Engineering Mathematics - III (Computer Engg. and IT Group) PDF

 ENGINEERING MATHEMATICS – III (With Large Number of Solved Examples and MCQ's) FOR SECOND YEAR DEGREE COURSES IN...

 ENGINEERING MATHEMATICS – III (With Large Number of Solved Examples and MCQ's) FOR SECOND YEAR DEGREE COURSES IN COMPUTER ENGINEERING AND INFORMATION TECHNOLOGY ACCORDING TO NEW REVISED CREDIT SYSTEM SYLLABUS (2019 PATTERN) OF SAVITRIBAI PHULE PUNE UNIVERSITY, PUNE (EFFECTIVE FROM ACADEMIC YEAR – 2020, SEMESTER - II) Dr. M. Y. GOKHALE Dr. N. S. MUJUMDAR M. Sc. (Pure Maths.), M. Sc. (App. Maths.) M. Sc., M. Phil., Ph. D. (Maths.) Ph. D. (I. I. T., Mumbai) Professor in Mathematics, Professor and Head, Deptt. of Mathematics, Jayawant Shikshan Prasarak Mandal's, Maharashtra Institute of Technology, Rajarshi Shahu College of Engineering, Kothrud, PUNE. Tathawade, PUNE. S. S. KULKARNI A. N. SINGH M. Sc. (Maths.) (I. I. T. Mumbai) M. A. (Mathematics Gold Medalist) Eminent Professor of Mathematics, Formerly, Head of Mathematics Deptt. Formerly, Lecturer, Mathematics Deptt. VIT. D. Y. Patil College of Engineering, PUNE. Pimpri, PUNE. K. R. ATAL M. Sc. (Mathematics), Formerly, Lecturer, (Selection Grade) Deptt. of Applied Sciences, Pune Institute of Computer Technology, (PICT), Dhankawdi, PUNE Price ` 550.00 N5696 ENGINEERING MATHEMATICS-III (Computer Engg. and IT Group) ISBN 978-93-5451-007-6 First Edition : February 2021 © : Authors The text of this publication, or any part thereof, should not be reproduced or transmitted in any form or stored in any computer storage system or device for distribution including photocopy, recording, taping or information retrieval system or reproduced on any disc, tape, perforated media or other information storage device etc., without the written permission of Author with whom the rights are reserved. Breach of this condition is liable for legal action. Every effort has been made to avoid errors or omissions in this publication. In spite of this, errors may have crept in. Any mistake, error or discrepancy so noted and shall be brought to our notice shall be taken care of in the next edition. It is notified that neither the publisher nor the author or seller shall be responsible for any damage or loss of action to any one, of any kind, in any manner, therefrom. Published By : Polyplate Printed By : NIRALI PRAKASHAN YOGIRAJ PRINTERS AND BINDERS Abhyudaya Pragati, 1312, Shivaji Nagar Survey No. 10/1A, Ghule Industrial Estate Off J.M. Road, PUNE – 411005 Nanded Gaon Road Tel - (020) 25512336/37/39, Fax - (020) 25511379 Nanded, Pune - 411041 Email : [email protected] Mobile No. 9404233041/9850046517
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Any type of error or mistake so noted, and shall be brought to our notice, shall be taken care of in the next edition. It is notified that neither the publisher, nor the author or book seller shall be responsible for any damage or loss of action to any one of any kind, in any manner, therefrom. The reader must cross check all the facts and contents with original Government notification or publications. [email protected] | www.pragationline.com Also find us on www.facebook.com/niralibooks  PREFACE Our text books on "Engineering Mathematics-III" have occupied place of pride among engineering student's community for more than twenty-five year now. All the teachers of this group of authors have been teaching mathematics in engineering colleges for the past several years. Difficulties of engineering students are well understood by the authors and that is reflected in the text material. As per the policy of the University, Engineering Syllabi is revised every five years. Last revision was in the year 2015. New revision is coming little earlier, as university has introduced Choice Based Credit System from Academic Year 2019-2020. As per the Choice Based Credit System, the theory examination shall be conducted by university in two phase. Phase-I will be In-Semester Examination of 30 marks theory examination based on first and second units. Phase-II will be End-Semester Examination of 70 marks theory examination based on third, fourth, fifth and sixths units. New text book on Engineering Mathematics III for second year Computer Engineering and Information Technology is written, taking in to account all the new features that have been introduced. All the entrants to the engineering field will definitely find this book, complete in all respect. Students will find the subject matter presentation quite lucid. There are large number of illustrative examples and well graded exercises. Multiple Choice Questions based on the syllabus are appended at the end of the book and will be useful across universities/colleges and various competitive exams. We take this opportunity to express our sincere thanks to Shri. Dineshbhai Furia of Nirali Prakashan, Pioneer in all fields of education. Thanks are also due to Shri. Jignesh Furia, whose dynamic leadership is helpful to all the authors of Nirali Prakashan. We specially appreciate the effects of Late M. P. Munde and entire team of Nirali Prakashan namely Mrs. Anagha Kawre (Co-Ordinator and Proof Reader), Late Mr. Santosh Bare, who really have taken keen interest and untiring efforts in publishing this text. We have no doubt that like our earlier texts, student's community will respond favourably to this new venture. The advice and suggestions of our esteemed readers to improve the text are most welcomed, and will be highly appreciated. 