ′ + Bv ′) + ( A′ u + B ′v) dx

(4)

Differential Equations

5.73

We choose A and B such that A′u + B′v = 0

(5)

Then (4) becomes dy = Au ′ + Bv ′ dx

(6)

Differentiating (6) with respect to x, we have d2 y = Au ′′ + Bv ′′ + A′ u ′ + B ′v ′ dx 2

(7)

Since (3) is a solution of Eq. (1), (3), (6) and (7) satisfy (1). (Au′′ + Bv″ + A′u′ + B′v′) + P(Au′ + Bv′) + Q(Au + Bv) = R

∴ i.e.

A (u″ + Pu′ + Qu) + B (v″ + Pv′ + Qv) + A′u′ + B′v′ = R

(8)

Since y = u is a solution of Equation (2) ∴

u′′+ Pu′ + Qu = 0

Similarly

v′′ + Pv′ + Qv = 0

Inserting these values in (8), it reduces to A′u′ + B′v′ = R

(9)

Solving (5) and (9), we get the values of A′ and B′ integrating which, we get the values of A and B as functions of x. Using these values in (3), we get the required general solution of Eq. (1).

Notes 1. To solve Eq. (1) by the method of variation of parameters, one should know the complementary function of (1) and remember (5) and (9), solving which the values of A and B are obtained. d2 y 2. Equations (5) and (9) hold good, only if the coefficient of in the given dx 2 differential equation is unity. 3. The method is known as variation of parameters, as we treat the parameters (arbitrary constants) A and B as varying functions of x.

WORKED EXAMPLE 5(e) Example 5.1 Solve the equation ( x 2 + 1) of variation of parameters.

dy 1 , by using the method + 4 xy = 2 dx x +1

5.74

Engineering Mathematics I

The homogeneous equation corresponding to the given equation is dy

( x 2 + 1) xy + 4 xy = 0

(1)

dy 4x dx = 0 + 2 y x +1

i.e.

Integrating, we get log y + 2 log (x2 + 1) = log c i.e.

y=

c

(x

2

+ 1)

2

is the solution of (1)

(2)

Treating c as a function of x and differentiating (2) with respect to x, we have 2 2 dy ( x + 1) c ' − c ⋅ 2 ( x + 1) ⋅ 2 x = 4 dx ( x 2 + 1) 2

(3)

Using (2) and (3) in the given equation, we have

( x 2 + 1)

2

c ' − 4cx ( x 2 + 1)

(x i.e.

2

+ 1)

3

+

4cx

(x

2

+ 1)

2

=

1 x2 + 1

(x2 + l)2 c′ – 4cx (x2 + l) + 4cx (x2 + 1) = (x2 + 1)2

i.e.

c′ = 1

∴ c=x+k

(4)

Using (4) in (2), the required general solution of the given equation is (x2 + 1)2 y = x + k, where k is an arbitrary constant. Example 5.2 Solve the equation

dy + x sin 2y = x3 cos2 y, by the method of dx

variation of parameters. The method of variation of parameters can be applied to solve only a linear differential equation. The given equation is not linear. We shall convert the given equation into a linear equation and then apply the method of variation of parameters. Dividing the given equation by cos2 y, we get sec 2 y

dy + 2 x tan y = x 3 dx

(1)

Putting tan y = z, Eq. (1) becomes dz + 2 x ⋅ z = x 3 , which is linear. dx The homogeneous equation corresponding to Eq. (2) is

(2)

Differential Equations dz + 2 xz = 0 dx

dz + 2 xdx = 0 z

or

5.75 (3)

Integrating, we get log z = log c – x2 i.e. the solution of Eq. (3) is z = ce− x

2

(4)

Treating c as a function of x and differentiating (4) with respect to x, 2 2 dz = − 2cxe− x + c ′e− x dx

(5)

Using (4) and (5) in (2), – 2cxe–x2 + c′e–x2 + 2cxe–x2 = x3 i.e.

c′ = x3 ex2

∴

c = ∫ x3 ex2 dx + k =

1 2

∫te

t

dt + k , on putting x2 = t

1 t (te − et ) + k 2 2 1 = ( x 2 − 1) e x + k 2

=

(6)

