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Advanced Engineering Mathematics, 6th Edition

Peter V. O'Neil

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                International Student Edition  ADVANCED ENGINEERING MATHEMATICS  This page intentionally left blank  ADVANCED ENGINEERING MATHEMATICS International Student Edition  PETER V. O'NEIL University of Alabama at Birmingham  Australia Canada Mexico Singapore Spain United Kingdom United States  Advanced Engineering Mathematics, International Student Edition by Peter V. O'Neil  Associate Vice-President and Editorial Director: Evelyn Veitch Publisher: Chris Carson Developmental Editor: Kamilah Reid Burrell/ Hilda Gowaus  Production Services: RPK Editorial Services  Creative Director: Angela Cluer  Copy Editor: Shelly Gerger-Knechtl/ Harlan James  Interior Design: Terri Wright  Proofreader: Erin Wagner/Harlan James  Cover Design: Andrew Adams  Indexer: RPK Editorial Services  Compositor: Integra  Permissions Coordinator: Vicki Gould  Production Manager: Renate McCloy  Printer: Quebecor World  COPYRIGHT © 2007 by Nelson, a division of Thomson Canada Limited.  ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transcribed, or used in any form or by any means—graphic, electronic, or mechanical, including photocopying, recording, taping, Web distribution, or information storage and retrieval systems—without the written permission of the publisher.  North America Nelson 1120 Birchmount Road Toronto, Ontario M1K 5G4 Canada  Printed and bound in the United States 1  2  3  4  07  06  For more information contact Nelson, 1120 Birchmount Road, Toronto, Ontario, Canada, M1K 5G4. Or you can visit our Internet site at http://www.nelson.com Library of Congress Control Number: 2006900028 ISBN: 0-495-08237-6 If you purchased this book within the United States or Canada you should be aware that it has been wrongfully imported without the approval of the Publisher or the Author.  For permission to use material from this text or product, submit a request online at www.thomsonrights.com Every effort has been made to trace ownership of all copyright material and to secure permission from copyr; ight holders. In the event of any question arising as to the use of any material, we will be pleased to make the necessary corrections in future printings.  Asia Thomson Learning 5 Shenton Way #01-01 UIC Building Singapore 068808 Australia/New Zealand Thomson Learning 102 Dodds Street Southbank, Victoria Australia 3006 Europe/Middle East/Africa Thomson Learning High Holborn House 50/51 Bedford Row London WC1R 4LR United Kingdom Latin America Thomson Learning Seneca, 53 Colonia Polanco 11560 Mexico D.F. Mexico Spain Paraninfo Calle/Magallanes, 25 28015 Madrid, Spain  Contents  PART  1  Chapter 1  Ordinary Differential Equations 1 First-Order Differential Equations  3  1.1 Preliminary Concepts 3 1.1.1 General and Particular Solutions 3 1.1.2 Implicitly Defined Solutions 4 1.1.3 Integral Curves 5 1.1.4 The Initial Value Problem 6 1.1.5 Direction Fields 7 1.2 Separable Equations 11 1.2.1 Some Applications of Separable Differential Equations 14 1.3 Linear Differential Equations 22 1.4 Exact Differential Equations 26 1.5 Integrating Factors 33 1.5.1 Separable Equations and Integrating Factors 37 1.5.2 Linear Equations and Integrating Factors 37 1.6 Homogeneous, Bernoulli, and Riccati Equations 38 1.6.1 Homogeneous Differential Equations 38 1.6.2 The Bernoulli Equation 42 1.6.3 The Riccati Equation 43 1.7 Applications to Mechanics, Electrical Circuits, and Orthogonal Trajectories 1.7.1 Mechanics 46 1.7.2 Electrical Circuits 51 1.7.3 Orthogonal Trajectories 53 1.8 Existence and Uniqueness for Solutions of Initial Value Problems 58  Chapter 2  Second-Order Differential Equations  46  61  2.1 Preliminary Concepts 61 2.2 Theory of Solutions of y 

 + p
xy 
 + q
xy = f
x 62 2.2.1 The Homogeneous Equation y 

 + p
xy 
 + q
x = 0 64 2.2.2 The Nonhomogeneous Equation y 

 + p
xy 
 + q
xy = f
x 2.3 Reduction of Order 69 2.4 The Constant Coefficient Homogeneous Linear Equation 73 2.4.1 Case 1: A 2 − 4B > 0 73 2.4.2 Case 2: A 2 − 4B = 0 74  68  v  vi  Contents 2.4.3 Case 3: A 2 − 4B < 0 74 2.4.4 An Alternative General Solution in the Complex Root Case 75 2.5 Euler's Equation 78 2.6 The Nonhomogeneous Equation y 

