Higher Engineering Mathematics

In memory of Elizabeth

Higher Engineering Mathematics Sixth Edition John Bird, BSc (Hons), CMath, CEng, CSci, FIMA, FIET, MIEE, FIIE, FCollT

AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Newnes is an imprint of Elsevier

Newnes is an imprint of Elsevier The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, UK 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA First edition 2010 Copyright © 2010, John Bird, Published by Elsevier Ltd. All rights reserved. The right of John Bird to be identiﬁed as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. No part of this publication 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 written permission of the publisher. Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: [emailprotected]. Alternatively you can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material. Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Because of rapid advances in the medical sciences, in particular, independent veriﬁcation of diagnoses and drug dosages should be made. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. Library of Congress Cataloging-in-Publication Data A catalogue record for this book is available from the Library of Congress. ISBN: 978-1-85-617767-2

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Typeset by: diacriTech, India Printed and bound in China 10 11 12 13 14 15 10 9 8 7 6 5 4 3 2 1

Contents Preface

xiii

Syllabus guidance

xv

1

1 1 1 3 6 8 10

2

3

4

Algebra 1.1 Introduction 1.2 Revision of basic laws 1.3 Revision of equations 1.4 Polynomial division 1.5 The factor theorem 1.6 The remainder theorem Partial fractions 2.1 Introduction to partial fractions 2.2 Worked problems on partial fractions with linear factors 2.3 Worked problems on partial fractions with repeated linear factors 2.4 Worked problems on partial fractions with quadratic factors

17

Logarithms 3.1 Introduction to logarithms 3.2 Laws of logarithms 3.3 Indicial equations 3.4 Graphs of logarithmic functions

20 20 22 24 25

Exponential functions 4.1 Introduction to exponential functions 4.2 The power series for e x 4.3 Graphs of exponential functions 4.4 Napierian logarithms 4.5 Laws of growth and decay 4.6 Reduction of exponential laws to linear form

27 27 28 29 31 34

Revision Test 1

5

Hyperbolic functions 5.1 Introduction to hyperbolic functions 5.2 Graphs of hyperbolic functions 5.3 Hyperbolic identities 5.4 Solving equations involving hyperbolic functions 5.5 Series expansions for cosh x and sinh x

13 13

6

7

13 16

37 40

41 41 43 45 47 49

Arithmetic and geometric progressions 6.1 Arithmetic progressions 6.2 Worked problems on arithmetic progressions 6.3 Further worked problems on arithmetic progressions 6.4 Geometric progressions 6.5 Worked problems on geometric progressions 6.6 Further worked problems on geometric progressions The binomial series 7.1 Pascal’s triangle 7.2 The binomial series 7.3 Worked problems on the binomial series 7.4 Further worked problems on the binomial series 7.5 Practical problems involving the binomial theorem Revision Test 2

8

Maclaurin’s series 8.1 Introduction 8.2 Derivation of Maclaurin’s theorem 8.3 Conditions of Maclaurin’s series 8.4 Worked problems on Maclaurin’s series 8.5 Numerical integration using Maclaurin’s series 8.6 Limiting values 9 Solving equations by iterative methods 9.1 Introduction to iterative methods 9.2 The bisection method 9.3 An algebraic method of successive approximations 9.4 The Newton-Raphson method

10 Binary, octal and hexadecimal 10.1 Introduction 10.2 Binary numbers 10.3 Octal numbers 10.4 Hexadecimal numbers Revision Test 3

51 51 51 52 54 55 56 58 58 59 59 62 64 67 68 68 68 69 69 73 74 77 77 77 81 84 87 87 87 90 92 96

vi Contents 11 Introduction to trigonometry 11.1 Trigonometry 11.2 The theorem of Pythagoras 11.3 Trigonometric ratios of acute angles 11.4 Evaluating trigonometric ratios 11.5 Solution of right-angled triangles 11.6 Angles of elevation and depression 11.7 Sine and cosine rules 11.8 Area of any triangle 11.9 Worked problems on the solution of triangles and ﬁnding their areas 11.10 Further worked problems on solving triangles and ﬁnding their areas 11.11 Practical situations involving trigonometry 11.12 Further practical situations involving trigonometry

97 97 97 98 100 105 106 108 108

12 Cartesian and polar co-ordinates 12.1 Introduction 12.2 Changing from Cartesian into polar co-ordinates 12.3 Changing from polar into Cartesian co-ordinates 12.4 Use of Pol/Rec functions on calculators

117 117

119 120

13 The circle and its properties 13.1 Introduction 13.2 Properties of circles 13.3 Radians and degrees 13.4 Arc length and area of circles and sectors 13.5 The equation of a circle 13.6 Linear and angular velocity 13.7 Centripetal force

122 122 122 123 124 127 129 130

Revision Test 4

109

15.5 Worked problems (ii) on trigonometric equations 15.6 Worked problems (iii) on trigonometric equations 15.7 Worked problems (iv) on trigonometric equations 16 The relationship between trigonometric and hyperbolic functions 16.1 The relationship between trigonometric and hyperbolic functions 16.2 Hyperbolic identities

156 157 157

159 159 160

110 111 113

117

133

14 Trigonometric waveforms 14.1 Graphs of trigonometric functions 14.2 Angles of any magnitude 14.3 The production of a sine and cosine wave 14.4 Sine and cosine curves 14.5 Sinusoidal form A sin(ωt ± α) 14.6 Harmonic synthesis with complex waveforms

134 134 135 137 138 143

15 Trigonometric identities and equations 15.1 Trigonometric identities 15.2 Worked problems on trigonometric identities 15.3 Trigonometric equations 15.4 Worked problems (i) on trigonometric equations

152 152

146

152 154 154

17 Compound angles 17.1 Compound angle formulae 17.2 Conversion of a sinωt + b cosωt into R sin(ωt + α) 17.3 Double angles 17.4 Changing products of sines and cosines into sums or differences 17.5 Changing sums or differences of sines and cosines into products 17.6 Power waveforms in a.c. circuits Revision Test 5

163 163 165 169 170 171 173 177

18 Functions and their curves 18.1 Standard curves 18.2 Simple transformations 18.3 Periodic functions 18.4 Continuous and discontinuous functions 18.5 Even and odd functions 18.6 Inverse functions 18.7 Asymptotes 18.8 Brief guide to curve sketching 18.9 Worked problems on curve sketching

178 178 181 186 186 186 188 190 196 197

19 Irregular areas, volumes and mean values of waveforms 19.1 Areas of irregular ﬁgures 19.2 Volumes of irregular solids 19.3 The mean or average value of a waveform

203 203 205 206

Revision Test 6

20 Complex numbers 20.1 Cartesian complex numbers 20.2 The Argand diagram 20.3 Addition and subtraction of complex numbers 20.4 Multiplication and division of complex numbers

212

213 213 214 214 216

vii

Contents 20.5 20.6 20.7 20.8

Complex equations The polar form of a complex number Multiplication and division in polar form Applications of complex numbers

217 218 220 221

21 De Moivre’s theorem 21.1 Introduction 21.2 Powers of complex numbers 21.3 Roots of complex numbers 21.4 The exponential form of a complex number

225 225 225 226

22 The theory of matrices and determinants 22.1 Matrix notation 22.2 Addition, subtraction and multiplication of matrices 22.3 The unit matrix 22.4 The determinant of a 2 by 2 matrix 22.5 The inverse or reciprocal of a 2 by 2 matrix 22.6 The determinant of a 3 by 3 matrix 22.7 The inverse or reciprocal of a 3 by 3 matrix

231 231

23 The solution of simultaneous equations by matrices and determinants 23.1 Solution of simultaneous equations by matrices 23.2 Solution of simultaneous equations by determinants 23.3 Solution of simultaneous equations using Cramers rule 23.4 Solution of simultaneous equations using the Gaussian elimination method Revision Test 7

24 Vectors 24.1 24.2 24.3 24.4 24.5 24.6 24.7 24.8 24.9

Introduction Scalars and vectors Drawing a vector Addition of vectors by drawing Resolving vectors into horizontal and vertical components Addition of vectors by calculation Vector subtraction Relative velocity i, j and k notation

25 Methods of adding alternating waveforms 25.1 Combination of two periodic functions 25.2 Plotting periodic functions 25.3 Determining resultant phasors by drawing

228

231 235 235 236 237 239

25.4 Determining resultant phasors by the sine and cosine rules 268 25.5 Determining resultant phasors by horizontal and vertical components 270 25.6 Determining resultant phasors by complex numbers 272 26 Scalar and vector products 26.1 The unit triad 26.2 The scalar product of two vectors 26.3 Vector products 26.4 Vector equation of a line Revision Test 8

275 275 276 280 283 286

27 Methods of differentiation 27.1 Introduction to calculus 27.2 The gradient of a curve 27.3 Differentiation from ﬁrst principles 27.4 Differentiation of common functions 27.5 Differentiation of a product 27.6 Differentiation of a quotient 27.7 Function of a function 27.8 Successive differentiation

287 287 287 288 289 292 293 295 296

28 Some applications of differentiation 28.1 Rates of change 28.2 Velocity and acceleration 28.3 Turning points 28.4 Practical problems involving maximum and minimum values 28.5 Tangents and normals 28.6 Small changes

299 299 300 303

29 Differentiation of parametric equations 29.1 Introduction to parametric equations 29.2 Some common parametric equations 29.3 Differentiation in parameters 29.4 Further worked problems on differentiation of parametric equations

315 315 315 315

254 255 260 262 263

30 Differentiation of implicit functions 30.1 Implicit functions 30.2 Differentiating implicit functions 30.3 Differentiating implicit functions containing products and quotients 30.4 Further implicit differentiation

320 320 320 321 322

265 265 265 267

31 Logarithmic differentiation 31.1 Introduction to logarithmic differentiation 31.2 Laws of logarithms 31.3 Differentiation of logarithmic functions

325 325 325 325

241 241 243 247 248

307 311 312

250

251 251 251 251 252

318

viii Contents 31.4 Differentiation of further logarithmic functions 31.5 Differentiation of [ f (x)]x Revision Test 9 32 Differentiation of hyperbolic functions 32.1 Standard differential coefﬁcients of hyperbolic functions 32.2 Further worked problems on differentiation of hyperbolic functions 33 Differentiation of inverse trigonometric and hyperbolic functions 33.1 Inverse functions 33.2 Differentiation of inverse trigonometric functions 33.3 Logarithmic forms of the inverse hyperbolic functions 33.4 Differentiation of inverse hyperbolic functions 34 Partial differentiation 34.1 Introduction to partial derivatives 34.2 First order partial derivatives 34.3 Second order partial derivatives 35 Total differential, rates of change and small changes 35.1 Total differential 35.2 Rates of change 35.3 Small changes 36 Maxima, minima and saddle points for functions of two variables 36.1 Functions of two independent variables 36.2 Maxima, minima and saddle points 36.3 Procedure to determine maxima, minima and saddle points for functions of two variables 36.4 Worked problems on maxima, minima and saddle points for functions of two variables 36.5 Further worked problems on maxima, minima and saddle points for functions of two variables Revision Test 10 37 Standard integration 37.1 The process of integration 37.2 The general solution of integrals of the form ax n 37.3 Standard integrals 37.4 Deﬁnite integrals

326 328 330 331 331 332 334 334 334

38 Some applications of integration 38.1 Introduction 38.2 Areas under and between curves 38.3 Mean and r.m.s. values 38.4 Volumes of solids of revolution 38.5 Centroids 38.6 Theorem of Pappus 38.7 Second moments of area of regular sections

375 375 375 377 378 380 381

39 Integration using algebraic substitutions 39.1 Introduction 39.2 Algebraic substitutions 39.3 Worked problems on integration using algebraic substitutions 39.4 Further worked problems on integration using algebraic substitutions 39.5 Change of limits

392 392 392

383

392 394 395

339 341 345 345 345 348 351 351 352 354 357 357 358

359

359

361 367 368 368 368 369 372

Revision Test 11 40 Integration using trigonometric and hyperbolic substitutions 40.1 Introduction 40.2 Worked problems on integration of sin2 x, cos2 x, tan2 x and cot2 x 40.3 Worked problems on powers of sines and cosines 40.4 Worked problems on integration of products of sines and cosines 40.5 Worked problems on integration using the sin θ substitution 40.6 Worked problems on integration using tan θ substitution 40.7 Worked problems on integration using the sinh θ substitution 40.8 Worked problems on integration using the cosh θ substitution

397

398 398 398 400 401 402 404 404 406

41 Integration using partial fractions 41.1 Introduction 41.2 Worked problems on integration using partial fractions with linear factors 41.3 Worked problems on integration using partial fractions with repeated linear factors 41.4 Worked problems on integration using partial fractions with quadratic factors

412

θ 42 The t = tan substitution 2 42.1 Introduction

414 414

θ 42.2 Worked problems on the t = tan 2 substitution

409 409 409

411

415

Contents θ 42.3 Further worked problems on the t = tan 2 substitution Revision Test 12

416 419

43 Integration by parts 43.1 Introduction 43.2 Worked problems on integration by parts 43.3 Further worked problems on integration by parts

420 420 420

44 Reduction formulae 44.1 Introduction 44.2 Using reduction formulae for integrals of the form x n e x dx 44.3 Using reduction formulae for integrals of the form x n cos x dx and x n sin x dx 44.4 Using reduction formulae for integrals of the form sinn x dx and cosn x dx 44.5 Further reduction formulae

426 426

429 432

45 Numerical integration 45.1 Introduction 45.2 The trapezoidal rule 45.3 The mid-ordinate rule 45.4 Simpson’s rule

435 435 435 437 439

Revision Test 13

46 Solution of ﬁrst order differential equations by separation of variables 46.1 Family of curves 46.2 Differential equations 46.3 The solution of equations of the form dy = f (x) dx 46.4 The solution of equations of the form dy = f (y) dx 46.5 The solution of equations of the form dy = f (x) · f (y) dx 47 hom*ogeneous ﬁrst order differential equations 47.1 Introduction 47.2 Procedure to solve differential equations dy =Q of the form P dx 47.3 Worked problems on hom*ogeneous ﬁrst order differential equations 47.4 Further worked problems on hom*ogeneous ﬁrst order differential equations

422

426 427

443

444 444 445 445 447 449 452 452 452 452 454

48 Linear ﬁrst order differential equations 48.1 Introduction 48.2 Procedure to solve differential equations dy + Py = Q of the form dx 48.3 Worked problems on linear ﬁrst order differential equations 48.4 Further worked problems on linear ﬁrst order differential equations 49 Numerical methods for ﬁrst order differential equations 49.1 Introduction 49.2 Euler’s method 49.3 Worked problems on Euler’s method 49.4 An improved Euler method 49.5 The Runge-Kutta method Revision Test 14 50 Second order differential equations of the form dy d2 y a 2 + b + cy= 0 dx dx 50.1 Introduction 50.2 Procedure to solve differential equations dy d2 y of the form a 2 + b + cy = 0 dx dx 50.3 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = 0 dx dx 50.4 Further worked problems on practical differential equations of the form dy d2 y a 2 + b + cy = 0 dx dx 51 Second order differential equations of the form dy d2 y a 2 + b + cy= f (x) dx dx 51.1 Complementary function and particular integral 51.2 Procedure to solve differential equations d2 y dy of the form a 2 + b + cy = f (x) dx dx 51.3 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is a constant or polynomial 51.4 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is an exponential function 51.5 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is a sine or cosine function

456 456 457 457 458

461 461 461 462 466 471 476

477 477 478

478

480

483

483 483

484

486

488

ix

x Contents 51.6 Worked problems on differential equations dy d2 y of the form a 2 + b + cy = f (x) dx dx where f (x) is a sum or a product 490 52 Power series methods of solving ordinary differential equations 52.1 Introduction 52.2 Higher order differential coefﬁcients as series 52.3 Leibniz’s theorem 52.4 Power series solution by the Leibniz–Maclaurin method 52.5 Power series solution by the Frobenius method 52.6 Bessel’s equation and Bessel’s functions 52.7 Legendre’s equation and Legendre polynomials 53 An introduction to partial differential equations 53.1 Introduction 53.2 Partial integration 53.3 Solution of partial differential equations by direct partial integration 53.4 Some important engineering partial differential equations 53.5 Separating the variables 53.6 The wave equation 53.7 The heat conduction equation 53.8 Laplace’s equation Revision Test 15

493 493 493 495 497 500 506 511 515 515 515

556 556 559

58 The normal distribution 58.1 Introduction to the normal distribution 58.2 Testing for a normal distribution

562 562 566

59 Linear correlation 59.1 Introduction to linear correlation 59.2 The product-moment formula for determining the linear correlation coefﬁcient 59.3 The signiﬁcance of a coefﬁcient of correlation 59.4 Worked problems on linear correlation

570 570

60 Linear regression 60.1 Introduction to linear regression 60.2 The least-squares regression lines 60.3 Worked problems on linear regression

575 575 575 576

Revision Test 17

570 571 571

581

516 518 518 519 523 525 528

54 Presentation of statistical data 54.1 Some statistical terminology 54.2 Presentation of ungrouped data 54.3 Presentation of grouped data

529 529 530 534

55 Measures of central tendency and dispersion 55.1 Measures of central tendency 55.2 Mean, median and mode for discrete data 55.3 Mean, median and mode for grouped data 55.4 Standard deviation 55.5 Quartiles, deciles and percentiles

541 541 541 542 544 546

56 Probability 56.1 Introduction to probability 56.2 Laws of probability 56.3 Worked problems on probability 56.4 Further worked problems on probability

548 548 549 549 551

Revision Test 16

57 The binomial and Poisson distributions 57.1 The binomial distribution 57.2 The Poisson distribution

554

61 Introduction to Laplace transforms 61.1 Introduction 61.2 Deﬁnition of a Laplace transform 61.3 Linearity property of the Laplace transform 61.4 Laplace transforms of elementary functions 61.5 Worked problems on standard Laplace transforms

582 582 582

62 Properties of Laplace transforms 62.1 The Laplace transform of eat f (t) 62.2 Laplace transforms of the form eat f (t) 62.3 The Laplace transforms of derivatives 62.4 The initial and ﬁnal value theorems

587 587 587 589 591

63 Inverse Laplace transforms 63.1 Deﬁnition of the inverse Laplace transform 63.2 Inverse Laplace transforms of simple functions 63.3 Inverse Laplace transforms using partial fractions 63.4 Poles and zeros

593 593

64 The solution of differential equations using Laplace transforms 64.1 Introduction 64.2 Procedure to solve differential equations by using Laplace transforms 64.3 Worked problems on solving differential equations using Laplace transforms

582 582 583

593 596 598 600 600 600 600

Contents 65 The solution of simultaneous differential equations using Laplace transforms 65.1 Introduction 65.2 Procedure to solve simultaneous differential equations using Laplace transforms 65.3 Worked problems on solving simultaneous differential equations by using Laplace transforms Revision Test 18 66 Fourier series for periodic functions of period 2π 66.1 Introduction 66.2 Periodic functions 66.3 Fourier series 66.4 Worked problems on Fourier series of periodic functions of period 2π 67 Fourier series for a non-periodic function over range 2π 67.1 Expansion of non-periodic functions 67.2 Worked problems on Fourier series of non-periodic functions over a range of 2π 68 Even and odd functions and half-range Fourier series 68.1 Even and odd functions

68.2 Fourier cosine and Fourier sine series 68.3 Half-range Fourier series

605 605

605

605 610

611 611 611 611 612 617 617 617 623 623

623 626

69 Fourier series over any range 69.1 Expansion of a periodic function of period L 69.2 Half-range Fourier series for functions deﬁned over range L

630

70 A numerical method of harmonic analysis 70.1 Introduction 70.2 Harmonic analysis on data given in tabular or graphical form 70.3 Complex waveform considerations

637 637 637 641

71 The complex or exponential form of a Fourier series 71.1 Introduction 71.2 Exponential or complex notation 71.3 The complex coefﬁcients 71.4 Symmetry relationships 71.5 The frequency spectrum 71.6 Phasors

644 644 644 645 649 652 653

Revision Test 19

630 634

658

Essential formulae

659

Index

675

xi

xii Contents

Website Chapters

72 Inequalities 72.1 Introduction to inequalities 72.2 Simple inequalities 72.3 Inequalities involving a modulus 72.4 Inequalities involving quotients 72.5 Inequalities involving square functions 72.6 Quadratic inequalities

1 1 1 2 3 4 5

73 Boolean algebra and logic circuits 73.1 Boolean algebra and switching circuits 73.2 Simplifying Boolean expressions 73.3 Laws and rules of Boolean algebra 73.4 De Morgan’s laws 73.5 Karnaugh maps 73.6 Logic circuits 73.7 Universal logic gates

7 7 12 12 14 15 19 23

Revision Test 20 74 Sampling and estimation theories 74.1 Introduction 74.2 Sampling distributions

28 29 29 29

74.3 The sampling distribution of the means 74.4 The estimation of population parameters based on a large sample size 74.5 Estimating the mean of a population based on a small sample size

29 33 38

75 Signiﬁcance testing 75.1 Hypotheses 75.2 Type I and Type II errors 75.3 Signiﬁcance tests for population means 75.4 Comparing two sample means

42 42 42 49 54

76 Chi-square and distribution-free tests 76.1 Chi-square values 76.2 Fitting data to theoretical distributions 76.3 Introduction to distribution-free tests 76.4 The sign test 76.5 Wilcoxon signed-rank test 76.6 The Mann-Whitney test

59 59 60 67 68 71 75

Revision Test 21

82

Preface This sixth edition of ‘Higher Engineering Mathematics’ covers essential mathematical material suitable for students studying Degrees, Foundation Degrees, Higher National Certiﬁcate and Diploma courses in Engineering disciplines. In this edition the material has been ordered into the following twelve convenient categories: number and algebra, geometry and trigonometry, graphs, complex numbers, matrices and determinants, vector geometry, differential calculus, integral calculus, differential equations, statistics and probability, Laplace transforms and Fourier series. New material has been added on logarithms and exponential functions, binary, octal and hexadecimal, vectors and methods of adding alternating waveforms. Another feature is that a free Internet download is available of a sample (over 1100) of the further problems contained in the book. The primary aim of the material in this text is to provide the fundamental analytical and underpinning knowledge and techniques needed to successfully complete scientiﬁc and engineering principles modules of Degree, Foundation Degree and Higher National Engineering programmes. The material has been designed to enable students to use techniques learned for the analysis, modelling and solution of realistic engineering problems at Degree and Higher National level. It also aims to provide some of the more advanced knowledge required for those wishing to pursue careers in mechanical engineering, aeronautical engineering, electronics, communications engineering, systems engineering and all variants of control engineering. In Higher Engineering Mathematics 6th Edition, theory is introduced in each chapter by a full outline of essential deﬁnitions, formulae, laws, procedures etc. The theory is kept to a minimum, for problem solving is extensively used to establish and exemplify the theory. It is intended that readers will gain real understanding through seeing problems solved and then through solving similar problems themselves. Access to software packages such as Maple, Mathematica and Derive, or a graphics calculator, will enhance understanding of some of the topics in this text.

Each topic considered in the text is presented in a way that assumes in the reader only knowledge attained in BTEC National Certiﬁcate/Diploma, or similar, in an Engineering discipline. ‘Higher Engineering Mathematics 6th Edition’ provides a follow-up to ‘Engineering Mathematics 6th Edition’. This textbook contains some 900 worked problems, followed by over 1760 further problems (with answers), arranged within 238 Exercises. Some 432 line diagrams further enhance understanding. A sample of worked solutions to over 1100 of the further problems has been prepared and can be accessed free via the Internet (see next page). At the end of the text, a list of Essential Formulae is included for convenience of reference. At intervals throughout the text are some 19 Revision Tests (plus two more in the website chapters) to check understanding. For example, Revision Test 1 covers the material in Chapters 1 to 4, Revision Test 2 covers the material in Chapters 5 to 7, Revision Test 3 covers the material in Chapters 8 to 10, and so on. An Instructor’s Manual, containing full solutions to the Revision Tests, is available free to lecturers adopting this text (see next page). Due to restriction of extent, ﬁve chapters that appeared in the ﬁfth edition have been removed from the text and placed on the website. For chapters on Inequalities, Boolean algebra and logic circuits, Sampling and estimation theories, Signiﬁcance testing and Chi-square and distribution-free tests (see next page). ‘Learning by example’ is at the heart of ‘Higher Engineering Mathematics 6th Edition’.

JOHN BIRD Royal Naval School of Marine Engineering, HMS Sultan, formerly University of Portsmouth and Highbury College, Portsmouth

xiv Preface Free web downloads Extra material available on the Internet at: www.booksite.elsevier.com/newnes/bird. It is recognised that the level of understanding of algebra on entry to higher courses is often inadequate. Since algebra provides the basis of so much of higher engineering studies, it is a situation that often needs urgent attention. Lack of space has prevented the inclusion of more basic algebra topics in this textbook; it is for this reason that some algebra topics – solution of simple, simultaneous and quadratic equations and transposition of formulae – have been made available to all via the Internet. Also included is a Remedial Algebra Revision Test to test understanding. To access the Algebra material visit the website. Five extra chapters Chapters on Inequalities, Boolean Algebra and logic circuits, Sampling and Estimation theories, Signiﬁcance testing, and Chi-square and distribution-free tests are available to download at the website.

Sample of worked Solutions to Exercises Within the text (plus the website chapters) are some 1900 further problems arranged within 260 Exercises. A sample of over 1100 worked solutions has been prepared and can be accessed free via the Internet. To access these worked solutions visit the website. Instructor’s manual This provides fully worked solutions and mark scheme for all the Revision Tests in this book (plus 2 from the website chapters), together with solutions to the Remedial Algebra Revision Test mentioned above. The material is available to lecturers only. To obtain a password please visit the website with the following details: course title, number of students, your job title and work postal address. To download the Instructor’s Manual visit the website and enter the book title in the search box.

Syllabus Guidance This textbook is written for undergraduate engineering degree and foundation degree courses; however, it is also most appropriate for HNC/D studies and three syllabuses are covered.

The appropriate chapters for these three syllabuses are shown in the table below. Chapter

Analytical Methods for Engineers

Further Analytical Methods for Engineers

1.

Algebra

×

2.

Partial fractions

×

3.

Logarithms

×

4.

Exponential functions

×

5.

Hyperbolic functions

×

6.

Arithmetic and geometric progressions

×

7.

The binomial series

×

8.

Maclaurin’s series

×

9.

Solving equations by iterative methods

×

10.

Binary, octal and hexadecimal

×

11.

Introduction to trigonometry

×

12.

Cartesian and polar co-ordinates

×

13.

The circle and its properties

×

14.

Trigonometric waveforms

×

15.

Trigonometric identities and equations

×

16.

The relationship between trigonometric and hyperbolic functions

×

17.

Compound angles

×

18.

Functions and their curves

×

19.

Irregular areas, volumes and mean values of waveforms

×

20.

Complex numbers

×

21.

De Moivre’s theorem

×

22.

The theory of matrices and determinants

×

23.

The solution of simultaneous equations by matrices and determinants

×

24.

Vectors

×

25.

Methods of adding alternating waveforms

×

Engineering Mathematics

(Continued )

xvi Syllabus Guidance Chapter

Analytical Methods for Engineers

Further Analytical Methods for Engineers

Engineering Mathematics

×

26.

Scalar and vector products

27.

Methods of differentiation

×

28.

Some applications of differentiation

×

29.

Differentiation of parametric equations

30.

Differentiation of implicit functions

×

31.

Logarithmic differentiation

×

32.

Differentiation of hyperbolic functions

×

33.

Differentiation of inverse trigonometric and hyperbolic functions

×

34.

Partial differentiation

×

35.

Total differential, rates of change and small changes

×

36.

Maxima, minima and saddle points for functions of two variables

×

37.

Standard integration

×

38.

Some applications of integration

×

39.

Integration using algebraic substitutions

×

40.

Integration using trigonometric and hyperbolic substitutions

×

41.

Integration using partial fractions

×

42.

The t = tan θ/2 substitution

43.

Integration by parts

×

44.

Reduction formulae

×

45.

Numerical integration

×

46.

Solution of ﬁrst order differential equations by separation of variables

×

47.

hom*ogeneous ﬁrst order differential equations

48.

Linear ﬁrst order differential equations

×

49.

Numerical methods for ﬁrst order differential equations

×

50.

Second order differential equations of the form d2 y dy + cy = 0 a 2 +b dx dx

×

51.

Second order differential equations of the form d2 y dy a 2 +b + cy = f (x) dx dx

×

52.

Power series methods of solving ordinary differential equations

×

53.

An introduction to partial differential equations

×

54.

Presentation of statistical data

×

× (Continued )

Syllabus Guidance Chapter

Analytical Methods for Engineers

Further Analytical Methods for Engineers

Engineering Mathematics

55.

Measures of central tendency and dispersion

×

56.

Probability

×

57.

The binomial and Poisson distributions

×

58.

The normal distribution

×

59.

Linear correlation

×

60.

Linear regression

×

61.

Introduction to Laplace transforms

×

62.

Properties of Laplace transforms

×

63.

Inverse Laplace transforms

×

64.

Solution of differential equations using Laplace transforms

×

65.

The solution of simultaneous differential equations using Laplace transforms

×

66.

Fourier series for periodic functions of period 2π

×

67.

Fourier series for non-periodic functions over range 2π

×

68.

Even and odd functions and half-range Fourier series

×

69.

Fourier series over any range

×

70.

A numerical method of harmonic analysis

×

71.

The complex or exponential form of a Fourier series

×

Website Chapters 72.

Inequalities

73.

Boolean algebra and logic circuits

74.

Sampling and estimation theories

×

75.

Signiﬁcance testing

×

76.

Chi-square and distribution-free tests

×

×

xvii

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Chapter 1

Algebra 1.1

3x + 2y x−y

Introduction

In this chapter, polynomial division and the factor and remainder theorems are explained (in Sections 1.4 to 1.6). However, before this, some essential algebra revision on basic laws and equations is included. For further Algebra revision, go to website: http://books.elsevier.com/companions/0750681527

1.2

Multiply by x → 3x 2 + 2x y Multiply by −y →

3x 2 − xy − 2y 2

Adding gives: Alternatively,

(3x + 2y)(x − y) = 3x 2 − 3x y + 2x y − 2y 2

Revision of basic laws

= 3x 2 − xy − 2y 2

(a) Basic operations and laws of indices The laws of indices are: (i) a m × a n = a m+n (iii)

(a m )n

(v)

a −n

=

a m×n

1 = n a

am (ii) = a m−n an √ m (iv) a n = n a m (vi)

a0

Problem 3. Simplify a = 3, b =

1 8

and c = 2.

When a = 3, b = Problem 1. Evaluate b = 12 and c = 1 12

when a = 2,

3 3 3 1 4a bc − 2ac = 4(2) − 2(2) 2 2 2 2

3

2

=

a 3 b 2 c4 and evaluate when abc−2

a 3 b 2 c4 = a 3−1b2−1c4−(−2) = a 2 bc6 abc−2

=1

4a 2 bc3−2ac

− 3x y − 2y 2

4 × 2 × 2 × 3 × 3 × 3 12 − 2×2×2×2 2

= 27 − 6 = 21 Problem 2. Multiply 3x + 2y by x − y.

and c = 2, a 2 bc6 = (3)2 18 (2)6 = (9) 18 (64) = 72 1 8

Problem 4. Simplify

x 2 y3 + x y2 xy

x 2 y3 + x y2 x 2 y3 x y2 = + xy xy xy = x 2−1 y 3−1 + x 1−1 y 2−1 = xy 2 + y or y(xy + 1)

2 Higher Engineering Mathematics

Problem 5.

Simplify

√ √ (x 2 y)( x 3 y 2 ) (x 5 y 3 )

1 2

√ √ (x 2 y)( x 3 y 2 )

(b) Brackets, factorization and precedence

1

(x 5 y 3 ) 2 1

1

5 2

3 2

Problem 6.

2

x2 y 2 x 2 y 3

=

x y

a 2 − (2a − ab) − a(3b + a) = a 2 − 2a + ab − 3ab − a 2 = −2a − 2ab or −2a(1 + b)

= x 2+ 2 − 2 y 2 + 3 − 2 1

5

1

2

3

= x 0 y− 3 1

Problem 7. expression:

1

= y − 3 or

Simplify a 2 − (2a − ab) − a(3b + a).

1 y

1 3

1 or √ 3 y

Remove the brackets and simplify the

2a − [3{2(4a − b) − 5(a + 2b)} + 4a]. Removing the innermost brackets gives: 2a − [3{8a − 2b − 5a − 10b} + 4a]

Now try the following exercise

Collecting together similar terms gives: Exercise 1 Revision of basic operations and laws of indices

2a − [3{3a − 12b} + 4a]

1. Evaluate 2ab + 3bc − abc when a = 2, b = −2 and c = 4. [−16] 2. Find the value of 5 pq 2r 3 when p = 25 , q = −2 and r = −1. [−8] 3. From 4x − 3y + 2z subtract x + 2y − 3z. [3x − 5y + 5z]

Removing the ‘curly’ brackets gives: 2a − [9a − 36b + 4a] Collecting together similar terms gives: 2a − [13a − 36b] Removing the square brackets gives:

4. Multiply 2a − 5b + c by 3a + b. [6a 2 − 13ab + 3ac − 5b 2 + bc] 5. Simplify (x y z)(x yz ) and evaluate when [x 5 y 4 z 3 , 13 12 ] x = 12 , y = 2 and z = 3. 2 3

3

3 2

1 2

8. Simplify

− 21

c) when a = 3, [±4 12 ]

a2b + a3b a 2 b2 1 2

− 12

(a 3 b c )(ab) √ √ ( a 3 b c)

11

36b − 11a

2

6. Evaluate (a bc−3)(a b b = 4 and c = 2. 7. Simplify

2a − 13a + 36b = −11a + 36b or

1

1+a b

Problem 8. Factorize (a) x y − 3x z (b) 4a 2 + 16ab3 (c) 3a 2 b − 6ab 2 + 15ab. (a)

x y − 3x z = x( y − 3z)

(b) 4a 2 + 16ab3 = 4a(a + 4b3 ) (c) 3a 2 b − 6ab 2 + 15ab = 3ab(a − 2b + 5)

1 3

3

a 6 b 3 c− 2

√ √ 6 11 3 a b or √ 3 c

Problem 9.

Simplify 3c + 2c × 4c + c ÷ 5c − 8c.

The order of precedence is division, multiplication, addition and subtraction (sometimes remembered by BODMAS). Hence

Algebra 3c + 2c × 4c + c ÷ 5c − 8c c = 3c + 2c × 4c + − 8c 5c 1 = 3c + 8c2 + − 8c 5 1 1 = 8c2 − 5c + or c(8c − 5) + 5 5

7. Simplify 3 ÷ y + 2 ÷ y − 1. 8. Simplify a 2 − 3ab × 2a ÷ 6b + ab.

1.3 Problem 10. Simplify (2a − 3) ÷ 4a + 5 × 6 −3a.

2a − 3 + 5 × 6 − 3a 4a

=

2a − 3 + 30 − 3a 4a

=

Revision of equations

Problem 11. Solve 4 − 3x = 2x − 11. Since 4 − 3x = 2x − 11 then 4 + 11 = 2x + 3x 15 i.e. 15 = 5x from which, x = =3 5 Problem 12. Solve

3 2a − + 30 − 3a = 4a 4a

4(2a − 3) − 2(a − 4) = 3(a − 3) − 1.

1 3 1 3 − + 30 − 3a = 30 − − 3a 2 4a 2 4a

Removing the brackets gives: 8a − 12 − 2a + 8 = 3a − 9 − 1 Rearranging gives: 8a − 2a − 3a = −9 − 1 + 12 − 8

Now try the following exercise Exercise 2 Further problems on brackets, factorization and precedence 1. Simplify 2( p + 3q − r) − 4(r − q + 2 p) + p. [−5 p + 10q − 6r] 2. Expand and simplify (x + y)(x − 2y). [x 2 − x y − 2y 2 ] 3. Remove the brackets and simplify: 24 p − [2{3(5 p − q) − 2( p + 2q)} + 3q]. [11q − 2 p]

i.e. and

3 4 = . x − 2 3x + 4

By ‘cross-multiplying’:

3(3x + 4) = 4(x − 2)

Removing brackets gives:

9x + 12 = 4x − 8

Rearranging gives:

9x − 4x = −8 − 12

i.e.

5. Factorize 2x y 2 + 6x 2 y + 8x 3 y. [2x y(y + 3x + 4x 2 )]

and

6. Simplify 2y + 4 ÷ 6y + 3 × 4− 5y. 2 − 3y + 12 3y

−6 = −2 3

Problem 13. Solve

[7ab(3ab − 4)]

4. Factorize 21a 2b2 − 28ab.

3a = −6 a=

5x = −20 −20 5 = −4

x=

√ t +3 Problem 14. Solve √ = 2. t

[ab]

(a) Simple equations

(2a − 3) ÷ 4a + 5 × 6 − 3a =

5 −1 y

3

4 Higher Engineering Mathematics

i.e.

√ √ t +3 √ =2 t t √ √ t +3= 2 t √ √ 3= 2 t − t √ 3= t

and

9= t

√ t i.e. and

Rearranging gives: d 2 p + D 2 p = D 2 f − d2 f Factorizing gives:

f (D 2 − d2 ) (d2 + D2 )

Now try the following exercise Exercise 3 Further problems on simple equations and transposition of formulae

Transpose the formula v = u +

to make f the subject.

ft m

In problems 1 to 4 solve the equations 1. 3x − 2 − 5x = 2x − 4.

ft ft = v from which, = v−u m m ft = m(v − u) and m m

u+

i.e.

f t = m(v − u)

and

m f = (v − u) t

X 2 = Z 2 − R 2 and reactance X =

3. 4.

Z2 − R2

f +p D , Problem 17. Given that = d f −p express p in terms of D, d and f . Rearranging gives: Squaring both sides gives:

2

[−3]

R 2 + X 2 = Z and squaring both sides gives R 2 + X 2 = Z 2 , from which,

1

2. 8 + 4(x − 1) − 5(x − 3) = 2(5 − 2x).

Problem 16. √The impedance of an a.c. circuit is given by Z = R 2 + X 2 . Make the reactance X the subject.

p=

and

(b) Transposition of formulae Problem 15.

p(d 2 + D 2 ) = f (D 2 − d2 )

D f +p = f −p d f +p D2 = 2 f −p d

‘Cross-multiplying’ gives: d2 ( f + p) = D 2 ( f − p) Removing brackets gives: d2 f + d2 p = D 2 f − D 2 p

1 1 + = 0. 3a − 2 5a + 3 √ 3 t √ = −6. 1− t

1 −8 [4]

3(F − f ) . for f . 5. Transpose y = L yL 3F − yL or f = F − f = 3 3 1 6. Make l the subject of t = 2π . g t 2g l= 4π 2 μL for L. 7. Transpose m = L + rC R mrC R L= μ−m 8. Make r the subject of the formula

x−y x 1 + r2 . r= = y 1 − r2 x+y

(c) Simultaneous equations Problem 18.

Solve the simultaneous equations: 7x − 2y = 26

(1)

6x + 5y = 29.

(2)

Algebra 5 × equation (1) gives: 35x − 10y = 130

The factors of −4 are +1 and −4 or −1 and +4, or −2 and +2. Remembering that the product of the two inner terms added to the product of the two outer terms must equal −11x, the only combination to give this is +1 and −4, i.e.,

(3)

2 × equation (2) gives: 12x + 10y = 58

(4)

equation (3) +equation (4) gives:

3x 2 − 11x − 4 = (3x + 1)(x − 4)

47x + 0 = 188 188 from which, x= =4 47 Substituting x = 4 in equation (1) gives:

(3x + 1)(x − 4) = 0 hence

Thus either

(x − 4) = 0 i.e. x = 4

or

28 − 2y = 26

(b) 4x 2 + 8x + 3 = (2x + 3)(2x + 1)

from which, 28 − 26 = 2y and y = 1

(2x + 3)(2x + 1) = 0 hence

Thus

Problem 19. Solve x 5 + =y 8 2 y 11 + = 3x. 3

(3x + 1) = 0 i.e. x = − 13

(1) (2)

either

(2x + 3) = 0 i.e. x = − 32

or

(2x + 1) = 0 i.e. x = − 12

Problem 21. The roots of a quadratic equation are 13 and −2. Determine the equation in x.

8 × equation (1) gives:

x + 20 = 8y

(3)

3 × equation (2) gives:

33 + y = 9x

(4)

i.e.

x − 8y = −20

(5)

and

9x − y = 33

(6)

i.e. x 2 + 2x − 13 x − 23 = 0

(7)

i.e.

x 2 + 53 x − 23 = 0

or

3x2 + 5x −2 = 0

8 × equation (6) gives: 72x − 8y = 264 Equation (7) − equation (5) gives: 71x = 284 284 =4 71 Substituting x = 4 in equation (5) gives: x=

from which,

4 − 8y = −20 4 + 20 = 8y and y = 3

from which,

(d) Quadratic equations Problem 20. Solve the following equations by factorization: (a) 3x 2 − 11x − 4 = 0 (b) 4x 2 + 8x + 3 = 0. (a)

The factors of 3x 2 are 3x and x and these are placed in brackets thus: (3x

)(x

)

If

1 3

and −2 are the roots of a quadratic equation then, (x − 13 )(x + 2) = 0

Problem 22. Solve 4x 2 + 7x + 2 = 0 giving the answer correct to 2 decimal places. From the quadratic formula if ax 2 + bx + c = 0 then, √ −b ± b2 − 4ac x= 2a Hence if 4x 2 + 7x + 2 = 0 −7 ± 72 − 4(4)(2) then x = 2(4) √ −7 ± 17 = 8 −7 ± 4.123 = 8 −7 + 4.123 −7 − 4.123 = or 8 8 i.e. x = −0.36 or −1.39

5

6 Higher Engineering Mathematics Now try the following exercise

For example,

Exercise 4 Further problems on simultaneous and quadratic equations In problems 1 to 3, solve the simultaneous equations

13 ——– 16 208 16 48 48 — ·· —

1. 8x − 3y = 51 3x + 4y = 14.

208 is achieved as follows: 16

[x = 6, y = −1]

(1) 16 divided into 2 won’t go 2. 5a = 1 − 3b 2b + a + 4 = 0. 3.

[a = 2, b = −3]

x 2y 49 + = 5 3 15

(2) 16 divided into 20 goes 1 (3) Put 1 above the zero (4) Multiply 16 by 1 giving 16 (5) Subtract 16 from 20 giving 4

3x y 5 − + = 0. 7 2 7

[x = 3, y = 4]

(6) Bring down the 8 (7) 16 divided into 48 goes 3 times

4. Solve the following quadratic equations by factorization: (a) x 2 + 4x − 32 = 0

[(a) 4, −8 (b) 54 , − 32 ] 5. Determine the quadratic equation in x whose roots are 2 and −5. [x 2 + 3x − 10 = 0] 6. Solve the following quadratic equations, correct to 3 decimal places: (a)

−4 = 0

(b) 4t 2 − 11t + 3 = 0.

(a) 0.637, −3.137 (b) 2.443, 0.307

1.4

(9) 3 × 16 = 48 (10) 48 − 48 = 0

(b) 8x 2 + 2x − 15 = 0.

2x 2 + 5x

(8) Put the 3 above the 8

Hence Similarly,

208 = 13 exactly 16

172 is laid out as follows: 15

11 ——– 15 172 15 22 15 — 7 — 7 7 172 = 11 remainder 7 or 11 + = 11 Hence 15 15 15 Below are some examples of division in algebra, which in some respects, is similar to long division with numbers. (Note that a polynomial is an expression of the form

Polynomial division f (x) = a + bx + cx 2 + d x 3 + · · ·

Before looking at long division in algebra let us revise long division with numbers (we may have forgotten, since calculators do the job for us!)

and polynomial division is sometimes required when resolving into partial fractions—see Chapter 2.)

Algebra Problem 23. Divide 2x 2 + x − 3 by x − 1.

(3) Subtract (4)

2x 2 + x − 3 is called the dividend and x − 1 the divisor. The usual layout is shown below with the dividend and divisor both arranged in descending powers of the symbols. 2x + 3 ——————– x − 1 2x 2 + x − 3 2x 2 − 2x 3x − 3 3x − 3 ——— · · ——— Dividing the ﬁrst term of the dividend by the ﬁrst term 2x 2 gives 2x, which is put above of the divisor, i.e. x the ﬁrst term of the dividend as shown. The divisor is then multiplied by 2x, i.e. 2x(x − 1) = 2x 2 − 2x, which is placed under the dividend as shown. Subtracting gives 3x − 3. The process is then repeated, i.e. the ﬁrst term of the divisor, x, is divided into 3x, giving +3, which is placed above the dividend as shown. Then 3(x − 1) = 3x − 3 which is placed under the 3x − 3. The remainder, on subtraction, is zero, which completes the process.

x into −2x 2 goes −2x. Put −2x above the dividend

(5) −2x(x + 1) = −2x 2 − 2x (6) Subtract (7)

x into 5x goes 5. Put 5 above the dividend

(8) 5(x + 1) = 5x + 5 (9) Subtract Thus

3x 3 + x 2 + 3x + 5 = 3x 2 − 2x + 5 x +1

Problem 25. Simplify

(1) (4) (7) x 2 − x y + y2 —————————– x + y x 3 + 0 + 0 + y3 x3 + x2 y − x2 y + y3 − x 2 y − x y2 ——————— x y2 + y3 x y2 + y3 ———– · · ———–

Thus (2x 2 + x − 3) ÷ (x − 1) = (2x + 3) [A check can be made on this answer by multiplying (2x + 3) by (x − 1) which equals 2x 2 + x − 3] Problem 24. Divide 3x 3 + x 2 + 3x + 5 by x + 1. (1) (4) (7) 3x 2 − 2x + 5 ————————— x + 1 3x 3 + x 2 + 3x + 5 3x 3 + 3x 2 − 2x 2 + 3x + 5 − 2x 2 − 2x ————– 5x + 5 5x + 5 ——— · · ——— (1)

x into 3x 3 goes 3x 2 . Put 3x 2 above 3x 3

(2) 3x 2 (x + 1) = 3x 3 + 3x 2

x 3 + y3 . x+y

(1)

x into x 3 goes x 2 . Put x 2 above x 3 of dividend

(2)

x 2 (x + y) = x 3 + x 2 y

(3) Subtract (4)

x into −x 2 y goes −x y. Put −x y above dividend

(5) −x y(x + y) = −x 2 y − x y 2 (6) Subtract (7)

x into x y 2 goes y 2 . Put y 2 above dividend

(8)

y 2 (x + y) = x y 2 + y 3

(9) Subtract Thus x 3 + y3 = x 2 − xy + y 2 x+y

7

8 Higher Engineering Mathematics The zero’s shown in the dividend are not normally shown, but are included to clarify the subtraction process and to keep similar terms in their respective columns. Problem 26.

Divide (x 2 + 3x − 2) by (x − 2).

x +5 ——————– x − 2 x 2 + 3x − 2 x 2 − 2x

14x 2 − 19x − 3 . 2x − 3

[7x + 1]

6. Find (5x 2 − x + 4) ÷ (x − 1). 5x + 4 +

8 x −1

7. Divide (3x 3 + 2x 2 − 5x + 4) by (x + 2). 2 2 3x − 4x + 3 − x +2

Hence 8 x 2 + 3x − 2 =x +5+ x −2 x−2 Divide 4a 3 − 6a 2 b + 5b 3 by

2a − 2ab − b ——————————— 2a − b 4a 3 − 6a 2 b + 5b 3 3 2 4a − 2a b 2

4. Find

5. Divide (x 3 + 3x 2 y + 3x y 2 + y 3 ) by (x + y). [x 2 + 2x y + y 2 ]

5x − 2 5x − 10 ——— 8 ———

Problem 27. 2a − b.

3. Determine (10x 2 + 11x − 6) ÷ (2x + 3). [5x − 2]

8. Determine (5x 4 + 3x 3 − 2x + 1)/(x − 3). 481 3 2 5x + 18x + 54x + 160 + x −3

2

1.5

−4a 2 b + 5b3 2 2 −4a b + 2ab ———— 2 −2ab + 5b 3 −2ab2 + b 3 —————– 4b 3 —————–

There is a simple relationship between the factors of a quadratic expression and the roots of the equation obtained by equating the expression to zero. For example, consider the quadratic equation x 2 + 2x − 8 = 0. To solve this we may factorize the quadratic expression x 2 + 2x − 8 giving (x − 2)(x + 4). Hence (x − 2)(x + 4) = 0. Then, if the product of two numbers is zero, one or both of those numbers must equal zero. Therefore,

Thus 4a 3 − 6a 2 b + 5b 3 2a − b = 2a 2 − 2ab − b2 +

either (x − 2) = 0, from which, x = 2 or (x + 4) = 0, from which, x = −4

4b3 2a − b

It is clear then that a factor of (x − 2) indicates a root of +2, while a factor of (x + 4) indicates a root of −4. In general, we can therefore say that:

Now try the following exercise Exercise 5 Further problems on polynomial division 1. Divide (2x 2 + x y − y 2 ) by (x + y). 2. Divide (3x 2 + 5x − 2) by (x + 2).

The factor theorem

[2x − y] [3x − 1]

a factor of (x − a) corresponds to a root of x = a In practice, we always deduce the roots of a simple quadratic equation from the factors of the quadratic expression, as in the above example. However, we could reverse this process. If, by trial and error, we could determine that x = 2 is a root of the equation x 2 + 2x − 8 = 0 we could deduce at once that (x − 2) is a factor of the

Algebra expression x 2 + 2x − 8. We wouldn’t normally solve quadratic equations this way — but suppose we have to factorize a cubic expression (i.e. one in which the highest power of the variable is 3). A cubic equation might have three simple linear factors and the difﬁculty of discovering all these factors by trial and error would be considerable. It is to deal with this kind of case that we use the factor theorem. This is just a generalized version of what we established above for the quadratic expression. The factor theorem provides a method of factorizing any polynomial, f (x), which has simple factors. A statement of the factor theorem says: ‘if x = a is a root of the equation f (x) = 0, then (x − a) is a factor of f (x)’ The following worked problems show the use of the factor theorem. Problem 28. Factorize x 3 − 7x − 6 and use it to solve the cubic equation x 3 − 7x − 6 = 0. Let

f (x) = x 3 − 7x − 6

If x = 1, then f (1) = 13 − 7(1) − 6 = −12 If x = 2, then f (2) = 23 − 7(2) − 6 = −12 If x = 3, then f (3)

= 33 − 7(3) − 6

=0

If f (3) = 0, then (x − 3) is a factor — from the factor theorem. We have a choice now. We can divide x 3 − 7x − 6 by (x − 3) or we could continue our ‘trial and error’ by substituting further values for x in the given expression — and hope to arrive at f (x) = 0. Let us do both ways. Firstly, dividing out gives: x + 3x + 2 ————————— x − 3 x 3 − 0 − 7x − 6 x 3 − 3x 2 2

3x 2 − 7x − 6 3x 2 − 9x ———— 2x − 6 2x − 6 ——— · · ——— x 3 − 7x − 6 = x 2 + 3x + 2 Hence x −3 i.e.

x 3 − 7x − 6 = (x − 3)(x 2 + 3x + 2)

x 2 + 3x + 2 factorizes ‘on sight’ as (x + 1)(x + 2). Therefore x 3 − 7x − 6 = (x − 3)(x + 1)(x + 2) A second method is to continue to substitute values of x into f (x). Our expression for f (3) was 33 − 7(3) − 6. We can see that if we continue with positive values of x the ﬁrst term will predominate such that f (x) will not be zero. Therefore let us try some negative values for x. Therefore f (−1) = (−1)3 − 7(−1) − 6 = 0; hence (x + 1) is a factor (as shown above). Also f (−2) = (−2)3 − 7(−2) − 6 = 0; hence (x + 2) is a factor (also as shown above). To solve x 3 − 7x − 6 = 0, we substitute the factors, i.e., (x − 3)(x + 1)(x + 2) = 0 from which, x = 3, x = −1 and x = −2. Note that the values of x, i.e. 3, −1 and −2, are all factors of the constant term, i.e. the 6. This can give us a clue as to what values of x we should consider. Problem 29. Solve the cubic equation x 3 − 2x 2 − 5x + 6 = 0 by using the factor theorem. Let f (x) = x 3 − 2x 2 − 5x + 6 and let us substitute simple values of x like 1, 2, 3, −1, −2, and so on. f (1) = 13 − 2(1)2 − 5(1) + 6 = 0, hence (x − 1) is a factor f (2) = 23 − 2(2)2 − 5(2) + 6 = 0 f (3) = 33 − 2(3)2 − 5(3) + 6 = 0, hence (x − 3) is a factor f (−1) = (−1)3 − 2(−1)2 − 5(−1) + 6 = 0 f (−2) = (−2)3 − 2(−2)2 − 5(−2) + 6 = 0, hence (x + 2) is a factor x 3 − 2x 2

− 5x + 6 = (x − 1)(x − 3)(x + 2) Hence Therefore if x 3 − 2x 2 − 5x + 6 = 0 then (x − 1)(x − 3)(x + 2) = 0 from which, x = 1, x = 3 and x = −2 Alternatively, having obtained one factor, i.e. (x − 1) we could divide this into (x 3 − 2x 2 − 5x + 6) as follows:

9

10 Higher Engineering Mathematics x − x −6 ————————– x − 1 x 3 − 2x 2 − 5x + 6 x3 − x2 2

1.6

Dividing a general quadratic expression (ax 2 + bx + c) by (x − p), where p is any whole number, by long division (see section 1.3) gives: ax + (b + ap) ————————————– x − p ax 2 + bx +c ax 2 − apx

− x 2 − 5x + 6 − x2 + x ————– − 6x + 6 − 6x + 6 ———– · · ———– Hence x 3 − 2x 2 − 5x + 6 = (x − 1)(x 2 − x − 6) = (x − 1)(x − 3)(x + 2) Summarizing, the factor theorem provides us with a method of factorizing simple expressions, and an alternative, in certain circ*mstances, to polynomial division.