Using (6) in (4), the required general solution of Eq. (2) is z=

2 1 2 x − 1) + ke− x ( 2

Therefore the general solution of the given equation is tan y =

2 1 2 ( x − 1) + ke−x 2

where k is an arbitrary constant d2 y Example 5.3 Solve the equation + y = x cos x, by the method of variation of dx 2 parameters. d2 y + y = x cos x dx 2

(1)

5.76

Engineering Mathematics I

The homogeneous equation corresponding to Eq. (1) is

d2 y + y= 0 dx 2

(2)

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

(3)

where A and B are parameters. Treating A and B as functions of x, A′ and B′ are given by – A′ sin x + B′ cos x = x cos x and

(4)

A′ cos x + B′ sin x = 0

(5)

by Eq. (5) and (9) of the discussion of the method of variation of parameters. Solving (4) and (5), we get A′ = − x sin x cos x or −

1 x sin 2 x and 2

(6)

1 x ( 1 + cos 2 x ) 2

(7)

1  − x cos 2 x sin 2 x  + + c1 2  2 4 

(8)

x 2 1  x sin 2 x cos 2 x  + + + c2 4 2  2 4 

(9)

B′ = x cos 2 x or

Integrating (6) and (7) with respect to x, we get A=−

and

B=

Using (8) and (9) in (3), the general solution of Eq. (1) is 1 x   y =  c1 + cos 2 x − sin 2 x  cos x   4 8   1 x2 x +  c2 + + sin 2 x + cos 2 x  sin x 4 4 8  

i.e. or

y = c1 cos x + c2 sin x −

x2 x 1 sin x + sin x + cos x 8 4 4

y = c1 cos x + c3 sin x +

x2 x sin x + cos x 4 4

 1 where c2 −  has been assumed as c3 .  8 

Differential Equations

5.77

d2 y Example 5.4 Solve the equation 2 + a 2 y = tan ax , by the method of variation of dx parameters. d2 y + a 2 y = tan ax dx 2 The homogeneous equation corresponding to Eq. (1) is

(1)

d2 y + a2 y = 0 dx 2

(2)

y = A cos ax + B sin ax

(3)

The solution of Eq. (2) is If we treat A and B as functions of x, A′ and B′are given by – aA′ sin ax + aB′ cos ax = tan ax

(4)

A′ cos ax + B′ sin ax =0

(5)

and Solving (4) and (5), we get

A′ = −

B′ =

and

1 sin 2 ax a cos ax

(6)

1 sin ax a

(7)

Integrating (6) and (7) with respect to x, we get A=

and

1 sin ax − log ( sec ax + tan ax )  + c1 a2 

B=−

(8)

1 cos ax + c2 a2

(9)

Using (8) and (9) in (3), the general solution of Eq. (1) is i.e.

1   y =  c1 + 2 sin ax − log ( sec ax + tan ax )  cos ax a  

{

}

1   +  c2 − 2 cos ax  sin ax   a i.e.

y = c1 cos ax + c2 sin ax −

1 cos ax ⋅ log ( sec ax + tan ax ) a2

5.78

Engineering Mathematics I

Example 5.5 Solve the equation (2D2 – D – 3) y = 25e–x, by the method of variation of parameters. As the formulae (5) and (9) given in the discussion of the procedure can be applied, only if the coefficient of D2y is unity, we rewrite the given equation as  2 1   D − D − 3  y = 25 e− x  2 2  2

(1)

The homogeneous equation corresponding to (1) is  2 1   D − D − 3  y = 0  2 2 

(2)

The auxiliary equation corresponding to Eq. (2) is

or

m2 −

1 3 m− = 0 2 2

3   m − 2 

( m + 1) = 0

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

3

x

y = Ae 2 + Be − x

(3)

Treating A and B as functions of x, A′ and B′ are given by 3 x 3 25 − x A′ ⋅ e 2 + B ′e− x = e 2 2

and

3

(5)

x

A′ e 2 + B ′e− x = 0

Solving (4) and (5), we get A′ = 5e

5 − x 2

and B ′ = − 5

Integrating, we get, 5 − x 2

A = − 2e

+ c1 and B = − 5 x + c2

Using these values in (3), the general solution of Eq. (1) is 5 3 − x  x  y =  c1 − 2e 2  e 2 + c2 − 5 x e − x  

(

(4)

)

Differential Equations 3

5.79

x

y = c1 e 2 + c2 e − x − 2e − x − 5 xe − x

i.e.

