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TUTORIAL WORKBOOK
ADVANCED ENGINEERING MATHEMATICS
Published By:
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Near Sardar Patel Statue Vallabh Vidyanagar-388120 Dist. Anand, Gujarat Phone: 02692-237474 Telefax: 02692-237373 Email: [email protected] Website: www.riapublishinghouse.com
Near Sardar Patel Statue Vallabh Vidyanagar-388120 Dist. Anand, Gujarat Phone: 02692-237273, 237171 Telefax: 02692-237373 Email: [email protected] Website: www.riapublishinghouse.com
Tutorial Workbook Advanced Engineering Mathematics Copyright © Reserved by the Authors Publication, Distribution and Promotion rights reserved by the Publisher Second Edition: 2015 ISBN Number: 978-93-84339-07-4 Price: Rs. 180/-
All Rights Reserved No part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher.
Disclaimer Every care has been taken by the authors, editors and publisher to give correct information. In case of any omission, printing mistake, or any other inadvertent error, none of the above, nor the distributers, take any legal responsibility. In case of any dispute, all matters are subject to the exclusive jurisdiction of the Hon’ble Courts in Anand only.
Technical content typeset using LaTeX Title Design: Prof. Kumar Trivedi
AS PER THE REVISED SYLLABUS OF G.T.U.
TUTORIAL WORKBOOK
ADVANCED ENGINEERING MATHEMATICS (Second Edition)
B.E. Semester III Subject Code: 2130002
A. H. Hasmani
T. S. Jani
Professor Department of Mathematics Sardar Patel University Vallabh Vidyanagar
Assistant Professor Department of Mathematics G. H. Patel College of Engg. & Technology Vallabh Vidyanagar
V. R. Shah Associate Professor Department of Mathematics G. H. Patel College of Engg. & Technology Vallabh Vidyanagar
Consulting Editors Dr. Naresh Jotwani
Dr. Subhash Bhatt
B.Tech.(IIT-Bombay), Ph.D. (Rice University, USA) Founder Principal G. H. Patel College of Engg. & Technology Vallabh Vidyanagar
M.Sc., Ph.D. Department of Mathematics Sardar Patel University Vallabh Vidyanagar
About the Authors Dr. Abdulvahid H. Hasmani, M.Sc., Ph.D., UGC-CSIR-NET-JRF is a Professor of Mathematics at Sardar Patel University, Vallabh Vidyanagar. His research interests are General Relativity, Algebraic Computations in GR, Differential Equations. He has published several research papers and attended several National and International seminars and conferences. He has taught Engineering Mathematics at S.S. Engineering College, Bhavnagar and B.V.M. Engineering College, Vallabh Vidyanagar. At present he is teaching courses in Applied Mathematics to M.Sc. and M.Phil. students.
Prof. Tejas S. Jani, M.Sc., M.Phil., UGC-CSIR-NET-JRF is an Assistant Professor at Department of Mathematics, G. H. Patel College of Engg. & Technology (GCET), Vallabh Vidyanagar. His areas of interests are Banach Algebras, Abstract Harmonic Analysis, Fractional Differential Equations.
Prof. Vipul R. Shah, M.Sc., M.Phil. is an Associate Professor at Department of Mathematics, G. H. Patel College of Engg. & Technology (GCET), Vallabh Vidyanagar. His research interests are in Applied Mathematics in particular Financial Mathematics. He is on the verge of completing his Ph.D. in Mathematics. He has published many research papers and attended several National and International conferences in India and abroad.
PREFACE
Problem solving is an essential key to learn mathematics and to enhance the student’s ability of logical thinking. This is the main reason for having a tutorial component in a course in Engineering Mathematics. Tutorials supplement the usual classroom teaching, and can be regarded as laboratory work in mathematics. Therefore this book is prepared in an innovative way to serve as a laboratory manual for the student. There exists rich literature and excellent text-books on the subject of Advanced Engineering Mathematics. These books can be consulted for rigorous mathematical concepts and the application of mathematics to engineering problems. Our aim is to provide a platform for the students to use their own ideas in solving problems, and thereby actively learn the basic concepts through their own practice. The book is divided into five chapters which cover the entire syllabus of Advanced Engineering Mathematics (BE Sem-III) of Gujarat Technological University (Subject Code 2130002). Sections in each chapter are labeled as tutorials. Each tutorial begins with basic concepts, and includes examples for understanding the relevant mathematical tools required for solving problems. Appropriate blank space is provided for solving each problem in the exercises. Also provided are additional exercises and sample questions for viva-voce examination. Problems are chosen with different difficulty levels, and repetitive problems are avoided. For ready reference, each section ends with answers to unsolved problems. This book has a dual role to play: As a tutorial book to solve problems in tutorial sessions, and also as important final preparation for examinations. Since the problems to be discussed in tutorial sessions are known beforehand, enthusiastic students can work on them in advance and discuss further aspects with the instructor. During the process of preparing the manuscript, we have referred to many standard text-books of Engineering Mathematics. We acknowledge all of them, without however including here the lengthy exhaustive list. This book is a revised edition of our earlier book. The revision is in view of feedback from the teachers and students who have used it. We are thankful to them for feedback. Also, some changes are made in pedagogic considerations. We express our gratitude to authorities and our colleagues respectively at the Department of Mathematics, Sardar Patel University, Vallabh Vidyanagar, the Department of Mathematics, G.H. Patel College of Engineering & Technology (GCET), Vallabh Vidyanagar, Dr. H.B. v
vi
Soni, Principal, GCET, and Charutar Vidya Mandal, Vallabh Vidyanagar. We are thankful to the consulting editors for their advices during the preparation of the manuscript for the enhancement of the quality. Our thanks are also due to Bipinbhai Panchal and Gaurang Panchal of Ria Publishing House for providing us the opportunity of writing this book. We hope students will find this book helpful in their journey towards a successful career in engineering. 21 July, 2015 Vallabh Vidyanagar
-Authors
CONTENTS
1 Fourier Series and Fourier Integrals 1.1 Tutorial : Fourier Series . . . . . . 1.2 Tutorial : Discontinuous Functions 1.3 Tutorial : Even and Odd Functions 1.4 Tutorial : Half-Range Expansions . 1.5 Tutorial : Fourier Integrals . . . . . 2 First Order 2.1 Tutorial 2.2 Tutorial 2.3 Tutorial 2.4 Tutorial 2.5 Tutorial 2.6 Tutorial
. . . . .
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Differential Equations : Basic Concepts . . . . . . . : Exact Differential Equations : Integrating Factor . . . . . . : Linear Differential Equations : Bernoulli Equation . . . . . : Orthogonal Trajectories . . .
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3 Higher Order Linear ODEs 3.1 Tutorial : Homogenous Linear ODEs . . . . . . . . . . . . . . . . . . 3.2 Tutorial : Homogeneous Linear ODEs with Constant Coefficients . . 3.3 Tutorial : Nonhomogeneous Linear ODEs with Constant Coefficients 3.4 Tutorial : Method of Undetermined Coefficients . . . . . . . . . . . . 3.5 Tutorial : Method of Variation of Parameters . . . . . . . . . . . . . 3.6 Tutorial : Euler-Cauchy Equations . . . . . . . . . . . . . . . . . . . 3.7 Tutorial : Series Solution of Differential Equations . . . . . . . . . . . 4 Laplace Transforms and Applications 4.1 Tutorial : Basic Definition and Examples . . . 4.2 Tutorial : Inverse Laplace Transforms . . . . . 4.3 Tutorial : First Shifting Theorem . . . . . . . 4.4 Tutorial : Method of Partial Fractions . . . . 4.5 Tutorial : Differentiation of Laplace Transform vii
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1 1 13 19 30 35
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43 43 46 51 61 65 70
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73 73 80 86 104 112 120 126
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133 133 138 142 146 150
viii
4.6 4.7 4.8 4.9 4.10 4.11
Tutorial Tutorial Tutorial Tutorial Tutorial Tutorial
: : : : : :
Integration of Laplace Transform . . . . . . . . . Unit Step Function and Second Shifting Theorem Laplace Transforms of Periodic Functions . . . . . Convolution . . . . . . . . . . . . . . . . . . . . . Laplace Transform of Integral . . . . . . . . . . . Solution of ODEs by Laplace Transforms . . . . .
5 Partial Differential Equations and Applications 5.1 Tutorial : Formation of PDEs . . . . . . . . . . . 5.2 Tutorial : Solution of PDEs by Direct Integration 5.3 Tutorial : First Order Linear PDEs . . . . . . . . 5.4 Tutorial : First Order Non-Linear PDEs . . . . . 5.5 Tutorial : Linear PDEs with Constant Coefficients 5.6 Tutorial : Method of Separation of Variables . . . 5.7 Tutorial : One Dimensional Wave Equation . . . 5.8 Tutorial : One Dimensional Heat Equation . . . .
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155 158 163 165 171 172
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177 178 184 187 195 202 210 216 222
Appendix A Multiple Choice Questions
229
Appendix B Differentiation and Integration Formulas
235
Chapter
1
Fourier Series and Fourier Integrals There are several periodic phenomena that occur frequently in engineering for instance, rotating parts of machine, alternating electric current etc. To study such phenomena, it is required to express the related complicated periodic functions in terms of simple ones. Fourier series is an important tool which represents the periodic functions in terms of trigonometric functions. The ideas and techniques of Fourier series can also be extended to nonperiodic functions. This leads to Fourier integrals.
1.1
Tutorial : Fourier Series
Periodic Function A function f (x) is called a periodic function if there exists a positive number p such that f (x + p) = f (x)
for all real x.
This number p is called a period of f (x). The smallest such p is called the fundamental period of f (x). For example, (1) The functions sin x and cos x are periodic functions with period 2π. (2) The function tan x is a periodic function with period π. (3) The functions sin 2x, cos 3x are periodic functions with periods π and 2π 3 respectively. x 3 (4) The functions e , x , ln x are not periodic functions.
Figure 1.1: Periodic Function
1
2
Chapter 1
Fourier Series and Fourier Integrals
Fourier Series Let f (x) be a periodic function with period p = 2L. Then the Fourier series of f (x) is given by ∞
nπx nπx i a0 X h f (x) = + an cos + bn sin , 2 L L n=1 where a0 , an and bn are called the Fourier coefficients of f (x), given by the Euler’s formulas a0 an bn
1 = L
Z
1 = L
Z
1 = L
Z
L
f (x)dx −L L
f (x) cos
nπx L
−L L
f (x) sin
nπx L
−L
dx
(n = 1, 2, . . .)
dx
(n = 1, 2, . . .).
In the formulas of Fourier coefficients, we can take the integration from 0 to 2L or over any other interval of length p = 2L, say interval (c, c + 2L). So, if the interval under consideration is (0, 2L), take the integration from 0 to 2L (See Example 1.1.3). Solved Examples Example 1.1.1. Find the Fourier series of the function defined by f (x) = x + x2 ,
x ∈ (−π, π) and f (x + 2π) = f (x).
Solution. The Fourier series of f (x) is given by ∞
nπx nπx i a0 X h f (x) = + an cos + bn sin . 2 L L n=1
(1.1.1)
Here, ⇒
p = 2L = 2π
L = π.
Hence, equation (1.1.1) becomes ∞
a0 X f (x) = + [an cos nx + bn sin nx] . 2 n=1
(1.1.2)
Now, 1 a0 = L
Z
L
1 f (x)dx = π −L
Z
π
Z
π
2 xdx + x dx = π −π −π 2
For n ≥ 1, we have an
1 = L
Z
L
f (x) cos −L
nπx L
dx
Z 0
π
x3 x dx = 3 2
π 0
2 π3 2π 2 = = . π 3 3
Tutorial 1.1
Fourier Series
= = = = = = =
3
Z 1 π (x + x2 ) cos nxdx π −π Z π Z π 1 2 x cos nxdx + x cos nxdx π −π −π Z 2 π 2 x cos nxdx π 0 π cos nx 2 sin nx sin nx 2 (x ) − (2x) − + (2) − 3 π n n2 n 0 cos nπ i 2h 0 + 2π − 0 − (0 + 0 − 0) π n2 2 2π(−1)n π n2 4(−1)n . n2
For n ≥ 1, we have bn = = = = = = = =
Z nπx 1 L f (x) sin dx L −L L Z 1 π (x + x2 ) sin nxdx π −π Z π Z π 1 2 x sin nxdx + x sin nxdx π −π −π Z 2 π x sin nxdx π 0 π 2 − cos nx sin nx (x) − (1) − 2 π n n 0 h i 2 cos nπ −π + 0 − (0 + 0) π n 2 π(−1)n − π n n 2(−1) − . n
Hence, by equation (1.1.2), we obtain ∞ π 2 X 4(−1)n 2(−1)n f (x) = + cos nx − sin nx . 3 n2 n n=1 Example 1.1.2. Find the Fourier series to represent ex in the interval (−π, π). [GTU, Dec. 2013] Solution. The given function is not periodic. So, to obtain its Fourier series representation, we have to extend it on the whole real line by 2π periodicity condition. Let f (x) = ex ,
x ∈ (−π, π) and f (x + 2π) = f (x).
4
Chapter 1
Fourier Series and Fourier Integrals
The Fourier series of f (x) is given by ∞
nπx nπx i a0 X h f (x) = + an cos + bn sin . 2 L L n=1
(1.1.3)
Here, p = 2L = 2π
⇒
L = π.
Hence, equation (1.1.3) becomes ∞
a0 X f (x) = + [an cos nx + bn sin nx] . 2 n=1 Now, 1 a0 = L
Z
L
1 f (x)dx = π −L
Z
π
−π
ex dx =
2 sinh π 1 h x iπ 1 e . = (eπ − e−π ) = π π π −π
For n ≥ 1, we have an = = = = = = =
Z nπx 1 L f (x) cos dx L −L L Z 1 π x e cos nxdx π −π π x 1 e (cos nx + n sin nx) π 1 + n2 −π π 1 e−π e (cos nπ + n sin nπ) − (cos nπ − n sin nπ) π 1 + n2 1 + n2 π 1 e−π e n n (−1) − (−1) π 1 + n2 1 + n2 (−1)n (eπ − e−π ) 2 π(1 + n ) 2(−1)n sinh π . π(1 + n2 )
For n ≥ 1, we have bn = = = = =
Z nπx 1 L f (x) sin dx L −L L Z 1 π x e sin nxdx π −π x π 1 e (sin nx − n cos nx) π 1 + n2 −π π 1 e e−π (sin nπ − n cos nπ) − (sin nπ − n cos nπ) π 1 + n2 1 + n2 π e e−π 1 n n {−n(−1) } − {−n(−1) } π 1 + n2 1 + n2
(1.1.4)
Tutorial 1.1
Fourier Series
5
n(−1)n π (e − e−π ) π(1 + n2 ) 2n(−1)n sinh π = − . π(1 + n2 )
= −
Hence, by equation (1.1.4), we obtain ∞
2(−1)n sinh π 2n(−1)n sinh π cos nx − sin nx . π(1 + n2 ) π(1 + n2 ) ∞ sinh π sinh π X 2(−1)n 2n(−1)n = + cos nx − sin nx . π π n=1 1 + n2 1 + n2
sinh π X + f (x) = π n=1
For x ∈ (−π, π), we obtain the Fourier series representation of ex as ∞ 2n(−1)n sinh π sinh π X 2(−1)n + cos nx − sin nx e = π π n=1 1 + n2 1 + n2 x
(−π < x < π).
Example 1.1.3. Express f (x) = x2 as a Fourier series for all values of x from 0 to 2π. [GTU, Jan. 2013] Solution. The given function is not periodic. So, to obtain its Fourier series representation, we have to extend it on the whole real line by 2π periodicity condition. Let f (x) = x2 ,
x ∈ (0, 2π) and f (x + 2π) = f (x).
The Fourier series of f (x) is given by ∞
nπx nπx i a0 X h f (x) = + an cos + bn sin . 2 L L n=1
(1.1.5)
Here, f (x) = x2
and p = 2L = 2π
⇒
L = π.
Hence, equation (1.1.5) becomes ∞
a0 X + [an cos nx + bn sin nx] . x = 2 n=1 2
Now, a0
1 = L
Z 0
2L
1 f (x)dx = π
Z 0
2π
2π 1 x3 1 8π 3 8π 2 x dx = = = . π 3 0 π 3 3 2
For n ≥ 1, we have an
Z nπx 1 2L = f (x) cos dx L 0 L Z 1 2π 2 = x cos nxdx π 0
(1.1.6)
6
Chapter 1
= = = =
Fourier Series and Fourier Integrals
2π cos nx sin nx sin nx 1 2 (x ) − (2x) − + (2) − 3 π n n2 n 0 1 1 0 + (4π) − 0 − (0 + 0 − 0) π n2 1 4π π n2 4 . n2
For n ≥ 1, we have bn = = = = = =
Z nπx 1 2L f (x) sin dx L 0 L Z 1 2π 2 x sin nxdx π 0 cos nx cos nx 2π sin nx 1 2 − (2x) − 2 (x ) − + (2) π n n n3 0 1 1 2 2 −(4π 2 ) + 0 + 3 − −0 + 0 + 3 π n n n 2 1 4π − π n 4π − . n
Hence, by equation (1.1.6), we obtain ∞ 4π 2 X 4 4π f (x) = + cos nx − sin nx 2 3 n n n=1 For x ∈ (0, 2π), we obtain the Fourier series of x2 as ∞ 4π 2 X 4 4π x = + cos nx − sin nx 2 3 n n n=1 2
(0 < x < 2π).
1 Example 1.1.4. Express f (x) = 2 (π − x) as a Fourier series with period 2π, to be valid in 1 1 1 π the interval 0 to 2π. Hence, prove that 1− 3 + 5 − 7 + · · · = 4 . [GTU, Dec. 2013] Solution. The Fourier series of f (x) is given by ∞
nπx nπx i a0 X h f (x) = + an cos + bn sin . 2 L L n=1 Here, 1 f (x) = (π − x) and p = 2L = 2π 2
⇒
L = π.