Now try the following exercise

Exercise 6 Further problems on the factor theorem Use the factor theorem to factorize the expressions given in problems 1 to 4. 1.

x 2 + 2x − 3

2.

x 3 + x 2 − 4x − 4

[(x − 1)(x + 3)]

3. 2x 3 + 5x 2 − 4x − 7

The remainder theorem

[(x + 1)(x + 2)(x − 2)] [(x + 1)(2x 2 + 3x − 7)]

4. 2x 3 − x 2 − 16x + 15 [(x − 1)(x + 3)(2x − 5)] 5. Use the factor theorem to factorize x 3 + 4x 2 + x − 6 and hence solve the cubic equation x 3 + 4x 2 + x − 6 = 0. ⎤ ⎡ 3 x + 4x 2 + x − 6 ⎥ ⎢ = (x − 1)(x + 3)(x + 2) ⎦ ⎣ x = 1, x = −3 and x = −2 6. Solve the equation x 3 − 2x 2 − x + 2 = 0. [x = 1, x = 2 and x = −1]

(b + ap)x + c (b + ap)x − (b + ap) p —————————– c + (b + ap) p —————————– The remainder, c + (b + ap) p = c + bp + ap 2 or ap2 + bp + c. This is, in fact, what the remainder theorem states, i.e., ‘if (ax 2 + bx + c) is divided by (x − p), the remainder will be ap 2 + bp + c’ If, in the dividend (ax 2 + bx + c), we substitute p for x we get the remainder ap2 + bp + c. For example, when (3x 2 − 4x + 5) is divided by (x − 2) the remainder is ap2 + bp + c (where a = 3, b = −4, c = 5 and p = 2), i.e. the remainder is 3(2)2 + (−4)(2) + 5 = 12 − 8 + 5 = 9 We can check this by dividing (3x 2 − 4x + 5) by (x − 2) by long division: 3x + 2 ——————– x − 2 3x 2 − 4x + 5 3x 2 − 6x 2x + 5 2x − 4 ——— 9 ——— Similarly, when (4x 2 − 7x + 9) is divided by (x + 3), the remainder is ap 2 + bp + c, (where a = 4, b = −7, c = 9 and p = −3) i.e. the remainder is 4(−3)2 + (−7)(−3) + 9 = 36 + 21 + 9 = 66. Also, when (x 2 + 3x − 2) is divided by (x − 1), the remainder is 1(1)2 + 3(1) − 2 = 2. It is not particularly useful, on its own, to know the remainder of an algebraic division. However, if the remainder should be zero then (x − p) is a factor. This is very useful therefore when factorizing expressions. For example, when (2x 2 + x − 3) is divided by (x − 1), the remainder is 2(1)2 + 1(1) − 3 = 0, which means that (x − 1) is a factor of (2x 2 + x − 3).

Algebra In this case the other factor is (2x + 3), i.e.,

i.e. the remainder = (1)(1)3 + (−2)(1)2 + (−5)(1) + 6

(2x 2 + x − 3) = (x − 1)(2x − 3) The remainder theorem may also be stated for a cubic equation as: ‘if (ax 3 + bx 2 + cx + d) is divided by (x − p), the remainder will be ap 3 + bp 2 + cp + d’ As before, the remainder may be obtained by substituting p for x in the dividend. For example, when (3x 3 + 2x 2 − x + 4) is divided by (x − 1), the remainder is ap 3 + bp2 + cp + d (where a = 3, b = 2, c = −1, d = 4 and p = 1), i.e. the remainder is 3(1)3 + 2(1)2 + (−1)(1) + 4 = 3 + 2 − 1 + 4 = 8. Similarly, when (x 3 − 7x − 6) is divided by (x − 3), the remainder is 1(3)3 + 0(3)2 − 7(3) − 6 = 0, which means that (x − 3) is a factor of (x 3 − 7x − 6). Here are some more examples on the remainder theorem. Problem 30. Without dividing out, ﬁnd the remainder when 2x 2 − 3x + 4 is divided by (x − 2). By the remainder theorem, the remainder is given by ap 2 + bp + c, where a = 2, b = −3, c = 4 and p = 2. Hence the remainder is: 2(2)2 + (−3)(2) + 4 = 8 − 6 + 4 = 6 Problem 31. Use the remainder theorem to determine the remainder when (3x 3 − 2x 2 + x − 5) is divided by (x + 2). By the remainder theorem, the remainder is given by ap 3 + bp2 + cp + d, where a = 3, b = −2, c = 1, d = −5 and p = −2. Hence the remainder is: 3(−2)3 + (−2)(−2)2 + (1)(−2) + (−5) = −24 − 8 − 2 − 5 = −39 Problem 32. Determine the remainder when (x 3 − 2x 2 − 5x + 6) is divided by (a) (x − 1) and (b) (x + 2). Hence factorize the cubic expression. (a)

When (x 3 − 2x 2 − 5x + 6) is divided by (x − 1), the remainder is given by ap 3 + bp2 + cp + d, where a = 1, b = −2, c = −5, d = 6 and p = 1,

11

= 1−2−5+6 = 0 Hence (x − 1) is a factor of (x 3 − 2x 2 − 5x + 6). (b) When (x 3 − 2x 2 − 5x + 6) is divided by (x + 2), the remainder is given by (1)(−2)3 + (−2)(−2)2 + (−5)(−2) + 6 = −8 − 8 + 10 + 6 = 0 Hence (x + 2) is also a factor of (x 3 − 2x 2 − 5x + 6). Therefore (x − 1)(x + 2)(x ) = x 3 − 2x 2 − 5x + 6. To determine the third factor (shown blank) we could (i) divide (x 3 − 2x 2 − 5x + 6) by (x − 1)(x + 2). or (ii) use the factor theorem where f (x) = x 3 − 2x 2 − 5x + 6 and hoping to choose a value of x which makes f (x) = 0. or (iii) use the remainder theorem, again hoping to choose a factor (x − p) which makes the remainder zero. (i) Dividing (x 3 − 2x 2 − 5x + 6) by (x 2 + x − 2) gives: x −3 ————————– x 2 + x − 2 x 3 − 2x 2 − 5x + 6 x 3 + x 2 − 2x —————— −3x 2 − 3x + 6 −3x 2 − 3x + 6 ——————– · · · ——————– Thus (x 3 − 2x 2 − 5x + 6) = (x − 1)(x + 2)(x − 3) (ii) Using the factor theorem, we let f (x) = x 3 − 2x 2 − 5x + 6 Then f (3) = 33 − 2(3)2 − 5(3) + 6 = 27 − 18 − 15 + 6 = 0 Hence (x − 3) is a factor. (iii) Using the remainder theorem, when (x 3 − 2x 2 − 5x + 6) is divided by (x − 3), the remainder is given by

12 Higher Engineering Mathematics ap3 + bp2 + cp + d, where a = 1, b = −2, c = −5, d = 6 and p = 3. Hence the remainder is: 1(3)3 + (−2)(3)2 + (−5)(3) + 6 = 27 − 18 − 15 + 6 = 0 Hence (x − 3) is a factor. Thus (x 3 − 2x 2 − 5x + 6) = (x − 1)(x + 2)(x − 3)

Now try the following exercise Exercise 7 Further problems on the remainder theorem 1. Find the remainder when 3x 2 − 4x + 2 is divided by (a) (x − 2) (b) (x + 1).

[(a) 6 (b) 9]

2. Determine the remainder when x 3 − 6x 2 + x − 5 is divided by (a) (x + 2) (b) (x − 3).

[(a) −39 (b) −29]

3. Use the remainder theorem to ﬁnd the factors of x 3 − 6x 2 + 11x − 6. [(x − 1)(x − 2)(x − 3)] 4. Determine the factors of x 3 + 7x 2 + 14x + 8 and hence solve the cubic equation x 3 + 7x 2 + 14x + 8 = 0. [x = −1, x = −2 and x = −4] 5. Determine the value of ‘a’ if (x + 2) is a factor of (x 3 − ax 2 + 7x + 10). [a = −3] 6. Using the remainder theorem, solve the equation 2x 3 − x 2 − 7x + 6 = 0. [x = 1, x = −2 and x = 1.5]

Chapter 2

Partial fractions 2.1

When the degree of the numerator is equal to or higher than the degree of the denominator, the numerator must be divided by the denominator until the remainder is of less degree than the denominator (see Problems 3 and 4). There are basically three types of partial fraction and the form of partial fraction used is summarized in Table 2.1, where f (x) is assumed to be of less degree than the relevant denominator and A, B and C are constants to be determined. (In the latter type in Table 2.1, ax 2 + bx + c is a quadratic expression which does not factorize without containing surds or imaginary terms.) Resolving an algebraic expression into partial fractions is used as a preliminary to integrating certain functions (see Chapter 41) and in determining inverse Laplace transforms (see Chapter 63).

Introduction to partial fractions

By algebraic addition, 1 3 (x + 1) + 3(x − 2) + = x −2 x +1 (x − 2)(x + 1) =

4x − 5 x2 − x − 2

The reverse process of moving from

4x − 5 −2

x2 − x

1 3 + is called resolving into partial x −2 x +1 fractions. In order to resolve an algebraic expression into partial fractions: to

(i) the denominator must factorize (in the above example, x 2 − x − 2 factorizes as (x − 2) (x + 1)), and

2.2 Worked problems on partial fractions with linear factors

(ii) the numerator must be at least one degree less than the denominator (in the above example (4x − 5) is of degree 1 since the highest powered x term is x 1 and (x 2 − x − 2) is of degree 2).

Problem 1. Resolve fractions.

11 − 3x into partial x 2 + 2x − 3

Table 2.1 Type

Denominator containing

1

Linear factors (see Problems 1 to 4)

2

Repeated linear factors (see Problems 5 to 7)

3

Quadratic factors (see Problems 8 and 9)

Expression

Form of partial fraction

f (x) (x + a)(x − b)(x + c)

A B C + + (x + a) (x − b) (x + c)

f (x) (x + a)3

A C B + + 2 (x + a) (x + a) (x + a)3

f (x) + c)(x + d)

(ax 2 + bx

Ax + B C + + bx + c) (x + d)

(ax 2

14 Higher Engineering Mathematics The denominator factorizes as (x − 1) (x + 3) and the numerator is of less degree than the denominator. Thus 11 − 3x may be resolved into partial fractions. x 2 + 2x − 3 Let

Let

2x 2 − 9x − 35 (x + 1)(x − 2)(x + 3) ≡

11 − 3x 11 − 3x ≡ − 3 (x − 1)(x + 3)

x 2 + 2x

≡

A B + (x − 1) (x + 3)

where A and B are constants to be determined, 11 − 3x A(x + 3) + B(x − 1) i.e. ≡ , (x − 1)(x + 3) (x − 1)(x + 3) by algebraic addition. Since the denominators are the same on each side of the identity then the numerators are equal to each other.

≡

When x = 1, then 11 −3(1) ≡ A(1 + 3) + B(0) 8 = 4A A =2

i.e. i.e.

When x = −3, then 11 −3(−3) ≡ A(0) + B(−3 −1) i.e.

20 = −4B

i.e.

B = −5

Thus

Check:

2 5 2(x + 3) − 5(x − 1) − = (x − 1) (x + 3) (x − 1)(x + 3) 11 − 3x = 2 x + 2x − 3

2x 2 − 9x − 35 into (x + 1)(x − 2)(x + 3) the sum of three partial fractions.

Problem 2.

Convert

2x 2 − 9x − 35 ≡ A(x − 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 2) Let x = − 1. Then 2(−1)2 − 9(−1) − 35 ≡ A(−3)(2) + B(0)(2) +C(0)(−3) −24 = −6 A

i.e.

A=

i.e.

−24 =4 −6

Let x = 2. Then 2(2)2 − 9(2) − 35 ≡ A(0)(5) + B(3)(5) + C(3)(0) −45 = 15B

i.e.

B=

i.e.

−45 = −3 15

Let x = − 3. Then

11 − 3x −5 2 + ≡ x2 + 2x − 3 (x − 1) (x + 3) 2 5 ≡ − (x − 1) (x + 3)

A(x − 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 2) (x + 1)(x − 2)(x + 3)

by algebraic addition. Equating the numerators gives:

Thus, 11 −3x ≡ A(x + 3) + B(x − 1) To determine constants A and B, values of x are chosen to make the term in A or B equal to zero.

A B C + + (x + 1) (x − 2) (x + 3)

2(−3)2 − 9(−3) − 35 ≡ A(−5)(0) + B(−2)(0) + C(−2)(−5) i.e.

10 = 10C

i.e.

C =1

Thus

2x 2 − 9x − 35 (x + 1)(x − 2)(x + 3) ≡

3 1 4 − + (x + 1) (x − 2) (x + 3)

Problem 3. fractions.

Resolve

x2

x2 + 1 into partial − 3x + 2

Partial fractions The denominator is of the same degree as the numerator. Thus dividing out gives: x 2 − 3x + 2

1 +1 x2 x 2 − 3x + 2 ————— 3x − 1 ———

Thus

x − 10 x 3 − 2x 2 − 4x − 4 ≡ x −3+ 2 2 x +x −2 x +x −2 ≡ x −3+

Let

x − 10 A B ≡ + (x + 2)(x − 1) (x + 2) (x − 1) ≡

For more on polynomial division, see Section 1.4, page 6. Hence

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

A B 3x − 1 ≡ + Let (x − 1)(x − 2) (x − 1) (x − 2) ≡

A(x − 2) + B(x − 1) (x − 1)(x − 2)

Equating numerators gives:

x − 10 ≡ A(x − 1) + B(x + 2) Let x = −2. Then

−12 = −3 A A= 4

i.e. Let x = 1. Then

−9 = 3B B = −3

i.e. Hence

x − 10 4 3 ≡ − (x + 2)(x − 1) (x + 2) (x − 1)

Thus

x3 − 2 x2 − 4x − 4 x2 + x − 2 ≡x−3+

Let x = 1. Then 2 = −A A = −2

A(x − 1) + B(x + 2) (x + 2)(x − 1)

Equating the numerators gives:

3x − 1 ≡ A(x − 2) + B(x − 1)

i.e.

x − 10 (x + 2)(x − 1)

4 3 − (x + 2) (x − 1)

Now try the following exercise

Let x = 2. Then 5 = B −2 5 3x − 1 ≡ + Hence (x − 1)(x − 2) (x − 1) (x − 2) Thus

2 5 x2 + 1 ≡ 1− + 2 x − 3x + 2 (x−1) (x−2)

Problem 4. Express fractions.

x 3 − 2x 2 − 4x − 4 in partial x2 + x − 2

The numerator is of higher degree than the denominator. Thus dividing out gives: x −3 x 2 + x − 2 x 3 − 2x 2 − 4x − 4 x 3 + x 2 − 2x —————— − 3x 2 − 2x − 4 − 3x 2 − 3x + 6 ——————— x − 10

Exercise 8 Further problems on partial fractions with linear factors Resolve the following into partial fractions. 2 2 12 − 1. x2 − 9 (x − 3) (x + 3)

2.

4(x − 4) 2 x − 2x − 3

3.

x 2 − 3x + 6 x(x − 2)(x − 1)

4.

3(2x 2 − 8x − 1) (x + 4)(x + 1)(2x − 1)

5 1 − (x + 1) (x − 3)

3 2 4 + − x (x − 2) (x − 1)

7 3 2 − − (x + 4) (x + 1) (2x − 1)

15

16 Higher Engineering Mathematics

5.

x 2 + 9x + 8 x2 + x − 6

6.

x 2 − x − 14 x 2 − 2x − 3

7.

2 6 1+ + (x + 3) (x − 2) 2 3 1− + (x − 3) (x + 1)

3x 3 − 2x 2 − 16x + 20 (x − 2)(x + 2) 3x − 2 +

When A = 2 and B = 7, R.H.S. = −2(2) + 7 = 3 = L.H.S.]

Hence

5x 2 − 2x − 19 as the sum (x + 3)(x − 1)2 of three partial fractions.

Problem 6. 5 1 − (x − 2) (x + 2)

Worked problems on partial fractions with repeated linear factors

Problem 5.

Resolve

fractions.

2x + 3 into partial (x − 2)2

The denominator contains a repeated linear factor, (x − 2)2 . A 2x + 3 B ≡ Let + (x − 2)2 (x − 2) (x − 2)2 A(x − 2) + B (x − 2)2

≡

Equating the numerators gives: 2x + 3 ≡ A(x − 2) + B Let x = 2. Then

7 = A(0) + B

i.e.

B =7

2x + 3 ≡ A(x − 2) + B ≡ Ax − 2 A + B Since an identity is true for all values of the unknown, the coefﬁcients of similar terms may be equated. Hence, equating the coefﬁcients of x gives: 2 = A. [Also, as a check, equating the constant terms gives:

Express

The denominator is a combination of a linear factor and a repeated linear factor. Let

2.3

2 7 2x + 3 ≡ + (x − 2)2 (x − 2) (x − 2)2

5x 2 − 2x − 19 (x + 3)(x − 1)2 ≡

A B C + + (x + 3) (x − 1) (x − 1)2

≡

A(x − 1)2 + B(x + 3)(x − 1) + C(x + 3) (x + 3)(x − 1)2

by algebraic addition. Equating the numerators gives: 5x 2 − 2x − 19 ≡ A(x − 1)2 + B(x + 3)(x − 1) + C(x + 3) Let x = −3. Then 5(−3)2 − 2(−3) − 19 ≡ A(−4)2 + B(0)(−4) + C(0) i.e. 32 = 16 A i.e. A= 2 Let x = 1. Then 5(1)2 − 2(1) − 19 ≡ A(0)2 + B(4)(0) + C(4) i.e. −16 = 4C i.e. C = −4 Without expanding the RHS of equation (1) it can be seen that equating the coefﬁcients of x 2 gives: 5 = A + B, and since A = 2, B = 3. [Check: Identity (1) may be expressed as: 5x 2 − 2x − 19 ≡ A(x 2 − 2x + 1) + B(x 2 + 2x − 3) + C(x + 3) i.e. 5x 2 − 2x − 19 ≡ Ax 2 − 2 Ax + A + Bx 2 + 2Bx

3 = −2 A + B

(1)

− 3B + Cx + 3C

Partial fractions Equating the x term coefﬁcients gives:

Equating the coefﬁcients of x terms gives: 16 = 6 A + B

−2 ≡ −2 A + 2B + C

Since A = 3, B = −2

When A = 2, B = 3 and C = −4 then

[Check: equating the constant terms gives:

−2 A + 2B + C = −2(2) + 2(3) − 4

15 = 9 A + 3B + C

= −2 = LHS

When A = 3, B = −2 and C = −6,

Equating the constant term gives:

9 A + 3B + C = 9(3) + 3(−2) + (−6)

−19 ≡ A − 3B + 3C

= 27 − 6 − 6 = 15 = LHS]

RHS = 2 − 3(3) + 3(−4) = 2 − 9 − 12 = −19 = LHS]

Hence

Thus

5x2 − 2x − 19 (x + 3)(x − 1)2 ≡

2 3 4 + − (x + 3) (x − 1) (x − 1)2

Now try the following exercise

3x 2 + 16x + 15 Problem 7. Resolve into partial (x + 3)3 fractions.

Let

3x 2 + 16x + 15 (x + 3)3

Exercise 9 Further problems on partial fractions with linear factors 4 4x − 3 7 1. − (x + 1)2 (x + 1) (x + 1)2 2.

≡

A C B + + (x + 3) (x + 3)2 (x + 3)3

≡

A(x + 3)2 + B(x + 3) + C (x + 3)3

3.

Equating the numerators gives: 3x 2 + 16x + 15 ≡ A(x + 3)2 + B(x + 3) + C Let x = −3. Then

3x2 + 16x + 15 (x + 3)3 3 6 2 ≡ − − 2 (x + 3) (x + 3) (x + 3)3

(1)

4.

x 2 + 7x + 3 x 2 (x + 3)

1 2 1 + − x 2 x (x + 3)

5x 2 − 30x + 44 (x − 2)3 5 4 10 + − (x − 2) (x − 2)2 (x − 2)3 18 + 21x − x 2 (x − 5)(x + 2)2

2 3 4 − + (x − 5) (x + 2) (x + 2)2

3(−3)2 + 16(−3) + 15 ≡ A(0)2 + B(0) + C i.e. −6 = C Identity (1) may be expanded as: 3x 2 + 16x + 15 ≡ A(x 2 + 6x + 9) + B(x + 3) + C

2.4 Worked problems on partial fractions with quadratic factors

i.e. 3x 2 + 16x + 15 ≡ Ax 2 + 6 Ax + 9 A + Bx + 3B + C Equating the coefﬁcients of x 2 terms gives: 3 = A

Problem 8. Express fractions.

7x 2 + 5x + 13 in partial (x 2 + 2)(x + 1)

17

18 Higher Engineering Mathematics The denominator is a combination of a quadratic factor, (x 2 + 2), which does not factorize without introducing imaginary surd terms, and a linear factor, (x + 1). Let,

Equating the numerators gives: 3 + 6x + 4x 2 − 2x 3 ≡ Ax(x 2 + 3) + B(x 2 + 3) + (Cx + D)x 2

7x 2 + 5x + 13 Ax + B C ≡ 2 + 2 (x + 2)(x + 1) (x + 2) (x + 1) ≡

(Ax + B)(x + 1) + C(x 2 + 2) (x 2 + 2)(x + 1)

≡ Ax 3 + 3 Ax + Bx 2 + 3B + Cx 3 + Dx 2 Let x = 0. Then 3 = 3B i.e.

Equating numerators gives: 7x 2 + 5x + 13 ≡ (Ax + B)(x + 1) + C(x 2 + 2) (1)

Equating the coefﬁcients of x 3 terms gives:

Let x = −1. Then

−2 = A + C

7(−1)2 + 5(−1) + 13 ≡ (Ax

+ B)(0) + C(1 + 2)

15 = 3C C= 5

i.e. i.e.

B=1

Equating the coefﬁcients of x 2 terms gives: 4= B+D Since B = 1, D = 3

Identity (1) may be expanded as: 7x 2 + 5x + 13 ≡ Ax 2 + Ax + Bx + B + Cx 2 + 2C

Equating the coefﬁcients of x terms gives:

Equating the coefﬁcients of x 2 terms gives: 7 = A + C, and since C = 5, A = 2 Equating the coefﬁcients of x terms gives: 5 = A + B, and since A = 2, B = 3

6 = 3A A=2

i.e.

From equation (1), since A = 2, C = −4 Hence

[Check: equating the constant terms gives:

3 + 6 x + 4x2 − 2 x3 −4x + 3 2 1 ≡ + 2+ 2 x2 (x2 + 3) x x x +3

13 = B + 2C

≡

When B = 3 and C = 5, B + 2C = 3 + 10 = 13 = LHS] Hence

7x2 + 5x + 13 (x2 + 2)(x + 1)

Problem 9.

≡

Resolve

partial fractions.

2x + 3 5 + ( x2 + 2) (x + 1)

3 + 6x + 4x 2 − 2x 3 into x 2 (x 2 + 3)

Terms such as x 2 may be treated as (x + 0)2 , i.e. they are repeated linear factors. Let

Ax(x 2 + 3) + B(x 2 + 3) + (Cx + D)x 2 x 2 (x 2 + 3)

2 3 − 4x 1 + 2+ 2 x x x +3

Now try the following exercise Exercise 10 Further problems on partial fractions with quadratic factors 2x + 3 1 x 2 − x − 13 − 1. (x 2 + 7)(x − 2) (x 2 + 7) (x − 2)

2.

6x − 5 (x − 4)(x 2 + 3)

3.

15 + 5x + 5x 2 − 4x 3 x 2 (x 2 + 5)

Cx + D A B 3 + 6x + 4x 2 − 2x 3 ≡ + 2+ 2 2 2 x (x + 3) x x (x + 3) ≡

(1)

1 2−x + (x − 4) (x 2 + 3) 1 2 − 5x 3 + + x x 2 (x 2 + 5)

Partial fractions

4.

following expression for L{θ} results:

x 3 + 4x 2 + 20x − 7 (x − 1)2 (x 2 + 8)

3 1 − 2x 2 + + (x − 1) (x − 1)2 (x 2 + 8)

5. When solving the differential equation d2θ dθ − 6 − 10θ = 20 − e2t by Laplace dt 2 dt transforms, for given boundary conditions, the

39 2 s + 42s − 40 2 L{θ} = s(s − 2)(s 2 − 6s + 10) 4s 3 −

Show that the expression can be resolved into partial fractions to give: L{θ} =

1 5s − 3 2 − + 2 s 2(s − 2) 2(s − 6s + 10)

19

Chapter 3

Logarithms 3.1

In another example, if we write down that 64 = 82 then the equivalent statement using logarithms is:

Introduction to logarithms

With the use of calculators ﬁrmly established, logarithmic tables are now rarely used for calculation. However, the theory of logarithms is important, for there are several scientiﬁc and engineering laws that involve the rules of logarithms. From the laws of indices:

16 = 2

4

The number 4 is called the power or the exponent or the index. In the expression 24 , the number 2 is called the base. In another example:

64 = 82

In this example, 2 is the power, or exponent, or index. The number 8 is the base. What is a logarithm? Consider the expression 16 = 24. An alternative, yet equivalent, way of writing this expression is: log2 16 = 4. This is stated as ‘log to the base 2 of 16 equals 4’. We see that the logarithm is the same as the power or index in the original expression. It is the base in the original expression which becomes the base of the logarithm. The two statements: 16 = 24 and log2 16 = 4 are equivalent. If we write either of them, we are automatically implying the other. In general, if a number y can be written in the form a x , then the index ‘x’ is called the ‘logarithm of y to the base of a’, i.e.

if y = a x then x = loga y

log8 64 = 2 In another example, if we write down that: log3 81 =4 then the equivalent statement using powers is: 34 = 81 So the two sets of statements, one involving powers and one involving logarithms, are equivalent. Common logarithms From above, if we write down that: 1000 = 103 , then 3 = log10 1000 This may be checked using the ‘log’ button on your calculator. Logarithms having a base of 10 are called common logarithms and log10 is often abbreviated to lg. The following values may be checked by using a calculator: lg 27.5 = 1.4393 . . ., lg 378.1 = 2.5776 . . . and lg 0.0204 = −1.6903 . . . Napierian logarithms Logarithms having a base of e (where ‘e’ is a mathematical constant approximately equal to 2.7183) are called hyperbolic, Napierian or natural logarithms, and loge is usually abbreviated to ln. The following values may be checked by using a calculator: ln 3.65 = 1.2947 . . ., ln 417.3 = 6.0338 . . . and ln 0.182 = −1.7037 . . . More on Napierian logarithms is explained in Chapter 4 following. Here are some worked problems to help understanding of logarithms.

Logarithms Problem 1. Evaluate log3 9.

Problem 6. Evaluate log3

Let x = log3 9 then 3 x = 9

from the deﬁnition of a logarithm,

3 x = 32

i.e.

1 . 81

Let x = log3

from which, x = 2

1 1 1 then 3 x = = 4 = 3−4 81 81 3 from which, x = −4

log3 9 = 2

Hence,

Hence,

log3

1 = −4 81

Problem 2. Evaluate log10 10. Problem 7. Solve the equation: lg x = 3. Let x = log10 10 then 10 x = 10

from the

deﬁnition of a logarithm, 10 = 10 x

i.e. Hence,

from which, x = 1

1

log10 10 = 1

(which may be checked by a calculator)

Problem 3. Evaluate log16 8.

If lg x = 3 then log10 x = 3 and

x = 103

i.e. x = 1000

Problem 8. Solve the equation: log2 x = 5. If log2 x = 5 then x = 25 = 32

Let x = log16 8 then 16 x = 8

from the deﬁnition

Problem 9. Solve the equation: log5 x = −2.

of a logarithm, i.e. (24 )x = 23 i.e. 24x = 23 from the laws of indices, from which, Hence,

4x = 3 and x = log16 8 =

If log5 x = −2 then x = 5−2 =

3 4

3 4

1 1 = 52 25

Now try the following exercise

Problem 4. Evaluate lg 0.001. then 10x = 0.001

Let x = lg 0.001 = log10 0.001 i.e. Hence,

10 x = 10−3

from which, x = −3

lg 0.001 = −3 (which may be checked

Exercise 11 logarithms

Further problems on laws of

In Problems 1 to 11, evaluate the given expressions: 1. log10 10000

[4]

2. log2 16

3. log5 125

[3] 1 3

4. log2 18

by a calculator) Problem 5. Evaluate ln e. 5. log8 2 Let x = ln e = loge e then ex = e i.e. Hence,

7. lg 100

ex = e1 from which, x = 1 ln e = 1 (which may be checked by a calculator)

9. log4 8 11. ln e2

[2] 1 1 2 [2]

6. log7 343 8. lg 0.01 10. log27 3

[4] [−3] [3] [−2] 1 3

21

22 Higher Engineering Mathematics The following may be checked using a calculator: In Problems 12 to 18 solve the equations: 12.

log10 x = 4

13.

lg x = 5

14.

log3 x = 2

15.

1 log4 x = −2 2

16.

lg x = −2

17.

log8 x = −

18.

ln x = 3

lg 52 = lg 25 = 1.39794. . .

[10000] [100000] [9] 1 32 [0.01] 1 16

4 3

[e3 ]

Also, 2 lg 5 = 2 × 0.69897. . . = 1.39794. . . lg 52 = 2 lg 5

Hence,

Here are some worked problems to help understanding of the laws of logarithms. Problem 10. Write log 4 + log 7 as the logarithm of a single number. log 4 + log 7 = log (7 × 4) by the ﬁrst law of logarithms = log 28

3.2

Laws of logarithms

There are three laws of logarithms, which apply to any base: (i) To multiply two numbers:

Problem 11. Write log 16 − log 2 as the logarithm of a single number.

16 log 16 − log 2 = log 2

by the second law of logarithms

log (A × B) = log A + log B The following may be checked by using a calculator: lg 10 = 1 Also, lg 5 + lg 2 = 0.69897. . . + 0.301029. . . = 1 Hence, lg (5 × 2) = lg 10 = lg 5 + lg 2 (ii) To divide two numbers: A = log A − log B log B The following may be checked using a calculator: 5 = ln 2.5 = 0.91629. . . ln 2 Also, Hence,

ln 5 − ln 2 = 1.60943. . . − 0.69314. . . = 0.91629. . . 5 = ln 5 − ln 2 ln 2

(iii) To raise a number to a power: log An = n log A

= log 8 Problem 12. Write 2 log 3 as the logarithm of a single number. 2 log 3 = log 32

by the third law of logarithms

= log 9 1 Problem 13. Write log 25 as the logarithm of a 2 single number. 1 1 log 25 = log 25 2 by the third law of logarithms 2 √ = log 25 = log 5

Problem 14.

Simplify: log 64 − log 128 + log32.

64 = 26, 128 = 27 and 32 = 25 Hence, log 64 − log 128 + log32 = log 26 − log 27 + log 25

Logarithms = 6 log2 − 7 log 2 + 5 log2 by the third law of logarithms = 4 log 2 1 1 Problem 15. Write log16 + log27 − 2 log5 2 3 as the logarithm of a single number. 1 1 log 16 + log 27 − 2 log5 2 3 1

1

= log 23 + log 5 4 − log 34 by the laws of indices √ 4 1 8× 5 = 3 log 2 + log 5 − 4 log 3 i.e. log 81 4 by the third law of logarithms Problem 18. Evaluate: log 25 − log125 + 12 log 625 . 3 log5

1

= log 16 2 + log 27 3 − log 52 by the third law of logarithms √ √ 3 = log 16 + log 27 − log 25 by the laws of indices

log 25 − log125 + 21 log 625 3 log5

= log4 + log 3 − log 25 4×3 = log 25 by the ﬁrst and second laws of logarithms 12 = log = log 0.48 25

=

log 52 − log 53 + 21 log 54 3 log5

=

2 log5 − 3 log 5 + 42 log 5 1 log5 1 = = 3 log5 3 log5 3

Problem 19. Solve the equation: log(x − 1) + log(x + 8) = 2 log(x + 2). LHS = log (x − 1) + log(x + 8)

Problem 16. Write (a) log30 (b) log 450 in terms of log 2, log3 and log 5 to any base.

= log (x − 1)(x + 8) from the ﬁrst law of logarithms

(a) log 30 = log(2 × 15) = log(2 × 3 × 5)

= log (x 2 + 7x − 8)

= log 2 + log 3 + log 5 by the ﬁrst law of logarithms

RHS = 2 log(x + 2) = log (x + 2)2

(b) log 450 = log(2 × 225) = log(2 × 3 × 75)

from the third law of logarithms

= log(2 × 3 × 3 × 25)

= log(x 2 + 4x + 4)

= log(2 × 32 × 52) = log2 + log 32 + log 52 by the ﬁrst law of logarithms

Hence,

log(x 2 + 7x − 8) = log (x 2 + 4x + 4) x 2 + 7x − 8 = x 2 + 4x + 4

i.e. log 450 = log 2 + 2 log 3 + 2 log 5 by the third law of logarithms

from which, i.e.

7x − 8 = 4x + 4

√ 4 8× 5 in terms of Problem 17. Write log 81 log 2, log3 and log 5 to any base.

i.e.

3x = 12

and

x=4

√ 4 √ 8× 5 4 = log 8 + log 5 − log 81 log 81 by the ﬁrst and second laws of logarithms

Problem 20. Solve the equation:

1 log 4 = log x. 2

1 1 log 4 = log4 2 from the third law of logarithms √ 2 = log 4 from the laws of indices

23

24 Higher Engineering Mathematics 1 log4 = log x 2 √ log 4 = log x

Hence, becomes

1 1 log 8 − log81 + log 27 3 2 1 log 4 − 2 log 3 + log45 9. 2 1 10. log 16 + 2 log3 − log 18 4 11. 2 log2 + log 5 − log 10 8.

log2 = log x

i.e.

2=x

from which,

i.e. the solution of the equation is: x = 2 Problem 21. Solve the equation: log x 2 − 3 − log x = log2.

x2 − 3 =2 x

Rearranging gives:

x 2 − 3 = 2x

13.

log 64 + log 32 − log 128 [log16 or log24 or 4 log2]

14.

log 8 − log4 + log 32 [log64 or log 26 or 6 log2]

16.

x = −1 is not a valid solution since the logarithm of a negative number has no real root. Hence, the solution of the equation is: x = 3 Now try the following exercise Exercise 12 logarithms

log 27 − log9 + log 81 [log 243 or log 35 or 5 log3]

15.

x = 3 or x = −1

from which,

[log 2]

12.

(x − 3)(x + 1) = 0

Factorizing gives:

[log 1 = 0]

Evaluate the expressions given in Problems 15 and 16:

x 2 − 2x − 3 = 0

and

[log 10]

Simplify the expressions given in Problems 12 to 14:

2 x −3 log x 2 − 3 − log x = log x from the second law of logarithms 2 x −3 Hence, = log 2 log x from which,

[log 6]

Further problems on laws of

In Problems 1 to 11, write as the logarithm of a single number:

1 1 2 log 16 − 3 log 8

log 4 log 9 − log3 + 12 log 81 2 log3

[0.5] [1.5]

Solve the equations given in Problems 17 to 22: 17.

log x 4 − log x 3 = log5x − log 2x

18.

log 2t 3 − log t = log 16 + logt

19.

2 logb 2 − 3 logb

20.

log (x + 1) + log(x − 1) = log 3

21. 22.

= log8b − log 4b

1 log 27 = log(0.5a) 3 log x 2 − 5 − log x = log 4

[x = 2.5] [t = 8] [b = 2] [x = 2] [a = 6] [x = 5]

1.

log 2 + log 3

[log 6]

2.

log 3 + log 5

[log 15]

3.

log 3 + log 4 − log 6

[log 2]

4.

log 7 + log 21 − log49

[log 3]

5.

2 log 2 + log 3

6.

2 log 2 + 3 log5

[log 500]

The laws of logarithms may be used to solve certain equations involving powers—called indicial equations. For example, to solve, say, 3 x = 27, logarithms to a base of 10 are taken of both sides,

7.

1 2 log 5 − log 81 + log 36 2

[log 100]

i.e. log10 3x = log10 27

[log 12]

3.3

Indicial equations

and x log10 3 = log10 27, by the third law of logarithms

Logarithms Rearranging gives x=

log10 27 1.43136 . . . = =3 log10 3 0.4771 . . .

which may be readily checked 8 log8 is not equal to lg Note, log2 2

log10 41.15 = 0.50449 3.2 Thus x = antilog 0.50449 =100.50449 = 3.195 correct to 4 signiﬁcant ﬁgures. Hence log10 x =

Now try the following exercise Exercise 13

Problem 22. Solve the equation 2 x = 3, correct to 4 signiﬁcant ﬁgures. Taking logarithms to base 10 of both sides of 2 x = 3 gives: log10 2x = log10 3 i.e.

x log10 2 = log10 3 log10 3 0.47712125 . . . = x= log10 2 0.30102999 . . . = 1.585, correct to 4 signiﬁcant ﬁgures

Indicial equations

Solve the following indicial equations for x, each correct to 4 signiﬁcant ﬁgures: 1. 3x = 6.4

[1.690]

2. 2 x = 9

[3.170]

3. 2 x−1 = 32x−1

[6.058]

5. 25.28 =4.2x

[2.251]

6. 42x−1 = 5x+2

[3.959]

7.

x −0.25 = 0.792

8. 0.027x = 3.26 equation 2 x+1 = 32x−5

Problem 23. Solve the correct to 2 decimal places.

Taking logarithms to base 10 of both sides gives:

[−0.3272]

where P1 is the power input and P2 is the P2 power output. Find the power gain when P1 n =25 decibels. [316.2]

(x + 1) log10 2 = (2x − 5) log10 3 x log10 2 + log10 2 = 2x log10 3 − 5 log10 3

x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771) i.e.

[2.542]

9. The decibel gain n of an ampliﬁer is given by: P2 n = 10 log10 P1

log10 2x+1 = log10 32x−5 i.e.

[0.2696]

x 1.5 = 14.91

4.

Rearranging gives:

25

0.3010x + 0.3010 = 0.9542x − 2.3855

Hence

3.4 2.3855 + 0.3010 = 0.9542x − 0.3010x 2.6865 = 0.6532x

from which x =

2.6865 = 4.11, correct to 0.6532 2 decimal places

Problem 24. Solve the equation x 3.2 = 41.15, correct to 4 signiﬁcant ﬁgures. Taking logarithms to base 10 of both sides gives: log10 x 3.2 = log10 41.15 3.2 log10 x = log10 41.15

Graphs of logarithmic functions

A graph of y = log10 x is shown in Fig. 3.1 and a graph of y = loge x is shown in Fig. 3.2. Both are seen to be of similar shape; in fact, the same general shape occurs for a logarithm to any base. In general, with a logarithm to any base a, it is noted that: (i) loga1 = 0 Let loga = x, then a x = 1 from the deﬁnition of the logarithm. If a x = 1 then x = 0 from the laws of indices. Hence loga 1 =0. In the above graphs it is seen that log10 1 = 0 and loge 1 = 0

26 Higher Engineering Mathematics y 2

y

1.0 1

0.5 0

1 x y 5 log10x

20.5

2 3

3 2

1

21

x 0.5

0.2

0.1

0.48 0.30 0 2 0.30 2 0.70 2 1.0

1

2

3

4

5

6

x

x 6 5 4 3 2 1 0.5 0.2 0.1 y 5 loge x 1.79 1.61 1.39 1.10 0.69 0 20.69 21.61 22.30

22

Figure 3.2

21.0

Figure 3.1

(ii) logaa = 1 Let loga a = x then a x = a from the deﬁnition of a logarithm. If a x = a then x = 1. Hence loga a = 1. (Check with a calculator that log10 10 = 1 and loge e = 1)

(iii) loga0 → −∞ Let loga 0 = x then a x = 0 from the deﬁnition of a logarithm. If a x = 0, and a is a positive real number, then x must approach minus inﬁnity. (For example, check with a calculator, 2−2 = 0.25, 2−20 = 9.54 × 10−7, 2−200 = 6.22 × 10−61, and so on) Hence loga 0 → −∞

Chapter 4

Exponential functions 4.1 Introduction to exponential functions An exponential function is one which contains ex , e being a constant called the exponent and having an approximate value of 2.7183. The exponent arises from the natural laws of growth and decay and is used as a base for natural or Napierian logarithms. The most common method of evaluating an exponential function is by using a scientiﬁc notation calculator. Use your calculator to check the following values: e = 2.7182818, correct to 8 signiﬁcant ﬁgures, 1

e−1.618 = 0.1982949, each correct to 7 signiﬁcant ﬁgures, e0.12 = 1.1275, correct to 5 signiﬁcant ﬁgures, e−1.47 = 0.22993, correct to 5 decimal places, e−0.431 = 0.6499, correct to 4 decimal places, e

9.32

0.0256 e5.21 − e2.49 = 0.0256 (183.094058 . . . − 12.0612761 . . .) = 4.3784, correct to 4 decimal places. Problem 2. Evaluate the following correct to 4 decimal places, using a calculator: e0.25 − e−0.25 5 0.25 e + e−0.25 5

e0.25 − e−0.25 e0.25 + e−0.25

1.28402541 . . . − 0.77880078 . . . 1.28402541 . . . + 0.77880078 . . . 0.5052246 . . . =5 2.0628262 . . .

=5

= 1.2246, correct to 4 decimal places.

= 11159, correct to 5 signiﬁcant ﬁgures,

e−2.785 = 0.0617291, correct to 7 decimal places.

Problem 1. Evaluate the following correct to 4 decimal places, using a calculator: 0.0256 e5.21 − e2.49

Problem 3. The instantaneous voltage v in a capacitive circuit is related to time t by the equation: v = V e−t /CR where V , C and R are constants. Determine v, correct to 4 signiﬁcant ﬁgures, when t = 50 ms, C = 10 μF, R = 47 k and V = 300 volts. v = V e−t /CR = 300e(−50×10

−3)/(10×10−6 ×47×103)

28 Higher Engineering Mathematics Using a calculator, v = 300e−0.1063829 ... = 300(0.89908025 . . .) = 269.7 volts Now try the following exercise Exercise 14 Further problems on evaluating exponential functions

(where 3! = 3 ×2 × 1 and is called ‘factorial 3’) The series is valid for all values of x. The series is said to converge, i.e. if all the terms are added, an actual value for e x (where x is a real number) is obtained. The more terms that are taken, the closer will be the value of ex to its actual value. The value of the exponent e, correct to say 4 decimal places, may be determined by substituting x = 1 in the power series of equation (1). Thus, e1 = 1 + 1 +

1. Evaluate the following, correct to 4 signiﬁcant ﬁgures: (a) e−1.8 (b) e−0.78 (c) e10 [(a) 0.1653 (b) 0.4584 (c) 22030]

+

2. Evaluate the following, correct to 5 signiﬁcant ﬁgures: (a) e1.629 (b) e−2.7483 (c) 0.62e4.178 [(a) 5.0988 (b) 0.064037 (c) 40.446] In Problems 3 and 4, evaluate correct to 5 decimal places: 5e2.6921 1 3. (a) e3.4629 (b) 8.52e−1.2651 (c) 1.1171 7 3e [(a) 4.55848 (b) 2.40444 (c) 8.05124] 5.6823 e2.1127 − e−2.1127 (b) e−2.1347 2 −1.7295 − 1) 4(e (c) e3.6817 [(a) 48.04106 (b) 4.07482 (c) −0.08286]

4. (a)

5. The length of a bar, l, at a temperature θ is given by l = l0 eαθ , where l0 and α are constants. Evaluate 1, correct to 4 signiﬁcant ﬁgures, where l0 = 2.587, θ = 321.7 and [2.739] α = 1.771 × 10−4. 6. When a chain of length 2L is suspended from two points, 2D metres hor apart, √on2the2 same L+ L +k . Evalizontal level: D = k ln k uate D when k = 75 m and L = 180 m. [120.7m]

4.2

The power series for ex

The value of e x can be calculated to any required degree of accuracy since it is deﬁned in terms of the following power series: ex = 1 + x +

x2 x3 x4 + + +··· 2! 3! 4!

(1)2 (1)3 (1)4 (1)5 + + + 2! 3! 4! 5!

(1)6 (1)7 (1)8 + + +··· 6! 7! 8!

= 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833 + 0.00139 + 0.00020 + 0.00002 + · · · i.e.

e = 2.71828 = 2.7183, correct to 4 decimal places

The value of e0.05, correct to say 8 signiﬁcant ﬁgures, is found by substituting x = 0.05 in the power series for e x . Thus e0.05 = 1 + 0.05 +

(0.05)2 (0.05)3 + 2! 3!

(0.05)4 (0.05)5 + +··· 4! 5! = 1 + 0.05 + 0.00125 + 0.000020833 +

+ 0.000000260 + 0.000000003 and by adding, e0.05 = 1.0512711, correct to 8 signiﬁcant ﬁgures In this example, successive terms in the series grow smaller very rapidly and it is relatively easy to determine the value of e0.05 to a high degree of accuracy. However, when x is nearer to unity or larger than unity, a very large number of terms are required for an accurate result. If in the series of equation (1), x is replaced by −x, then, e−x = 1 + (−x) + i.e. e−x = 1 − x +

(−x)2 (−x)3 + +··· 2! 3!

x2 x3 − +··· 2! 3!

In a similar manner the power series for e x may be used to evaluate any exponential function of the form a ekx ,

Exponential functions where a and k are constants. In the series of equation (1), let x be replaced by kx. Then, (kx)2 (kx)3 kx a e = a 1 + (kx) + + +··· 2! 3! (2x)2 (2x)3 2x Thus 5 e = 5 1 + (2x) + + +··· 2! 3! 4x 2 8x 3 = 5 1 + 2x + + +··· 2 6 4 i.e. 5 e2x = 5 1 + 2x + 2x 2 + x 3 + · · · 3 Problem 4. Determine the value of 5 e0.5 , correct to 5 signiﬁcant ﬁgures by using the power series for ex . ex = 1 + x + Hence

x2 2!

+

x3 3!

+

x4 4!

+···

(0.5)2 (0.5)3 e0.5 = 1 + 0.5 + + (2)(1) (3)(2)(1) (0.5)4

(0.5)5

+ (4)(3)(2)(1) (5)(4)(3)(2)(1) (0.5)6 + (6)(5)(4)(3)(2)(1)

+

= 1 + 0.5 + 0.125 + 0.020833 + 0.0026042 + 0.0002604 + 0.0000217 i.e.

e0.5

= 1.64872, correct to 6 signiﬁcant ﬁgures

Hence 5e0.5 = 5(1.64872) = 8.2436, correct to 5 signiﬁcant ﬁgures Problem 5. Expand ex (x 2 − 1) as far as the term in x 5 . The power series for ex is, ex = 1 + x +

x2 x3 x4 x5 + + + +··· 2! 3! 4! 5!

Hence e x (x 2 − 1) x2 x3 x4 x5 = 1+x + + + + + · · · (x 2 − 1) 2! 3! 4! 5!

=

x2 + x3 +

x4 x5 + +··· 2! 3!

29

x2 x3 x4 x5 + + + +··· − 1+x + 2! 3! 4! 5!

Grouping like terms gives: ex (x 2 − 1)

x2 x3 = −1 − x + x 2 − + x3 − 2! 3! 4 x x4 x5 x5 +··· + − + − 2! 4! 3! 5!

1 5 11 19 5 = − 1 −x + x2 + x3 + x4 + x 2 6 24 120 when expanded as far as the term in x 5 . Now try the following exercise Exercise 15 series for ex

Further problems on the power

1. Evaluate 5.6 e−1 , correct to 4 decimal places, [2.0601] using the power series for e x . 2. Use the power series for ex to determine, correct to 4 signiﬁcant ﬁgures, (a) e2 (b) e−0.3 and check your result by using a calculator. [(a) 7.389 (b) 0.7408] 3. Expand (1 − 2x) e2x as far as the term in x 4 . 8x 3 1 − 2x 2 − − 2x 4 3 2 1 4. Expand 2 ex x 2 to six terms. ⎤ ⎡ 1 5 9 1 13 2x 2 + 2x 2 + x 2 + x 2 ⎥ ⎢ 3 ⎥ ⎢ ⎦ ⎣ 17 21 1 1 + x2 + x2 12 60

4.3

Graphs of exponential functions

Values of ex and e−x obtained from a calculator, correct to 2 decimal places, over a range x = −3 to x = 3, are shown in the following table.

30 Higher Engineering Mathematics x ex

y

−3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0 0.05

0.08

0.14

0.22

0.37

0.61 1.00

e−x 20.09 12.18

7.39

4.48

2.72

1.65 1.00

5 3.87

y5 2e0.3x

4 3

x

0.5

1.0

1.5

2.0

2.5

3.0

2

ex

1.65

2.72

4.48

7.39

12.18

20.09

1

e−x

0.61

0.37

0.22

0.14

0.08

0.05

1.6

Figure 4.1 shows graphs of y = ex and y = e−x

y 5 ex 16

−1.5 −1.0 −0.5

x 8

−2x e−2x

4

22

21

1

2 2.2

3

x

A table of values is drawn up as shown below.

12

23

21 0 20.74

Problem 7. Plot a graph of y = 13 e−2x over the range x = −1.5 to x = 1.5. Determine from the graph the value of y when x = −1.2 and the value of x when y = 1.4.