Example 5.6 Solve the equation x 2

d2 y dy − 4x + 6 y = sin ( log x ) , by the method dx dx 2

of variation of parameters. d2 y as unity, we rewrite the given equation as dx 2

To make the coefficient of

d 2 y 4 dy 6 1 − + 2 y = 2 sin ( log x ) 2 x dx x dx x

(1)

The homogeneous equation corresponding to (1) is d 2 y 4 dy 6 y = 0 or − + dx 2 x dx x 2 d2 y dy − 4x + 6y = 0 dx dx 2

x2

Putting x = e′ or log x = t and denoting

(2)

d by θ, Eq. (2) becomes dt

θ (θ −1) − 4θ + 6 y = 0  

(θ

i.e.

2

− 5θ + 6) y = 0

(3)

The auxiliary equation corresponding to (3) is m2 –5m + 6 = 0 ∴

m = 2, 3

Therefore the solution of Eq. (2) is y = Ae 2t + Be3t y = Ax 2 + Bx 3

or

(4)

Treating A and B as functions of x, A′ and B′are given by 2 A′x + 3B ′x 2 =

1 sin (log x) x2

A′ x 2 + B′ x3 = 0

and

(6)

Solving (5) and (6), we get A′ =

1 sin log x and x3

B′ =

(5)

1 sin log x. x4

5.80

Engineering Mathematics I 1 sin log x dx + c1 and B = x3

∫

1 sin log x dx + c2 x4

∴

A=−∫

i.e.

A = − ∫ e −2t sin t dt + c1 and B = ∫ e −3t sin t dt + c2 ,

on putting log x = t or x = et i.e.

i.e.

A = c1 −

e−2t ( − 2 sin t − cos t ) and 5

B = c2 +

e− 3t ( − 3 sin t − cos t ) 10

A = c1 +

1 ( 2 sin log x + cos log x ) and 5x2

B = c2 −

1 ( 3 sin log x + cos log x ) 10 x 3

Using these values of A and B in (4), the required solution of Eq. (1) is 1   y = c1 + 2 ( 2 sin log x + cos log x )  x 2 5 x   1  + c2 − ( 3 sin log x + cos log x ) x3 10 x 3  

i.e.

2 3  1 1  y = c1 x 2 + c2 x 3 +  −  sin log x +  −  cos log x  5 10   5 10 

i.e.

y = c1 x 2 + c2 x 3 +

1 {sin log x + cos log x} . 10

d2 y dy Example 5.7 Solve the equation x 2 −x + y = x log x , by the method of 2 dx dx variation of parameters.

To make the coefficient of

d2 y as unity, we rewrite the given equation as d x2

d2 y 1 d y 1 1 − + 2 y = log x 2 x x dx dx x The homogeneous equation corresponding to Eq. (1) is

(1)

Differential Equations d2 y 1 d y 1 − + 2 y = 0 or 2 x dx x dx 2 y d dy x2 −x + y=0 2 dx dx Putting x = et or t = log x and denoting

5.81

(2)

d by θ , Eq. (2) becomes dt

θ (θ −1)− θ +1 y = 0  

(θ − 1) 2 y = 0

i.e.

(3)

Therefore the solution of Eq. (3) is y = ( At + B ) et or y = Ax log x + Bx

(4)

Treating A and B as functions of x, A′ and B′ are given by A′ (1 + log x ) + B ′ =

1 log x x

A′ x log x + B ′ x = 0

and

(6)

Solving (5) and (6), we get A′ =

1 1 log x and B ′ = − (log x ) 2 x x 1 1 log x dx and B = − ∫ (log x ) 2 dx x x

∴

A= ∫

i.e.