26th January 2021 Pune Authors  SYLLABUS Unit I : Linear Differential Equations (LDE) (08 Hrs.) LDE of nth order with constant coefficients, Complementary function, Particular integral, General method, Short methods, Method of variation of parameters, Cauchy's and Legendre's DE, Simultaneous and Symmetric simultaneous DE. Unit II : Transforms (08 Hrs.) Fourier Transform (FT): Complex exponential form of Fourier series, Fourier integral theorem, Fourier transform, Fourier sine and cosine integrals, Fourier transforms, Fourier Sine and Cosine transforms and their inverses, Discrete Fourier Transform. Z-Transform (ZT): Introduction, Definition, Standard properties, ZT of standard sequences and their inverses. Solution of difference equations. Unit III : Statistics (07 Hrs.) Measures of central tendency, Measure of dispersion, Coefficient of variation, Moments, Skewness and Kurtosis, Curve fitting: fitting of straight line, parabola and related curves, Correlation and Regression, Reliability of Regression Estimates. Unit IV : Probability and Probability Distributions (07 Hrs.) Probability, Theorems on Probability, Bayes Theorem, Random variables, Mathematical Expectation, Probability density function, Probability distributions : Binomial, Poisson, Normal and Hypergeometric, Sampling distributions, Test of Hypothesis : Chi-square test, t-test. Unit V : Numerical Methods (08 Hrs.) Numerical solution of Algebraic and Transcendental equations: Bisection, Secant, Regula-Falsi, Newton- Raphson and Successive Approximation Methods, Convergence and Stability. Numerical Solution of System of linear equations: Gauss elimination, LU Decomposition, Cholesky, Jacobi and Gauss-Seidel Methods. Unit VI : Numerical Methods (08 Hrs.) Interpolation: Finite Differences, Newton's and Lagrange's Interpolation formulae, Numerical Differentiation. Numerical Integration: Trapezoidal and Simpson's rules, Bound of truncation error. Numerical Solution of Ordinary differential equations: Euler's, Modified Euler's, Runge-Kutta 4th order methods and Predictor-Corrector methods.  CONTENTS Unit-I : Linear Differential Equations (LDE) 1. Linear Differential Equations with Constant Coefficients 1.1. − 1.48 2. Simultaneous Linear Differential Equations, Symmetrical Simultaneous D.E. 2.1 − 2.16 Unit-II : Transforms 3. Fourier Transform 3.1 − 3.40 4. The Z-Transform 4.1 − 4.64 Unit-III: Statistics 5. Statistics, Correlation and Regression 5.1 − 5.60 Unit-IV : Probability and Probability Distributions 6. Probability and Probability Distributions 6.1 − 6.56 Unit-V : Numerical Methods 7. Numerical Solutions of Algebraic and Transcendental Equations 7.1 − 7.44 8. System of Linear Equations 8.1 − 8.44 Unit-VI : Numerical Methods 9. Interpolation, Numerical Differentiation and Integration 9.1 − 9.42 10. Numerical Solutions of Ordinary Differential Equations 10.1 − 10.48 Appendix : Multiple Choice Questions (Chapter wise) A.1 − A.70 Model Question Paper : Phase I : In Semester Examination (ISE) P.1 − P.1 Model Question Paper : Phase II : End Semester Examination (ESE) P.2 − P.4 ❒❒❒  UNIT I : LINEAR DIFFERENTIAL EQUATIONS (LDE) CHAPTER-1 LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 1.1 INTRODUCTION Differential equations are widely used in fields of Engineering and Applied Sciences. Mathematical formulations of most of the physical problems are in the forms of differential equations. Use of differential equations is most prominent in subjects like Circuit Analysis, Theory of Structures, Vibrations, Heat Transfer, Fluid Mechanics etc. Differential equations are of two types : Ordinary and Partial Differential Equations. In ordinary equations, there is one dependent variable depending for its value on one independent variable. Partial differential equations will have more than one independent variables. In what follows, we shall discuss ordinary and partial differential equations, which are of common occurrence in engineering fields. Applications to some areas will also be dealt. 1.2 PRELIMINARIES I. Second Degree Polynomials and Their Factorization (a) (i) D2 – 2D – 3 = (D + 1) (D – 3) (ii) D2 + 5D + 6 = (D + 2) (D + 3) (iii) D2 + 2D + 1 = (D + 1)2 (iv) D2 – 5D + 6 = (D – 2) (D – 3) 2 (v) D + 3D + 2 = (D + 2) (D + 1) (vi) D2 – D – 2 = (D – 2) (D + 1) (vii) D2 – 4D + 4 = (D – 2)2 (viii) D2 – a2 = (D – a) (D + a) 2 2 (ix) D + a = (D + ia) (D – ia) –b± b2 – 4ac (b) The roots of ax2 + bx + c = 0 are x = , these roots are imaginary if b2 – 4ac < 0. 