20 y

22

Figure 4.2

y

5 e2x

23

1

2

3

3

2

0.5

1.0

1.5

−1

−2

−3

20.086 7.389 2.718 1.00 0.368 0.135 0.050

1 −2x 6.70 e 3

x

1

2.46 0.91 0.33 0.12 0.05 0.02

A graph of 13 e−2x is shown in Fig. 4.3. Figure 4.1

y 1 e22x

y 53

7 6

Problem 6. Plot a graph of y = 2 e0.3x over a range of x = − 2 to x = 3. Hence determine the value of y when x = 2.2 and the value of x when y = 1.6.

5 4

3.67

3 2

A table of values is drawn up as shown below.

1.4

1

x

−3

−2

−1

1

2

3

21.5 21.0 20.5

0.5

1.0

1.5

x

21.2 20.72

0.3x

−0.9 −0.6 −0.3

e0.3x

0.407 0.549 0.741 1.000 1.350 1.822 2.460

0.3

0.6

0.9

2 e0.3x 0.81 1.10 1.48 2.00 2.70 3.64 4.92 A graph of y = 2 e0.3x is shown plotted in Fig. 4.2. From the graph, when x = 2.2, y = 3.87 and when y = 1.6, x = −0.74.

Figure 4.3

From the graph, when x = −1.2, y = 3.67 and when y = 1.4, x = −0.72. Problem 8. The decay of voltage, v volts, across a capacitor at time t seconds is given by v = 250 e

−t 3 .

Draw a graph showing the natural

Exponential functions decay curve over the ﬁrst 6 seconds. From the graph, ﬁnd (a) the voltage after 3.4 s, and (b) the time when the voltage is 150 V.

of y when x = 1.4 and the value of x when y = 4.5. [3.95, 2.05] 2. Plot a graph of y = 12 e−1.5x over a range x = −1.5 to x = 1.5 and hence determine the value of y when x = −0.8 and the value of x when y = 3.5. [1.65, −1.30]

A table of values is drawn up as shown below. t

e

−t 3

−t v = 250 e 3

2

3

1.00

0.7165 0.5134 0.3679

250.0

179.1

t e

1

−t 3

−t v = 250 e 3

128.4 5

6

0.2636

0.1889

0.1353

65.90

47.22

33.83

The natural decay curve of v = 250 e Fig. 4.4.

−t 3

3. In a chemical reaction the amount of starting material C cm3 left after t minutes is given by C = 40 e−0.006t . Plot a graph of C against t and determine (a) the concentration C after 1 hour, and (b) the time taken for the concentration to decrease by half. [(a) 28 cm3 (b) 116 min]

91.97

4

is shown in

250 t

y 5 250e2 3

Voltage v (volts)

200

150

4. The rate at which a body cools is given by θ = 250 e−0.05t where the excess of temperature of a body above its surroundings at time t minutes is θ ◦ C. Plot a graph showing the natural decay curve for the ﬁrst hour of cooling. Hence determine (a) the temperature after 25 minutes, and (b) the time when the temperature is 195◦C. [(a) 70◦C (b) 5 min]

4.4

100 80 50

1 1.5 2 3 3.4 4 Time t(seconds)

5

6

Figure 4.4

Napierian logarithms

Logarithms having a base of ‘e’ are called hyperbolic, Napierian or natural logarithms and the Napierian logarithm of x is written as loge x, or more commonly as ln x. Logarithms were invented by John Napier, a Scotsman (1550–1617). The most common method of evaluating a Napierian logarithm is by a scientiﬁc notation calculator. Use your calculator to check the following values:

From the graph: (a)

31

when time t = 3.4 s, voltage v = 80 V and

(b) when voltage v = 150 V, time t = 1.5 s. Now try the following exercise

ln 4.328 = 1.46510554 . . . = 1.4651, correct to 4 decimal places ln 1.812 = 0.59443, correct to 5 signiﬁcant ﬁgures ln 1 = 0 ln 527 = 6.2672, correct to 5 signiﬁcant ﬁgures ln 0.17 = −1.772, correct to 4 signiﬁcant ﬁgures

Exercise 16 Further problems on exponential graphs

ln 0.00042 = −7.77526, correct to 6 signiﬁcant ﬁgures

1. Plot a graph of y = 3 e0.2x over the range x = −3 to x = 3. Hence determine the value

ln e3 = 3 ln e1 = 1

32 Higher Engineering Mathematics From the last two examples we can conclude that: loge ex = x This is useful when solving equations involving exponential functions. For example, to solve e3x = 7, take Napierian logarithms of both sides, which gives: ln e3x = ln 7 i.e. from which

3x = ln 7 1 x = ln 7 = 0.6486, correct to 4 3 decimal places.

Problem 9. Evaluate the following, each correct to 5 signiﬁcant ﬁgures: (a) (a)

(b)

(c)

1 ln 7.8693 3.17 ln 24.07 . ln 4.7291 (b) (c) 2 7.8693 e−0.1762

1 1 ln 4.7291 = (1.5537349 . . .) = 0.77687, 2 2 correct to 5 signiﬁcant ﬁgures ln 7.8693 2.06296911 . . . = = 0.26215, 7.8693 7.8693 correct to 5 signiﬁcant ﬁgures 3.17 ln 24.07 3.17(3.18096625 . . .) = e−0.1762 0.83845027 . . . = 12.027, correct to 5 signiﬁcant ﬁgures.

Problem 10. (a)

(a)

(b)

Evaluate the following:

ln e2.5 5e2.23 lg 2.23 (b) (correct to 3 0.5 ln 2.23 lg 10 decimal places).

t Problem 12. Given 32 = 70 1 − e− 2 determine the value of t , correct to 3 signiﬁcant ﬁgures. t

Rearranging 32 = 70(1 − e− 2 ) gives: t 32 = 1 − e− 2 70 t 32 38 and e− 2 = 1 − = 70 70 Taking the reciprocal of both sides gives: t 70 e2 = 38 Taking Napierian logarithms of both sides gives: t 70 ln e 2 = ln 38 t 70 i.e. = ln 2 38 70 from which, t = 2 ln = 1.22, correct to 3 signiﬁ38 cant ﬁgures.

Problem 13.

ln e2.5 2.5 = =5 lg 100.5 0.5

4.87 Solve the equation: 2.68 = ln x

to ﬁnd x.

5e2.23 lg 2.23 ln 2.23 5(9.29986607 . . .)(0.34830486 . . .) = 0.80200158 . . . = 20.194, correct to 3 decimal places.

Problem 11. Solve the equation: 9 = 4e−3x to ﬁnd x, correct to 4 signiﬁcant ﬁgures. Rearranging 9 = 4e−3x gives:

Taking the reciprocal of both sides gives: 4 1 = e3x = 9 e−3x Taking Napierian logarithms of both sides gives: 4 = ln(e3x ) ln 9 4 α = 3x Since loge e = α, then ln 9 1 4 1 Hence, x = ln = (−0.81093) = −0.2703, 3 9 3 correct to 4 signiﬁcant ﬁgures.

9 = e−3x 4

From thedeﬁnition of a logarithm, since 4.87 4.87 then e2.68 = 2.68 = ln x x 4.87 Rearranging gives: x = 2.68 = 4.87e−2.68 e i.e. x = 0.3339, correct to 4 signiﬁcant ﬁgures. 7 Problem 14. Solve = e3x correct to 4 signi4 ﬁcant ﬁgures.

Exponential functions Taking natural logs of both sides gives:

Since ln e = 1

ln

7 = ln e3x 4

ln

7 = 3x ln e 4

ln

7 = 3x 4

−1 ±

12 − 4(1)(−10.953) 2 √ −1 ± 44.812 −1 ± 6.6942 = = 2 2

x = 2.847 or −3.8471

i.e.

x = 0.1865, correct to 4 signiﬁcant ﬁgures.

i.e.

Using the quadratic formula, x=

0.55962 = 3x

i.e.

Problem 15. Solve: e x−1 = 2e3x−4 correct to 4 signiﬁcant ﬁgures.

x = −3.8471 is not valid since the logarithm of a negative number has no real root. Hence, the solution of the equation is: x = 2.847 Now try the following exercise

Taking natural logarithms of both sides gives: ln e x−1 = ln 2e3x−4

Exercise 17 Further problems on evaluating Napierian logarithms

and by the ﬁrst law of logarithms, ln e x−1 = ln 2 + ln e3x−4

In Problems 1 and 2, evaluate correct to 5 signiﬁcant ﬁgures:

x − 1 = ln 2 + 3x − 4

i.e.

Rearranging gives: 4 − 1 − ln 2 = 3x − x 3 − ln 2 = 2x

i.e.

3 − ln 2 2 = 1.153

x=

from which,

Problem 16. Solve, correct to 4 signiﬁcant ﬁgures: ln(x − 2)2 = ln(x − 2) − ln(x + 3) + 1.6 Rearranging gives: ln(x − 2)2 − ln(x − 2) + ln(x + 3) = 1.6

5e−0.1629 1.786 ln e1.76 (b) lg 101.41 2 ln 0.00165 ln 4.8629 − ln 2.4711 (c) 5.173 [(a) 2.2293 (b) −0.33154 (c) 0.13087]

2. (a)

3. ln x = 2.10

[8.166]

4. 24 + e2x = 45

[1.522]

5. 5 =

and by the laws of logarithms,

Cancelling gives:

1 ln 82.473 ln 5.2932 (b) 3 4.829 5.62 ln 321.62 (c) e1.2942 [(a) 0.55547 (b) 0.91374 (c) 8.8941]

1. (a)

In Problems 3 to 7 solve the given equations, each correct to 4 signiﬁcant ﬁgures.

ln

33

(x − 2)2 (x + 3) = 1.6 (x − 2)

ln {(x − 2)(x + 3)} = 1.6

and

(x − 2)(x + 3) = e1.6

i.e.

x 2 + x − 6 = e1.6

or

x 2 + x − 6 − e1.6 = 0

i.e.

x 2 + x − 10.953 = 0

e x+1 − 7

6. 1.5 = 4e2t 7. 7.83 =

2.91e−1.7x

t −2 8. 16 = 24 1 − e x 9. 5.17 = ln 4.64 1.59 = 2.43 10. 3.72 ln x

[1.485] [−0.4904] [−0.5822] [2.197] [816.2] [0.8274]

34 Higher Engineering Mathematics

11. 12.

y

−x 5 = 8 1−e 2

[1.962]

ln(x + 3) − ln x = ln(x − 1) − 1)2

− ln 3 = ln(x − 1)

y 5 Ae2kx

[3] [4]

13.

ln(x

14.

ln(x + 3) + 2 = 12 − ln(x − 2)

[147.9]

15.

e(x+1)

[4.901]

16.

ln(x + 1)2 = 1.5 − ln(x − 2) + ln(x + 1)

17.

18.

=

3e(2x−5)

19.

If U2 = U1 e

W PV

formula. 20.

A

y 5 A(12e2kx )

makeW the subject of the U2 W = PV ln U1

Laws of growth and decay

The laws of exponential growth and decay are of the form y = A e−kx and y = A(1 − e−kx ), where A and k are constants. When plotted, the form of each of these equations is as shown in Fig. 4.5. The laws occur frequently in engineering and science and examples of quantities related by a natural law include. l = l0 eαθ

(ii) Change in electrical resistance with temperature Rθ = R0 eαθ (iii) Tension in belts

y

The work done in an isothermal expansion of a gas from pressure p1 to p2 is given by: p1 w = w0 ln p2

(i) Linear expansion

x

[3.095]

If the initial pressure p1 = 7.0 kPa, calculate the ﬁnal pressure p2 if w = 3 w0 . [ p2 = 348.5 Pa]

4.5

0 (a)

Transpose: b = ln t − a ln D to make t the subject. a [t = eb+a ln D = eb ea ln D = eb eln D i.e. t = eb D a ] R1 P = 10 log10 ﬁnd the value of R1 If Q R2 when P = 160, Q = 8 and R2 = 5. [500]

A

Figure 4.5

(v) Biological growth

y = y0 ekt

(vi) Discharge of a capacitor q = Q e−t/CR (vii) Atmospheric pressure

p = p0 e−h/c

(viii) Radioactive decay

N = N0 e−λt

(ix) Decay of current in an inductive circuit

i = I e− Rt /L

(x) Growth of current in a capacitive circuit

i = I (1 − e−t/CR )

Problem 17. The resistance R of an electrical conductor at temperature θ ◦ C is given by R = R0 eαθ , where α is a constant and R0 = 5 × 103 ohms. Determine the value of α, correct to 4 signiﬁcant ﬁgures, when R = 6 ×103 ohms and θ = 1500◦C. Also, ﬁnd the temperature, correct to the nearest degree, when the resistance R is 5.4 ×103 ohms. R = eαθ . R0 Taking Napierian logarithms of both sides gives:

Transposing R = R0 eαθ gives

T1 = T0 eμθ

(iv) Newton’s law of cooling θ = θ0 e−kt

x

(b)

ln

R = ln eαθ = αθ R0

Exponential functions 6 × 103 1 1 R Hence α = ln = ln θ R0 1500 5 × 103 =

1 (0.1823215 . . .) 1500

= 1.215477 · · ·× 10−4 Hence α = 1.215 × 10−4 , correct to 4 signiﬁcant ﬁgures. R = αθ From above, ln R0 θ=

hence

1 R ln α R0

When R = 5.4 × 103, α = 1.215477 . . . × 10−4 and R0 = 5 ×103 5.4 × 103 1 θ= ln 1.215477 . . . × 10−4 5 × 103 =

104 (7.696104 . . . × 10−2) 1.215477 . . . ◦

= 633 C, correct to the nearest degree.

35

Problem 19. The current i amperes ﬂowing in a capacitor at time t seconds is given by −t

i = 8.0(1 − e CR ), where the circuit resistance R is 25 ×103 ohms and capacitance C is 16 ×10−6 farads. Determine (a) the current i after 0.5 seconds and (b) the time, to the nearest millisecond, for the current to reach 6.0 A. Sketch the graph of current against time. (a)

−t

Current i = 8.0(1 − e CR ) −0.5

= 8.0[1 − e (16 ×10−6 )(25 ×103 ) ] =8.0(1 − e−1.25) = 8.0(1 − 0.2865047 . . .) = 8.0(0.7134952 . . .) = 5.71 amperes −t

(b) Transposing i = 8.0(1 − e CR )

gives

−t i = 1 −e CR 8.0 −t

from which, e CR = 1 −

i 8.0 − i = 8.0 8.0

Taking the reciprocal of both sides gives: Problem 18. In an experiment involving Newton’s law of cooling, the temperature θ(◦ C) is given by θ = θ0 e−kt . Find the value of constant k when θ0 = 56.6◦ C, θ = 16.5◦ C and t = 83.0 seconds. Transposing

θ = θ0 e−kt gives θ = e−kt θ0

from which

θ0 1 = −kt = ekt θ e

Taking Napierian logarithms of both sides gives: θ0 ln = kt θ from which, 56.6 1 1 θ0 ln k = ln = t θ 83.0 16.5 1 = (1.2326486 . . .) 83.0 Hence k = 1.485 × 10−2

t

e CR =

8.0 8.0 − i

Taking Napierian logarithms of both sides gives: t 8.0 = ln CR 8.0 − i Hence

8.0 t = CRln 8.0 − i

= (16 × 10−6)(25 × 103 ) ln

8.0 8.0 − 6.0

when i = 6.0 amperes, 8.0 400 i.e. t = 3 ln = 0.4 ln 4.0 10 2.0 = 0.4(1.3862943 . . .) = 0.5545 s = 555 ms, to the nearest millisecond. A graph of current against time is shown in Fig. 4.6.

36 Higher Engineering Mathematics i (A)

Hence the time for the temperature θ 2 to be one half of the value of θ 1 is 41.6 s, correct to 1 decimal place.

8 6 5.71

i 5 8.0 (12e2t/CR)

4

Now try the following exercise

2

Exercise 18 Further problems on the laws of growth and decay

0.5 0.555

1.0

1.5

t(s)

Figure 4.6

Problem 20. The temperature θ2 of a winding which is being heated electrically at time t is given −t θ2 = θ1 (1 − e τ )

by: where θ1 is the temperature (in degrees Celsius) at time t = 0 and τ is a constant. Calculate, (a)

θ1 , correct to the nearest degree, when θ2 is 50◦ C, t is 30 s and τ is 60 s

(b) the time t , correct to 1 decimal place, for θ2 to be half the value of θ1 . (a) Transposing the formula to make θ1 the subject gives: θ1 =

θ2

−t (1 − e T )

=

50 1−e

−30 60

50 50 = = 1 − e−0.5 0.393469 . . . i.e. θ 1 = 127◦ C, correct to the nearest degree. (b) Transposing to make t the subject of the formula gives: −t θ2 =1−e τ θ1 −t θ2 from which, e τ = 1 − θ 1 t θ2 Hence − = ln 1 − τ θ 1 θ2 i.e. t = −τ ln 1 − θ1 1 Since θ2 = θ1 2 1 t = −60 ln 1 − 2 = −60 ln 0.5 = 41.59 s

1. The temperature, T ◦C, of a cooling object varies with time, t minutes, according to the equation: T = 150e−0.04t . Determine the temperature when (a) t = 0, (b) t = 10 minutes. [(a) 150◦ C (b) 100.5◦ C ] 2. The pressure p pascals at height h metres −h

above ground level is given by p = p0 e C , where p0 is the pressure at ground level and C is a constant. Find pressure p when p0 = 1.012 × 105 Pa, height h = 1420 m, and C = 71500. [99210] 3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given − Rt

by v = 200 e L , where R = 150 and L =12.5 × 10−3 H. Determine (a) the voltage when t = 160 ×10−6 s, and (b) the time for the voltage to reach 85 V. [(a) 29.32 volts (b) 71.31 × 10−6 s] 4. The length l metres of a metal bar at temperature t ◦ C is given by l = l0 eαt , where l0 and α are constants. Determine (a) the value of α when l = 1.993 m, l0 = 1.894 m and t = 250◦C, and (b) the value of l0 when l = 2.416, t = 310◦C and α = 1.682 ×10−4. [(a) 2.038 × 10−4 (b) 2.293 m] 5. The temperature θ2◦ C of an electrical conductor at time t seconds is given by: θ2 = θ1 (1 − e−t / T ), where θ1 is the initial temperature and T seconds is a constant. Determine: (a) θ2 when θ1 = 159.9◦C, t = 30 s and T = 80 s, and (b) the time t for θ2 to fall to half the value of θ1 if T remains at 80 s. [(a) 50◦ C (b) 55.45 s ] 6. A belt is in contact with a pulley for a sector of θ = 1.12 radians and the coefﬁcient

Exponential functions

of friction between these two surfaces is μ = 0.26. Determine the tension on the taut side of the belt, T newtons, when tension on the slack side T0 = 22.7 newtons, given that these quantities are related by the law T = T0 eμθ .Determine also the value of θ when T = 28.0 newtons. [30.4 N, 0.807 rad] 7. The instantaneous current i at time t is given −t

by: i = 10 e CR when a capacitor is being charged. The capacitance C is 7 ×10−6 farads and the resistance R is 0.3 × 106 ohms. Determine: (a) the instantaneous current when t is 2.5 seconds, and (b) the time for the instantaneous current to fall to 5 amperes Sketch a curve of current against time from t = 0 to t = 6 seconds. [(a) 3.04 A (b) 1.46 s] 8. The amount of product x (in mol/cm3) found in a chemical reaction starting with 2.5 mol/cm3 of reactant is given by x = 2.5(1 − e−4t ) where t is the time, in minutes, to form product x. Plot a graph at 30 second intervals up to 2.5 minutes and determine x after 1 minute. [2.45 mol/cm3] 9. The current i ﬂowing in a capacitor at time t is given by: −t

be determined. This technique is called ‘determination of law’. Graph paper is available where the scale markings along the horizontal and vertical axes are proportional to the logarithms of the numbers. Such graph paper is called log-log graph paper. A logarithmic scale is shown in Fig. 4.7 where the distance between, say 1 and 2, is proportional to lg 2 − lg 1, i.e. 0.3010 of the total distance from 1 to 10. Similarly, the distance between 7 and 8 is proportional to lg 8 − lg 7, i.e. 0.05799 of the total distance from 1 to 10. Thus the distance between markings progressively decreases as the numbers increase from 1 to 10. 1

2

3

4

5

6 7 8 910

Figure 4.7

With log-log graph paper the scale markings are from 1 to 9, and this pattern can be repeated several times. The number of times the pattern of markings is repeated on an axis signiﬁes the number of cycles. When the vertical axis has, say, 3 sets of values from 1 to 9, and the horizontal axis has, say, 2 sets of values from 1 to 9, then this log-log graph paper is called ‘log 3 cycle × 2 cycle’. Many different arrangements are available ranging from ‘log 1 cycle × 1 cycle’ through to ‘log 5 cycle × 5 cycle’. To depict a set of values, say, from 0.4 to 161, on an axis of log-log graph paper, 4 cycles are required, from 0.1 to 1, 1 to 10, 10 to 100 and 100 to 1000. Graphs of the form y = a ekx

i = 12.5(1 − e CR )

Taking logarithms to a base of e of both sides of y = a ekx gives:

where resistance R is 30 kilohms and the capacitance C is 20 micro-farads. Determine:

ln y = ln(a ekx ) = ln a + ln ekx = ln a + kx ln e

(a)

the current ﬂowing after 0.5 seconds, and

(b) the time for the current to reach 10 amperes. [(a) 7.07 A (b) 0.966 s]

4.6 Reduction of exponential laws to linear form Frequently, the relationship between two variables, say x and y, is not a linear one, i.e. when x is plotted against y a curve results. In such cases the non-linear equation may be modiﬁed to the linear form, y = mx + c, so that the constants, and thus the law relating the variables can

37

i.e. ln y = kx + ln a

(since ln e = 1)

which compares with Y = m X + c Thus, by plotting ln y vertically against x horizontally, a straight line results, i.e. the equation y = a ekx is reduced to linear form. In this case, graph paper having a linear horizontal scale and a logarithmic vertical scale may be used. This type of graph paper is called log-linear graph paper, and is speciﬁed by the number of cycles on the logarithmic scale. Problem 21. The data given below is believed to be related by a law of the form y = a ekx , where a and b are constants. Verify that the law is true and

38 Higher Engineering Mathematics The law of the graph is thus y = 18 e0.55x

determine approximate values of a and b. Also determine the value of y when x is 3.8 and the value of x when y is 85. x −1.2 0.38 y

9.3

1.2

2.5

3.4

4.2

When x is 3.8, y = 18 e0.55(3.8) = 18 e2.09 = 18(8.0849) = 146

5.3

When y is 85, 85 = 18 e0.55x

22.2 34.8 71.2 117 181 332

Since y = a ekx then ln y = kx + ln a (from above), which is of the form Y = m X + c, showing that to produce a straight line graph ln y is plotted vertically against x horizontally. The value of y ranges from 9.3 to 332 hence ‘log 3 cycle × linear’ graph paper is used. The plotted co-ordinates are shown in Fig. 4.8 and since a straight line passes through the points the law y = a ekx is veriﬁed. Gradient of straight line, k=

AB ln 100 − ln 10 2.3026 = = BC 3.12 − (−1.08) 4.20

e0.55x =

and

0.55x = ln 4.7222 = 1.5523 x=

Hence

Since ln y = kx + ln a, when x = 0, ln y = ln a, i.e. y = a The vertical axis intercept value at x = 0 is 18, hence a = 18 1000 y

1.5523 = 2.82 0.55

Problem 22. The voltage, v volts, across an inductor is believed to be related to time, t ms, by t

the law v = V e T , where V and T are constants. Experimental results obtained are: v volts 883

= 0.55, correct to 2 signiﬁcant ﬁgures.

85 = 4.7222 18

Hence,

t ms

347

90

55.5 18.6

5.2

10.4 21.6 37.8 43.6 56.7 72.0

Show that the law relating voltage and time is as stated and determine the approximate values of V and T . Find also the value of voltage after 25 ms and the time when the voltage is 30.0 V. t

Since v = V e T then ln v = T1 t + ln V which is of the form Y = m X + c. Using ‘log3 cycle × linear’ graph paper, the points are plotted as shown in Fig. 4.9. Since the points are joined by a straight line the law

y 5a e kx

100

10

t

v = Ve T is veriﬁed. Gradient of straight line, 1 AB = T BC ln 100 − ln 10 = 36.5 − 64.2

A

B

C

=

2.3026 −27.7

Hence T =

−27.7 2.3026

= −12.0, correct to 3 signiﬁcant ﬁgures. 1 22

21

Figure 4.8

1

2

3

4

5

6

x

Since the straight line does not cross the vertical axis at t = 0 in Fig. 4.9, the value of V is determined by selecting any point, say A, having co-ordinates t

(36.5,100) and substituting these values into v = V e T .

Exponential functions Now try the following exercise

1000

v 5Ve

Exercise 19 Further problems on reducing exponential laws to linear form

t T

1. Atmospheric pressure p is measured at varying altitudes h and the results are as shown below:

(36.5, 100)

100

A

Voltage, v volts

Altitude, h m

10

B

C

1 0

10

20

30

40 50 Time, t ms

60

70

80

Figure 4.9

−36.5

correct to 3 signiﬁcant ﬁgures.

−t

Hence the law of the graph is v = 2090 e 12.0 . When time t = 25 ms, −25

v = 2090 e 12.0 = 260 V −t

When the voltage is 30.0 volts, 30.0 = 2090 e 12.0 , hence

−t

e 12.0 =

30.0 2090

t

2090 = 69.67 30.0 Taking Napierian logarithms gives: and

1500

68.42

3000

61.60

5000

53.56

8000

43.41

a = 76, k = −7 × 10−5, −5 h

, 37.74 cm

2. At particular times, t minutes, measurements are made of the temperature, θ ◦ C, of a cooling liquid and the following results are obtained:

e 12.0 = 2090 volts,

voltage

73.39

p = 76 e−7×10

100

V =

500

Show that the quantities are related by the law p =a ekh , where a and k are constants. Determine the values of a and k and state the law. Find also the atmospheric pressure at 10 000 m.

36.5

Thus 100 = V e −12.0 i.e.

90

pressure, p cm

e 12.0 =

t = ln 69.67 = 4.2438 12.0 from which, time t = (12.0)(4.2438) = 50.9 ms

Temperature θ ◦ C

Time t minutes

92.2

10

55.9

20

33.9

30

20.6

40

12.5

50

Prove that the quantities follow a law of the form θ = θ0 ekt , where θ0 and k are constants, and determine the approximate value of θ0 and k. [θ0 = 152, k = − 0.05]

39

Revision Test 1 This Revision Test covers the material contained in Chapters 1 to 4. The marks for each question are shown in brackets at the end of each question. 1.

Factorise x 3 + 4x 2 + x − 6 using the factor theorem. Hence solve the equation x 3 + 4x 2 + x − 6 =0

2.

(6)

Use the remainder theorem to ﬁnd the remainder when 2x 3 + x 2 − 7x − 6 is divided by (a) (x − 2) (b) (x + 1) Hence factorise the cubic expression 6x 2 + 7x − 5 by dividing out 2x − 1

3.

Simplify

4.

Resolve the following into partial fractions (a) (c)

5.

x − 11 −2

x2 − x

(b)

(x 2

(4)

3−x + 3)(x + 3)

x 3 − 6x + 9 x2 + x − 2

8. (24)

Evaluate, correct to 3 decimal places, 5 e−0.982 3 ln0.0173

6.

(7)

(2)

Solve the following equations, each correct to 4 signiﬁcant ﬁgures: x

Solve the following equations: (a) log x 2 + 8 − log(2x) = log 3

(b) ln x + ln(x – 3) = ln 6x – ln(x – 2) (13) R U2 9. If θ f − θi = ln ﬁnd the value of U2 J U1 given that θ f = 3.5, θi = 2.5, R = 0.315, J = 0.4, (6) U1 = 50 10.

Solve, correct to 4 signiﬁcant ﬁgures: (a) 13e2x−1 = 7ex

(a) ln x = 2.40 (b) 3x−1 = 5x−2 (c) 5 = 8(1 − e− 2 )

7. (a) The pressure p at height h above ground level is given by: p = p0 e−kh where p0 is the pressure at ground level and k is a constant. When p0 is 101 kilopascals and the pressure at a height of 1500 m is 100 kilopascals, determine the value of k. (b) Sketch a graph of p against h ( p the vertical axis and h the horizontal axis) for values of height from zero to 12 000 m when p0 is 101 kilopascals. (c) If pressure p = 95 kPa, ground level pressure p0 = 101 kPa, constant k = 5 × 10−6, determine the height above ground level, h, in kilometres correct to 2 decimal places. (13)

(10)

(b) ln (x + 1)2 = ln(x + 1) – ln(x + 2) + 2

(15)

Chapter 5

Hyperbolic functions (v) Hyperbolic secant of x,

5.1 Introduction to hyperbolic functions

sech x =

Functions which are associated with the geometry of the conic section called a hyperbola are called hyperbolic functions. Applications of hyperbolic functions include transmission line theory and catenary problems. By deﬁnition: (i) Hyperbolic sine of x, ex − e−x sinh x = 2

1 2 = cosh x e x + e−x

(5)

‘sech x’ is pronounced as ‘shec x’ (vi) Hyperbolic cotangent of x, coth x =

e x + e−x 1 = x −x tanh x e − e

(6)

‘coth x’ is pronounced as ‘koth x’ (1)

Some properties of hyperbolic functions Replacing x by 0 in equation (1) gives:

‘sinh x’ is often abbreviated to ‘sh x’ and is pronounced as ‘shine x’

sinh 0 =

(ii) Hyperbolic cosine of x, e x + e−x cosh x = 2

Replacing x by 0 in equation (2) gives: (2)

‘cosh x’ is often abbreviated to ‘ch x’ and is pronounced as ‘kosh x’ (iii) Hyperbolic tangent of x, sinh x e x − e−x = tanh x = cosh x e x + e−x

(3)

‘tanh x’ is often abbreviated to ‘th x’ and is pronounced as ‘than x’ (iv) Hyperbolic cosecant of x, cosech x =

1 2 = sinh x e x − e−x

‘cosech x’ is pronounced as ‘coshec x’

e0 − e−0 1−1 = =0 2 2

(4)

cosh 0 =

e0 + e−0 1 + 1 = =1 2 2

If a function of x, f (−x) = − f (x), then f (x) is called an odd function of x. Replacing x by −x in equation (1) gives: e−x − e x e−x − e−(−x) = 2 2 x −x e −e =− = −sinh x 2

sinh(−x) =

Replacing x by −x in equation (3) gives: e−x − e−(−x) e−x − e x = e−x + e−(−x) e−x + e x x e − e−x =− x = −tanh x e + e−x

tanh(−x) =

42 Higher Engineering Mathematics Hence sinh x and tanh x are both odd functions 1 and (see Section 5.1), as also are cosech x = sinh x 1 coth x = tanh x If a function of x, f (−x) = f (x), then f (x) is called an even function of x. Replacing x by −x in equation (2) gives: e−x + e−(−x) e−x + e x = 2 2 = cosh x

cosh(−x) =

Hence cosh xis an evenfunction (see Section 5.2), as 1 also is sech x = cosh x Hyperbolic functions may be evaluated easiest using a calculator. Many scientiﬁc notation calculators actually possess sinh and cosh functions; however, if a calculator does not contain these functions, then the deﬁnitions given above may be used. Problem 1. Evaluate sinh 5.4, correct to 4 signiﬁcant ﬁgures.

Problem 3. Evaluate th 0.52, correct to 4 signiﬁcant ﬁgures. Using a calculator with the procedure similar to that used in Worked Problem 1, th 0.52 = 0.4777, correct to 4 signiﬁcant ﬁgures. Problem 4. Evaluate cosech 1.4, correct to 4 signiﬁcant ﬁgures. cosech 1.4 =

1 sinh 1.4

Using a calculator, (i) press hyp (ii) press 1 and sinh( appears (iii) type in 1.4 (iv) press ) to close the brackets (v) press = and 1.904301501 appears (vi) press x −1

Using a calculator, (i) press hyp (ii) press 1 and sinh( appears

(vii) press = and 0.5251269293 appears Hence, cosech 1.4 = 0.5251, correct to 4 signiﬁcant ﬁgures.

(iii) type in 5.4 (iv) press ) to close the brackets (v) press = and 110.7009498 appears Hence, sinh 5.4 = 110.7, correct to 4 signiﬁcant ﬁgures. 1 Alternatively, sinh 5.4 = (e5.4 − e−5.4 ) 2 1 = (221.406416 . . . − 0.00451658 . . .) 2 1 = (221.401899 . . .) 2 = 110.7, correct to 4 signiﬁcant ﬁgures. Problem 2. Evaluate cosh 1.86, correct to 3 decimal places. Using a calculator with the procedure similar to that used in Worked Problem 1, cosh 1.86 = 3.290, correct to 3 decimal places.

Problem 5. Evaluate sech 0.86, correct to 4 signiﬁcant ﬁgures. sech 0.86 =

1 cosh 0.86

Using a calculator with the procedure similar to that used in Worked Problem 4, sech 0.86 = 0.7178, correct to 4 signiﬁcant ﬁgures. Problem 6. Evaluate coth 0.38, correct to 3 decimal places. coth 0.38 =

1 tanh 0.38

Using a calculator with the procedure similar to that used in Worked Problem 4, coth 0.38 = 2.757, correct to 3 decimal places.

Hyperbolic functions

43

y

Now try the following exercise

10 8 6

Exercise 20 Further problems on evaluating hyperbolic functions

y 5sinh x

4 2

In Problems 1 to 6, evaluate correct to 4 signiﬁcant ﬁgures.

23 22 21 0 1 2 22

1. (a) sh 0.64 (b) sh 2.182

3 x

24 26

[(a) 0.6846 (b) 4.376]

28

2. (a) ch 0.72 (b) ch 2.4625

210

[(a) 1.271 (b) 5.910] Figure 5.1

3. (a) th 0.65 (b) th 1.81 [(a) 0.5717 (b) 0.9478] 4. (a) cosech 0.543 (b) cosech 3.12 [(a) 1.754 (b) 0.08849] 5. (a) sech 0.39 (b) sech 2.367 [(a) 0.9285 (b) 0.1859]

cosh x is an even function (as stated in Section 5.1). The shape of y = cosh x is that of a heavy rope or chain hanging freely under gravity and is called a catenary. Examples include transmission lines, a telegraph wire or a ﬁsherman’s line, and is used in the design of roofs and arches. Graphs of y = tanh x, y = cosech x, y = sech x and y = coth x are deduced in Problems 7 and 8. y

6. (a) coth 0.444 (b) coth 1.843 [(a) 2.398 (b) 1.051]

10

7. A telegraph wire hangs so that its shape is x described by y = 50 ch . Evaluate, correct 50 to 4 signiﬁcant ﬁgures, the value of y when x = 25. [56.38] 8. The length l of a heavy cable hanging under gravity is given by l = 2c sh (L/2c). Find the value of l when c = 40 and L =30. [30.71] 9.

V 2 = 0.55L tanh (6.3 d/L) is a formula for velocity V of waves over the bottom of shallow water, where d is the depth and L is the wavelength. If d = 8.0 and L =96, calculate the value of V . [5.042]

6 4 2 23 22 21 0

Graphs of hyperbolic functions

A graph of y = sinhx may be plotted using calculator values of hyperbolic functions. The curve is shown in Fig. 5.1. Since the graph is symmetrical about the origin, sinh x is an odd function (as stated in Section 5.1). A graph of y = cosh x may be plotted using calculator values of hyperbolic functions. The curve is shown in Fig. 5.2. Since the graph is symmetrical about the y-axis,

1 2

3

x

Figure 5.2

Problem 7. Sketch graphs of (a) y = tanh x and (b) y = coth x for values of x between −3 and 3. A table of values is drawn up as shown below −3

x

5.2

y 5cosh x

8

sh x

−10.02

ch x

10.07

y = th x =

sh x ch x

y = coth x =

ch x sh x

−2

−1

−3.63 −1.18 3.76

1.54

−0.995 −0.97 −0.77 −1.005 −1.04 −1.31

44 Higher Engineering Mathematics x

1

2

3

sh x

1.18 3.63 10.02

ch x

1

1.54 3.76 10.07

0.77 0.97

A table of values is drawn up as shown below −4

x

sh x ch x

0.995

cosech x =

1 sh x

ch x ch x y = coth x = sh x

±∞ 1.31 1.04

1.005

A graph of y = tanh x is shown in Fig. 5.3(a)

sech x =

−0.10 −0.28 −0.85

27.31

10.07

3.76

1.54

0.04

0.10

0.27

0.65

3

4

1 ch x 0

(b) A graph of y = coth x is shown in Fig. 5.3(b)

sh x

Both graphs are symmetrical about the origin thus tanh x and coth x are odd functions.

cosech x =

1 sh x

ch x Problem 8. Sketch graphs of (a) y = cosech x and (b) y = sech x from x = −4 to x = 4, and, from the graphs, determine whether they are odd or even functions.

y 5 tanh x

y 1 23 22 21

sech x =

1 ch x

1

2

1.18 3.63 10.02 27.29

±∞ 0.85 0.28

0.10

2 3

1.54 3.76 10.07 27.31

1

0.65 0.27

0.10

3

x

2

y 5 cosech x

1

(a) 232221

y

01 2 3 21

y 5 cosech x

3

x

22 y 5coth x

23

1 (a) 23 22 21 0

1

2 3

x

y

21 y 5 coth x

1

22 23

232221 0 (b)

(b)

Figure 5.4 Figure 5.3

0.04

A graph of y = cosech x is shown in Fig. 5.4(a). The graph is symmetrical about the origin and is thus an odd function. (b) A graph of y = sech x is shown in Fig. 5.4(b). The graph is symmetrical about the y-axis and is thus an even function. (a)

21

2

0.04

1

y 0 1

−1

−0.04

x

(a)

−2

−22.29 −10.02 −3.63 −1.18

sh x

y = th x =

−3

y 5 sech x 1 2 3

x

45

Hyperbolic functions 5.3

Hyperbolic identities

For every trigonometric identity there is a corresponding hyperbolic identity. Hyperbolic identities may be proved by either (i) replacing sh x

by

e x + e−x

e x − e−x 2

Problem 9. Prove the hyperbolic identities (a) ch 2 x − sh2 x = 1 (b) 1 − th2 x = sech2 x (c) coth 2 x − 1 =cosech2 x.

(a) and ch x

by

, or 2 (ii) by using Osborne’s rule, which states: ‘the six trigonometric ratios used in trigonometrical identities relating general angles may be replaced by their corresponding hyperbolic functions, but the sign of any direct or implied product of two sines must be changed’. For example, since cos2 x + sin2 x = 1 then, by Osborne’s rule, ch2 x − sh2 x = 1, i.e. the trigonometric functions have been changed to their corresponding hyperbolic functions and since sin2 x is a product of two sines the sign is changed from + to −. Table 5.1 shows some trigonometric identities and their corresponding hyperbolic identities.

x e − e−x e x + e−x + = ex ch x + sh x = 2 2 x x e + e−x e − e−x ch x − sh x = − 2 2

= e+−x (ch x + sh x)(ch x − sh x) = (e x )(e−x ) = e0 = 1 i.e. ch2 x − sh2 x = 1

(b) Dividing each term in equation (1) by ch2 x gives: ch2 x sh2 x 1 − = 2 ch2 x ch2 x ch x i.e. 1 −th2 x = sech2 x

Table 5.1 Trigonometric identity

Corresponding hyperbolic identity

cos2 x + sin2 x = 1

ch2 x − sh2 x = 1

1 + tan2 x = sec2 x

1 −th2 x = sech2 x

cot 2 x + 1 =cosec 2 x

coth2 x − 1 = cosech2 x Compound angle formulae

sin (A ± B) = sin A cos B ± cos A sin B

sh (A ± B) = sh A ch B ± ch A sh B

cos (A ± B) = cos A cos B ∓ sin A sin B

ch (A ± B) = ch A ch B ± sh A sh B

tan (A ± B) =

tan A ± tan B 1 ∓ tan A tan B

th (A ± B) =

th A ± th B 1 ±th A th B

Double angles sin 2x = 2 sin x cos x

sh 2x = 2 sh x ch x

cos 2x = cos2 x − sin2 x

ch 2x =ch2 x + sh2 x

= 2 cos2 x − 1

= 2 ch2 x − 1

= 1 − 2 sin2 x

= 1 + 2sh2 x

tan 2x =

2 tan x 1 − tan2 x

(1)

th 2x =

2 th x 1 + th2 x

46 Higher Engineering Mathematics (c)

Dividing each term in equation (1) by sh2 x gives: ch2 x sh2 x 1 − = 2 sh2 x sh2 x sh x

Problem 12.

Show that th2 x + sech2 x = 1.

L.H.S. = th2 x + sech2 x =

i.e. coth2 x − 1 =cosech2 x

=

Problem 10. Prove, using Osborne’s rule (a) ch 2 A = ch2 A + sh2 A (b) 1 −th2 x = sech2 x. From trigonometric ratios, cos 2 A = cos2 A − sin 2 A

(1)

Osborne’s rule states that trigonometric ratios may be replaced by their corresponding hyperbolic functions but the sign of any product of two sines has to be changed. In this case, sin2 A = (sin A)(sin A), i.e. a product of two sines, thus the sign of the corresponding hyperbolic function, sh2 A, is changed from + to −. Hence, from (1), ch 2A = ch2 A + sh2 A (b) From trigonometric ratios, 1 + tan2 x

= sec2 x

and tan2 x =

sin2 x cos2 x

(2) =

(sin x)(sin x) cos2 x

i.e. a product of two sines. Hence, in equation (2), the trigonometric ratios are changed to their equivalent hyperbolic function and the sign of th2 x changed + to −, i.e. 1 −th2 x = sech2 x Problem 11.

Prove that 1 + 2 sh2 x = ch 2x.

Left hand side (L.H.S.)

2 e x − e−x = 1 + 2 sh x = 1 + 2 2 2x e − 2e x e−x + e−2x = 1+2 4

e2x − 2 + e−2x 2 2x e + e−2x 2 =1+ − 2 2 =1+

=

+ e−2x 2

sh2 x + 1 ch2 x = 2 = 1 = R.H.S. ch2 x ch x

Problem 13. Given Ae x + Be−x ≡ 4ch x−5 sh x, determine the values of A and B. Ae x + Be−x ≡ 4 ch x − 5 sh x x x e + e−x e − e−x −5 =4 2 2 5 5 = 2e x + 2e−x − e x + e−x 2 2 1 9 = − e x + e−x 2 2 Equating coefﬁcients gives: A = −

1 1 and B = 4 2 2

Problem 14. If 4e x − 3e−x ≡ Psh x + Qch x, determine the values of P and Q. 4e x − 3e−x ≡ P sh x + Q ch x x x e − e−x e + e−x +Q =P 2 2 P x P −x Q x Q −x e − e + e + e 2 2 2 2 P+Q x Q − P −x e + e = 2 2 =

2

e2x

sh2 x + 1 ch2 x

Since ch2 x − sh2 x = 1 then 1 + sh2 x = ch2 x Thus

(a)

1 sh2 x + 2 2 ch x ch x

= ch 2x = R.H.S.

Equating coefﬁcients gives: 4=

P+Q Q−P and −3 = 2 2

i.e. P + Q = 8 −P + Q = −6

(1) (2)

Adding equations (1) and (2) gives: 2Q = 2, i.e. Q = 1 Substituting in equation (1) gives: P = 7.

Hyperbolic functions Now try the following exercise Exercise 21 Further problems on hyperbolic identities In Problems 1 to 4, prove the given identities. 1. (a) ch (P − Q) ≡ ch P ch Q − sh P sh Q (b) ch 2x ≡ ch2 x + sh2 x 2. (a) coth x ≡ 2 cosech 2x + th x (b) ch 2θ − 1 ≡2 sh2 θ th A − th B 1 −th A th B (b) sh 2 A ≡ 2 sh A ch A

3. (a) th (A − B) ≡

4. (a) sh (A + B) ≡ sh A ch B + ch A sh B (b)

sh2 x + ch2 x − 1 ≡ tanh4 x 2ch2 x coth2 x

5. Given Pe x − Qe−x ≡ 6 ch x − 2 sh x, ﬁnd P and Q [P = 2, Q =−4] 6. If 5e x − 4e−x ≡ A sh x + B ch x, ﬁnd A and B. [A = 9, B = 1]

5.4 Solving equations involving hyperbolic functions Equations such as sinh x = 3.25 or coth x = 3.478 may be determined using a calculator. This is demonstrated in Worked Problems 15 to 21. Problem 15. Solve the equation sh x = 3, correct to 4 signiﬁcant ﬁgures. If sinh x = 3, then x = sinh−1 3 This can be determined by calculator. (i) Press hyp (ii) Choose 4, which is sinh−1 (iii) Type in 3 (iv) Close bracket ) (v) Press = and the answer is 1.818448459 i.e. the solution of sh x = 3 is: x = 1.818, correct to 4 signiﬁcant ﬁgures. Problem 16. Solve the equation ch x = 1.52, correct to 3 decimal places.

47

Using a calculator with a similar procedure as in Worked Problem 15, check that: x = 0.980, correct to 3 decimal places. With reference to Fig. 5.2, it can be seen that there will be two values corresponding to y = cosh x = 1.52. Hence, x = ±0.980 Problem 17. Solve the equation tanh θ = 0.256, correct to 4 signiﬁcant ﬁgures. Using a calculator with a similar procedure as in Worked Problem 15, check that gives θ = 0.2618, correct to 4 signiﬁcant ﬁgures. Problem 18. Solve the equation sech x = 0.4562, correct to 3 decimal places. sech then x = sech −10.4562 = x = 0.4562, 1 1 cosh−1 since cosh = 0.4562 sech

If

i.e. x = 1.421, correct to 3 decimal places. With reference to the graph of y = sech x in Fig. 5.4, it can be seen that there will be two values corresponding to y = sech x = 0.4562 Hence, x = ±1.421 Problem 19. Solve the equation cosech y = −0.4458, correct to 4 signiﬁcant ﬁgures. −1 If cosechy = − 0.4458, then y = cosech (−0.4458) 1 1 since sinh = = sinh−1 − 0.4458 cosech i.e. y = −1.547, correct to 4 signiﬁcant ﬁgures.

Problem 20. Solve the equation coth A = 2.431, correct to 3 decimal places. coth 2.431, then A = coth−1 2.431 = A= 1 1 tanh−1 since tanh = 2.431 coth i.e. A= 0.437, correct to 3 decimal places. If

Problem 21. A chain hangs in the form given by x y = 40 ch . Determine, correct to 4 signiﬁcant 40 ﬁgures, (a) the value of y when x is 25, and (b) the value of x when y = 54.30

48 Higher Engineering Mathematics (a)

x , and when x = 25, 40 25 y = 40 ch = 40 ch 0.625 40

y = 40 ch

= 40(1.2017536 . . .) = 48.07 x (b) When y = 54.30, 54.30 =40 ch , from which 40 x 54.30 ch = = 1.3575 40 40 x Hence, = cosh−1 1.3575 =±0.822219 . . .. 40 (see Fig. 5.2 for the reason as to why the answer is ±) from which, x = 40(±0.822219 . . ..) = ±32.89

Following the above procedure: (i) 2.6 ch x + 5.1 sh x = 8.73 x x e + e−x e − e−x i.e. 2.6 + 5.1 = 8.73 2 2 (ii) 1.3e x + 1.3e−x + 2.55e x − 2.55e−x = 8.73 i.e. 3.85e x − 1.25e−x − 8.73 =0 (iii) 3.85(e x )2 − 8.73e x − 1.25 =0 (iv) e x

−(−8.73) ± [(−8.73)2 − 4(3.85)(−1.25)] = 2(3.85) √ 8.73 ± 95.463 8.73 ±9.7705 = = 7.70 7.70 Hence e x = 2.4027 or e x = −0.1351

Equations of the form a ch x + b sh x = c, where a, b and c are constants may be solved either by: (a)

plotting graphs of y = a ch x + b sh x and y = c and noting the points of intersection, or more accurately,

(b) by adopting the following procedure: x e − e−x and ch x to (i) Change sh x to 2 x e + e−x 2 (ii) Rearrange the equation into the form pe x + qe−x +r = 0, where p, q and r are constants.

(v)

Now try the following exercise Exercise 22 Further problems on hyperbolic equations In Problems 1 to 8, solve the given equations correct to 4 decimal places. 1.

2.

(iv) Solve the quadratic equation p(e x )2 +re x + q = 0 for e x by factorising or by using the quadratic formula.

3.

(b) sh A = −2.43

(a) cosh B = 1.87 (b) 2 ch x = 3 [(a) ±1.2384 (b) ±0.9624] (a) tanh y = −0.76 (b) 3 th x = 2.4 [(a) −0.9962 (b) 1.0986]

4.

(a) sech B = 0.235 (b) sech Z = 0.889 [(a) ±2.1272 (b) ±0.4947]

5.

This procedure is demonstrated in Problem 22.

(a) cosech θ = 1.45 (b) 5 cosech x = 4.35 [(a) 0.6442 (b) 0.5401]

6. Problem 22. Solve the equation 2.6 ch x + 5.1 sh x = 8.73, correct to 4 decimal places.

(a) sinh x = 1

[(a) 0.8814 (b) −1.6209]

(iii) Multiply each term by e x , which produces an equation of the form p(e x )2 +re x + q = 0 (since (e−x )(e x ) = e0 = 1)

(v) Given e x = a constant (obtained by solving the equation in (iv)), take Napierian logarithms of both sides to give x = ln (constant)

x = ln 2.4027 or x = ln(−0.1351) which has no real solution. Hence x = 0.8766, correct to 4 decimal places.

(a) coth x = 2.54 (b) 2 coth y = −3.64 [(a) 0.4162 (b) −0.6176]

7.

3.5 sh x + 2.5 ch x = 0

[−0.8959]

Hyperbolic functions 8. 2 sh x + 3 ch x = 5 9. 4 th x − 1 = 0

[0.6389 or −2.2484] [0.2554]

10. A chain hangs so its shape is of the xthat . Determine, correct to form y = 56 cosh 56 4 signiﬁcant ﬁgures, (a) the value of y when x is 35, and (b) the value of x when y is 62.35 [(a) 67.30 (b) ±26.42]

x3 x5 i.e. sinh x = x + + + · · · (which is valid for all 3! 5! values of x). sinh x is an odd function and contains only odd powers of x in its series expansion. Problem 23. Using the series expansion for ch x evaluate ch 1 correct to 4 decimal places. ch x = 1 + Let

5.5 Series expansions for cosh x and sinh x

x = 1,

then ch 1 = 1 +

By deﬁnition, x2 x3 x4 x5 + + + +··· 2! 3! 4! 5!

from Chapter 4. Replacing x by −x gives: e−x = 1 − x +

x2 x4 + + · · ·from above 2! 4!

x2 x3 x4 x5 − + − +··· . 2! 3! 4! 5!

1 cosh x = (e x + e−x ) 2

x2 x3 x4 x5 1 1+x + = + + + +··· 2 2! 3! 4! 5! x2 x3 x4 x5 − + − +··· + 1−x + 2! 3! 4! 5! 2x 2 2x 4 1 2+ + +··· = 2 2! 4! x2 x4 i.e. cosh x = 1 + + + · · · (which is valid for all 2! 4! values of x). cosh x is an even function and contains only even powers of x in its expansion. 1 sinh x = (e x − e−x ) 2

x2 x3 x4 x5 1 1+x + = + + + +··· 2 2! 3! 4! 5! x2 x3 x4 x5 − + − +··· − 1−x + 2! 3! 4! 5!

2x 3 2x 5 1 2x + + + ··· = 2 3! 5!