A = ∫ t dt and B = − ∫ t 2 dt on putting log x = t

∴

A=

1 1 (log x) 2 + c1 and B = − (log x) 3 + c2 2 3

Using these values of A and B in (4), the required solution of Eq. (1) is 1   1  y =  ( log x ) 2 + c1  x log x +  − ( log x ) 3 + c2  x 2   3  i.e.

y = c1 x log x + c2 x +

1 x ( log x ) 6

(5)

3

5.82

Engineering Mathematics I EXERCISE 5(e)

Solve the following equations by the method of variation of parameters.

1.

dy − y tan x = e x sec x dx

2. x

dy + (1 + x ) y = e− x dx

3.

d2 y + y = x sin x d2 x

4.

d2 y + a 2 y = sec ax d2 x

4.

d2 y dy − 2 + 2 y = e x tan x dx dx 2

6.

d2 y dy 1 − 6 + 9 y = 2 e3 x 2 dx dx x

2 7. x 2 d y + 4 x dy + 2 y = x 2 + 1 dx dx 2 x2 2 d y dy 2 8. x 2 2 − 2 x − 4 y = 32 (log x ) dx dx

ANSWERS Exercise 5(a) (7) ( y − 2x − C )( y − 3 x − C ) = 0 (9) (y – ex – C) (ey – x – C) = 0 (11) y = px +

p C ; y = Cx + p −1 C −1

  x2 (8)  y − − C (log y − x − C ) = 0  2   y2  (10) ( y − log x − C ) − x − C  = 0   2 (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

   1 x2 (17)  y − − C  x + + C  = 0  y 4   

(18) (y – Cx2) (3x2 – y2 – C) = 0

Differential Equations (19) (y – Cx) (x2 – y2 – C) = 0

5.83

(20) (xy – C) (x2 – y2 – C) = 0

(21) [(y (1 + cos x) – C] [y (1 – cos x) – C] = 0   −1 y  y + log x − C sin −1 − log x − C  = 0 (22) sin      x x (23) y +

1 C = C2; y + 2 = 0 x 4x

(24) Eliminant of p between 4y = x2 + p2 and log ( p − x ) = (25) Eliminant of p between x = − (26) Eliminant of p between x =

x + C. p−x

n C p n−1 + 2 and the given equation. n +1 p

p 2 2C 2p C . + 2 and y = + p 3 3 p p −1

(27) Eliminant of p between x + C = log

p +1 2

−p (28) Eliminant of p between x = Ce −

− tan −1 p and the given equation.

ep 1 and y = C (1 + p ) e− p + (1− p ) e p 2 2

 3   (29) y = 3 x + log  1− Ce3 x  (30) 64y = C(C – 4x)2; 4x3 = 27y (31) Eliminant of p between y = C – [2 log (p – 1) + 2p + p2] and x = C – [2 log(p – l) + 2p] (32) y2 = 2Cx + 4C2 (33) y2 = 2Cx + C3 (34) log y = Cx + C2 −1 (35) y = − x (1− x ) + tan

(37) y2 =Cx2+ C2 (39) xy= Cx – C2

x 1− x

(36) y = 1 + log x (38) y2 = 2Cx + C2 (40) ey = Cex + C3

Exercise 5(b)

x  3 3  2 x + (C3 x + C4 ) sin x. (1) y = e (C1 x + C2 ) cos 2 2  

(2)

x3 x 1 − x e − e 6 8

(3)

(4)

x 3 −2 x e 6

(5) (x – 2) e2x

(6) – e–x cos x

x (b sin a x − c cos a x) . 2a

1 (7) − xe x cos 2 x 4

5.84

Engineering Mathematics I

1 −2 x x e 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 = C1 cos 3 x + C2 sin 3 x +

1 x2 2 − + cosh x 9 81 10

1 −x 3 2 (13) y = ( Ax + B ) e + x − 6 x + 18 x − 24 + 25 (4 sin 2 x − 3 cos 2 x)