2a –2± 4–8 (i) D2 + 2D + 2 = 0 ⇒ D = = –1±i 2 –1± 1–4 –1 3 (ii) D2 + D + 1 = 0 ⇒ D = = ± i 2 2 2 –1 3 1 3 If D = ± i = α ± iβ then α = – , β = , β is always positive; α may be positive, negative or zero. 2 2 2 2 (iii) D2 + 1 = 0 ⇒ D2 = – 1 i.e. D=±i ∴ α = 0, β = 1. 2 (iv) D + 4 = 0 ⇒ D 2 = –4 i.e. D = ± 2i ∴ α = 0, β = 2. II. Third Degree Polynomials and Their Factorization (a) (i) D3 – a3 = (D – a) (D2 + aD + a2) (iii) D3 + a3 = (D + a) (D2 – aD + a2) 3 2 3 (ii) D + 3D + 3D + 1 = (D + 1) (iv) D3 – 3D2 + 3D – 1 = (D – 1)3 (b) Use of Synthetic Division : (i) f(D) = D3 – 7D – 6 = 0; for D = – 1, f(–1) = 0 ∴ (D + 1) is one of the factors. –1 1 0 –7 –6 ∴ D3 – 7D – 6 = 0 ⇒ (D + 1) (D2 – D – 6) = 0 (D + 1) (D – 3) (D + 2) = 0 ⇒ D = – 1, – 2, 3. –1 1 6 1 –1 –6 0 (ii) For D3 – 2D + 4 = 0; D = – 2 ∴ f(–2) = 0 ∴ (D + 2) is one of the factors. –2 1 0 –2 4 ∴ D3 – 2D + 4 = 0 ⇒ (D + 2) (D2 – 2D + 2) = 0 –2 4 –4 D = – 2 and D = 1 ± i, α = 1, β = 1. 1 –2 2 0 (1.1) ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.2) LINEAR DIFFERENTIAL EQUATIONS… … III. Fourth Degree Polynomials and Their Factorization (a) D4 – a4 = (D2 – a2) (D2 + a2) = (D – a) (D + a) (D + ia) (D – ia) (b) Making a Perfect Square by Introducing a Middle Term : (i) For D4 + a4 = 0 ; consider (D2 + a2)2 = D4 + 2a2 D2 + a4 2 D4 + a4 = (D4 + 2a2 D2 + a4) – (2a2 D2) = (D2 + a2)2 – ( 2 a D) D4 + a4 = (D 2 – 2 a D + a2) (D 2 + 2 a D + a2) 2 (ii) For D4 + 1 = D4 + 2D2 + 1 – 2D2 = (D2 + 1)2 – ( 2 D) D4 + 1 = (D 2 – 2 D + 1) (D 2 + 2 D + 1) (c) D4 + 8D2 + 16 = (D2 + 4)2, D4 + 2D2 + 1 = (D2 + 1)2 = (D + i)2 (D – i)2 D4 + 10D2 + 9 = (D2 + 9) (D2 + 1) = (D + 3i) (D – 3i) (D + i) (D – i) (d) (i) f(D) = D4 – 2D3 – 3D2 + 4D + 4 = 0, for D = – 1, f (–1) = 0 –1 1 –2 –3 4 4 2 2 ∴ Factors are (D + 1) (D – 2) = 0. –1 3 0 –4 –1 1 –3 0 4 0 –1 4 –4 2 1 –4 4 0 2 –4 1 –2 0 On a similar line, (ii) D4 – D3 – 9D2 – 11 D – 4 = (D + 1)3 (D – 4) (e) Perfect square of the type (a + b + c)2 (i) D4 + 2D3 + 3D2 + 2D + 1 = (D2)2 + 2 · D2 · D + D2 + 2D2 + 2D + 1 = (D2 + D)2 + 2 (D2 + D) + 1 = [(D2 + D) + 1]2 = (D2 + D + 1)2 (ii) D4 – 4D3 + 8D2 – 8D + 4 = (D2)2 – 2D2 · 2D + (2D)2 + 4D2 – 8D + 4 = (D2 – 2D)2 + 4 (D2 – 2D) + 4 = [(D2 – 2D) + 2]2 = (D2 – 2D + 2)2 IV. Fifth Degree Polynomials and Their Factorization (i) D5 – D4 + 2D3 – 2D2 + D – 1 = D4 (D – 1) + 2D2 (D – 1) + 1 (D – 1) = (D4 + 2D2 + 1) (D – 1) = (D – 1) (D2 + 1)2 = (D – 1) (D + i)2 (D – i)2 1.3 THE nth ORDER LINEAR DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS A differential equation which contains the differential coefficients and the dependent variable in the first degree, does not involve the product of a derivative with another derivative or with dependent variable, and in which the coefficients are constants is called a linear differential equation with constant coefficients. The general form of such a differential equation of order "n" is dny dn–1 y dn–2 y dy a0 n + a1 n–1 + a2 + … + an–1 + an y = f(x) … (1) dx dx dxn–2 dx Here a0, a1, a2 … are constants. Equation (1) is a nth order linear differential equation with constant coefficients. ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.3) LINEAR DIFFERENTIAL EQUATIONS… … e.g. Put n = 3 in equation (1), we get d3y d2y dy a0 3 + a1 + a2 + a3 y = f(x) which is a 3rd order linear differential equation with constant coefficients. dx dx2 dx d dy 2 d2y dn y Using the differential operator D to stand for i.e. Dy = ; D y = 2 , … Dny = , the equation (1) will take the form dx dx dx dx a0 Dn y + a1 Dn–1 y + a2 Dn–2 y + … + an–1 Dy + an y = f(x) OR (a0 Dn + a1 Dn–1 + a2 Dn–2 + … + an-1 D + an) y = f(x) … (2) in which each term in the parenthesis is operating on y and the results are added. Let φ(D) ≡ a0 Dn + a1 Dn–1 + a2 Dn–2 + … + an–1 D + an , φ(D) is called as nth order polynomial in D. ∴ Equation (2) can be written as φ(D) y = f(x) … (3) Note : In equation (1), if a0, a1, … an are functions of x then it is called nth order linear differential equation. 1.4 THE NATURE OF DIFFERENTIAL OPERATOR "D" d It is convenient to introduce the symbol D to represent the operation of differentiation with respect to x i.e. D ≡ , so that dx dy d2 y 2 d3 y 3 dn y dy = Dy; 2 = D y; 3 = D y; ……; = Dny and + ay = (D + a) y dx dx dx dxn dx The differential operator D or (Dn) obeys the laws of Algebra. Properties of the Operator D : If y1 and y2 are differentiable functions of x and "a" is a constant and m, n are positive integer then (i) Dm (Dn) y = Dn (Dm) y = Dm+n y (ii) (D – m1) (D – m2) y = (D – m2) (D – m1) y 2 (iii) (D – m1) (D – m2) y = [D – (m1 + m2) D + m1 m2] y (iv) D (au) = a · D(u); Dn (au) = a · Dn (u) (v) D (y1 + y2) = D (y1) + D(y2); Dn (y1 + y2) = Dn (y1) + Dn (y2). 1.5 LINEAR DIFFERENTIAL EQUATION φ (D) y = 0 Consider, (D) y = 0 … (4) n n–1 n–2 n–3 th where, φ(D) = a0 D + a1 D + a2 D + a3 D + … + an–1 D + an is n order polynomial in D and D obeys the laws of algebra, we can in general factorise φ(D) in n linear factors as φ(D) = (D – m1) (D – m2) (D – m3) … (D – mn) where m1, m2, m3, … mn are the roots of the algebraic equation φ(D) = 0 ∴ Equation (4) can be written as; φ(D) y = (D – m1) (D – m2) (D – m3) … (D – mn) y = 0 … (5) Note : These factors can be taken in any sequence. 