14 12 + 2 × 1 4 ×3 × 2 × 1

16 + ··· 6 ×5 × 4 × 3 ×2 × 1

+

ex = 1 + x +

49

= 1 + 0.5 + 0.04167 + 0.001389 + · · · i.e. ch 1 = 1.5431, correct to 4 decimal places, which may be checked by using a calculator. Problem 24. Determine, correct to 3 decimal places, the value of sh 3 using the series expansion for sh x. sh x = x +

x3 x5 + + · · · from above 3! 5!

Let x = 3, then 33 35 37 39 311 + + + + +··· 3! 5! 7! 9! 11! = 3 + 4.5 + 2.025 + 0.43393 + 0.05424

sh 3 = 3 +

+ 0.00444 + · · · i.e. sh 3 = 10.018, correct to 3 decimal places. Problem 25. Determine the power series for θ − sh 2θ as far as the term in θ 5 . 2 ch 2 In the series expansion for ch x, let x = 2 ch

θ then: 2

θ (θ/2)2 (θ/2)4 =2 1+ + +··· 2 2! 4! =2+

θ2 θ4 + +··· 4 192

50 Higher Engineering Mathematics In the series expansion for sh x, let x = 2θ, then: (2θ)3 (2θ)5 + +··· 3! 5! 4 4 = 2θ + θ 3 + θ 5 + · · · 3 15

sh 2θ = 2θ +

Hence θ θ2 θ4 ch − sh 2θ = 2 + + +··· 2 4 192 4 4 − 2θ + θ 3 + θ 5 + · · · 3 15 = 2 −2θ + −

θ2 4 3 θ4 − θ + 4 3 192

4 5 θ + · · · as far the term in θ 5 15

Now try the following exercise Exercise 23 Further problems on series expansions for cosh x and sinh x 1. Use the series expansion for ch x to evaluate, correct to 4 decimal places: (a) ch 1.5 (b) ch 0.8 [(a) 2.3524 (b) 1.3374]

2. Use the series expansion for sh x to evaluate, correct to 4 decimal places: (a) sh 0.5 (b) sh 2 [(a) 0.5211 (b) 3.6269] 3. Expand the following as a power series as far as the term in x 5 : (a) sh 3x (b) ch 2x ⎡ ⎤ 9 3 81 5 (a) 3x + + x x ⎢ 2 40 ⎥ ⎣ ⎦ 2 (b) 1 + 2x 2 + x 4 3 In Problems 4 and 5, prove the given identities, the series being taken as far as the term in θ 5 only. 4. sh 2θ − sh θ ≡ θ +

5. 2 sh

31 5 7 3 θ + θ 6 120

θ θ θ2 θ3 θ4 − ch ≡ − 1 + θ − + − 2 2 8 24 384 +

θ5 1920

Chapter 6

Arithmetic and geometric progressions 6.1

Arithmetic progressions

When a sequence has a constant difference between successive terms it is called an arithmetic progression (often abbreviated to AP). Examples include:

i.e.

For example, the sum of the ﬁrst 7 terms of the series 1, 4, 7, 10, 13, . . . is given by 7 S7 = [2(1) + (7 − 1)3], since a = 1 and d = 3 2

(i) 1, 4, 7, 10, 13, . . . where the common difference is 3 and

7 7 = [2 + 18] = [20] = 70 2 2

(ii) a, a + d, a + 2d, a + 3d,. . .where the common difference is d. General expression for the n’th term of an AP If the ﬁrst term of an AP is ‘a’ and the common difference is ‘d’ then the n’th term is: a + (n − 1)d In example (i) above, the 7th term is given by 1 + (7 − 1)3 = 19, which may be readily checked. Sum of n terms of an AP The sum S of an AP can be obtained by multiplying the average of all the terms by the number of terms. a +l , where ‘a’ is the The average of all the terms = 2 ﬁrst term and l is the last term, i.e. l = a + (n − 1)d, for n terms. Hence the sum of n terms, a +l Sn = n 2 n = {a + [a + (n − 1)d]} 2

n S n = [2a + (n − 1)d] 2

6.2 Worked problems on arithmetic progressions Problem 1. Determine (a) the ninth, and (b) the sixteenth term of the series 2, 7, 12, 17, . . . 2, 7, 12, 17, . . . is an arithmetic progression with a common difference, d, of 5. (a)

The n’th term of an AP is given by a + (n −1)d Since the ﬁrst term a = 2, d = 5 and n =9 then the 9th term is: 2 + (9 −1)5 = 2 + (8)(5) = 2 + 40 =42

(b) The 16th term is: 2 + (16 −1)5 = 2 +(15)(5) = 2 + 75 =77. Problem 2. The 6th term of an AP is 17 and the 13th term is 38. Determine the 19th term.

52 Higher Engineering Mathematics The n’th term of an AP is a + (n −1)d

The sum of the ﬁrst 21 terms,

a + 5d = 17

(1)

The 13th term is: a + 12d= 38

(2)

The 6th term is:

Equation (2) −equation (1) gives: 7d = 21, from which, 21 d = = 3. 7 Substituting in equation (1) gives: a + 15 =17, from which, a = 2. Hence the 19th term is: a + (n − 1)d = 2 + (19 − 1)3 = 2 + (18)(3) = 2 + 54 = 56.

is

an

AP

where

a = 2 12

1. Find the 11th term of the series 8, 14, 20, 26, . . . [68] 2. Find the 17th term of the series 11, 10.7, 10.4, 10.1, . . . [6.2] and

Hence if the n’th term is 22 then: a + (n − 1)d = 22 i.e. 2 12 + (n − 1) 1 12 = 22 (n − 1) 1 12 = 22 − 2 12 = 19 12 . n −1 =

19 12 1 12

= 13 and n = 13 + 1 = 14

i.e. the 14th term of the AP is 22. Problem 4. Find the sum of the ﬁrst 12 terms of the series 5, 9, 13, 17, . . . 5, 9, 13, 17, . . . is an AP where a = 5 and d = 4. The sum of n terms of an AP, n Sn = [2a + (n − 1)d] 2 Hence the sum of the ﬁrst 12 terms, S12 =

Now try the following exercise Exercise 24 Further problems on arithmetic progressions

Problem 3. Determine the number of the term whose value is 22 in the series 2 12 , 4, 5 12 , 7, . . . 2 12 , 4, 5 12 , 7, . . . d = 1 12 .

21 [2a + (n − 1)d] 2 21 21 = [2(3.5) + (21 − 1)0.6] = [7 + 12] 2 2 399 21 = 199.5 = (19) = 2 2

S21 =

12 [2(5) + (12 − 1)4] 2

= 6[10 + 44] = 6(54) = 324 Problem 5. Find the sum of the ﬁrst 21 terms of the series 3.5, 4.1, 4.7, 5.3, . . . 3.5, 4.1, 4.7, 5.3, . . . is an AP where a = 3.5 and d = 0.6

3. The seventh term of a series is 29 and the eleventh term is 54. Determine the sixteenth term. [85.25] 4. Find the 15th term of an arithmetic progression of which the ﬁrst term is 2.5 and the tenth term is 16. [23.5] 5. Determine the number of the term which is 29 in the series 7, 9.2, 11.4, 13.6, . . . [11th ] 6. Find the sum of the ﬁrst 11 terms of the series 4, 7, 10, 13, . . . [209] 7. Determine the sum of the series 6.5, 8.0, 9.5, 11.0, . . . , 32 [346.5]

6.3 Further worked problems on arithmetic progressions Problem 6. The sum of 7 terms of an AP is 35 and the common difference is 1.2. Determine the ﬁrst term of the series. n = 7, d = 1.2 and S7 = 35 Since the sum of n terms of an AP is given by Sn =

n [2a + (n − 1)d], then 2

7 7 35 = [2a + (7 − 1)1.2] = [2a + 7.2] 2 2

Arithmetic and geometric progressions 35 × 2 = 2a + 7.2 7 10 = 2a + 7.2 2a = 10 − 7.2 = 2.8, 2.8 a= = 1.4 2

Hence

Thus from which

i.e. the ﬁrst term, a = 1.4

Problem 9. The ﬁrst, twelfth and last term of an arithmetic progression are 4, 31 12 , and 376 12 respectively. Determine (a) the number of terms in the series, (b) the sum of all the terms and (c) the ‘80’th term. (a)

Problem 7. Three numbers are in arithmetic progression. Their sum is 15 and their product is 80. Determine the three numbers.

Let the AP be a, a +d, a +2d, . . . , a + (n − 1)d, where a = 4 The 12th term is: a + (12 −1)d = 31 12 4 + 11d = 31 12 ,

i.e.

Let the three numbers be (a − d), a and (a + d)

from which, 11d = 31 12 − 4 = 27 12

Then (a − d) + a + (a + d) = 15, i.e. 3a = 15, from which, a = 5

Hence d =

27 12 = 2 12 11 The last term is a + (n − 1)d i.e. 4 + (n − 1) 2 12 = 376 12

Also, a(a − d)(a + d) = 80, i.e. a(a 2 − d 2 ) = 80 Since a = 5, 5(52 − d 2 ) = 80 125 − 5d 2 = 80 125 − 80 = 5d 2

(n − 1) =

376 12 − 4 2 12

45 = 5d 2 √ 45 from which, d 2 = = 9. Hence d = 9 = ±3. 5 The three numbers are thus (5 − 3), 5 and (5 + 3), i.e. 2, 5 and 8. Problem 8. Find the sum of all the numbers between 0 and 207 which are exactly divisible by 3.

=

a + (n − 1)d = 207

i.e.

3 + (n − 1)3 = 207,

n [2a + (n − 1)d] 2 150 1 = 2(4) + (150 − 1) 2 2 2 1 = 75 8 + (149) 2 2

= 85[8 + 372.5] = 75(380.5) = 28537

The sum of all 69 terms is given by n [2a + (n − 1)d] 2 69 = [2(3) + (69 − 1)3] 2 69 69 = [6 + 204] = (210) = 7245 2 2

S69 =

= 149

S150 =

207 − 3 = 68 3 n = 68 + 1 = 69

Hence

2 12

(b) Sum of all the terms,

(n − 1) =

from which

372 12

Hence the number of terms in the series, n = 149 +1 =150

The series 3, 6, 9, 12, . . ., 207 is an AP whose ﬁrst term a = 3 and common difference d = 3 The last term is

53

(c)

1 2

The 80th term is: a + (n − 1)d = 4 + (80 − 1) 2 12 = 4 + (79) 2 12 = 4 + 197.5 = 201 12

54 Higher Engineering Mathematics Problem 10. An oil company bores a hole 80 m deep. Estimate the cost of boring if the cost is £30 for drilling the ﬁrst metre with an increase in cost of £2 per metre for each succeeding metre.

8. An oil company bores a hole 120 m deep. Estimate the cost of boring if the cost is £70 for drilling the ﬁrst metre with an increase in cost of £3 per metre for each succeeding metre. [£29820]

The series is: 30, 32, 34, . . . to 80 terms, i.e. a = 30, d = 2 and n = 80 Thus, total cost, n Sn = 2a + (n − 1)d 2 =

80 [2(30) + (80 − 1)(2)] 2

= 40[60 + 158] = 40(218) = £8720

6.4

Geometric progressions

When a sequence has a constant ratio between successive terms it is called a geometric progression (often abbreviated to GP). The constant is called the common ratio, r. Examples include (i) 1, 2, 4, 8, . . . where the common ratio is 2 and

Now try the following exercise

(ii) a, ar, ar 2 , ar 3 , . . . where the common ratio is r. General expression for the n’th term of a GP

Exercise 25 Further problems on arithmetic progressions

If the ﬁrst term of a GP is ‘a’ and the common ratio is r, then

1. The sum of 15 terms of an arithmetic progression is 202.5 and the common difference is 2. Find the ﬁrst term of the series. [−0.5]

the n’th term is: ar n−1

2. Three numbers are in arithmetic progression. Their sum is 9 and their product is 20.25. Determine the three numbers. [1.5, 3, 4.5] 3. Find the sum of all the numbers between 5 and 250 which are exactly divisible by 4. [7808] 4. Find the number of terms of the series 5, 8, 11, . . . of which the sum is 1025. [25] 5. Insert four terms between 5 and 22.5 to form an arithmetic progression. [8.5, 12, 15.5, 19] 6. The ﬁrst, tenth and last terms of an arithmetic progression are 9, 40.5, and 425.5 respectively. Find (a) the number of terms, (b) the sum of all the terms and (c) the 70th term. [(a) 120 (b) 26070 (c) 250.5] 7. On commencing employment a man is paid a salary of £16000 per annum and receives annual increments of £480. Determine his salary in the 9th year and calculate the total he will have received in the ﬁrst 12 years. [£19840, £223,680]

which can be readily checked from the above examples. For example, the 8th term of the GP 1, 2, 4, 8, . . . is (1)(2)7 = 128, since a = 1 and r = 2. Sum of n terms of a GP Let a GP be a, ar, ar 2 , ar 3 , . . . , ar n−1 then the sum of n terms, Sn = a + ar + ar 2 + ar 3 + · · · + ar n−1 · · ·

(1)

Multiplying throughout by r gives: r Sn = ar + ar 2 + ar 3 + ar 4 + · · · + ar n−1 + ar n + · · ·

(2)

Subtracting equation (2) from equation (1) gives: Sn − r Sn = a − ar n i.e. Sn (1 − r) = a(1 − r n ) n

−r ) Thus the sum of n terms, S n = a(1 (1 − r ) which is valid when r < 1.

Arithmetic and geometric progressions Subtracting equation (1) from equation (2) gives a(r n − 1) Sn = which is valid when r > 1. (r − 1) For example, the sum of the ﬁrst 8 terms of the GP 1, 2, 1(28 − 1) 4, 8, 16, . . . is given by S8 = , since a = 1 and (2 − 1) r =2 i.e. S8 =

1(256 − 1) = 255 1

Sum to inﬁnity of a GP When the common ratio r of a GP is less than unity, the a(1 −r n ) , which may be written sum of n terms, Sn = (1 −r) a ar n − as Sn = (1 −r) (1 −r) Since r < 1, r n becomes less as n increases, i.e. r n → 0 as n →∞. n a ar Hence → 0 as n →∞. Thus Sn → as (1 −r) (1 −r) n →∞. a is called the sum to inﬁnity, S∞, The quantity (1 −r) and is the limiting value of the sum of an inﬁnite number of terms, a i.e. S ∞ = which is valid when −1 0, then the stationary ∂x2 point is a minimum point. if < 0 and

∂2z ∂z = 2(x − 1) = 2x − 2, 2 = 2 ∂x ∂x ∂z ∂2z = 2(y − 2) = 2y − 4, 2 = 2 ∂y ∂y

∂2z ∂ and = ∂x∂ y ∂x

∂2z ∂2 z ∂2 z , and ∂x 2 ∂ y 2 ∂x∂ y

2

(c)

Following the above procedure:

for each stationary point,

and evaluate, (viii) (a) if > 0 then the stationary point is a saddle point. (b)

Problem 1. Show that the function z =(x − 1)2 + (y − 2)2 has one stationary point only and determine its nature. Sketch the surface represented by z and produce a contour map in the x-y plane.

and since

into the equation

36.4 Worked problems on maxima, minima and saddle points for functions of two variables

2

(vii) substitute the values of

359

(v)

∂z ∂ = (2y − 4) = 0 ∂y ∂x

∂2 z ∂2z ∂2z = = 2 and =0 ∂x 2 ∂ y 2 ∂x∂ y

(vi)

2 ∂2 z =0 ∂x∂ y

(vii) = (0)2 − (2)(2) = −4 ∂2z (viii) Since < 0 and 2 > 0, the stationary point ∂x (1, 2) is a minimum. The surface z = (x − 1)2 + (y − 2)2 is shown in three dimensions in Fig. 36.7. Looking down towards the x-y plane from above, it is possible to produce a contour map. A contour is a line on a map which gives places having the same vertical height above a datum line (usually the mean sea-level on a geographical map).

360 Higher Engineering Mathematics z

Problem 2. Find the stationary points of the surface f (x, y) = x 3 − 6x y + y 3 and determine their nature. y 1

Let z = f (x, y) = x 3 − 6x y + y 3

2

Following the procedure: (i)

o 1

∂z ∂z = 3x 2 − 6y and = −6x + 3y 2 ∂x ∂y

(ii) for stationary points, 3x 2 − 6y = 0 x

−6x + 3y 2 = 0

and

Figure 36.7

(iii) from equation (1), 3x 2 = 6y

A contour map for z =(x − 1)2 + (y − 2)2 is shown in Fig. 36.8. The values of z are shown on the map and these give an indication of the rise and fall to a stationary point.

and

y=

3x 2 1 2 = x 6 2

y

z51

2

z54

z59

z 5 16

1

1

Figure 36.8

2

x

(1) (2)

Maxima, minima and saddle points for functions of two variables and substituting in equation (2) gives: 1 2 2 =0 x −6x + 3 2 3 −6x + x 4 = 0 4 3 x 3x −2 = 0 4 from which, x = 0 or

x3 − 2 =0 4

i.e. x 3 = 8 and x = 2 When x = 0, y = 0 and when x = 2, y = 2 from equations (1) and (2). Thus stationary points occur at (0, 0) and (2, 2). ∂2z ∂2z ∂ ∂z ∂2z = 6x, = 6y and = (iv) ∂x 2 ∂ y2 ∂x∂ y ∂x ∂ y =

∂ (−6x + 3y 2 ) = −6 ∂x

∂2 z ∂2 z = 0, 2 = 0 2 ∂x ∂y ∂2 z and = −6 ∂x∂ y ∂2 z ∂2 z = 12, 2 = 12 for (2, 2), 2 ∂x ∂y ∂2 z and = −6 ∂x∂ y 2 2 ∂ z = (−6)2 = 36 (vi) for (0, 0), ∂x∂ y 2 2 ∂ z for (2, 2), = (−6)2 = 36 ∂x∂ y (v)

for (0, 0)

∂2z (vii) (0, 0) = ∂x∂ y

2 −

∂2z ∂x 2

∂2 z ∂ y2

Now try the following exercise Exercise 143 Further problems on maxima, minima and saddle points for functions of two variables 1. Find the stationary point of the surface f (x, y) = x 2 + y 2 and determine its nature. Sketch the surface represented by z. [Minimum at (0, 0)] 2. Find the maxima, minima and saddle points for the following functions: (a) f (x, y) = x 2 + y 2 − 2x + 4y + 8 (b) f (x, y) = x 2 − y 2 − 2x + 4y + 8 (c) f (x, y) = 2x⎡+ 2y − 2x y − 2x 2 − y 2 + 4.⎤ (a) Minimum at (1, −2) ⎣ (b) Saddle point at (1, 2) ⎦ (c) Maximum at (0, 1) 3. Determine the stationary values of the function f (x, y) = x 3 − 6x 2 − 8y 2 and distinguish between them. Sketch an approximate contour map to representthe surface f (x, y). Maximum point at (0, 0), saddle point at (4, 0) 4. Locate the stationary point of the function z =12x 2 + 6x y + 15y 2 . [Minimum at (0, 0)] 5. Find the stationary points of the surface z = x 3 − x y + y 3 and distinguish between them. saddle point at 1(0,1 0), minimum at 3 , 3

= 36 − (0)(0) = 36 (2, 2) = 36 − (12)(12) = −108 (viii) Since (0, 0) > 0 then (0, 0) is a saddle point. ∂2 z > 0, then (2, 2) is a Since (2, 2) < 0 and ∂x 2 minimum point.

36.5 Further worked problems on maxima, minima and saddle points for functions of two variables Problem 3. Find the co-ordinates of the stationary points on the surface z = (x 2 + y 2 )2 − 8(x 2 − y 2 ) and distinguish between them. Sketch the approximate contour map associated with z.

361

362 Higher Engineering Mathematics (vii) (0, 0) = (0)2 − (−16)(16) = 256

Following the procedure:

(2, 0) = (0)2 − (32)(32) = −1024

∂z (i) = 2(x 2 + y 2 )2x − 16x and ∂x ∂z = 2(x 2 + y 2 )2y + 16y ∂y

(−2, 0) = (0)2 − (32)(32) = −1024

(ii) for stationary points, 2(x 2 + y 2 )2x − 16x = 0 i.e.

4x 3 + 4x y 2 − 16x = 0

and

2(x 2 + y 2 )2y + 16y = 0

i.e.

4y(x 2 + y 2 + 4) = 0

(iii) From equation (1), y 2 = Substituting

y2 = 4 − x 2

(1) (2)

16x − 4x 3 =4 − x2 4x in equation (2) gives

4y(x 2 + 4 − x 2 + 4) = 0 i.e. 32y = 0 and y = 0 When y = 0 in equation (1),

4x 3 − 16x = 0

i.e.

4x(x 2 − 4) = 0

from which, x = 0 or x = ±2 The co-ordinates of the stationary points are (0, 0), (2, 0) and (−2, 0). ∂2z = 12x 2 + 4y 2 − 16, (iv) ∂x 2 ∂2 z ∂ y2

= 4x 2 + 12y 2 + 16 and

∂2z ∂x∂ y

= 8x y

(v) For the point (0, 0), ∂2z ∂2z ∂2z = −16, = 16 and =0 ∂x 2 ∂ y2 ∂x∂ y For the point (2, 0), ∂2z ∂2z ∂2z = 32, = 32 and =0 ∂x 2 ∂ y2 ∂x∂ y For the point (−2, 0), ∂2z ∂2z ∂2z = 32, 2 = 32 and =0 2 ∂x ∂y ∂x∂ y (vi)

∂2 z ∂x∂ y

2 = 0 for each stationary point

(viii) Since (0, 0) > 0, the point (0, 0) is a saddle point. 2 ∂ z > 0, the point Since (0, 0) < 0 and ∂x 2 (2, 0) (2, 0) is a minimum point. 2 ∂ z Since (−2, 0) < 0 and > 0, the ∂x 2 (−2, 0) point (−2, 0) is a minimum point. Looking down towards the x-y plane from above, an approximate contour map can be constructed to represent the value of z. Such a map is shown in Fig. 36.9. To produce a contour map requires a large number of x-y co-ordinates to be chosen and the values of z at each co-ordinate calculated. Here are a few examples of points used to construct the contour map. When z = 0, 0 =(x 2 + y 2 )2 − 8(x 2 − y)2 In addition, when, say, y = 0 (i.e. on the x-axis) 0 = x 4 − 8x 2 , i.e. x 2 (x 2 − 8) = 0 √ from which, x = 0 or x = ± 8 √ Hence the contour z = 0 crosses the x-axis at 0 and ± 8, i.e. at co-ordinates (0, 0), (2.83, 0) and (−2.83, 0) shown as points, S, a and b respectively. When z = 0 and x =2 then 0 = (4 + y 2 )2 − 8(4 − y 2 ) i.e. 0 = 16 + 8y 2 + y 4 − 32 + 8y 2 i.e. 0 = y 4 + 16y 2 − 16 Let y 2 = p, then p2 + 16 p − 16 = 0 and −16 ± 162 − 4(1)(−16) p= 2 −16 ± 17.89 = 2 = 0.945 or −16.945 Hence y =

√

p=

(0.945) or (−16.945)

= ± 0.97 or complex roots.

Maxima, minima and saddle points for functions of two variables

363

y 4

i

z5

128

2

z59 c

g

0 z5

S f

b

3 22

3 2

a

e

x

d h 22

j 24

Figure 36.9

Hence the z =0 contour passes through the co-ordinates (2, 0.97) and (2, −0.97) shown as a c and d in Fig. 36.9. Similarly, for the z = 9 contour, when y = 0, 9=

(x 2

+ 02 )2

i.e.

9 = x 4 − 8x 2

i.e.

x 4 − 8x 2 − 9 =0

− 8(x 2 − 02 )

Hence (x 2 − 9)(x 2 + 1) = 0. from which, x = ±3 or complex roots. Thus the z = 9 contour passes through (3, 0) and (−3, 0), shown as e and f in Fig. 36.9. If z = 9 and x = 0, 9 = y 4 + 8y 2 i.e.

y 4 + 8y 2 − 9 = 0

i.e.

(y 2 + 9)(y 2 − 1) = 0

from which, y = ±1 or complex roots. Thus the z = 9 contour also passes through (0, 1) and (0, −1), shown as g and h in Fig. 36.9.

When, say, x = 4 and y = 0, z = (42 )2 − 8(42 ) = 128. when z = 128 and x = 0, 128 = y 4 + 8y 2 i.e.

y 4 + 8y 2 − 128 = 0

i.e. (y 2 + 16)(y 2 − 8) = 0 √ from which, y = ± 8 or complex roots. Thus the z = 128 contour passes through (0, 2.83) and (0, −2.83), shown as i and j in Fig. 36.9. In a similar manner many other points may be calculated with the resulting approximate contour map shown in Fig. 36.9. It is seen that two ‘hollows’ occur at the minimum points, and a ‘cross-over’ occurs at the saddle point S, which is typical of such contour maps. Problem 4. Show that the function f (x, y) = x 3 − 3x 2 − 4y 2 + 2 has one saddle point and one maximum point. Determine the maximum value.

364 Higher Engineering Mathematics

Let z = f (x, y) = x 3 − 3x 2 − 4y 2 + 2.

(vi)

Following the procedure: (i)

(ii) for stationary points, 3x 2 −6x = 0

(1)

−8y = 0

(2)

(iii) From equation (1), 3x(x − 2) = 0 from which, x = 0 and x = 2.

∂2z (viii) Since (0, 0) < 0 and < 0, the ∂x 2 (0, 0) point (0, 0) is a maximum point and hence the maximum value is 0. Since (2, 0) > 0, the point (2, 0) is a saddle point.

Hence the stationary points are (0, 0) and (2, 0). (iv)

= 6x − 6,

∂x 2

∂2z ∂ y2

= −8 and

= (0)2 = 0

From equation (2), y = 0.

∂2z

2

(vii) (0, 0) = 0 −(−6)(−8) = −48 (2, 0) = 0 −(6)(−8) = 48

∂z ∂z = 3x 2 − 6x and = − 8y ∂x ∂y

and

∂2 z ∂x∂ y

∂2z ∂x∂ y

The value of z at the saddle point is 23 − 3(2)2 − 4(0)2 + 2 =−2.

=0

An approximate contour map representing the surface f (x, y) is shown in Fig. 36.10 where a ‘hollow effect’ is seen surrounding the maximum point and a ‘cross-over’ occurs at the saddle point S.

(v) For the point (0, 0), ∂2 z ∂2 z ∂2 z = −6, = −8 and =0 ∂x 2 ∂ y2 ∂x∂ y For the point (2, 0),

Problem 5. An open rectangular container is to have a volume of 62.5 m3 . Determine the least surface area of material required.

∂2z ∂2z ∂2 z = 6, 2 = −8 and =0 2 ∂x ∂y ∂x∂ y

y 2

z5

MAX

S 2 2 52

z

z5

22

Figure 36.10

24

21

21

z5

z5

22

3

2

4

x

Maxima, minima and saddle points for functions of two variables From equation (1),

(5) (5) z =62.5 z=

from which,

365

62.5 = 2.5 m 25

∂ 2 S 250 ∂ 2 S 250 ∂2 S = 3 , 2 = 3 and =1 2 ∂x x ∂y y ∂x∂ y When x = y = 5,

y z

∂2 S ∂2 S ∂2 S = 2, = 2 and =1 ∂x 2 ∂ y2 ∂x∂ y

= (1)2 − (2)(2) = −3 ∂2 S > 0, then the surface area S is a Since < 0 and ∂x 2 minimum.

x

Figure 36.11

Hence the minimum dimensions of the container to have a volume of 62.5 m3 are 5 m by 5 m by 2.5 m. Let the dimensions of the container be x, y and z as shown in Fig. 36.11.

= (5)(5) + 2(5)(2.5) + 2(5)(2.5)

Volume

V = x yz = 62.5

(1)

Surface area,

S = x y + 2yz + 2x z

(2)

Exercise 144 Further problems on maxima, minima and saddle points for functions of two variables

Substituting in equation (2) gives:

i.e.

S=xy +

62.5 62.5 + 2x xy xy

1. The function z = x 2 + y 2 + x y + 4x − 4y + 3 has one stationary value. Determine its co-ordinates and its nature. [Minimum at (−4, 4)]

125 125 + x y

which is a function of two variables ∂s 125 = y − 2 = 0 for a stationary point, ∂x x hence x 2 y =125 ∂s 125 = x − 2 = 0 for a stationary point, ∂y y hence x y 2 = 125

(3)

(4)

Dividing equation (3) by (4) gives: x2 y x = 1, i.e. = 1, i.e. x = y x y2 y Substituting y = x in equation (3) gives x 3 = 125, from which, x = 5 m. Hence y = 5 m also

= 75 m2 Now try the following exercise

62.5 From equation (1), z = xy

S = x y + 2y

From equation (2), minimum surface area, S

2. An open rectangular container is to have a volume of 32 m3 . Determine the dimensions and the total surface area such that the total surface area is a minimum. 4 m by 4 m by 2 m, surface area = 48m2 3. Determine the stationary values of the function f (x, y) = x 4 + 4x 2 y 2 − 2x 2 + 2y 2 − 1 and distinguish between them. ⎡ ⎤ Minimum at (1, 0), ⎣ minimum at (−1, 0), ⎦ saddle point at (0, 0)

366 Higher Engineering Mathematics 4. Determine the stationary points of the surface f (x, y) = x 3 − 6x 2 − y 2 . Maximum at (0, 0), saddle point at (4, 0) 5. Locate the stationary points on the surface f (x, y) = 2x 3 + 2y 3 − 6x − 24y + 16 and determine their nature. ⎡ ⎤ Minimum at (1, 2), ⎣ maximum at (−1, −2), ⎦ saddle points at (1, −2) and (−1, 2)

6. A large marquee is to be made in the form of a rectangular box-like shape with canvas covering on the top, back and sides. Determine the minimum surface area of canvas necessary if the volume of the marquee is to the 250 m3. [150 m2 ]

Revision Test 10 This Revision Test covers the material contained in Chapters 32 to 36. The marks for each question are shown in brackets at the end of each question. 1.

(a) 5 ln (shx) (b) 3 ch3 2x 2x

(c) e 2.

6.

sech 2x

(7)

Differentiate the following functions with respect to the variable: x 1 (a) y = cos−1 5 2 (b) y = 3esin

2 sec−1 5x x (d) y = 3 sinh−1 (2x 2 − 1)

4.

∂z ∂z , , , , and . ∂x ∂ y ∂x 2 ∂ y 2 ∂x∂ y ∂ y∂x ∂2z

5.

8.

The volume V of a liquid of viscosity coefﬁcient η delivered after time t when passed through a tube of length L and diameter d by a pressure p pd 4t . If the errors in V , p and is given by V = 128ηL L are 1%, 2% and 3% respectively, determine the error in η. (8)

9.

Determine and distinugish between the stationary values of the function

Evaluate the following, each correct to 3 decimal places: (6)

If z = f (x, y) and z = x cos(x + y) determine ∂2z

∂2 z

∂2 z

(12)

The magnetic ﬁeld vector H due to a steady current I ﬂowing around a circular wire of radius r and at a distance x from its centre is given by x I ∂ √ H =± 2 ∂x r2 + x2

(6)

An engineering function z = f (x, y) and y z = e 2 ln(2x + 3y). Determine the rate of increase of z, correct to 4 signiﬁcant ﬁgures, when x = 2 cm, y = 3 cm, x is increasing at 5 cm/s and y is increasing at 4 cm/s. (8)

(14)

(a) sinh−1 3 (b) cosh−1 2.5 (c) tanh−1 0.8

If x yz = c, where c is constant, show that dx d y + dz = −z x y

(7)

7.

−1 t

(c) y =

3.

r2 I Show that H = ± 2 (r 2 + x 2 )3

Differentiate the following functions with respect to x:

f (x, y) = x 3 − 6x 2 − 8y 2 and sketch an approximate contour map to represent the surface f (x, y). (20) 10. An open, rectangular ﬁsh tank is to have a volume of 13.5 m3 . Determine the least surface area of glass required. (12)

Chapter 37

Standard integration 37.1

The process of integration

The process of integration reverses the process of differentiation. In differentiation, if f (x) = 2x 2 then f (x) = 4x. Thus the integral of 4x is 2x 2 , i.e. integration is the process of moving from f (x) to f (x). By similar reasoning, the integral of 2t is t 2. Integration is a process of summation oradding parts together and an elongated S, shown as , is used to replace the words ‘the integral of’. Hence, from above, 4x = 2x 2 and 2t is t 2. dy In differentiation, the differential coefﬁcient indidx cates that a function of x is being differentiated with respect to x, the dx indicating that it is ‘with respect to x’. In integration the variable of integration is shown by adding d (the variable) after the function to be integrated.

37.2 The general solution of integrals of the form ax n The general solution of integrals of the form ax n dx, where a and n are constants is given by: ! ax n dx =

This rule is true when n is fractional, zero, or a positive or negative integer, with the exception of n = −1. Using this rule gives: ! 3x 4+1 3 (i) 3x 4 dx = + c = x5 + c 4+1 5 ! ! 2x −2+1 2 −2 dx = 2x dx = (ii) +c x2 −2 +1

! Thus

and

2t dt means ‘the integral of 2t with respect to t ’.

As stated above, the differential coefﬁcient of 2x 2 is 4x, hence 4x dx = 2x 2 . However, the differential coefﬁcient of 2x 2 + 7 is also 4x. Hence 4x dx is also equal to 2x 2 + 7. To allow for the possible presence of a constant, whenever the process of integration is performed, a constant ‘c’ is added to the result. ! Thus

=

4x dx means ‘the integral of 4x with respect to x’, !

! 4x dx = 2x 2 + c and

2t dt = t 2 + c

‘c’ is called the arbitrary constant of integration.

ax n+1 +c n+1

! (iii)

√

2x −1 −2 +c= + c, and −1 x !

x dx =

1

1 x2

3

x 2 +1 x2 dx = +c= +c 1 3 +1 2 2

2√ 3 x +c 3 Each of these three results may be checked by differentiation. =

(a)

The integral of a constant k is kx + c. For example, ! 8 dx = 8x + c

(b) When a sum of several terms is integrated the result is the sum of the integrals of the separate terms.

Standard integration For example, ! (3x + 2x 2 − 5) dx ! ! ! 2 = 3x dx + 2x dx − 5 dx =

37.3

3x 2 2x 3 + − 5x + c 2 3

Standard integrals

Since integration is the reverse process of differentiation the standard integrals listed in Table 37.1 may be deduced and readily checked by differentiation. Table 37.1 Standard integrals !

ax n+1 +c n +1 (except when n =−1)

ax n dx =

(i) !

cos ax dx =

(ii) ! (iii) !

1 sin ax dx = − cos ax + c a sec 2 ax dx =

(iv) ! (v) ! (vi) ! (vii) ! !

(b) When a = 2 and n = 3 then ! 2t 3 dt =

1 cosec ax cot ax dx = − cosec ax + c a 1 sec ax tan ax dx = sec ax + c a 1 ax e +c a

1 dx = ln x + c x

Problem 1. Determine (a) 5x 2 dx (b) 2t 3 dt . ax n+1 The standard integral, ax n dx = +c n +1 (a) When a = 5 and n =2 then ! 5x 2+1 5x 3 5x 2 dx = +c= +c 2+1 3

2t 3+1 2t 4 1 +c= +c= t4 +c 3+1 4 2

Each of these results may be checked by differentiating them. Problem 2. Determine ! 3 4 + x − 6x 2 dx. 7 (4 + 37 x − 6x 2 ) dx may be written as 4 dx + 37 x dx − 6x 2 dx, i.e. each term is integrated separately. (This splitting up of terms only applies, however, for addition and subtraction.) ! 3 2 Hence 4 + x − 6x dx 7 1+1 3 x x 2+1 = 4x + − (6) +c 7 1+1 2+1 2 3 x x3 = 4x + − (6) + c 7 2 3

1 tan ax + c a

1 cosec 2 ax dx = − cot ax + c a

eax dx =

(viii) (ix)

1 sin ax + c a

369

= 4x +

3 2 x − 2x 3 + c 14

Note that when an integral contains more than one term there is no need to have an arbitrary constant for each; just a single constant at the end is sufﬁcient. Problem 3. Determine ! ! 2x 3 − 3x (a) dx (b) (1 − t )2 dt. 4x (a)

Rearranging into standard integral form gives: ! 2x 3 − 3x dx 4x ! ! 2 2x 3 3x x 3 = − dx = − dx 4x 4x 2 4 2+1 1 x 3 = − x +c 2 2+1 4 3 1 x 3 3 1 − x + c = x3 − x + c = 2 3 4 6 4

370 Higher Engineering Mathematics ! !

2t 1+1 t 2+1 + +c 1+1 2+1

(1 − 2t + t 2) dt = t −

2t 2 t 3 + +c 2 3

=t−

1 = t −t 2 + t 3 +c 3 This problem shows that functions often have to be rearranged into the standard form of ax n dx before it is possible to integrate them. ! Problem 4.

1 5 4 1 5 t4 +c = − = − t4 +c 9 14 9 1 20 √ 4 t+c =− 9

(1 − t )2 dt gives:

(b) Rearranging

Determine

3 dx. x2

! Problem 7. !

Determine

(1 + θ)2 dθ = √ θ

!

3x −2+1 3x −1 3x −2 dx = +c = +c −2 + 1 −1 −3 = −3x −1 + c = +c x √ Problem 5. Determine 3 x dx. For fractional powers it is necessary to appreciate √ m n m a =a n !

√

!

3 x dx =

3x 2 +1 dx = +c 1 +1 2

3

−5 dt = √ 4 9 t3

!

=

!

=

θ

θ

−1 2

− 12

θ2 1 2

+1

+

+1

1

=

1 3 + 2θ 2 + θ 2 dθ

−1 2

2θ 1 2

3

+

2θ 2 3 2

1 2

+1

+1

+

θ

3 2 +1

3 2

+1

5

+

θ2 5 2

+c

1 4 3 2 5 = 2θ 2 + θ 2 + θ 2 + c 3 5 √ 4 3 2 5 θ + θ +c = 2 θ+ 3 5

1

1 3x 2

3 3x 2 + c = 2x 2 + c = 2 x 3 + c = 3 2 ! −5 Problem 6. Determine √ dt . 4 9 t3 !

!

(1 + 2θ + θ 2 ) dθ √ θ ! 1 2θ θ2 = + + dθ 1 1 1 θ2 θ2 θ2 ! −1 1− 1 2− 1 = θ 2 + 2θ 2 + θ 2 dθ

!

! 3 dx = 3x −2 dx. Using the standard integral, 2 x ! ax n dx when a = 3 and n =−2 gives:

(1 + θ)2 dθ. √ θ

−5 3

9t 4

dt =

! 5 −3 t 4 dt − 9

3 − +1 5 t 4 +c = − 3 9 − +1 4

Problem 8. Determine (a) 4 cos3x dx (b) 5 sin 2θ dθ. (a) From Table 37.1(ii), ! 1 4 cos3x dx = (4) sin 3x + c 3 4 = sin 3x + c 3 (b) From Table 37.1(iii), ! 1 5 sin 2θ dθ = (5) − cos 2θ + c 2 5 = − cos 2θ + c 2

+c

371

Standard integration Problem 9. Determine (a) 7 sec2 4t dt (b) 3 cosec 2 2θ dθ.

=

2m 2 + ln m + c 2

= m 2 + ln m + c (a)

From Table 37.1(iv), ! 7 sec2 4t dt = (7)

1 tan 4t + c 4

7 = tan 4t + c 4 (b) From Table 37.1(v), ! 3

1 cot 2θ + c cosec 2θ dθ = (3) − 2 2

Now try the following exercise Exercise 145 integrals

Further problems on standard

In Problems 1 to 12, determine the indeﬁnite integrals. ! ! 1. (a) 4 dx (b) 7x dx

3 = − cot 2θ + c 2 !

Problem 10. Determine ! ! 2 (a) 5 e3x dx (b) dt. 3 e4t (a)

2.

From Table 37.1(viii), ! 1 3x 5 3x 5 e dx = (5) e + c = e3x + c 3 3 !

(b)

2 dt = 3 e4t

!

(a) 4x + c (b)

3.

2 1 −4t 2 −4t − e +c e dt = 3 3 4

1 1 = − e−4t + c = − 4t + c 6 6e

(a)

(a)

Problem 11. Determine ! ! 2 3 2m + 1 dm. (a) dx (b) 5x m ! (a)

3 dx = 5x

! 1 3 3 dx = ln x +c 5 x 5

(b)

! 2 2m 1 2m 2 + 1 dm = + dm m m m =

! 1 dm 2m + m

5 3 x dx 6

(b)

5 4 2 3 x + c (b) x +c (a) 15 24

4 dx (b) 3x 2

!

3 dx 4x 4

(a)

5.

(a) 2

!

! x 3 dx (b) (a)

(from Table 37.1(ix)) !

!

! 6.

! ! 2 3x − 5x dx (b) (2 + θ)2 dθ (a) x ⎡ ⎤ 3x 2 (a) − 5x + c ⎢ ⎥ 2 ⎢ ⎥ ⎣ ⎦ 3 θ 2 (b) 4θ + 2θ + + c 3 !

4.

2 2 x dx 5

7x 2 +c 2

(a)

−1 −4 + c (b) 3 + c 3x 4x 1 4 x 5 dx 4

1√ 4√ 5 4 9 x + c (b) x +c 5 9

! −5 3 √ dt (b) √ dx 5 t3 7 x4 10 15 √ 5 (a) √ + c (b) x +c 7 t

372 Higher Engineering Mathematics ! 7.

(a)

! 3 cos2x dx (b)

7 sin 3θ dθ ⎡

⎤ 3 ⎢ (a) 2 sin 2x + c ⎥ ⎢ ⎥ ⎣ ⎦ 7 (b) − cos 3θ + c 3 ! ! 3 sec2 3x dx (b) 2 cosec 2 4θ dθ 8. (a) 4 1 1 (a) tan 3x +c (b) − cot 4θ +c 4 2 ! 9. (a) 5 cot 2t cosec 2t dt ! 4 sec 4t tan 4t dt (b) 3 ⎡ ⎤ 5 cosec 2t + c (a) − ⎢ ⎥ 2 ⎢ ⎥ ⎣ ⎦ 1 (b) sec 4t + c 3 ! ! 2 dx 3 2x e dx (b) 10. (a) 4 3 e5x −2 3 + c (a) e2x + c (b) 8 15 e5x ! ! 2 2 u −1 11. (a) du dx (b) 3x u 2 u2 (a) ln x + c (b) − ln u + c 3 2 ! 12.

(a)

(2+3x)2 √ dx (b) x ⎡

!

!

x3 +c 3

3

3 33 1 +c − +c 3 3 1 1 2 = (9 + c) − + c =8 3 3 =

Note that the ‘c’ term always cancels out when limits are applied and it need not be shown with deﬁnite integrals. Problem 12. Evaluate 3 2 (a) 1 3x dx (b) −2 (4 − x 2 ) dx. !

2

(a) 1

3x 2 3x dx = 2

2

3 2 3 2 = (2) − (1) 2 2 1

1 1 =6 − 1 =4 2 2 ! (b)

3 x3 (4 − x ) dx = 4x − 3 −2 −2 3

2

(3)3 (−2)3 = 4(3) − − 4(−2) − 3 3 −8 = {12 − 9} − −8 − 3 1 1 = {3} − −5 =8 3 3 4

!

√

Deﬁnite integrals

x 2 dx =

1

Problem 13.

Evaluate

positive square roots only. 4

! 1

37.4

3

2 1 + 2t dt t

⎤ 18 √ 5 (a) 8 x + 8 x 3 + x +c ⎢ ⎥ 5 ⎢ ⎥ ⎣ ⎦ 3 1 4t (b) − + 4t + +c t 3 √

limit and ‘a’ the lower limit. The operation of applying the limits is deﬁned as [x]ba = (b) − (a). The increase in the value of the integral x 2 as x increases 3 from 1 to 3 is written as 1 x 2 dx. Applying the limits gives:

1

θ +2 √ dθ, taking θ

! 4 θ +2 θ 2 √ + dθ dθ = 1 1 θ 1 θ2 θ2 ! 4 1 −1 θ 2 + 2θ 2 dθ = 1

Integrals containing an arbitrary constant c in their results are called indeﬁnite integrals since their precise value cannot be determined without further information. Deﬁnite integrals are those in which limits are applied. If an expression is written as [x]ba, ‘b’ is called the upper

⎡ ⎢θ =⎣

1 2 +1

1 +1 2

+

⎤4 −1 2 +1

⎥ ⎦ 1 − +1 2 1

2θ

Standard integration ⎡ =⎣

3

θ2 3 2

1

+

2θ 2 1 2

⎤4 ⎦ = 1

√ 4 2 3 θ +4 θ 3 1

√ 2 2 3 3 = (4) + 4 4 − (1) + 4 (1) 3 3 16 2 = +8 − +4 3 3

Problem 16. Evaluate ! ! 2 4 e2x dx (b) (a) 1

π 2

Problem 14. Evaluate

!

2

(a)

4 e2x dx =

1

π 2

!

4

(b)

2 = 2[ e2x ]21 = 2[ e4 − e2 ] 1

1

4 3 3 3 du = ln u = [ln 4 − ln 1] 4u 4 4 1

3 sin 2x dx.

3 sin 2x dx

4 2x e 2

= 2[54.5982 −7.3891] =94.42

3 = [1.3863 −0] =1.040 4

!

1

3 du, 4u

each correct to 4 signiﬁcant ﬁgures.

2 2 1 = 5 +8− −4 = 8 3 3 3 !

4

Now try the following exercise

π π 2 2 3 1 = − cos 2x = (3) − cos 2x 2 2 0 0 π 3 3 − − cos 2(0) = − cos 2 2 2 2 3 3 = − cos π − − cos 0 2 2 3 3 3 3 = − (−1) − − (1) = + = 3 2 2 2 2

!

2

Problem 15. Evaluate

4 cos 3t dt.

Exercise 146 integrals

Further problems on deﬁnite

In problems 1 to 8, evaluate the deﬁnite integrals (where necessary, correct to 4 signiﬁcant ﬁgures). ! 1 ! 4 3 2 5x dx (b) − t 2 dt 1. (a) 1 −1 4 1 (a) 105 (b) − 2 ! 2 ! 3 2. (a) (3 − x 2 ) dx (b) (x 2 − 4x + 3) dx −1

1

1 (a) 6 (b) −1 3

1

!

2 2 2 1 4 4 cos3t dt = (4) sin 3t = sin 3t 3 3 1 1 1 4 4 = sin 6 − sin 3 3 3

Note that limits of trigonometric functions are always expressed in radians—thus, for example, sin 6 means the sine of 6 radians= −0.279415 . . . ! 2 4 cos 3t dt Hence 1

4 4 = (−0.279415 . . .) − (0.141120 . . .) 3 3 = (−0.37255) − (0.18816) = −0.5607

!

π

3 cos θ dθ 2

3. (a) 0

!

π 2

(b)

4 cos θ dθ

[(a) 0 (b) 4] ! 4. (a)

π 3 π 6

!

2

2 sin 2θ dθ (b)

3 sin t dt 0

[(a) 1 (b) 4.248] ! 5. (a)

!

1

π 6

5 cos3x dx (b) 0

3 sec2 2x dx

[(a) 0.2352 (b) 2.598]

373

374 Higher Engineering Mathematics !

2

6. (a)

1 litre to 3 litres for a temperature rise from 100 K to 400 K given that:

cosec 2 4t dt

1

! (b)

π 2

π 4

(3 sin 2x − 2 cos3x) dx [(a) 0.2527 (b) 2.638]

!

1

7. (a)

! 3 e3t dt (b)

2

2 dx 2x 3 e −1 [(a) 19.09 (b) 2.457]

!

3

8. (a) 2

2 dx (b) 3x

!

3 1

2x 2 + 1 dx x [(a) 0.2703 (b) 9.099]

9. The entropy change S, for an ideal gas is given by: ! V2 ! T2 dT dV Cv −R S = T T1 V1 V where T is the thermodynamic temperature, V is the volume and R = 8.314. Determine the entropy change when a gas expands from

Cv = 45 + 6 × 10−3 T + 8 × 10−6 T 2 . [55.65] 10. The p.d. between boundaries a and b of an ! b Q electric ﬁeld is given by: V = dr 2πrε 0 εr a If a = 10, b = 20, Q =2 × 10−6 coulombs, ε0 = 8.85 ×10−12 and εr = 2.77, show that V = 9 kV. 11. The average value of a complex voltage waveform is given by: ! 1 π (10 sin ωt + 3 sin 3ωt V AV = π 0 + 2 sin 5ωt) d(ωt) Evaluate V AV correct to 2 decimal places. [7.26]

Chapter 38

Some applications of integration 38.1

Introduction

There are a number of applications of integral calculus in engineering. The determination of areas, mean and r.m.s. values, volumes, centroids and second moments of area and radius of gyration are included in this chapter.

38.2

Areas under and between curves

When y = 0, x = 0 or (x + 2) = 0 or (x − 4) = 0, i.e. when y = 0, x = 0 or −2 or 4, which means that the curve crosses the x-axis at 0, −2, and 4. Since the curve is a continuous function, only one other co-ordinate value needs to be calculated before a sketch of the curve can be produced. When x = 1, y = −9, showing that the part of the curve between x = 0 and x = 4 is negative. A sketch of y = x 3 − 2x 2 − 8x is shown in Fig. 38.2. (Another method of sketching Fig. 38.2 would have been to draw up a table of values.) y

In Fig. 38.1, ! total shaded area =

b

10

!

c

f (x)dx −

a

b

f (x)dx ! +

22

d

21

y 5 x 3 2 2x 2 2 8x

1

2

3

4

f (x)dx 210

c

y 220 y 5 f (x) G

Figure 38.2

E 0

a

b

F

c

d

x

Figure 38.1

Problem 1. Determine the area between the curve y = x 3 − 2x 2 − 8x and the x-axis. y = x 3 −2x 2 − 8x = x(x 2 −2x − 8) = x(x + 2)(x − 4)

Shaded area ! 0 ! 4 = (x 3 − 2x 2 − 8x)dx − (x 3 − 2x 2 − 8x)dx −2

x4

2x 3

0 8x 2

x 4 2x 3 8x 2 = − − − − − 4 3 2 −2 4 3 2 2 2 1 = 6 − −42 = 49 square units 3 3 3

4 0

x

376 Higher Engineering Mathematics 1 1 − −13 = 7 3 2

Problem 2. Determine the area enclosed between the curves y = x 2 + 1 and y = 7 − x. At the points of intersection the curves are equal. Thus, equating the y values of each curve gives: x2 + 1 = 7 − x x2 + x − 6 = 0

from which,

Factorizing gives (x − 2)(x + 3) = 0 from which x = 2 and x = −3 By ﬁrstly determining the points of intersection the range of x-values has been found. Tables of values are produced as shown below. x

−3 −2 −1 0 1 2

y = x2 + 1

10

5

2 1 2

x

−3

0 2

y = 7−x

10

7

21

!

2 −3

! =

2 −3

! =

2 −3

y542x

y 5 3x

3y 5 x (or y 5 x3 )

2

1

2

3

4

x

Figure 38.4

Shaded area ! 1 ! 3 x x = dx + 3x − (4 − x) − dx 3 3 0 1

y 5 x 2 11

1 3 3x 2 x 2 x2 x2 + 4x − − − 2 6 0 2 6 1 3 1 9 9 − (0) + 12 − − = − 2 6 2 6 1 1 − 4− − 2 6 1 1 + 6−3 = 4 square units = 1 3 3

=

y572x

1

2

x

Figure 38.3

Shaded area =

y

5

5

22

Each of the straight lines are shown sketched in Fig. 38.4.