(1.1.7)
Tutorial 1.1
Fourier Series
7
Hence, equation (1.1.7) becomes ∞
a0 X 1 (π − x) = + [an cos nx + bn sin nx] . 2 2 n=1
(1.1.8)
Now, 1 a0 = L
Z 0
2L
1 f (x)dx = π
Z 0
2π
2π 1 1 x2 1 (π − x)dx = πx − = (2π 2 − 2π 2 ) = 0. 2 2π 2 0 2π
For n ≥ 1, we have an = = = = = =
Z nπx 1 2L f (x) cos dx L 0 L Z 1 2π 1 (π − x) cos nxdx π 0 2 cos nx 2π 1 sin nx (π − x) − (−1) − 2π n n2 0 1 cos 2nπ cos 0 0− − 0− 2 2 2π n n 1 1 1 − 2+ 2 2π n n 0.
For n ≥ 1, we have bn = = = = = = =
Z nπx 1 2L f (x) sin dx L 0 L Z 1 2π 1 (π − x) sin nxdx π 0 2 2π cos nx 1 sin nx (π − x) − − (−1) − 2 2π n n 0 1 cos 2nπ cos 0 (−π) − − 0 − (π) − −0 2π n n 1 hπ π i + 2π n n 1 2π 2π n 1 . n
Hence, by equation (1.1.8), the required Fourier series is ∞ X 1 1 1 1 1 (π − x) = sin nx = sin x + sin 2x + sin 3x + · · · . 2 n 1 2 3 n=1
8
Chapter 1
Fourier Series and Fourier Integrals
π For x = 2 , we obtain 1 π 1 1 3π 1 π sin + sin π + sin + ··· = π− 1 2 2 3 2 2 2 1 1 1 π ⇒ 1 − + − + ··· = . 3 5 7 4 Exercises Exercise 1.1.1. Find the Fourier series of the 2π-periodic function f (x) = x + |x|, Solution.
−π < x < π.
[GTU- March 2010, Jan. 2015]
Tutorial 1.1
Fourier Series
Exercise 1.1.2. Find the Fourier series for f (x) = e−x , 0 < x < 2π. Solution.
9
[GTU, June 2014]
10
Chapter 1
Exercise 1.1.3. Obtain the Fourier series of f (x) = π2 1 1 1 Hence, deduce that 12 = 2 − 2 + 2 + · · · . 1 2 3 Solution.
Fourier Series and Fourier Integrals
π−x 2 in the interval 0 ≤ x ≤ 2π. 2 [GTU, Jan. 2015]
Tutorial 1.1
Fourier Series
11
Exercise 1.1.4. Find the Fourier series of the periodic function f (x) = π sin πx, Solution.
0 < x < 1 with p = 2L = 1.
[GTU, Dec. 2010]
12
Chapter 1
Fourier Series and Fourier Integrals
Viva Questions Question 1.1.5. Define periodic function. Give one example. Question 1.1.6. What is Fourier series ? Answers 1.1.1 1.1.2
π 2
+
2 π
1−e−2π 2π
∞ P n=1
+
(−1)n −1 n2 ∞ P
n=1
cos nx + 2
1−e−2π π(n2 +1)
∞ P
1 (−1)n+1 n
n=1 ∞ P
cos nx +
n=1
sin nx
1−e−2π n sin nx π(n2 +1)
zzzzzzz
1.1.3
π2 12
+
1.1.4 2 − 4
∞ P n=1 ∞ P n=1
cos nx n2 1 4n2 −1
cos(2nπx)
Tutorial 1.2
1.2
Discontinuous Functions
13
Tutorial : Discontinuous Functions
In the last section, we have seen that a continuous function f (x) on the interval (c, c + 2L) can be represented by a Fourier series. But the Fourier series of a discontinuous function does not converge to the function for all points. However, if the function f (x) has finitely many points of jump discontinuity, then its Fourier series converges to f (x) except at points x0 where f (x) is discontinuous. At the points x0 , the series converges to
+ f (x− 0 )+f (x0 ) . 2
Solved Examples Example 1.2.1. Find the Fourier series of the periodic function x, −1 < x < 0 f (x) = 2, 0 < x < 1.
[GTU, Jan. 2013]
Solution. The Fourier series of f (x) is given by ∞
nπx nπx i a0 X h an cos + + bn sin . f (x) = 2 L L n=1
(1.2.1)
Here, p = 2L = 2
⇒
L = 1.
Therefore, equation (1.2.1) becomes ∞
a0 X [an cos nπx + bn sin nπx] . f (x) = + 2 n=1
(1.2.2)
Now, 1 a0 = L
Z
L
1 f (x)dx = 1 −L
Z
1
1 f (x)dx = 1 −1
Z
0
Z xdx +
−1
0
1
x2 2dx = 2
0
1 1 3 + 2x 0 = + 2 = . 2 2 −1
For n ≥ 1, we have an
Z nπx 1 L = f (x) cos dx L −L L Z 1 1 = f (x) cos nπxdx 1 −1 Z 0 Z 1 = x cos nπxdx + 2 cos nπxdx −1
= = = =
0
1 cos nπx 0 sin nπx sin nπx (x) − (1) − 2 2 +2 nπ nπ nπ −1 0 cos(−nπ) 1 0+ 2 2 − 0+ + 2[0 − 0] nπ n2 π 2 1 cos nπ − 2 2 n2 π 2 nπ n 1 − (−1) . n2 π 2
14
Chapter 1
Fourier Series and Fourier Integrals
For n ≥ 1, we have bn
Z nπx 1 L f (x) sin dx = L −L L Z 1 1 = f (x) sin nπxdx 1 −1 Z 1 Z 0 x sin nπxdx + 2 sin nπxdx = −1
0
0 cos nπx h cos nπx i1 sin nπx (x) − − (1) − 2 2 +2 − nπ nπ nπ 0 −1 cos nπ 1 cos(−nπ) +0 −2 − {−0 + 0} − nπ nπ nπ n n (−1) 2(−1) 2 − − + nπ nπ nπ 2 − 3(−1)n . nπ
= = = =
Hence, by equation (1.2.2), we obtain ∞ 3 X 1 − (−1)n 2 − 3(−1)n f (x) = + cos nπx + sin nπx . 4 n=1 n2 π 2 nπ Example 1.2.2. Find the Fourier series of the periodic function f (x) with period 2, where π, 0≤x≤1 f (x) = π(2 − x), 1 ≤ x ≤ 2. [GTU, June 2013] Solution. The Fourier series of f (x) is given by ∞
f (x) =
nπx nπx i a0 X h an cos + + bn sin . 2 L L n=1
(1.2.3)
Here, p = 2L = 2
⇒
L = 1.
Therefore, equation (1.2.3) becomes ∞
f (x) =
a0 X + [an cos nπx + bn sin nπx] . 2 n=1
Now, a0
Z 1 2L = f (x)dx L 0 Z 1 2 = f (x)dx 1 0 Z Z 2 1 1 = πdx + π(2 − x)dx 1 0 1
(1.2.4)
Tutorial 1.2
Discontinuous Functions
15
2 1 x2 = π x 0 + π 2x − 2 1 1 = π + π (4 − 2) − 2 − 2 π = π+ 2 3π . = 2 For n ≥ 1, we have Z nπx 1 2L f (x) cos dx L 0 L Z 1 2 f (x) cos nπxdx 1 0 Z 1 Z 2 π cos nπxdx + π(2 − x) cos nπxdx 0 1 1 cos nπx 2 sin nπx sin nπx + π (2 − x) − (−1) − 2 2 π nπ nπ nπ 0 1 n o cos 2nπ cos nπ π(0 − 0) + π 0 − 2 2 − 0− 2 2 nπ nπ n 1 (−1) π − 2 2+ 2 2 nπ nπ 1 [(−1)n − 1] n2 π 0, if n is even 2 − n2 π , if n is odd.
an = = = = = = = = Thus
a1 = −
2 12 π
,
a2 = 0,
a3 = −
2 32 π
,
a4 = 0,
a5 = −
2 52 π
,...
For n ≥ 1, we have bn = = = = = =
Z nπx 1 2L f (x) sin dx L 0 L Z 1 2 f (x) sin nπxdx 1 0 Z 1 Z 2 π sin nπxdx + π (2 − x) sin nπxdx 0 1 2 h cos nπx i1 cos nπx sin nπx π − + π (2 − x) − − (−1) − 2 2 nπ nπ nπ 0 1 h n oi cos nπ cos nπ 1 −π − + π {0 − 0} − − −0 nπ nπ nπ (−1)n 1 (−1)n 1 + + = . − n n n n
16
Chapter 1
Fourier Series and Fourier Integrals
Hence, by equation (1.2.4), we obtain a0 + a1 cos πx + a3 cos 3πx + · · · + b1 sin πx + b2 sin 2πx + · · · 2 2 1 1 3π 1 1 − sin πx + sin 2πx + · · · . = cos πx + 2 cos 3πx + · · · + 4 π 12 3 1 2
f (x) =
Exercises Exercise 1.2.1. Find the Fourier series of the function 2 x , x ∈ (0, π) f (x) = 0, x ∈ (π, 2π). Solution.
[GTU, Jan. 2013]
Tutorial 1.2
Discontinuous Functions
17
Exercise 1.2.2. Find the Fourier series expansion for f (x), if f (x) = 1 1 π2 1 Deduce that 2 + 2 + 2 + · · · = 8 . 1 3 5 Solution.
−π, −π < x < 0 x, 0 < x < π.
[GTU- May 2012, June 2014]
18
Chapter 1
Fourier Series and Fourier Integrals
Viva Questions Question 1.2.3. If x0 is a point of discontinuity for the function f (x), then the Fourier series of f (x) at x0 converges to . Question 1.2.4. Where does the Fourier series in Exercise 1.2.1 converge to at x = π ? Answers 1.2.1
π2 6
+2
1.2.2 − π4 −
∞ P
(−1)n n2
n=1 2 cos x π 12
+
cos nx + cos 3x 32
+
1 π
∞ h P
n=1 cos 5x + 2 5
2 [(−1)n n3
+
− 1] −
π2 (−1)n n
3
i
sin nx
sin x − 21 sin 2x + 33 sin 3x − · · · 1
zzzzzzz
Tutorial 1.3
1.3
Even and Odd Functions
19
Tutorial : Even and Odd Functions
A function f (x) is said to be even if f (−x) = f (x) for all x. The graph of an even function is symmetric about Y -axis (Examples in Figure 1.2). The Fourier series of an even function of period 2L is given by ∞ nπx a0 X an cos + , f (x) = 2 L n=1 where 2 a0 = L
Z 0
L
2 f (x)dx and an = L
L
Z
f (x) cos 0
nπx L
dx.
Figure 1.2: Even Functions A function f (x) is said to be odd if f (−x) = −f (x) for all x. The graph of an odd function is symmetric about origin (Examples in Figure 1.3). The Fourier series of an odd function of period 2L is given by ∞ X nπx , f (x) = bn sin L n=1 where 2 bn = L
Z
L
f (x) sin 0
nπx L
dx.
Figure 1.3: Odd Functions
20
Chapter 1
Fourier Series and Fourier Integrals
Solved Examples Example 1.3.1. Obtain the Fourier series of the periodic function f (x) = 2x,
−1 < x < 1,
p = 2L = 2.
[GTU, Dec. 2009]
Solution. Since f (x) is an odd function on (−1, 1), its Fourier series is given by f (x) =
∞ X
bn sin
n=1
nπx L
.
(1.3.1)
Now, bn = = = = = =
Z nπx 2 L f (x) sin dx L 0 L Z 2 1 2x sin nπxdx 1 0 Z 1 4 x sin nπxdx 0 1 sin nπx cos nπx − (1) − 2 2 4 (x) − nπ nπ 0 h cos nπ i 4 − nπ 4(−1)n − . nπ
Hence, by equation (1.3.1), we obtain ∞ 4 X (−1)n+1 f (x) = sin nπx. π n=1 n
Example 1.3.2. Find the Fourier series of the function f (x) = cos πx,
1 1 − x>0 0 > x > −π.
22
Chapter 1
Fourier Series and Fourier Integrals
−1, −π < x < 0 1, 0 < x < π. = −f (x). = −
Thus f (x) is an odd function, so that a0 and an are zero for all n. Therefore, the Fourier series of f (x) is given by ∞ nπx X . (1.3.4) f (x) = bn sin L n=1 Here, ⇒
p = 2L = 2
L = 1,
So, equation (1.3.4) becomes f (x) =
∞ X
bn sin nπx.
(1.3.5)
n=1
Now, for n ≥ 1, we have bn = = = = = = =
Z 2 L nπx f (x) sin dx L 0 L Z π 2 sin nxdx π 0 2 h cos nx iπ − π n 0 2 cos nπ 1 − − π n n n 1 2 (−1) − − π n n 2 [1 − (−1)n ] nπ 0, if n is even 4 , if n is odd. nπ
Thus
4 , b2 = 0, π Hence, by equation (1.3.5), we obtain b1 =
b3 =
4 , 3π
b4 = 0,
b5 =
4 ,... 5π
f (x) = b1 sin x + b3 sin 3x + b5 sin 5x + · · · 4 1 1 sin x + sin 3x + sin 5x + · · · . = π 3 5 Example 1.3.4. Find the Fourier series of the function defined by 1 + x, −1 ≤ x ≤ 0 f (x) = p = 2. 1 − x, 0 ≤ x ≤ 1, Hence, deduce that 1 1 1 π2 + + + · · · = . 12 32 52 8
Tutorial 1.3
Even and Odd Functions
23
Solution. Observe that
1 − x, −1 ≤ −x ≤ 0 1 + x, 0 ≤ −x ≤ 1.
1 − x, 1 + x,
f (−x) = =
1≥x≥0 0 ≥ x ≥ −1.
1 + x, −1 ≤ x ≤ 0 1 − x, 0 ≤ x ≤ 1. = f (x). =
Thus f (x) is an even function, so that bn = 0 for all n. Therefore, the Fourier series of f (x) is given by ∞ nπx a0 X + an cos . (1.3.6) f (x) = 2 L n=1 Here, p = 2L = 2
⇒
L = 1.
Therefore, equation (1.3.6) becomes ∞
a0 X f (x) = + an cos nπx. 2 n=1
(1.3.7)
Now, 2 a0 = L
Z
L
1
Z f (x)dx = 2
0
0
x2 (1 − x)dx = 2 x − 2
1 0
1 = 2 1− 2
1 = 2 = 1. 2
For n ≥ 1, we have Z 2 L nπx f (x) cos dx L 0 L Z 1 2 (1 − x) cos nπxdx 0 cos nπx 1 sin nπx 2 (1 − x) − (−1) − 2 2 nπ nπ 0 cos nπ 1 2 0− 2 2 − 0− 2 2 nπ nπ n (−1) 1 2 − 2 2 + 2 2 nπ nπ 2 1 − (−1)n 2 2 nπ 0, if n is even 4 , if n is odd. n2 π 2
an = = = = = = = Thus a1 =
4 12 π 2
,
a2 = 0,
a3 =
4 32 π 2
,
a4 = 0,
a5 =
4 52 π 2
,...
24
Chapter 1
Fourier Series and Fourier Integrals
Hence, by equation (1.3.7), we obtain a0 + a1 cos πx + a3 cos 3πx + a5 cos 5πx + · · · 2 4 cos πx cos 3πx cos 5πx 1 + = + + + ··· . 2 π2 12 32 52
f (x) =
For x = 0, we obtain f (0) ⇒ ⇒ ⇒
1 1 2 1 12
1 4 1 1 1 = + 2 2 + 2 + 2 + ··· 2 π 1 3 5 4 1 1 1 1 = + 2 2 + 2 + 2 + ··· 2 π 1 3 5 4 1 1 1 = 2 2 + 2 + 2 + ··· π 1 3 5 1 1 π2 + 2 + 2 + ··· = . 3 5 8
Exercises Exercise 1.3.1. Find the Fourier expansion for the function f (x) = x − x3 in the interval −1 < x < 1. [GTU, Dec. 2013] Solution.
Tutorial 1.3
Even and Odd Functions
25
Exercise 1.3.2. Find the Fourier series of the function f (x) = 1 − x2
(−1 < x < 1),
p = 2.
Solution.
Exercise 1.3.3. Find the Fourier series of the function given by ( , −π ≤ x ≤ 0 1 + 2x π f (x) = 1 − 2x , 0 ≤ x ≤ π. π 1 1 1 π2 Hence, show that 2 + 2 + 2 + · · · = 8 . 1 3 5 Solution.
[GTU, May 2011]
26
Chapter 1
Fourier Series and Fourier Integrals
Tutorial 1.3
Even and Odd Functions
Exercise 1.3.4. Find the Fourier series of the function −k, −2 < x < 0 f (x) = k, 0 < x < 2. 1 1 1 π Hence, deduce that 1 − 3 + 5 − 7 + · · · = 4 . Solution.
27
p = 4.
28
Chapter 1
Fourier Series and Fourier Integrals
Exercise 1.3.5. Find the Fourier series of the periodic function with period 2 given by 0, −1 ≤ x ≤ 0 f (x) = x, 0 ≤ x ≤ 1. [GTU, June 2013] Solution.
Tutorial 1.3
Even and Odd Functions
29
Additional Exercises Exercise 1.3.6. Sketch the function f (x) = x + π, (−π < x < π) where f (x + 2π) = f (x) and find its Fourier series. [GTU, March 2010] Exercise 1.3.7. Find the Fourier series expansion of f (x) = x2 , −2 < x < 2. [GTU, Dec. 2011] Exercise 1.3.8. Find the Fourier series for f (x) = | sin x| in −π < x < π. [GTU, May 2011] Exercise 1.3.9. Find the Fourier series of the function defined by −x, − 21 < x < 0 p = 1. f (x) = x, 0 < x < 12 , Viva Questions Question 1.3.10. Sketch the function f (x) = whether it is even or odd.