1.6 AUXILIARY EQUATION (A.E.) The equation φ(D) = 0 is called as an auxiliary equation (A.E.) for equations (3), (4). d2 y dy e.g. –5 + 6y = 0 dx2 dx d By using operator D for , we have (D2 – 5D + 6) y = 0 dx φ(D) = D2 – 5D + 6 = 0 is the A.E. ∴ (D2 – 5D + 6) y = (D – 3) (D – 2) y = (D – 2) (D – 3) y. 1.7 SOLUTION OF φ(D) y = 0 Being nth order DE, equation (4) or (5) will have exactly n arbitrary constants in its general solution. The equation (5) will be satisfied by the solution of the equation (D – mn) y = 0 dy i.e. – mn y = 0 dx ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.4) LINEAR DIFFERENTIAL EQUATIONS… … On solving this 1st order 1st degree DE by separting variables, we get y = cn emnx, where, cn is an arbitrary constant. Similarly, since the factors in equation (5) can be taken in any order, the equation will be satisfied by the solution of each of the mx mx equations (D – m1) y = 0, (D – m2) y = 0 … etc., that is by y = c1 e 1 , y = c2 e 2 ………… etc. It can, therefore, easily be proved that the sum of these individual solutions, i.e. mx mx mx y = c1 e 1 + c2 e 2 + … + cn e n … (6) also satisfies the equation (5) and as it contains n arbitrary constants, and the equation (4) is of the nth order, (6) constitutes the general solution of the equation (4). ∴ The general solution of the equation φ(D) y = 0 is mx mx mx y = c1 e 1 + c2 e 2 + … + cn e n where m1, m2, … mn are the roots of the auxiliary equation φ(D) = 0. d3 y d2 y dy Ex. 1 : Solve 3 – 6 + 11 – 6y = 0. dx dx2 dx d Sol. : Let D stand for and the given equation can be written as dx (D3 – 6D2 + 11D – 6) y = 0. Here auxiliary equation is D3 – 6D2 + 11D – 6 = 0 i.e. (D – 1) (D – 2) (D – 3) = 0 ⇒ m1 = 1, m2 = 2, m3 = 3, are roots of AE. ∴ The general solution is y = c1 ex + c2 e2x + c3 e3x. Ex. 2 : For (4D2 – 8D + 1) y = 0. 3 Sol. : Here auxiliary equation is 4D2 − 8D + 1 = 0 i.e. D = 1 ± 2  3  3 1 + x 1 – 2  x y = c1 e 2  + c2 e . 1.8 DIFFERENT CASES DEPENDING UPON THE NATURE OF ROOTS OF THE AUXILIARY EQUATION φ(D) = 0 A. The Case of Real and Different Roots If roots of φ (D) = 0 be m1, m2, m3 … mn, all are real and different, then the solution of φ (D) y = 0 will be m1x mx mx mx y = c1 e + c2 e 2 + c3 e 3 + … + cn e n B. The Case of Real and Repeated Roots (The Case of Multiple Roots) Let m1 = m2, m3, m4 … mn be the roots of φ(D) = 0, then the part of solution corresponding to m1 and m2 will look like mx mx mx mx c1 e 1 + c2 e 1 (m1 = m2) = (c1 + c2) e 1 = c'e 1 But this means that number of arbitrary constants now in the solution will be n – 1 instead of n. Hence it is no longer the general solution. The anomaly can be rectified as under. Pertaining to m1 = m2, the part of the equation will be (D – m1) (D – m1) y = 0 Put (D – m1) y = z, temporarily, then we have (D – m1) z = 0 ∴ z = c1 em1x Hence putting value of z in (D – m1) y = z, we have mx dy mx (D – m1) y = c1 e 1 or – m1 y = c1 e 1 dx – ∫ m1dx – m1x which is a linear differential equation. Its I.F. = e =e and hence solution is ⌠ ( y e – m1 x ) = ⌡ c1 e m1x ·e –m1x dx + c2 = c1 x + c2 mx ∴ y = (c1 x + c2) e 1 If m1 = m2 are real, and the remaining roots m3, m4, m5, …, mn are real and different then solution of φ(D) y = 0 is mx mx mx mx y = (c1 x + c2) e 1 + c3 e 3 + c4 e 4 + … + cn e n ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.5) LINEAR DIFFERENTIAL EQUATIONS… … Similarly, when three roots are repeated. i.e. if m1 = m2 = m3 are real, and the remaining roots m4, m5, … mn are real and different then solution of φ(D) y = 0 is mx mx mx y = (c1 x2 + c2 x + c3) e 1 + c4 e 4 + … + cn e n If m1 = m2 = m3 = … = mn i.e. n roots are real and equal then solution of φ(D) y = 0 is m1x y = (c1 xn–1 + c2 xn–2 + … + cn–1 x + cn) e Ex. 1. For (D2 – 6D + 9) y = 0 A.E. = (D – 3)2 = 0 and solution is y = (c1 x + c2) e3x 2. For (D – 1)3 (D + 1) y = 0, solution is y = (c1 x2 + c2 x + c3) ex + c4 e–x 3. For (D – 1)2 (D + 1)2 y = 0, solution is y = (c1 x + c2) ex + (c3 x + c4) e–x. C. The Case of Imaginary (Complex) Roots For practical problems in engineering, this case has special importance. Since the coefficients of the auxiliary equation are real, the imaginary roots (if exists) will occur in conjugate pairs. Let α ± iβ be one such pair. Therefore m1 = α + iβ, m2 = α – iβ The corresponding part of the solution of the equation φ(D) y = 0, then takes the form (α + iβ) x (α – iβ) x y = Ae + Be y = eαx [Aeiβx + Be−iβx] y = eαx [A (cos βx + i sin βx) + B (cos βx – i sin βx)] y = eαx [(A + B) cos βx + i (A – B) sin βx] y = eαx [c1 cos β x + c2 sin β x] where, c1 = A + B and c2 = i (A – B) are arbitrary constants. Using c1 = C cos θ, c2 = – sin θ, this can also be put sometimes into the form as given below (recall SHM). y = C eαx cos (β β x + θ) where C,, θ are arbitrary constants. ILLUSTRATIONS 2 Ex. 1 : Solve (D + 2D + 