5

y

23

Problem 3. Determine by integration the area bounded by the three straight lines y = 4 − x, y = 3x and 3y = x.

4

A sketch of the two curves is shown in Fig. 38.3.

10

5 = 20 square units 6

! (7 − x)dx −

2 −3

(x 2 + 1)dx

[(7 − x) − (x 2 + 1)]dx Now try the following exercise (6 − x − x 2 )dx

x2 x3 = 6x − − 2 3

2 −3

9 8 − −18 − + 9 = 12 − 2 − 3 2

Exercise 147 Further problems on areas under and between curves 1. Find the area enclosed by the curve y = 4 cos 3x, the x-axis and ordinates x = 0 π [1 13 square units] and x = 6

Some applications of integration

377

[Note that for a sine wave, 2. Sketch the curves y = x 2 + 3 and y = 7 − 3x and determine the area enclosed by them. [20 56 square units] 3. Determine the area enclosed by the three straight lines y = 3x, 2y = x and y + 2x = 5. [2 12 square units]

In this case, mean value = (b) r.m.s. value =

38.3

Mean and r.m.s. values =

With reference to Fig. 38.5, ! b 1 mean value, y = y dx b −a a 75 6 8 ! b 8 1 y2 dx and r.m.s. value = 9 b−a a y

2 × maximum value π

mean value=

=

!

1 π −0 !

1 π

π

π

2 × 100 = 63.66 V] π

v 2 d(ωt )

(100 sin ωt )2 d(ωt )

10000 π

!

π

sin2 ωt d(ωt ) ,

which is not a ‘standard’ integral. It is shown in Chapter 17 that cos 2 A = 1 − 2 sin2 A and this formula is used whenever sin2 A needs to be integrated.

y 5 f(x)

Rearranging cos 2 A = 1 − 2 sin2 A gives

y

Hence 0

x5a

x5b

Figure 38.5

Problem 4. A sinusoidal voltage v = 100 sin ωt volts. Use integration to determine over half a cycle (a) the mean value, and (b) the r.m.s. value. (a)

x

Half a cycle means the limits are 0 to π radians. ! π 1 Mean value, y = v d(ωt ) π −0 0 ! 1 π = 100 sinωt d(ωt ) π 0 100 = [−cos ωt ]π0 π 100 = [(−cos π) − (−cos 0)] π 200 100 [(+1) − (−1)] = = π π = 63.66 volts

= =

1 sin2 A = (1 − cos 2 A) 2 10000 π

10000 π

!

!

sin2 ωt d(ωt )

0 π

π

1 (1 − cos 2ωt ) d(ωt ) 2

10000 1 sin 2ωt π ωt − π 2 2 0

7⎧ 8 10000 1 sin 2π 8⎪ ⎪ 8⎨ π− 8 π 2 2 =8 sin 0 9⎪ ⎪ − 0− ⎩ 2 = =

10000 1 [π] π 2

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

100 10000 = √ = 70.71 volts 2 2

[Note that for a sine wave, 1 r.m.s. value= √ × maximum value. 2

378 Higher Engineering Mathematics y

In this case,

y 5 f (x)

1 r.m.s. value = √ × 100 = 70.71 V] 2 A

Now try the following exercise x5a

Exercise 148 Further problems on mean and r.m.s. values 1. The vertical height h km of a missile varies with the horizontal distance d km, and is given by h = 4d − d 2 . Determine the mean height of the missile from d = 0 to d = 4 km. [2 23 km]. 2. The distances of points y from the mean value of a frequency distribution are related to the 1 variate x by the equation y = x + . Deterx mine the standard deviation (i.e. the r.m.s. value), correct to 4 signiﬁcant ﬁgures for values of x from 1 to 2. [2.198] 3. A current i = 25 sin 100πt mA ﬂows in an electrical circuit. Determine, using integral calculus, its mean and r.m.s. values each correct to 2 decimal places over the range t = 0 to t = 10 ms. [15.92 mA, 17.68 mA]

generated, V , is given by: ! d πx2 dy V= c

Problem 5. The curve y = x 2 + 4 is rotated one revolution about the x-axis between the limits x = 1 and x = 4. Determine the volume of solid of revolution produced. Revolving the shaded area shown in Fig. 38.7, 360◦ about the x-axis produces a solid of revolution given by: ! 4 ! 4 π y 2 dx = π(x 2 + 4)2 dx Volume = !

1

1 4

=

π(x 4 + 8x 2 + 16) dx

1

x 5 8x 3 =π + + 16x 5 3

v = E 1 sin ωt + E 3 sin 3ωt

4 1

= π[(204.8 + 170.67 + 64)

where E 1 , E 3 and ω are constants. Determine the r.m.s. value of v over the π interval 0 ≤ t ≤ . ω ⎤ ⎡ E 12 + E 32 ⎦ ⎣ 2

− (0.2 + 2.67 + 16)] = 420.6π cubic units y 30

20

Volumes of solids of revolution

With reference to Fig. 38.6, the volume of revolution, V , obtained by rotating area A through one revolution about the x-axis is given by: ! b πy2 dx V=

A

10 5 D 4 0

a

If a curve x = f ( y) is rotated 360◦ about the y-axis between the limits y = c and y = d then the volume

x

Figure 38.6

4. A wave is deﬁned by the equation:

38.4

x5b

Figure 38.7

y 5 x21 4

B

C

1

2

3

4

5

x

Some applications of integration Problem 6. Determine the area enclosed by the two curves y = x 2 and y 2 = 8x. If this area is rotated 360◦ about the x-axis determine the volume of the solid of revolution produced.

{(volume produced by revolving y 2 = 8x) − (volume produced by revolving y = x 2 )} !

2

i.e. volume =

!

x 4 − 8x = 0

Hence, at the points of intersection, x = 0 and x = 2. When x = 0, y = 0 and when x = 2, y = 4. The points of intersection of the curves y = x 2 and y 2 = 8x are therefore at (0,0) and (2,4).√A sketch is shown in Fig. 38.8. If y 2 = 8x then y = 8x.

Shaded area !

2

=

1 8 x 2 − x 2 dx

! √ 8x − x 2 dx =

2 √ 0

⎤2 5 √ √ ⎡ 6 √ x 32 3 8 8 8 x − {0} − =⎣ 8 3 − ⎦ = 3 3 3 2 2 0

=

16 8 8 2 − = = 2 square units 3 3 3 3 y5x2

y

y 2 5 8x (or y 5Œ(8x)

4

2

2

8x 2 x 5 = π (8x − x )dx = π − 2 5 0

2

4

= 9.6π cubic units

x(x 3 − 8) = 0

and

π(x 4 )dx

32 − (0) = π 16 − 5

x 4 = 8x from which,

!

2

π(8x)dx −

At the points of intersection the co-ordinates of the curves are equal. Since y = x 2 then y 2 = x 4 . Hence equating the y 2 values at the points of intersection:

Now try the following exercise Exercise 149

Further problems on volumes

1. The curve x y = 3 is revolved one revolution about the x-axis between the limits x = 2 and x = 3. Determine the volume of the solid produced. [1.5π cubic units] y 2. The area between 2 = 1 and y + x 2 = 8 is x rotated 360◦ about the x-axis. Find the volume produced. [170 23 π cubic units] 3. The curve y = 2x 2 + 3 is rotated about (a) the x-axis between the limits x = 0 and x = 3, and (b) the y-axis, between the same limits. Determine the volume generated in each case. [(a) 329.4π (b) 81π] 4. The proﬁle of a rotor blade is bounded by the lines x = 0.2, y = 2x, y = e−x , x = 1 and the x-axis. The blade thickness t varies linearly with x and is given by: t = (1.1 − x)K, where K is a constant. (a) Sketch the rotor blade, labelling the limits. (b) Determine, using an iterative method, the value of x, correct to 3 decimal places, where 2x = e−x

1

2

x

Figure 38.8

The volume produced by revolving the shaded area about the x-axis is given by:

379

(c) Calculate the cross-sectional area of the blade, correct to 3 decimal places. (d) Calculate the volume of the blade in terms of K, correct to 3 decimal places. [(b) 0.352 (c) 0.419 square units (d) 0.222 K]

380 Higher Engineering Mathematics 38.5

Centroids

A lamina is a thin ﬂat sheet having uniform thickness. The centre of gravity of a lamina is the point where it balances perfectly, i.e. the lamina’s centre of mass. When dealing with an area (i.e. a lamina of negligible thickness and mass) the term centre of area or centroid is used for the point where the centre of gravity of a lamina of that shape would lie. If x and y denote the co-ordinates of the centroid C of area A of Fig. 38.9, then: !

!

b

1 2

xy dx x = !a

b

y2 dx

y dx

!

2

=

0 ! 2

!

1 2

y 2 dx

2

(3x 2 )2 dx

8

y dx 0

=

=

!

1 2

2

9x 4 dx =

8

32 5 8

9 2

2 9 x5 2 5 0

8

=

18 = 3.6 5

Hence the centroid lies at (1.5, 3.6)

and y = ! ab

b

y=

1 2

y dx

a

Problem 8. Determine the co-ordinates of the centroid of the area lying between the curve y = 5x − x 2 and the x-axis.

a

y y 5 f(x)

y = 5x − x 2 = x(5 − x). When y = 0, x = 0 or x = 5. Hence the curve cuts the x-axis at 0 and 5 as shown in Fig. 38.10. Let the co-ordinates of the centroid be (x , y) then, by integration,

Area A C

!

x x5b

x

x= !

= !

x(5x − x 2 ) dx

5

5

y dx

Figure 38.9

Problem 7. Find the position of the centroid of the area bounded by the curve y = 3x 2 , the x-axis and the ordinates x = 0 and x = 2. If (x , y) are co-ordinates of the centroid of the given area then: !

!

2

2

x y dx x = !0

=

2

x(3x 2 ) dx

2

y dx 2

3x dx 0

5

= !0 5

= (5x − x ) dx 2

12 = 1.5 8

5x 3 3

−

5x 2 2

−

5 x4 4 0 5 x3 3 0

y

y 5 5x 2 x 2

6 4

3

(5x 2 − x 3 ) dx

2

2 3x 4 3x dx 4 0 = = !0 2 [x 3 ]20 3x 2 dx !

!

(5x − x 2 ) dx

8

!

=

5

x y dx

x5a

!

5

y

C

x 2

y 0

Figure 38.10

1

2

3

4

5

x

Some applications of integration 625 − = 3 125 − 2 =

y=

1 2

625 12 !

5

625 625 4 = 12 125 125 3 6

6 125

2

y dx =

0 ! 5

!

=

=

=

5

0 ! 5

y dx 0

1 2

4. Find the co-ordinates of the centroid of the area which lies between the curve y/x = x − 2 and the x-axis. [(1, −0.4)] 5. Sketch the curve y 2 = 9x between the limits x = 0 and x = 4. Determine the position of the centroid of this area. [(2.4, 0)]

5 = = 2.5 2

1 2

(5x − x 2 )2 dx

(5x − x 2 ) dx

!

5

(25x − 10x + x ) dx 3

Theorem of Pappus

4

‘If a plane area is rotated about an axis in its own plane but not intersecting it, the volume of the solid formed is given by the product of the area and the distance moved by the centroid of the area’. With reference to Fig. 38.11, when the curve y = f (x) is rotated one revolution about the x-axis between the limits x = a and x = b, the volume V generated is given by:

125 6

5 1 25x 3 10x 4 x 5 − + 2 3 4 5 0

125 6 1 2

38.6

A theorem of Pappus states: 2

381

25(125) 6250 − + 625 3 4 125 6

volume V = (A)(2π y ), from which, y =

V 2π A

y

= 2.5

y 5 f(x) Area A

Hence the centroid of the area lies at (2.5, 2.5).

C

(Note from Fig. 38.10 that the curve is symmetrical about x = 2.5 and thus x could have been determined ‘on sight’.)

y x5a

x5b x

Figure 38.11

Now try the following exercise Exercise 150 Further problems on centroids In Problems 1 and 2, ﬁnd the position of the centroids of the areas bounded by the given curves, the x-axis and the given ordinates. 1.

y = 3x + 2 x = 0, x = 4

2.

y=

5x 2

x = 1, x = 4

[(2.5, 4.75)] [(3.036, 24.36)]

3. Determine the position of the centroid of a sheet of metal formed by the curve y = 4x − x 2 which lies above the x-axis. [(2, 1.6)]

Problem 9. (a) Calculate the area bounded by the curve y = 2x 2 , the x-axis and ordinates x = 0 and x = 3. (b) If this area is revolved (i) about the x-axis and (ii) about the y-axis, ﬁnd the volumes of the solids produced. (c) Locate the position of the centroid using (i) integration, and (ii) the theorem of Pappus. (a)

The required area is shown shaded in Fig. 38.12. ! 3 ! 3 y dx = 2x 2 dx Area = 0

=

3 2x 3 3

= 18 square units

382 Higher Engineering Mathematics y

y 5 2x 2

y=

18

1 2

!

2

y dx =

0 ! 3

1 2

!

3

18

x

6

y 0

1

2

=

x

3

Figure 38.12

1 2

!

3

4x 4 dx =

18

(2x 2 )2 dx

y dx

12

(b)

3

3 1 4x 5 2 5 0

18

= 5.4

(ii) using the theorem of Pappus:

(i) When the shaded area of Fig. 38.12 is revolved 360◦ about the x-axis, the volume generated !

3

=

!

3

π y 2 dx =

π(2x 2 )2 dx

3 x5 4π x dx = 4π = 5 0 0 243 = 194.4πcubic units = 4π 5 !

3

4

Volume generated when shaded area is revolved about OY= (area)(2π x ). 81π = (18)(2π x ),

i.e. from which,

x=

Volume generated when shaded area is revolved about OX = (area)(2π y). 194.4π = (18)(2π y),

i.e.

y=

from which, (ii) When the shaded area of Fig. 38.12 is revolved 360◦ about the y-axis, the volume generated = (volume generated by x = 3) − (volume generated by y = 2x 2 ) ! 18 ! 18 y 2 = π(3) dy − π dy 2 0 0 18 ! 18 y2 y =π dy = π 9y − 9− 2 4 0 0 = 81π cubic units (c) If the co-ordinates of the centroid of the shaded area in Fig. 38.12 are (x, y) then: (i) by integration, !

!

3

3

x y dx x = !0

=

3

= =

3 0

18 81 = 2.25 36

=

Hence the centroid of the shaded area in Fig. 38.12 is at (2.25, 5.4).

Problem 10. A metal disc has a radius of 5.0 cm and is of thickness 2.0 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine, using Pappus’ theorem, the volume and mass of metal removed and the volume and mass of the pulley if the density of the metal is 8000 kg m−3. A side view of the rim of the disc is shown in Fig. 38.13. 2.0 cm P

x(2x 2 ) dx

Q

18

2x 3 dx

194.4π = 5.4 36π

y dx !

81π = 2.25 36π

5.0 cm S

3 2x 4 4 18

X

Figure 38.13

R X

Some applications of integration When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid of the 4r semicircular area removed is at a distance of from its 3π diameter (see ‘Engineering Mathematics 6th edition’, 4(1.0) Chapter 58), i.e. , i.e. 0.424 cm from PQ. Thus 3π the distance of the centroid from XX is 5.0 − 0.424, i.e. 4.576 cm. The distance moved through in one revolution by the centroid is 2π(4.576) cm. π(1.0)2 π πr 2 = = cm2 Area of semicircle = 2 2 2 By the theorem of Pappus, volume generated = area × distance moved by π (2π)(4.576). centroid = 2 i.e. volume of metal removed = 45.16 cm3 Mass of metal removed = density × volume 45.16 3 m 106 = 0.3613 kg or 361.3 g

= 8000 kg m−3×

volume of pulley = volume of cylindrical disc − volume of metal removed = π(5.0)2 (2.0) − 45.16 = 111.9 cm3 Mass of pulley = density× volume = 8000 kg m−3 ×

111.9 3 m 106

= 0.8952 kg or 895.2 g

Now try the following exercise Exercise 151 Further problems on the theorem of Pappus 1. A right angled isosceles triangle having a hypotenuse of 8 cm is revolved one revolution about one of its equal sides as axis. Determine the volume of the solid generated using Pappus’ theorem. [189.6 cm3 ] 2. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid of a metal template in the form of a quadrant

383

of a circle of radius 4 cm. (The equation of a circle, centre 0, radius r is x 2 + y 2 = r 2 ). ⎡ ⎤ On the centre line, distance ⎢ 2.40 cm from the centre, ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ i.e. at co-ordinates ⎦ (1.70, 1.70) 3.

(a) Determine the area bounded by the curve y = 5x 2 , the x-axis and the ordinates x = 0 and x = 3. (b) If this area is revolved 360◦ about (i) the x-axis, and (ii) the y-axis, ﬁnd the volumes of the solids of revolution produced in each case. (c) Determine the co-ordinates of the centroid of the area using (i) integral calculus, and (ii) the theorem of Pappus. ⎡ ⎤ (a) 45 square units ⎢(b) (i) 1215π cubic units ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ (ii) 202.5π cubic units⎦ (c) (2.25, 13.5)

4. A metal disc has a radius of 7.0 cm and is of thickness 2.5 cm. A semicircular groove of diameter 2.0 cm is machined centrally around the rim to form a pulley. Determine the volume of metal removed using Pappus’ theorem and express this as a percentage of the original volume of the disc. Find also the mass of metal removed if the density of the metal is 7800 kg m−3. [64.90 cm3 , 16.86%, 506.2 g] For more on areas, mean and r.m.s. values, volumes and centroids, see ‘Engineering Mathematics 6th edition’, Chapters 55 to 58.

38.7 Second moments of area of regular sections The ﬁrst moment of area about a ﬁxed axis of a lamina of area A, perpendicular distance y from the centroid of the lamina is deﬁned as Ay cubic units. The second moment of area of the same lamina as above is given by Ay 2 , i.e. the perpendicular distance from the centroid of the area to the ﬁxed axis is squared.

384 Higher Engineering Mathematics Second moments of areas are usually denoted by I and have units of mm4 , cm4 , and so on.

limit

Radius of gyration

δx→0

Several areas, a1 , a2, a3 , . . . at distances y1 , y2, y3 , . . . from a ﬁxed axis, may be replaced by a single area + a3 + · · · at distance k from the A, where A = a1 + a2 ; axis, such that Ak 2 = ay 2 . k is called the radius of ; gyration of area A about the given axis. Since Ak 2 = ay 2 = I then the radius of gyration, k=

It is a fundamental theorem of integration that

I A

The second moment of area is a quantity much used in the theory of bending of beams, in the torsion of shafts, and in calculations involving water planes and centres of pressure. The procedure to determine the second moment of area of regular sections about a given axis is (i) to ﬁnd the second moment of area of a typical element and (ii) to sum all such second moments of area by integrating between appropriate limits. For example, the second moment of area of the rectangle shown in Fig. 38.14 about axis PP is found by initially considering an elemental strip of width δx, parallel to and distance x from axis PP. Area of shaded strip = bδx.

x=l

4ac) or (ii) two equal real roots (when b 2 = 4ac) or (iii) two complex roots (when b2 < 4ac).

478 Higher Engineering Mathematics Using the above procedure:

50.2

Procedure to solve differential equations of the form d2 y dy a 2 + b + cy = 0 dx dx

(a) Rewrite the differential equation dy d2 y a 2 +b + cy = 0 dx dx as

(aD2 + bD + c)y = 0

(b) Substitute m for D and solve the auxiliary equation am 2 + bm + c = 0 for m. (c) If the roots of the auxiliary equation are: (i) real and different, say m = α and m = β, then the general solution is y = Aeαx + Beβx (ii) real and equal, say m = α twice, then the general solution is y = (Ax + B)eαx (iii) complex, say m = α ± jβ, then the general solution is y = eαx {A cosβx + B sinβx} (d) Given boundary conditions, constants A and B, may be determined and the particular solution of the differential equation obtained. The particular solutions obtained in the worked problems of Section 50.3 may each be veriﬁed by substid2 y dy and 2 into the original tuting expressions for y, dx dx equation.

dy d2 y + 5 − 3y = 0 in D-operator form is dx 2 dx d (2D2 + 5D − 3)y = 0, where D ≡ dx (b) Substituting m for D gives the auxiliary equation (a) 2

2m 2 + 5m − 3 = 0. Factorising gives: (2m − 1)(m + 3) = 0, from which, m = 12 or m = −3. (c) Since the roots are real and different the general 1 solution is y = Ae 2 x + Be−3x . (d) When x = 0, y = 4, hence

4= A+ B

Since

y = Ae 2 x + Be−3x

(1)

1

dy 1 1 x = Ae 2 − 3Be−3x dx 2 dy When x = 0, =9 dx 1 (2) thus 9 = A − 3B 2 Solving the simultaneous equations (1) and (2) gives A = 6 and B = −2. then

Hence the particular solution is y = 6e 2 x − 2e−3x 1

Problem 2. Find the general solution of d2 y dy 9 2 − 24 + 16y = 0 and also the particular dt dt solution given the boundary conditions that when dy t = 0, y = = 3. dt Using the procedure of Section 50.2:

50.3

Worked problems on differential equations of d2 y dy the form a 2 + b + cy = 0 dx dx

Problem 1. Determine the general solution of d2 y dy 2 2 + 5 − 3y = 0. Find also the particular dx dx dy solution given that when x = 0, y = 4 and = 9. dx

d2 y dy − 24 + 16y = 0 in D-operator form is dt 2 dt d 2 (9D − 24D +16)y = 0 where D ≡ dt (b) Substituting m for D gives the auxiliary equation 9m 2 − 24m + 16 =0. (a) 9

Factorizing gives: (3m − 4)(3m − 4) = 0, i.e. m = 43 twice. (c) Since the roots are real and equal, the general 4 solution is y = (At +B)e 3 t .

2

479

Second order differential equations of the form a ddxy2 + b dy dx + cy = 0 (d) When t = 0, y = 3 hence 3 = (0 + B)e0, i.e. B = 3.

then

4

Since y = (At + B)e 3 t 4 dy 4 4t 3 then = (At + B) e + Ae 3 t , by the dt 3 product rule. dy When t = 0, =3 dt 4 thus 3= (0 + B) e0 + Ae0 3

4

y = (−t + 3)e 3 t or y = (3 − t)e 3 t Problem 3. Solve the differential equation d2 y dy + 6 + 13y = 0, given that when x = 0, y = 3 2 dx dx dy and = 7. dx Using the procedure of Section 50.2: (a)

− 3e−3x (A cos 2x + B sin 2x), by the product rule, −3x

=e

dy d2 y + 6 + 13y = 0 in D-operator form is dx 2 dx d (D2 + 6D + 13)y = 0, where D ≡ dx

[(2B − 3 A) cos 2x − (2 A + 3B) sin 2x]

When x = 0,

4 i.e. 3 = B + A from which, A = −1, since 3 B = 3. Hence the particular solution is 4

dy = e−3x (−2 A sin 2x + 2B cos 2x) dx

dy = 7, dx

hence 7 =e0 [(2B − 3 A) cos 0 − (2 A + 3B) sin 0] i.e. 7 =2B − 3 A, from which, B = 8, since A = 3. Hence the particular solution is y = e−3x(3 cos 2x + 8 sin 2x) Since, from Chapter 17, page 165, a cos ωt + b sin ωt = R sin(ωt + α), where a R = (a 2 + b2) and α = tan −1 then b 3 cos2x + 8 sin 2x = (32 + 82 ) sin(2x + tan−1 38 ) √ = 73 sin(2x + 20.56◦ ) √ = 73 sin(2x + 0.359) Thus the particular solution may also be expressed as √ y = 73 e−3x sin(2x + 0.359)

(b) Substituting m for D gives the auxiliary equation m 2 + 6m + 13 =0. Now try the following exercise Using the quadratic formula: −6 ± [(6)2 − 4(1)(13)] m= 2(1) √ −6 ± (−16) = 2 −6 ± j 4 = −3 ± j 2 i.e. m= 2 (c)

Since the roots are complex, the general solution is

Exercise 187 Further problems on differential equations of the form dy d2 y a 2 + b + cy = 0 dx dx In Problems 1 to 3, determine the general solution of the given differential equations. 1. 6

y = e−3x (A cos 2x + B sin 2x) (d) When x = 0, y = 3, hence 3 =e0 (A cos 0 + B sin 0), i.e. A = 3. Since y = e−3x (A cos 2x + B sin 2x)

2. 4

d2 y d y − − 2y = 0 dt 2 dt

d2θ dθ +4 +θ =0 2 dt dt

y = Ae 3 t + Be− 2 t 2

1

1 θ = (At + B)e− 2 t

480 Higher Engineering Mathematics

3.

d2 y dy + 2 + 5y = 0 2 dx dx [y = e−x (A cos 2x + B sin 2x)]

In Problems 4 to 9, ﬁnd the particular solution of the given differential equations for the stated boundary conditions. dy d2 y 4. 6 2 + 5 − 6y = 0; when x = 0, y = 5 and dx dx 2 3 dy = −1. y = 3e 3 x + 2e− 2 x dx dy d2 y 5. 4 2 − 5 + y = 0; when t = 0, y = 1 and dt dt 1 dy = −2. y = 4e 4 t − 3et dt d 6. (9D2 + 30D +25)y = 0, where D ≡ ; when dx dy x = 0, y = 0 and = 2. dx 5 y = 2xe− 3 x 7.

8.

d2 x dx − 6 + 9x = 0; when t = 0, x = 2 and dt 2 dt dx = 0. [x = 2(1 − 3t )e3t ] dt dy d2 y + 6 + 13y = 0; when x = 0, y = 4 and 2 dx dx dy = 0. [y = 2e−3x (2 cos 2x + 3 sin 2x)] dx

d 9. (4D2 + 20D + 125)θ = 0, where D ≡ ; when dt dθ t = 0, θ = 3 and = 2.5. dt [θ = e−2.5t (3 cos 5t + 2 sin 5t )]

where x is the displacement in metres of the body from its equilibrium position after time t seconds. Determine x in terms of t given that at time t = 0, dx x = 2m and = 0. dt d2 x + m 2 x = 0 is a differAn equation of the form dt 2 ential equation representing simple harmonic motion (S.H.M.). Using the procedure of Section 50.2: (a)

d2 x + 100x = 0 in D-operator form is dt 2 (D2 + 100)x = 0.

(b) The auxiliary equation is m 2 + 100 = 0, i.e. √ 2 m = −100 and m = (−100), i.e. m = ± j 10. (c) Since the roots are complex, the general solution is x = e0 (A cos 10t + B sin 10t ), i.e. x =(A cos 10t +B sin10t) metres (d) When t = 0, x = 2, thus 2 = A dx = −10 A sin 10t + 10B cos 10t dt dx When t = 0, =0 dt thus 0 = −10 A sin 0 + 10B cos 0, i.e. B = 0 Hence the particular solution is x = 2 cos 10t metres Problem 5. Given the differential equation d2 V = ω2 V , where ω is a constant, show that its dt 2 solution may be expressed as: V = 7 cosh ωt + 3 sinh ωt given the boundary conditions that when

50.4

Further worked problems on practical differential equations d2 y dy of the form a 2 + b + cy = 0 dx dx

Problem 4. The equation of motion of a body oscillating on the end of a spring is d2 x + 100x = 0, dt 2

t = 0, V = 7 and

dV = 3ω. dt

Using the procedure of Section 50.2: (a)

d2 V d2 V 2 = ω V , i.e. − ω2 V = 0 in D-operator dt 2 dt 2 d form is (D2 − ω2 )v = 0, where D ≡ . dx

(b) The auxiliary equation is m 2 − ω2 = 0, from which, m 2 = ω2 and m = ±ω.

2

Second order differential equations of the form a ddxy2 + b dy dx + cy = 0 (c)

Since the roots are real and different, the general solution is

(d) When t = 0, V = 7 hence 7 = A + B

t = 0,

d2 i R di 1 + + i = 0 in D-operator form is dt 2 L dt LC d R 1 2 i = 0 where D ≡ D + D+ L LC dt

(1)

dV = Aωeωt − Bωe−ωt dt When

Using the procedure of Section 50.2: (a)

V = Aeωt + Be−ωt

(b) The auxiliary equation is m 2 +

dV = 3ω, dt Hence m =

3 = A− B

i.e.

(2)

Hence the particular solution is V = 5eωt + 2e−ωt

and

m=

sinh ωt = 12 (eωt − e−ωt ) cosh ωt =

1 ωt 2 (e

+ e−ωt )

then sinh ωt + cosh ωt = eωt and

cosh ωt − sinh ωt = e−ωt from Chapter 5.

Hence the particular solution may also be written as V = 5(sinh ωt + cosh ωt ) + 2(cosh ωt − sinh ωt ) i.e. V = (5 + 2) cosh ωt + (5 − 2) sinh ωt i.e. V = 7 cosh ωt + 3 sinh ωt Problem 6. The equation d2i R di 1 + + i =0 dt 2 L dt LC represents a current i ﬂowing in an electrical circuit containing resistance R, inductance L and capacitance C connected in series. If R = 200 ohms, L =0.20 henry and C = 20 ×10−6 farads, solve the equation for i given the boundary conditions that di when t = 0, i = 0 and = 100. dt

2

When R = 200, L =0.20 and C = 20 ×10−6, then

From equations (1) and (2), A = 5 and B = 2

Since

R 1 m+ =0 L LC

7

8 2 1 R R 8 9 − 4(1) − ± L L LC

3ω = Aω − Bω,

thus

481

=

(c)

7

8 200 8 4 200 2 9 − − ± 0.20 0.20 (0.20)(20 × 10−6 ) 2 −1000 ± 2

√

= −500

Since the two roots are real and equal (i.e. −500 twice, since for a second order differential equation there must be two solutions), the general solution is i = (At +B)e−500t .

(d) When t = 0, i = 0, hence B = 0 di = (At + B)(−500e−500t ) + (e−500t )(A), dt by the product rule di = 100, thus 100 =−500B + A dt i.e. A = 100, since B = 0 When t = 0,

Hence the particular solution is i = 100te−500t Problem 7. The oscillations of a heavily damped pendulum satisfy the differential equation dx d2 x + 6 + 8x = 0, where x cm is the dt 2 dt displacement of the bob at time t seconds. The initial displacement is equal to +4 cm and the dx is 8 cm/s. Solve the initial velocity i.e. dt equation for x.

482 Higher Engineering Mathematics Using the procedure of Section 50.2:

from

2. A body moves in a straight line so that its distance s metres from the origin after time d2 s t seconds is given by 2 + a2 s = 0, where a dt is a constant. Solve the equation for s given ds 2π that s = c and = 0 when t = . dt a [s = c cos at ]

(c) Since the roots are real and different, the general solution is x =Ae−2t + Be−4t .

3. The motion of the pointer of a galvanometer about its position of equilibrium is represented by the equation

dx d2 x + 6 + 8x = 0 in D-operator form is (a) 2 dt dt d 2 (D + 6D + 8)x = 0, where D ≡ . dt (b) The auxiliary equation is m 2 + 6m + 8 =0. Factorising gives: (m + 2)(m + 4) = 0, which, m = −2 or m = −4.

(d) Initial displacement means that time t = 0. At this instant, x = 4. Thus 4 = A + B

I

If I , the moment of inertia of the pointer about its pivot, is 5 ×10−3, K , the resistance due to friction at unit angular velocity, is 2 × 10−2 and F, the force on the spring necessary to produce unit displacement, is 0.20, solve the equation for θ in terms of t given that when dθ t = 0, θ = 0.3 and = 0. dt [θ = e−2t (0.3 cos 6t + 0.1 sin 6t )]

(1)

Velocity, dx = −2 Ae−2t − 4Be−4t dt dx = 8 cm/s when t = 0, dt thus

8 = −2 A − 4B

(2)

From equations (1) and (2), A = 12 and B = −8 Hence the particular solution is x = 12e−2t − 8e−4t

4. Determine an expression for x for a differential dx d2 x equation 2 + 2n + n 2 x = 0 which repredt dt sents a critically damped oscillator, given that dx at time t = 0, x = s and = u. dt [x = {s + (u + ns)t }e−nt ] 5.

i.e. displacement, x = 4(3e−2t − 2e−4t ) cm

Now try the following exercise Exercise 188 Further problems on second order differential equations of the form dy d2 y a 2 + b + cy = 0 dx dx 1. The charge, q, on a capacitor in a certain electrical circuit satisﬁes the differential equadq d2 q tion 2 + 4 + 5q = 0. Initially (i.e. when dt dt dq t = 0), q = Q and = 0. Show that the dt charge √ in the circuit can be expressed as: q = 5 Qe−2t sin(t + 0.464).

dθ d2θ +K + Fθ = 0. 2 dt dt

di 1 d2i L 2 + R + i = 0 is an equation repredt dt C senting current i in an electric circuit. If inductance L is 0.25 henry, capacitance C is 29.76 ×10−6 farads and R is 250 ohms, solve the equation for i given the boundary di conditions that when t = 0, i = 0 and = 34. dt 1 −160t − e−840t e i= 20

6. The displacement s of a body in a damped mechanical system, with no external forces, satisﬁes the following differential equation: 2

ds d2 s + 6 + 4.5s = 0 2 dt dt

where t represents time. If initially, when ds t = 0, s = 0 and = 4, solve the differential dt 3 equation for s in terms of t . [s = 4t e− 2 t ]

Chapter 51

Second order differential equations of the form d2 y dy a dx 2 + b dx 51.1 Complementary function and particular integral If in the differential equation a

d2 y dy +b + cy = f (x) 2 dx dx

(1)

the substitution y = u + v is made then: a

+ cy = f (x) The general solution, u, of equation (3) will contain two unknown constants, as required for the general solution of equation (1). The method of solution of equation (3) is shown in Chapter 50. The function u is called the complementary function (C.F.). If the particular solution, v, of equation (2) can be determined without containing any unknown constants then y = u +v will give the general solution of equation (1). The function v is called the particular integral (P.I.). Hence the general solution of equation (1) is given by:

d(u + v) d2(u + v) +b + c(u + v) = f (x) dx 2 dx

y = C.F. + P.I.

Rearranging gives: 2 2 du dv d u d v a 2 +b + cu + a 2 + b +cv dx dx dx dx

51.2

= f (x) If we let a

d2 v dx 2

+b

dv + cv = f (x) dx

(i) Rewrite the given differential equation as (aD2 + bD+ c)y = f (x). (2)

then du d2 u a 2 +b + cu = 0 dx dx

Procedure to solve differential equations of the form d2 y dy a 2 + b + cy = f (x) dx dx

(3)

(ii) Substitute m for D, and solve the auxiliary equation am 2 + bm +c = 0 for m. (iii) Obtain the complementary function, u, which is achieved using the same procedure as in Section 50.2(c), page 478.

484 Higher Engineering Mathematics Table 51.1 Form of particular integral for different functions Type

Straightforward cases ‘Snag’ cases Try as particular integral: Try as particular integral:

(a) f (x) = a constant

v=k

(b) f (x) = polynomial (i.e.

v = a + bx + cx 2 + · · ·

f (x) = L + M x + N x 2 +

v = kx (used when C.F. contains a constant)

See problem 1, 2 3

···

where any of the coefﬁcients may be zero) (c) f (x) = an exponential function (i.e. f (x) =

v = keax

(i) v = kxeax (used when eax

Aeax )

4, 5

appears in the C.F.) (ii) v = kx 2 eax (used when eax and xeax both appear in the C.F.)

(d) f (x) = a sine or cosine function v = A sin px + B cos px

v = x(A sin px + B cos px)

(i.e. f (x) = a sin px + b cos px,

(used when sin px and/or

where a or b may be zero)

cos px appears in the C.F.)

6

7, 8

(e) f (x) = a sum e.g. (i)

f (x) = 4x 2 − 3 sin 2x

(ii)

f (x) = 2 − x + e3x

9 (i)

v = ax 2 + bx + c + d sin 2x + e cos 2x

(f ) f (x) = a product e.g. f (x) = 2ex

(ii) v = ax + b + ce3x v = ex (A sin 2x + B cos 2x)

10

cos 2x

(iv) To determine the particular integral, v, ﬁrstly assume a particular integral which is suggested by f (x), but which contains undetermined coefﬁcients. Table 51.1 gives some suggested substitutions for different functions f (x). (v) Substitute the suggested P.I. into the differential equation (aD2 + bD +c)v = f (x) and equate relevant coefﬁcients to ﬁnd the constants introduced. (vi) The general solution is given by y = C.F. + P.I., i.e. y = u +v. (vii) Given boundary conditions, arbitrary constants in the C.F. may be determined and the particular solution of the differential equation obtained.

51.3

Worked problems on differential equations of the d2 y dy form a 2 + b + cy = f (x) dx dx where f (x) is a constant or polynomial

Problem 1. Solve the differential equation d2 y d y + − 2y = 4. dx 2 dx Using the procedure of Section 51.2: (i)

d2 y d y + − 2y = 4 in D-operator form is dx 2 dx (D2 + D − 2)y = 4.

2

Second order differential equations of the form a ddxy2 + b dy dx + cy = f (x) (ii) Substituting m for D gives the auxiliary equation m 2 + m − 2 = 0. Factorizing gives: (m − 1) (m + 2) = 0, from which m = 1 or m = −2. (iii) Since the roots are real and different, the C.F., u = Aex + Be−2x . (iv) Since the term on the right hand side of the given equation is a constant, i.e. f (x) = 4, let the P.I. also be a constant, say v = k (see Table 51.1(a)). (v) Substituting v = k into (D2 + D − 2)v = 4 gives (D2 + D − 2)k = 4. Since D(k) = 0 and D2 (k) = 0 then −2k = 4, from which, k = −2. Hence the P.I., v = −2. (vi) The general solution is given by y = u + v, i.e. y = Aex + Be−2x − 2. Problem 2. Determine the particular solution of d2 y dy the equation 2 − 3 = 9, given the boundary dx dx dy conditions that when x = 0, y = 0 and = 0. dx

d2 y dy − 3 =9 2 dx dx (D2 − 3D)y = 9.

in

D-operator

Hence the particular solution is y = −1 + 1e3x − 3x, i.e. y = e3x − 3x − 1

Problem 3. Solve the differential equation d2 y dy 2 2 − 11 + 12y = 3x − 2. dx dx Using the procedure of Section 51.2: dy d2 y (i) 2 2 − 11 + 12y = 3x − 2 dx dx form is

in

D-operator

(2D2 − 11D + 12)y = 3x − 2. (ii) Substituting m for D gives the auxiliary equation 2m 2 − 11m + 12 =0. Factorizing gives: (2m − 3)(m − 4) = 0, from which, m = 32 or m = 4. (iii) Since the roots are real and different, the C.F., 3

u =Ae 2 x + Be4x (iv) Since f (x) = 3x − 2 is a polynomial, let the P.I., v = ax + b (see Table 51.1(b)).

Using the procedure of Section 51.2: (i)

485

form

is

(ii) Substituting m for D gives the auxiliary equation m 2 − 3m =0. Factorizing gives: m(m − 3) = 0, from which, m = 0 or m = 3.

(v) Substituting v = ax + b into (2D2 − 11D +12)v = 3x − 2 gives: (2D2 − 11D + 12)(ax + b) = 3x − 2, i.e. 2D2 (ax + b) − 11D(ax + b) + 12(ax + b) = 3x − 2

(iii) Since the roots are real and different, the C.F., u = Ae0 + Be3x , i.e. u = A +Be3x .

i.e.

(iv) Since the C.F. contains a constant (i.e. A) then let the P.I., v = kx (see Table 51.1(a)).

Equating the coefﬁcients of x gives: 12a = 3, from which, a = 14 .

(v) Substituting v = kx into (D2 − 3D)v = 9 gives (D2 − 3D)kx = 9. D(kx) = k and D2 (kx) = 0. Hence (D2 − 3D)kx = 0 −3k = 9, from which, k = −3. Hence the P.I., v = −3x. (vi) The general solution is given by y = u + v, i.e. y = A +Be3x −3x. (vii) When x = 0, y = 0, thus 0 = A + Be0 − 0, i.e. 0= A+ B (1) dy d y = 3Be3x − 3; = 0 when x = 0, thus dx dx 0 = 3Be0 − 3 from which, B = 1. From equation (1), A = −1.

0 − 11a + 12ax + 12b = 3x − 2

Equating the constant terms gives: −11a + 12b = −2. i.e. −11 14 + 12b = −2 from which, 1 11 3 = i.e. b = 4 4 16 1 1 Hence the P.I., v = ax + b = x + 4 16 (vi) The general solution is given by y = u + v, i.e. 12b = −2 +

3 1 1 y = Ae 2 x + Be4x + x + 4 16

486 Higher Engineering Mathematics Now try the following exercise Exercise 189 Further problems on differential equations of the form dy d2 y a 2 +b + cy = f (x) where f (x) is a dx dx constant or polynomial. In Problems 1 and 2, ﬁnd the general solutions of the given differential equations. 1. 2

2. 6

6. In a galvanometer the deﬂection θ satisﬁes d2θ dθ the differential equation 2 + 4 + 4 θ = 8. dt dt Solve the equation for θ given that when t = 0, dθ θ= = 2. [θ = 2(t e−2t + 1)] dt

51.4

dy d2 y + 5 − 3y = 6 dx 2 dx 1 y = Ae 2 x + Be−3x − 2 d2 y dy + 4 − 2y = 3x − 2 dx 2 dx 1 y = Ae 3 x + Be−x − 2 − 32 x

In Problems 3 and 4 ﬁnd the particular solutions of the given differential equations. d2 y d y 3. 3 2 + − 4y = 8; when x = 0, y = 0 and dx dx dy = 0. dx 4 −3 x 2 x y = 7 (3e + 4e ) − 2 d2 y dy − 12 + 4y = 3x − 1; when x = 0, 2 dx dx dy 4 y = 0 and =− dx 3 2 y = − 2 + 34 x e 3 x + 2 + 34 x

4. 9

5. The charge q in an electric circuit at time t satd2 q dq 1 isﬁes the equation L 2 + R + q = E, dt dt C where L, R, C and E are constants. Solve the equation given L = 2H , C = 200 ×10−6 F and E = 250 V, when (a) R = 200 and (b) R is negligible. Assume that when t = 0, q = 0 and dq =0 dt ⎡ ⎤ 5 1 1 −50t e − t + (a) q = ⎢ ⎥ 20 2 20 ⎢ ⎥ ⎣ ⎦ 1 (b) q = (1 − cos 50t ) 20

Worked problems on differential equations of the form d2 y dy a 2 +b + cy = f (x) where dx dx f (x) is an exponential function

Problem 4. Solve the equation d2 y dy − 2 + y = 3e4x given the boundary 2 dx dx dy conditions that when x = 0, y = − 23 and = 4 13 dx Using the procedure of Section 51.2: (i)

d2 y dy − 2 + y = 3e4x in D-operator form is dx 2 dx (D2 − 2D + 1)y = 3e4x .

(ii) Substituting m for D gives the auxiliary equation m 2 − 2m + 1 =0. Factorizing gives: (m − 1)(m − 1) = 0, from which, m = 1 twice. (iii) Since the roots are real and equal the C.F., u = (Ax + B)ex . (iv) Let the particular integral, v = ke4x Table 51.1(c)).

(see

(v) Substituting v = ke4x into (D2 − 2D + 1)v = 3e4x gives: (D2 − 2D + 1)ke4x = 3e4x i.e. D2 (ke4x ) − 2D(ke4x ) + 1(ke4x ) = 3e4x i.e.

16ke4x − 8ke4x + ke4x = 3e4x

Hence 9ke4x = 3e4x , from which, k = 13 Hence the P.I., v = ke4x = 13 e4x . (vi) The general solution is given by y = u + v, i.e. y = (Ax + B)ex + 13 e4x . (vii) When x = 0, y = − 23 thus

2

487

Second order differential equations of the form a ddxy2 + b dy dx + cy = f (x) 3 3 = 2 ke 2 x 94 x + 3 − ke 2 x 32 x + 1

− 23 = (0 + B)e0 + 13 e0 , from which, B = −1. dy = (Ax + B)ex + ex (A) + 43 e4x . dx dy 1 13 4 When x = 0, = 4 , thus = B + A+ dx 3 3 3 from which, A = 4, since B = −1. Hence the particular solution is:

3 3 x 2 − 3 kxe = 5e 2 x i.e.

3

(v) The

3

the C.F., u =Ae 2 x + Be−x . 3

(see Table 51.1(c), snag case (i)). 3

(iv) Substituting v = kxe 2 x into (2D2 − D − 3)v = 3 3 (2D2 − D − 3)kxe 2 x = 5e 2 x .

gives: 3 3 3 x x 3 2x 2 2 D kxe = (kx) 2 e + e (k),

+1

3

9

4x

d2 y dy − 4 + 4y = 3e2x . dx 2 dx

Using the procedure of Section 51.2: dy d2 y − 4 + 4y = 3e2x in D-operator form is dx 2 dx (D2 − 4D +4)y = 3e2x .

(ii) Substituting m for D gives the auxiliary equation m 2 − 4m + 4 = 0. Factorizing gives: (m − 2)(m − 2) = 0, from which, m = 2 twice. (iii) Since the roots are real and equal, the C.F., u =(Ax + B)e2x .

(v) Substituting v = kx 2 e2x into (D2 − 4D + 4)v = 3e2x gives: (D2 − 4D + 4)(kx 2 e2x ) = 3e2x

D2 (kx 2 e2x ) = D[2ke2x (x 2 + x)] = (2ke2x )(2x + 1) + (x 2 + x)(4ke2x )

3

+3

i.e.

= 2ke2x (x 2 + x)

3 x 3 2 = ke 2

= ke 2 x

y = u + v,

D(kx 2 e2x ) = (kx 2 )(2e2x ) + (e2x )(2kx)

3 3 x x 2 3 D kxe 2 = D ke 2 2 x + 1

+

is

(iv) Since e2x and xe2x both appear in the C.F. let the P.I., v = kx 2 e2x (see Table 51.1(c), snag case (ii)).

by the product rule, =

solution

Problem 6. Solve

(i)

(iii) Since e 2 x appears in the C.F. and in the right hand side of the differential equation, let the

general

3 3 y = Ae 2 x + Be−x + xe 2 x .

3 − 3y = 5e 2 x

3 ke 2 x 32 x

3

Hence the P.I., v = kxe 2 x = xe 2 x .

(ii) Substituting m for D gives the auxiliary equation 2m 2 − m − 3 = 0. Factorizing gives: (2m − 3)(m + 1) = 0, from which, m = 32 or m = −1. Since the roots are real and different then

3 5e 2 x

3

3

dy − in D-operator form is dx 2 dx 3 (2D2 − D − 3)y = 5e 2 x .

P.I.,

3

Equating coefﬁcients of e 2 x gives: 5k = 5, from which, k = 1.

Using the procedure of Section 51.2:

3 v = kxe 2 x

3

− 3kxe 2 x = 5e 2 x

Problem 5. Solve the differential equation 3 d2 y d y 2 2− − 3y = 5e 2 x . dx dx

(i) 2

3

+ 6ke 2 x − 32 xke 2 x − ke 2 x 3

y = (4x − 1)ex + 13 e4x

d2 y

3 9 2x 2 kxe

2x +1

3 Hence (2D2 − D − 3) kxe 2 x

3 3 2x 2 ke

= 2ke2x (4x + 1 + 2x 2 ) Hence (D2 − 4D + 4)(kx 2 e2x ) = [2ke2x (4x + 1 + 2x 2 )] − 4[2ke2x (x 2 + x)] + 4[kx 2 e2x ] = 3e2x

488 Higher Engineering Mathematics from which, 2ke2x = 3e2x and k = 32 Hence the P.I., v = kx2 e2x = 32 x2 e2x .

51.5

(vi) The general solution, y = u + v, i.e. y = (Ax + B)e2x + 23 x2e2x Now try the following exercise Exercise 190 Further problems on differential equations of the form dy d2 y a 2 + b +cy = f (x) where f (x) is an dx dx exponential function

Worked problems on differential equations of the d2 y dy form a 2 + b + cy= f (x) dx dx where f (x) is a sine or cosine function

Problem 7. Solve the differential equation d2 y dy 2 2 + 3 − 5y = 6 sin 2x. dx dx Using the procedure of Section 51.2:

In Problems 1 to 4, ﬁnd the general solutions of the given differential equations. 1.

2.

3.

d2 y d y − − 6y = 2ex dx 2 dx

y = Ae3x + Be−2x − 13 ex dy d2 y − 3 − 4y = 3e−x 2 dx dx

y = Ae4x + Be−x − 35 xe−x d2 y + 9y = 26e2x dx 2 [ y = A cos 3x + B sin 3x + 2e2x ]

4. 9

t dy d2 y 3 − 6 + y = 12e dt 2 dt 1 1 t 2 2 3t 3 y = (At + B)e + 3 t e

In problems 5 and 6 ﬁnd the particular solutions of the given differential equations. dy 1 d2 y + 9 − 2y = 3ex ; when x = 0, y = 2 dx dx 4 dy and = 0. dx 1 1 x 5 x −2x 5 e −e + e y= 44 4

5. 5

6.

dy d2 y − 6 + 9y = 4e3t ; when t = 0, y = 2 dt 2 dt dy and =0 [ y = 2e3t (1 − 3t + t 2)] dt

d2 y dy + 3 − 5y = 6 sin 2x in D-operator form dx 2 dx is (2D2 + 3D − 5)y = 6 sin 2x

(i) 2

(ii) The auxiliary equation is 2m 2 + 3m −5 = 0, from which, (m − 1)(2m + 5) = 0, i.e. m = 1 or m = −52 (iii) Since the roots are real and different the C.F., 5 u = Aex + Be− 2 x. (iv) Let the P.I., Table 51.1(d)).

v = A sin 2x + B cos 2x

(see

(v) Substituting v = A sin 2x + B cos 2x into (2D2 + 3D −5)v = 6 sin 2x gives: (2D2 + 3D−5)(A sin 2x + B cos 2x) = 6 sin 2x. D(A sin 2x + B cos 2x) = 2 A cos 2x − 2B sin 2x D2 (A sin 2x + B cos 2x) = D(2 A cos 2x − 2B sin 2x) = −4 A sin 2x − 4B cos 2x Hence (2D2 + 3D −5)(A sin 2x + B cos 2x) = −8 A sin 2x − 8B cos 2x + 6 A cos 2x − 6B sin 2x − 5 A sin 2x − 5B cos 2x = 6 sin 2x Equating coefﬁcient of sin 2x gives: −13 A − 6B = 6

(1)

2

Second order differential equations of the form a ddxy2 + b dy dx + cy = f (x) D[x(C sin 4x + D cos 4x)]

Equating coefﬁcients of cos 2x gives: 6 A − 13B = 0

(2)

6 × (1)gives : −78 A − 36B = 36 13 × (2)gives :

by the product rule D2 [x(C sin 4x + D cos 4x)]

−36 into equation (1) or (2) Substituting B = 205 −78 gives A = 205 −78 36 Hence the P.I., v = sin 2x − cos 2x. 205 205 (vi) The general solution, y = u +v, i.e. x

+ (C sin 4x + D cos 4x)(1),

(4)

− 205B = 36 −36 B= 205

from which,

= x(4C cos 4x − 4D sin 4x)

(3)

78 A − 169B = 0

(3) + (4)gives :

489

− 52 x

y = Ae + Be 2 − (39 sin 2x + 18 cos 2x) 205

= x(−16C sin 4x − 16D cos 4x) + (4C cos 4x − 4D sin 4x) + (4C cos 4x − 4D sin 4x) Hence (D2 + 16)[x(C sin 4x + D cos 4x)] = −16Cx sin 4x −16Dx cos 4x + 4C cos 4x − 4D sin 4x + 4C cos 4x − 4D sin 4x + 16Cx sin 4x + 16Dx cos 4x = 10 cos4x,

d2 y Problem 8. Solve 2 + 16y = 10 cos4x given dx dy y = 3 and = 4 when x = 0. dx

i.e. −8D sin 4x + 8C cos 4x = 10 cos4x Equating coefﬁcients of cos 4x gives: 10 5 8C = 10, from which, C = = 8 4

Using the procedure of Section 51.2: (i)

Equating coefﬁcients of sin 4x gives: −8D = 0, from which, D = 0. Hence the P.I., v = x 45 sin 4x .

d2 y + 16y = 10 cos 4x in D-operator form is dx 2 (D2 + 16)y = 10 cos4x

(ii) The auxiliary √ equation is which m = −16 = ± j 4.

m 2 + 16 = 0,

(vi) The general solution, y = u +v, i.e. from

(vii) When x = 0, y = 3, thus 3 = A cos 0 + B sin 0 + 0, i.e. A = 3.