 + p
xy 
 + q
xy = f
x 82 2.6.1 The Method of Variation of Parameters 82 2.6.2 The Method of Undetermined Coefficients 85 2.6.3 The Principle of Superposition 91 2.6.4 Higher-Order Differential Equations 91 2.7 Application of Second-Order Differential Equations to a Mechanical System 2.7.1 Unforced Motion 95 2.7.2 Forced Motion 98 2.7.3 Resonance 100 2.7.4 Beats 102 2.7.5 Analogy with an Electrical Circuit 103  Chapter 3  The Laplace Transform  107  3.1 Definition and Basic Properties 107 3.2 Solution of Initial Value Problems Using the Laplace Transform 3.3 Shifting Theorems and the Heaviside Function 120 3.3.1 The First Shifting Theorem 120 3.3.2 The Heaviside Function and Pulses 122 3.3.3 The Second Shifting Theorem 125 3.3.4 Analysis of Electrical Circuits 129 3.4 Convolution 134 3.5 Unit Impulses and the Dirac Delta Function 139 3.6 Laplace Transform Solution of Systems 144 3.7 Differential Equations with Polynomial Coefficients 150  Chapter 4  Series Solutions 4.1 4.2 4.3 4.4  Chapter 5  155  Power Series Solutions of Initial Value Problems 156 Power Series Solutions Using Recurrence Relations 161 Singular Points and the Method of Frobenius 166 Second Solutions and Logarithm Factors 173  Numerical Approximation of Solutions  181  5.1 Euler's Method 182 5.1.1 A Problem in Radioactive Waste Disposal 5.2 One-Step Methods 190 5.2.1 The Second-Order Taylor Method 190 5.2.2 The Modified Euler Method 193 5.2.3 Runge-Kutta Methods 195 5.3 Multistep Methods 197 5.3.1 Case 1 r = 0 198 5.3.2 Case 2 r = 1 198 5.3.3 Case 3 r = 3 199 5.3.4 Case 4 r = 4 199  187  116  93  Contents  PART  2  Chapter 6  Vectors and Linear Algebra 201 Vectors and Vector Spaces 6.1 6.2 6.3 6.4 6.5  Chapter 7  203  The Algebra and Geometry of Vectors 203 The Dot Product 211 The Cross Product 217 The Vector Space R n 223 Linear Independence, Spanning Sets, and Dimension in R n  Matrices and Systems of Linear Equations  228  237  7.1 Matrices 238 7.1.1 Matrix Algebra 239 7.1.2 Matrix Notation for Systems of Linear Equations 242 7.1.3 Some Special Matrices 243 7.1.4 Another Rationale for the Definition of Matrix Multiplication 246 7.1.5 Random Walks in Crystals 247 7.2 Elementary Row Operations and Elementary Matrices 251 7.3 The Row Echelon Form of a Matrix 258 7.4 The Row and Column Spaces of a Matrix and Rank of a Matrix 266 7.5 Solution of Homogeneous Systems of Linear Equations 272 7.6 The Solution Space of AX = O 280 7.7 Nonhomogeneous Systems of Linear Equations 283 7.7.1 The Structure of Solutions of AX = B 284 7.7.2 Existence and Uniqueness of Solutions of AX = B 285 7.8 Matrix Inverses 293 7.8.1 A Method for Finding A −1 295  Chapter 8  Determinants 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9  Chapter 9  299  Permutations 299 Definition of the Determinant 301 Properties of Determinants 303 Evaluation of Determinants by Elementary Row and Column Operations Cofactor Expansions 311 Determinants of Triangular Matrices 314 A Determinant Formula for a Matrix Inverse 315 Cramer's Rule 318 The Matrix Tree Theorem 320  Eigenvalues, Diagonalization, and Special Matrices 9.1 Eigenvalues and Eigenvectors 324 9.1.1 Gerschgorin's Theorem 328 9.2 Diagonalization of Matrices 330 9.3 Orthogonal and Symmetric Matrices  339  323  307  vii  viii  Contents 9.4 Quadratic Forms 347 9.5 Unitary, Hermitian, and Skew Hermitian Matrices  PART  3  Chapter 10  352  Systems of Differential Equations and Qualitative Methods 359 Systems of Linear Differential Equations  361  10.1 Theory of Systems of Linear First-Order Differential Equations 361 10.1.1 Theory of the Homogeneous System X
 = AX 365 10.1.2 General Solution of the Nonhomogeneous System X
 = AX + G 372 10.2 Solution of X
 = AX when A is Constant 374 10.2.1 Solution of X
 = AX when A has Complex Eigenvalues 377 10.2.2 Solution of X
 = AX when A does not have n Linearly Independent Eigenvectors 379 10.2.3 Solution of X
 = AX by Diagonalizing A 384 10.2.4 Exponential Matrix Solutions of X
 = AX 386 10.3 Solution of X
 = AX + G 394 10.3.1 Variation of Parameters 394 10.3.2 Solution of X
 = AX + G by Diagonalizing A 398  Chapter 11  Qualitative Methods and Systems of Nonlinear Differential Equations 11.1 11.2 11.3 11.4 11.5 11.6 11.7  PART  4  Chapter 12  Nonlinear Systems and Existence of Solutions 403 The Phase Plane, Phase Portraits and Direction Fields Phase Portraits of Linear Systems 413 Critical Points and Stability 424 Almost Linear Systems 431 Lyapunov's Stability Criteria 451 Limit Cycles and Periodic Solutions 461  406  Vector Analysis 473 Vector Differential Calculus  475  12.1 Vector Functions of One Variable 475 12.2 Velocity, Acceleration, Curvature and Torsion 481 12.2.1 Tangential and Normal Components of Acceleration 488 12.2.2 Curvature as a Function of t 491 12.2.3 The Frenet Formulas 492 12.3 Vector Fields and Streamlines 493 12.4 The Gradient Field and Directional Derivatives 499 12.4.1 Level Surfaces, Tangent Planes and Normal Lines 503 12.5 Divergence and Curl 510 12.5.1 A Physical Interpretation of Divergence 512 12.5.2 A Physical Interpretation of Curl 513  403  Contents  Chapter 13  Vector Integral Calculus  517  13.1 Line Integrals 517 13.1.1 Line Integral with Respect to Arc Length 525 13.2 Green's Theorem 528 13.2.1 An Extension of Green's Theorem 532 13.3 Independence of Path and Potential Theory in the Plane 536 13.3.1 A More Critical Look at Theorem 13.5 539 13.4 Surfaces in 3-Space and Surface Integrals 545 13.4.1 Normal Vector to a Surface 548 13.4.2 The Tangent Plane to a Surface 551 13.4.3 Smooth and Piecewise Smooth Surfaces 552 13.4.4 Surface Integrals 553 13.5 Applications of Surface Integrals 557 13.5.1 Surface Area 557 13.5.2 Mass and Center of Mass of a Shell 557 13.5.3 Flux of a Vector Field Across a Surface 560 13.6 Preparation for the Integral Theorems of Gauss and Stokes 562 13.7 The Divergence Theorem of Gauss 564 13.7.1 Archimedes's Principle 567 13.7.2 The Heat Equation 568 13.7.3 The Divergence Theorem as a Conservation of Mass Principle 13.8 The Integral Theorem of Stokes 572 13.8.1 An Interpretation of Curl 576 13.8.2 Potential Theory in 3-Space 576  PART  5  Chapter 14  570  Fourier Analysis, Orthogonal Expansions, and Wavelets 581 Fourier Series  583  14.1 Why Fourier Series? 583 14.2 The Fourier Series of a Function 586 14.2.1 Even and Odd Functions 589 14.3 Convergence of Fourier Series 593 14.3.1 Convergence at the End Points 599 14.3.2 A Second Convergence Theorem 601 14.3.3 Partial Sums of Fourier Series 604 14.3.4 The Gibbs Phenomenon 606 14.4 Fourier Cosine and Sine Series 609 14.4.1 The Fourier Cosine Series of a Function 610 14.4.2 The Fourier Sine Series of a Function 612 14.5 Integration and Differentiation of Fourier Series 614 14.6 The Phase Angle Form of a Fourier Series 623 14.7 Complex Fourier Series and the Frequency Spectrum 630 14.7.1 Review of Complex Numbers 630 14.7.2 Complex Fourier Series 631  ix  x  Contents  Chapter 15  The Fourier Integral and Fourier Transforms 15.1 15.2 15.3 15.4  15.5 15.6 15.7  15.8  15.9  Chapter 16  637  The Fourier Integral 637 Fourier Cosine and Sine Integrals 640 The Complex Fourier Integral and the Fourier Transform 642 Additional Properties and Applications of the Fourier Transform 652 15.4.1 The Fourier Transform of a Derivative 652 15.4.2 Frequency Differentiation 655 15.4.3 The Fourier Transform of an Integral 656 15.4.4 Convolution 657 15.4.5 Filtering and the Dirac Delta Function 660 15.4.6 The Windowed Fourier Transform 661 15.4.7 The Shannon Sampling Theorem 665 15.4.8 Lowpass and Bandpass Filters 667 The Fourier Cosine and Sine Transforms 670 The Finite Fourier Cosine and Sine Transforms 673 The Discrete Fourier Transform 675 15.7.1 Linearity and Periodicity 678 15.7.2 The Inverse N -Point DFT 678 15.7.3 DFT Approximation of Fourier Coefficients 679 Sampled Fourier Series 681 15.8.1 Approximation of a Fourier Transform by an N -Point DFT 685 15.8.2 Filtering 689 The Fast Fourier Transform 694 15.9.1 Use of the FFT in Analyzing Power Spectral Densities of Signals 15.9.2 Filtering Noise From a Signal 696 15.9.3 Analysis of the Tides in Morro Bay 697  Special Functions, Orthogonal Expansions, and Wavelets  695  701  16.1 Legendre Polynomials 701 16.1.1 A Generating Function for the Legendre Polynomials 704 16.1.2 A Recurrence Relation for the Legendre Polynomials 706 16.1.3 Orthogonality of the Legendre Polynomials 708 16.1.4 Fourier–Legendre Series 709 16.1.5 Computation of Fourier–Legendre Coefficients 711 16.1.6 Zeros of the Legendre Polynomials 713 16.1.7 Derivative and Integral Formulas for Pn 
x 715 16.2 Bessel Functions 719 16.2.1 The Gamma Function 719 16.2.2 Bessel Functions of the First Kind and Solutions of Bessel's Equation 16.2.3 Bessel Functions of the Second Kind 722 16.2.4 Modified Bessel Functions 725 16.2.5 Some Applications of Bessel Functions 727 16.2.6 A Generating Function for Jn 
x 732 16.2.7 An Integral Formula for Jn 
x 733 16.2.8 A Recurrence Relation for Jv 
x 735 16.2.9 Zeros of Jv 