5) y = 0. Sol. : The auxiliary equation is D2 + 2D + 5 = 0 whose roots are D = – 1 ± 2i which are both imaginary. Here α = – 1, β = 2. Hence the solution is y = e–x [A cos 2x + B sin 2x] 4 2 dy dy dy Ex. 2 : Solve – 5 2 + 12 + 28y = 0. dx4 dx dx Sol. : The auxiliary equation is D4 – 5D2 + 12D + 28 = 0 having roots D = – 2, –2, 2 ± 3 i. (Here α = 2, β = 3). Hence the solution is y = (c1 x + c2 ) e–2x + e2x [A cos 3 x + B sin 3 x] 2 Ex. 3 : For (D + 4)y = 0. Sol. : The auxiliary equation is D2 + 4 = 0 having roots D = 0 ± 2i (Here α = 0, β = 2). Hence the solution is y = A cos 2x + B sin 2x. D. The Case of Repeated Imaginary Roots If the imaginary roots m1 = α + iβ and m2 = α – iβ occur twice, then the part of solution of φ (D) y = 0 will be mx mx y = (A x + B) e 1 + (C x + D ) e 2 … (by using case B) (α + iβ) x = (A x + B) e + (C x + D) e(α – iβ) x = eαx [(A x + B) eiβx + (C x + D) e–iβx] = eαx [(A x + B) {cos βx + i sin βx} + (C x + D) {cos βx – i sin βx}] = eαx [(A x + B + C x + D) cos βx + i (A x + B – C x – D) sin βx] ∴ y = eαx [(c1 x + c2) cos β x + (c3 x + c4) sin β x] with proper changes in the constants c1, c2, c3 and c4. ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.6) LINEAR DIFFERENTIAL EQUATIONS… … ILLUSTRATIONS 6 4 2 dy dy dy Ex. 1 : Solve + 6 4 +9 = 0. dx6 dx dx2 Sol. : The auxiliary equation D6 + 6D4 + 9D2 = 0 has roots D = 0, 0, ± i 3,±i 3 where the imaginary roots ± i 3 are repeated. Hence the solution is y = c1 x + c2 + (c3 x + c4) cos 3 x + (c5 x + c6) sin 3 x 4 2 Ex. 2 : (D + 2D + 1) y = 0. Sol. : The auxiliary equation D4 + 2D2 + 1 = 0 has roots D = ± i, ± i, repeated imaginary roots. Hence the solution is y = (c1x + c2) cos x + (c3x + c4) sin x Now we will summarise the four cases for ready reference. Case 1 : Real & Distinct Roots : A.E. ⇒ (D – m1) (D – m2) (D – m3) …(D – mn) = 0 mx mx mx Solution is y = c1 e 1 + c2 e 2 + c3 e 3 + … + cn emnx Case 2 : Repeated Real Roots (i) For m1 = m2 ⇒ A.E. ⇒ (D – m1) (D – m1) (D – m3) …… (D – mn) = 0 m1x mx Solution is y = (c1 x + c2) e + c3 e 3 + … + cn emnx (ii) For m1 = m2 = m3 ⇒ A.E. ⇒ (D – m1) (D – m1) (D – m1) (D – m4) … (D – mn) = 0 m1x mx mx Solution is y = (c1 x2 + c2 x + c3) e + c4 e 4 + … + cn e n Case 3 : Imaginary Roots : For D = α ± i β Solution is y = eαx [c1 cos β x + c2 sin β x] Case 4 : Repeated Imaginary Roots : For D = α ± iβ be repeated twice Solution is y = eαx [(c1 x + c2) cos β x + (c3 x + c4) sin β x] ILLUSTRATIONS 2 d x Ex.1 : Solve + 4x = 0. dt2 d Sol. : Let D stand for. dt ∴ A.E. : D2 + 4 = 0 ⇒ D = 0 ± 2i ∴ The solution is x = c1 cos 2t + c2 sin 2t. d4 y Ex. 2 : Solve – 16y = 0. dz2 d Sol. : Let D stand for. dz ∴ A.E. : D4 – 16 = 0, (D – 2) (D + 2) (D2 + 4) = 0. ∴ The solution is y = c1 e2z + c2 e–2z + c3 cos 2z + c4 sin 2z. Special Case : If the two real roots of φ(D) y = 0 be m and – m [e.g. D2 – m2 = 0], then the corresponding part of the solution is y = A emx + B e–mx OR y = A (cosh mx + sinh mx) + B (cosh mx – sinh mx) OR y = (A + B) cosh mx + (A – B) sinh mx ∴ y = c1 cosh mx + c2 sinh mx We note here that (in some particular cases) solution of D2 – m2 = 0 can be written as y = c1 emx + c2 e–mx or y = c1 cosh mx + c2 sinh mx. e.g. 1. (D – 1) y = 0 ⇒ y = c1 cosh x + c2 sinh x. 2 2. (D2 – 4) y = 0 ⇒ y = c1 cosh 2x + c2 sinh 2x. ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.7) LINEAR DIFFERENTIAL EQUATIONS… … EXERCISE 1.1 Solve the following Differential Equations d2 y dy d2y dy 1. –5 – 6y = 0. 2. 2 – – 10y = 0. dx2 dx dx2 dx Ans. y = c1 e–x + c2 e6x Ans. y = c1 e–2x + c2 e(5/2) x d3 y d2y dy 3. 3 + 2 + = 0. 4. (D4 – 2D3 + D2) y = 0. dx dx2 dx Ans. y = c1 + e–x (c2 x + c3) Ans. y = c1 x + c2 + (c3 x + c4) ex 5. (D6 – 6D5 + 12D4 – 6D3 – 9D2 + 12D – 4) y = 0. 6. (D3 + 6D2 + 11D + 6) y = 0. Ans. y = (c1x2 + c2x + c3) ex + (c4x + c5) e2x + c6e–x Ans. y = c1e–x + c2e–2x + c3e–3x d2x dx 7. 4y" – 8y' + 7y = 0. 8. +2 + 5x = 0, x (0) = 2, x'(0) = 0. dt2 dt   3   3  Ans. y = ex A cos  x + B sin  x Ans. x = e–t (2 cos 2t + sin 2t)   2   2  d2 s ds ds 9. = – 16 – 64 s, s = 0, = – 4 when t = 0. 10. (D3 + D2 – 2D + 12) y = 0. dt2 dt dt Ans. s = – 4e8t t Ans. y = c1 e–3x + ex [A cos 3 x + B sin 3x ] 11. (D2 + 1)3 (D2 + D + 1)2 y = 0.   3   3  Ans. y = (c1 + c2 x + c3 x2) cos x + (c4 + c5 x + c6 x2) sin x + e–x/2 (c7 + c8 x) cos  x + (c9 + c10 x) sin  x   2   2  d4 y 12. + m4y = 0. dx4 (mx/ 2) A cos mx + B sin mx + e– (mx/ 2) C cos mx + D sin mx Ans. y = e   2  2   2  2           d2s 3t 3t 13. 4 = – 9s. Ans. s = c1 sin + c2 cos dt2 2 2 d2y 14. The equation for the bending of a strut is EI + Py = 0. If y = 0 when dx2 P a sin x l EI x = 0 and y = a when x = , find y. Ans. y = 2 P l sin · EI 2 1.9 THE GENERAL SOLUTION OF THE LINEAR DIFFERENTIAL EQUATION φ(D) y = f(x) The general solution of the equation φ(D)y = f(x) can be written as y = yc + yp where, 1. yc is the solution of the given equation with f(x) = 0, that is of equation φ(D) y = 0 (which is known as Associated equation or Reduced equation) and is called the complimentary function (C.F.). It involves n arbitrary constants and is denoted by C.F. then φ(D) yc = 0. 