(iii) Since the roots are complex the C.F., u =e0 (A cos 4x + B sin 4x) i.e. u =Acos 4x + B sin4x (iv) Since sin 4x occurs in the C.F. and in the right hand side of the given differential equation, let the P.I., v = x(C sin 4x + D cos 4x) (see Table 51.1(d), snag case—constants C and D are used since A and B have already been used in the C.F.). (v) Substituting v = x(C sin 4x + D cos 4x) (D2 + 16)v = 10 cos 4x gives: (D2 + 16)[x(C sin 4x + D cos 4x)] = 10 cos 4x

y = A cos 4x + B sin 4x + 45 x sin 4x

into

dy = −4 A sin 4x + 4B cos 4x dx + 54 x(4 cos 4x) + 54 sin 4x When x = 0,

dy = 4, thus dx

4 = −4 A sin 0 + 4B cos 0 + 0 + 54 sin 0 i.e. 4 =4B, from which, B = 1 Hence the particular solution is y = 3 cos 4x + sin 4x + 54 x sin 4x

490 Higher Engineering Mathematics Now try the following exercise given by: Exercise 191 Further problems on differential equations of the form dy d2 y a 2 + b + cy = f (x) where f (x) is a sine or dx dx cosine function

y = e−4t (A cos 2t + B sin 2t ) 15 + (sin 4t − 8 cos4t ) 13 7.

In Problems 1 to 3, ﬁnd the general solutions of the given differential equations. 1. 2

2.

3.

d2 y d y − − 3y = 25 sin 2x dx 2 dx

3 y = Ae 2 x + Be−x − 15 (11 sin 2x − 2 cos 2x)

dq 1 d2q L 2 + R + q = V0 sin ωt represents the dt dt C variation of capacitor charge in an electric circuit. Determine an expression for q at time t seconds given that R = 40 , L =0.02 H, C = 50 × 10−6 F, V0 = 540.8 V and ω = 200 rad/s and given the boundary dq conditions that when t = 0, q = 0 and = 4.8 dt

q = (10t + 0.01)e−1000t + 0.024 sin 200t − 0.010 cos 200t

d2 y dy − 4 + 4y = 5 cos x dx 2 dx

y = (Ax + B)e2x − 45 sin x + 35 cos x d2 y + y = 4 cos x dx 2

51.6

[ y = A cos x + B sin x + 2x sin x] 4. Find the particular solution of the differend2 y dy tial equation 2 − 3 − 4y = 3 sin x; when dx dx dy x = 0, y = 0 and = 0. dx ⎤ ⎡ 1 4x − 51e−x ) (6e y = ⎥ ⎢ 170 ⎥ ⎢ ⎦ ⎣ 1 − (15 sin x − 9 cos x) 34 5. A differential equation representing the d2 y + n 2 y = k sin pt , motion of a body is dt 2 where k, n and p are constants. Solve the equation (given n = 0 and p2 = n 2) given that when dy t = 0, y = = 0. dt k p y= 2 sin nt sin pt − n − p2 n 6. The motion of a vibrating mass is given by d2 y dy + 8 + 20y = 300 sin4t . Show that the 2 dt dt general solution of the differential equation is

Worked problems on differential equations of the d2 y dy form a 2 + b + cy = f (x) dx dx where f (x) is a sum or a product

Problem 9. Solve d2 y d y + − 6y = 12x − 50 sin x. dx 2 dx Using the procedure of Section 51.2: (i)

d2 y d y + − 6y = 12x − 50 sin x in D-operator dx 2 dx form is (D2 + D − 6)y = 12x − 50 sin x

(ii) The auxiliary equation is (m 2 + m − 6) = 0, from which, (m − 2)(m + 3) = 0, i.e. m = 2 or m = −3 (iii) Since the roots are real and different, the C.F., u = Ae2x + Be−3x . (iv) Since the right hand side of the given differential equation is the sum of a polynomial and a sine function let the P.I. v = ax + b + c sin x + d cos x (see Table 51.1(e)).

2

Second order differential equations of the form a ddxy2 + b dy dx + cy = f (x) (v) Substituting v into

Using the procedure of Section 51.2:

(D2 + D −6)v = 12x − 50 sin x gives:

(i)

(D + D − 6)(ax + b + c sin x + d cos x) 2

= 12x − 50 sin x

D2 (ax + b + c sin x + d cos x) = −c sin x − d cos x (D2 + D − 6)(v)

= (−c sin x − d cos x) + (a + c cos x − d sin x) − 6(ax + b + c sin x + d cos x) = 12x − 50 sin x Equating constant terms gives: a − 6b = 0

(1)

Equating coefﬁcients of x gives: −6a = 12, from which, a = −2. Hence, from (1), b = − 13

D-operator

(iii) Since the roots are complex, the C.F., u = ex (A cos x + B sin x). (iv) Since the right hand side of the given differential equation is a product of an exponential and a cosine function, let the P.I., v = ex (C sin 2x + D cos 2x) (see Table 51.1(f) — again, constants C and D are used since A and B have already been used for the C.F.). (v) Substituting v into (D2 − 2D +2)v = 3ex cos 2x gives: (D2 − 2D + 2)[ex (C sin 2x + D cos 2x)]

Equating the coefﬁcients of cos x gives:

= 3ex cos 2x

−d + c − 6d = 0

(2)

c − 7d = 0

−c − d − 6c = −50

(≡ex {(2C + D) cos 2x

Solving equations (2) and (3) gives: c = 7 and d = 1. Hence the P.I., υ = −2x − 13 + 7 sin x + cos x (vi) The general solution, y = u +v, i.e. y = Ae + Be

+ (C − 2D) sin 2x})

(3)

i.e. − 7c − d = −50

−3x

D(v) = ex (2C cos 2x − 2D sin 2x) + ex (C sin 2x + D cos 2x)

Equating the coefﬁcients of sin x gives:

2x

in

(ii) The auxiliary equation is m 2 − 2m + 2 = 0 Using the quadratic formula, √ 2 ± [4 − 4(1)(2)] m= 2 √ 2 ± −4 2 ± j 2 = = i.e. m = 1 ± j 1. 2 2

= a + c cos x − d sin x

i.e.

dy d2 y − 2 + 2y = 3ex cos 2x dx 2 dx form is

(D2 − 2D + 2)y = 3ex cos 2x

D(ax + b + c sin x + d cos x)

Hence

491

− 2x

− 13 + 7 sin x + cos x Problem 10. Solve the differential equation d2 y dy − 2 + 2y = 3ex cos 2x, given that when 2 dx dx dy x = 0, y = 2 and = 3. dx

D2 (v) = ex (−4C sin 2x − 4D cos 2x) + ex (2C cos 2x − 2D sin 2x) + ex (2C cos 2x − 2D sin 2x) + ex (C sin 2x + D cos 2x) ≡ ex {(−3C − 4D) sin 2x + (4C − 3D) cos 2x} Hence (D2 − 2D + 2)v = ex {(−3C − 4D) sin 2x + (4C − 3D) cos 2x} − 2ex {(2C + D) cos 2x + (C − 2D) sin 2x} + 2ex (C sin 2x + D cos 2x) = 3ex cos 2x

492 Higher Engineering Mathematics Equating coefﬁcients of ex sin 2x gives: −3C − 4D − 2C + 4D + 2C = 0

1. 8

i.e. −3C = 0, from which, C = 0. Equating coefﬁcients of ex cos 2x gives: 4C − 3D − 4C − 2D + 2D = 3 i.e. −3D = 3, from which, D = −1. Hence the P.I., υ = ex (−cos 2x). (vi) The general solution, y = u + v, i.e.

2.

y = ex (A cos x +B sinx) − ex cos 2x (vii) When x = 0, y = 2 thus

d2 y dy − 6 + y = 2x + 40 sin x dx 2 dx ⎤ ⎡ x x y = Ae 4 + Be 2 + 2x + 12 ⎦ ⎣ 8 + (6 cos x − 7 sin x) 17

d2 y dy − 3 + 2y = 2 sin 2 θ − 4 cos 2 θ dθ 2 dθ

y = Ae2θ + Beθ + 12 (sin 2 θ + cos 2 θ)

2 = e0 (A cos 0 + B sin 0) − e0 cos 0 i.e.

When thus

2 = A − 1, from which, A = 3 dy = ex (− A sin x + B cos x) dx + ex (A cos x + B sin x) − [ex (−2 sin 2x) + ex cos 2x] dy x = 0, =3 dx 0 3 = e (− A sin 0 + B cos 0) + e0 (A cos 0 + B sin 0) − e0 (−2 sin 0) − e0 cos 0

i.e.

3.

− 12 x − 12 x 2 + 14 e2x

4.

Hence the particular solution is y = ex (3 cos x + sin x) − ex cos 2x Now try the following exercise Exercise 192 Further problems on second order differential equations of the form dy d2 y a 2 +b + cy = f (x) where f (x) is a sum or dx dx product In Problems 1 to 4, ﬁnd the general solutions of the given differential equations.

d2 y dy − 2 + 2y = et sin t 2 dt dt

y = et (A cos t + B sin t ) − 2t et cos t

In Problems 5 to 6 ﬁnd the particular solutions of the given differential equations.

3 = B + A − 1, from which, B = 1, since A = 3

d2 y d y + − 2y = x 2 + e2x dx 2 dx

y = Aex + Be−2x − 34

5.

d2 y dy − 7 + 10y = e2x + 20; when x = 0, 2 dx dx dy 1 y = 0 and =− dx 3 4 5x 10 2x 1 2x y = e − e − xe + 2 3 3 3 d2 y d y − − 6y = 6ex cos x; when x = 0, dx 2 dx 21 dy 20 y = − and = −6 29 dx 29 ⎡ ⎤ 3 y = 2e− 2 x − 2e2x ⎣ ⎦ 3ex + (3 sin x − 7 cos x) 29

6. 2

Chapter 52

Power series methods of solving ordinary differential equations 52.1

Introduction

Second order ordinary differential equations that cannot be solved by analytical methods (as shown in Chapters 50 and 51), i.e. those involving variable coefﬁcients, can often be solved in the form of an inﬁnite series of powers of the variable. This chapter looks at some of the methods that make this possible—by the Leibniz– Maclaurin and Frobinius methods, involving Bessel’s and Legendre’s equations, Bessel and gamma functions and Legendre’s polynomials. Before introducing Leibniz’s theorem, some trends with higher differential coefﬁcients are considered. To better understand this chapter it is necessary to be able to: (i) differentiate standard functions (as explained in Chapters 27 and 32), (ii) appreciate the binomial theorem (as explained in Chapters 7), and (iii) use Maclaurins theorem (as explained in Chapter 8).

52.2 Higher order differential coefﬁcients as series The following is an extension of successive differentiation (see page 296), but looking for trends, or series,

as the differential coefﬁcient of common functions rises. dy d2 y = a 2 eax , and so = aeax , (i) If y = eax , then 2 dx dx on. If we abbreviate

dy d2 y as y

, … and as y , dx dx 2

dn y as y (n) , then y = aeax , y

= a 2eax , and the dx n emerging pattern gives:

y(n) = an eax

For example, if y = 3e2x , then d7 y = y (7) = 3(27 ) e2x = 384e2x dx 7 (ii) If y = sin ax, π y = a cos ax = a sin ax + 2 y

= −a 2 sin ax = a 2 sin(ax + π) 2π 2 = a sin ax + 2 y

= −a 3 cos x 3π and so on. = a 3 sin ax + 2

(1)

494 Higher Engineering Mathematics In general, y(n) = an sin ax + nπ 2

(2)

For example, if d5 y y = sin 3x, then 5 = y (5) dx 5π π 5 = 3 sin 3x + = 35 sin 3x + 2 2

y(n) =

y = sinh 2x, then

(iii) If y = cos ax, π y = −a sin ax = a cos ax + 2 2π y

= −a 2 cos ax = a 2 cos ax + 2 3π

3 3 and so on. y = a sin ax = a cos ax + 2 nπ y(n) = an cos ax + 2

(3)

(5)

(iv) If y = x a, y = a x a−1 , y

= a(a − 1)x a−2 , y

= a(a − 1)(a − 2)x a−3 , and y(n) = a(a − 1)(a − 2) . . . . . (a − n + 1) x a−n

(v) If y = sinh ax, y = a cosh ax

y = a sinh ax 2

y

= a 3 cosh ax, and so on

25 {[0] sinh 2x + [2] cosh 2x} 2 = 32 cosh 2x

=

(vi) If y = cosh ax,

Since cosh ax is not periodic (see graph on page 43), again it is more difﬁcult to ﬁnd a general statement for y (n) . However, this is achieved with the following general series:

= −256 cos 2x

d4 y For example, if y = 2x6 , then 4 = y (4) dx 6! = (2) x 6−4 (6 − 4)! 6 × 5 × 4× 3 × 2× 1 2 = (2) x 2×1 = 720x2

+ [1 − (−1)5 ] cosh 2x}

y

= a 3 sinh ax, and so on

= 4(26 ) cos (2x + π)

where a is a positive integer.

25 {[1 + (−1)5 ] sinh 2x 2

y

= a 2 cosh ax

6π (6) 6 then 6 = y = 4(2 ) cos 2x + dx 2 6 = 4(2 ) cos (2x + 3π) d6 y

a! xa−n (a − n)!

=

d5 y = y (5) dx 5

y = a sinh ax

For example, if y = 4 cos 2x,

or y(n) =

an {[1 +(−1)n ] sinh ax 2 + [1 −(−1)n ] cosh ax}

For example, if

= 243 cos 3x

In general,

Since sinh ax is not periodic (see graph on page 43), it is more difﬁcult to ﬁnd a general statement for y (n) . However, this is achieved with the following general series:

(4)

y(n) =

an {[1 − (−1)n ] sinh ax 2 + [1 + (−1)n ] cosh ax}

(6)

1 For example, if y = cosh 3x, 9 7 1 3 d7 y (7) (2 sinh 3x) then 7 = y = dx 9 2 = 243 sinh 3x 1 1 2 (vii) If y = ln ax, y = , y

= − 2 , y

= 3 , and so x x x on. In general, y(n) = (−1)n−1

(n − 1)! xn

For example, if y = ln 5x, then d6 y (6) = (−1)6−1 5! = − 120 = y dx 6 x6 x6

(7)

Power series methods of solving ordinary differential equations 1 Note that if y = ln x, y = ; if in equation (7), x

0 (0)! n = 1 then y = (−1) 1 x 1 (−1)0 = 1 and if y = then (0)!= 1 (Check that x (−1)0 = 1 and (0)! = 1 on a calculator).

52.3

Leibniz’s theorem y = uv

If

495

(8)

where u and v are each functions of x, then by using the product rule, y = uv + vu

(9)

y

= uv

+ v u + vu

+ u v

Now try the following exercise

= u

v + 2u v + uv

(10)

(11)

y (4) = u (4)v + 4u (3)v (1) + 6u (2)v (2)

1. (a) y (4) when y = e2x (b) y (5) when y

t = 8e2

1 t (b) e 2 ] 4

+ 4u (1)v (3) + uv (4)

[(a) 81 sin3t (b) −1562.5 cos5θ] 3. (a) y (8) when y = cos 2x 2 (b) y (9) when y = 3 cos t 3 2 29 (a) 256 cos2x (b) − 8 sin t 3 3 t7 8 (b) 630 t ]

4. (a) y (7) when y = 2x 9 (b) y (6) when y =

1 5. (a) y (7) when y = sinh 2x 4 (b) y (6) when y = 2 sinh 3x [(a) 32 cosh 2x (b) 1458 sinh 3x] 6. (a) y (7) when y = cosh 2x 1 (b) y (8) when y = cosh 3x 9 [(a) 128 sinh 2x (b) 729 cosh 3x]

(12)

From equations (8) to (12) it is seen that (a)

2. (a) y (4) when y = sin 3t 1 (b) y (7) when y = sin 5θ 50

7. (a) y (4) when y = 2ln 3θ 1 (b) y (7) when y = ln 2t 3

= u

v + 3u

v + 3u v

+ uv

Determine the following derivatives:

[(a) (9! )x 2

y = u v + vu + 2u v + 2v u + uv + v u

Exercise 193 Further problems on higher order differential coefﬁcients as series

[(a) 16 e2x

the n’th derivative of u decreases by 1 moving from left to right,

(b) the n’th derivative of v increases by 1 moving from left to right, (c)

the coefﬁcients 1, 4, 6, 4, 1 are the normal binomial coefﬁcients (see page 58).

In fact, (uv)(n) may be obtained by expanding (u + v)(n) using the binomial theorem (see page 59), where the ‘powers’ are interpreted as derivatives. Thus, expanding (u + v)(n) gives: y(n) = (uv)(n) = u(n) v + nu(n−1) v (1) n(n− 1) (n−2) (2) v u 2! n(n− 1)(n −2) (n−3) (3) + v +··· u 3! +

(13)

Equation (13) is a statement of Leibniz’s theorem, which can be used to differentiate a product n times. The theorem is demonstrated in the following worked problems. Problem 1. Determine y (n) when y = x 2 e3x . For a product y = uv, the function taken as

(a) −

240 6 (b) 7 θ4 t

(i) u is the one whose nth derivative can readily be determined (from equations (1) to (7)), (ii) v is the one whose derivative reduces to zero after a few stages of differentiation.

496 Higher Engineering Mathematics Thus, when y = x 2 e3x , v = x 2 , since its third derivative is zero, and u = e3x since the nth derivative is known from equation (1), i.e. 3n eax Using Leinbiz’s theorem (equation (13), y

(n)

=u

(n)

n(n − 1) (n−2) (2) v + nu v + v u 2! n(n − 1)(n − 2) (n−3) (3) + v + ··· u 3!

By Leibniz’s equation, equation (13), n(n − 1) (n) y (2)+ 0 y (n+2)(1 + x 2 ) + n y (n+1)(2x)+ 2! + 2{y (n+1) (x) + n y (n) (1) + 0} − 3{y (n) } = 0

(n−1) (1)

i.e. (1 + x 2 )y (n+2) + 2n x y (n+1) + n(n − 1)y (n) + 2x y (n+1) + 2 ny (n) − 3y (n) = 0 (1 + x 2 )y (n+2) + 2(n + 1)x y (n+1)

where in this case v = x 2 , v (1) = 2x, v (2) = 2 and v (3) = 0

or

Hence, y (n) = (3n e3x )(x 2 ) + n(3n−1 e3x )(2x)

i.e. (1 + x2 )y(n+2) + 2(n + 1)xy(n+1)

n(n − 1) n−2 3x (3 e )(2) 2! n(n − 1)(n − 2) n−3 3x (3 e )(0) + 3! = 3n−2 e3x (32 x 2 + n(3)(2x)

+ (n 2 − n + 2n − 3)y (n) = 0

+ (n2 + n − 3)y(n) = 0

+

Problem 4.

+ n(n − 1) + 0) i.e.

y(n) = e3x 3n−2 (9x2 + 6nx + n(n− 1))

Problem 2. If x 2 y

+ 2x y + y = 0 show that: x y (n+2) + 2(n + 1)x y (n+1) + (n 2 + n + 1)y (n) = 0 Differentiating each term of x 2 y

+ 2x y + y = 0 n times, using Leibniz’s theorem of equation (13), gives:

y (n+2) x 2 + n y (n+1) (2x) +

n(n − 1) (n) y (2) + 0 2!

+ {y (n+1) (2x) + n y (n) (2) + 0} + {y (n) } = 0 i.e. x 2 y (n+2) + 2n x y (n+1) + n(n − 1)y (n) + 2x y (n+1) + 2n y (n) + y (n) = 0 i.e. x 2 y (n+2) + 2(n + 1)x y (n+1) + (n 2 − n + 2n + 1)y (n) = 0 or

Find the 5th derivative of y = x 4 sin x.

If y = x 4 sin x, then using Leibniz’s equation with u = sin x and v = x 4 gives: nπ 4 y (n) = sin x + x 2 (n − 1)π 3 + n sin x + 4x 2 (n − 2)π n(n − 1) 2 sin x + 12x + 2! 2 n(n − 1)(n − 2) (n − 3)π + sin x + 24x 3! 2 n(n − 1)(n − 2)(n − 3) sin x + 4! (n − 4)π + 24 2 5π + 20x 3 sin(x + 2π) and y (5) = x 4 sin x + 2 (5)(4) 3π 2 + (12x ) sin x + 2 2

x2 y(n+2) + 2(n + 1) x y(n+1) + (n + n + 1)y 2

(n)

+

=0

π (5)(4)(3)(2) (24) sin x + (4)(3)(2) 2 5π π sin x + ≡ sin x + ≡ cos x, 2 2 +

Problem 3. Differentiate the following differential equation n times: (1 + x 2 )y

+ 2x y − 3y = 0.

Since

(5)(4)(3) (24x) sin (x + π) (3)(2)

497

Power series methods of solving ordinary differential equations 3π sin(x + 2π) ≡ sin x, sin x + ≡ −cos x, 2 and

52.4 Power series solution by the Leibniz–Maclaurin method

sin (x + π) ≡ −sin x,

then y (5) = x 4 cos x + 20x 3 sin x + 120x 2 (−cos x) + 240x(−sin x) + 120 cos x i.e. y(5) = (x4 − 120x2 + 120)cos x + (20x3 − 240x) sin x

(i) Differentiate the given equation n times, using the Leibniz theorem of equation (13),

Now try the following exercise

(ii) rearrange the result to obtain the recurrence relation at x = 0,

Exercise 194 Further problems on Leibniz’s theorem Use the theorem of Leibniz in the following problems: 1. Obtain the n’th derivative of: x 2 y.

2 (n) x y + 2n x y (n−1) + n(n − 1)y (n−2) 2. If ⎡ ⎢ ⎣

y = x 3 e2x

ﬁnd

y (n)

and hence

y (3) .

⎤

y (n) = e2x 2n−3 {8x 3 + 12nx 2

⎥ + n(n − 1)(6x) + n(n − 1)(n − 2)} ⎦

y (3) = e2x (8x 3 + 36x 2 + 36x + 6) 3. Determine the 4th derivative of: y = 2x 3 e−x . [ y (4) = 2e−x (x 3 − 12x 2 + 36x − 24)] 4. If y = x 3 cos x determine the 5th derivative. [ y (5) = (60x − x 3 ) sin x + (15x 2 − 60) cos x] 5. Find an expression for y (4) if y = e−t sin t . [ y (4)

=

−4 e−t sin t ]

6. If y = x 5 ln 2x ﬁnd y (3) . [ y (3) = x 2 (47 + 60 ln 2x)] 7. Given 2x 2 y

+ x y + 3y = 0 show that 2x 2 y (n+2) + (4n + 1)x y (n+1) + (2n 2 − n + 3)y (n) = 0. 8. If y = (x 3 + 2x 2 )e2x determine an expansion for y (5). [ y (5)

=

e2x 24 (2x 3

For second order differential equations that cannot be solved by algebraic methods, the Leibniz–Maclaurin method produces a solution in the form of inﬁnite series of powers of the unknown variable. The following simple 5-step procedure may be used in the Leibniz–Maclaurin method:

+ 19x 2 + 50x

+ 35)]

(iii) determine the values of the derivatives at x = 0, i.e. ﬁnd ( y)0 and ( y )0 , (iv) substitute in the Maclaurin expansion for y = f (x) (see page 69, equation (5)), (v) simplify the result where possible and apply boundary condition (if given). The Leibniz–Maclaurin method is demonstrated, using the above procedure, in the following worked problems. Problem 5. Determine the power series solution of the differential equation: dy d2 y + x + 2y = 0 using Leibniz–Maclaurin’s 2 dx dx method, given the boundary conditions that at dy = 2. x = 0, y = 1 and dx Following the above procedure: (i) The differential equation is rewritten as: y

+ x y + 2y = 0 and from the Leibniz theorem of equation (13), each term is differentiated n times, which gives: y (n+2) +{y (n+1) (x)+n y (n) (1)+0}+2 y (n) = 0 i.e.

y (n+2) + x y (n+1) + (n + 2) y (n) = 0 (14)

(ii) At x = 0, equation (14) becomes: y (n+2) + (n + 2) y (n) = 0 from which, y (n+2) = −(n +2) y (n)

498 Higher Engineering Mathematics This equation is called a recurrence relation or recurrence formula, because each recurring term depends on a previous term. (iii) Substituting n =0, 1, 2, 3, … will produce a set of relationships between the various coefﬁcients. For n =0,

( y

)0 = −2( y)0

n =1, ( y

)0 = −3( y )0

(v) Collecting similar terms together gives: 2x 2 2 × 4x 4 y = ( y)0 1 − + 2! 4! 2 × 4 × 6x 6 2 × 4 × 6 × 8x 8 + 6! 8! 5 3x 3 3 × 5x 5 + − · · · + ( y )0 x − 3! 5!

−

n =2, ( y (4) )0 = −4( y

)0 = −4{−2( y)0 } = 2 × 4( y)0 n =3,

( y (5) )0 = −5( y

)0 = −5{−3( y )0 } = 3 × 5( y )0

−

n =5, ( y (7) )0 = −7( y (5) )0 = −7{3×5( y )0 }

+ 5

+ ( y )0 ×

x7 − +··· 2×4×6

n =6, ( y (8) )0 = −8( y (6) )0 =

(iv) Maclaurin’s theorem from page 69 may be written as: y = ( y)0 + x( y )0 +

x 2

x3 ( y )0 + ( y

)0 2! 3! +

x 4 (4) ( y )0 + · · · 4!

Substituting the above values into Maclaurin’s theorem gives: y = ( y)0 + x( y )0 +

x2 {−2( y)0 } 2!

x4 x3 + {−3( y )0 } + {2 × 4( y)0 } 3! 4! +

x6 x5 {3 × 5( y )0 } + {−2 × 4 ×6( y)0 } 5! 6!

+

x7 {−3 × 5 × 7( y )0 } 7! +

x8 8!

{2 × 4 × 6 × 8( y)0 }

x8 − ··· 3×5×7

x x3 x5 − + 1 1×2 2×4

= −3 × 5 × 7( y )0

−8{−2 × 4 × 6( y)0}= 2 × 4 × 6×8(y)0

x4 x6 x2 i.e. y = ( y)0 1 − + − 1 1×3 3×5

n =4, ( y (6) )0 = −6( y (4) )0 = −6{2 × 4( y)0 } = −2 × 4 × 6( y)0

3 × 5 × 7x 7 + ··· 7!

6

The boundary conditions are that at x = 0, y = 1 dy = 2, i.e. ( y)0 = 1 and ( y )0 = 2. and dx Hence, the power series solution of the differendy d2 y tial equation: 2 + x + 2y = 0 is: dx dx x2 x4 x6 y = 1− + − 1 1 ×3 3 ×5 x x8 x3 + −··· +2 − 3 ×5 × 7 1 1×2 5 7 x x + − +··· 2×4 2×4×6 Problem 6. Determine the power series solution of the differential equation: d2 y d y + + x y = 0 given the boundary conditions dx 2 dx dy that at x = 0, y = 0 and = 1, using dx Leibniz–Maclaurin’s method. Following the above procedure: (i) The differential equation is rewritten as: y

+ y + x y = 0 and from the Leibniz theorem of

Power series methods of solving ordinary differential equations equation (13), each term is differentiated n times, which gives: y

(n+2)

i.e.

+y

(n+1)

+y

(n)

(x) + n y

(n−1)

(1) + 0 = 0

y (n+2) + y (n+1) + x y (n) + n y (n−1) = 0 (15)

(ii) At x = 0, equation (15) becomes: y (n+2) + y (n+1) + n y (n−1) = 0 from which, y (n+2) = −{y (n+1) + n y (n−1) } This is the recurrence relation and applies for n ≥1 (iii) Substituting n = 1, 2, 3, . . . will produce a set of relationships between the various coefﬁcients. For n = 1, ( y

)0 = −{( y

)0 + ( y)0 } n = 2, ( y (4) )0 = −{( y

)0 + 2( y )0 } n = 3, ( y (5) )0 = −{( y (4) )0 + 3( y

)0 } n = 4, ( y (6) )0 = −{( y (5) )0 + 4( y

)0 } n = 5, ( y (7) )0 = −{( y (6) )0 + 5( y (4) )0 } n = 6, ( y (8) )0 = −{( y (7) )0 + 6( y (5) )0 } From the given boundary conditions, at x = 0, dy y = 0, thus ( y)0 = 0, and at x = 0, = 1, thus dx

( y )0 = 1 From the given differential equation, y

+ y + x y = 0, and, at x = 0, ( y

)0 + ( y )0 + (0)y = 0 from which, ( y

)0 = −( y )0 = −1 Thus, ( y)0 = 0, ( y )0 = 1, ( y

)0 = −1, ( y

)0 = −{( y

)0 + ( y)0 } = −(−1 +0) = 1 ( y (4) )0 = −{( y

)0 + 2( y )0 } = −[1 + 2(1)] = −3 ( y (5) )0 = −{( y (4) )0 + 3( y

)0 } = −[−3 +3(−1)] =6 ( y (6) )0 = −{( y (5) )0 + 4( y

)0 } = −[6 + 4(1)] = −10 ( y (7) )0 = −{( y (6) )0 + 5( y (4) )0 } = −[−10 +5(−3)] =25

499

( y (8) )0 = −{( y (7) )0 + 6( y (5) )0 } = −[25 +6(6)] = −61 (iv) Maclaurin’s theorem states: x2 x3 y = ( y)0 + x( y )0 + ( y

)0 + ( y

)0 2! 3! x 4 (4) ( y )0 + · · · 4! and substituting the above values into Maclaurin’s theorem gives: +

y = 0 + x(1) +

x2 x3 x4 {−1} + {1} + {−3} 2! 3! 4!

+

x6 x7 x5 {6} + {−10} + {25} 5! 6! 7!

x8 {−61} + · · · 8! (v) Simplifying, the power series solution of d2 y d y + the differential equation: + x y = 0 is dx 2 dx given by: +

y = x−

x2 x3 3x4 6x5 10x6 + − + − 2! 3! 4! 5! 6! +

25x7 61x8 − +··· 7! 8!

Now try the following exercise Exercise 195 Further problems on power series solutions by the Leibniz–Maclaurin method 1. Determine the power series solution of the difdy d2 y ferential equation: 2 + 2x + y = 0 using dx dx the Leibniz–Maclaurin method, given that at dy x = 0, y = 1 and = 2. dx ⎤ ⎡ x 2 5x 4 5 × 9x 6 ⎥ ⎢ y = 1 − 2! + 4! − 6! ⎥ ⎢ ⎥ ⎢ 3 ⎢ 5 × 9 × 13x 8 3x ⎥ ⎥ ⎢ + −··· +2 x − ⎢ 8! 3! ⎥ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎦ ⎣ 3 × 7x 5 3 × 7 × 11x 7 + − +··· 5! 7!

500 Higher Engineering Mathematics 2. Show that the power series solution of the difd2 y dy ferential equation: (x + 1) 2 + (x − 1) − dx dx 2y = 0, using the Leibniz–Maclaurin method, is given by: y = 1 + x 2 + ex given the boundary dy conditions that at x = 0, y = = 1. dx 3. Find the particular solution of the differd2 y dy − 4y = 0 ential equation: (x 2 + 1) 2 + x dx dx using the Leibniz–Maclaurin method, given the boundary conditions that at x = 0, y = 1 dy and = 1.

dx 3 5 7 x x x y = 1 + x + 2x 2 + − + +··· 2 8 16 4. Use the Leibniz–Maclaurin method to determine the power series solution for the differend2 y d y + x y = 1 given that tial equation: x 2 + dx dx dy at x = 0, y = 1 and = 2. dx ⎤ ⎡ x4 x6 x2 ⎢ y = 1 − 22 + 22 × 42 − 22 × 42 × 62 ⎥ ⎥ ⎢ 5 ⎥ ⎢ x3 x5 ⎥ ⎢ ⎥ ⎢ + ··· +2 x − 2 + 2 2 ⎥ ⎢ 3 3 ×5 ⎥ ⎢ ⎥ ⎢ 7 x ⎦ ⎣ − 2 + · · · 3 × 52 × 72

(iv) equate coefﬁcients of corresponding powers of the variable on each side of the equation; this enables index c and coefﬁcients a1 , a2 , a3 , … from the trial solution, to be determined. This introductory treatment of the Frobenius method covering the simplest cases is demonstrated, using the above procedure, in the following worked problems. Problem 7. Determine, using the Frobenius method, the general power series solution of the d2 y d y differential equation: 3x 2 + − y = 0. dx dx The differential equation may be rewritten as: 3x y

+ y − y = 0. (i) Let a trial solution be of the form y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · · 4 + ar x r + · · ·

(16)

where a0 = 0, i.e. y = a0 x + a1 x c

c+1

+ a2 x

c+2

+ a3 x c+3

+ · · · + ar x c+r + · · ·

(17)

(ii) Differentiating equation (17) gives: y = a0cx c−1 + a1 (c + 1)x c + a2(c + 2)x c+1 + · · · + ar (c + r)x c+r−1 + · · · and

y

= a0c(c − 1)x c−2 + a1 c(c + 1)x c−1 + a2 (c + 1)(c + 2)x c + · · ·

52.5 Power series solution by the Frobenius method A differential equation of the form y

+ P y + Qy = 0, where P and Q are both functions of x, such that the equation can be represented by a power series, may be solved by the Frobenius method. The following 4-step procedure may be used in the Frobenius method: (i) Assume a trial solution of the form y4 = : xc a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · ·

+ ar (c + r − 1)(c + r)x c+r−2 + · · · (iii) Substituting y, y and y

into each term of the given equation 3x y

+ y − y = 0 gives: 3x y

= 3a0 c(c − 1)x c−1 + 3a1 c(c + 1)x c + 3a2(c + 1)(c + 2)x c+1 + · · · + 3ar (c + r − 1)(c+r)x c+r−1 +· · · (a) y = a0 cx c−1 +a1 (c + 1)x c +a2 (c + 2)x c+1 + · · · + ar (c + r)x c+r−1 + · · ·

(b)

(ii) differentiate the trial series, (iii) substitute the results in the given differential equation,

−y = −a0 x c − a1 x c+1 − a2 x c+2 − a3 x c+3 − · · · − ar x c+r − · · ·

(c)

Power series methods of solving ordinary differential equations (iv) The sum of these three terms forms the left-hand side of the equation. Since the right-hand side is zero, the coefﬁcients of each power of x can be equated to zero. For example, the coefﬁcient of x c−1 is equated to zero giving: 3a0 c(c − 1) + a0 c = 0 or a0 c[3c − 3 + 1] = a0 c(3c − 2) = 0

(18)

The coefﬁcient of x c is equated to zero giving: 3a1c(c + 1) + a1 (c + 1) − a0 = 0 i.e.

a1 (3c2 + 3c + c + 1) − a0 = a1(3c2 + 4c + 1) − a0 = 0

or

a1 (3c + 1)(c + 1) − a0 = 0

501

a1 a0 = (2 × 4) (2 × 4) since a1 = a0 a2 a0 when r = 2, a3 = = (3 × 7) (2 × 4)(3 × 7) a0 or (2 × 3)(4 × 7) a3 when r = 3, a4 = (4 × 10) a0 = (2 × 3 × 4)(4 × 7 × 10) and so on.

Thus, when r = 1, a2 =

From equation (16), the trial solution was: (19)

In each of series (a), (b) and (c) an x c term is involved, after which, a general relationship can be obtained for x c+r , where r ≥ 0. In series (a) and (b), terms in x c+r−1 are present; replacing r by (r + 1) will give the corresponding terms in x c+r , which occurs in all three equations, i.e. in series (a), 3ar+1 (c + r)(c + r + 1)x c+r in series (b), ar+1 (c + r + 1)x c+r in series (c), −ar x c+r Equating the total coefﬁcients of x c+r to zero gives: 3ar+1 (c + r)(c + r + 1) + ar+1 (c + r + 1) − ar = 0

y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·+ ar x r + · · ·} Substituting c = 0 and the above values of a1 , a2 , a3, … into the trial solution gives: y = x a0 + a0 x +

a0 x2 (2 × 4) a0 x3 + (2 × 3)(4 × 7) a0 x4 + · · · + (2 × 3 × 4)(4 × 7 × 10) x3 x2 i.e. y = a0 1 + x + + (2 × 4) (2 × 3) (4 × 7) x4 + +··· (21) (2 × 3 × 4)(4 × 7 × 10) 0

which simpliﬁes to: ar+1 {(c + r + 1)(3c + 3r +1)} − ar = 0

(20)

Equation (18), which was formed from the coefﬁcients of the lowest power of x, i.e. x c−1, is called the indicial equation, from which, the value of c is obtained. From equation (18), since a0 = 0, 2 then c = 0 or c = 3

(a) When c = 0: From equation (19), if c = 0, a1 (1 × 1) − a0 = 0, i.e. a1 = a0 From equation (20), if c = 0, ar+1 (r + 1)(3r + 1) − ar = 0, ar i.e. ar+1 = r ≥0 (r + 1)(3r + 1)

2 (b) When c = : 3

5 2 − a0 = 0, i.e. From equation (19), if c = , a1(3) 3 3 a0 a1 = 5 2 From equation (20), if c = 3 2 ar+1 + r + 1 (2 + 3r + 1) − ar = 0, 3 5 (3r + 3) − ar i.e. ar+1 r + 3 = ar+1 (3r 2 + 8r + 5) − ar = 0, ar i.e. ar+1 = r ≥0 (r + 1)(3r + 5)

502 Higher Engineering Mathematics a1 a0 = (2 × 8) (2 × 5 × 8) a0 since a1 = 5 a2 when r = 2, a3 = (3 × 11) a0 = (2 × 3)(5 × 8 × 11) a3 when r = 3, a4 = (4 × 14) a0 = (2×3×4)(5×8×11×14) and so on.

Thus, when r = 1, a2 =

y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·+ ar x r + · · ·} 2 Substituting c = and the above values of a1 , a2 , 3 a3 , … into the trial solution gives: a 2 a0 0 y = x 3 a0 + x+ x2 5 2×5×8 a0 + x3 (2 × 3)(5 × 8 × 11) a0 + x4 + · · · (2 × 3 × 4)(5 × 8 × 11 × 14) 2 x2 x 3 i.e. y = a0 x 1 + + 5 (2 × 5 × 8) x3 (2 × 3)(5 × 8 × 11)

+

x4 + ··· (2 × 3 × 4)(5 × 8 × 11 × 14)

x3 (2 × 3)(5 × 8 × 11)

+

x4 +··· (2 × 3 × 4)(5 × 8 × 11 × 14)

Problem 8. Use the Frobenius method to determine the general power series solution of the differential equation: d2 y dy 2x 2 2 − x + (1 − x)y = 0. dx dx The differential equation may be rewritten as: 2x 2 y

− x y + (1 − x)y = 0. (i) Let a trial solution be of the form

From equation (16), the trial solution was:

+

+

y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · · + ar x r + · · ·}

(23)

where a0 = 0, i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3 + · · · + ar x c+r + · · ·

(24)

(ii) Differentiating equation (24) gives: y = a0 cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1 + · · · + ar (c + r)x c+r−1 + · · · and y

= a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1 + a2(c + 1)(c + 2)x c + · · · + ar (c + r − 1)(c + r)x c+r−2 + · · · (iii) Substituting y, y and y

into each term of the given equation 2x 2 y

− x y + (1 − x)y = 0 gives:

(22)

Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different. Let a0 = A in equation (21), and a0 = B in equation (22). Also, if the ﬁrst solution is denoted by u(x) and the second by v(x), then the general solution of the given differential equation is y = u(x) + v(x). Hence, x3 x2 + y = A 1 +x + (2 × 4) (2 × 3)(4 × 7) x4 + +··· (2 ×3 × 4)(4 × 7 × 10) 2 x2 x +Bx3 1+ + 5 (2 × 5 ×8)

2x 2 y

= 2a0 c(c − 1)x c + 2a1 c(c + 1)x c+1 + 2a2 (c + 1)(c + 2)x c+2 + · · · + 2ar (c + r − 1)(c + r)x c+r + · · · (a) −x y = −a0 cx c − a1 (c + 1)x c+1 − a2 (c + 2)x c+2 − · · · − ar (c + r)x c+r − · · ·

(b)

(1 − x)y = (1 − x)(a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3 + · · · + ar x c+r + · · ·) = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3 + · · · + ar x c+r + · · ·

Power series methods of solving ordinary differential equations − a0 x c+1 − a1 x c+2 − a2 x c+3 − a3 x

c+4

− · · · − ar x

c+r+1

−···

(c)

(iv) The indicial equation, which is obtained by equating the coefﬁcient of the lowest power of x to zero, gives the value(s) of c. Equating the total coefﬁcients of x c (from equations (a) to (c)) to zero gives: i.e.

2a0c(c − 1) − a0 c + a0 = 0 a0 [2c(c − 1) − c + 1] = 0

i.e.

a0 [2c2 − 2c − c + 1] = 0

i.e. i.e.

a0 [2c2 − 3c + 1] = 0 a0 [(2c − 1)(c − 1)] = 0 c = 1 or c =

2ar (c + r − 1)(c + r) − ar (c + r) + ar − ar−1 = 0 from which, ar [2(c + r − 1)(c + r) − (c + r) + 1] = ar−1 and ar =

ar−1 2(c +r −1)(c +r)−(c +r) +1

when r = 4, a3 a3 = a4 = 4(8 + 1) 4 × 9 a0 = (1 × 2 × 3 × 4) × (3 × 5 × 7 × 9) and so on. From equation (23), the trial solution was: y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · · + ar x r + · · ·

1 2 The coefﬁcient of the general term, i.e. x c+r , gives (from equations (a) to (c)): from which,

i.e. y = a0 x 1 1+

(25)

+

ar−1 2(r)(1 + r) − (1 + r ) +1 ar−1 = 2 2r + 2r − 1 − r + 1 ar−1 ar−1 = 2 = 2r + r r (2r + 1)

when r = 2, a1 a1 = 2(4 + 1) (2 × 5) a0 a0 = or (1 × 3)(2 × 5) (1 × 2) × (3 × 5)

a2 =

when r = 3, a2 a2 a3 = = 3(6 + 1) 3 × 7 a0 = (1 × 2 × 3) × (3 × 5 × 7)

4

Substituting c = 1 and the above values of a1 , a2 , a3, … into the trial solution gives: a0 a0 1 y = x a0 + x+ x2 (1×3) (1×2)×(3×5) a0 + x3 (1 × 2 × 3) × (3 × 5 × 7) a0 + x4 (1×2×3×4)×(3×5×7×9) + ···

(a) With c = 1, ar =

Thus, when r = 1, a0 a0 a1 = = 1(2 + 1) 1 × 3

503

+

(b) With c =

x2 x + (1×3) (1 × 2) × (3 × 5)

x3 (1 × 2 × 3) × (3 × 5 × 7) x4 (1×2×3×4)×(3×5×7×9) + ··· (26)

1 2

ar−1 2(c + r − 1)(c + r) − (c + r) + 1 from equation (25) ar−1 i.e. ar = 1 1 1 2 +r −1 +r − + r +1 2 2 2 ar−1 = 1 1 1 2 r− r+ − −r +1 2 2 2 ar−1 = 1 1 2 r2 − − −r +1 4 2 ar =

504 Higher Engineering Mathematics ar−1 ar−1 = 2 1 1 2r −r 2r 2 − − − r + 1 2 2 ar−1 = r(2r − 1)

solution of the given differential equation is y = u(x) + v(x),

=

i.e. y = A x 1 +

a0 a0 = 1(2 − 1) 1 × 1 a1 a1 when r = 2, a2 = = 2(4 − 1) (2 × 3) a0 = (2 × 3) a2 a2 when r = 3, a3 = = 3(6 − 1) 3 × 5 a0 = (2 × 3) × (3 × 5) a3 a3 when r = 4, a4 = = 4(8 − 1) 4 × 7 a0 = (2×3×4)×(3×5×7)

Thus, when r = 1, a1 =

and so on. From equation (23), the trial solution was: y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · · 4 + ar x r + · · ·

x4 (2 × 3 × 4) × (3 × 5 × 7) + ···

x4 (1 × 2 × 3×4)×(3×5×7×9) 1 x2 +··· +Bx2 1+x+ (2 × 3)

+

+

x3 (2 × 3) × (3 × 5)

+

x4 +··· (2 × 3 × 4) ×(3 × 5 ×7)

Problem 9. Use the Frobenius method to determine the general power series solution of the d2 y differential equation: 2 − 2y = 0. dx

(i) Let a trial solution be of the form y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · · + ar x r + · · ·

4

(28)

where a0 = 0, i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3 + · · · + ar x c+r + · · ·

x2 (2 × 3)

(29)

(ii) Differentiating equation (29) gives:

x3 + (2 × 3) × (3 × 5) +

x3 (1 × 2 × 3) × (3 × 5 ×7)

The differential equation may be rewritten as: y

− 2y = 0.

1 Substituting c = and the above values of a1 , a2 , 2 a3 , … into the trial solution gives: 1 a0 2 a0 2 x + x3 y=x a0 +a0 x + (2×3) (2×3)×(3×5) a0 + x4 + · · · (2 × 3 × 4) × (3 × 5 × 7) 1 i.e. y = a0 x 2 1 + x +

+

x2 x + (1 × 3) (1 × 2) × (3 × 5)

y = a0cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1 + · · · + ar (c + r)x c+r−1 + · · · (27)

Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different. Let a0 = A in equation (26), and a0 = B in equation (27). Also, if the ﬁrst solution is denoted by u(x) and the second by v(x), then the general

and y

= a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1 + a2(c + 1)(c + 2)x c + · · · + ar (c + r − 1)(c + r)x c+r−2 + · · · (iii) Replacing r by (r + 2) in ar (c + r − 1)(c + r) x c+r−2 gives: ar+2 (c + r + 1)(c + r + 2)x c+r

Power series methods of solving ordinary differential equations

505

2x 2 4x 4 = a0 1 + + +··· 2! 4! 5 6 2x 3 4x 5 + a1 x + + +··· 3! 5!

Substituting y and y

into each term of the given equation y

− 2y = 0 gives: y

− 2y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1 + [a2(c+1)(c + 2)−2a0 ]x c +· · ·

Hence, a0 c(c − 1) =0 from which, c = 0 or c = 1 since a0 = 0

Since a0 and a1 are arbitrary constants depending on boundary conditions, let a0 = P and a1 = Q, then: 2x2 4x4 + +··· y=P 1 + 2! 4! 3 4x5 2x + +··· (33) +Q x+ 3! 5!

For the term in x c−1 , i.e. a1c(c + 1) = 0

(b) When c =1: a1 = 0, and from equation (31),

+ [ar+2 (c + r + 1)(c + r + 2) − 2ar ] x c+r + · · · = 0

(30)

(iv) The indicial equation is obtained by equating the coefﬁcient of the lowest power of x to zero.

With c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined with the zero value of c would make the product zero.

a2 = Since

For the term in x c , a2 (c + 1)(c + 2) − 2a0 = 0 from which, 2a0 a2 = (31) (c + 1)(c + 2) For the term in x c+r ,

(32)

(a) When c = 0: a1 is indeterminate, and from equation (31) 2a0 2a0 a2 = = (1 × 2) 2! 2ar and (r + 1)(r + 2) 2a1 2a1 2a1 = = when r = 1, a3 = (2 × 3) (1 × 2 × 3) 3! 2a2 4a0 = when r = 2, a4 = 3×4 4! 2a0 2 2a1 3 Hence, y = x 0 a0 + a1 x + x + x 2! 3! 4a0 4 + x + ··· 4! In general, ar + 2 =

from equation (28)

2ar (c + r + 1)(c + r + 2) 2ar = (r + 2)(r + 3)

ar+2 =

from equation (32) and when r = 1, a3 =

a4 =

from which, 2ar (c + r + 1)(c + r + 2)

c = 1,

2a1 = 0 since a1 = 0 (3 × 4)

when r = 2,

ar+2 (c + r + 1)(c + r + 2) − 2ar = 0

ar+2 =

2a0 2a0 = (2 × 3) 3!

2a2 2 2a0 4a0 = × = (4 × 5) (4 × 5) 3! 5!

when r = 3, a5 =

2a3 =0 (5 × 6)

Hence, when c = 1, 2a0 2 4a0 4 x + x +··· y = x 1 a0 + 3! 5! from equation (28) 5 6 2x 3 4x 5 i.e. y = a0 x + + + ... 3! 5! Again, a0 is an arbitrary constant; let a0 = K , 2x3 4x5 + +··· then y=K x+ 3! 5! However, this latter solution is not a separate solution, for it is the same form as the second series in equation (33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solution. This is always

506 Higher Engineering Mathematics the case when the two values of c differ by an integer (i.e. whole number). From the above three worked problems, the following can be deduced, and in future assumed: (i) if two solutions of the indicial equation differ by a quantity not an integer, then two independent solutions y = u(x) + v(x) result, the general solution of which is y = Au + Bv (note: Problem 7 1 2 had c = 0 and and Problem 8 had c = 1 and ; 3 2 in neither case did c differ by an integer) (ii) if two solutions of the indicial equation do differ by an integer, as in Problem 9 where c = 0 and 1, and if one coefﬁcient is indeterminate, as with when c = 0, then the complete solution is always given by using this value of c. Using the second value of c, i.e. c = 1 in Problem 9, always gives a series which is one of the series in the ﬁrst solution. Now try the following exercise Exercise 196 Further problems on power series solution by the Frobenius method 1. Produce, using Frobenius’ method, a power series solution for the differential equation: d2 y d y 2x 2 + − y = 0. dx dx ⎤ ⎡ 2 x ⎥ ⎢y = A 1 + x + ⎥ ⎢ (2 × 3) ⎥ ⎢ ⎢ ⎥ 3 ⎥ ⎢ x ⎢ + +··· ⎥ ⎥ ⎢ (2 × 3)(3 × 5) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 ⎥ x2 x ⎢ ⎥ ⎢ +Bx2 1+ + ⎥ ⎢ (1 × 3) (1 × 2)(3 × 5) ⎥ ⎢ ⎥ ⎢ 3 ⎥ ⎢ x ⎣ + + ··· ⎦ (1 × 2 × 3)(3 × 5 × 7) 2. Use the Frobenius method to determine the general power series solution of the differend2 y tial equation: 2 + y = 0. dx ⎡ ⎤ x2 x4 ⎢ y = A 1 − 2! + 4! − · · · ⎥ ⎢ ⎥ ⎥ ⎢ 3 5 ⎢ ⎥ x x ⎢ + B x − + − ··· ⎥ ⎢ ⎥ 3! 5! ⎣ ⎦ = P cos x + Q sin x

3. Determine the power series solution of the d2 y dy differential equation: 3x 2 + 4 − y = 0 dx dx using the Frobenius method. ⎡ ⎤ x x2 ⎢y = A 1 + (1 × 4) + (1 × 2)(4 × 7) ⎥ ⎢ ⎥ ⎢ ⎥ x3 ⎢ ⎥ ⎢ + +··· ⎥ ⎢ ⎥ (1 × 2 × 3)(4 × 7 × 10) ⎢ ⎥ ⎢ ⎥ 2 1 x x ⎢ ⎥ −3 + 1+ ⎢ + Bx ⎥ ⎢ (1 × 2) (1 × 2)(2 × 5)⎥ ⎢ ⎥ ⎢ ⎥ x3 ⎣ + + ··· ⎦ (1 × 2 × 3)(2 × 5 × 8) 4. Show, using the Frobenius method, that the power series solution of the differential d2 y − y = 0 may be expressed as equation: dx 2 y = P cosh x + Q sinh x, where P and Q are constants. [Hint: check the series expansions for cosh x and sinh x on page 47]

52.6 Bessel’s equation and Bessel’s functions One of the most important differential equations in applied mathematics is Bessel’s equation and is of the form: d2 y dy + (x 2 − v 2 )y = 0 x2 2 + x dx dx where v is a real constant. The equation, which has applications in electric ﬁelds, vibrations and heat conduction, may be solved using Frobenius’ method of the previous section. Problem 10. Determine the general power series solution of Bessels equation. d2 y dy +x + (x 2 − v 2 )y = 0 may 2 dx dx be rewritten as: x 2 y

+ x y + (x 2 − v 2 )y = 0 Bessel’s equation x 2

Using the Frobenius method from page 500: (i) Let a trial solution be of the form y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · · + ar x r + · · ·} where a0 = 0,

(34)

507

Power series methods of solving ordinary differential equations i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3 + · · · + ar x

c+r

+···

(35)

Similarly, if c = −va1[1 − 2v] = 0 The terms (2v + 1) and (1 − 2v) cannot both be zero since v is a real constant, hence a1 = 0.