(

4x (14) y = e Ae

7x

+ Be−

7x



) + 291 (5 cos 5x − 2 sin 5x) + 19  x

2

+

16 110  x+  9 81 

1 −x −2 x 2 (15) y = C1e + C2 e + 20 (cos 2 x − 3 sin 2 x ) + x − 3 x + 4 −

(16) y = (C1 x + C2 ) + e

x 2

  C cos 3 x + C sin 3 x + x 4 − 4 x 3 +  4  3 2 2  

1 (8 cos 3x − 3 sin 3x) − 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 −x (19) y = C1 + (C2 x + C3 ) e +

e2 x 18

 2 7   x − x + 11  3 6 

2 x 2x −2 x (20) y = C1e + C2 e − sinh x − cosh x 3 9 (21) y = (C1x + C2) cos x + (C3x + C4) sin x + (x2 + 4x + 4) e–x  1 2 7  5 21 (22) y = C1e x + C2 e 4 x + e−2 x + e− x  x −  +  x 2 − x +   18 5 10   2 8 x x 3x (23) y = C1e + C2 e − e (cox 2 x + sin 2 x )

1 x −x (24) y = C1e + C2 e + C3 cos x + C4 sin x − 5 sin x cosh x 1 x 1 x x (25) y = e (C1 cos 2 x + C2 sin 2 x) + e − xe cos 2 x 4 4 x  3 3  1 − x 1 −x 2 x + C3 sin x + xe + e− x (2 siin x (26) y = C1e + e C2 cos  2 2  6 26  + 3 cos x ) 3 (27) y = C1 cos 2x + C2 sin 2x + e 2 x  (7 sin x − 4 cos x )  65 1 + (12 cos 3x + sin 3x) 145

Differential Equations (28) y = C1e x + C2 e3 x + 1

884

1 ( sin x + 2 cos x ) . 20

( 10 cos 5 x − 11 sin 5 x ) +

x x (29) y = ( Ax + B ) e − e ( x sin x + 2 cos x )

(30) y = A + Be− x +

5.85

sin x x ( sin x − cos x ) + cos x + 2 2

(31) y = ( Ax + B ) e 2 x +

2 x ( 3 sin x + 4 cos x ) + ( 11 cos x + 2 sin x ) 25 125

(32) y = Ae x + Be− x +

1 1 − x 2 cos x + x sin x. 2

(

)

(33) y = ( C1 x + C2 ) cos 2 x + ( C3 x + C4 ) sin 2 x +

x ( sin 2 x − 2 x cos 2 x ) 64

(

)

1  24 x cos 2 x + 9 x 2 − 26 sin 2 x  .  27  (35) y = A cos 2 x + B sin 2 x − cos 2 x log ( sec 2 x + tan 2 x ) . (34) y = A cos x + B sin x −

Exercise 5(c) (1) y = A log x + B − x (3) (θ − 4θ + 1) y = e 2

(2) y = x ( A log x + B )

3 2 −t (4) (θ − 4θ + 4θ ) y = e

3t

n n+1 (5) y = Ax + Bx .

  3   3   x  B cos  log x  + C sin  log x    2   2    (7) x = A cos t + B sin t + 2 (6) y =

A + x

1 1 B 1 4 . + x log x − x 2 − 5 9 7 20 x 1 (9) y = A + x 2 ( B + C log x ) − 9x (8) y = Ax 4 +

(10) y = ( A log x + B ) x + Cx 2 + (11) y = Ax 3 + (12) y =

1 { sin ( 2 log x ) + 7 cos ( 2 log x ) } . 100

3 1 B sin ( log x ) + sin ( 3 log x ) . − 3 40 72 x

1 1  16 78  A cos ( 3 log x ) + B sin ( 3 log x )  + ( log x ) 2 − log x +  x4  25  25 625 

2 (13) y = ( A log x + B ) cos ( log x ) + ( C log x + D ) sin ( log x ) + ( log x ) + 2 log x − 3

(14) y = x

2

( A log x + B ) +

x

{ ( log x )

2

− 4 log x + 6

}

(15) y =  A ( log x ) 3 + B ( log x ) 2 + C log x + D  x + x 2 ( log x − 4 ) 1 B 3 (16) y = Ax + x − 20 x { 2 sin ( 2 log x ) + cos ( 2 log x ) }

5.86

Engineering Mathematics I

 1 x  (17) y =  A cos (2 log x ) + B sin (2 log x ) + {4 sin (log x ) + 7 cos (log x )} 65 x  1  2  (3 x + 2) log (3 x + 2) +1  108  (19) y = A cos log (x + 1) + B sin log (x + 1) + 2 log (x + 1) sin log (x + 1) −2