x 737  721  Contents 16.2.10 Fourier–Bessel Expansions 739 16.2.11 Fourier–Bessel Coefficients 741 16.3 Sturm–Liouville Theory and Eigenfunction Expansions 745 16.3.1 The Sturm–Liouville Problem 745 16.3.2 The Sturm–Liouville Theorem 752 16.3.3 Eigenfunction Expansions 755 16.3.4 Approximation in the Mean and Bessel's Inequality 759 16.3.5 Convergence in the Mean and Parseval's Theorem 762 16.3.6 Completeness of the Eigenfunctions 763 16.4 Wavelets 765 16.4.1 The Idea Behind Wavelets 765 16.4.2 The Haar Wavelets 767 16.4.3 A Wavelet Expansion 774 16.4.4 Multiresolution Analysis with Haar Wavelets 774 16.4.5 General Construction of Wavelets and Multiresolution Analysis 16.4.6 Shannon Wavelets 776  PART  6  Chapter 17  xi  775  Partial Differential Equations 779 The Wave Equation  781  17.1 The Wave Equation and Initial and Boundary Conditions 781 17.2 Fourier Series Solutions of the Wave Equation 786 17.2.1 Vibrating String with Zero Initial Velocity 786 17.2.2 Vibrating String with Given Initial Velocity and Zero Initial Displacement 791 17.2.3 Vibrating String with Initial Displacement and Velocity 793 17.2.4 Verification of Solutions 794 17.2.5 Transformation of Boundary Value Problems Involving the Wave Equation 796 17.2.6 Effects of Initial Conditions and Constants on the Motion 798 17.2.7 Numerical Solution of the Wave Equation 801 17.3 Wave Motion Along Infinite and Semi-Infinite Strings 808 17.3.1 Wave Motion Along an Infinite String 808 17.3.2 Wave Motion Along a Semi-Infinite String 813 17.3.3 Fourier Transform Solution of Problems on Unbounded Domains 815 17.4 Characteristics and d'Alembert's Solution 822 17.4.1 A Nonhomogeneous Wave Equation 825 17.4.2 Forward and Backward Waves 828 17.5 Normal Modes of Vibration of a Circular Elastic Membrane 831 17.6 Vibrations of a Circular Elastic Membrane, Revisited 834 17.7 Vibrations of a Rectangular Membrane 837  Chapter 18  The Heat Equation  841  18.1 The Heat Equation and Initial and Boundary Conditions 18.2 Fourier Series Solutions of the Heat Equation 844  841  xii  Contents 18.2.1 Ends of the Bar Kept at Temperature Zero 844 18.2.2 Temperature in a Bar with Insulated Ends 847 18.2.3 Temperature Distribution in a Bar with Radiating End 848 18.2.4 Transformations of Boundary Value Problems Involving the Heat Equation 851 18.2.5 A Nonhomogeneous Heat Equation 854 18.2.6 Effects of Boundary Conditions and Constants on Heat Conduction 857 18.2.7 Numerical Approximation of Solutions 859 18.3 Heat Conduction in Infinite Media 865 18.3.1 Heat Conduction in an Infinite Bar 865 18.3.2 Heat Conduction in a Semi-Infinite Bar 868 18.3.3 Integral Transform Methods for the Heat Equation in an Infinite Medium 869 18.4 Heat Conduction in an Infinite Cylinder 873 18.5 Heat Conduction in a Rectangular Plate 877  Chapter 19  The Potential Equation  879  19.1 19.2 19.3 19.4 19.5  Harmonic Functions and the Dirichlet Problem 879 Dirichlet Problem for a Rectangle 881 Dirichlet Problem for a Disk 883 Poisson's Integral Formula for the Disk 886 Dirichlet Problems in Unbounded Regions 888 19.5.1 Dirichlet Problem for the Upper Half Plane 889 19.5.2 Dirichlet Problem for the Right Quarter Plane 891 19.5.3 An Electrostatic Potential Problem 893 19.6 A Dirichlet Problem for a Cube 896 19.7 The Steady-State Heat Equation for a Solid Sphere 898 19.8 The Neumann Problem 902 19.8.1 A Neumann Problem for a Rectangle 904 19.8.2 A Neumann Problem for a Disk 906 19.8.3 A Neumann Problem for the Upper Half Plane 908  PART  7  Chapter 20  Complex Analysis 911 Geometry and Arithmetic of Complex Numbers  913  20.1 Complex Numbers 913 20.1.1 The Complex Plane 914 20.1.2 Magnitude and Conjugate 915 20.1.3 Complex Division 916 20.1.4 Inequalities 917 20.1.5 Argument and Polar Form of a Complex Number 918 20.1.6 Ordering 920 20.2 Loci and Sets of Points in the Complex Plane 921 20.2.1 Distance 922 20.2.2 Circles and Disks 922 20.2.3 The Equation z − a = z − b 923 20.2.4 Other Loci 925 20.2.5 Interior Points, Boundary Points, and Open and Closed Sets  925  Contents 20.2.6 20.2.7 20.2.8 20.2.9  Chapter 21  Limit Points 929 Complex Sequences 931 Subsequences 934 Compactness and the Bolzano-Weierstrass Theorem  Complex Functions  935  939  21.1 Limits, Continuity, and Derivatives 939 21.1.1 Limits 939 21.1.2 Continuity 941 21.1.3 The Derivative of a Complex Function 943 21.1.4 The Cauchy–Riemann Equations 945 21.2 Power Series 950 21.2.1 Series of Complex Numbers 951 21.2.2 Power Series 952 21.3 The Exponential and Trigonometric Functions 957 21.4 The Complex Logarithm 966 21.5 Powers 969 21.5.1 Integer Powers 969 21.5.2 z 1/n for Positive Integer n 969 21.5.3 Rational Powers 971 21.5.4 Powers z w 972  Chapter 22  Complex Integration  975  22.1 Curves in the Plane 975 22.2 The Integral of a Complex Function 980 22.2.1 The Complex Integral in Terms of Real Integrals 983 22.2.2 Properties of Complex Integrals 985 22.2.3 Integrals of Series of Functions 988 22.3 Cauchy's Theorem 990 22.3.1 Proof of Cauchy's Theorem for a Special Case 993 22.4 Consequences of Cauchy's Theorem 994 22.4.1 Independence of Path 994 22.4.2 The Deformation Theorem 995 22.4.3 Cauchy's Integral Formula 997 22.4.4 Cauchy's Integral Formula for Higher Derivatives 1000 22.4.5 Bounds on Derivatives and Liouville's Theorem 1001 22.4.6 An Extended Deformation Theorem 1002  Chapter 23  Series Representations of Functions  1007  23.1 Power Series Representations 1007 23.1.1 Isolated Zeros and the Identity Theorem 1012 23.1.2 The Maximum Modulus Theorem 1016 23.2 The Laurent Expansion 1019  Chapter 24  Singularities and the Residue Theorem  1023  24.1 Singularities 1023 24.2 The Residue Theorem 1030 24.3 Some Applications of the Residue Theorem  1037  xiii  xiv  Contents 24.3.1 The Argument Principle 1037 24.3.2 An Inversion for the Laplace Transform 24.3.3 Evaluation of Real Integrals 1040  Chapter 25  Conformal Mappings 25.1 25.2 25.3 25.4 25.5  PART  8  Chapter 26  1039  1055  Functions as Mappings 1055 Conformal Mappings 1062 25.2.1 Linear Fractional Transformations 1064 Construction of Conformal Mappings Between Domains 1072 25.3.1 Schwarz–Christoffel Transformation 1077 Harmonic Functions and the Dirichlet Problem 1080 25.4.1 Solution of Dirichlet Problems by Conformal Mapping 1083 Complex Function Models of Plane Fluid Flow 1087  Probability and Statistics 1097 Counting and Probability  1099  26.1 26.2 26.3  The Multiplication Principle 1099 Permutations 1102 Choosing r Objects from n Objects 1104 26.3.1 r Objects from n Objects, with Order 1104 26.3.2 r Objects from n Objects, without Order 1106 26.3.3 Tree Diagrams 1107 26.4 Events and Sample Spaces 1112 26.5 The Probability of an Event 1116 26.6 Complementary Events 1121 26.7 Conditional Probability 1122 26.8 Independent Events 1126 26.8.1 The Product Rule 1128 26.9 Tree Diagrams in Computing Probabilities 1130 26.10 Bayes' Theorem 1134 26.11 Expected Value 1139  Chapter 27  Statistics 27.1  27.2 27.3  27.4  1143  Measures of Center and Variation 1143 27.1.1 Measures of Center 1143 27.1.2 Measures of Variation 1146 Random Variables and Probability Distributions 1150 The Binomial and Poisson Distributions 1154 27.3.1 The Binomial Distribution 1154 27.3.2 The Poisson Distribution 1157 A Coin Tossing Experiment, Normally Distributed Data, and the Bell Curve 27.4.1 The Standard Bell Curve 1174 27.4.2 The 68, 95, 99.7 Rule 1176  1159  Contents 27.5 27.6 27.7 27.8  Sampling Distributions and the Central Limit Theorem 1178 Confidence Intervals and Estimating Population Proportion 1185 Estimating Population Mean and the Student t Distribution 1190 Correlation and Regression 1194  Answers and Solutions to Selected Problems Index  I1  A1  xv  This page intentionally left blank  Preface  This Sixth Edition of Advanced Engineering Mathematics maintains the primary goal of previous editions—to engage much of the post-calculus mathematics needed and used by scientists, engineers, and applied mathematicians, in a setting that is helpful to both students and faculty. The format used throughout begins with the correct developments of concepts such as Fourier series and integrals, conformal mappings, and special functions. These ideas are then brought to bear on applications and models of important phenomena, such as wave and heat propagation and filtering of signals. This edition differs from the previous one primarily in the inclusion of statistics and numerical methods. The statistics part treats random variables, normally distributed data, bell curves, the binomial, Poisson, and student t-distributions, the central limit theorem, confidence intervals, correlation, and regression. This is preceded by prerequisite topics from probability and techniques of enumeration. The numerical methods are applied to initial value problems in ordinary differential equations, including a proposal for radioactive waste disposal, and to boundary value problems involving the heat and wave equations. Finally, in order to include these topics without lengthening the book, some items from the fifth edition have been moved to a website, located at http://engineering.thomsonlearning.com. I hope that this provides convenient accessibility. Material selected for this move includes some biographies and historical notes, predator/prey and competing species models, the theory underlying the efficiency of the FFT, and some selected examples and problems. The chart on the following page offers a complete organizational overview.  Acknowledgments This book is the result of a team effort involving much more than an author. Among those to whom I owe a debt of appreciation are Chris Carson, Joanne Woods, Hilda Gowans and Kamilah Reid-Burrell of Thomson Engineering, and Rose Kernan and the professionals at RPK Editorial Services, Inc. I also want to thank Dr. Thomas O'Neil of the California Polytechnic State University for material he contributed, and Rich Jones, who had the vision for the first edition of this book many years ago. Finally, I want to acknowledge the reviewers, whose suggestions for improvements and clarifications are much appreciated: Preliminary Review  Panagiotis Dimitrakopoulos, University of Maryland Mohamed M. Hafez, University of California, Davis Jennifer Hopwood, University of Western Australia Nun Kwan Yip, Purdue University xvii  xviii Organizational Overview Ordinary Differential Equations  Laplace Transforms  Series Solutions  Systems of Ordinary Differential Equations  Special Functions  Vectors, Matrices, Determinants  Statistical Analysis Systems of Algebraic Equations  Eigenfunction Expansions, Completeness Haar Wavelets  Vector Analysis Qualitative Methods, Stability, Analysis of Critical Points Probability  Partial Differential Equations Statistics  Mathematical Models  Fourier Analysis Complex Analysis Fourier Series, Integrals  Fourier Transforms  Discrete Fourier Transform  P r eface  xix  Draft Review  Sabri Abou-Ward, University of Toronto Craig Hildebrand, California State University – Fresno Seiichi Nomura, University of Texas, Arlington David L. Russell, Virginia Polytechnic Institute and State University Y.Q. Sheng, McMaster University  Peter V. O'neil University of Alabama at Birmingham  This page intentionally left blank  PA RT  1  CHAPTER 1 First-Order Differential Equations  CHAPTER 2 Second-Order Differential Equations  CHAPTER 3  Ordinary Differential Equations  The Laplace Transform  CHAPTER 4 Series Solutions  CHAPTER 5 Numerical Approximation of Solutions  A differential equation is an equation that contains one or more derivatives. For example, y