(ii) Differentiating equation (35) gives: y = a0cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1 + · · · + ar (c + r)x c+r−1 + · · · and y

= a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1 + a2 (c + 1)(c + 2)x c + · · · + ar (c + r − 1)(c + r)x c+r−2 + · · · y

y

(iii) Substituting y, and into each term of the given equation: x 2 y

+ x y + (x 2 − v 2 )y = 0 gives:

Since a1 = 0, then from a3 = a5 = a7 = . . . = 0

+ a2(c + 1)(c + 2)x c+2 + · · ·

a2 =

a0 v 2 − (c + 2)2

a4 =

a0 2 2 [v − (c + 2) ][v 2 − (c + 4)2 ]

a6 =

a0 [v 2 − (c + 2)2 ][v 2 −(c + 4)2 ][v 2 − (c + 6)2 ] and so on.

a2 =

+ a1(c + 1)x c+1 + a2 (c + 2)x c+2 + · · · + ar (c + r)x c+r + · · · + a0 x c+2 + a1 x c+3 + a2 x

− a1 v x

+ · · · + ar x

2 c+1

2 c

+ · · · − a0 v x

− · · · − ar v x

2 c+r

+··· = 0 (36)

(iv) The indicial equation is obtained by equating the coefﬁcient of the lowest power of x to zero. Hence,

a0 [c2 − c + c − v 2 ] = 0

from which, c = +v or c = −v since a0 = 0 For the term in x c+r , − ar v 2 = 0

ar [(c + r)2 − v 2 ] =−ar−2

For the term in x c+1 , a1[c(c + 1) + (c + 1) − v 2 ] = 0 i.e.

a1 [(c + 1)2 − v 2 ] = 0

but if c = v

a1 [(v + 1)2 − v 2 ] = 0

a6 =

=

ar [(c + r − 1)(c + r) + (c + r) − v 2 ] =−ar−2 i.e. ar [(c + r)(c + r − 1 + 1) − v 2 ] =−ar−2

i.e. the recurrence relation is: ar−2 ar = for r ≥ 2 v2 − (c + r)2

=

=

ar (c + r − 1)(c + r) + ar (c + r) + ar−2

i.e.

a4 =

=

a0 [c2 − v 2 ] = 0

i.e.

=

=

a0c(c − 1) + a0 c − a0 v 2 = 0

from which,

(37)

When c = +v,

+ ar (c + r − 1)(c + r)x c+r + · · · + a0 cx c

c+r+2

equation

and

a0 c(c − 1)x c + a1 c(c + 1)x c+1

c+4

a1[2v + 1] = 0

i.e.

(37)

=

a0 2 v − (v + 2)2

=

a0 2 2 v − v − 4v − 4

−a0 −a0 = 2 4 + 4v 2 (v + 1) a0

v 2 − (v + 2)2 v 2 − (v + 4)2 a0 [−22 (v + 1)][−23(v + 2)] a0 5 2 (v + 1)(v + 2) a0 24 × 2(v + 1)(v + 2) a0 2 2 2 [v −(v+2) ][v −(v+4)2 ][v 2−(v+6)2 ] a0 4 [2 × 2(v + 1)(v + 2)][−12(v + 3)] −a0 24 × 2(v + 1)(v + 2) × 22 × 3(v + 3) −a0 and so on. 26 × 3! (v + 1)(v + 2)(v + 3)

The resulting solution for c = +v is given by: y=u= A x v 1−

x4 x2 + 22 (v +1) 24 × 2! (v +1)(v +2) x6 − 6 +··· 2 × 3! (v +1)(v + 2)(v + 3) (38)

508 Higher Engineering Mathematics which is valid provided v is not a negative integer and where A is an arbitrary constant. When c = −v, a0 a0 = 2 a2 = 2 v − (−v + 2)2 v − (v 2 − 4v + 4) −a0 −a0 = = 2 4 − 4v 2 (v − 1) a0 a4 = 2 [2 (v − 1)][v 2 − (−v + 4)2 ] a0 = 2 [2 (v − 1)][23 (v − 2)] a0 = 4 2 × 2(v − 1)(v − 2) a0 Similarly, a6 = 6 2 × 3! (v−1)(v−2)(v−3)

upper case Greek letter gamma, and the gamma function (x) is deﬁned by the integral ! ∞ t x−1 e−t dt (40) (x) = 0

and is convergent for x > 0

∞

t x e−t dt

and by using integration by parts (see page 420): x e−t ∞ (x + 1) = t −1 0 ! ∞ −t e x t x−1 dx − −1 0 ! ∞ e−t t x−1 dt = (0 − 0) + x 0

Hence,

= x(x) from equation (40)

y =w=

x4 x2 B x −v 1 + 2 + 4 2 (v−1) 2 ×2! (v−1)(v−2) x6 + 6 +··· 2 × 3! (v − 1)(v − 2)(v − 3) which is valid provided v is not a positive integer and where B is an arbitrary constant. The complete solution of Bessel’s equation: x2

!

From equation (40), (x + 1) =

d2 y dy 2 +x + x − v 2 y = 0 is: dx 2 dx

y= u +w = x4 x2 + 4 A xv 1 − 2 2 (v + 1) 2 × 2!(v + 1)(v + 2) x6 − 6 +··· 2 × 3!(v + 1)(v + 2)(v + 3) x2 −v +Bx 1+ 2 2 (v − 1) x4 + 4 2 × 2!(v − 1)(v − 2) +

x6 +· · · 6 2 × 3!(v−1)(v−2)(v−3)

(39)

The gamma function The solution of the Bessel equation of Problem 10 may be expressed in terms of gamma functions. is the

This is an important recurrence relation for gamma functions. Thus, since then similarly,

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

and

(41)

(x + 3) = (x + 2)(x + 2) = (x + 2)(x + 1)x(x), and so on.

These relationships involving gamma functions are used with Bessel functions.

Bessel functions The power series solution of the Bessel equation may be written in terms of gamma functions as shown in worked problem 11 below. Problem 11. Show that the power series solution of the Bessel equation of worked problem 10 may be written in terms of the Bessel functions Jv (x) and J−v (x) as: AJv (x) + BJ −v (x) x v 1 x2 = − 2 2 (v + 1) 2 (1! )(v + 2) x4 + 4 −··· 2 (2! )(v + 4)

Power series methods of solving ordinary differential equations

+

x −v 2

This is called the Bessel function of the ﬁrst order kind, of order v, and is denoted by Jv (x), x v 1 x2 i.e. Jv (x) = − 2 2 (v + 1) 2 (1!)(v + 2) x4 + 4 −··· 2 (2!)(v + 3)

1 x2 − 2 (1 − v) 2 (1! )(2 − v) x4 + 4 −··· 2 (2! )(3 − v)

From Problem 10 above, when c = +v, −a0 a2 = 2 2 (v + 1) If we let a0 =

provided v is not a negative integer.

1 2v (v + 1)

For the second solution, when c = −v, replacing v by −v in equation (42) above gives:

then −1 −1 = 22 (v + 1) 2v (v + 1) 2v+2 (v + 1)(v + 1) −1 = v+2 from equation (41) 2 (v + 2)

a2k =

a2 =

Similarly, a4 =

a2 2 v − (c + 4)2

from equation (37)

from =

(−1)k 22k−v (k! ) (k − v + 1)

which,

when

a2 a2 = (v − c − 4)(v + c + 4) −4(2v + 4) since c = v −1 −a2 −1 = 3 = 2 (v + 2) 23 (v + 2) 2v+2 (v + 2)

= when k = 2, a4 =

1

=

2v+4 (2! )(v + 3)

when k = 3, a6 = =

The recurrence relation is: ar =

(−1)r/2 r r 2v+r ! v + +1 2 2

(−1)1 −1 22−v (1! )(2 − v) (−1)2 24−v (2! )(2 − v + 1) 1 24−v (2! )(3 − v)

a2k =

(−1)k 2v+2k (k!)(v + k + 1)

(42) for k = 1, 2, 3, . . .

Hence, it is possible to write the new form for equation (38) as: 1 x2 v y = Ax − 2v (v + 1) 2v+2 (1! )(v + 2) x4 + v+4 −··· 2 (2! )(v + 3)

(−1)3 26−v (3! )(3 − v + 1) 1 26−v (3! )(4 − v)

and so on.

1 x2 − 2−v (1 − v) 22−v (1! )(2 − v) x4 + 4−v −··· 2 (2! )(3 − v) x −v 1 x2 − 2 i.e. J−v (x)= 2 (1 −v) 2 (1!)(2 − v) x4 −· · · + 4 2 (2!)(3 −v) Hence, y = Bx −v

And if we let r = 2k, then

(−1)0 2−v (0! )(1 − v)

22−v (1! )(1 − v + 1)

since (v + 2)(v + 2) = (v + 3) −1 and a6 = v+6 and so on. 2 (3! )(v + 4)

k = 0, a0 =

1 since 0! = 1 (see page 495) 2−v (1 − v)

when k = 1, a2 =

=

=

509

provided v is not a positive integer. Jv (x) and J−v (x) are two independent solutions of the Bessel equation; the complete solution is: y = AJ v (x) + B J −v (x) where A and B are constants

510 Higher Engineering Mathematics i.e. y = AJ v (x)+ BJ −v (x) x v 1 x2 − 2 =A 2 (v + 1) 2 (1!)(v + 2) + +B

x −v 2

From this series two commonly used function are derived,

x4 − ··· 24 (2!)(v + 4)

1 x2 − 2 (1 −v) 2 (1!)(2 − v) x4 + 4 − ··· 2 (2!)(3 −v)

In general terms: Jv (x) =

x v ; ∞

i.e. J0(x) =

1 1 x 4 1 x 2 + − (0! ) (1! )2 2 (2! )2 2 1 x 6 − +··· (3! )2 2

= 1−

(−1)k x 2k 22k (k! )(v+k+1)

2 k=0 x −v ; ∞ (−1)k x 2k and J−v (x) = 2k 2 k=0 2 (k! )(k − v + 1)

x2 22 (1!)2

x4 24 (2!)

− 2

=

x 2 x n 1 1 − 2 n! (n + 1)! 2

x 4 1 + − ··· (2! )(n + 2)! 2

26 (3!)2

+···

x x3 x5 − 3 + 5 2 2 (1!)(2!) 2 (2!)(3!) −

It may be shown that another series for Jn(x) is given by:

x6

x 2 1 x 1 and J1(x) = − 2 (1! ) (1! )(2! ) 2 x 4 1 + −··· (2! )(3! ) 2

Another Bessel function

Jn (x) =

+

x7 +··· 27 (3!)(4!)

Tables of Bessel functions are available for a range of values of n and x, and in these, J0 (x) and J1(x) are most commonly used. Graphs of J0 (x), which looks similar to a cosine, and J1 (x), which looks similar to a sine, are shown in Figure 52.1.

y 1 y ⫽ J0(x)

0.5 y ⫽ J1(x)

⫺0.5

Figure 52.1

2

4

6

8

10

12

14

x

511

Power series methods of solving ordinary differential equations i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3 Now try the following exercise Exercise 197 Further problems on Bessel’s equation and Bessel’s functions 1. Determine the power series solution of Besd2 y dy sel’s equation: x 2 2 + x + (x 2 −v 2 )y = 0 dx dx 6. when v = 2,up to and including the term in x 2 4 x x − ··· y = Ax 2 1 − + 12 384

+ · · · + ar x c+r + · · ·

(44)

(ii) Differentiating equation (44) gives: y = a0 cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1 + · · · + ar (c + r)x c+r−1 + · · · and y

= a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1 + a2 (c + 1)(c + 2)x c + · · ·

2. Find the power series solution of the Bessel function: x 2 y

+ x y + x 2 − v 2 y = 0 in terms of the Bessel function J3(x) when v = 3. Give the answer up to and including the term in x 7 . ⎡ ⎤ x 3 1 x2 − ⎢ y = AJ3 (x) = 2 4 22 5 ⎥ ⎢ ⎥ ⎣ ⎦ x4 + 5 −··· 2 6 3. Evaluate the Bessel functions J0 (x) and J1 (x) when x = 1, correct to 3 decimal places. [J0(x) = 0.765, J1(x) = 0.440]

+ ar (c + r − 1)(c + r)x c+r−2 + · · · (iii) Substituting y, y and y

into each term of the given equation: 1 − x 2 y

− 2x y + k(k + 1)y = 0 gives: a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1 + a2 (c + 1)(c + 2)x c + · · · + ar (c + r − 1)(c + r)x c+r−2 + · · · − a0 c(c − 1)x c − a1 c(c + 1)x c+1 − a2 (c + 1)(c + 2)x c+2 − · · ·

52.7 Legendre’s equation and Legendre polynomials Another important differential equation in physics and engineering applications is Legendre’s equation d2 y dy of the form: (1 − x 2 ) 2 − 2x + k(k + 1)y = 0 or dx dx 2

(1 − x )y − 2x y + k(k + 1)y = 0 where k is a real constant. Problem 12. Determine the general power series solution of Legendre’s equation. To solve Legendre’s equation (1 − x 2 )y

− 2x y + k(k + 1)y = 0 using the Frobenius method: (i) Let a trial solution be of the form y = x c a0 + a1 x + a2 x 2 + a3 x 3

4 + · · · + ar x r + · · · (43) where a0 = 0,

− ar (c + r − 1)(c + r)x c+r − · · · − 2a0 cx c − 2a1 (c + 1)x c+1 − 2a2 (c + 2)x c+2 − · · · − 2ar (c + r)x c+r − · · · + k 2 a0 x c + k 2 a1 x c+1 + k 2 a2 x c+2 + · · · + k 2 ar x c+r + · · · + ka0 x c + ka1 x c+1 + · · · + kar x c+r + · · · = 0

(45)

(iv) The indicial equation is obtained by equating the coefﬁcient of the lowest power of x (i.e. x c−2 ) to zero. Hence, a0c(c − 1) = 0 from which, c = 0 or c = 1 since a0 = 0. For the term in x c−1 , i.e. a1 c(c + 1) = 0 With c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined with the zero value of c would make the product zero. For the term in x c+r , ar+2 (c + r + 1)(c + r + 2) −ar (c + r − 1) (c + r) − 2ar (c + r) + k 2 ar + kar = 0

512 Higher Engineering Mathematics from which,

ar (c+r−1)(c+r)+2(c+r)−k 2 −k ar+2 = (c+r+1)(c+r +2) ar [(c + r)(c + r + 1) − k(k + 1)] = (c + r + 1)(c + r + 2) (46) When c = 0, ar+2 =

ar [r(r + 1) − k(k + 1)] (r + 1)(r + 2)

For r = 0, a2 =

a0 [−k(k + 1)] (1)(2)

For r = 1, a3 = =

a1[(1)(2) − k(k + 1)] (2)(3) −a1 [k 2 + k − 2] −a1 (k − 1)(k + 2) = 3! 3!

a2 [(2)(3) − k(k + 1)] −a2 k 2 + k − 6 = a4 = (3)(4) (3)(4) =

−a2 (k + 3)(k − 2) (3)(4)

=

−(k + 3)(k − 2) a0 [−k(k + 1)] . (3)(4) (1)(2)

a0 k(k + 1)(k + 3)(k − 2) = 4! For r = 3,

=

a3[(3)(4) − k(k + 1)] −a3 [k 2 + k − 12] = (4)(5) (4)(5) −a3 (k + 4)(k − 3) (4)(5)

−(k + 4)(k − 3) −a1 (k − 1)(k + 2) = . (4)(5) (2)(3) =

a1(k − 1)(k − 3)(k + 2)(k + 4) and so on. 5!

Substituting values into equation (43) gives: a0 k(k + 1) 2 0 y = x a0 + a1 x − x 2! −

a1 (k − 1)(k + 2) 3 x 3!

+ ··· k(k + 1) 2 i.e. y = a0 1 − x 2! k(k +1)(k − 2)(k + 3) 4 x −··· 4! (k − 1)(k + 2) 3 + a1 x − x 3!

+

+

For r = 2,

a5 =

a0k(k + 1)(k − 2)(k + 3) 4 x 4! a1 (k − 1)(k − 3)(k + 2)(k + 4) 5 + x 5!

+

(k − 1)(k − 3)(k + 2)(k + 4) 5 x − · · · (47) 5!

From page 506, it was stated that if two solutions of the indicial equation differ by an integer, as in this case, where c = 0 and 1, and if one coefﬁcient is indeterminate, as with when c = 0, then the complete solution is always given by using this value of c. Using the second value of c, i.e. c = 1 in this problem, will give a series which is one of the series in the ﬁrst solution. (This may be checked for c = 1 and where a1 = 0; the result will be the ﬁrst part of equation (47) above).

Legendre’s polynomials (A polynomial is an expression of the form: f (x) = a + bx + cx 2 + d x 3 + · · ·). When k in equation (47) above is an integer, say, n, one of the solution series terminates after a ﬁnite number of terms. For example, if k = 2, then the ﬁrst series terminates after the term in x 2 . The resulting polynomial in x, denoted by Pn (x), is called a Legendre polynomial. Constants a0 and a1 are chosen so that y = 1 when x = 1. This is demonstrated in the following worked problems. Problem 13. P2 (x).

Determine the Legendre polynomial

Since in P2 (x), n =k = 2, then from the ﬁrst part of equation (47), i.e. the even powers of x: 2(3) 2 y = a0 1 − x + 0 = a0 {1 − 3x 2 } 2! a0 is chosen to make y = 1 when x = 1 i.e. 1 = a0 {1 −3(1)2 } = −2a0 , from which, a0 = −

1 2

Power series methods of solving ordinary differential equations Hence, P2 (x)= −

1 1 1 − 3x 2 = (3x2 − 1) 2 2

Problem 14. Determine the Legendre polynomial P3 (x). Since in P3 (x), n =k = 3, then from the second part of equation (47), i.e. the odd powers of x: (k − 1)(k + 2) 3 x y = a1 x − 3! (k − 1)(k − 3)(k + 2)(k + 4) 5 + x − ··· 5! (2)(5) 3 (2)(0)(5)(7) 5 i.e. y = a1 x − x + x 3! 5! 5 3 = a1 x − x + 0 3 a1 is chosen to make y = 1 when x = 1. 3 5 2 from which, a1 = − i.e. 1 = a1 1 − = a1 − 3 3 2 5 3 1 Hence, P3 (x) =− x− x 3 or P3 (x) = (5x3− 3x) 2 3 2

Rodrigue’s formula An alternative method of determining Legendre polynomials is by using Rodrigue’s formula, which states: n 1 dn x2 − 1 Pn (x)= n (48) 2 n! dxn This is demonstrated in the following worked problems. Problem 15. Determine the Legendre polynomial P2 (x) using Rodrigue’s formula. n 1 dn x 2 − 1 In Rodrigue’s formula, Pn (x) = n 2 n! dx n and when n =2, P2 (x) = =

1 d 2 (x 2 − 1)2 22 2! dx 2 1 d2 (x 4 − 2x 2 + 1) 23 dx 2 d 4 (x − 2x 2 + 1) dx

= 4x 3 − 4x

513

d2 x 4 − 2x 2 + 1 d(4x 3 − 4x) = and = 12x 2 − 4 dx 2 dx 1 d2 x 4 −2x 2 +1 1 = 12x 2 − 4 Hence, P2 (x) = 3 2 2 dx 8 1 2 i.e. P2 (x) = 3x − 1 the same as in Problem 13. 2 Problem 16. Determine the Legendre polynomial P3 (x) using Rodrigue’s formula. n 1 dn x 2 − 1 and In Rodrigue’s formula, Pn (x) = n 2 n! dx n when n = 3, 3 1 d3 x 2 − 1 P3 (x) = 3 2 3! dx 3 1 d3 x 2 − 1 x 4 − 2x 2 + 1 = 3 2 (6) dx 3 1 d3 x 6 − 3x 4 + 3x 2 − 1 = (8)(6) dx 3 d x 6 −3x 4 +3x 2 −1 = 6x 5 − 12x 3 + 6x dx d 6x 5 −12x 3 +6x = 30x 4 − 36x 2 + 6 dx d 30x 4 − 36x 2 + 6 and = 120x 3 − 72x dx 1 d3 x 6 − 3x 4 + 3x 2 − 1 Hence, P3 (x) = (8)(6) dx 3 1 1 = 120x 3 − 72x = 20x 3 − 12x (8)(6) 8 1 i.e. P3 (x)= 5x3 − 3x the same as in Problem 14. 2 Now try the following exercise Exercise 198 Legendre’s equation and Legendre polynomials 1. Determine the power series solution of the Legendre equation: 1 − x 2 y

− 2x y + k(k + 1)y = 0 when (a) k = 0 (b) k = 2, up to and including the

514 Higher Engineering Mathematics term in x 5 . ⎡

x3 x5 ⎢(a) y = a0 + a1 x + 3 + 5 + · · · ⎢ : 4 ⎢ ⎢(b) y = a0 1 − 3x 2 ⎢ ⎣ 1 2 + a1 x − x 3 − x 5 3 5

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

2. Find the following Legendre polynomials: (a) P1 (x) (b) P4 (x) (c) P5 (x). ⎤ 1 4 2 ⎢ (a) x (b) 8 35x − 30x + 3 ⎥ ⎣ ⎦ 1 (c) 63x 5 − 70x 3 + 15x 8 ⎡

Chapter 53

An introduction to partial differential equations 53.1

Introduction

A partial differential equation is an equation that contains one or more partial derivatives. Examples include: ∂u ∂u +b =c (i) a ∂x ∂y (ii)

(iii)

∂ 2u 1 ∂u = 2 2 ∂x c ∂t (known as the heat conduction equation) ∂ 2u ∂ 2 u + =0 ∂x 2 ∂ y 2 (known as Laplace’s equation)

Equation (i) is a ﬁrst order partial differential equation, and equations (ii) and (iii) are second order partial differential equations since the highest power of the differential is 2. Partial differential equations occur in many areas of engineering and technology; electrostatics, heat conduction, magnetism, wave motion, hydrodynamics and aerodynamics all use models that involve partial differential equations. Such equations are difﬁcult to solve, but techniques have been developed for the simpler types. In fact, for all but for the simplest cases, there are a number of numerical methods of solutions of partial differential equations available. To be able to solve simple partial differential equations knowledge of the following is required: (a)

partial integration,

(b) ﬁrst and second order partial differentiation — as explained in Chapter 34, and (c)

the solution of ordinary differential equations — as explained in Chapters 46–51.

It should be appreciated that whole books have been written on partial differential equations and their solutions. This chapter does no more than introduce the topic.

53.2

Partial integration

Integration is the reverse process of differentiation. ∂u Thus, if, for example, = 5 cos x sin t is integrated par∂t tially with respect to t , then the 5 cosx term is considered as a constant, ! ! and u = 5 cos x sin t dt = (5 cos x) sin t dt = (5 cos x)(−cos t ) + c = −5 cos x cos t + f (x) ∂2u = 6x 2 cos 2y is integrated partially ∂x∂ y with respect to y, ! ! ∂u then cos 2y d y = 6x 2 cos 2y d y = 6x 2 ∂x 1 2 = 6x sin 2y + f (x) 2

Similarly, if

= 3x 2 sin 2y + f (x)

516 Higher Engineering Mathematics and integrating ! u=

∂u partially with respect to x gives: ∂x

[3x 2 sin 2y + f (x)] dx

= x3 sin 2y + (x)f (x) + g(y) f (x) and g(y) are functions that may be determined if extra information, called boundary conditions or initial conditions, are known.

53.3

From the boundary conditions, when x = 0, u = cos y, hence cos y =

from which, F( y) = cos y ∂2u Hence, the solution of 2 = 6x 2 (2y − 1) for the given ∂x boundary conditions is: u=

Solution of partial differential equations by direct partial integration

The simplest form of partial differential equations occurs when a solution can be determined by direct partial integration. This is demonstrated in the following worked problems. Problem 1. Solve the differential equation ∂2u = 6x 2 (2y − 1) given the boundary conditions ∂x 2 ∂u that at x = 0, = sin 2y and u =cos y. ∂x ∂ 2u Since 2 = 6x 2 (2y − 1) then integrating partially with ∂x respect to x gives: ! ! ∂u = 6x 2 (2y − 1)dx = (2y − 1) 6x 2 dx ∂x = (2y − 1)

6x 3 + f ( y) 3

= 2x 3 (2y − 1) + f ( y) where f (y) is an arbitrary function. From the boundary conditions, when x = 0, ∂u = sin 2y. ∂x sin 2y = 2(0)3 (2y − 1) + f ( y)

Hence, from which,

f ( y) = sin 2y

∂u = 2x 3 (2y − 1) + sin 2y ∂x Integrating partially with respect to x gives: ! u = [2x 3 (2y − 1) + sin 2y]dx Now

=

2x 4 (2y − 1) + x(sin 2y) + F( y) 4

(0)4 (2y − 1) + (0)sin 2y + F( y) 2

x4 (2y − 1) + x sin y + cos y 2

Problem 2. Solve the differential equation: ∂2u ∂u = cos(x + y) given that = 2 when y = 0, ∂x∂ y ∂x and u = y 2 when x = 0. ∂ 2u = cos(x + y) then integrating partially with ∂x∂ y respect to y gives: ! ∂u = cos(x + y)d y = sin(x + y) + f (x) ∂x Since

∂u = 2 when y = 0, From the boundary conditions, ∂x hence 2 = sin x + f (x) from which, f (x) = 2 − sin x i.e.

∂u = sin(x + y) + 2 − sin x ∂x

Integrating partially with respect to x gives: ! u = [sin(x + y) + 2 − sin x]dx = −cos(x + y) + 2x + cos x + f (y) From the boundary conditions, u = y 2 when x = 0, hence y 2 = −cos y + 0 + cos 0 + f ( y) = 1 − cos y + f ( y) from which, f (y) = y 2 − 1 + cos y Hence, the solution of

∂ 2u = cos(x + y) is given by: ∂x∂ y

u = −cos(x + y) + 2x + cos x + y2 − 1 + cos y

517

An introduction to partial differential equations ⎞ ⎛ 2 3x − (x 2 + y 2 + z 2 ) ⎟ ⎜ ⎜ + 3y 2 − (x 2 + y 2 + z 2 )⎟ ⎠ ⎝

Problem 3. Verify that 1 φ(x, y, z) = satisﬁes the partial x 2 + y2 + z2 ∂ 2φ ∂ 2φ ∂ 2φ differential equation: 2 + 2 + 2 = 0. ∂x ∂y ∂z

= Thus,

The partial differential equation ∂ 2φ ∂ 2φ ∂ 2φ + + = 0 is called Laplace’s equation. ∂x 2 ∂ y 2 ∂z 2 If φ(x, y, z) =

1

1

then differentiating partially with respect to x gives: 3 ∂φ 1 = − (x 2 + y 2 + z 2 )− 2 (2x) ∂x 2

= −x(x 2 + y 2 + z 2 )− 2 ∂ 2φ 3 2 2 2 − 52 = (−x) − (x + y + z ) (2x ) ∂x 2 2 2

2

=

3x 2

−

2 − 32

(3x 2 ) − (x 2

y2 + z2 )

+

(x 2 + y 2 + z 2 ) 2

(3y 2 ) − (x 2 + y 2 + z 2 ) ∂ 2φ = 5 ∂ y2 (x 2 + y 2 + z 2 ) 2

of

[u =2t y 2 +

f (t )]

∂u = 2t cos θ given that u = 2t when ∂t θ = 0. [u =t 2 (cos θ − 1) + 2t ]

Verify that u(θ, t ) =θ 2 + θt is a solution of ∂u ∂u −2 =t. ∂θ ∂t

4.

Verify that u = e−y cos x is a solution of ∂ 2u ∂ 2u + = 0. ∂x 2 ∂ y 2

5.

Solve

6.

Solve

∂ 2φ (3z 2 ) − (x 2 + y 2 + z 2 ) = 5 ∂z 2 (x 2 + y 2 + z 2 ) 2

Thus, ∂ 2 φ ∂ 2 φ ∂ 2 φ (3x 2 ) − (x 2 + y 2 + z 2 ) + 2+ 2 = 5 ∂x 2 ∂y ∂z (x 2 + y 2 + z 2 ) 2

+

solution

3.

5

+

general

Solve

y2 + z2 ) 2

Similarly, it may be shown that

and

the

3

(x 2 + y 2 + z 2 ) +

Determine ∂u = 4t y. ∂y

2.

1

5 2

(x 2

satisﬁes the Laplace equation

Now try the following exercise

1.

+ (x + y + z ) (−1) by the product rule =

=0

Exercise 199 Further problems on the solution of partial differential equations by direct partial integration

3

and

1 x 2 + y2 + z2

5

(x 2 + y 2 + z 2 ) 2

∂ 2φ ∂ 2φ ∂ 2φ + + =0 ∂x 2 ∂ y 2 ∂z 2

= (x 2 + y 2 + z 2 )− 2

x 2 + y2 + z2

+ 3z 2 − (x 2 + y 2 + z 2 )

(3y 2 ) − (x 2 + y 2 + z 2 ) 5

(x 2 + y 2 + z 2 ) 2 (3z 2 ) − (x 2 + y 2 + z 2 ) 5

(x 2 + y 2 + z 2 ) 2

∂2u = 8e y sin 2x given that at y = 0, ∂x∂ y ∂u π = sin x, and at x = , u =2y 2 . ∂x 2

u = −4e y cos 2x − cos x + 4 cos 2x + 2y 2 − 4e y + 4

∂2u = y(4x 2 − 1) given that at x = 0, ∂x 2 ∂u u =sin y and = cos 2y. ∂x 4 x x2 + x cos 2y + sin y − u=y 3 2

518 Higher Engineering Mathematics (c) 7.

8.

9.

10.

∂2u ∂u Solve = sin(x + t) given that =1 ∂x∂t ∂x when t = 0, and when u =2t when x = 0. [u =−sin(x + t) + x + sin x + 2t + sin t ] x Show that u(x, y) = x y + is a solution of y ∂2u ∂2u 2x + y 2 = 2x. ∂x∂ y ∂y Find the particular solution of the differential ∂ 2u equation = cos x cos y given the ini∂x∂ y ∂u = x, and tial conditions that when y =π, ∂x when x = π, u =2 cos y. π2 x2 u = sin x sin y + + 2 cos y − 2 2 Verify that φ(x, y) = x cos y + e x sin y satisﬁes the differential equation ∂2φ ∂2φ + + x cos y = 0. ∂x 2 ∂ y 2

53.4 Some important engineering partial differential equations There are many types of partial differential equations. Some typically found in engineering and science include: (a)

Laplace’s equation, used extensively with electrostatic ﬁelds is of the form: ∂ 2u ∂ 2u ∂ 2u + + = 0. ∂x 2 ∂ y 2 ∂z 2

(d) The transmission equation, where the potential u in a transmission cable is of the form: ∂ 2u ∂2 u ∂u = A +B + Cu where A, B and C are 2 2 ∂x ∂t ∂t constants. Some of these equations are used in the next sections.

53.5

Let u(x, t ) = X (x)T (t ), where X (x) is a function of x only and T (t ) is a function of t only, be a trial solution to ∂ 2u 1 ∂ 2u the wave equation 2 = 2 2 . If the trial solution is ∂x c ∂t ∂ 2u ∂u = X

T . simpliﬁed to u = XT, then = X T and ∂x ∂x 2 ∂ 2u ∂u Also = XT and 2 = XT

. ∂t ∂t ∂ 2u Substituting into the partial differential equation 2 = ∂x 1 ∂ 2u gives: c2 ∂t 2 1 X

T = 2 XT

c Separating the variables gives: X 1 T = 2 X c T

The wave equation, where the equation of motion is given by: 1 ∂2u ∂ 2u = ∂x 2 c2 ∂t 2 T , with T being the tension in a string ρ and ρ being the mass/unit length of the string. where c2 =

(b) The heat conduction equation is of the form: ∂ 2 u 1 ∂u = ∂x 2 c2 ∂t h , with h being the thermal conducσρ tivity of the material, σ the speciﬁc heat of the material, and ρ the mass/unit length of material. where c2 =

Separating the variables

X

1 T

= 2 where μ is a constant. X c T X

Thus, since μ = (a function of x only), it must be X 1 T

independent of t ; and, since μ = 2 (a function of t c T only), it must be independent of x. Let μ =

If μ is independent of x and t , it can only be a conX

then X

= μX or X

− μX = 0 and if stant. If μ = X 1 T

μ= 2 then T

= c2 μT or T

− c2 μT = 0. c T Such ordinary differential equations are of the form found in Chapter 50, and their solutions will depend on whether μ > 0, μ = 0 or μ < 0.

An introduction to partial differential equations

Problem 4. Find the general solution of the following differential equations: (a) X

− 4X =0

y u 5 f (x, t )

Worked Problem 4 will be a reminder of solving ordinary differential equations of this type.

P u(x, t )

(b) T

+ 4T = 0. 0

(a)

m 2 − 4 = 0 i.e. m 2 = 4 from which, m = +2 or m = −2 Thus, the general solution is: X = Ae2x + Be−2x (b) If T

+ 4T = 0 then the auxiliary equation is: m 2 + 4 = 0 i.e. m 2 = −4 from which, √ m = −4 = ± j 2 Thus, the general solution is: T = e0 {A cos 2t + B sin 2t } =A cos2t + B sin2t Now try the following exercise

1. Solve T

= c2 μT given c = 3 and μ = 1. [T = Ae3t + Be−3t ] 2. Solve T

− c2 μT = 0 given c = 3 and μ = −1. [T = A cos 3t + B sin 3t] 3. Solve X

= μX given μ = 1.

X = Aex + Be−x 4. Solve

x

Figure 53.1

ﬁxed. The position of any point P on the string depends on its distance from one end, and on the instant in time. Its displacement u at any time t can be expressed as u = f (x, t ), where x is its distance from 0. The equation of motion is as stated in Section 53.4 (a), 1 ∂2u ∂2u i.e. 2 = 2 2 . ∂x c ∂t The boundary and initial conditions are: (i) The string is ﬁxed at both ends, i.e. x = 0 and x = L for all values of time t . Hence, u(x, t ) becomes: u(0, t ) = 0 for all values of t ≥ 0 u(L , t ) = 0 (ii) If the initial deﬂection of P at t = 0 is denoted by f (x) then u(x, 0) = f (x)

Exercise 200 Further problems on revising the solution of ordinary differential equation

X

− μX

L x

X

− 4X

If =0 then the auxiliary equation (see Chapter 50) is:

519

= 0 given μ = −1. [X = A cos x + B sin x]

(iii) Let the initial velocity of P be g(x), then ∂u = g(x) ∂t t =0 Initially a trial solution of the form u(x, t ) = X (x)T (t ) is assumed, where X (x) is a function of x only and T (t ) is a function of t only. The trial solution may be simpliﬁed to u = XT and the variables separated as explained in the previous section to give: X

1 T

= 2 X c T When both sides are equated to a constant μ this results in two ordinary differential equations: T

− c2 μT = 0 and X

− μX =0

53.6

The wave equation

An elastic string is a string with elastic properties, i.e. the string satisﬁes Hooke’s law. Figure 53.1 shows a ﬂexible elastic string stretched between two points at x = 0 and x = L with uniform tension T . The string will vibrate if the string is displace slightly from its initial position of rest and released, the end points remaining

Three cases are possible, depending on the value of μ.

Case 1: μ > 0 For convenience, let μ = p2 , where p is a real constant. Then the equations X

− p 2 X = 0 and T

− c2 p2 T = 0

520 Higher Engineering Mathematics have solutions: X = Ae px + Be− px and T = Cecpt + De−cpt where A, B, C and D are constants. But X =0 at x = 0, hence 0 = A + B i.e. B = −A and X = 0 at x = L , hence 0 = Ae p L + Be− p L = A(e p L − e− p L ). Assuming (e p L – e− p L ) is not zero, then A = 0 and since B = −A, then B = 0 also. This corresponds to the string being stationary; since it is non-oscillatory, this solution will be disregarded.

nπ Thus sin pL =0 i.e. pL =nπ or p = for inteL ger values of n. Substituting in equation (4) gives: cnπt cnπt nπ x C cos + D sin u = B sin L L L nπ x cnπt cnπt i.e. u = sin An cos + Bn sin L L L

In this case, since μ = p2 = 0, T

= 0 and X

= 0. We will assume that T (t ) = 0. Since X

= 0, X = a and X = ax + b where a and b are constants. But X =0 at x = 0, hence b = 0 and X = ax and X =0 at x = L, hence a = 0. Thus, again, the solution is non-oscillatory and is also disregarded.

(where constant An = BC and Bn = B D). There will be many solutions, depending on the value of n. Thus, more generally, ∞ < cnπ t nπx un (x, t) = An cos sin L L n=1 cnπt + Bn sin (5) L

Case 3: μ < 0

To ﬁnd An and Bn we put in the initial conditions not yet taken into account.

Case 2: μ = 0

For convenience, let μ = − p2 then X

+ p 2 X =0 from which, X = A cos px + B sin px and T

+ c2 p2 T = 0

(i) At t = 0, u(x, 0) = f (x) for 0 ≤ x ≤ L Hence, from equation (5), (1) u(x, 0) = f (x) =

from which,

n=1

T = C cos cpt + D sin cpt

(2)

(ii) Also at t = 0,

(see worked Problem 4 above). Thus, the suggested solution u = XT now becomes: u = {A cos px + B sin px}{C cos cpt + D sin cpt } (3) Applying the boundary conditions: (i) u = 0 when x = 0 for all values of t , thus 0 = {A cos 0 + B sin 0}{C cos cpt i.e.

∂u ∂t

An sin

nπx L

(6)

t =0

= g(x) for 0 ≤ x ≤ L

Differentiating equation (5) with respect to t gives: ∞ ∂u < cnπt nπ x cnπ An − = sin sin ∂t L L L n=1 cnπ cnπt + Bn cos L L and when t = 0,

+ D sin cpt }

0 = A{C cos cpt + D sin cpt }

i.e. g(x) =

+ D sin cpt } = 0)

∞ cπ < nπx Bn n sin L L

(7)

n=1

u = {B sin px}{C cos cpt + D sin cpt }

∞ < nπ x cnπ sin g(x) = Bn L L n=1

from which, A = 0, (since {C cos cpt Hence,

∞

0) s2 + a2

=

!

∞

L{t } =

(s 2 + 4) − s 2 4 = 2 2 2s(s + 4) 2s(s + 4) 2 = 2 s(s + 4)

e−st t 2 dt

t 2 e−st 2t e−st 2e−st = − 3 − −s s2 s

∞ 0

by integration by parts twice, 2 = (0 − 0 − 0) − 0 − 0 − 3 s =

(b) Since cosh 2x = 2 cosh2 x − 1 then 1 cosh2 x = (1 + cosh 2x) from Chapter 5. 2 1 Hence cosh2 3x = (1 + cosh 6x) 2 1 Thus L{cosh 2 3x} = L (1 + cosh 6x) 2

2 (provided s > 0) s3

1 1 = L{1} + L{cosh 6x} 2 2 1 s 1 1 + = 2 s 2 s 2 − 62

(c) From equation (1), L{cosh at } = L

1 at (e + e−at ) , 2

= from Chapter 5

1 1 = L{eat } + L{e−at }, 2 2 equations (2) and (3) 1 1 1 1 = + 2 s −a 2 s − (−a) from (iii) of Table 61.1 1 1 1 + = 2 s −a s +a 1 (s + a) + (s − a) = 2 (s − a)(s + a) =

s s2 − a2

(provided s > a)

Problem 4. Determine the Laplace transforms of: (a) sin2 t (b) cosh2 3x. (a)

1 s 1 − s 2 s 2 + 22 from (i) and (v) of Table 61.1

=

(b) From equation (1), 2

1 2

585

Since cos 2t = 1 −2sin2 t then 1 sin2 t = (1 − cos2t ). Hence, 2 1 2 L{sin t } = L (1 − cos 2t ) 2 1 1 = L{1} − L{cos 2t } 2 2

2s 2 − 36 s2 − 18 = 2s(s 2 − 36) s(s2 − 36)

Problem 5. Find the Laplace transform of 3 sin(ωt + α), where ω and α are constants. Using the compound angle formula for sin(A + B), from Chapter 17, sin(ωt + α) may be expanded to (sin ωt cos α + cos ωt sin α). Hence, L{3sin (ωt + α)} = L{3(sin ωt cos α + cos ωt sin α)} = 3 cosαL{sin ωt } + 3 sin αL{cosωt }, since α is a constant s ω + 3 sin α 2 = 3 cosα 2 s + ω2 s + ω2 from (iv) and (v) of Table 61.1 3 = 2 (ω cos α + s sin α) (s + ω 2 ) Now try the following exercise Exercise 219 Further problems on an introduction to Laplace transforms Determine the Laplace transforms in Problems 1 to 9.

586 Higher Engineering Mathematics 1.

(a) 2t − 3 (b) 5t 2+ 4t − 3 10 4 3 2 3 (a) 2 − (b) 3 + 2 − s s s s s

6. (a) 2 cos2 t (b) 3 sin2 2x 24 2(s 2 + 2) (b) (a) s(s 2 + 4) s(s 2 + 16)

2.

(a)

t3 t2 t5 − 3t + 2 (b) − 2t 4 + 24 15 2 3 2 8 48 1 1 (a) 4 − 2 + (b) 6 − 5 + 3 4s s s s s s

7. (a) cosh2 t (b) 2 sinh2 2θ 16 s2 − 2 (b) (a) s(s 2 − 4) s(s 2 − 16)

3.

(a) 5e3t (b) 2e−2t

4.

(a) 4 sin 3t (b) 3 cos2t (a)

5.

(a) 7 cosh 2x (b)

(a)

1 sinh 3t 3

2 5 (b) s −3 s +2

3s 12 (b) 2 2 s +9 s +4

1 7s (b) 2 (a) 2 s −4 s −9

8. 4 sin(at + b), where a and b are constants. 4 (a cos b + s sin b) s2 + a2 9. 3 cos(ωt − α), where ω and α are constants. 3 (s cos α + ω sin α) s 2 + ω2

10. Show that L(cos2 3t − sin2 3t ) =

s s 2 + 36

Chapter 62

Properties of Laplace transforms 62.1

The Laplace transform of eat f (t)

From Chapter 61, the deﬁnition of the Laplace transform of f (t ) is: ! ∞ L{ f (t )} = e−st f (t ) dt (1) 0

!

∞

Thus L{eat f (t )} =

∞

=

e−st (eat f (t )) dt e

f (t ) dt

(2)

Hence the substitution of (s − a) for s in the transform shown in equation (1) corresponds to the multiplication of the original function f (t ) by eat . This is known as a shift theorem.

62.2 Laplace transforms of the form eat f(t) From equation (2), Laplace transforms of the form eat f (t ) may be deduced. For example: (i) L{eat t n }

page 584.

n!

then L{eat t n } =

ω from (iv) of Table s 2 + ω2

ω from equa(s −a)2 + ω2 tion (2) (provided s > a). then L{eat sin ωt} =

61.1, page 584. −(s−a)

(where a is a real constant)

s n+1

61.1, page 584.

Since L{cosh ωt } =

Since L{t n } =

Since L{sin ωt } =

(iii) L{eat cosh ωt}

!

(ii) L{eat sin ωt}

from (viii) of Table 61.1, n!

from equation (2)

from (ix) of Table

s−a from equa(s − a)2 − ω2 tion (2) (provided s > a). then L{eat cosh ωt} =

A summary of Laplace transforms of the form eat f (t ) is shown in Table 62.1. Table 62.1 Laplace transforms of the form eat f (t ) Function eat f (t ) (a is a real constant) (i) eat t n (ii) eat sin ωt (iii) eat cos ωt (iv) eat sinh ωt

(s − a)n+1 above (provided s > a).

s s 2 − ω2

(v) eat cosh ωt

Laplace transform L{eat f (t )} n! (s − a)n+1 ω (s − a)2 + ω2 s −a (s − a)2 + ω2 ω (s − a)2 − ω2 s −a (s − a)2 − ω2

588 Higher Engineering Mathematics =

Problem 1. Determine (a) L{2t 4e3t } (b) L{4e3t cos 5t }.

= (a) From (i) of Table 62.1,

4! L{2t e } = 2L{t e } = 2 (s − 3)4+1 4 3t

4 3t

= 8L{e3t cos 2t } − 10L{e3t sin 2t } =

(b) From (iii) of Table 62.1,

=

4(s − 3) s2 − 6s +34

=

1 Since cos 2x = 1 −2 sin2 x, sin2 x = (1 − cos 2x). 2 Hence, 1 L 3e− 2 x sin2 x

(a) From (ii) of Table 62.1, L{e

3 3 sin 3t }= = 2 2 (s − (−2)) + 3 (s +2)2 + 9 =

3 3 = s 2 + 4s + 4 + 9 s2 + 4s + 13

(b) From (v) of Table 62.1, L{3eθ cosh 4θ} = 3L{eθ cosh 4θ}= =

3(s − 1) s 2 −2s +1−16

8s − 44 8(s − 3) − 10(2) = 2 (s − 3)2 + 22 s − 6s + 13

Problem 4. Show that 1 48 −2x 2 sin x = L 3e (2s + 1)(4s 2 + 4s + 17)

Problem 2. Determine (a) L{e−2t sin 3t } (b) L{3eθ cosh 4θ}.

−2t

10(2) 8(s − 3) − 2 2 (s − 3) + 2 (s − 3)2 + 22 from (iii) and (ii) of Table 62.1

L{4e3t cos 5t } = 4L{e3t cos 5t } s −3 =4 (s − 3)2 + 52 4(s − 3) s 2 − 6s + 9 + 25

10 s2 + 6s + 5

(b) L{2e3t (4 cos 2t − 5 sin 2t )}

2(4)(3)(2) 48 = = 5 (s − 3) (s − 3)5

=

10 10 = (s + 3)2 − 22 s 2 + 6s+9 − 4

3(s − 1) (s − 1)2 − 42

=

1 1 = L 3e− 2 x (1 − cos 2x) 2 1 1 3 3 = L e− 2 x − L e− 2 x cos 2x 2 2 ⎛ ⎛ ⎞

⎞ 1 s− − ⎟ ⎟ 3⎜ 3⎜ 1 2 ⎟ ⎟− ⎜ = ⎜ ⎜ ⎟ ⎝ ⎠ 2 1 2 2⎝ ⎠ 1 s− − +22 s− − 2 2

3(s − 1) s2 − 2s −15

Problem 3. Determine the Laplace transforms of (a) 5e−3t sinh 2t (b) 2e3t (4 cos 2t − 5 sin 2t ). (a) From (iv) of Table 62.1, L{5e−3t sinh 2t } = 5L{e−3t sinh 2t } 2 =5 (s − (−3))2 − 22

from (iii) of Table 61.1 (page 584) and (iii) of Table 62.1 above, 1 3 s+ 3 2 − = 1 1 2 2 s+ 2 2 s+ +2 2 2 =

3 6s + 3 − 1 2s + 1 4 s2 + s + + 4 4

Properties of Laplace transforms =

3 6s + 3 − 2 2s + 1 4s + 4s + 17

=

3(4s 2 + 4s + 17) − (6s + 3)(2s + 1) (2s + 1)(4s 2 + 4s + 17)

1 7. (a) 2e−t sinh 3t (b) e−3t cosh 2t 4 s +3 6 (b) (a) 2 s + 2s − 8 4(s 2 + 6s + 5) 8. (a) 2et (cos 3t − 3 sin 3t )

12s 2 + 12s + 51 − 12s 2 − 6s − 6s − 3 = (2s + 1)(4s 2 + 4s + 17) =

(b) 3e−2t (sinh 2t − 2 cosh 2t ) −6(s + 1) 2(s − 10) (b) (a) 2 s − 2s + 10 s(s + 4)

48 (2s + 1)(4s2 + 4s + 17)

62.3 The Laplace transforms of derivatives

Now try the following exercise Exercise 220 Further problems on Laplace transforms of the form eat f (t) Determine the Laplace transforms of the following functions: 1. (a) 2t e2t (b) t 2et

2 2 (b) (a) 2 (s − 2) (s − 1)3

1 2. (a) 4t 3e−2t (b) t 4e−3t 2 12 24 (b) (a) (s + 2)4 (s + 3)5 3. (a) et cos t (b) 3e2t sin 2t s −1 6 (a) 2 (b) 2 s − 2s + 2 s − 4s + 8 4. (a) 5e−2t cos 3t (b) 4e−5t sin t 5(s + 2) 4 (a) 2 (b) 2 s + 4s + 13 s + 10s + 26 1 5. (a) 2et sin2 t (b) e3t cos2 t 2 ⎡ 1 s −1 (a) − 2 ⎢ s − 1 s − 2s + 5 ⎢ ⎣ 1 1 s −3 (b) + 2 4 s − 3 s − 6s + 13

589

(a) First derivative Let the ﬁrst derivative of f (t ) be f (t ) then, from equation (1), ! ∞ e−st f (t ) dt L{ f (t )} = 0

From Chapter 43, when integrating by parts ! ! dv du u dt = uv − v dt dt dt ∞ −st

When evaluating 0 e f (t ) dt , let u = e−st and

dv = f (t ) dt

from which,

!

du = −se −st and v = dt ! ∞ e−st f (t ) dt Hence 0

∞

= e−st f (t ) 0 −

!

! ⎤ ⎥ ⎥ ⎦

6. (a) et sinh t (b) 3e2t cosh 4t 1 3(s − 2) (a) (b) 2 s(s − 2) s − 4s − 12

= [0 − f (0)] + s

f (t ) dt = f (t )

∞ 0 ∞

f (t )(−se −st ) dt

e−st f (t ) dt

= − f (0) + sL{ f (t )} assuming e−st f (t ) → 0 as t → ∞, and f (0) is the value of f (t ) at t = 0. Hence, ⎫ L{ f (t)} = sL{ f (t)} − f (0) ⎬ (3) dy or L = sL{ y} − y(0) ⎭ dx where y(0) is the value of y at x = 0.