1 2 1 2
+ x, − 21 < x < 0 p = 1. Also, Check − x, 0 < x < 12 ,
Question 1.3.11. The Fourier series expansion of f (x) = x cos x in the interval (−1, 1) contains only sine terms or cosine terms or both ? Question 1.3.12. Find bn for the Fourier series of the function f (x) = sin2 x in (−π, π). Answers 12 π3
∞ P
(−1)n+1 n3
1.3.2 23 + π42 cos12πx − cos22πx + 2 n=1 x cos 3x cos 5x 1.3.3 π82 cos + 32 + 52 + · · · 12 π 1 3π 1 5π 1.3.4 4k sin x + sin x + sin x + · · · π 2 3 2 5 2 ∞ ∞ cos nπx 1 P P 1 1 n 1.3.5 4 + π2 (−1) − 1 n2 − π (−1)n sin nnπx 1.3.1
sin nπx
n=1
1.3.8
2 π
−
4 π
n=1
− ···
n=1
1.3.6 π + 2 sin x − 12 sin 2x + 13 sin 3x − · · · 1.3.7 ∞ P
cos 3πx 32
cos 2nx 4n2 −1
1.3.9
1 4
+
∞ P n=1
1.3.10 even 1.3.11 only sine terms
2 − nπ sin nπ + 2
4 3
+
4 n2 π 2
1.3.12 0 zzzzzzz
16 π2
∞ P n=1
(−1)n n2
cos nπ − 2
cos
4 n2 π 2
nπx 2
cos 2nπx
30
1.4
Chapter 1
Fourier Series and Fourier Integrals
Tutorial : Half-Range Expansions
Half Range Cosine Series Let f : (0, L) → R be a given function. Extending f (x) as an even function on (−L, L), we obtain the Fourier series of f (x) as ∞
nπx a0 X + an cos , f (x) = 2 L n=1 where
Z Z nπx 2 L 2 L a0 = f (x)dx and an = f (x) cos dx. L 0 L 0 L This series is called the half range cosine series or Fourier cosine series. Half Range Sine Series Let f : (0, L) → R be a given function. Extending f (x) as an odd function on (−L, L), we obtain the Fourier series of f (x) as f (x) =
∞ X
bn sin
n=1
nπx , L
where
Z nπx 2 L bn = f (x) sin dx. L 0 L This series is called the half range sine series or Fourier sine series. Remark. A function can be extended as an even or odd function, hence Fourier cosine or Fourier sine series can be obtained for any function. Solved Examples Example 1.4.1. Find the half range cosine series for f (x) = x, 0 < x < 3. [GTU, Jan. 2013] Solution. The half range cosine series of f (x) is given by ∞
f (x) =
nπx a0 X . + an cos 2 L n=1
Comparing 0 < x < 3 with 0 < x < L, we get L = 3. Now, 3 Z Z 2 L 2 3 2 x2 2 9 a0 = f (x)dx = xdx = = = 3. L 0 3 0 3 2 0 3 2 For n ≥ 1, we have an
2 = L
Z
L
f (x) cos 0
nπx L
dx
(1.4.1)
Tutorial 1.4
Half-Range Expansions 2 = 3
Z
3
x cos
31 nπx
0
3
dx !
" !#3 nπx sin nπx cos 2 3 = (x) − (1) − n2 π23 nπ 3 3 9 0 2 9 9 = 0 + 2 2 cos nπ − 0 + 2 2 cos 0 3 nπ nπ 9 9 2 n = (−1) − 3 n2 π 2 n2 π 2 6 = (−1)n − 1 2 2 nπ 0, if n is even = 12 − n2 π2 , if n is odd. Thus
12 12 12 , a2 = 0, a3 = − 2 2 , a4 = 0, a5 = − 2 2 , . . . . 2 2 1π 3π 5π Hence, by equation (1.4.1), we obtain πx a0 3πx 5πx f (x) = + a1 cos + a3 cos + a5 cos + ··· 2 3 3 3 πx 5πx 1 1 3 12 1 − cos + 2 cos (πx) + 2 cos = + ··· 2 π 2 12 3 3 5 3 a1 = −
Example 1.4.2. If f (x) =
mx, 0 < x < π2 then show that m(π − x), π2 < x < π,
4m sin x sin 3x sin 5x f (x) = − + − ··· . π 12 32 52
[GTU, May 2012]
Solution. The half range sine series of f (x) is given by f (x) =
∞ X n=1
bn sin
nπx L
(1.4.2)
Comparing 0 < x < π with 0 < x < L, we get L = π. Now, Z 2 π f (x) sin nxdx bn = π 0 (Z π ) Z π 2 2 = mx sin nxdx + m(π − x) sin nxdx π π 0 2 (Z π ) Z π 2 2m = x sin nxdx + (π − x) sin nxdx π π 0 2 ( ) cos nx sin nx π2 cos nx sin nx π 2m = x − − − 2 + (π − x) − + − 2 π n n n n π 0 2
32
Chapter 1
Fourier Series and Fourier Integrals
nπ nπ nπ nπ π 1 π 1 2m − cos + 2 sin + cos + 2 sin = π 2n 2 n 2 2n 2 n 2 2m 2 nπ = sin 2 π n 2 4m 1 nπ = sin π n2 2 Hence, by equation (1.4.2), we obtain ∞ nπx nπ 4m X 1 f (x) = sin sin π n=1 n2 2 L 4m sin x sin 3x sin 5x − + − ··· . = π 12 32 52
Exercises Exercise 1.4.1. Find the Fourier cosine series of f (x) = ex , 0 < x < l. [GTU, March 2010] Solution.
Tutorial 1.4
Half-Range Expansions
33
Exercise 1.4.2. Find the Fourier sine series of a function f (x) = π − x, 0 < x < π. [GTU, March 2010] Solution.
Exercise 1.4.3. Find the Fourier cosine series for f (x) = x2 , 0 < x ≤ c. Also, sketch f (x) and its periodic extension. [GTU- Dec. 2010, June 2013] Solution.
34
Chapter 1
Fourier Series and Fourier Integrals
Additional Exercises Exercise 1.4.4. Find the half range cosine series for f (x) = sin x in (0, π) and show that 1 1 1 π 1− + − +··· = . [GTU- May 2012, Jan. 2015] 3 5 7 4 x, 0 < x < π2 Exercise 1.4.5. If f (x) = then show that π − x, π2 < x < π, π 2 cos 2x cos 6x cos 10x + + + ··· . f (x) = − 4 π 12 32 52 Viva Questions Question 1.4.6. Suppose you are given a function that is defined on the interval (0, L). How to obtain its Fourier cosine series expansion and Fourier sine series expansion ? Answers 1.4.1
el −1 l
1.4.3
c2 3
+
∞ P
+
2l l2 +n2 π 2
n=1 ∞ P
4c2 π2
n=1
(−1)n n2
el (−1)n − 1 cos nπx l
cos
nπx c
1.4.2 2
zzzzzzz
sin x 1
+
sin 2x 2
+
sin 3x 3
+ ···
Tutorial 1.5
1.5
Fourier Integrals
35
Tutorial : Fourier Integrals
In this section, we will express nonperiodic functions in terms of integrals. These integrals are called Fourier integrals. Precisely, we have the following definitions. Fourier Integral Let f (x) be absolutely integrable on R, i.e.,
R∞
|f (x)|dx < ∞. Then the representation
−∞
Z
∞
f (x) =
A(w) cos wx + B(w) sin wx dw,
0
where
1 A(w) = π
Z
∞
f (v) cos wv dv −∞
1 and B(w) = π
Z
∞
f (v) sin wv dv −∞
is called the Fourier integral of f (x). Fourier Cosine Integral If f (x) is an even function, then the representation Z ∞ Z 2 ∞ A(w) cos wx dw, where A(w) = f (x) = f (v) cos wv dv π 0 0 is called the Fourier cosine integral of f (x). Fourier Sine Integral If f (x) is an odd function, then the representation Z ∞ Z 2 ∞ f (x) = B(w) sin wx dw, where B(w) = f (v) sin wv dv π 0 0 is called the Fourier sine integral of f (x). Fourier Integral Theorem If a function f (x) has finitely many points of jump discontinuity over the interval (−∞, ∞), then the value of its Fourier integral is f (x) except at points x0 where f (x) is discontinuous. At the points x0 , the value of the integral is
+ f (x− 0 )+f (x0 ) . 2
Solved Examples Example 1.5.1. Find the Fourier integral representation of the function 2, |x| < 2 f (x) = 0, |x| > 2. [GTU- Jan. 2013, June 2014]
36
Chapter 1
Fourier Series and Fourier Integrals
Solution. Observe that f (−x) =
2, | − x| < 2 0, | − x| > 2
2, |x| < 2 0, |x| > 2 = f (x). =
Thus f (x) is an even function. Hence, Fourier integral of f (x) becomes Fourier cosine integral of f (x) and is given by Z ∞ A(w) cos wx dw. (1.5.1) f (x) = 0
Now, A(w) = = = = = =
2 π 2 π 4 π 4 π 4 π 4 π
Z
∞
f (v) cos wv dv Z 2 Z ∞ 2 cos wv dv + 0 cos wv dv 0 2 Z 2 cos wv dv 0 2 sin wv w 0 sin 2w sin 0 − w w sin 2w . w 0
Hence, by (1.5.1), we obtain 4 f (x) = π
∞
Z 0
sin 2w cos wx dw. w
Example 1.5.2. Find the Fourier cosine integral of f (x) = e−kx , where x > 0, k > 0. [GTU, March 2010] Solution. The Fourier cosine integral of f (x) is given by Z ∞ f (x) = A(w) cos wxdw. 0
Now, Z 2 ∞ A(w) = f (v) cos wvdv π 0 Z ∞ 2 = e−kv cos wvdv π 0 −kv ∞ 2 e = (−k cos wv + w sin wv) π k2 + w2 0
(1.5.2)
Tutorial 1.5
Fourier Integrals
37
1 2 0− 2 = (−k + 0) π k + w2 2k . = π(k 2 + w2 ) Hence, by (1.5.2), we obtain 2k f (x) = π
Z
∞
(
x, 0 < x < 1 0, x > 1.
cos wx dw. k2 + w2 0 ∞ R sin w−w cos w πx/2, 0 < x < 1 π/4, x = 1 Example 1.5.3. Prove that sin xw dw = w2 0 0, x > 1. Solution. Consider the function f (x) =
The Fourier sine integral of f (x) is given by Z ∞ f (x) = B(w) sin wx dw.
(1.5.3)
0
Now, B(w) = = = = = =
2 π 2 π 2 π 2 π 2 π 2 π
Z
∞
f (v) sin wv dv 0
Z
1
Z
0 sin wv dv
v sin wv dv + Z
∞
1
0 1
v sin wv dv 0
1 cos wv sin wv (v) − − (1) − w w2 0 cos w sin w − + w w2 sin w − w cos w . w2
Hence, by (1.5.3), we obtain 2 f (x) = π
Z 0
∞
sin w − w cos w sin wx dw. w2
The function f (x) is discontinuous at the point x = 1. We find that f (1− ) + f (1+ ) 1+0 1 = = . 2 2 2
38
Chapter 1
Fourier Series and Fourier Integrals
Hence, by Fourier integral theorem, we obtain 2 π
Z 0
∞
x, 0 < x < 1 sin w − w cos w 1 , x=1 sin wx dw = 2 w2 0, x > 1,
which gives Z∞ 0
πx/2, 0 < x < 1 sin w − w cos w π/4, x = 1 sin xw dw = w2 0, x > 1.
Exercises Exercise 1.5.1. Find the Fourier integral representation of the function 1 if |x| < 1 f (x) = 0 if |x| > 1
and hence evaluate
R∞ sin λ cos λx 0
Solution.
λ
dλ.
[GTU- Dec. 2013, Jan. 2015]
Tutorial 1.5
Fourier Integrals
Exercise 1.5.2. Show that
R∞ cos wx+w sin wx 0
Solution.
39
1+w2
dw =
0
π 2
πe−x
if x < 0 if x = 0 if x > 0. [GTU, Dec. 2010]
40
Chapter 1
Fourier Series and Fourier Integrals
Exercise 1.5.3. Using Fourier sine integral, show that Z ∞ w sin wx π dw = e−x (x > 0). 2 1+w 2 0 Solution.
Exercise 1.5.4. Prove that
R∞ sin w cos xw 0
Solution.
w
π 2 π 4
if 0 ≤ x < 1 if x = 1 dw = 0 if x > 1.
Tutorial 1.5
Fourier Integrals
Exercise 1.5.5. Prove that
41
R∞ 1−cos πw 0
w
sin wx dw =
π , 2
0,
if 0 < x < π if x > π. [GTU, March 2010]
Solution.
42
Chapter 1
Fourier Series and Fourier Integrals
Additional Exercises Exercise 1.5.6. Find the Fourier sine integral of the function sin x, 0 < x < π f (x) = 0, x > π. Viva Questions Question 1.5.7. Define Fourier integral, Fourier cosine integral and Fourier sine integral. Question 1.5.8. If x0 is a point of discontinuity for the function f (x), then the value of the Fourier integral of f (x) at x0 is Answers 1.5.1 f (x) =
1.5.6
2 π
2 π
R∞ 0
cos λx sin λ wλ
R∞ sin πw sin xw 0
1−w2
dλ;
R∞ 0
cos λx sin λ wλ
π 2 π 4
if 0 ≤ x < 1 if x = 1 dλ = 0 if x > 1
dw zzzzzzz
Chapter
2
First Order Differential Equations Differential equations are of great importance in engineering because many engineering problems when expressed mathematically, appear in the form of differential equations. For instance, the current I in an RLC circuit is governed by the differential equation L
dI 1 dE d2 I +R + I = 2 dt dt C dt
where L is the inductance, R is the resistance, C is the capacitance and E is the electromotive force. In this chapter, we shall introduce the relevant concepts in order to be able to discuss differential equations and we shall discuss some methods to solve first order ordinary differential equations.
2.1
Tutorial : Basic Concepts
Definition An equation involving the derivatives of a dependent variable with respect to one or more independent variables is called a differential equation. • Ordinary Differential Equation: A differential equation that contains derivatives with respect to only one independent variable is called an ordinary differential equation (ODE). For example, (i)
dy + y = sin x, dx
(ii)
d2 y dy − 2x + 2y = 0, 2 dx dx
(iii) x3
d3 y dy + 2x + y = x log x. 3 dx dx
• Partial Differential Equation: A differential equation that contains derivatives with respect to more than one independent variable is called a partial differential equation (PDE). For example, (i) x
∂u ∂u +y = u, ∂x ∂y
(ii)
∂u ∂ 2u = c2 2 , ∂t ∂x 43
(iii)
∂ 2u ∂ 2u + = 0. ∂x2 ∂y 2
44
Chapter 2
First Order Differential Equations
Order and Degree • The order of a differential equation is the order of the highest derivative occurring in the equation. • The degree of a differential equation is the power of the highest ordered derivative occurring in the equation provided all the derivatives are made free from radicals and fractions. For example, (1) The differential equation
dy + 4y = ex is of order 1 and degree 1. dx
∂ 2u ∂ 2u + 4 = cos x is of order 2 and degree 1. ∂x2 ∂y 2 3 2 2 dy dy + y (3) The differential equation + y = cos x is of order 2 and degree 2. dx2 dx (2) The differential equation
d3 y d2 y dy + 3 + 3 + y = ex is of order 3 and degree 1. 3 2 dx dx dx " 3 2 # 25 dy dy (5) The differential equation 1 + = is of order 3 and degree 4. dx3 dx
(4) The differential equation
Concept of Solution A solution of a differential equation is a relation between the variables which does not involve any derivatives and satisfies the given differential equation. • General Solution: The solution of a differential equation is intitutively integrations, hence the solution will contain arbitrary constants. Such a solution is called a general solution. The number of arbitrary constants in the general solution is equal to the order of the differential equation. • Particular Solution: A solution of a differential equation obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution. • Singular Solution: A solution which cannot be obtained from the general solution is called a singular solution. Furthermore, it does not involve any arbitrary constant. Initial and Boundary Value Problems A differential equation together with some conditions specified at one value of the independent variable is called an initial value problem (IVP). These conditions are called initial conditions. For Example, d2 y dy + + 4y = sin 2x; y(0) = 1, y 0 (0) = 4. 2 dx dx
Tutorial 2.1
Basic Concepts
45
A differential equation together with some conditions specified at more than one value of the independent variable is called a boundary value problem (BVP). These conditions are called boundary conditions. For Example, d2 y + 9y = 0; dx2
y(0) = 2, y(1) = 3.
First Order ODEs The standard form of a first order ordinary differential equation in the unknown function y(x) is dy = f (x, y). (2.1.1) dx The differential form of a first order ordinary differential equation in the unknown function y(x) is M (x, y)dx + N (x, y)dy = 0. (2.1.2) The first order ordinary differential equations can be classified as follows: (1) Variable Separable Equations;
(3) Exact Differential Equations;
(2) Homogeneous Differential Equations;
(4) Linear Differential Equations.
Variable Separable Equations A differential equation which can be expressed so that the coefficients of dx is only a function of x and that of dy is only a function of y is called a variable separable equation. Thus the general form of such an equation is g(y)dy = f (x)dx
(2.1.3)
To solve (2.1.3), integrate both sides, we obtain Z Z g(y)dy = f (x)dx + c. The result is an equation involving x and y that determines y as a function of x provided above integrals exist. Homogeneous Differential Equations A differential equation of the form M (x, y)dx + N (x, y)dy = 0 is said to be a homogeneous differential equation if all the terms in M (x, y) and N (x, y) have same degree. For example, (i) (x2 − 2xy)dx + (y 2 − x2 )dy = 0,
(ii) (x3 − 3x2 y)dx + (x2 y − y 3 )dy = 0.