 
x + y
x = 4 sin
3x and d4 w − 
w
t2 = e−t dt4 are differential equations. These are ordinary differential equations because they involve only total derivatives, rather than partial derivatives. Differential equations are interesting and important because they express relationships involving rates of change. Such relationships form the basis for developing ideas and studying phenomena in the sciences, engineering, economics, and increasingly in other areas, such as the business world and the stock market. We will see examples of applications as we learn more about differential equations. 1  The order of a differential equation is the order of its highest derivative. The first example given above is of second order, while the second is of fourth order. The equation xy
 − y2 = ex is of first order. A solution of a differential equation is any function that satisfies it. A solution may be defined on the entire real line, or on only part of it, often an interval. For example, y = sin
2x is a solution of y

 + 4y = 0 because, by direct differentiation, y

 + 4y = −4 sin
2x + 4 sin
2x = 0 This solution is defined for all x (that is, on the whole real line). By contrast, y = x ln
x − x is a solution of y
 =  y + 1 x  but this solution is defined only for x > 0. Indeed, the coefficient 1/x of y in this equation means that x = 0 is disallowed from the start. We now begin a systematic development of ordinary differential equations, starting with the first order case.  2  CHAPTER  1  PRELIMINARY CONCEPTS SEPARABLE EQUATIONS HOMOGENEOUS, BERNOULLI, AND RICCATI EQUATIONS APPLICATIONS TO MECHANICS, ELECTRICAL CIRCUITS, AND ORTHOGONAL TRAJECTORIES EXI  First-Order Differential Equations  1.1  Preliminary Concepts Before developing techniques for solving various kinds of differential equations, we will develop some terminology and geometric insight.  1.1.1  General and Particular Solutions  A first-order differential equation is any equation involving a first derivative, but no higher derivative. In its most general form, it has the appearance F
x y y
  = 0  (1.1)  in which y
x is the function of interest and x is the independent variable. Examples are y
 − y2 − ey = 0 y
 − 2 = 0 and y
 − cos
x = 0 Note that y
 must be present for an equation to qualify as a first-order differential equation, but x and/or y need not occur explicitly. A solution of equation (1.1) on an interval I is a function  that satisfies the equation for all x in I. That is, F
x 
x 
 
x = 0  for all x in I.  For example, 
x = 2 + ke−x 3  4  CHAPTER 1  First-Order Differential Equations  is a solution of y
 + y = 2 for all real x, and for any number k. Here I can be chosen as the entire real line. And 
x = x ln
x + cx is a solution of y
 =  y +1 x  for all x > 0, and for any number c. In both of these examples, the solution contained an arbitrary constant. This is a symbol independent of x and y that can be assigned any numerical value. Such a solution is called the general solution of the differential equation. Thus 
x = 2 + ke−x is the general solution of y
 + y = 2. Each choice of the constant in the general solution yields a particular solution. For example, f
x = 2 + e−x  and  g
x = 2 − e−x  √ h
x = 2 − 53e−x  are√all particular solutions of y
 + y = 2, obtained by choosing, respectively, k = 1, −1 and − 53 in the general solution.  1.1.2  Implicitly Defined Solutions  Sometimes we can write a solution explicitly giving y as a function of x. For example, y = ke−x is the general solution of y
 = −y as can be verified by substitution. This general solution is explicit, with y isolated on one side of an equation, and a function of x on the other. By contrast, consider y
 = −  2xy3 + 2  3x2 y2 + 8e4y  We claim that the general solution is the function y
x implicitly defined by the equation x2 y3 + 2x + 2e4y = k  (1.2)  in which k can be any number. To verify this, implicitly differentiate equation (1.2) with respect to x, remembering that y is a function of x. We obtain 2xy3 + 3x2 y2 y
 + 2 + 8e4y y
 = 0 and solving for y
 yields the differential equation. In this example we are unable to solve equation (1.2) explicitly for y as a function of x, isolating y on one side. Equation (1.2), implicitly defining the general solution, was obtained by a technique we will develop shortly, but this technique cannot guarantee an explicit solution.  1.1 Preliminary Concepts  1.1.3  5  Integral Curves  A graph of a solution of a first-order differential equation is called an integral curve of the equation. If we know the general solution, we obtain an infinite family of integral curves, one for each choice of the arbitrary constant.  EXAMPLE 1.1  We have seen that the general solution of y
 + y = 2 is y = 2 + ke−x for all x. The integral curves of y
 + y = 2 are graphs of y = 2 + ke−x for different choices of k. Some of these are shown in Figure 1.1. y 30  k6  20 k3 10 k  0 ( y  2) 
2 k  
3  k  
6  
1  0  1  2  3  
10  
20  FIGURE 1.1 Integral curves of y
 + y = 2 for k = 0 3 −3 6, and −6.  EXAMPLE 1.2  It is routine to verify that the general solution of y
 +  y = ex x  is y=  1 
xex − ex + c x  4  5  6  x  6  CHAPTER 1  First-Order Differential Equations  for x = 0. Graphs of some of these integral curves, obtained by making choices for c, are shown in Figure 1.2. y 40 30  c  20  20 10  c5 c0  0.5 
10 
20  1.5  1.0  2.0  2.5  x 3.0  c  
6 c  
10  FIGURE 1.2  −10.  Integral curves of y
 + x1 y = ex for c = 0 5 20 −6, and  We will see shortly how these general solutions are obtained. For the moment, we simply want to illustrate integral curves. Although in simple cases integral curves can be sketched by hand, generally we need computer assistance. Computer packages such as MAPLE, MATHEMATICA and MATLAB are widely available. Here is an example in which the need for computing assistance is clear.  EXAMPLE 1.3  The differential equation y
 + xy = 2 has general solution y
x = e−x  2 /2  
  x  2e  0  2 /2  d + ke−x /2  2  Figure 1.3 shows computer-generated integral curves corresponding to k = 0, 4, 13, −7, −15 and −11.  1.1.4  The Initial Value Problem  The general solution of a first-order differential equation F
x y y
  = 0 contains an arbitrary constant, hence there is an infinite family of integral curves, one for each choice of the constant. If we specify that a solution is to pass through a particular point 
x0  y0 , then we must find that particular integral curve (or curves) passing through this point. This is called an initial value problem. Thus, a first order initial value problem has the form F
x y y
  = 0  y
x0  = y0   in which x0 and y0 are given numbers. The condition y
x0  = y0 is called an initial condition.  1.1 Preliminary Concepts  7  y k  13 10 k4  5 
2  k0  0  k  
7  
4  2  4  x  
5 k  
11  
10  k  
15 
15 FIGURE 1.3 Integral curves of y
 + xy = 2 for k = 0 4 13 −7 −15, and  −11.  EXAMPLE 1.4  Consider the initial value problem y
 + y = 2  y
1 = −5  
  From Example 1.1, the general solution of y + y = 2 is y = 2 + ke−x  Graphs of this equation are the integral curves. We want the one passing through 
1 −5. Solve for k so that y
1 = 2 + ke−1 = −5 obtaining k = −7e The solution of this initial value problem is y = 2 − 7ee−x = 2 − 7e−
x−1  As a check, y
1 = 2 − 7 = −5 The effect of the initial condition in this example was to pick out one special integral curve as the solution sought. This suggests that an initial value problem may be expected to have a unique solution. We will see later that this is the case, under mild conditions on the coefficients in the differential equation.  1.1.5  Direction Fields  Imagine a curve, as in Figure 1.4. If we choose some points on the curve and, at each point, draw a segment of the tangent to the curve there, then these segments give a rough outline of the shape of the curve. This simple observation is the key to a powerful device for envisioning integral curves of a differential equation.  8  CHAPTER 1  First-Order Differential Equations y  x Short tangent segments suggest the shape of the curve.  FIGURE 1.4  The general first-order differential equation has the form F
x y y
  = 0 Suppose we can solve for y
 and write the differential equation as y
 = f
x y Here f is a known function. Suppose f
x y is defined for all points 
x y in some region R of the plane. The slope of the integral curve through a given point 
x0  y0  of R is y
 