2. yp is any function of x, which satisfies the equation φ(D) y = f(x), so that φ(D) yp = f(x) yp is called the particular integral and is denoted by P.I. It does not contain any arbitrary constant. Thus, on substituting y = yc + yp in φ(D) y, φ(D) [yc + yp] = φ(D) yc + φ(D) yp = 0 + f(x) = f(x) ∴ y = yc + yp satisfies the equation φ(D) y = f(x) and as it contains exactly n arbitrary constants, is the general (or complete) solution of the equation. Note : 1. The complete solution of φ(D) y = f(x) is y = C.F. + P.I. = yc + yp. 2. The general solution of φ(D) y = f(x) has arbitrary constants equal in number to the order of the differential equation. ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.8) LINEAR DIFFERENTIAL EQUATIONS… … 1 1.10 THE INVERSE OPERATOR AND THE SYMBOLIC EXPRESSION FOR THE PARTICULAR INTEGRAL φ(D) 1 We define f(x) as that function of x which, when acted upon by the differential operator φ(D) gives f(x). φ(D)  1   1  Thus by this definition, φ(D)  f(x) = f(x) and so  f(x) satisfies the equation φ(D) y = f(x) and so is the P.I. of the φ(D)  φ(D)  equation φ(D) y = f(x). Thus the P. I. of the equation φ(D) y = f(x) is symbolically given by; 1 P.I. = yp = f(x) φ(D) 1 e.g. 1. (D2 – 1) y = x2 ∴ yp = x2 D2 – 1 1 2. (D2 – 3D + 2) y = sin ex ∴ yp = sin ex. D2 – 3D + 2 1.11 METHODS OF OBTAINING PARTICULAR INTEGRAL 1 There are three methods to evaluate the particular integral yp = f(x). φ(D) (A) General method (B) Short-cut methods (C) Method of variation of parameters. Now we will discuss these methods in detail. (A) General Method This method is useful when the short-cut methods given in (B) are not applicable. This method involves integration. 1 1 (i) f(x) : By definition of the P. I., f(x) will be the P.I. of the equation (D – m) y = f(x) i.e. the part in the solution of D–m D–m dy this equation which does not contain the arbitrary constant. We have, – my = f(x) (linear) dx I.F. = e–mx and the general solution is y e–mx = ⌠ ⌡ f(x) · e–mx dx + c1 emx ⌠  ∴ y = (c1 emx) +  ⌡ e–mx f(x) · dx  i.e. y = yc + yp Here c1 emx is the C.F. and emx ⌠ –mx ⌡ e f(x) dx must be the P.I. ∴ yp = P. I. = 1 ⌠ f(x) = emx ⌡ e–mx f(x) dx D–m 1 f(x) = e–mx ⌠ emx f(x) dx Similarly, yp = P. I. = D+m ⌡ Put m = 0 yp = 1 ⌠ f(x) = ⌡ f(x) dx D ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.9) LINEAR DIFFERENTIAL EQUATIONS… … Also, yp = 1 f(x) = 1  1 f(x) D2 D D  ⌠  ⌠  ⌡ f(x) dx = ⌠ 1 = ⌡ ⌡ f(x) dx dx D     1 ⌠   ∴ yp = D2 f(x) = ⌡ ⌠ ⌡ f(x) dx dx  1 ⌠ ⌠ ⌠   Similarly, yp = f(x) = ⌡ ⌡ ⌡ f(x) dx dx dx … and so on. D3     1 (ii) f(x) : (D – m1) (D – m2) 1 1 mx ⌠ – m2x yp = (D – m1) (D – m2) f(x) = (D – m1) e 2 ⌡ e f(x) dx m1x ⌠ –m1x  mx⌠ – m2 x  yp = e ⌡ e e 2 ⌡ e f(x) dx dx   (iii) Use of Partial Fraction : 1 yp = f(x) (D – m1) (D – m2) = 1  1 – 1  f(x) (m1 – m2) D – m1 D – m2 1  1 1  =  f(x) – f(x) m1 – m2  D – m1 D – m2  1  mx⌠ m x ⌠ –m x  yp = e 1 ⌡ e 1 f(x) dx – e 2 ⌡ e 2 f(x) dx –m x m1 – m2   ILLUSTRATIONS ON GENERAL METHOD d2y dy x Ex. 1 : Solve 2 + 3 + 2y = ee (Dec. 2010, 2016, 2017; May 2011) dx dx Sol. : For C.F., A.E. is D2 + 3D + 2 = 0 ⇒ (D + 2) (D + 1) = 0. Hence D = –1, –2 and C.F. = c1 e–x + c2 e–2x 1 x Here P. I. = yp = (ee ) (D + 2) (D + 1) = 1  1 eex D + 2 D + 1   ⌠  1 e–x  x  ex ee dx = D+2  ⌡  [put ex = t ∴ ex dx = dt]  ⌠  = 1 e–x  t e dt  D+2  ⌡  x 1 = D+2 [e –x ee]=e x –2x ⌠ 2x –x e ⌡e e e dx x x ⌠ ex ee dx = e–2x ee P. I. = e–2x ⌡ Hence the complete solution will be x y = c1 e–x + c2 e–2x + e–2x ee ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.10) LINEAR DIFFERENTIAL EQUATIONS… … d2y dy 1 Ex. 2 : Solve + =. (Dec. 2008, 2010, 2012; May 2012) dx2 dx 1 + ex 1 d Sol. : We have (D2 + D) y = , here D ≡ 1 + ex dx For C.F., A.E. ⇒ D (D + 1) = 0 ∴ D = 0, – 1. ∴ C.F. = yc = c1 + c2 e–x P.I. = 1  1  D (D + 1) 1 + ex =  1 – 1   1 x by partial fraction D D + 1 1 + e  1  1  1  1  = – D 1 + ex D + 1 1 + ex ⌠ 1 dx ⌠ ex 1 + ex dx – e–x ⌡ = ⌡ 1 + ex ⌠ ex dx x – e–x ⌠ x dx putx 1 + e = t = ⌡ e (1 + ex) x ⌡ e 1 + ex  e dx = dt  ⌠ dt – e–x ⌠ dt = ⌡ t (t – 1) ⌡ t ⌠  1 – 1 dt – e–x log (ex + 1) = ⌡ t – 1 t  = log (t – 1) – log t – e–x log (ex + 1) = log (ex) – log (1 + ex) – e–x log (ex + 1) = x – log (1 + ex) – e–x log (ex + 1) Hence the complete solution is y = c1 + c2 e–x + x – log (1 + ex) – e–x log (1 + ex) Ex. 3 : Solve (D + 5D + 6) y = e sec2 x (1 + 2 tan x) 2 –2x (Dec. 2005, May 2008, 2011) Sol. : A.E. is D2 + 5D + 6 = 0 gives (D + 2) (D + 3) = 0 ⇒ D = – 2, – 3 C.F. = c1 e–2x + c2 e–3x 1 P.I. = [e–2x sec2 x (1 + 2 tan x)] (D + 3) (D + 2) 1  –2x ⌠  = e ⌡ e2x · e–2x sec2 x (1 + 2 tan x) dx D+3   1  –2x ⌠  = e ⌡ sec2 x (1 + 2 tan x) dx put tan x = t, sec2 x dx = dt D+3   1  –2x ⌠  1 = e ⌡ (1 + 2t) dt = [e–2x (t + t2)] D+3   D+3 1 = [e–2x (tan x + tan2 x)] D+3 = e–3x ⌠ ⌡ e3x · e–2x [(tan x – 1) + sec2 x] dx ⌠ ex [(tan x – 1) + sec2 x] dx = e–3x ⌡ = e–3x [ex (tan x – 1)] ⌠ ‡⌡ ex [f(x) + f'(x)] dx = ex f(x) = e–2x (tan x – 