This type of differential equations can be solved by substituting y = vx. Above two types of differential equations are easy to solve and have been already studied earlier. So, we will start with the exact differential equations. zzzzzzz
46
2.2
Chapter 2
First Order Differential Equations
Tutorial : Exact Differential Equations
Definition A first order differential equation of the form M (x, y)dx + N (x, y)dy = 0
(2.2.1)
is said to be exact if there exists a function U (x, y) such that M dx + N dy = dU . Condition For Exactness The necessary and sufficient condition for the differential equation (2.2.1) to be exact is ∂M ∂N = , ∂y ∂x ∂N where ∂M ∂y and ∂x denote partial derivatives of M and N w.r.t. y and x respectively.
Method of Solution For a given differential equation M (x, y)dx + N (x, y)dy = 0, ∂N • Test the condition of exactness, i.e. ∂M ∂y = ∂x .
• If the condition is satisfied, then the given differential equation is exact and its solution is given by Z Z M dx + (Terms in N not involving x) dy = c. y const.
∂N • If the condition of exactness is not satisfied, i.e. ∂M ∂y 6= ∂x , then the differential equation is not exact.
Solved Examples Example 2.2.1. Solve (x2 + 2xy)dx + (x2 − y 2 )dy = 0. Solution. The given equation is of the form M dx + N dy = 0 with M = x2 + 2xy
and N = x2 − y 2 .
First we check for exactness. It can be seen that ∂M = 2x and ∂y Thus
∂N = 2x. ∂x
∂M ∂N = . ∂y ∂x
Tutorial 2.2
Exact Differential Equations
47
Hence, the equation is exact and its solution is given by Z Z M dx + (Terms in N not involving x) dy = c y const.
Z ⇒
Z
2
(x + 2xy)dx −
y 2 dy = c
y const.
⇒
y3 x3 + x2 y − = c. 3 3
Example 2.2.2. Find the solution of the differential equation yex dx + (2y + ex )dy = 0, where y(0) = −1.
[GTU, March 2010]
Solution. The given equation is of the form M dx + N dy = 0 with M = yex
and N = 2y + ex .
First we check for exactness. It can be seen that ∂M = ex ∂y Thus
and
∂N = ex . ∂x
∂M ∂N = . ∂y ∂x
Hence, the equation is exact and its solution is given by Z Z M dx + (Terms in N not involving x) dy = c y const.
Z ⇒
x
Z
(ye )dx +
2ydy = c
y const.
⇒ yex + y 2 = c. Now, y(0) = −1
⇒
(−1)(e0 ) + (−1)2 = c
⇒
c = 0.
So, the required particular solution is yex + y 2 = 0. Example 2.2.3. Solve the initial value problem (cosh x cos y)dx − (sinh x sin y)dy = 0,
y(0) = π.
Solution. The given equation is of the form M (x, y)dx + N (x, y)dy = 0 with M = cosh x cos y
and N = − sinh x sin y.
48
Chapter 2
First Order Differential Equations
First we check for exactness. It can be seen that ∂M = − cosh x sin y ∂y Thus
and
∂N = − cosh x sin y. ∂x
∂M ∂N = . ∂y ∂x
Hence, the equation is exact and its solution is given by Z Z M dx + (Terms in N not involving x) dy = c y const.
Z ⇒
(cosh x cos y)dx = c y const.
⇒ cos y sinh x = c. Now, y(0) = π
⇒
cos π sinh 0 = c
⇒
Hence, the required particular solution is cos y sinh x = 0. Exercises Exercise 2.2.1. Solve (2ey − yex )dx + (2xey − ex + y 3 )dy = 0. Solution.
c = 0.
Tutorial 2.2
Exact Differential Equations
49
Exercise 2.2.2. Solve (1 + 2xy)dx + (1 + x2 + 2y)dy = 0. Solution.
Exercise 2.2.3. Test for the exactness and solve (x + 1)ex − ey dx − xey dy = 0, y(1) = 0. [GTU- Dec. 2011, June 2014] Solution.
50
Chapter 2
First Order Differential Equations
Exercise 2.2.4. Solve the differential equation yexy dx + (xexy + 2y)dy = 0. Solution.
Additional Exercises Exercise 2.2.5. Find the general solution of the differential equation y 1 − 2 + 2 cos 2x dx + − 2 sin 2y dy = 0. x x y cos x+sin y+y dy Exercise 2.2.6. Solve dx + sin x+x cos y+x = 0.
[GTU, June 2013]
Viva Questions Question 2.2.7. When the equation M dx + N dy = 0 is said to be exact ? Question 2.2.8. How to give the general solution if the equation is exact ? Question 2.2.9. Is the differential equation (ex+y − y)dx + (xex+y + 1)dy = 0 exact ? Question 2.2.10. Is the differential equation (ex − y)dx + (ey − x)dy = 0 exact ? Answers 2.2.1 2.2.3 2.2.5 2.2.9
y4
2xey − yex + 4 = c 2.2.2 x(1 + xy) + y(1 + y) = c x(ex − ey ) = e − 1 2.2.4 exy + y 2 = c y 2.2.6 y sin x + x sin y + xy = c x + sin 2x + cos 2y = c no 2.2.10 yes zzzzzzz
Tutorial 2.3
2.3
Integrating Factor
51
Tutorial : Integrating Factor
Definition Sometimes a differential equation which is not exact, can be made so by multiplying a suitable factor called an integrating factor (I.F.). Rules For Finding an Integrating Factor Consider a differential equation of the form M (x, y)dx + N (x, y)dy = 0.
(2.3.1)
Rule-1. If (2.3.1) is a homogeneous equation, then I.F. =
1 , provided M x + N y 6= 0. Mx + Ny
Rule-2. If (2.3.1) is of the form f (xy)ydx + g(xy)xdy = 0, then I.F. =
Rule-3. For (2.3.1), if
1 N
∂M ∂y
1 , provided M x − N y 6= 0. Mx − Ny
−
∂N ∂x
is a function of x only, say f (x), then R
I.F. = e Rule-4. For (2.3.1), if
1 M
∂M ∂y
−
∂N ∂x
f (x)dx
.
is a function of y only, say g(y), then
I.F. = e−
R
g(y)dy
.
Solved Examples Example 2.3.1. Solve (x2 y − 2xy 2 )dx − (x3 − 3x2 y)dy = 0. Solution. Here, M = x2 y − 2xy 2 ,
and N = −x3 + 3x2 y.
Therefore, ∂M = x2 − 4xy ∂y Thus
and
∂N = −3x2 + 6xy. ∂x
∂M ∂N 6= . ∂y ∂x
Hence, the given equation is not exact. But the equation is homogeneous in x and y. Also, M x + N y = (x2 y − 2xy 2 )x + (−x3 + 3x2 y)y
52
Chapter 2
First Order Differential Equations
= x3 y − 2x2 y 2 − x3 y + 3x2 y 2 = x2 y 2 6= 0. Therefore, I.F. =
1 1 = 2 2. Mx + Ny xy
Multiplying the equation throughout by x21y2 , we obtain 1 2 x 3 − dx − dy = 0 − y x y2 y which is exact (verify!). Therefore, the solution is given by x − 2 ln x + 3 ln y = c. y Example 2.3.2. Solve (1 + xy)ydx + (1 − xy)xdy = 0. Solution. Here, M = y + xy 2
and N = x − x2 y.
Therefore, ∂M = 1 + 2xy ∂y Thus
and
∂N = 1 − 2xy. ∂x
∂N ∂M 6= . ∂y ∂x
Hence, the given equation is not exact. But the equation is of the form f (xy)ydx + g(xy)xdy = 0. Also, M x − N y = (y + xy 2 )x − (x − x2 y)y = xy + x2 y 2 − xy + x2 y 2 = 2x2 y 2 6= 0. Therefore, I.F. =
1 1 = 2 2. Mx − Ny 2x y
Multiplying the equation throughout by 2x12 y2 , we obtain 1 1 1 1 + dx + − dy = 0 2x2 y 2x 2xy 2 2y which is exact (verify!). Therefore, the solution is given by 1 1 1 1 x 1 − + ln x − ln y = c0 or ln − = c, 2y x 2 2 y xy
where c = 2c0 .
Tutorial 2.3
Integrating Factor
53
Example 2.3.3. Find the general solution of the differential equation 2 sin y 2 dx + xy cos y 2 dy = 0. Solution. Here, M = 2 sin y 2
and N = xy cos y 2 .
Therefore, ∂M = 4y cos y 2 ∂y Thus
and
∂N = y cos y 2 . ∂x
∂M ∂N 6= . ∂y ∂x
Hence, the equation is not exact. But 3y ∂N 1 3 1 ∂M − = 4y cos y 2 − y cos y 2 = = 2 N ∂y ∂x xy cos y xy x which is a function of x only. Therefore, I.F. = e
R
3 dx x
3
= e3 ln x = eln x = x3 .
Multiplying the given equation by x3 , we obtain 2x3 sin y 2 dx + x4 y cos y 2 dy = 0 which is exact (verify!). Therefore, its solution is given by x4 sin y 2 = c. 2 Example 2.3.4. Solve (xy 3 + y)dx + 2(x2 y 2 + x + y 4 )dy = 0. Solution. Here, M = xy 3 + y
and N = 2(x2 y 2 + x + y 4 )dy = 0.
Therefore, ∂M = 3xy 2 + 1 and ∂y Thus
∂N = 4xy 2 + 2. ∂x
∂M ∂N 6= . ∂y ∂x
Hence, the equation is not exact. But 1 ∂M ∂N 1 −xy 2 − 1 1 2 2 − = (3xy + 1 − 4xy − 2) = =− 2 2 M ∂y ∂x y(xy + 1) y(xy + 1) y which is a function of y only. Therefore, I.F. = e−
R
− y1 dy
= eln y = y.
Multiplying the given equation by y, we obtain (xy 4 + y 2 )dx + (2x2 y 3 + 2xy + 2y 5 )dy = 0 which is exact (verify!). Therefore, its solution is given by Z Z x2 y 4 y6 4 2 5 2 (xy + y )dx + 2y dy = c ⇒ + xy + = c. 2 3 y const.
54
Chapter 2
First Order Differential Equations
Exercises Exercise 2.3.1. Solve (x2 y 2 + 2)ydx + (2 − x2 y 2 )xdy = 0. [GTU- May 2012, Jan. 2015] Solution.
Tutorial 2.3
Integrating Factor
Exercise 2.3.2. Solve (ex+y − y)dx + (xex+y + 1)dy = 0. Solution.
55
56
Chapter 2
Exercise 2.3.3. Solve x2 ydx − (x3 + xy 2 )dy = 0. Solution.
First Order Differential Equations
[GTU- Jan. 2013, Jan. 2015]
Tutorial 2.3
Integrating Factor
Exercise 2.3.4. Solve (y 4 + 2y)dx + (xy 3 + 2y 4 − 4x)dy = 0. Solution.
57
58
Chapter 2
First Order Differential Equations
Exercise 2.3.5. Solve (xy − 2y 2 )dx − (x2 − 3xy)dy = 0.
[GTU, Dec. 2013]
Solution.
Tutorial 2.3
Integrating Factor
Exercise 2.3.6. Solve (1 + 2x2 y 2 )ydx + (1 + xy + 2x2 y 2 )xdy = 0. Solution.
59
60
Chapter 2
First Order Differential Equations
Additional Exercises Exercise 2.3.7. Solve y(2xy + 1)dx + x(1 + 2xy − x3 y 3 )dy = 0. Exercise 2.3.8. Solve 2xydx + 3x2 dy = 0. Exercise 2.3.9. Solve (x3 + y 2 )dx − xydy = 0. Viva Questions Question 2.3.10. What is meant by an integrating factor ? 1 Question 2.3.11. Is xy an integrating factor for the equation (2y + xy)dx + 2xdy = 0 ?
Question 2.3.12. What is the form an integrating factor if the equation is homogeneous ? Answers x2 2.3.1 21 log xy − 2x12 y2 = c 2.3.2 xey + ye−x = c 2.3.3 − 2y 2 + log y = c 1 2 2.3.4 xy+ 2x 2.3.5 xy −2 log x + 3 log y = c 2.3.6 xy −2xy − log y = c y 2 +y = c
2.3.7 x21y2 + 3x13 y3 + ln y = c
2.3.8 x2 y 3 = c
y2
2.3.9 x− 2x2 = c
zzzzzzz
2.3.11 Yes
Tutorial 2.4
2.4
Linear Differential Equations
61
Tutorial : Linear Differential Equations
Definition A first order differential equation of the form dy + p(x)y = q(x) (where p and q are the functions of x) dx is called a linear differential equation.
(2.4.1)
Method of Solution • Compare the given equation with the standard form dy + p(x)y = q(x) dx to determine p(x) and q(x). R
• Find I.F. using the formula e p(x)dx . • Write the general solution as y(I.F.) =
R
q(x)(I.F.)dx + c.
Solved Examples dy [GTU, Jan. 2013] Example 2.4.1. Solve dx + y tan x = sin 2x, y(0) = 1. Solution. The given equation is linear with p(x) = tan x and q(x) = sin 2x. Therefore, R
I.F. = e
tan xdx
= eln sec x = sec x.
Hence, the general solution is Z y sec x =
sin 2x sec xxdx + c Z
⇒ y sec x =
2 sin x cos x sec xdx + c Z
⇒ y sec x = 2
sin xdx + c
⇒ y sec x = −2 cos x + c ⇒ y = c cos x − 2 cos2 x. Now, y(0) = 1
⇒
c cos 0 − 2 cos2 0 = 1
⇒
c(1) − 2(1)2 = 1
Hence, the required particular solution is y = 3 cos x − 2 cos2 x.
⇒
c=3
62
Chapter 2
First Order Differential Equations
x2 dy Example 2.4.2. Solve dx + xy = e− 2 . 2
x Solution. The given equation is linear with p(x) = x and q(x) = e− 2 . Therefore,
R
p(x)dx
Z
ye
x2 2
⇒ ye
x2 2
I.F. = e
R
=e
xdx
x2
=e2.
Hence, the general solution is =
x2
x2
e− 2 · e 2 dx + c
Z =
1 dx + c
x2
⇒ ye 2 = x + c x2
⇒ y = (x + c)e− 2 . Exercises dy 1 4x y= 2 . Exercise 2.4.1. Solve dx + 2 x +1 (x +1)3 Solution.
[GTU, Dec. 2013]
Tutorial 2.4
Linear Differential Equations
63
dy Exercise 2.4.2. Solve the linear equation dx − y cos x = esin x . Solution.
3
−2x Exercise 2.4.3. Solve the initial value problem y 0 + 6x2 y = e x2 , where y(1) = 0. [GTU March 2010]
Solution.
64
Chapter 2
First Order Differential Equations
y dy Exercise 2.4.4. Solve the differential equation dx + x = ex . Solution.
Additional Exercises dy Exercise 2.4.5. Solve (x + 1) dx −y = e3x (x + 1)2 . dx Exercise 2.4.6. Solve (1 + y 2 ) dy = tan−1 y − x.
[GTU, June 2014] [GTU, June 2013]
Exercise 2.4.7. Solve the initial value problem y 0 − (1 + 3x−1 )y = x + 2, y(1) = e − 1. [GTU, Dec. 2010] Viva Questions Question 2.4.8. What is the standard form of a linear equation ? Question 2.4.9. What is the integrating factor for a linear equation ? Question 2.4.10. Which of the following equations are linear ? (i)
dy + x3 y = sin x dx
(ii)
dy + xy 2 = ex dx
(iii)
dy 1 = dx 6x + ey
(iv)
dy x +y = dx y
Answers 3
2.4.1 y(x + 1)2 = tan−1 x + c 2.4.2 y = (x + c)esin x 2.4.3 ye2x = 1 − x1 3x −1 2.4.4 y = 1 + x1 ex + xc 2.4.5 y = (x + 1) e3 + c 2.4.6 x = tan−1 y − 1 + ce− tan y 2.4.7 y = x(ex x2 − 1)
2.4.10 (i)-linear in y, (iii)-linear in x zzzzzzz
Tutorial 2.5
2.5
Bernoulli Equation
65
Tutorial : Bernoulli Equation
Definition An equation of the form dy + p(x)y = q(x)y n dx
(where p and q are functions of x)
(2.5.1)
is called the Bernoulli equation. Method of Solution • Divide equation (2.5.1) by y n to get y −n
dy + p(x)y 1−n = q(x). dx
dy • Put y 1−n = z, so that (1 − n)y −n dx =
dz . dx
(2.5.2)
Thus equation (2.5.2) becomes
1 dz + p(x)z = q(x) or 1 − n dx
dz + p(x)(1 − n)z = q(x)(1 − n) dx
(2.5.3)
which is a linear equation in z. • Solve equation (2.5.3) by the usual method of linear differential equation. • Replace z by y 1−n in the solution of equation (2.5.3) which will give the general solution of equation (2.5.1). Solved Examples Example 2.5.1. Find the general solution of the differential equation dy + 2y = y 2 . dx Solution. The given equation is a Bernoulli equation. Dividing the equation throughout by y 2 , we get dy y −2 + 2y −1 = 1. (2.5.4) dx dy dz Put y −1 = z, so that −y −2 dx = dx . Therefore, equation (2.5.4) becomes −
dz + 2z = 1 or dx
dz − 2z = −1. dx
which is a linear equation in z with p(x) = −2 and q(x) = −1. So, we have I.F. = e
R
p(x)dx
= e−
R
2dx
= e−2x .