x0 , which equals f
x0  y0 . If we compute f
x y at selected points in R, and draw a small line segment having slope f
x y at each 
x y, we obtain a collection of segments which trace out the shapes of the integral curves. This enables us to obtain important insight into the behavior of the solutions (such as where solutions are increasing or decreasing, limits they might have at various points, or behavior as x increases). A drawing of the plane, with short line segments of slope f
x y drawn at selected points 
x y, is called a direction field of the differential equation y
 = f
x y. The name derives from the fact that at each point the line segment gives the direction of the integral curve through that point. The line segments are called lineal elements.  EXAMPLE 1.5  Consider the equation y
 = y2  Here f
x y = y2 , so the slope of the integral curve through 
x y is y2 . Select some points and, through each, draw a short line segment having slope y2 . A computer generated direction field is shown in Figure 1.5(a). The lineal elements form a profile of some integral curves and give us some insight into the behavior of solutions, at least in this part of the plane. Figure 1.5(b) reproduces this direction field, with graphs of the integral curves through 
0 1, 
0 2, 
0 3, 
0 −1, 
0 −2 and 
0 −3. By a method we will develop, the general solution of y
 = y2 is y=−  1  x+k  so the integral curves form a family of hyperbolas, as suggested by the curves sketched in Figure 1.5(b).  1.1 Preliminary Concepts  9  y 4  2  
4  
2  2  0  4  x  
2  
4 FIGURE 1.5(a) A direction field for y
 = y2 .  y 4  2  2 
4  
2  4  0  
2  
4 FIGURE 1.5(b) Direction field for y
 = y2 and integral curves through 
0 1, 
0 2,  
0 3
0 −1, 
0 −2, and 
0 −3.  x  10  CHAPTER 1  First-Order Differential Equations  EXAMPLE 1.6  Figure 1.6 shows a direction field for y
 = sin
xy together with the integral curves through 
0 1, 
0 2, 
0 3, 
0 −1, 
0 −2 and 
0 −3. In this case, we cannot write a simple expression for the general solution, and the direction field provides information about the behavior of solutions that is not otherwise readily apparent. y 4  2  
4  
2  0  2  4  x  
2  
4 Direction field for y
 = sin
xy and integral curves through 
0 1, 
0 2, 
0 3, 
0 −1, 
0 −2, and 
0 −3.  FIGURE 1.6  With this as background, we will begin a program of identifying special classes of firstorder differential equations for which there are techniques for writing the general solution. This will occupy the next five sections.  SECTION 1.1  PROBLEMS  In each of Problems 1 through 6, determine whether the given function is a solution of the differential equation. √ 1. 2yy
 = 1 
x = x − 1 for x > 1 2. y
 + y = 0 
x = Ce−x C −e 2y + e for x > 0 
x = 2x 2x √ 2xy C for x = ± 2 
x = 2 4. y
 = 2 − x2 x −2 3. y
 = −  x  x  5. xy
 = x − y 
x =  x2 − 3 for x = 0 2x  6. y
 + y = 1 
x = 1 + Ce−x In each of Problems 7 through 11, verify by implicit differentiation that the given equation implicitly defines a solution of the differential equation. 7. y2 + xy − 2x2 − 3x − 2y = C y − 4x − 2 + 
x + 2y − 2y
 = 0  1.2 Separable Equations 8. xy3 − y = C y3 + 
3xy2 − 1y
 = 0  17. y
 = x + y y
2 = 2 
  18. y
 = x − xy y
0 = −1  9. y − 4x + e = C 8x − ye − 
2y + xe y = 0 2  2  xy  xy  xy  19. y
 = xy y
0 = 2  10. 8 lnx − 2y + 4 − 2x + 6y = C; y
 =  20. y
 = x − y + 1 y
0 = 1  x − 2y 3x − 6y + 4  2x3 + 2xy2 − y x + 2 y
 = 0 11. tan 
y/x + x = C 2 2 x +y x + y2 −1  2  In each of Problems 12 through 16, solve the initial value problem and graph the solution. Hint: Each of these differential equations can be solved by direct integration. Use the initial condition to solve for the constant of integration. 12. y
 = 2x y
2 = 1 
  In each of Problems 21 through 26, generate a direction field and some integral curves for the differential equation. Also draw the integral curve representing the solution of the initial value problem. These problems should be done by a software package. 21. y
 = sin
y y
1 = /2 22. y
 = x cos
2x − y y
1 = 0 23. y
 = y sin
x − 3x2  y
0 = 1  −x  13. y = e  y
0 = 2  24. y
 = ex − y y
−2 = 1  
  14. y = 2x + 2 y
−1 = 1  25. y
 − y cos
x = 1 − x2  y
2 = 2  15. y
 = 4 cos
xsin
x y
/2 = 0  26. y
 = 2y + 3 y
0 = 1  
  16. y = 8x + cos
2x y
0 = −3 In each of Problems 17 through 20 draw some lineal elements of the differential equation for −4 ≤ x ≤ 4, −4 ≤ y ≤ 4. Use the resulting direction field to sketch a graph of the solution of the initial value problem. (These problems can be done by hand.)  1.2  11  27. Show that, for the differential equation y
 + p
xy = q
x, the lineal elements on any vertical line x = x0 , with p
x0  = 0, all pass through the single point 
 , where  = x0 +  1 p
x0   and  =  q
x0   p
x0   Separable Equations DEFINITION 1.1  Separable Differential Equation  A differential equation is called separable if it can be written y
 = A
xB
y  In this event, we can separate the variables and write, in differential form, 1 dy = A
x dx B
y wherever B
y = 0. We attempt to integrate this equation, writing 
  