590 Higher Engineering Mathematics (b) Second derivative

Substituting into equation (3) gives:

Let the second derivative of f (t ) be f

(t ), then from equation (1), L{ f

(t )} =

!

∞

i.e.

e−st f

(t ) dt

Hence

(c) Let f (t ) = e−at then f (t ) = −ae−at and f (0) = 1.

Integrating by parts gives: !

∞

−st

e 0

∞ f (t ) dt = e−st f (t ) 0 + s

!

∞

e

Substituting into equation (3) gives: −st

f (t ) dt

L{−ae−at } = sL{e−at } − 1

−aL{e−at } = sL{e−at } − 1

= [0 − f (0)] + sL{ f (t )}

1 = sL{e−at } + aL{e−at }

assuming e−st f (t ) → 0 as t → ∞, and f (0) is the value of f (t ) at t = 0. Hence { f

(t )} = − f (0) + s[s( f (t )) − f (0)], from equation (3), ⎫ ⎪ ⎪ ⎪ ⎪ = s2 L{ f (t)} − sf (0) − f (0) ⎪ ⎪ ⎬ 2 d y ⎪ or L ⎪ ⎪ dx2 ⎪ ⎪ ⎪ ⎭ 2 = s L{ y} − sy(0) − y (0) L{ f (t)}

i.e.

(4)

dy at x = 0. where y (0) is the value of dx Equations (3) and (4) are important and are used in the solution of differential equations (see Chapter 64) and simultaneous differential equations (Chapter 65). Problem 5. Use the Laplace transform of the ﬁrst derivative to derive: (a) L{k} =

k 2 (b) L{2t } = 2 s s

1 (c) L{e−at } = s +a From equation (3), L{ f (t )} = sL{ f (t )} − f (0). (a) Let f (t ) = k, then f (t ) = 0 and f (0) = k. Substituting into equation (3) gives: L{0} = sL{k} − k k = sL{k} k Hence L{k} = s (b) Let f (t ) = 2t then f (t ) = 2 and f (0) = 0. i.e.

L{2} = sL{2t } − 0 2 = sL{2t } s 2 L{2t}= 2 s

1 = (s + a)L{e−at } Hence L{e−at } =

1 s+a

Problem 6. Use the Laplace transform of the second derivative to derive s L{cos at } = 2 s + a2 From equation (4), L{ f

(t )} = s 2 L{ f (t )} − s f (0) − f (0) Let f (t ) = cos at , then f (t ) = −a sin at and f

(t ) = −a 2 cosat , f (0) = 1 and f (0) = 0 Substituting into equation (4) gives: L{−a 2 cos at } = s 2 {cos at } − s(1) − 0 i.e.

−a 2 L{cos at } = s 2 L{cos at } − s s = (s 2 + a 2 )L{cos at }

Hence

from which, L{cos at } =

s s2 + a2

Now try the following exercise Exercise 221 Further problems on the Laplace transforms of derivatives 1. Derive the Laplace transform of the ﬁrst derivative from the deﬁnition of a Laplace transform. Hence derive the transform L{1} =

1 s

Properties of Laplace transforms Let 2. Use the Laplace transform of the ﬁrst derivative to derive the transforms: 1 6 (b) L{3t 2} = 3 (a) L{eat } = s −a s 3. Derive the Laplace transform of the second derivative from the deﬁnition of a Laplace transform. Hence derive the transform a L{sin at } = 2 s + a2 4. Use the Laplace transform of the second derivative to derive the transforms: a (a) L{sinh at } = 2 s − a2 s (b) L{cosh at } = 2 s − a2

f (t ) = 5 + 2 cos3t

L{ f (t )} = L{5 + 2 cos3t } =

5 2s + 2 s s +9

from (ii) and (v) of Table 61.1, page 584. By the initial value theorem, limit[ f (t )] = limit [sL{ f (t )}] t →0

s→∞

5 2s i.e. limit[5 + 2 cos 3t ]= limit s + s→∞ t →0 s s2 + 9 2s 2 = limit 5 + 2 s→∞ s +9 2∞2 = 5+2 ∞2 + 9 i.e. 7 = 7, which veriﬁes the theorem in this case. 5 + 2(1) = 5 +

i.e.

The initial value of the voltage is thus 7 V.

62.4 The initial and ﬁnal value theorems There are several Laplace transform theorems used to simplify and interpret the solution of certain problems. Two such theorems are the initial value theorem and the ﬁnal value theorem.

(a) The initial value theorem states:

Problem 8. Verify the initial value theorem for the function (2t − 3)2 and state its initial value. Let Let

f (t ) = (2t − 3)2 = 4t 2 − 12t + 9 L{ f (t )} = L(4t 2 − 12t + 9) 2 12 9 =4 3 − 2 + s s s

from (vii), (vi) and (ii) of Table 61.1, page 584. limit [ f (t)]= limit [sL{ f (t)}] s→∞

t→0

For example, if f (t ) = 3e4t then L{3e4t } =

3 s −4

from (iii) of Table 61.1, page 584. By the initial value theorem, 3 limit[3e4t ] = limit s s→∞ t →0 s −4 3 i.e. 3e0 = ∞ ∞−4 i.e.

3 =3, which illustrates the theorem.

Problem 7. Verify the initial value theorem for the voltage function (5 + 2 cos3t ) volts, and state its initial value.

By the initial value theorem, 8 12 9 limit[(2t − 3)2 ] = limit s 3 − 2 + s→∞ t →0 s s s 8 12 = limit 2 − +9 s→∞ s s 8 12 +9 i.e. (0 − 3)2 = 2 − ∞ ∞ i.e. 9 = 9, which veriﬁes the theorem in this case. The initial value of the given function is thus 9.

(b) The ﬁnal value theorem states: limit [f (t)]= limit [sL{ f (t)}] t→∞

s→0

For example, if f (t ) = 3e−4t then: 3 limit[3e−4t ] = limit s t →∞ s→0 s +4

591

592 Higher Engineering Mathematics i.e.

3e−∞ = (0)

3 0+4

i.e. 0 = 0, which illustrates the theorem. Problem 9. Verify the ﬁnal value theorem for the function (2 + 3e−2t sin 4t ) cm, which represents the displacement of a particle. State its ﬁnal steady value. f (t ) = 2 + 3e−2t sin 4t

Let

L{ f (t )} = L{2 + 3e−2t sin 4t } 2 4 = +3 s (s − (−2))2 + 42 12 2 = + s (s + 2)2 + 16 from (ii) of Table 61.1, page 584 and (ii) of Table 62.1 on page 587. By the ﬁnal value theorem, t →∞

s→0

limit[2 + 3e−2t sin 4t ] t →∞

2 12 + s→0 s (s + 2)2 + 16 12s = limit 2 + s→0 (s + 2)2 + 16

Now try the following exercise Exercise 222 Further problems on initial and ﬁnal value theorems 1. State the initial value theorem. Verify the theorem for the functions (a) 3 −4 sin t (b) (t − 4)2 and state their initial values. [(a) 3 (b) 16] 2. Verify the initial value theorem for the voltage functions: (a) 4 +2 cos t (b) t − cos 3t and state their initial values. [(a) 6 (b) −1]

limit[ f (t )] = limit[sL{ f (t )}] i.e.

The initial and ﬁnal value theorems are used in pulse circuit applications where the response of the circuit for small periods of time, or the behaviour immediately after the switch is closed, are of interest. The ﬁnal value theorem is particularly useful in investigating the stability of systems (such as in automatic aircraft-landing systems) and is concerned with the steady state response for large values of time t , i.e. after all transient effects have died away.

= limit s

i.e. 2 + 0 = 2 +0 i.e. 2 = 2, which veriﬁes the theorem in this case. The ﬁnal value of the displacement is thus 2 cm.

3. State the ﬁnal value theorem and state a practical application where it is of use. Verify the theorem for the function 4 +e−2t (sin t + cos t ) representing a displacement and state its ﬁnal value. [4] 4. Verify the ﬁnal value theorem for the function 3t 2e−4t and determine its steady state value. [0]

Chapter 63

Inverse Laplace transforms 63.1 Deﬁnition of the inverse Laplace transform If the Laplace transform of a function f (t ) is F(s), i.e. L{ f (t )} = F(s), then f (t ) is called the inverse Laplace transform of F(s) and is written as f (t ) = L−1{F(s)}. 1 −1 1 = 1. For example, since L{1} = then L s s a Similarly, since L{sin at } = 2 then s + a2 −1

L

a s2 + a2

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

= sin at, and so on.

Tables of Laplace transforms, such as the tables in Chapters 61 and 62 (see pages 584 and 587) may be used to ﬁnd inverse Laplace transforms. However, for convenience, a summary of inverse Laplace transforms is shown in Table 63.1.

Problem 1. Find the following inverse Laplace transforms: 1 5 (a) L−1 2 (b) L−1 s +9 3s − 1 From (iv) of Table 63.1, a −1 L = sin at, s2 + a2

L−1 {F(s)} = f (t)

(i)

1 s

1

(ii)

k s

k

(iii)

1 s −a

eat

(iv)

a s 2 +a 2

sin at

(v)

s s 2 +a 2

cosat

(vi)

1 s2

t

(vii)

2! s3

t2

(viii)

n! s n+1

tn

(ix)

a s 2 −a 2

sinh at

(x)

s s 2 −a 2

cosh at

(xi)

n! (s − a)n+1

eat t n

(xii)

ω (s − a)2 + ω2

eat sinωt

(xiii)

s−a (s − a)2 + ω2

eat cosωt

(xiv)

ω (s − a)2 − ω2

eat sinhωt

(xv)

s−a (s − a)2 − ω2

eat coshωt

63.2 Inverse Laplace transforms of simple functions

(a)

Table 63.1 Inverse Laplace transforms

594 Higher Engineering Mathematics Hence L−1

1 1 −1 = L s2 + 9 s 2 + 32 3 1 −1 = L 3 s 2 + 32 =

(b) L−1

1 sin 3t 3

⎧ ⎪ ⎪ ⎨

⎫ ⎪ ⎪ ⎬

5 5 = L−1 1 ⎪ ⎪ 3s − 1 ⎪ ⎪ ⎩3 s − ⎭ 3 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎬ 5 1 1 5 −1 ⎨ = e3t = L ⎪ 1 ⎪ 3 3 ⎪ ⎪ ⎭ ⎩ s− 3

(b) L−1

4s s −1 = 4L s 2 − 16 s 2 − 42 = 4 cosh 4t, from (x) of Table 63.1

Problem 4. Find 3 2 (b) L−1 (a) L−1 2 s −7 (s − 3)5 (a) From (ix) of Table 63.1, a −1 L = sinh at s2 − a2 Thus L−1

from (iii) of Table 63.1 Problem 2. Find the following inverse Laplace transforms: 6 3 −1 −1 (a) L (b) L 3 s s4 2 (a) From (vii) of Table 63.1, =t2 s3 6 2 −1 −1 Hence L = 3L = 3t 2 . 3 s s3 L−1

(b) From (viii) of Table 63.1, if s is to have a power of 4 then n = 3. 3! 6 −1 3 −1 Thus L = t i.e. L = t3 s4 s4 Hence

L−1

3 1 −1 6 1 = L = t3 . 4 4 s 2 s 2

Problem 3. Determine 7s 4s (a) L−1 2 (b) L−1 2 s +4 s − 16 (a) L−1

7s s −1 = 7L = 7 cos 2t, s2 + 4 s 2 + 22 from (v) of Table 63.1

3 1 −1 = 3L √ s2 − 7 s 2 − ( 7)2 5 6 √ 7 3 −1 √ =√ L 7 s 2 − ( 7)2 √ 3 = √ sinh 7t 7

(b) From (xi) of Table 63.1, n! −1 L = eat t n (s − a)n+1 1 1 Thus L−1 = eat t n n+1 (s − a) n! 2 −1 and comparing with L shows that (s − 3)5 n = 4 and a = 3. Hence L

−1

2 (s − 3)5

1 = 2L (s − 3)5 1 3t 4 1 =2 e t = e3t t 4 4! 12

Problem 5. Determine 3 −1 (a) L s 2 − 4s + 13 2(s + 1) −1 (b) L s 2 + 2s + 10

−1

595

Inverse Laplace transforms (a) L

−1

3 3 −1 =L s 2 − 4s + 13 (s − 2)2 + 32 = e2t sin 3t,

(b) L−1

from (xii) of Table 63.1 2(s + 1) 2(s + 1) −1 =L s 2 + 2s + 10 (s + 1)2 + 32

= 2e−t cos 3t,

Now try the following exercise Exercise 223 Further problems on inverse Laplace transforms of simple functions Determine the inverse Laplace transforms of the following: 1. (a)

7 2 (b) s s −5

[(a) 7 (b) 2e5t ]

from (xiii) of Table 63.1 2. (a)

Problem 6. Determine 5 (a) L−1 2 s + 2s − 3 4s − 3 −1 (b) L s 2 − 4s − 5 (a) L

−1

3. (a)

5 5 −1 =L s 2 + 2s − 3 (s + 1)2 − 22 ⎧ ⎫ 5 ⎪ ⎪ ⎨ ⎬ (2) −1 2 =L 2 2 ⎪ ⎩ (s + 1) − 2 ⎪ ⎭

4. (a)

2s 3 (b) 2 2s + 1 s +4 3 −1t 2 (b) 2 cos2t (a) e 2 1 s 2 + 25

(b)

5s 2s 2 + 18

5 = e−t sinh 2t, 2 from (xiv) of Table 63.1 4s − 3 4s − 3 −1 −1 =L (b) L s 2 − 4s − 5 (s − 2)2 − 32 4(s − 2) + 5 = L−1 (s − 2)2 − 32 4(s − 2) = L−1 (s − 2)2 − 32 5 + L−1 (s − 2)2 − 32 ⎧ ⎫ 5 ⎪ ⎪ ⎨ ⎬ (3) 2t −1 3 = 4e cosh 3t + L 2 2 ⎪ ⎩ (s − 2) − 3 ⎪ ⎭

from (xv) of Table 63.1 = 4e2t cosh 3t +

5 2t e sinh 3t, 3 from (xiv) of Table 63.1

5. (a)

6. (a)

7. (a)

8. (a)

5 8 (b) 4 s3 s

4 s2 + 9

(b)

1 4 (a) sin 5t (b) sin 3t 5 3

6 s2

5 (a) cos 3t (b) 6t 2 (a)

4 5 2 t (b) t 3 2 3

3s 7 (b) 2 1 2 s − 16 s −8 2 7 (a) 6 cosh 4t (b) sinh 4t 4 4 15 (b) 3s 2 − 27 (s − 1)3 5 (a) sinh 3t (b) 2 et t 2 3 1 3 (b) (s + 2)4 (s − 3)5 1 1 (a) e−2t t 3 (b) e3t t 4 6 8

596 Higher Engineering Mathematics s +1 3 9. (a) 2 (b) 2 s + 2s + 10 s + 6s + 13 3 (a) e−t cos 3t (b) e−3t sin 2t 2 10. (a)

2(s − 3) s 2 − 6s + 13

(b)

7 s 2 − 8s + 12

7 4t 3t (a) 2e cos 2t (b) e sinh 2t 2

11. (a)

2s + 5 3s + 2 (b) 2 s 2 + 4s − 5 s − 8s + 25 ⎤ ⎡ 1 (a) 2e−2t cosh 3t + e−2t sinh 3t ⎥ ⎢ 3 ⎦ ⎣ 14 4t 4t (b) 3e cos 3t + e sin 3t 3

When s = 2, 3 =3 A, from which, A = 1. When s = −1, −9 = −3B, from which, B = 3. 4s − 5 −1 Hence L s2 − s − 2 1 3 −1 ≡L + s −2 s +1 1 3 = L−1 + L−1 s −2 s +1 = e2t + 3e−t , from (iii) of Table 63.1 Problem 8.

Find L−1

3s 3 + s 2 + 12s + 2 (s − 3)(s + 1)3

3s 3 + s 2 + 12s + 2 (s − 3)(s + 1)3 A D B C + + + 2 s − 3 s + 1 (s + 1) (s + 1)3 A(s + 1)3 + B(s − 3)(s + 1)2 + C(s − 3)(s + 1) + D(s − 3) ≡ (s − 3)(s + 1)3 ≡

63.3 Inverse Laplace transforms using partial fractions Sometimes the function whose inverse is required is not recognisable as a standard type, such as those listed in Table 63.1. In such cases it may be possible, by using partial fractions, to resolve the function into simpler fractions which may be inverted on sight. For example, the function, F(s) =

2s − 3 s(s − 3)

Problem 7.

Determine L−1

4s − 5 2 s −s −2

4s − 5 4s − 5 A B ≡ ≡ + s2 − s − 2 (s − 2)(s + 1) (s − 2) (s + 1) A(s +1) + B(s −2) ≡ (s − 2)(s + 1) Hence 4s − 5 ≡ A(s + 1) + B(s − 2).

3s 3 + s 2 + 12s + 2 ≡ A(s + 1)3 + B(s − 3)(s + 1)2 + C(s − 3)(s + 1) + D(s − 3) When s = 3, 128 =64 A, from which, A = 2. When s = −1, −12 =−4D, from which, D = 3.

cannot be inverted on sight from Table 63.1. However, 2s − 3 1 1 by using partial fractions, ≡ + which s(s − 3) s s − 3 may be inverted as 1 + e3t from (i) and (iii) of Table 61.1. Partial fractions are discussed in Chapter 2, and a summary of the forms of partial fractions is given in Table 2.1 on page 13.

Hence

Equating s 3 terms gives: 3 = A + B, from which, B = 1. Equating constant terms gives: 2 = A − 3B − 3C − 3D, i.e.

2 = 2 − 3 − 3C − 9,

from which, 3C = −12 and C = − 4 Hence 3 2 3s + s + 12s + 2 L−1 (s − 3)(s + 1)3 2 3 1 4 + ≡ L−1 + − s − 3 s + 1 (s + 1)2 (s + 1)3 3 = 2e3t + e−t − 4e−t t + e−t t 2 , 2 from (iii) and (xi) of Table 63.1

Inverse Laplace transforms

7s + 13 s(s 2 + 4s + 13) −s + 3 −1 1 ≡L + s s 2 + 4s + 13 1 −s + 3 + L−1 ≡ L−1 s (s + 2)2 + 32 −1 1 −1 −(s + 2) + 5 ≡L +L s (s + 2)2 + 32 s +2 −1 1 −1 −L ≡L s (s + 2)2 + 32 5 + L−1 (s + 2)2 + 32

Hence L−1

Problem 9. Determine 5s 2 + 8s − 1 L−1 (s + 3)(s 2 + 1) 5s 2 + 8s − 1 A Bs + C ≡ + 2 2 (s + 3)(s + 1) s + 3 (s + 1) ≡

A(s 2 + 1) + (Bs + C)(s + 3) (s + 3)(s 2 + 1)

Hence 5s 2 + 8s − 1 ≡ A(s 2 + 1) + (Bs + C)(s + 3). When s = −3, 20 =10 A, from which, A = 2. Equating s 2 terms gives: 5 = A + B, from which, B = 3, since A = 2. Equating s terms gives: 8 = 3B + C, from which, C = −1, since B = 3. 5s 2 + 8s − 1 −1 Hence L (s + 3)(s 2 + 1)

2 3s − 1 + 2 s +3 s +1 2 3s + L−1 2 ≡ L−1 s +3 s +1 ≡ L−1

− L−1

5 ≡ 1 − e−2t cos 3t + e−2t sin 3t 3 from (i), (xiii) and (xii) of Table 63.1

Now try the following exercise

1

s2 + 1

Use partial fractions to ﬁnd the inverse Laplace transforms of the following functions: 1.

11 −3s s 2 + 2s − 3

2.

2s 2 − 9s − 35 (s + 1)(s − 2)(s + 3)

[4e−t − 3e2t + e−3t ]

3.

5s 2 − 2s − 19 (s + 3)(s − 1)2

[2e−3t + 3et − 4et t ]

4.

3s 2 + 16s + 15 (s + 3)3

[e−3t (3 − 2t − 3t 2)]

= 2e−3t + 3 cost − sin t, from (iii), (v) and (iv) of Table 63.1 Problem 10. Find

L−1

7s + 13 s(s 2 + 4s + 13)

7s + 13 A Bs + C ≡ + 2 s(s 2 + 4s + 13) s s + 4s + 13 A(s 2 + 4s + 13) + (Bs + C)(s) ≡ s(s 2 + 4s + 13)

5.

Hence 7s + 13 ≡ A(s 2 + 4s + 13) + (Bs + C)(s). When s = 0, 13 =13 A, from which, A = 1. Equating B = −1.

s2

Exercise 224 Further problems on inverse Laplace transforms using partial fractions

terms gives: 0 = A + B, from which,

Equating s terms gives: 7 =4 A + C, from which, C = 3.

597

6.

[2et − 5e−3t ]

7s 2 + 5s + 13 (s 2 + 2)(s + 1) √ √ 3 2 cos 2t + √ sin 2t + 5e−t 2 3 +6s + 4s 2 − 2s 3 s 2 (s 2 + 3)

√ √ √ [2 + t + 3 sin 3t − 4 cos 3t ]

598 Higher Engineering Mathematics

7.

26 −s 2 s(s 2 + 4s + 13) 2 2 − 3e−2t cos 3t − e−2t sin 3t 3

63.4

Poles and zeros

It was seen in the previous section that Laplace transφ(s) forms, in general, have the form f (s) = . This is θ(s) the same form as most transfer functions for engineering systems, a transfer function being one that relates the response at a given pair of terminals to a source or stimulus at another pair of terminals. Let a function in the s domain be given by: φ(s) f (s) = where φ(s) is of less (s − a)(s − b)(s − c) degree than the denominator. Poles: The values a, b, c, … that makes the denominator zero, and hence f (s) inﬁnite, are called the system poles of f (s). If there are no repeated factors, the poles are simple poles. If there are repeated factors, the poles are multiple poles. Zeros: Values of s that make the numerator φ(s) zero, and hence f (s) zero, are called the system zeros of f (s). s −4 has simple poles at s = −1 (s + 1)(s − 2) s +3 has a and s = +2, and a zero at s = 4 (s + 1)2 (2s + 5) 5 simple pole at s = − and double poles at s = −1, and 2 s +2 a zero at s = −3 and has simple s(s − 1)(s + 4)(2s + 1) 1 poles at s = 0, +1, −4, and − and a zero at s = −2 2 For example:

The location of a pole in the s-plane is denoted by a cross (×) and the location of a zero by a small circle (o). This is demonstrated in the following examples. From the pole-zero diagram it may be determined that the magnitude of the transfer function will be larger when it is closer to the poles and smaller when it is close to the zeros. This is important in understanding what the system does at various frequencies and is crucial in the study of stability and control theory in general. Problem 11. R(s) =

Determine for the transfer function:

400 (s + 10) s (s + 25)(s 2 + 10s + 125)

(a) the zero and (b) the poles. Show the poles and zero on a pole-zero diagram. (a) For the numerator to be zero, (s + 10) = 0. Hence, s = −10 is a zero of R(s). (b) For the denominator to be zero, s = 0 or s = −25 or s 2 + 10s + 125 =0. Using the quadratic formula. √ −10 ± 102 −4(1)(125) −10 ± −400 = s= 2 2 =

−10 ± j 20 2

= (−5 ± j 10) Hence, poles occur at s = 0, s =−25, (−5 + j10) and (−5 −j10) The pole-zero diagram is shown in Figure 63.1. j

j10

225

220

215

210

25

Pole-zero diagram The poles and zeros of a function are values of complex frequency s and can therefore be plotted on the complex frequency or s-plane. The resulting plot is the pole-zero diagram or pole-zero map. On the rectangular axes, the real part is labelled the σ -axis and the imaginary part the jω-axis.

2j10

Figure 63.1

Inverse Laplace transforms Now try the following exercise

Problem 12. Determine the poles and zeros for the function: F(s) =

(s + 3)(s − 2) (s + 4)(s 2 + 2s + 2)

Exercise 225 and zeros

Further problems on poles

and plot them on a pole-zero map. 1. Determine For the numerator to be zero, (s + 3) =0 and (s − 2) = 0, hence zeros occur at s = −3 and at s = +2 Poles occur when the denominator is zero, i.e. when (s + 4) = 0, i.e. s = −4, and when s 2 + 2s + 2 = 0, i.e. s =

−2 ±

√ 22 − 4(1)(2) − 2 ± −4 = 2 2

−2 ± j2 = = (−1 +j) or (−1 −j) 2 The poles and zeros are shown on the pole-zero map of F(s) in Figure 63.2.

j

23

22

21

function:

(a) the zero and (b) the poles. Show the poles and zeros on a pole-zero diagram. (a) s = −4 (b) s = 0, s = −2, s = 4 + j 3, s= 4 − j 3 2. Determine the poles and zeros for the function: (s − 1)(s + 2) F(s) = and plot them on (s + 3)(s 2 − 2s + 5) a pole-zero map. poles at s = −3, s = 1 + j 2, s = 1 − j 2, zeros at s = +1, s = −2 s −1 (s + 2)(s 2 + 2s + 5) determine the poles and zeros and show them on a pole-zero diagram. ⎡ ⎤ poles at s = −2, s = −1 + j 2, ⎣ ⎦ s = −1 − j 2, zero at s = 1

3. For the function G(s) =

j

24

for the transfer 50 (s + 4) R(s) = s (s + 2)(s 2 − 8s + 25)

1

2

3

2j

Figure 63.2

It is seen from these problems that poles and zeros are always real or complex conjugate.

4. Find the poles and zeros for the transfer funcs 2 − 5s − 6 tion: H (s) = and plot the results in s(s 2 + 4) the s-plane. poles at s = 0, s = + j 2, s = − j 2, zeros at s = −1, s = 6

599

Chapter 64

The solution of differential equations using Laplace transforms 64.1

Introduction

An alternative method of solving differential equations to that used in Chapters 46 to 51 is possible by using Laplace transforms.

64.2

Procedure to solve differential equations by using Laplace transforms

(i) Take the Laplace transform of both sides of the differential equation by applying the formulae for the Laplace transforms of derivatives (i.e. equations (3) and (4) of Chapter 62) and, where necessary, using a list of standard Laplace transforms, such as Tables 61.1 and 62.1 on pages 584 and 587. (ii) Put in the given initial conditions, i.e. y(0) and y (0). (iii) Rearrange the equation to make L{y} the subject. (iv) Determine y by using, where necessary, partial fractions, and taking the inverse of each term by using Table 63.1 on page 593.

64.3

Worked problems on solving differential equations using Laplace transforms

Problem 1. Use Laplace transforms to solve the differential equation dy d2 y − 3y = 0, given that when x = 0, 2 2 +5 dx dx dy y = 4 and = 9. dx This is the same problem as Problem 1 of Chapter 50, page 478 and a comparison of methods can be made. Using the above procedure: dy d2 y − 3L{y} = L{0} + 5L (i) 2L dx 2 dx

2[s 2 L{y} − sy(0) − y (0)] + 5[sL{y} − y(0)] − 3L{y} = 0, from equations (3) and (4) of Chapter 62.

The solution of differential equations using Laplace transforms (ii)

y(0) = 4 and y (0) = 9 Thus 2[s 2 L{y} − 4s − 9] + 5[sL{y} − 4] − 3L{y} = 0 i.e.

This is the same as Problem 3 of Chapter 50, page 479. Using the above procedure: 2 d x dy (i) L + 13L{y} = L{0} + 6L d y2 dx

2s 2 L{y} − 8s − 18 + 5sL{y} − 20

+ 6[sL{y} − y(0)] + 13L{y} = 0,

(iii) Rearranging gives:

from equations (3) and (4) of Chapter 62.

(2s 2 + 5s − 3)L{y} = 8s + 38 8s + 38 i.e. L{y} = 2 2s + 5s − 3 y = L−1

8s + 38 2s 2 + 5s − 3

[s 2 L{y} − sy(0) − y (0)]

Hence

− 3L{y} = 0

(iv)

601

(ii)

y(0) = 3 and y (0) = 7 Thus s 2 L{y} − 3s − 7 + 6sL{y}

− 18 + 13L{y} = 0 (iii) Rearranging gives:

8s + 38 8s + 38 ≡ + 5s − 3 (2s − 1)(s + 3)

(s 2 + 6s + 13)L{y} = 3s + 25

2s 2

L{y} =

A B ≡ + 2s − 1 s + 3

i.e.

A(s + 3) + B(2s − 1) ≡ (2s − 1)(s + 3)

y = L−1

Hence 8s + 38 = A(s + 3) + B(2s − 1). 1 1 When s = , 42 =3 A, from which, A = 12. 2 2 When s = −3, 14 =−7B, from which, B = −2. 8s + 38 −1 Hence y = L 2s 2 + 5s − 3 12 2 −1 =L − 2s − 1 s + 3 6 5 12 2 −1 −1 =L −L s +3 2 s − 12 1

Hence y = 6e 2 x − 2e−3x , from (iii) of Table 63.1. Problem 2. Use Laplace transforms to solve the differential equation: dy d2 y +6 + 13y = 0, given that when x = 0, y = 3 2 dx dx dy and = 7. dx

(iv)

= L−1 =L

−1

3s + 25 s 2 + 6s + 13

3s + 25 + 6s + 13

3s + 25 (s + 3)2 + 22 3(s + 3) + 16 (s + 3)2 + 22

3(s + 3) =L (s + 3)2 + 22 8(2) + L−1 (s + 3)2 + 22 −1

s2

= 3e−3t cos2t + 8e−3t sin 2t, from (xiii) and (xii) of Table 63.1 Hence y = e−3t (3 cos 2t + 8 sin 2t) Problem 3. Use Laplace transforms to solve the differential equation: d2 y dy −3 = 9, given that when x = 0, y = 0 and 2 dx dx dy = 0. dx This is the same problem as Problem 2 of Chapter 51, page 485. Using the procedure:

602 Higher Engineering Mathematics

d2 y dy (i) L = L{9} − 3L dx 2 dx Hence [s 2 L{y} − sy(0) − y (0)]

Using the procedure: 2 dy d y (i) L + 10L{y} = L{ e2x + 20} − 7L dx 2 dx Hence [s 2 L{y} − sy(0) − y (0)] − 7[sL{y} 1 20 − y(0)] + 10L{y} = + s −2 s

9 −3[sL{y} − y(0)] = s y(0) = 0 and y (0) = 0

(ii)

(ii) Hence s 2 L{y} − 3sL{y} =

9 s

(iii) Rearranging gives: 9 (s 2 − 3s)L{y} = s 9 9 i.e. L{y} = = s(s 2 − 3s) s 2 (s − 3) y = L−1

(iv)

9 2 s (s − 3)

+ 10L{y} = (iii) (s 2 − 7s + 10)L{y} =

9 C A B ≡ + 2+ − 3) s s s −3

s 2 (s

≡

A(s)(s − 3) + B(s s 2 (s − 3)

When s = 0, 9 =−3B, from which, B = −3. When s = 3, 9 =9C, from which, C = 1. Equating s 2 terms gives: 0 = A + C, from which, A = −1, since C = 1. Hence, L−1

9 1 3 1 −1 =L − − 2+ s 2 (s − 3) s s s −3 = −1 − 3x + e3x , from (i),

(iv)

≡

3(21s − 40) − s(s − 2) 3s(s − 2)

=

−s 2 + 65s − 120 3s(s − 2)

A B C D + + + s s − 5 s − 2 (s − 2)2 A(s − 5)(s − 2)2 + B(s)(s − 2)2

≡

+ C(s)(s − 5)(s − 2) + D(s)(s − 5) s(s − 5)(s − 2)2

Hence

d2 y

−s 2 + 65s − 120

dy + 10y = e2x + 20, given that when dx dx dy 1 x = 0, y = 0 and =− dx 3

=

−s 2 + 65s − 120 s(s − 5)(s − 2)2

Problem 4. Use Laplace transforms to solve the differential equation: −7 2

21s − 40 1 − s(s − 2) 3

−s 2 + 65s − 120 3s(s − 2)(s 2 − 7s + 10) −s 2 + 65s − 120 1 = 3 s(s − 2)(s − 2)(s − 5) 1 −s 2 + 65s − 120 = 3 s(s − 5)(s − 2)2 2 −s + 65s − 120 1 y = L−1 3 s(s − 5)(s − 2)2

(vi) and (iii) of Table 63.1. i.e. y = e3x − 3x −1

21s − 40 s(s − 2)

Hence L{y} =

− 3) + Cs 2

Hence 9 ≡ A(s)(s − 3) + B(s − 3) + Cs 2 .

1 3 1 2 Hence s L{y} − 0 − − − 7sL{y} + 0 3 y(0) = 0 and y (0) = −

≡A(s − 5)(s − 2)2 + B(s)(s − 2)2 + C(s)(s − 5)(s − 2) + D(s)(s − 5)

The solution of differential equations using Laplace transforms When s = 0, −120 = − 20 A, from which, A = 6.

Hence

E = A(R + Ls) + Bs

When s = 5, 180 =45B, from which, B = 4.

When

s = 0, E = AR,

When s = 2, 6 =−6D, from which, D = −1.

from which,

A=

Equating s 3 terms gives: 0 = A + B + C, from which, C = −10. 2 −s + 65s − 120 1 Hence L−1 3 s(s − 5)(s − 2)2 1 −1 6 4 10 1 = L + − − 3 s s − 5 s − 2 (s − 2)2

Problem 5. The current ﬂowing in an electrical circuit is given by the differential equation Ri + L(di/dt ) = E, where E, L and R are constants. Use Laplace transforms to solve the equation for current i given that when t = 0, i = 0. Using the procedure: di = L{E} (i) L{Ri} + L L dt

from which,

B =−

L−1

Hence current i = E s

(ii) i(0) = 0, hence RL{i} + LsL{i} =

E s

(iii) Rearranging gives: E s

E s(R + Ls) E (iv) i = L−1 s(R + Ls) i.e. L{i} =

E A B ≡ + s(R + Ls) s R + Ls ≡

R R s =− , E = B − L L

A(R + Ls) + Bs s(R + Ls)

EL R

E s(R + Ls) −E L/R −1 E/R =L + s R + Ls E EL = L−1 − Rs R(R + Ls) ⎧ ⎛ ⎞⎫ ⎪ ⎪ ⎨ E 1 E ⎬ ⎜ 1 ⎟ = L−1 − ⎝ ⎠ ⎪ ⎪ R R ⎩R s +s ⎭ L ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ E −1 ⎨ 1 1 = L − ⎪ R ⎪ R s ⎪ ⎪ ⎩ ⎭ s+ L

10 x 4 Thus y = 2 + e5x − e2x − e2x 3 3 3

(R + Ls)L{i} =

When

Hence

1 = [6 + 4 e5x − 10 e2x − x e2x ] 3

i.e. RL{i} + L[sL{i} − i(0)] =

E R

Rt E 1 − e− L R

Now try the following exercise Exercise 226 Further problems on solving differential equations using Laplace transforms 1.

A ﬁrst order differential equation involving current i in a series R − L circuit is given by: di E + 5i = and i = 0 at time t = 0. dt 2 Use Laplace transforms to solve for i when (a) E = 20 (b) E = 40 e−3t and (c) E = 50 sin 5t . ⎤ ⎡ (a) i = 2(1 − e−5t ) ⎥ ⎢(b) i = 10( e−3t − e−5t ) ⎥ ⎢ ⎦ ⎣ 5 −5t (c) i = ( e − cos 5t + sin 5t ) 2

603

604 Higher Engineering Mathematics In Problems 2 to 9, use Laplace transforms to solve the given differential equations. 2.

9

7.

dy d2 y − 24 + 16y = 0, given y(0) = 3 and dt 2 dt

y (0) = 3.

4

y = (3 − t ) e 3 t 8.

3.

d2 x + 100x = 0, given x(0) = 2 and dt 2 [x = 2 cos10t ] x (0) = 0.

4.

d2 i di + 1000 + 250000i = 0, given 2 dt dt i(0) = 0 and i (0) = 100. [i = 100t e−500t ]

5.

d2 x dx +6 + 8x = 0, given x(0) = 4 and dt 2 dt

x (0) = 8. [x = 4(3e−2t − 2e−4t )]

6.

dy 2 d2 y −2 + y = 3 e4x , given y(0) = − dx 2 dx 3 1 and y (0) = 4 3 1 4x x y = (4x − 1) e + e 3

d2 y + 16y = 10 cos4x, given y(0) = 3 and dx 2

y (0) = 4. 5 y = 3 cos4x + sin 4x + x sin 4x 4 d2 y dy + − 2y = 3 cos3x − 11 sin 3x, given dx 2 dx y(0) = 0 and y (0) = 6 [ y = ex − e−2x + sin 3x]

9.

d2 y dy −2 + 2y = 3 e x cos 2x, given 2 dx dx y(0) = 2 and y (0) = 5

y = 3e x (cos x + sin x) − ex cos 2x

10. Solve, using Laplace transforms, Problems 4 to 9 of Exercise 187, page 480 and Problems 1 to 5 of Exercise 188, page 482. 11. Solve, using Laplace transforms, Problems 3 to 6 of Exercise 189, page 486, Problems 5 and 6 of Exercise 190, page 488, Problems 4 and 7 of Exercise 191, page 490 and Problems 5 and 6 of Exercise 192, page 492.

Chapter 65

The solution of simultaneous differential equations using Laplace transforms 65.1

Introduction

It is sometimes necessary to solve simultaneous differential equations. An example occurs when two electrical circuits are coupled magnetically where the equations relating the two currents i1 and i2 are typically:

L1

di1 di2 +M + R1 i1 = E 1 dt dt

L2

di2 di1 +M + R2 i2 = 0 dt dt

where L represents inductance, R resistance, M mutual inductance and E 1 the p.d. applied to one of the circuits.

65.2 Procedure to solve simultaneous differential equations using Laplace transforms (i) Take the Laplace transform of both sides of each simultaneous equation by applying the formulae for the Laplace transforms of derivatives (i.e. equations (3) and (4) of Chapter 62, page 589) and using a list of standard Laplace transforms, as in Table 61.1, page 584 and Table 62.1, page 587.

(ii) Put in the initial conditions, i.e. x(0), y(0), x (0), y (0). (iii) Solve the simultaneous equations for L{y} and L{x} by the normal algebraic method. (iv) Determine y and x by using, where necessary, partial fractions, and taking the inverse of each term.

65.3 Worked problems on solving simultaneous differential equations by using Laplace transforms Problem 1. Solve the following pair of simultaneous differential equations dy +x =1 dt dx − y + 4et = 0 dt given that at t = 0, x = 0 and y = 0. Using the above procedure: dy + L{x} = L{1} (i) L dt

(1)

606 Higher Engineering Mathematics L

dx − L{y} + 4L{et } = 0 dt

(2)

Hence −4s 2 + s − 1 = A(s − 1)(s 2 + 1) + Bs(s 2 + 1)

Equation (1) becomes:

+ (Cs + D)s(s − 1)

1 [sL{y} − y(0)] + L{x} = s

(1 )

from equation (3), page 589 and Table 61.1, page 584.

When s = 0, −1 = −A

hence A = 1

When s = 1, −4 = 2B

hence B =−2

Equating s 3 coefﬁcients:

Equation (2) becomes: [sL{x} − x(0)] − L{y} = − (ii)

4 s −1

0 = A + B + C hence C = 1

(2 )

x(0) = 0 and y(0) = 0 hence

(since A = 1 and B = −2) Equating s 2

Equation (1 ) becomes: sL{y} + L{x} =

−4 = −A + D − C hence D =−2

1 s

(since A = 1 and C = 1)

(1

)

and equation (2 ) becomes:

Thus L{x} =

4 s −1 4 or −L{y} + sL{x} = − s −1 sL{x} − L{y} = −

(iii)

1 × equation (1

)

and

= (2

)

s × equation (2

)

gives:

1 sL{y} + L{x} = s −sL{y} + s 2 L{x} = −

4s s −1

(4)

=L

−1

+s −1 s(s − 1)(s 2 + 1)

(5)

−4s 2 + s − 1 A B Cs + D ≡ + + 2 2 s(s − 1)(s + 1) s (s − 1) (s + 1) A(s − 1)(s 2 + 1) + Bs(s 2 + 1) s(s − 1)(s 2 + 1)

1 2 s −2 − + s (s − 1) (s 2 + 1)

1 2 s 2 − + − s (s − 1) (s 2 + 1) (s 2 + 1)

x = 1 −2et + cos t − 2 sin t,

y=

Using partial fractions

+ (Cs + D)s(s − 1)

2 s −2 1 − + s (s − 1) (s 2 + 1)

dx − y + 4 et = 0 dt from which,

−4s 2 + s − 1 s(s − 1) −4s 2

−4s 2 + s − 1 s(s − 1)(s 2 + 1)

from Table 63.1, page 593 From the second equation given in the question,

1 4s − s s −1 (s − 1) − s(4s) = s(s − 1)

=

x =L

−1

i.e.

(s 2 + 1)L{x} =

from which, L{x} =

(iv) Hence

(3)

Adding equations (3) and (4) gives:

=

coefﬁcients:

=

dx + 4 et dt d (1 − 2 et + cos t − 2 sin t ) + 4 et dt

= −2 et − sin t − 2 cos t + 4 et i.e. y = 2et − sin t − 2 cos t [Alternatively, to determine equations (1

) and (2

)]

y,

return

to

607

The solution of simultaneous differential equations using Laplace transforms and equation (2 ) becomes

Problem 2. Solve the following pair of simultaneous differential equations

(2s − 1)L{y} − 2(3) − sL{x }

dx dy − 5 + 2x = 6 dt dt dy dx − − y = −1 2 dt dt

+8=−

3

i.e. (3s + 2)L{x} − 5sL{y} =

given that at t = 0, x = 8 and y = 3.

(3s + 2)L{x} − 5sL{y}

Using the above procedure: dy dx (i) 3L − 5L + 2L{x} = L{6} dt dt dy dx 2L −L − L{y} = L{−1} dt dt

6 +9 s − sL{x} + (2s − 1)L{y} =

(1)

1 = − −2 s

(2)

3[sL{x} − x(0)] − 5[sL{y} − y(0)] 6 s

from equation (3), page 589, and Table 61.1, page 584.

i.e. (3s + 2)L{x} − 3x(0) − 5sL{y} 6 + 5y(0) = s Equation (2) becomes:

6 s (1 )

(1

)

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

(2 ) ⎭

(A)

1 = (3s + 2) − − 2 (4) s i.e. s(3s + 2)L{x} − 5s 2 L{y} = 6 + 9s

(3 )

−s(3s + 2)L{x} + (6s 2 + s − 2)L{y} = −6s −

2 −7 s

(4 )

Adding equations (3 ) and (4 ) gives:

2[sL{y} − y(0)] − [sL{x } − x(0)] − L{y} = −

1 s

(s 2 + s − 2)L{y} = −1 + 3s −

from equation (3), page 589, and Table 61.1, page 584,

+ x(0) − L{y} = −

1 s

i.e. (2s − 1)L{y} − 2y(0) − sL{x}

2 s

=

−s + 3s 2 − 2 s

from which, L{y} =

3s 2 − s − 2 s(s 2 + s − 2)

i.e. 2sL{y} − 2y(0) − sL{x}

+ x(0) = −

6 +9 s ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪

(1 ) ⎪ ⎬

−s(3s + 2)L{x} + (3s + 2)(2s − 1)L{y}

i.e. 3sL{x} − 3x(0) − 5sL{y} + 5y(0) + 2L{x} =

(2

)

(iii) s × equation (1

) and (3s + 2) × equation (2

) gives: 6 +9 (3) s(3s + 2)L{x} − 5s 2 L{y} = s s

Equation (1) becomes: + 2L{x} =

1 s

Using partial fractions 1 s

(2 )

(ii) x(0) = 8 and y(0) = 3, hence equation (1 ) becomes (3s + 2)L{x} − 3(8) − 5sL{y} + 5(3) =

6 s

(1

)

3s 2 − s − 2 s(s 2 + s − 2) ≡

A B C + + s (s + 2) (s − 1)

=

A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2) s(s + 2)(s − 1)

608 Higher Engineering Mathematics i.e. 3s 2 − s − 2 = A(s + 2)(s − 1)

Using partial fractions

+ Bs(s − 1) + Cs(s + 2)

8s 2 − 2s − 6 s(s + 2)(s − 1)

When s = 0, −2 = −2 A, hence A =1 When s = 1, 0 = 3C, hence C = 0

≡

A B C + + s (s + 2) (s − 1)

=

A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2) s(s + 2)(s − 1)

When s = −2, 12 =6B, hence B =2 Thus L{y} =

3s 2 − s − 2 1 2 = + s(s 2 + s − 2) s (s + 2)

(iv) Hence y = L−1

i.e. 8s 2 − 2s − 6 = A(s + 2)(s − 1)

1 2 = 1 +2e−2t + s s +2

+ Bs(s − 1) + Cs(s + 2)

Returning to equations (A) to determine L{x} and hence x: (2s − 1) × equation (1

) and 5s × (2

) gives: (2s − 1)(3s + 2)L{x} − 5s(2s − 1)L{y} 6 = (2s − 1) +9 s and −s(5s)L{x} + 5s(2s − 1)L{y} 1 = 5s − − 2 s

6 −9 s

(5) Thus L{x} =

(6)

and − 5s 2 L{x} + 5s(2s − 1)L{y} (6 )

Adding equations (5 ) and (6 ) gives: (s 2 + s − 2)L{x} = −2 + 8s − =

6 s

−2s + 8s 2 − 6 s

from which, L{x} = =

8s 2 − 2s − 6 s(s 2 + s − 2) 8s 2 − 2s − 6 s(s + 2)(s − 1)

8s 2 − 2s − 6 3 5 = + s(s + 2)(s − 1) s (s + 2)

Hence x = L−1

3 5 = 3 + 5e−2t + s s +2

Therefore the solutions of the given simultaneous differential equations are (5 )

= −5 − 10s

When s = 1, 0 =3C, hence C = 0 When s = −2, 30 = 6B, hence B = 5

i.e. (6s 2 + s − 2)L{x} − 5s(2s − 1)L{y} = 12 + 18s −

When s = 0, −6 = −2 A, hence A = 3

y = 1 +2e−2t and x = 3 +5e−2t (These solutions may be checked by substituting the expressions for x and y into the original equations.) Problem 3. Solve the following pair of simultaneous differential equations d2 x −x = y dt 2 d2 y + y = −x dt 2 dx =0 given that at t = 0, x = 2, y = −1, dt dy and = 0. dt

The solution of simultaneous differential equations using Laplace transforms Equation (7) −equation (8) gives:

Using the procedure: (i)

[s 2 L{x} − sx(0) − x (0)] − L{x} = L{y}

[−1 − (s 2 − 1)(s 2 + 1)]L{y}

(1)

= 2s + s(s 2 − 1)

[s 2 L{y} − sy(0) − y (0)] + L{y} = −L{x} (2)

i.e.

−s 4 L{y} = s 3 + s

and

L{y} =

from equation (4), page 590 (ii) x(0) = 2, y(0) = −1, x (0) = 0 and y (0) = 0 hence s 2 L{x} − 2s − L{x} = L{y} s 2 L{y} + s + L{y} = −L{x}

(1 )

1 1 s3 + s =− − 3 4 −s s s 1 1 −1 y=L − − 3 s s

from which,

(2 )

1 y = −1 − t 2 2

i.e.

(iii) Rearranging gives: (s 2 − 1)L{x} − L{y} = 2s 2

L{x} + (s + 1)L{y} = −s

(3) (4)

Equation (3) ×(s 2 + 1) and equation (4) ×1 gives: (s 2 + 1)(s 2 − 1)L{x} − (s 2 + 1)L{y} = (s 2 + 1)2s L{x} + (s 2 + 1)L{y} = −s

Now try the following exercise Exercise 227 Further problems on solving simultaneous differential equations using Laplace transforms

(5)

Solve the following pairs of simultaneous differential equations:

(6)

1.

Adding equations (5) and (6) gives: [(s 2 + 1)(s 2 − 1) + 1]L{x} = (s 2 + 1)2s − s i.e. s 4 L{x} = 2s 3 + s = s(2s 2 + 1) 2.

s(2s 2 + 1) 2s 2 + 1 = from which, L{x} = s4 s3 =

(iv)

Hence x = L−1

2s 2 1 2 1 + 3 = + 3 s3 s s s

2 1 + 3 s s

3.

Returning to equations (3) and (4) to determine y: 1 × equation (3) and (s 2 − 1) × equation (4) gives: 2

(s − 1)L{x} − L{y} = 2s

(7)

(s − 1)L{x} + (s − 1)(s + 1)L{y} = −s(s 2 − 1)

(8)

2

2

dy dx −y+x + − 5 sin t = 0 dt dt dx dy 3 + x − y + 2 − et = 0 dt dt given that at t = 0, x = 0 and y = 0.

x = 5 cos t + 5 sin t − e2t − et − 3 and 2

y = e2t + 2et − 3 − 5 sin t

1 x = 2 + t2 2

i.e.

dx dy + = 5 et dt dt dy dx −3 =5 dt dt given that when t = 0, x = 0 and y = 0. [x = et − t − 1 and y = 2t − 3 + 3et ] 2

2

d2 x + 2x = y dt 2 d2 y + 2y = x dt 2 given that at t = 0, x = 4, y = 2, and

dy = 0. dt

dx =0 dt

√ x = 3 cos t + cos(√3 t ) and y = 3 cos t − cos( 3 t )

609

Revision Test 18 This Revision Test covers the material contained in Chapters 61 to 65. The marks for each question are shown in brackets at the end of each question. 1.

Find the Laplace transforms of the following functions: (a) 2t 3 − 4t + 5 (b) 3e−2t − 4 sin 2t

2.

(c) 3 cosh 2t

(d) 2t 4e−3t

(e) 5e2t cos 3t

(f) 2e3t sinh 4t

4. (16)

(c) (e) (g)

12 5 (b) 5 2s + 1 s 4s 5 (d) 2 2 s +9 s −9 s −4 3 (f) 2 (s + 2)4 s − 8s − 20 8 s 2 − 4s + 3

13 − s 2 s(s 2 + 4s + 13)

5.

(24)

In a galvanometer the deﬂection θ satisﬁes the differential equation:

Use Laplace transforms to solve the equation for θ dθ = 0. (13) given that when t = 0, θ = 0 and dt Solve the following pair of simultaneous differential equations: 3

dx = 3x + 2y dt

2

dy + 3x = 6y dt

(17)

Use partial fractions to determine the following: 5s − 1 (a) L−1 2 s −s −2 2 2s + 11s − 9 (b) L−1 s(s − 1)(s + 3)

d2 θ dθ +2 +θ = 4 2 dt dt

Find the inverse Laplace transforms of the following functions: (a)

3.

(c) L−1

given that when t = 0, x = 1 and y = 3. 6.

(20)

Determine the poles and zeros for the transfer func(s + 2)(s − 3) tion: F(s) = and plot them on (s + 3)(s 2 + 2s + 5) a pole-zero diagram. (10)

Chapter 66

Fourier series for periodic functions of period 2π f (x)

66.1

Introduction 1

Fourier series provides a method of analysing periodic functions into their constituent components. Alternating currents and voltages, displacement, velocity and acceleration of slider-crank mechanisms and acoustic waves are typical practical examples in engineering and science where periodic functions are involved and often requiring analysis.

66.2

Periodic functions

A function f (x) is said to be periodic if f (x + T ) = f (x) for all values of x, where T is some positive number. T is the interval between two successive repetitions and is called the period of the functions f (x). For example, y = sin x is periodic in x with period 2π since sin x = sin(x + 2π) = sin(x + 4π), and so on. In general, if y = sin ωt then the period of the waveform is 2π/ω. The function shown in Fig. 66.1 is also periodic of period 2π and is deﬁned by: −1, when −π < x < 0 f (x) = 1, when 0<x