(2.5.5)
66
Chapter 2
First Order Differential Equations
Thus the general solution of equation (2.5.5) is given by Z e−2x −2x ze = − e−2x dx + c ⇒ ze−2x = + c. 2 Replacing z by y −1 , we get the general solution of the given equation as y −1 e−2x =
e−2x +c 2
1 1 = + ce2x . y 2
⇒
Example 2.5.2. Find the solution of the differential equation dy x +y =− . dx y
[GTU, Dec. 2009]
Solution. The given equation is a Bernoulli equation. Multiplying the equation throughout by y, we get dy + y 2 = −x. (2.5.6) y dx dy dz Let y 2 = z. Then 2y dx = dx . Therefore, equation (2.5.6) becomes 1 dz + z = −x or 2 dx
dz + 2z = −2x. dx
which is a linear equation in z with p(x) = 2 and q(x) = −2. So, we have I.F. = e
R
p(x)dx
R
=e
2dx
= e2x .
Thus the general solution of equation (2.5.7) is given by Z 2x ze = −2xe2x dx + c 2x 2x e e 2x ⇒ ze = −2 (x) − (1) +c 2 4 1 ⇒ ze2x = − (2xe2x − e2x ) + c 2 1 ⇒ z = − (2x − 1) + ce−2x . 2 Replacing z by y 2 , we get the general solution of the given equation as 1 y 2 = − (2x − 1) + ce−2x 2
⇒
2y 2 = 1 − 2x + 2ce−2x .
Exercises Exercise 2.5.1. Find the general solution of the differential equation dy + 3y = e2x y 3 . dx Solution.
(2.5.7)
Tutorial 2.5
Bernoulli Equation
Exercise 2.5.2. Solve xy 0 = y 2 + y. Solution.
67
[GTU, Dec. 2010]
68
dy 1 ey Exercise 2.5.3. Solve dx + x = 2 . x Solution.
Chapter 2
First Order Differential Equations
[GTU, May 2011]
Tutorial 2.5
Bernoulli Equation
69
Additional Exercises Exercise 2.5.4. Find the general solution of the differential equation dy 2 + (x + 1)y = ex y 3 . dx Exercise 2.5.5. Solve the Bernoulli equation y 0 + 31 y = 13 (1 − 2x)y 4 . Exercise 2.5.6. Solve the differential equation y 0 + xy = xy −1 . Viva Questions Question 2.5.7. What is the standard form of a Bernoulli equation ? Question 2.5.8. How to reduce a Bernoulli equation into a linear equation ? Question 2.5.9. Which of the following are Bernoulli equations ? (i)
dy + 2xy = y 4 dx
(ii)
dy + y sin x = xy −1 dx
(iii)
dy + x2 y 3 = x2 dx
(iv)
dy + x2 y 3 = x2 y dx
Answers 1 2.5.1 y22 = e2x (1 + ce4x ) 2.5.2 x(1 + y) = cy 2.5.3 e−y = 2x + cx 2.5.5 y13 = cex − 2x − 1 √ 2 2.5.4 y12 = ex (1 + ce2x ) 2.5.6 y = 1 + ce−x2 2.5.9 (i), (ii), (iv)
zzzzzzz
70
2.6
Chapter 2
First Order Differential Equations
Tutorial : Orthogonal Trajectories
Definition Given a family of curves there exists another family of curves such that each member of the either family cuts each member of the other family at a right angle. These curves are called the orthogonal trajectories of the given curves.
Figure 2.1: Orthogonal Trajectories Working Rules • For the given family of curves, find the differential equation y 0 = f (x, y) by eliminating the arbitrary constant. 1 . • Write the differential equation of the orthogonal trajectories as y 0 = − f (x,y)
• Solve the differential equation to obtain the orthogonal trajectories. Solved Examples Example 2.6.1. Find the orthogonal trajectories of the curves y = cx2 . Solution. The given curves are y = cx2 Differentiating with respect to x, we get 2 1 0 y +y − 3 =0 ⇒ x2 x
y = c. x2
⇒
y0 −
2y =0 x
⇒
y0 =
2y . x
which is the differential equation of the given curves, so that the differential equation of the orthogonal trajectories is x y0 = − . (2.6.1) 2y We find the orthogonal trajectories by solving (2.6.1). Now from (2.6.1), dy x =− dx 2y
⇒
2ydy = −xdx.
x2 + c? 2
⇒
x2 + y 2 = c? . 2
Integrating both sides, we get y2 = −
Tutorial 2.6
Orthogonal Trajectories
71
Exercises Exercise 2.6.1. Find the orthogonal trajectories of the curves y = x2 + c. [GTU, Dec. 2010] Solution.
c Exercise 2.6.2. Find the orthogonal trajectories of the curves y = x . Solution.
72
Chapter 2
First Order Differential Equations
Exercise 2.6.3. Find the orthogonal trajectories of the curves (x − c)2 + y 2 = c2 . Solution.
Viva Questions Question 2.6.4. What is meant by orthogonal trajectories ? Question 2.6.5. How to find orthogonal trajectories of given curves ? Answers 2.6.1 log x + 2y = c?
2.6.2 x2 − y 2 = c?
2.6.3 x2 + (y − c? )2 = c? 2
zzzzzzz
Chapter
3
Higher Order Linear ODEs The differential equation in which dependent variable and its derivatives occur only in first degree and are not multiplied together is called a linear differential equation. The standard form of an nth -order linear ODE is dn y dn−1 y dn−2 y dy + p (x) + p (x) + · · · + p1 (x) + p0 (x)y = r(x), n−1 n−2 n n−1 n−2 dx dx dx dx
(3.0.1)
where the coefficients p0 (x), p1 (x), . . . , pn−1 (x) and r(x) are functions of x. If r(x) = 0 for all x under consideration (usually in some open interval I), then equation (3.0.1) is called homogeneous. If r(x) 6= 0 for at least one x under consideration, then equation (3.0.1) is called nonhomogeneous.
3.1
Tutorial : Homogenous Linear ODEs
The standard form of an nth -order homogeneous linear ODE is dn−1 y dn−2 y dy dn y + p (x) + p (x) + · · · + p1 (x) + p0 (x)y = 0, n−1 n−2 n n−1 n−2 dx dx dx dx
(3.1.1)
where p0 (x), p1 (x), . . . , pn−1 (x) are functions of x. Basis, General Solution, Particular Solution • A basis of solutions of equation (3.1.1) on an open interval I is a set of n linearly independent solutions of the equation on I. • If {y1 , y2 , . . . , yn } is a basis of solutions of equation (3.1.1) on an open interval I then the general solution of the equation on I is given by y = c1 y 1 + c2 y 2 + · · · + cn y n ,
(3.1.2)
where c1 , . . . , cn are arbitrary constants. • A particular solution of equation (3.1.1) on an open interval I is obtained by assigning specific values to the constants c1 , c2 , . . . , cn in the general solution (3.1.2). 73
74
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Wronskian The Wronskian of a set of n-functions {y1 , y2 , . . . , yn } on an open interval I having the property that each function possess n − 1 derivatives on I, is defined by y1 y2 ··· yn y10 y20 ··· yn0 W (y1 , y2 , . . . , yn ) = .. .. .. .. . . . . (n−1) (n−1) (n−1) y y · · · y n 1 2 Wronskian and Linear Independence If the Wronskian of a set of n functions defined on an open interval I is nonzero for at least one point in this interval, then the set of functions is linearly independent there. Finding a Basis if One Solution Is Known, Reduction of Order This method is exclusively for second order homogeneous linear ODEs. The method can be described as follows: If one solution y1 of the differential equation y 00 + p(x)y 0 + q(x)y = 0 is known, then a second linearly independent solution y2 of the equation can be obtained as Z 1 − R p(x)dx e dx. y2 = y1 y12 Solved Examples Example 3.1.1. Verify that the functions cos 3x and sin 3x form a basis of solutions of the differential equation y 00 + 9y = 0 and solve the equation when y(0) = 4, y 0 (0) = −6. Solution. Let y1 = cos 3x and y2 = sin 3x. Then y10 = −3 sin 3x,
y20 = 3 cos 3x,
y100 = −9 cos 3x,
y200 = −9 sin 3x.
Observe that y100 + 9y1 = −9 cos 3x + 9 cos 3x = 0 and y200 + y2 = −9 sin 3x + 9 sin 3x = 0, so that y1 and y2 are solutions of the given equation. Also, cos 3x sin 3x = 3 cos2 3x + 3 sin2 3x = 3 6= 0. W (y1 , y2 ) = −3 sin 3x 3 cos 3x
Tutorial 3.1
Homogenous Linear ODEs
75
Thus y1 and y2 are linearly independent and hence form a basis of solution. Therefore, the general solution of the given equation is ⇒
y(x) = c1 y1 (x) + c2 y2 (x)
y(x) = c1 cos 3x + c2 sin 3x
(3.1.3)
Differentiating (3.1.3) w.r.t. x, we get y 0 (x) = −3c1 sin 3x + 3c2 cos 3x.
(3.1.4)
Using the condition y(0) = 4, we obtain c1 cos 0 + c2 sin 0 = 4
⇒
c1 = 4.
Using the condition y 0 (0) = −6, we obtain −3c1 sin 0 + 3c2 cos 0 = −6
⇒
3c2 = −6
⇒
c2 = −2.
Hence, the required particular solution is y(x) = 4 cos 3x − 2 sin 3x. Example 3.1.2. Find a basis of solutions for the following second order homogeneous linear equation x > 0 : x2 y 00 − xy 0 + y = 0.
[GTU, Dec. 2010]
Solution. By inspection y1 = x is a solution. Dividing the equation throughout by x2 , we get y 00 −
1 0 1 y + 2 y = 0. x x
Comparing with the standard form y 00 + p(x)y 0 + q(x)y = 0, 1 we get p(x) = − x . Therefore, e−
R
p(x)dx
R
=e
1 dx x
= eln x = x.
By reduction of order, we obtain Z Z Z 1 − R p(x)dx 1 1 y2 = y1 e dx = x · xdx = x dx = x log x. 2 2 y1 x x Observe that y1 y2 x x log x = W (y1 , y2 ) = 0 y1 y20 1 1 + log x
= x + x log x − x log x = x 6= 0 for x > 0.
Thus y1 = x and y2 = x log x are linearly independent and hence form a basis for the given differential equation.
76
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Linear ODEs of Higher Order
Example 3.1.3. Find the general solution of the differential equation xy 00 + 2y 0 + xy = 0
if y1 =
sin x . x
Solution. Dividing the equation throughout by x, we get y 00 +
2 0 y + y = 0. x
Comparing with the standard form y 00 + p(x)y 0 + q(x)y = 0, 2 we get p(x) = x . Therefore, e−
R
p(x)dx
= e−
R
2 dx x
= e−2 ln x = eln x
−2
= x−2 .
By the Reduction of order, we obtain Z
1 − R p(x)dx e dx y12 Z sin x x2 −2 x dx = x sin2 x Z sin x = cosec2 xdx x
y2 = y1
sin x (− cot x) x sin x cos x = − x sin x
=
= −
cos x . x
Observe that y1 = y2
sin x x
x − = − tan x cos x
⇒
y1 6= k (const.) y2
⇒
y1 6= ky2 .
Thus y1 and y2 are linearly independent and hence form a basis for the given differential equation. Therefore, the general solution of the equation is given by cos x sin x y = c1 y1 + c2 y2 ⇒ y = c1 + c2 . x x Exercises Exercise 3.1.1. Verify that the functions e−x and xe−x form a basis of solutions of the differential equation y 00 + 2y 0 + y = 0 and solve the equation when y(0) = 1, y 0 (0) = 0. Solution.
Tutorial 3.1
Homogenous Linear ODEs
77
1
3
Exercise 3.1.2. Verify that the functions x− 2 and x 2 form a basis of solutions of the equation 4x2 y 00 − 3y = 0 and solve it when y(1) = 3, y 0 (1) = 2.5. [GTU, Dec. 2011] Solution.
78
Chapter 3
Linear ODEs of Higher Order
Exercise 3.1.3. If one solution of x2 y 00 − 4xy 0 + 6y = 0 is y1 = x2 , x > 0, then determine its second solution. [GTU, Jan. 2013] Solution.
Exercise 3.1.4. Find the general solution of the differential equation 1 1 2 00 0 2 y=0 if y1 = x− 2 cos x. x y + xy + x − 4
Tutorial 3.1
Homogenous Linear ODEs
79
Solution.
Additional Exercises Exercise 3.1.5. Verify that the functions 1, cos x and sin x form a basis of solutions of the differential equation y 000 + y 0 = 0 and solve the when y(0) = 15, y 0 (0) = 0, y 00 (0) = −3. Exercise 3.1.6. Solve (1 − x2 )y 00 − 2xy 0 + 2y = 0 if y1 = x. Viva Questions Question 3.1.7. What is the standard form of an nth order linear differential equation ? When it is called homogeneous? Question 3.1.8. What is Wronskian ? How to check whether the given functions are linearly independent ? Question 3.1.9. Define a basis of solutions for the differential equation. How to find it if one solution is known ? Answers 3.1.1 y = (1 + x)e−x 3.1.5 12 + 3 cos x
5
3.1.2 y = x− 2 + 2x 2 3.1.3 x3 3.1.4 y = c1 x−1/2 cos x + c2 x−1/2 sin x q 3.1.6 y = c1 x + c2 1 + x ln x−1 x+1 1
3
zzzzzzz
80
Chapter 3
3.2
Higher Order Linear ODEs
Tutorial : Homogeneous Linear ODEs with Constant Coefficients
The standard form of an nth order homogeneous linear ODE with constant coefficients is dn−1 y dn−2 y dy dn y + a0 y = 0, + a + a + · · · + a1 n−1 n−2 n n−1 n−2 dx dx dx dx
(3.2.1)
where a0 , a1 , . . . , an−1 are constants. Method of Solution • Write the given homogeneous linear ODE dn y dn−1 y dn−2 y dy + a + a + · · · + a1 + a0 y = 0, n−1 n−2 n n−1 n−2 dx dx dx dx
(3.2.2)
in symbolic form as (Dn + an−1 Dn−1 + an−2 Dn−2 + · · · + a0 )y = 0,
where D =
d dx
• Write the auxiliary equation for (3.2.2) as mn + an−1 mn−1 + an−2 mn−2 + · · · + a0 = 0
(3.2.3)
• Find the roots m1 , m2 , . . . , mn of equation (3.2.3). The general solution of the differential equation (3.2.2) depends on the nature of these roots. We have following four possibilities for the roots: (1) Distinct real roots: If all the roots are real and distinct, then the general solution of the differential equation (3.2.2) is given by y(x) = c1 em1 x + c2 em2 x + · · · + cn emn x . (2) Equal real roots: If two roots are equal, say m1 = m2 , then the general solution of the differential equation (3.2.2) is given by y(x) = (c1 + c2 x)em1 x + c3 em3 x + · · · + cn emn x . Similarly, if three roots are equal, say m1 = m2 = m3 , then the general solution of the differential equation (3.2.2) is given by y(x) = (c1 + c2 x + c3 x2 )em1 x + c4 em4 x + · · · + cn emn x . (3) One pair of roots is complex: If one pair of roots is complex, say m1 = α + iβ and m2 = α − iβ, then the general solution of the differential equation (3.2.2) is given by y(x) = eαx (c1 cos βx + c2 sin βx) + c3 em3 x + · · · + cn emn x .
Tutorial 3.2
Homogeneous Linear ODEs with Constant Coefficients
81
(4) Two pairs of complex roots are equal: If two pairs of roots are complex and equal, say m1 = m2 = α + iβ and m3 = m4 = α − iβ, then the general solution of the differential equation (3.2.2) is given by y(x) = eαx [(c1 + c2 x) cos βx + (c3 + c4 x) sin βx] + c5 em5 x + · · · + cn emn x . Solved Examples Example 3.2.1. Solve the initial value problem y 00 + y 0 − 2y = 0; y(0) = 4 and y 0 (0) = −5. [GTU, Dec. 2009] Solution. The symbolic form of the given equation is (D2 + D − 2)y = 0,
where D =
d . dx
Therefore, the auxiliary equation is m2 + m − 2 = 0
⇒
(m + 2)(m − 1) = 0
⇒
m = −2, 1 (distinct real roots).
Thus the general solution is y(x) = c1 e−2x + c2 ex .
(3.2.4)
Differentiating equation (3.2.4), we get y 0 (x) = −2c1 e−2x + c2 ex .
(3.2.5)
Since y(0) = 4, from (3.2.4) we obtain c1 + c2 = 4.
(3.2.6)
−2c1 + c2 = −5.
(3.2.7)
Since y 0 (0) = −5, from (3.2.5) we obtain
Solving equations (3.2.6) and (3.2.7), we get c1 = 3 and c2 = 1. Hence, the required particular solution is y(x) = 3e−2x + ex . Example 3.2.2. Solve the initial value problem dy d2 y − 6 + 9y = 0, 2 dx dx
y(0) = 1, y 0 (0) = 0.
Solution. The symbolic form of the given equation is (D2 − 6D + 9)y = 0,
where D =
d . dx
Therefore, the auxiliary equation is m2 − 6m + 9 = 0
⇒
(m − 3)2 = 0
⇒
m = 3, 3 (equal real roots).
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Chapter 3
Higher Order Linear ODEs
Thus the general solution of the given equation is y(x) = (c1 + c2 x)e3x .
(3.2.8)
Differentiating equation (3.2.8), we get y 0 (x) = (c1 + c2 x)3e3x + c2 e3x .
(3.2.9)
Since y(0) = 1, from (3.2.8) we obtain c1 = 1. Since y 0 (0) = 0, from (3.2.9) we obtain ⇒
3c1 + c2 = 0
c2 = −3c1 = −3 (∵ c1 = 1).
Hence, the required particular solution is y(x) = (1 − 3x)e3x . Example 3.2.3. Find the general solution of 16y 00 − 8y 0 + 5y = 0. Solution. The symbolic form of the given equation is (D2 − 8D + 5)y = 0,
where D =
d . dx
Therefore, the auxiliary equation is 16m2 − 8m + 5 = 0 ⇒ m= ⇒ m=
8± 8±
√
64 − 320 32
√
−256 32
⇒ m=
8 ± 16i 32
⇒ m=
1 1 ±i 4 2
(pair of complex roots).