 1 dy = A
x dx B
y  This yields an equation in x, y, and a constant of integration. This equation implicitly defines the general solution y
x. It may or may not be possible to solve explicitly for y
x.  12  CHAPTER 1  First-Order Differential Equations  EXAMPLE 1.7  y
 = y2 e−x is separable. Write dy = y2 e−x dx as 1 dx = e−x dx y2 for y = 0. Integrate this equation to obtain 1 − = −e−x + k y an equation that implicitly defines the general solution. In this example we can explicitly solve for y, obtaining the general solution y=  1  −k  e−x  Now recall that we required that y = 0 in order to separate the variables by dividing by y2 . In fact, the zero function y
x = 0 is a solution of y
 = y2 ex , although it cannot be obtained from the general solution by any choice of k. For this reason, y
x = 0 is called a singular solution of this equation. Figure 1.7 shows graphs of particular solutions obtained by choosing k as 0, 3, −3, 6 and −6. y 3  2 k0 1 k  
3 k  
6 
2 k6  
1 k3  1  2  x  
1  Integral curves of y
 = y2 e−x for k = 0 3 −3 6, and −6. FIGURE 1.7  Whenever we use separation of variables, we must be alert to solutions potentially lost through conditions imposed by the algebra used to make the separation.  1.2 Separable Equations  13  EXAMPLE 1.8  x2 y
 = 1 + y is separable, and we can write 1 1 dy = 2 dx 1+y x The algebra of separation has required that x = 0 and y = −1, even though we can put x = 0 and y = −1 into the differential equation to obtain the correct equation 0 = 0. Now integrate the separated equation to obtain 1 ln1 + y = − + k x This implicitly defines the general solution. In this case, we can solve for y
x explicitly. Begin by taking the exponential of both sides to obtain 1 + y = ek e−1/x = Ae−1/x  in which we have written A = ek . Since k could be any number, A can be any positive number. Then 1 + y = ±Ae−1/x = Be−1/x  in which B = ±A can be any nonzero number. The general solution is y = −1 + Be−1/x  in which B is any nonzero number. Now revisit the assumption that x = 0 and y = −1. In the general solution, we actually obtain y = −1 if we allow B = 0. Further, the constant function y
x = −1 does satisfy x2 y
 = 1 + y. Thus, by allowing B to be any number, including 0, the general solution y
x = −1 + Be−1/x contains all the solutions we have found. In this example, y = −1 is a solution, but not a singular solution, since it occurs as a special case of the general solution. Figure 1.8 shows graphs of solutions corresponding to B = −8 −5 0 4 and 7. y  B7  4  B4  2 0 
2  1  2  
4 
6  3  4 5 B0  x  B  
5 B  
8  
8 FIGURE 1.8 Integral curves of x2 y
 = 1 + y for  B = 0 4 7 −5, and −8.  We often solve an initial value problem by finding the general solution of the differential equation, then solving for the appropriate choice of the constant.  14  CHAPTER 1  First-Order Differential Equations  EXAMPLE 1.9  Solve the initial value problem y
 = y2 e−x  y
1 = 4 We know from Example 1.7 that the general solution of y
 = y2 e−x is y
x =  1  −k  e−x  Now we need to choose k so that y
1 =  1 = 4 e−1 − k  from which we get 1 k = e−1 −  4 The solution of the initial value problem is y
x =  1 e−x  + 41 − e−1    EXAMPLE 1.10  The general solution of y
 = y  
x − 12 y+3  is implicitly defined by 1 y + 3 lny = 
x − 13 + k 3  (1.3)  To obtain the solution satisfying y
3 = −1, put x = 3 and y = −1 into equation (1.3) to obtain 1 −1 = 
23 + k 3 hence k=−  11  3  The solution of this initial value problem is implicitly defined by 1 11 y + 3 lny = 
x − 13 −  3 3  1.2.1  Some Applications of Separable Differential Equations  Separable equations arise in many contexts, of which we will discuss three.  1.2 Separable Equations  15  EXAMPLE 1.11  (The Mathematical Policewoman) A murder victim is discovered, and a lieutenant from the forensic science laboratory is summoned to estimate the time of death. The body is located in a room that is kept at a constant 68 degrees Fahrenheit. For some time after the death, the body will radiate heat into the cooler room, causing the body's temperature to decrease. Assuming (for want of better information) that the victim's temperature was a "normal" 986 at the time of death, the lieutenant will try to estimate this time by observing the body's current temperature and calculating how long it would have had to lose heat to reach this point. According to Newton's law of cooling, the body will radiate heat energy into the room at a rate proportional to the difference in temperature between the body and the room. If T
t is the body temperature at time t, then for some constant of proportionality k, T 
 