(18) y = A(3 x + 2) + B (3 x + 2) 2

+

5 31 1 2t t− − e 14 196 12 2 1 9 1 y =− Ae−2t + Be−7t − t + + e 2t 3 7 98 6

−2 t −7 t (20) x = Ae + Be +

5 4B t 1 t e + e (21) x =− Ae−2t − 3 3 3 y = Ae−2t + Bet (22) x = 2 cosh t; y = sin t – 2 sinh t −

(23) x = ( A + Bt ) et + Ce

3t 2

−

t 2

y = (−2 A + 6 B − 2 Bt ) et − (24) x = A + B cos t + C sin t + y = B cos t + C sin t −

3t

C −2 1 e − 3 3

9 2t e 10

6 2t e 10

1 (1+ cos 2t ) 4 y =1− 2 A cos 2 t − 2 B sin 2 t − C cos 3 t − D sin 3 t

(25) x = A cos 2 t + B sin 2 t + C cos 3 t + D sin 3 t −

1 t (26) x = ( At + B ) cos t + (Ct + D ) sin t + 25 e (4 sin t − 3 cos t ) y =−( At + B ) sin t + (Ct + D ) cos t −

1 t e (3 sin t + 4 cos t ) 25

4 5 37 (27) x = Aet + Be−t + C cos 3t + D sin 3t − t 2 + t − . 9 9 81 5 4 44 y =− Aet − Be−t + C cos 3t + D sin 3t + t 2 − t + . 9 9 81 Exercise 5(d) x x 5 x x (11) y = A( x +1) e − e + Be . 4 (12) y = ex (c1 log x + c2 + x).

Differential Equations 1 x −x x (13) y = Ae (2 sin x + cos x ) + Be − e cos x 2 3 –x (14) y = (Ax + B)e (15) y = Axe–2x + Be–x (16) y = e–x (c1 log x + c2) 1+ x  c 3 3 − c1 + c2 x + x 2 (17) y =  1 +  x log   2 4 2 1− x  (18) y = c1e2x + c2x (19) y = c1 xex + c2x + x (x – 1)ex (20) y = c1 x3 + c2x + x3 log 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 log cos x (24) y= c1 cos 2x + c2 sin 2x – cos 2x log (sec 2x + tan 2x) (25) y =

1 x

cos x  A + B ( x tan x + log cos x )

2 (26) y = c1 (1+ x ) + c2 ⋅

1 x

1+ x  1 . (27) y = c1 ( x − x3 ) + c2 (4 − 3 x − 6 x 2 + 3 x3 ) log   1− x − 6 (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 + B ) x −

x2 2

(31) y = c1x – c2 cos x (32) 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 =  A cos 2 x + B sin 2 x + sin x  3 x (36) y = 2 sin x – sin x cos x – x – 1 (33) y =

2 1 π sin −1 x ) + sin −1 x ( 2 4  1 1 (38) y =− x + 1+ 2  log (1+ c1 x ) + c2  c  c

(37) y =

1

1

5.87

5.88

Engineering Mathematics I

(39) x= c1 sin (y – c2) (40) y + c1 log 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 − x 2 + c2 / 1 − x 2 + 1 2 2 -1 (49) y 1 + x = 1 + x + c1 sinh x + c2 (50) xyex = c1 ex + c2 – x.

Exercise 5(e) x sin 2x + +c 2 4 (2) xy ex = x + c (1) y cos x =

1 1 x sin x − x 2 cos x 4 4 1 1 y = c1 cos ax + c2 sin ax + 2 cos ax log cos ax + x sin ax a a y = ex (c1 cos x + c2 sin x) – ex cos x log (sec x + tan x) y = (c1x + c2) e3x – e3x log x c c 1 2 1 y = 1 + 22 + x − 2 log x 12 x x x 1 2 y = c1 x 4 + c2 ⋅ − 8 (log x ) + 12 log x + 13 x

(3) y = c1 cos x + c2 sin x + (4) (5) (6) (7) (8)