1) Hence the complete solution is y = c2 e–3x + e–2x [c1 + tan x – 1] = c2 e–3x + e–2x [c3 + tan x] ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.11) LINEAR DIFFERENTIAL EQUATIONS… … d2y Ex. 4 : Solve + 9y = sec 3x (May 2008) dx2 Sol. : A.E. is D2 + 9 = 0, or D = ± 3i C.F. = c1 cos 3x + c2 sin 3x and 1 1 P.I. = (sec 3x) = sec 3x D2 + 9 (D + 3i) (D – 3i) = 1  1 – 1  sec 3x 6i D – 3i D + 3i 1 1 1 1 = sec 3x – sec 3x … (1) 6i D – 3i 6i D + 3i 1 sec 3x = e3ix ⌠ e–3ix sec 3x dx = e3ix ⌠ cos 3x – i sin 3x Now, D – 3i ⌡ ⌡ cos 3x dx = e3ix ⌠ ⌡ [1 – i tan 3x] dx = e3ix x +  i log (cos 3x)   3  Changing i to – i in this, we have (sec 3x) = e–3ix x – log (cos 3x) 1 i D + 3i  3  Putting values in (1), we have P.I. = 1 e3ix x + i log (cos 3x) – e–3ix x – i log (cos 3x) 6i   3   3  x 3ix e3ix log (cos 3x) x e–3ix e–3ix log (cos 3x) = ·e + – + 6i 18 6i 18 Combining the like terms, we get e 3ix x – e–3ix 1 e 3ix + e–3ix = + log (cos 3x) 3  2i  9  2  x 1 P.I. = sin 3x + cos 3x log (cos 3x) 3 9 Hence the general solution will be x 1 y = c1 cos 3x + c2 sin 3x + sin 3x + cos 3x · log (cos 3x) 3 9 d2 y dy 1 1 Ex. 5 : Solve – – 2y = 2 log x + + 2 (May 2006, Dec. 2012) dx2 dx x x 1 1 Sol. : (D2 – D – 2) y = 2 log x + + 2 x x A.E. : D2 – D – 2 = 0 ∴ (D – 2) (D + 1) = 0 ∴ yc = c1 e2x + c2 e–x yp = 1 2 log x + 1 + 12 (D – 2) (D + 1)  x x 1 e–x ⌠  1 1   ⌡ = ex 2 log x + + dx D–2  x x2  1 e–x ⌠  2 1 1   ⌡ = ex 2 log x + – +  dx D–2  x x x2  ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.12) LINEAR DIFFERENTIAL EQUATIONS… …  –x ⌠ x   e ⌡ e 2 log x –  +  + 2 dx 1 1 2 1 = D–2   x   x x   = 1  e–x · ex 2 log x – 1  = 1 2 log x – 1 D–2  x  D – 2  x –2x  1 = e2x ⌠ ⌡ e 2 log x – x  dx  1  = e2x ⌠ ⌡ 2 log x · e dx – ⌠ –2x ⌡ e –2x dx  x   e–2x 2 e–2x  = e2x 2 log x   – ⌠ 1 ⌡ · dx – ⌠ ⌡ e–2x · dx   –2  x  –2  x   1 1  ⌠ e–2x = e2x – log x · e–2x + ⌡ x ⌠ e–2x x dx dx – ⌡   = e2x {– log x e–2x} = – log x ∴ y = C.F. + P.I. = yc + yp y = c1 e2x + c2 e–x – log x Ex. 6 : Solve (D2 – 1) y = e–x sin e–x + cos e–x. Sol. : AE : D2 – 1 = 0 or (D – 1) (D + 1) = 0 ⇒ D = – 1, + 1 ∴ yc = c1 ex + c2 e–x 1 yp = (e–x sin e–x + cos e–x) (D – 1) (D + 1) 1  –x ⌠    = e ⌡ ex (cos e–x + e–x sin e–x) dx Use ⌠ x x ⌡ e [f + f'] dx = e ·f D–1     1 1 = {e–x · ex cos e–x} = · cos e–x D–1 D–1 = ex ⌠ ⌡ e cos e dx = – e ⌠ –x –x x –x –x ⌡ cos e (– e dx) {Use e–x = t} = – ex sin e–x ∴ y = c1 ex + c2 e–x – ex sin e–x. Ex. 7 : Solve (D2 – 1) y = (1 + e–x)–2. Sol. : A.E. : D2 – 1 = 0 C.F. = c1 ex + c2 e–x 1 P.I. = (1 + e–x)–2 (D + 1) (D – 1) 1 = D+1 ex ⌠ –x –x –2 ⌡ e (1 + e ) dx 1 –1 = (– ex) ⌠ –x –2 –x x –x –1 ⌡ (1 + e ) (– e dx) = D + 1 [– e (1 + e ) ] D+1 ex · ex = e–x ⌠ ⌡ 1 + e–x dx e2x · (ex dx) = e–x ⌠ ⌡ 1 + ex (1 + ex = t) (t – 1)2 2 t – 2t + log t = e–x ⌠ ⌡ t dt = e–x 2  (1 + e ) x 2 = e–x – 2 (1 + ex) + log (1 + ex)   2  ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.13) LINEAR DIFFERENTIAL EQUATIONS… … e–x ∴ y = c1 ex + c2 e–x + (1 + ex)2 + e–x log (1 + ex) – 2e–x – 2 2 x 2 or y = A ex + B e–x + e–x (1 + e ) + log (1 + ex) – 2  2  x Ex. 8 : Solve (D2 + 3D + 2) y = e + cos ex. e (Dec. 2007) 2 Sol. : A.E. : D + 3D + 2 = (D + 2) (D + 1) = 0 C.F. = c1 e–2x + c2 e–x 1 x P.I. = (ee + cos ex) (D + 2) (D + 1) 1 x = e–x ⌠ x e x ⌡ e (e + cos e ) dx D+2 1 x = e–x (ee + sin ex) D+2 x = e–2x ⌠ 2x –x e x ⌡ e e (e + sin e ) dx x = e–2x ⌠ x e x ⌡ e (e + sin e ) dx x = e–2x (ee – cos ex) x ∴ y = c1 e–2x + c2 e–x + e–2x (ee – cos ex) Ex. 9 : Solve (D2 + 3D + 2) y = sin ex. (May 2012, Dec. 2012) Sol. : A.E. : D2 + 3D + 2 = (D + 2) (D + 1) = 0 ⇒ D = – 2, – 1. C.F. = c1 e–2x + c2 e–x 1 1 P.I. = sin ex = e–x ⌠ ⌡ ex sin ex dx (D + 2) (D + 1) D+2 1 = e–x (– cos ex) = – e–2x ⌠ ⌡ e cos e dx x x D+2 = – e–2x sin ex ∴ y = c1 e–2x + c2 e–x – e–2x sin ex d2y Ex. 10 : Solve + y = cosec x. dx2 Sol. : A.E. : D2 + 1 = (D + i) (D – i) = 0 ⇒ D = ± i. C.F. = c1 cos x + c2 sin x 1 1  1 1  P.I. = 2 cosec x = – cosec x D +1 2i D – i D + i 1  1 cosec x 1 = cosec x – 2i D – i D+i  1  ix  = e ⌠ ⌡ e–ix cosec x dx – e–ix ⌠ ⌡ e cosec x dx ix 2i   1  ix  = e ⌠ ⌡ ⌠ (cos x + i sin x) cosec x dx (cos x – i sin x) cosec x dx – e–ix ⌡ 2i   1  ix  e ⌠ (cot x – i) dx – e–ix ⌠ 2i  ⌡ ⌡ (cot x + i) dx = 1 = [eix (log sin x – ix) – e–ix (log sin x + ix)] 2i 1 = [log sin x) (eix – e–ix) – ix (eix + e–ix)] 2i = sin x log sin x – x cos x ∴ y = c1 cos x + c2 sin x + sin x log sin x – x cos x ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.14) LINEAR DIFFERENTIAL EQUATIONS… … (B) Short-cut Methods for Finding P.I. in Certain Standard Cases Although the general method (A) discussed in the previous article will always work in the theory, it many a times leads to laborious and difficult integration. To avoid this, short methods of finding P.I. without actual integration are developed depending upon the particular form of function f(x). Case I : P.I. when f(x) = eax, a is any constant. 