t = k T
t − 68  The lieutenant recognizes this as a separable differential equation and writes 1 dT = k dt T − 68 Upon integrating, she gets lnT − 68 = kt + C Taking exponentials, she gets T − 68 = ekt+C = Aekt  in which A = eC . Then T − 68 = ±Aekt = Bekt  Then T
t = 68 + Bekt  Now the constants k and B must be determined, and this requires information. The lieutenant arrived at 9:40 p.m. and immediately measured the body temperature, obtaining 944 degrees. Letting 9:40 be time zero for convenience, this means that T
0 = 944 = 68 + B and so B = 264. Thus far, T
t = 68 + 264ekt  To determine k, the lieutenant makes another measurement. At 11:00 she finds that the body temperature is 892 degrees. Since 11:00 is 80 minutes past 9:40, this means that T
80 = 892 = 68 + 264e80k  Then e80k = so  212  264   80k = ln  212 264    16  CHAPTER 1  First-Order Differential Equations  and    1 212 k= ln  80 264  The lieutenant now has the temperature function: T
t = 68 + 264eln
212/264t/80  In order to find when last time when the body was 986 (presumably the time of death), solve for the time in T
t = 986 = 68 + 264eln
212/264t/80  To do this, the lieutenant writes 306 = eln
212/264t/80 264 and takes the logarithm of both sides to obtain     306 t 212 ln = ln  264 80 264 Therefore the time of death, according to this mathematical model, was 80 ln
306/264  ln
212/264  t=  which is approximately −538 minutes. Death occurred approximately 538 minutes before (because of the negative sign) the first measurement at 9:40, which was chosen as time zero. This puts the murder at about 8:46 p.m.  EXAMPLE 1.12  (Radioactive Decay and Carbon Dating) In radioactive decay, mass is converted to energy by radiation. It has been observed that the rate of change of the mass of a radioactive substance is proportional to the mass itself. This means that, if m
t is the mass at time t, then for some constant of proportionality k that depends on the substance, dm = km dt This is a separable differential equation. Write it as 1 dm = k dt m and integrate to obtain lnm = kt + c Since mass is positive, m = m and ln
m = kt + c Then m
t = ekt+c = Aekt  in which A can be any positive number.  1.2 Separable Equations  17  Determination of A and k for a given element requires two measurements. Suppose at some time, designated as time zero, there are M grams present. This is called the initial mass. Then m
0 = A = M so m
t = Mekt  If at some later time T we find that there are MT grams, then m
T = MT = MekT  Then   ln  hence  MT M   = kT    1 MT k = ln  T M  This gives us k and determines the mass at any time: m
t = Meln
MT /Mt/T  We obtain a more convenient formula for the mass if we choose the time of the second measurement more carefully. Suppose we make the second measurement at that time T = H at which exactly half of the mass has radiated away. At this time, half of the mass remains, so MT = M/2 and MT /M = 1/2. Now the expression for the mass becomes m
t = Meln
1/2t/H  or m
t = Me− ln
2t/H   (1.4)  This number H is called the half-life of the element. Although we took it to be the time needed for half of the original amount M to decay, in fact, between any times t1 and t1 + H, exactly half of the mass of the element present at t1 will radiate away. To see this, write m
t1 + H = Me− ln
2
t1 +H/H = Me− ln
2t1 /H e− ln
2H/H = e− ln
2 m
t1  = 21 m
t1  Equation (1.4) is the basis for an important technique used to estimate the ages of certain ancient artifacts. The earth's upper atmosphere is constantly bombarded by high-energy cosmic rays, producing large numbers of neutrons, which collide with nitrogen in the air, changing some of it into radioactive carbon-14, or 14 C. This element has a half-life of about 5,730 years. Over the relatively recent period of the history of this planet in which life has evolved, the fraction of 14 C in the atmosphere, compared to regular carbon, has been essentially constant. This means that living matter (plant or animal) has injested 14 C at about the same rate over a long historical period, and objects living, say, two million years ago would have had the same ratio of carbon-14 to carbon in their bodies as objects alive today. When an organism dies, it ceases its intake of 14 C, which then begins to decay. By measuring the ratio of 14 C to carbon in an artifact, we can estimate the amount of the decay, and hence the time it took, giving an  18  CHAPTER 1  First-Order Differential Equations  estimate of the time the organism was alive. This process of estimating the age of an artifact is called carbon dating. Of course, in reality the ratio of 14 C in the atmosphere has only been approximately constant, and in addition a sample may have been contaminated by exposure to other living organisms, or even to the air, so carbon dating is a sensitive process that can lead to controversial results. Nevertheless, when applied rigorously and combined with other tests and information, it has proved a valuable tool in historical and archeological studies. To apply equation (1.4) to carbon dating, use H = 5730 and compute ln
2 ln
2 = ≈ 0000120968 H 5730 in which ≈ means "approximately equal" (not all decimal places are listed). Equation (1.4) becomes m
t = Me−0000120968t  Now suppose we have an artifact, say a piece of fossilized wood, and measurements show that the ratio of 14 C to carbon in the sample is 37 percent of the current ratio. If we say that the wood died at time 0, then we want to compute the time T it would take for one gram of the radioactive carbon to decay this amount. Thus, solve for T in 037 = e−0000120968T  We find that T =−  ln
037 ≈ 8,219 0000120968  years. This is a little less than one and one-half half-lives, a reasonable estimate if nearly the 14 C has decayed.  2 3  of  EXAMPLE 1.13  (Torricelli's Law) Suppose we want to estimate how long it will take for a container to empty by discharging fluid through a drain hole. This is a simple enough problem for, say, a soda can, but not quite so easy for a large oil storage tank or chemical facility. We need two principles from physics. The first is that the rate of discharge of a fluid flowing through an opening at the bottom of a container is given by dV = −kAv dt in which V
t is the volume of fluid in the container at time t, v
t is the discharge velocity of fluid through the opening, A is the cross sectional area of the opening (assumed constant), and k is a constant determined by the viscosity of the fluid, the shape of the opening, and the fact that the cross-sectional area of fluid pouring out of the opening is slightly less than that of the opening itself. In practice, k must be determined for the particular fluid, container, and opening, and is a number between 0 and 1. We also need Torricelli's law, which states that v
t is equal to the velocity of a free-falling particle released from a height equal to the depth of the fluid at time t. (Free-falling means that the particle is influenced by gravity only). Now the work done by gravity in moving the particle from its initial point by a distance h
t is mgh
t, and this must equal the change in the kinetic energy, 
 21 mv2 . Therefore,  v
t = 2gh
t  1.2 Separable Equations 18 
 h  r  19  18  h  FIGURE 1.9  Putting these two equations together yields  dV = −kA 2gh
t dt  (1.5)  We will apply equation (1.5) to a specific case to illustrate its use. Suppose we have a hemispherical tank of water, as in Figure 1.9. The tank has radius 18 feet, and water drains through a circular hole of radius 3 inches at the bottom. How long will it take the tank to empty? Equation (1.5) contains two unknown functions, V
t and h
t, so one must be eliminated. Let r
t be the radius of the surface of the fluid at time t and consider an interval of time from t0 to t1 = t0 + t. The volume V of water draining from the tank in this time equals the volume of a disk of thickness h (the change in depth) and radius r
t∗ , for some t∗ between t0 and t1 . Therefore V =  r
t∗   2  h  so V =  r
t∗  t  2  h  t  In the limit as t → 0, dV dh = r 2  dt dt Putting this into equation (1.5) yields r 2   dh = −kA 2gh dt  Now V has been eliminated, but at the cost of introducing r
t. However, from Figure 1.9, r 2 = 182 − 
18 − h2 = 36h − h2 so    dh  36h − h2 = −kA 2gh dt This is a separable differential equation, which we write as    36h − h2 dh = −kA 2g dt 1/2 h  Take g to be 32 feet per second per second. The radius of the circular opening is 3 inches, or 1 feet, so its area is A = /16 square feet. For water, and an opening of this shape and size, 4 the experiment gives k = 08. The last equation becomes  √   1 1/2 3/2 dh = −
08 64 dt 36h − h 16  CHAPTER 1  20  First-Order Differential Equations  or     36h1/2 − h3/2 dh = −04 dt  A routine integration yields 2 2 24h3/2 − h5/2 = − t + c 5 5 or 60h3/2 − h5/2 = −t + k Now h
0 = 18, so √  60
183/2 − 
185/2 = k  Thus k = 2268 2 and h
t is implicitly determined by the equation √ 60h3/2 − h5/2 = 2268 2 − t √ The tank is empty when h = 0, and this occurs when t = 2268 2 seconds, or about 53 minutes, 28 seconds. The last three examples contain an important message. Differential equations can be used to solve a variety of problems, but a problem usually does not present itself as a differential equation. Normally we have some event or process, and we must use whatever information we have about it to derive a differential equation and initial conditions. This process is called mathematical modeling. The model consists of the differential equation and other relevant information, such as initial conditions. We look for a function satisfying the differential equation and the other information, in the hope of being able to predict future behavior, or perhaps better understand the process being considered.  SECTION 1.2  PROBLEMS  In each of Problems 1 through 10, determine if the differential equation is separable. If it is, find the general solution (perhaps implicitly defined). If it is not separable, do not attempt a solution at this time.  In each of Problems 11 through 15, solve the initial value problem. 11. xy2 y
 = y + 1 y
3e2  = 2 12. y
 = 3x2 
y + 2 y
2 = 8  1. 3y
 = 4x/y2  13. ln
yx y
 = 3x2 y y
2 = e3  2. y + xy
 = 0  14. 2yy
 = ex−y  y
4 = −2  3. cos
yy
 = sin
x + y  15. yy
 = 2x sec
3y y
2/3 = /3  4. e  x+y 
  y = 3x  
  5. xy + y = y 6. y
 =  2  
x + 12 − 2y 2y  7. x sin
yy
 = cos
y 8.  x 
 2y2 + 1 y = y x+1  9. y + y
 = ex − sin
y 10. cos
x + y + sin
x − y y
 = cos
2x  2  16. An object having a temperature of 90 degrees Fahrenheit is placed into an environment kept at 60 degrees. Ten minutes later the object has cooled to 88 degrees. What will be the temperature of the object after it has been in this environment for 20 minutes? How long will it take for the object to cool to 65 degrees? 17. A thermometer is carried outside a house whose ambient temperature is 70 degrees Fahrenheit. After five minutes the thermometer reads 60 degrees, and fifteen minutes after this, 504 degrees. What is the outside temperature (which is assumed to be constant)?  1.2 Separable Equations 18. Assume that the population of bacteria in a petri dish changes at a rate proportional to the population at that time. This means that, if P
t is the population at time t, then dP = kP dt for some constant k. A particular culture has a population density of 100,000 bacteria per square inch. A culture that covered an area of 1 square inch at 10:00 a.m. on Tuesday was found to have grown to cover 3 square inches by noon the following Thursday. How many bacteria will be present at 3:00 p.m. the following Sunday? How many will be present on Monday at 4:00 p.m.? When will the world be overrun by these bacteria, assuming that they can live anywhere on the earth's surface? (Here you need to look up the land area of the earth.) 19. Assume that a sphere of ice melts at a rate proportional to its surface area, retaining a spherical shape. Interpret melting as a reduction of volume with respect to time. Determine an expression for the volume of the ice at any time t. 20. A radioactive element has a half-life of ln
2 weeks. If e3 tons are present at a given time, how much will be left 3 weeks later? 21. The half-life of uranium-238 is approximately 45 · 109 years. How much of a 10-kilogram block of U-238 will be present 1 billion years from now? 22. Given that 12 grams of a radioactive element decays to 91 grams in 4 minutes, what is the half-life of this element? 23. Evaluate 
  e−t  2 −9/t 2  dt  0  Hint: Let I
x =  
  e−t  2 −
x/t2  dt  0  Calculate I 
 
x by differentiating under the integral sign, then let u = x/t. Show that I 
 
x = −2I
x and solve for I
x. Evaluate the constant by using the stan √ 2 dard result that 0 e−t dt = /2. Finally, evaluate I
3. 24. Derive the fact used in Example 1.13 that v
t =  2gh
t. Hint: Consider a free-falling particle having height h
t at time t. The work done by gravity in moving the particle from its starting point to a given point is mgh
t, and this must equal the change in the kinetic energy, which is 
1/2mv2 .  21  25. Calculate the time required to empty the hemispherical tank of Example 1.13 if the tank is positioned with its flat side down. 26. (Draining a Hot Tub) Consider a cylindrical hot tub with a 5-foot radius and height of 4 feet, placed on one of its circular ends. Water is draining from the tub through a circular hole 58 inches in diameter located in the base of the tub. (a) Assume a value k = 06 to determine the rate at which the depth of the water is changing. Here it is useful to write dh dh dV dV/dt = =  dt dV dt dV/dh (b) Calculate the time T required to drain the hot tub if it is initially full. Hint: One way to do this is to write T=  
  0 H  dt dh dh  (c) Determine how much longer it takes to drain the lower half than the upper half of the tub. Hint: Use the integral suggested in (b), with different limits for the two halves. 27. (Draining a Cone) A tank shaped like a right circular cone, with its vertex down, is 9 feet high and has a diameter of 8 feet. It is initially full of water. (a) Determine the time required to drain the tank through a circular hole of diameter 2 inches at the vertex. Take k = 06. (b) Determine the time it takes to drain the tank if it is inverted and the drain hole is of the same size and shape as in (a), but now located in the new base. 28. (Drain Hole at Unknown Depth) Determine the rate of change of the depth of water in the tank of Problem 27 (vertex at the bottom) if the drain hole is located in the side of the cone 2 feet above the bottom of the tank. What is the rate of change in the depth of the water when the drain hole is located in the bottom of the tank? Is it possible to determine the location of the drain hole if we are told the rate of change of the depth and the depth of the water in the tank? Can this be done without knowing the size of the drain opening? 29. Suppose the conical tank of Problem 27, vertex at the bottom, is initially empty and water is added at the constant rate of /10 cubic feet per second. Does the tank ever overflow? 30. (Draining a Sphere) Determine the time it takes to completely drain a spherical tank of radius 18 feet if it is initially full of water and the water drains through a circular hole of radius 3 inches located in the bottom of the tank. Use k = 08.  22  1.3  CHAPTER 1  First-Order Differential Equations  Linear Differential Equations DEFINITION 1.2  Linear Differential Equation  A first-order differential equation is linear if it has the form y
 