Hence, the general solution is given by y(x) = e
x 4
x x c1 cos + c2 sin . 2 2
Example 3.2.4. Find the general solution of the differential equation d4 y d2 y + 4 + 4y = 0. dx4 dx2
[GTU, Dec. 2011]
Tutorial 3.2
Homogeneous Linear ODEs with Constant Coefficients
83
Solution. The symbolic form of the given equation is (D4 + 4D2 + 4)y = 0,
where D =
d . dx
Therefore, the auxiliary equation is √ √ m4 + 4m2 + 4 = 0 ⇒ (m2 + 2)2 = 0 ⇒ m = ± 2i, ± 2i
(equal pairs of complex roots).
Hence, the general solution is √ √ y(x) = (c1 + c2 x) cos 2x + (c3 + c4 x) sin 2x. Exercises Exercise 3.2.1. Find the general solution of the differential equation y 00 + 4y 0 − 12y = 0. Solution.
Exercise 3.2.2. Solve the initial value problem y 00 − 4y 0 + 4y = 0; y(0) = 3, y 0 (0) = 1. [GTU, Jan. 2015] Solution.
84
Exercise 3.2.3. Find the general solution of
Chapter 3
Higher Order Linear ODEs
d4 y d2 y −18 2 + 81y = 0. 4 dx dx [GTU, May 2012]
Solution.
Exercise 3.2.4. Solve (D3 − 3D2 + 3D − 1)y = 0. Solution.
Exercise 3.2.5. Solve y 00 + 2y 0 + 2y = 0, y(0) = 1, y(π/2) = 0. Solution.
[GTU, Dec. 2010]
Tutorial 3.2
Homogeneous Linear ODEs with Constant Coefficients
85
Exercise 3.2.6. Solve y 000 − y 00 + 100y 0 − 100y = 0, y(0) = 4, y 0 (0) = 11, y 00 (0) = −299. [GTU- Dec. 2011, Jan. 2013] Solution.
Additional Exercises Exercise 3.2.7. Find the general solution of (D2 − 2D + 4)y = 0. Exercise 3.2.8. Find the solution of differential equation y 00 −5y 0 +6y = 0 with initial condition y(1) = e2 and y 0 (1) = 3e2 . [GTU, May 2012] Viva Questions Question 3.2.9. What is meant by D ? Question 3.2.10. Define auxiliary equation. Question 3.2.11. Find the general solution of the following differential equation: (i) y 00 + 5y 0 + 4y = 0;
(ii) y 00 − y = 0;
(iii) (D2 + 1)y = 0.
Answers 3.2.1 c1 e−6x + c2 e2x 3.2.2 (3 − 5x)e2x 3.2.3 (c1 + c2 x)e−3x + (c3 + c4 x)e3x 3.2.4 (c1 + c2 x + c3 x2 )ex 3.2.5 e−x cos x 3.2.6 ex + 3 cos 10x + sin 10x √ √ 3.2.7 ex (c1 cos 3x + c2 sin 3x) 3.2.8 e3x−1 3.2.11 (i) c1 e−x + c2 e−4x , (ii) c1 ex + c2 e−x , (iii) c1 cos x + c2 sin x zzzzzzz
86
Chapter 3
3.3
Higher Order Linear ODEs
Tutorial : Nonhomogeneous Linear ODEs with Constant Coefficients
The standard form of an nth order nonhomogeneous linear ODE with constant coefficients is dn y dn−1 y dn−2 y dy + a0 y = r(x), + a + a + · · · + a1 n−1 n−2 n n−1 n−2 dx dx dx dx
(3.3.1)
where a0 , a1 , . . . , an−1 are constants and r(x) 6= 0 for at least one x under consideration. Method of Solution Consider a nonhomogeneous linear ODE with constant coefficients of the form y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = r(x).
(3.3.2)
• First find the general solution of the corresponding homogeneous equation y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = 0
(3.3.3)
by the usual method described in the section 3.2. This solution is called the complementary function (C.F.) of (3.3.2). It is denoted by YC . • The symbolic form of (3.3.2) is (Dn + an−1 Dn−1 + . . . + a1 D + a0 )y = r(x)
⇒
f (D)y = r(x).
1 Applying f (D) (inverse of f (D)) on both sides, we obtain 1 1 f (D)y = r(x) f (D) f (D)
⇒
y=
1 r(x). f (D)
This solution is called the particular integral (P.I.) of (3.3.2). It is denoted by YP . • The general solution of (3.3.2) is given by y = C.F. + P.I. = YC + YP . Direct Method For Finding Particular Integral Consider the nonhomogeneous equation of the form (Dn + an−1 Dn−1 + . . . + a1 D + a0 )y = r(x) or f (D)y = r(x).
(3.3.4)
Then the particular integral (P.I.) is given by P.I. = YP =
1 r(x). f (D)
The expression of YP depends on the nature of r(x). The following are some special cases for r(x):
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
87
Case-1. r(x) = eax In this case, YP =
1 ax 1 ax e = e , f (D) f (a)
provided f (a) 6= 0.
If f (a) = 0, then YP =
1 ax x e = 0 eax , f (D) f (a)
provided f 0 (a) 6= 0.
YP =
x2 1 ax e = 00 eax , f (D) f (a)
provided f 00 (a) 6= 0
If f 0 (a) = 0, then
and so on. If f (D) = (D − a)r eax , then YP = Case-2. r(x) = cos(ax + b) or
1 xr ax ax e = e . (D − a)r r!
sin(ax + b)
In this case, YP =
1 1 cos(ax + b) = cos(ax + b), f (D2 ) f (−a2 )
provided f (−a2 ) 6= 0.
If f (−a2 ) = 0, then YP =
1 x cos(ax + b) = 0 cos(ax + b), 2 f (D ) f (−a2 )
provided f 0 (−a2 ) 6= 0.
If f 00 (−a2 ) = 0, then YP =
x2 1 cos(ax + b) = cos(ax + b), f (D2 ) f 00 (−a2 )
provided f 00 (−a2 ) 6= 0
and so on. If f (D2 ) = (D2 + a2 )2 , then YP =
1 1 x2 cos ax = − · cos ax. (D2 + a2 )2 4a2 2!
The method for r(x) = sin(ax + b) is similar. Case-3. r(x) = xn In this case, YP =
1 xn . f (D)
Take the constant, if not, then the lowest powered D (with sign) common from f (D) and then 1 expand f (D) by either of the following binomial expansions: 1 = 1 + D + D2 + . . . 1−D
or
1 = 1 − D + D2 − . . . . 1+D
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Chapter 3
Higher Order Linear ODEs
Operate the resulting expansion on xn . We need to expand up to power Dn as higher derivatives vanish. Case-4. r(x) = eax φ(x), where φ(x) is any function of x In this case, YP = Case-5. r(x) = x cos ax
1 1 eax φ(x) = eax φ(x). f (D) f (D + a)
or x sin ax
In this case, 1 f 0 (D) 1 YP = x cos ax = x − cos ax. f (D) f (D) f (D) The method for r(x) = x sin ax is similar. Solved Examples Example 3.3.1. Solve the differential equation y 00 + 7y 0 + 10y = e−x . Solution. The symbolic form of given equation is (D2 + 7D + 10)y = e−x . First we find YC by solving the corresponding homogeneous equation (D2 + 7D + 10)y = 0. The auxiliary equation is m2 + 7m + 10 = 0
⇒
(m + 2)(m + 5) = 0
⇒
m = −2, −5.
Thus YC = c1 e−2x + c2 e−5x . Now YP = = = =
1 r(x) f (D) D2
1 e−x + 7D + 10
(−1)2
1 e−x + 7(−1) + 10
1 −x e . 4
Hence, the general solution is given by y = YC + YP
⇒
y = c1 e−2x + c2 e−5x +
e−x . 4
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
Example 3.3.2. Find the general solution of the differential equation d2 y dy + 2y = e2x . − 3 dx2 dx Solution. The symbolic form of given equation is (D2 − 3D + 2)y = e2x . Therefore, the auxiliary equation is m2 − 3m + 2 = 0
⇒
(m − 1)(m − 2) = 0
⇒
m = 1, 2.
Thus YC = c1 ex + c2 e2x . Now YP =
1 r(x) f (D)
=
1 e2x D2 − 3D + 2
=
x e2x 2D − 3
=
x e2x 2(2) − 3
(∵ f (2) = 0)
= xe2x . Hence, the general solution is given by y = YC + YP
⇒
y = c1 ex + c2 e2x + xe2x .
Example 3.3.3. Find the general solution of the differential equation d3 y d2 y dy + 2− − y = sin 2x. 3 dx dx dx Solution. The symbolic form of given equation is (D3 + D2 − D − 1)y = sin 2x. Therefore, the auxiliary equation is
⇒ ⇒ ⇒ ⇒
m3 + m2 − m − 1 = 0 m2 (m + 1) − (m + 1) = 0 (m + 1)(m2 − 1) = 0 (m + 1)(m + 1)(m − 1) = 0 m = −1, −1, 1.
89
90
Chapter 3
Higher Order Linear ODEs
Thus YC = (c1 + c2 x)e−x + c3 ex . Now, YP = = = = = = = = =
1 r(x) f (D2 ) 1 sin 2x DD2 + D2 − D − 1 D(−22 )
1 sin 2x + (−22 ) − D − 1
1 sin 2x −4D − 4 − D − 1 1 sin 2x −5D − 5 1 1 sin 2x − 5 D+1 1 D−1 − sin 2x 5 (D − 1)(D + 1) 1 D−1 − sin 2x 5 D2 − 1 1 D−1 − sin 2x 5 −22 − 1
=
1 (D − 1) sin 2x 25
=
1 (D sin 2x − sin 2x) 25
=
1 (2 cos 2x − sin 2x). 25
Hence, the general solution is given by y = YC + YP
⇒
y = (c1 + c2 x)e−x + c3 ex +
1 (2 cos 2x − sin 2x). 25
Example 3.3.4. Solve (D4 + 2a2 D2 + a4 )y = cos ax.
[GTU, May 2011]
Solution. The auxiliary equation equation is m4 + 2a2 m2 + a4 = 0
⇒
(m2 + a2 )2 = 0
⇒
m = ±ia, ±ia.
Thus YC = (c1 + c2 x) cos ax + (c3 + c4 x) sin ax.
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
91
Now, YP =
x2 1 1 1 x2 cos ax = − cos ax = cos ax = − · cos ax. D4 + 2a2 D2 + a4 (D2 + a2 )2 4a2 2! 8a2
Hence, the general solution is given by y = YC + YP = (c1 + c2 x) cos ax + (c3 + c4 x) sin ax −
x2 cos ax. 8a2
Example 3.3.5. Solve y 00 + 2y 0 + 3y = 2x2 .
[GTU, Jan. 2015]
Solution. The symbolic form of the given equation is (D2 + 2D + 3)y = 2x2 . Therefore, the auxiliary equation is 2
m + 2m + 3 = 0
⇒
Thus
m=
−2 ±
√
4 − 12
2
√ −2 ± 2 2i m= 2
⇒
⇒
m = −1 ±
√
2i.
√ √ YC = e−x (c1 cos 2x + c2 sin 2x).
Now, YP = =
= = = =
(D2
1 2x2 + 2D + 3)
1 2x2 D2 +2D 3 1+ 3 " # 2 2 2 2 D + 2D D + 2D 1− + − · · · x2 3 3 3 2 2 1 2 1 4 2 3 2 2 x − (D + 2D)x + (D + 4D + 4D )x 3 3 9 2 2 1 1 x − (2 + 4x) + (0 + 0 + 8) 3 3 9 2 2 4 2 x − x+ 3 3 9
Hence, the general solution is given by −x
y = YC + YP = e (c1 cos
√
2x + c2 sin
√
2 2 4 2 2x) + x − x+ . 3 3 9
Example 3.3.6. Solve the initial value problem y 00 + 4y = 8e−2x + 4x2 + 2,
y(0) = 2, y 0 (0) = 2.
[GTU, Dec. 2010]
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Higher Order Linear ODEs
Solution. The symbolic form of the given equation is (D2 + 4)y = 8e−2x + 4x2 + 2. Therefore, the auxiliary equation is m2 + 4 = 0
⇒
m2 = −4
⇒
m = ±2i.
Thus YC = c1 cos 2x + c2 sin 2x. Now, YP = =
= = = =
1 (8e−2x + 4x2 + 2) +4 1 1 1 −2x 2 8 e x +2 1 +4 D2 + 4 D2 + 4 D2 + 4 # " 1 1 1 −2x 2 0x e e 8 + +2 2 x (−2)2 + 4 D2 + 4 1 + D4 1 −2x D2 D4 1 2 0x e 8 e + 1− + − ··· x + 2 2 8 4 16 0 +4 1 1 2 2 1 4 2 −2x 2 e + x − D (x ) + D (x ) + 4 16 2 1 1 −2x 2 e + x − + 2 2 D2
= e−2x + x2 . Hence, the general solution is given by y = YC + YP = c1 cos 2x + c2 sin 2x + e−2x + x2 .
(3.3.5)
Using the condition y(0) = 2, we get c1 cos 0 + c2 sin 0 + e0 + 0 = 2
⇒
c1 + 1 = 2
⇒
c1 = 1.
Differentiating (3.3.5) w. r. t. x, we get y 0 (x) = −2c1 sin 2x + 2c2 cos 2x − 2e−2x + 2x. Using the condition y 0 (0) = 2, we get −2c1 sin 0 + 2c2 cos 0 − 2e0 + 2(0) = 2
⇒
2c2 − 2 = 2
Thus the required particular solution is cos 2x + 2 sin 2x + e−2x + x2 .
⇒
c2 = 2.
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
93
Example 3.3.7. Solve (D3 − 3D + 2)y = xex . Solution. The auxiliary equation is m3 −3m+2 = 0 ⇒ (m−1)(m2 +m−2) = 0 ⇒ (m−1)(m−1)(m+2) = 0 ⇒ m = 1, 1, −2. Thus YC = (c1 + c2 x)ex + c3 e−2x . Now YP =
1 r(x) f (D)
=
1 xex D3 − 3D + 2
=
1 xex (D − 1)(D − 1)(D + 2)
= ex = ex = = = = =
1 x (D)(D)(D + 1 + 2) 1 D2 (D
+ 3)
x
1 1 1 e 2 x D 3 1 + D/3 ex 1 D 1 − + ··· x 3 D2 3 ex 1 1 x − (1) 3 D2 3 ex 1 x2 x − 3 D 2 3 ex x3 x2 − 3 6 6 x
x2 ex = (x − 1). 18 Hence, the general solution is y = YC + YP
⇒
y = (c1 + c2 x)ex + c3 e−2x +
Example 3.3.8. Solve (D2 + 1)y = x sin 2x. Solution. The auxiliary equation is m2 + 1 = 0
⇒
m = ±i.
x2 ex (x − 1). 18
94
Chapter 3
Higher Order Linear ODEs
Thus YC = c1 cos x + c2 sin x. Now, YP = = = = = = = = =
1 x sin 2x f (D) 1 f 0 (D) sin 2x x− f (D) f (D) 2D 1 x− 2 sin 2x 2 D +1 D +1 1 2D sin 2x x− 2 2 D + 1 −2 + 1 2D 1 x− 2 − sin 2x D +1 3 1 1 − x sin 2x − 2D 2 sin 2x 3 D +1 1 1 sin 2x − x sin 2x − 2D 2 3 −2 + 1 1 2 d − x sin 2x + sin 2x 3 3 dx 1 4 − x sin 2x + cos 2x 3 3
1 = − [3x sin 2x + 4 cos 2x] . 9 Hence, the general solution is y = YC + YP
⇒
y = c1 cos x + c2 sin x −
1 (3x sin 2x + 4 cos 2x) . 9
Exercises Exercise 3.3.1. Solve Solution.
d2 y dy + dx −12y = e6x . 2 dx
[GTU, Jan. 2013]
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
95
Exercise 3.3.2. Solve the non-homogeneous equation y 00 − 3y 0 + 2y = ex . [GTU, Dec. 2011] Solution.
Exercise 3.3.3. Find the particular solution of y =
Solution.
d 1 cosh x, where D = dx . 2 (D+1) [GTU, Dec. 2012]
96
Chapter 3
Exercise 3.3.4. Solve y 000 − 3y 00 + 3y 0 − y = 4et .
Higher Order Linear ODEs
[GTU- Dec. 2011, Jan. 2015]
Solution.
Exercise 3.3.5. Find the general solution of
Solution.
d4 y d2 y −2 +y = cos t + e2t + et . dt4 dt2 [GTU, May 2012]
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
Exercise 3.3.6. Solve (D2 + 1)y = sin x sin 2x. Solution.
97
98
Chapter 3
Higher Order Linear ODEs
Exercise 3.3.7. Find the general solution of the differential equation d2 y − 9y = x2 . dx2 Solution.
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
Exercise 3.3.8. Solve (D3 − D2 − 6D)y = x2 + 1. Solution.
99
[GTU, June 2013]
100
Exercise 3.3.9. Solve Solution.
Chapter 3 d3 y d2 y dy − 2 + 3 dx + 5y = ex cos 3x. 3 dx dx
Higher Order Linear ODEs
[GTU, May 2012]
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
Exercise 3.3.10. Solve Solution.
d2 y dy e2x − 4 dx + 4y = 5 . 2 dx x
101
[GTU, May 2012]
102 Exercise 3.3.11. Solve (D2 + 4)y = x sin x. Solution.
Chapter 3
Higher Order Linear ODEs
Tutorial 3.3
Nonhomogeneous Linear ODEs with Constant Coefficients
103
Additional Exercises Exercise 3.3.12. Solve (D2 − 4D + 3)y = sin 3x cos 2x.