1 To obtain yp = eax, we have D eax = a eax, D2 eax = a2 eax …… Dn ean = an eax φ(D) ∴ (a0 Dn + a1 Dn–1 + …… + an) eax = (a0 an + a1 an–1 + …… + an) eax or φ (D) eax = φ (a) eax 1 Operating on both sides by , we have φ (D) 1 1 [φ (D) eax] = [φ (a) eax] φ (D) φ (D) 1 1 or eax = φ (a) (eax) , (‡ is a linear operator) φ (D) φ (D) Dividing by φ (a), we have the formula 1 1 eax = eax provided φ (a) ≠ 0 … (A) φ (D) φ (a) Case of Failure : If φ (a) = 0, above rule fails and we proceed as under. Since φ (a) = 0, D – a must be a factor of φ (D) (by Factor Theorem). Let φ (D) = (D – a) ψ (D) where, ψ (a) ≠ 0. Then 1 1 1 (eax) = eax φ (D) D – a ψ (D) 1 eax = … from (A) D – a ψ (a) 1 1 = eax ψ (a) D – a 1 = ψ (a) ⌠ e–ax eax dx eax ⌡ … (refer 1.11-A (i)) 1 = eax ⌠ ⌡ dx ψ (a) 1 = x· eax, where ψ (a) = φ'(a) ≠ 0. ψ (a) 1 1 i.e. eax = x · eax provided φ'(a) ≠ 0 … (B) φ (D) φ' (a) If φ' (a) = 0 then we shall apply (B) again to get 1 1 (eax) = x2 eax, provided φ"(a) ≠ 0 , and so on. φ (D) φ" (a) Remark 1 : Since φ (D) = (D – a) ψ (D) φ'(D) = (D – a) ψ'(D) + ψ(D) ∴ φ'(a) = 0 + ψ (a) or φ'(a) = ψ (a) ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.15) LINEAR DIFFERENTIAL EQUATIONS… … Remark 2 : It can also be established that 1 1 xr ax r eax = e , provided ψ(a) ≠ 0. (D – a) ψ(D) ψ(a) r! Remark 3 : Any constant k can be expressed as k = k · e0x 1 1 1 ∴ yp = (k) = k · e0x = k · e0x φ(D) φ(D) φ(D) 1 = k· , φ(0) ≠ 0 φ(0) Remark 4 : If f(x) = ax then we use ax = ex log a 1 1 ∴ yp = ax = ex log a φ(D) φ(D) 1 = ax Replace D with log a. φ (log a) –x –x x log 1/a x (–log a) If f(x) = a then we use a = e =e 1 1 ∴ yp = a–x = ex (–log a) φ(D) φ(D) 1 = a–x. Replace D with – log a. φ(– log a) Formulae for Ready Reference : 1 1 x2 ax 1 x3 ax 1 xr ax 1. eax = x · eax 2. 2 e ax = e 3. 3 e ax = e 4. r e ax = e D–a (D – a) 2! (D – a) 3! (D – a) r! 1 1 1 1 xr ax 5. r eax = r e ax = e , ψ(a) ≠ 0 (D – a) ψ (D) ψ(a) (D – a) ψ(a) r! ILLUSTRATIONS Ex. 1 : Find the Particular Integral of (D2 – 5D + 6) y = 3 e5x. 3 3 e5x e5x Sol. : P.I. = 2 (e5x) = 2 = D – 5D + 6 5 – 5.5 + 6 2 d2y dy Ex. 2 : Find the Particular Integral of +4 + 3y = e–3x dx2 dx 1 Sol. : Here P.I. = (e–3x) , φ (D) = D2 + 4D + 3 D2 + 4D + 3 But φ (– 3) = 9 – 12 + 3 = 0 hence φ (–3) = 0 and case I fails. x e–3x 1 ∴ P.I. = = x· · e–3x, D → a = – 3 φ' (a) 2D + 4 x e–3x x e–3x = = 2 (–3) + 4 –2 3 Ex. 3 : Find the Particular Integral of (D – 1)3 y = ex + 2x –. 2 1 1 3 1 Sol. : yp = ex + 2x – e0x (D – 1)3 (D – 1)3 2 (D – 1)3 x3 x 1 3 1 = e + (log 2 – 1)3 2x – 3! 2 (0 – 1)3 x3 x 1 3 = e + 2x + 6 (log 2 – 1)3 2 ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.16) LINEAR DIFFERENTIAL EQUATIONS… … Ex. 4 : Find the Particular Integral of (D – 2)2 (D + 1) y = e2x + 2–x 1 1 Sol. : yp = e2x + 2–x (D – 2)2 (D + 1) (D – 2)2 (D + 1) 1 1 2–x = 2 · e2x + (D – 2) (2 + 1) (– log 2 – 2)2 (– log 2 + 1) 1 x2 2x 2–x = · e + 3 2! (– log 2 – 2)2 (– log 2 + 1) Case II : P.I. when f(x) = sin (ax + b) or cos (ax + b). 1 1 To obtain yp = sin (ax + b) or cos (ax + b), we have φ(D2) φ(D2) D sin (ax + b) = a cos (ax + b) D2 sin (ax + b) = – a2 sin (ax + b) D3 sin (ax + b) = – a3 cos (ax + b) D4 sin (ax + b) = a4 sin (ax + b) or (D2)2 sin (ax + b) = (–a2)2 sin (ax + b) Similarly (D2)p sin (ax + b) = (– a2)p sin (ax + b) and we may generalise that φ (D2) sin (ax + b) = φ (– a2) sin (ax + b) 1 Operating on both sides by , we have φ (D2) 1 1 [φ (D2) sin (ax + b)] = [φ (–a2) sin (ax + b)] φ (D2) φ (D2) 1 sin (ax + b) = φ (–a2) sin (ax + b) φ (D2) Dividing now by φ (–a2), we have 1 1 sin (ax + b) = sin (ax + b), provided φ (–a2) ≠ 0 φ (D2) φ (–a2) Case of Failure : But if φ (–a2) = 0, above rule fails and we proceed as under : We know by Euler's Theorem that cos (ax + b) + i sin (ax + b) = ei (ax + b) hence 1 1 sin (ax + b) = Imag. Part of ei (ax + b) φ (D2) φ (D2) 1 = I.P. of ei (ax + b) φ (D2) 1 = I.P. of x ei (ax + b) , (D2 = – a2) φ'(D2) 1 1 Hence sin (ax + b) = x sin (ax + b) provided φ' (–a2) ≠ 0 φ (D2) φ' (–a2) Add if φ' (– a2) ≠ 0, we have 1 1 sin (ax + b) = x2 sin (ax + b), provided φ"(–a2) ≠ 0 φ(D2) φ"(–a )2 Similarly formulae for cos (ax + b) viz. 1 1 cos (ax + b) = cos (ax + b), provided φ (–a2) ≠ 0 φ (D2) φ (–a2) ENGINEERING MATHEMATICS – III (Comp. Engg. and IT Group) (S-II) (1.17) LINEAR DIFFERENTIAL EQUATIONS… … But if φ (–a2) = 0, we have 1 1 cos (ax + b) = x cos (ax + b), provided φ' (–a2 ) ≠ 0 φ (D2) φ' (–a2) And if φ' (–a2) = 0, we have 1 1 cos (ax + b) = x2 cos (ax + b), provided φ" (–a2) ≠ 0 φ (D2) φ" (–a2) and so on and so forth. Additional Results : 1 1 sin ax = sin ax , φ(–a2) ≠ 0 (Replace D2 with – a2) φ(D2) φ(–a2) 1 1 cos ax = cos ax , φ(– a2) ≠ 0, (Replace D2 with – a2) φ(D2) φ(–a2) For the case of failure, it can also be established that 1 x sin (ax + b) = – cos (ax + b) D2 + a2 2a 1 x cos (ax + b) = sin (ax + b) D2 + a2 2a 1 sin (ax + b) =  –x r 1  sin ax + b + rπ (D2 + a2)r 2a r!  2 r 1 cos (ax + b) = – x 1 cos ax + b +  rπ (D + a2)r 2  2a  r!  2 Useful Formulae : 1 – cos 2x eox cos 2x sin2 x = = – 2 2 2 1 + cos 2x eox cos 2x cos2 x = = + 2 2 2 1 sin A sin B = [cos (A – B) – cos (A + B)] 2 1 sin A cos B = [sin (A + B) + sin (A – B)]