[GTU, Dec. 2013]
Exercise 3.3.13. Find the particular solution of y 00 − 2y 0 + 5y = 5x3 − 6x2 + 6x. [GTU, Dec. 2010] Exercise 3.3.14. Find the general solution of (D2 − 4)y = x3 e2x . Exercise 3.3.15. Solve the differential equation
Exercise 3.3.16. Solve (D2 − 4D + 4)y =
d2 y dy x − 3 dx + 2y = 2ex cos 2 . 2 dx [GTU, May 2012]
e2x d , where D = dx . 1+x2
[GTU, May 2012]
Viva Questions Question 3.3.17. When a linear differential equation is nonhomogeneous ? Question 3.3.18. How to give the general solution of a non-homogeneous equation ? Question 3.3.19. What will be P.I. for (D2 + 2D + 1)y = e−x sin x ? Question 3.3.20. How to find P.I. when r(x) = eax ? Question 3.3.21. How to find P.I. when r(x) = sin ax ? Answers 6x
3.3.1 c1 e−4x + c2 e3x + e30 3.3.2 c1 e2x + c2 ex − xex 3.3.3 81 ex + 14 x2 e−x 3.3.4 (c1 + c2 t + c3 t2 )et + 23 t3 et 2t
3.3.5 (c1 + c2 t)e−t + (c3 + c4 t)et + cos4 t + e9 + x + cos83x 3.3.6 c1 cos x + c2 sin x + 21 x sin 2
t2 et 8
2
2 3.3.7 c1 e3x + c2 e−3x − x9 − 81 1 3 1 2 25 3.3.8 c1 + c2 e3x + c3 e−2x − 18 x + 36 x − 108 x ex −x x 3.3.9 c1 e + e (c2 cos 2x + c3 sin 2x) − 65 (3 sin 3x + 2 cos 3x)
3.3.10 3.3.11 3.3.12 3.3.13 3.3.14 3.3.15 3.3.16 3.3.19
2x
1 e (c1 + c2 x)e2x + 12 x3 y = c1 cos 2x + c2 sin 2x + 19 (3x sin x − 2 cos x) 1 1 c1 ex + c2 e3x + 884 (10 cos 5x − 11 sin 5x) + 20 (2 cos x + sin x) x3 e2x c1 e2x + c2 e−2x + 128 (8x4 − 8x3 + 6x2 − 3x) c1 ex + c2 e2x − 85 ex cos x2 + 2 sin x2 (c1 + c2 x)e2x + e2x x tan−1 x − 21 log(x2 + 1) −e−x sin x
zzzzzzz
104
3.4
Chapter 3
Higher Order Linear ODEs
Tutorial : Method of Undetermined Coefficients
This method can be applied when the function r(x) is an exponential function, a power of x, a cosine or sine, or sums or products of such functions. These functions have derivatives similar to r(x) itself. In this method we will assume a form of YP similar to r(x), but with unknown coefficients to be determined by substituting that YP and its derivatives into the given differential equation. The choice of YP depending on r(x) and corresponding rules are as follows: Term in r(x)
Set of Solutions
Choice for YP
Keax
{eax }
Ceax
Kxn
{xn , xn−1 , . . . , x, 1}
Kn xn + Kn−1 xn−1 + · · · + K1 x + K0
K cos ax
{cos ax, sin ax}
K1 cos ax + K2 sin ax
K sin ax
{cos ax, sin ax}
K1 cos ax + K2 sin ax
Keax cos ax
{eax cos ax, eax sin ax}
eax (K1 cos ax + K2 sin ax)
Keax sin ax
{eax cos ax, eax sin ax}
eax (K1 cos ax + K2 sin ax)
Basic Rule If r(x) is one of the functions in the first column of above table, choose YP in the same line and determine its unknown coefficients by substituting YP and its derivatives into the given equation. Modification Rule If any member from the solution set of r(x) occurs in YC corresponding to a simple root, then multiply each member of the set by x (or by x2 if the member is corresponding to a double root and so on.) Sum Rule If r(x) is a sum of functions in the first column of above table, choose for YP the sum of the functions in the corresponding lines of the third column. Solved Examples Example 3.4.1. Using the method of undetermined coefficients, solve the differential equation y 00 + 4y = 8x2 .
[GTU, Dec. 2009]
Tutorial 3.4
Method of Undetermined Coefficients
105
Solution. The symbolic form of the given equation is (D2 + 4)y = 8x2 . First we find YC by solving (D2 + 4)y = 0. The auxiliary equation is m2 + 4 = 0
⇒
m2 = −4
⇒
m = ±2i.
Thus YC = c1 cos 2x + c2 sin 2x. Now, the solution set of 8x2 is {x2 , x, 1}. Therefore, Y P = K 2 x2 + K 1 x + K 0 YP0 = 2K2 x + K1 YP00 = 2K2 . Substituting all these values in the given equation, we get 2K2 + 4K2 x2 + 4K1 x + 4K0 = 8x2 ⇒ 4K2 x2 + 4K1 x + (4K0 + 2K2 ) = 8x2 . Equating the corresponding coefficients on both the sides, we get 4K2 = 8,
4K1 = 0,
4K0 + 2K2 = 0.
Solving, we get K2 = 2. K1 = 0,
K0 = −1.
Thus YP = 2x2 − 1. Hence, the general solution is given by y = YC + YP
y = c1 cos 2x + c2 sin 2x + 2x2 − 1.
⇒
Example 3.4.2. By the method of undetermined coefficients, find the general solution of y 00 + y = 6 cos 2x. Solution. The symbolic form of the given equation is (D2 + 1)y = 6 cos 2x. Therefore, the auxiliary equation is m2 + 1 = 0
⇒
m2 = −1
⇒
Thus YC = c1 cos x + c2 sin x.
m = ±i.
106
Chapter 3
Higher Order Linear ODEs
Now, the solution set of 6 cos 2x is {cos 2x, sin 2x}. Therefore, YP = K1 cos 2x + K2 sin 2x YP0 = −2K1 sin 2x + 2K2 cos 2x YP00 = −4K1 cos 2x − 4K2 sin 2x. Substituting all these values in the given equation, we get −4K1 cos 2x − 4K2 sin 2x + K1 cos 2x + K2 sin 2x = 6 cos 2x ⇒ −3K1 cos 2x − 4K2 sin 2x = 6 cos 2x. Equating the corresponding coefficients on both the sides, we get −3K1 = 6 and
− 4K2 = 0
⇒
K1 = −2 and K2 = 0.
Thus YP = −2 cos 2x. Hence, the general solution is given by y = YC + YP
⇒
y = c1 cos 2x + c2 sin 2x − 2 cos 2x.
Example 3.4.3. Using the method of undetermined coefficients solve the differential equation d2 y − 4y = e−2x − 2x. 2 dx Solution. The symbolic form of the given equation is (D2 − 4)y = e−2x − 2x. Therefore, the auxiliary equation is m2 − 4 = 0
⇒
(m − 2)(m + 2) = 0
⇒
m = 2, −2.
Thus YC = c1 e2x + c2 e−2x . Now, the solution sets of e−2x and −2x are {e−2x } and {x, 1} respectively. Since e−2x occurs in YC corresponding to a simple root, we have to modify the first solution set as {xe−2x }. Thus YP = Axe−2x + Bx + C YP0 = −2Axe−2x + Ae−2x + B YP00 = 4Axe−2x − 2Ae−2x − 2Ae−2x . Substituting all these values in the given equation, we get 4Axe−2x − 4Ae−2x − 4Axe−2x − 4Bx − 4C = e−2x − 2x ⇒ −4Ae−2x − 4Bx − 4C = e−2x − 2x. Equating the corresponding coefficients on both the sides, we get −4A = 1,
−4B = −2,
−4C = 0.
Tutorial 3.4
Method of Undetermined Coefficients
107
Solving, we get 1 A=− , 4
1 B= , 2
C = 0.
Therefore, 1 1 YP = − xe−2x + x. 4 2 Hence, the general solution is given by y = YC + YP
⇒
1 1 y = c1 e2x + c2 e−2x − xe−2x + x. 4 2
Exercises Exercise 3.4.1. Find the solution of differential equation y 00 + 4y = 2 sin 3x by the method of undetermined coefficients. [GTU, March 2010] Solution.
108
Chapter 3
Higher Order Linear ODEs
Exercise 3.4.2. Using the method of undetermined coefficients, find the general solution of the differential equation y 00 + 2y 0 + 10y = 25x2 + 3. [GTU, Dec. 2010] Solution.
Tutorial 3.4
Method of Undetermined Coefficients
109
Exercise 3.4.3. Using the method of undetermined coefficients, find the general solution of y 00 + 8y 0 + 16y = 64 cosh 4x. Hint: Use the formula cosh ax = Solution.
eax +e−ax 2
110
Chapter 3
Higher Order Linear ODEs
Exercise 3.4.4. Use the method of undetermined coefficients to solve the initial value problem y 00 + 4y = 16 cos 2x, Solution.
y(0) = 0, y 0 (0) = 0.
Tutorial 3.4
Method of Undetermined Coefficients
111
Additional Exercises Exercise 3.4.5. Solve the initial value problem y 00 + y 0 = 2 + 2x + x2 ,
y(0) = 8, y 0 (0) = −1
using the method of undetermined coefficients. Viva Questions Question 3.4.6. For which r(x), the method of undetermined coefficients is applicable ? Question 3.4.7. For the differential equation y 00 + y = x2 + 4, give the form of YP . Question 3.4.8. How to modify the form of YP if one of the terms in YP is a solution of the corresponding homogeneous equation (i.e., the term occurs in YC ) ? Question 3.4.9. For the differential equation y 00 + 4y = e4x + sin 2x, give the form of YP . Answers 3.4.1 3.4.3 3.4.4 3.4.5 3.4.7 3.4.9
c1 cos 2x + c2 sin 2x − 25 sin 3x (c1 + c2 x)e−4x + 12 e4x + 16x2 e−4x 4x sin 2x 3e−x + 5 + 2x + 13 x3 Ax2 + Bx + C Ae2x + Bx cos 2x + Cx sin 2x zzzzzzz
112
3.5
Chapter 3
Higher Order Linear ODEs
Tutorial : Method of Variation of Parameters
The method of undetermined coefficients is restricted to functions r(x) whose derivatives are of a form similar to r(x) itself. Of course direct method is applicable almost in all cases but some times turns out to be complicated. So, it is required to establish a more general method. One such method, called the method of variation of parameters. We now discuss this method. Method for Second Order Linear ODEs Consider a nonhomogeneous linear ODEs of the form y 00 + p(x)y 0 + q(x)y = r(x) (where p(x) and q(x) are continuous). Suppose that YC (x) = c1 y1 (x) + c2 y2 (x). Then
Z YP (x) = −y1 (x)
y2 (x) r(x)dx + y2 (x) W (x)
Z
y1 (x) r(x)dx, W (x)
where W is the Wronskian of y1 , y2 , i.e., y1 y2 . W (x) = 0 y1 y20 Method for Third Order Linear ODEs Consider a nonhomogeneous linear ODE of the form y 000 + p2 (x)y 00 + p1 (x)y 0 + p0 (x)y = r(x) (where p0 (x), p1 (x) and p2 (x) are continuous). Suppose that YC (x) = c1 y1 (x) + c2 y2 (x) + c3 y3 (x). Then Z YP (x) = y1 (x)
W1 (x) r(x)dx + y2 (x) W (x)
Z
W2 (x) r(x)dx + y3 (x) W (x)
Z
W3 (x) r(x)dx, W (x)
where y1 y2 y3 W = y10 y20 y30 , y100 y200 y300
W1 = y2 y30 − y3 y20 ,
W2 = y3 y10 − y1 y30 ,
W3 = y1 y20 − y2 y10 .
Solved Examples Example 3.5.1. Using the method of variation of parameters solve the differential equation y 00 + y = cosec x.
Tutorial 3.5
Method of Variation of Parameters
113
Solution. The symbolic form of the given equation is (D2 + 1)y = cosec x. First we find YC by solving the corresponding homogeneous equation (D2 + 1)y = 0. The auxiliary equation is m2 + 1 = 0
⇒
m2 = −1
⇒
m = ±i.
Thus YC = c1 cos x + c2 sin x. Comparing this with YC = c1 y1 + c2 y2 , we obtain y1 = cos x, Now
y2 = sin x.
y1 y2 cos x sin x = W (x) = 0 y1 y20 − sin x cos x
= cos2 x + sin2 x = 1.
By the method of variation of parameters, we have Z Z y2 (x) y1 (x) YP = −y1 (x) r(x)dx + y2 (x) r(x)dx W (x) W (x) Z Z sin x cos x = − cos x cosec xdx + sin x cosec xdx 1 1 Z Z = − cos x dx + sin x cot xdx = −x cos x + sin x · ln | sin x|. Hence, the general solution is y = YC + YP
⇒
y = c1 cos x + c2 sin x − x cos x + sin x · ln | sin x|.
Example 3.5.2. Find the general solution of the differential equation y 000 − 6y 00 + 11y 0 − 6y = e−x using the method of variation of parameters. Solution. The symbolic form of the given equation is (D3 − 6D2 + 11D − 6)y = e−x . First we find YC by solving the corresponding homogeneous equation (D3 − 6D2 + 11D − 6)y = 0.
[GTU, May 2012]
114
Chapter 3
Higher Order Linear ODEs
The auxiliary equation is m3 − 6m2 + 11m − 6 = 0 ⇒ (m − 1)(m2 − 5m + 6) = 0 ⇒ (m − 1)(m − 2)(m − 3) = 0 ⇒ m = 1, 2, 3. Thus YC = c1 ex + c2 e2x + c3 e3x . Comparing this with YC = c1 y1 + c2 y2 + c3 y3 , we obtain y1 = ex , ⇒ y10 = ex , ⇒ y100 = ex ,
y2 = e2x , y20 = 2e2x , y200 = 4e2x ,
y3 = e3x y30 = 3e3x y300 = 9e3x .
By the method of variation of parameters, we have Z Z Z W2 (x) W3 (x) W1 (x) YP (x) = y1 (x) r(x)dx + y2 (x) r(x)dx + y3 (x) r(x)dx. W (x) W (x) W (x)
(3.5.1)
Now y1 y2 y3 ex e2x 0 W = y1 y20 y30 = ex 2e2x y100 y200 y300 ex 4e2x W1 = y2 y30 − y3 y20 = e2x 3e3x −
e3x ex e2x e3x 3e3x = 0 e2x 2e3x = ex (8e5x − 6e5x ) = 2e6x ; 9e3x 0 3e2x 8e3x e3x 2e2x = 3e5x − 2e5x = e5x ;
W2 = y3 y10 − y1 y30 = e3x (ex ) − (ex ) 3e3x = e4x − 3e4x = −2e4x ; W3 = y1 y20 − y2 y10 = (ex ) 2e2x − e2x (ex ) = 2e3x − e3x = e3x . Substituting all these values in Equation (3.5.1), we get Z Z 3x Z 5x −2e4x −x e e −x 2x 3x x e dx + e e dx + e e−x dx YP = e 6x 6x 2e 2e 2e6x Z Z Z 1 3x 1 x −2x 2x −3x = e e dx − e e dx + e e−4x dx 2 2 −2x −3x 1 x e e 1 3x e−4x 2x = e −e + e 2 −2 −3 2 −4 1 1 1 = − e−x + e−x − e−x 4 3 8
Tutorial 3.5
Method of Variation of Parameters
= −
115
1 −x e . 24
Hence, the general solution is y = YC + YP
⇒
y = c1 ex + c2 e2x + c3 e3x −
1 −x e . 24
Exercises Exercise 3.5.1. Solve y 00 + 9y = sec 3x by the method of variation of parameters. [GTU, March 2010] Solution.
116
Chapter 3
Higher Order Linear ODEs
ex Exercise 3.5.2. Solve (D2 − 3D + 2)y = 1+ex by method of variation of parameters. [GTU, Jan. 2013] Solution.
Exercise 3.5.3. Use the method of variation of parameters to find the general solution of y 00 − 4y 0 + 4y = Solution.
e2x . x
Tutorial 3.5
Method of Variation of Parameters
Exercise 3.5.4. Solve
Solution.
117
d2 y + 4y = tan 2x by method of variation of parameters. dx2 [GTU, June 2014]
118
Chapter 3
Higher Order Linear ODEs
d3 y dy Exercise 3.5.5. Solve the differential equation 3 + dx = cosec x by method of variation of dx parameters. [GTU- May 2012, Jan. 2013]
Tutorial 3.5
Method of Variation of Parameters
119
Additional Exercises Exercise 3.5.6. Using the method of variation of parameters find the general solution of the differential equation (D2 + 2D + 1)y = 3x3/2 ex . [GTU, Dec. 2010] Exercise 3.5.7. Solve (D2 +2D+2)y = 4e−x sec3 x using the method of variation of parameters. Viva Questions Question 3.5.8. When should we use the method of variation of parameters ? Question 3.5.9. In the method of variation of parameters, what is the form of YP for a second order linear ODE ? Question 3.5.10. In the method of variation of parameters, what is the form of YP for a third order linear ODE ? Answers 3.5.1 3.5.2 3.5.3 3.5.4 3.5.5 3.5.6 3.5.7
c1 cos 3x + c2 sin 3x + 19 cos 3x log cos 3x + 13 x sin 3x c1 ex + c2 e2x + ex log(1 + e−x ) − ex + e2x log(1 + e−x ) (c1 + c2 x)e2x + (ln x − 1)xe2x c1 cos 2x + c2 sin 2x − 41 cos 2x log(sec 2x + tan 2x) c1 + c2 cos x + c3 sin x + log(cosec x − cot x) − cos x log sin x − x sin x 12 7/2 x (c1 + c2 x)ex + 35 x e −x e (c1 cos x + c2 sin x) + 2e−x sin2 x sec x zzzzzzz
120
3.6
Chapter 3
Higher Order Linear ODEs
Tutorial : Euler-Cauchy Equations
Definition An equation of the form xn y (n) + an−1 xn−1 y (n−1) + · · · + a1 xy (1) + a0 y = r(x),
(3.6.1)
where a0 , a1 , · · · , an−1 are constants, is called an Euler-Cauchy Equation of order n. Method of Solution • For a given Euler-Cauchy equation, substitute x = ez or z = ln x. Then we obtain d 0 2 00 3 000 xy = Dy, x y = D(D − 1)y, x y = D(D − 1)(D − 2)y where D = dz and so on. • Substitute all these values in the given equation. Now the equation becomes a linear differential equation with constant coefficients. Solve it by the usual methods discussed in the previous chapter. Here the solution will be in the form y ≡ y(z). • Replace z by ln x or ez by x to obtain the general solution of the given equation. Solved Examples d3 y dy + 7x dx −27y = 0. dx3 Solution. Let x = ez or z = ln x. Then 2 dy 3d y x = Dy, x = D(D − 1)(D − 2)y dx dx3
Example 3.6.1. Solve x3
d where D = . dz
Substituting all these values in the given equation, we get [D(D − 1)(D − 2) + 7D − 27]y = 0 ⇒ [D(D2 − 3D + 2) + 7D − 27]y = 0 ⇒ (D3 − 3D2 + 9D − 27)y = 0 which is a linear differential equation with constant coefficients. The auxiliary equation is m3 − 3m2 + 9m − 27 = 0 ⇒ m2 (m − 3) + 9(m − 3) = 0 ⇒ (m − 3)(m2 + 9) = 0 ⇒ m = 3, ±3i. Hence, the general solution is y = c1 e3z + c2 cos 3z + c3 sin 3z. Replacing z by ln x or ez by x, we obtain y = c1 x3 + c2 cos(3 ln x) + c3 sin(3 ln x).
Tutorial 3.6
Euler-Cauchy Equations
121
Example 3.6.2. Solve x2 y 00 + 3xy 0 + y = x2 log x. Solution. Let x = ez or z = ln x. Then 0
2 00
x y = D(D − 1)y
xy = Dy,
d where D = dz
.
Substituting all these values in the given equation, we get [D(D − 1) + 3D + 1]y = e2z z ⇒ (D2 + 2D + 1)y = e2z z. which is a linear differential equation with constant coefficients. The auxiliary equation is m2 + 2m + 1 = 0
(m + 1)2 = 0
⇒
⇒
m = −1, −1.
Thus YC = (c1 + c2 z)e−z . Now YP = =
1 e2z z 2 2 (D + 2D + 1) 1 e2z z (D + 1)2
= e2z
1 z (D + 3)2
1 z D2 + 6D + 9 ! 1 e2z = z 2 9 1 + D +6D 9 e2z D2 + 6D = + ··· z 1− 9 9 e2z 2 = z− 9 3 = e2z
e2z = (3z − 2). 27 Hence, the general solution is given by y = YC + YP
⇒
y = (c1 + c2 z)e−z +
e2z (3z − 2). 27
Replacing z by ln x or ez by x, we obtain y=
c1 + c2 ln x x2 + (3 ln x − 2). x 27
Exercises Exercise 3.6.1. Solve the Euler-Cauchy equation x2 y 00 − 7xy 0 + 16y = 0. Solution.
122
Chapter 3
Higher Order Linear ODEs
Exercise 3.6.2. Solve (x2 D2 − 3xD + 4)y = 0, y(1) = 0, y 0 (1) = 3. Solution.
[GTU, Dec. 2011]
Tutorial 3.6
Euler-Cauchy Equations
Exercise 3.6.3. Solve (x2 D2 + xD − 9)y = 48x5 . Solution.
123
124
Chapter 3
Higher Order Linear ODEs
Exercise 3.6.4. Find the general solution of the equation (x2 D2 − 2xD + 2)y = x3 cos x. [GTU, Dec. 2010] Solution.
Tutorial 3.6
Euler-Cauchy Equations
125
Additional Exercises Exercise 3.6.5. Find the general solution of the differential equation 3 2 1 3d y 2d y . [GTU- May 2011, Jan. 2013] x + 2x + 2y = 10 x + dx3 dx2 x Exercise 3.6.6. Solve x2
d2 y dy − 3x dx + 3y = 3 log x − 4. dx2
[GTU, March 2010]
Viva Questions Question 3.6.7. What is the standard form of an Euler-Cauchy equation of nth order ? Question 3.6.8. How to solve an Euler-Cauchy equation ? Question 3.6.9. Which of the following are Euler-Cauchy equations ? (i) x3 y 000 − 5xy 0 + 6y = sin x
(iii) xy 000 + y 0 = 0
(ii) xy 00 + 10y 0 = 12x7
(iv) xy 000 + 3y 00 = ex
Answers 3.6.1 3.6.2 3.6.3 3.6.4 3.6.5 3.6.6 3.6.9
(c1 + c2 ln x)x4 3x2 log x c1 x3 + c2 x−3 + 3x5 c1 x + c2 x2 − x cos x c1 + x[c2 cos(log x) + c3 sin(log x)] + 5x + x2 log x x c1 x + c2 x3 + log x (i), (ii), (iv) zzzzzzz
126
3.7
Chapter 3
Higher Order Linear ODEs
Tutorial : Series Solution of Differential Equations
Power Series A series of the form ∞ X
an (x − x0 )n = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · · .
n=0
is called a power series in powers of x − x0 or near x0 . In particular, for x0 = 0, we obtain a power series in powers of x as ∞ X
an xn = a0 + a1 x + a2 x2 + · · · .
n=0
Analytic Function A function is said to be analytic at a point x0 if it can be expressed in a power series in powers of x − x0 . Ordinary and Singular Points Consider the differential equation y 00 + p(x)y 0 + q(x)y = 0.
(3.7.1)
• A point x0 is called an ordinary point of (3.7.1) if the functions p(x) and q(x) both are analytic at x0 . • A point x0 is called a singular point of (3.7.1) if either p(x) or q(x) is not analytic at x0 . • A singular point x0 is called a regular singular point if (x − x0 )p(x) and (x − x0 )2 q(x) both are analytic at x0 , otherwise it is called an irregular singular point. Power Series Method This method is used to obtain a power series solution of the differential equation y 00 + p(x)y 0 + q(x)y = 0.
(3.7.2)
at an ordinary point x0 . The method can be described as follows: • Assume that y =
∞ P
ak (x − x0 )k is a solution of (3.7.2). Differentiating with respect to x, we
k=0
obtain 0
y =
∞ X k=1
k−1
kak (x − x0 )
00
and y =
∞ X k=2
• Substitute the expressions of y, y 0 and y 00 in (3.7.2).
k(k − 1)ak (x − x0 )k−2 .
Tutorial 3.7
Series Solution of Differential Equations
127
• Shift the summation index so that powers of x become same in all the summations. • Make the starting index same of all the summations by writing first few terms, if necessary and merge all the summations. • Equate to zero the coefficients of various powers of x and find an 0 s. • Substitute the expressions of an 0 s in y =
∞ P
ak (x − x0 )k which gives the required solution.
k=0
Solved Examples Example 3.7.1. Solve the equation
Solution. Let y =
∞ P
d2 y + y = 0 by the power series method. dx2 [GTU- Dec. 2009, Dec. 2011]
ak xk be a solution of the given equation. Then
k=0 ∞
dy X = kak xk−1 dx k=1
∞
and
d2 y X = k(k − 1)ak xk−2 . dx2 k=2
Substituting all these values in the given equation, we obtain ∞ X
⇒ ⇒
k(k − 1)ak x
k−2
k=2 ∞ X
+
∞ X
ak x k = 0
k=0 k
(k + 2)(k + 1)ak+2 x +
k=0 ∞ X
∞ X
ak x k = 0
k=0
(k + 2)(k + 1)ak+2 + ak xk = 0.
k=0
Equating the corresponding coefficients of xk on both the sides, we obtain ⇒
(k + 2)(k + 1)ak+2 + ak = 0
ak+2 = −
For k = 0, a2 = −
a0 . 1·2
a3 = −
a1 . 2·3
For k = 1,
For k = 2,
ak (k + 1)(k + 2)
a4 = −
a2 a0 = . 3·4 1·2·3·4
a5 = −
a3 a1 = . 4·5 2·3·4·5
For k = 3,
(k ≥ 0).
128
Chapter 3
Higher Order Linear ODEs
Hence, the required solution is y = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 a5 x 5 + · · · a1 3 a0 a1 a0 2 x − x + x4 + x5 − · · · 1·2 2·3 1·2·3·4 2·3·4·5 x2 x4 x3 x5 = a0 1 − + − · · · + a1 x − + − ··· 2! 4! 3! 5! = a0 + a1 x −
= a0 cos x + a1 sin x. Example 3.7.2. Find the series solution of (x2 + 1)y 00 + xy 0 − xy = 0 near x = 0. [GTU- May 2011, Jan. 2013, June 2013] Solution. Dividing the equation by x2 + 1, we get y 00 +
x2
x x y0 − 2 y = 0. +1 x +1
Here,
x x and q(x) = − x2 + 1 x2 + 1 which both are analytic at x = 0. Thus x = 0 is an ordinary point and we apply the power series method. Now, the given equation is p(x) =
x2 y 00 + y 00 + xy 0 − xy = 0. Let y =
∞ P
(3.7.3)
ak xk be a solution of the given equation. Then
k=0
0
y =
∞ X
kak x
k−1
00
and y =
∞ X
k(k − 1)ak xk−2 .
k=2
k=1
Substituting all these values in (3.7.3), we obtain ∞ X
⇒
k=2 ∞ X
k
∞ X
k
k=2 ∞ X
k(k − 1)ak x + k(k − 1)ak x +
k=2
k(k − 1)ak x
k−2
+
∞ X
k
kak x −
k=1 k
(k + 2)(k + 1)ak+2 x +
k=0
∞ X
ak xk+1 = 0
k=0 ∞ X
k
kak x −
k=1
∞ X
ak−1 xk = 0
k=1
∞ X ⇒ 2a2 + 6a3 x + a1 x − a0 x + k(k − 1)ak + (k + 1)(k + 2)ak+2 + kak − ak−1 xk = 0. k=2
Equating the corresponding coefficients of 1, x, . . . , xk on both the sides, we obtain 2a2 = 0
⇒
a2 = 0
6a3 + a1 − a0 = 0 .. .
⇒
1 a3 = (a0 − a1 ) 6
Tutorial 3.7
Series Solution of Differential Equations
129
k(k − 1) + k ak + (k + 1)(k + 2) ak+2 − ak−1 = 0 ⇒ k 2 ak + (k + 1)(k + 2) ak+2 − ak−1 = 0 ak−1 − k 2 ak ⇒ ak+2 = (k ≥ 2). (k + 1)(k + 2) For k = 2, a4 =
a1 − 4a2 3·4
⇒
a4 =
a1 12
(∵ a2 = 0).
For k = 3, a2 − 9a3 a5 = 4·5
⇒
9 1 a5 = − (a0 − a1 ) 20 6
⇒
a5 = −
3 (a0 − a1 ). 40
Hence, the required solution is y = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 a5 x 5 + · · · 1 3 = a0 + a1 x + (a0 − a1 )x3 − (a0 − a1 )x5 + · · · 6 40 3 5 1 4 3 5 1 3 1 3 = a0 1 + x − x + · · · + a1 x − x + x − x + · · · . 6 40 6 12 40 Exercises Exercise 3.7.1. Find a power series solution in powers of x of y 0 + 2xy = 0. [GTU, Dec. 2011] Solution.
130
Exercise 3.7.2. Solve in series the equation Solution.
Chapter 3
d2 y + x2 y = 0. dx2
Higher Order Linear ODEs
[GTU, Dec. 2013]
Tutorial 3.7
Series Solution of Differential Equations
131
Exercise 3.7.3. Find the series solution of (1 − x2 )y 00 − 2xy 0 + 2y = 0. [GTU- March 2010, Dec. 2013, Jan. 2015] Solution.
132
Chapter 3
Higher Order Linear ODEs
Additional Exercises Exercise 3.7.4. Find the series solution of y 00 = 2y 0 in powers of x.
[GTU, Jan. 2013]
Exercise 3.7.5. Find the series solution of (1 + x2 )y 00 + xy 0 − 9y = 0.
[GTU, May 2012]
Viva Questions Question 3.7.6. Define ordinary point, singular point, regular singular point. Question 3.7.7. Check whether the point x = 0 is an ordinary point of the differential equation (1 − x2 )y 00 − 2xy 0 + 2y = 0. Question 3.7.8. Determine the singular points of the differential equation 2x(x − 2)2 + 3xy 0 + (x − 2)y = 0 and classify them as regular or irregular. Answers h i h i h i 4 x4 x8 x5 x9 3.7.1 a0 1 − x2 + x2 − · · · 3.7.2 a0 1 − 3·4 + 3·4·7·8 − · · · + a1 x − 4·5 + 4·5·8·9 − ··· i h 6 4 3.7.3 a0 1 − x2 − x3 − x5 − · · · + a1 x 3.7.4 a0 + a1 x 1 + x + 32 x2 + 13 x3 + · · · h i h i 15x4 7x6 9x2 4x3 3.7.5 a0 1 + 2 + 8 − 16 · · · + a1 x + 3 · · · zzzzzzz
Chapter
4
Laplace Transforms and Applications The Laplace transform is very useful tool for solving differential equations. It greatly simplifies the method of solution by transforming differential equations into algebraic equations. So, it is frequently used in solving differential equations arising in many engineering problems.
4.1
Tutorial : Basic Definition and Examples
Definition Let f (t) be a function that is defined for all t ≥ 0. Then the Laplace transform of f (t) is denoted by L[f (t)] or F (s) or f (s) and is defined as Z ∞ e−st f (t)dt (4.1.1) L[f (t)] = F (s) = 0
provided the improper integral exists, where s is a parameter real or complex. The following table shows Laplace transforms of some standard functions: Function
Laplace Transform
1
1
1 s
2
t
3
t2
4
tn (n = 0, 1, 2, . . .)
5 ta (a is positive real )
Function
Laplace Transform
6
eat
1 s−a
1 s2
7
cos ωt
s s2 +ω 2
2 s3
8
sin ωt
ω s2 +ω 2
sn+1
9
cosh ωt
s2 −ω 2
Γ(a+1) sa+1
10
sinh ωt
ω s2 −ω 2
n!
133
s
134
Chapter 4
Laplace Transforms and Applications
Linearity of Laplace Transform Let f (t) and g(t) be any functions whose Laplace transforms exist and a and b be any constants. Then L[af (t) + bg(t)] = aL[f (t)] + bL[g(t)]. Solved Examples Example 4.1.1. Find the Laplace transform of 0, 0 < t < 2 f (t) = 3, t ≥ 2.
[GTU, Dec. 2011]
Solution. By the definition, Z
L[f (t)] = = = = = =
∞
e−st f (t)dt Z0 2 Z ∞ 0dt + e−st 3dt 0 2 Z ∞ e−st dt 3 2 −st ∞ e 3 −s 2 e−2s 3 −0 + s −2s 3e . s
Example 4.1.2. Find the Laplace transforms of (i) 1 + 2t + 3t2 , (ii) ea−bt . Solution. Using linearity of Laplace transform, we obtain 1 1 2! 1 2 6 2 2 (i) L[1 + 2t + 3t ] = L[1] + 2L[t] + 3L[t ] = + 2 2 + 3 3 = + 2 + 3 . s s s s s s a e (ii) L[ea−bt ] = L ea e−bt = ea L e−bt = . s+b Example 4.1.3. Find L[sin 2t cos 2t].
[GTU, Dec. 2009]
Solution. We have 1 1 1 4 2 L[sin 2t cos 2t] = L[2 sin 2t cos 2t] = L[sin 4t] = = 2 . 2 2 2 2 s + 16 s + 16 Example 4.1.4. Find the Laplace transforms of (i) sin2 at, (ii) cos 4t cos 3t. Solution. (i) We have
1 − cos 2at L[sin at] = L 2 2
Tutorial 4.1
Basic Definition and Examples
= = = = =
135
i 1h L[1] − L[cos 2at] 2 s 1 1 − 2 s s2 + 4a2 1 s2 + 4a2 − s2 2 s(s2 + 4a2 ) 1 4a2 2 s(s2 + 4a2 ) 2a2 . s(s2 + 4a2 )
(ii) Observe that 1 1 1 cos 4t cos 3t = [2 cos 4t cos 3t] = cos 7t + cos t = [cos 7t + cos t]. 2 2 2 Thus L[cos 4t cos 3t] = = = = =
i 1h L[cos 7t] + L[cos t] 2 s s 1 + 2 s2 + 49 s2 + 1 s 1 1 + 2 s2 + 49 s2 + 1 2s2 + 50 s 2 (s2 + 49)(s2 + 1) s(s2 + 25) . (s2 + 49)(s2 + 1)
Example 4.1.5. Find the Laplace transform of cos3 t. Solution. Observe that cos 3t = 4 cos3 t − 3 cos t
⇒
4 cos3 t = cos 3t + 3 cos t
⇒
Thus [cos3 t] = = = = =
i 1h L[cos 3t] + 3L[cos t] 4 1 s s +3 2 4 s2 + 9 s +1 s 1 3 + 4 s2 + 9 s2 + 1 s 4s2 + 28 4 (s2 + 9)(s2 + 1) s(s2 + 7) . (s2 + 9)(s2 + 1)
1 cos3 t = [cos 3t + 3 cos t]. 4
136
Chapter 4
Laplace Transforms and Applications
Exercises Exercise 4.1.1. Find the Laplace transform of f (t) =
0, 0