x + p
xy = q
x Assume that p and q are continuous on an interval I (possibly the whole real line). Because of the special form of the linear equation, we can obtain the general solution on I by a clever  observation. Multiply the differential equation by e p
x dx to get   e  p
x dx 
  y 
x + p
xe    p
x dx  y = q
xe    p
x dx  The left side of this equation is the derivative of the product y
xe      p
x dx  , enabling us to write    d  y
xe p
x dx = q
xe p
x dx  dx  Now integrate to obtain   y
xe  p
x dx  Finally, solve for y
x: y
x = e−    p
x dx  
  =    q
xe  
    q
xe  p
x dx  p
x dx  dx + C  dx + Ce−    p
x dx    (1.6)    The function e p
x dx is called an integrating factor for the differential equation, because multiplication of the differential equation by this factor results in an equation that can be integrated to obtain the general solution. We do not recommend memorizing equation (1.6). Instead, recognize  the form of the linear equation and understand the technique of solving it by multiplying by e p
x dx . EXAMPLE 1.14  The equation y
 + y = sin
x is linear. Here p
x = 1 and q
x = sin
x, both continuous for all x. An integrating factor is   e  dx    or ex . Multiply the differential equation by ex to get y
 ex + yex = ex sin
x or 
yex 
 = ex sin
x Integrate to get  
 1 yex = ex sin
x dx = ex sin
x − cos
x + C 2 The general solution is y
x =  1 sin
x − cos
x + Ce−x  2  1.3 Linear Differential Equations  23  EXAMPLE 1.15  Solve the initial value problem y y
 = 3x2 −  x  y
1 = 5  First recognize that the differential equation can be written in linear form: 1 y
 + y = 3x2  x An integrating factor is e to get    
1/x dx  = eln
x = x, for x > 0. Multiply the differential equation by x xy
 + y = 3x3   or 
xy
 = 3x3  Integrate to get 3 xy = x4 + C 4 Then C 3 y
x = x3 + 4 x for x > 0. For the initial condition, we need y
1 = 5 =  3 +C 4  so C = 17/4 and the solution of the initial value problem is 17 3 y
x = x3 + 4 4x for x > 0. Depending on p and q, it may not be possible to evaluate all of the integrals in the general solution 1.6 in closed form (as a finite algebraic combination of elementary functions). This occurs with y
 + xy = 2 whose general solution is y
x = 2e−x  2 /2  
  ex  2 /2  dx + Ce−x /2  2   2 We cannot write ex /2 dx in elementary terms. However, we could still use a software package to generate a direction field and integral curves, as is done in Figure 1.10. This provides some idea of the behavior of solutions, at least within the range of the diagram.  24  CHAPTER 1  First-Order Differential Equations y 4  2  
4  
2  x 0  2  4  
2  
4  Integral curves of y
 + xy = 2 passing through 
0 2, 
0 4, 
0 −2, and 
0 −5. FIGURE 1.10  Linear differential equations arise in many contexts. Example 1.11, involving estimation of time of death, involved a separable differential equation which is also linear and could have been solved using an integrating factor.  EXAMPLE 1.16  (A Mixing Problem) Sometimes we want to know how much of a given substance is present in a container in which various substances are being added, mixed, and removed. Such problems are called mixing problems, and they are frequently encountered in the chemical industry and in manufacturing processes. As an example, suppose a tank contains 200 gallons of brine (salt mixed with water), in which 100 pounds of salt are dissolved. A mixture consisting of 18 pound of salt per gallon is flowing into the tank at a rate of 3 gallons per minute, and the mixture is continuously stirred. Meanwhile, brine is allowed to empty out of the tank at the same rate of 3 gallons per minute (Figure 1.11). How much salt is in the tank at any time?  1  8 lb/gal; 3 gal/min  3 gal/min  FIGURE 1.11  Before constructing a mathematical model, notice that the initial ratio of salt to brine in the tank is 100 pounds per 200 gallons, or 21 pound per gallon. Since the mixture pumped in has a constant ratio of 18 pound per gallon, we expect the brine mixture to dilute toward the incoming ratio, with a "terminal" amount of salt in the tank of 18 pound per gallon, times 200 gallons. This leads to the expectation that in the long term (as t → ) the amount of salt in the tank should approach 25 pounds.  1.3 Linear Differential Equations  25  Now let Q
t be the amount of salt in the tank at time t. The rate of change of Q
t with time must equal the rate at which salt is pumped in, minus the rate at which it is pumped out. Thus dQ = (rate in) − (rate out) dt       1 pounds gallons Q
t pounds gallons = 3 − 3 8 gallon minute 200 gallon minute =  3 3 − Q
t 8 200  This is the linear equation Q
 
t + An integrating factor is e obtain    
3/200 dt  3 3 Q=  200 8  = e3t/200 . Multiply the differential equation by this factor to  Q
 e3t/200 +  3 3t/200 3 Q = e3t/200  e 200 8  or   
 3 Qe3t/200 = e3t/200  8  Then Qe3t/200 =  3 200 3t/200 e + C 8 3  so Q
t = 25 + Ce−3t/200  Now Q
0 = 100 = 25 + C so C = 75 and Q
t = 25 + 75e−3t/200  As we expected, as t increases, the amount of salt approaches the limiting value of 25 pounds. From the derivation of the differential equation for Q
t, it is apparent that this limiting value depends on the rate at which salt is poured into the tank, but not on the initial amount of salt in the tank. The term 25 in the solution is called the steady-state part of the solution because it is independent of time, and the term 75e−3t/200 is the transient part. As t increases, the transient part exerts less influence on the amount of salt in the tank, and in the limit the solution approaches its steady-state part.  CHAPTER 1  26  First-Order Differential Equations  SECTION 1.3  PROBLEMS  In each of Problems 1 through 8, find the general solution. Not all integrals can be done in closed form. 3 1. y
 − y = 2x2 x 2. y
 − y = sinh
x 3. y
 + 2y = x 4. sin
2xy
 + 2y sin2 
x = 2 sin
x 5. y
 − 2y = −8x2 6. 
x2 − x − 2y
 + 3xy = x2 − 4x + 4 7. y
 + y =  x−1 x2  8. y
 + sec
xy = cos
x In each of Problems 9 through 14, solve the initial value problem. 9. y
 +  1 y = 3x y
3 = 4 x−2  10. y
 + 3y = 5e2x − 6 y
0 = 2 11. y
 +  2 y = 3 y
0 = 5 x+1  16. A 500-gallon tank initially contains 50 gallons of brine solution in which 28 pounds of salt have been dissolved. Beginning at time zero, brine containing 2 pounds of salt per gallon is added at the rate of 3 gallons per minute, and the mixture is poured out of the tank at the rate of 2 gallons per minute. How much salt is in the tank when it contains 100 gallons of brine? Hint: The amount of brine in the tank at time t is 50 + t. 17. Two tanks are cascaded as in Figure 1.12. Tank 1 initially contains 20 pounds of salt dissolved in 100 gallons of brine, while tank 2 contains 150 gallons of brine in which 90 pounds of salt are dissolved. At time zero a brine solution containing 21 pound of salt per gallon is added to tank 1 at the rate of 5 gallons per minute. Tank 1 has an output that discharges brine into tank 2 at the rate of 5 gallons per minute, and tank 2 also has an output of 5 gallons per minute. Determine the amount of salt in each tank at any time t. Also determine when the concentration of salt in tank 2 is a minimum and how much salt is in the tank at that time. Hint: Solve for the amount of salt in tank 1 at time t first and then use this solution to determine the amount in tank 2.  5 gal/min  12. 
x2 − 2xy
 + 
x2 − 5x + 4y = 
x4 − 2x3 e−x  y
3 = 18e−3  1 2  lb/gal Tank 1  13. y
 − y = 2e4x  y
0 = −3 14. y
 +  5y = 3x3 + x y
−1 = 4 9x  Tank 2 5 gal/min  15. Find all functions with the property that the y-intercept of the tangent to the graph at 
x y is 2x2 .  1.4  5 gal/min  FIGURE 1.12  Mixing between tanks in Problem 17.  Exact Differential Equations We continue the theme of identifying certain kinds of first-order differential equations for which there is a method leading to a solution. We can write any first order equation y
 = f
x y in the form M
x y + N
x yy
 = 0 For example, put M
x y = −f
x y and N
x y = 1. An interesting thing happens if there is a function  such that  = M
x y and x   = N
x y y  (1.7)  1.4 Exact Differential Equations  27  In this event, the differential equation becomes   dy + = 0 x y dx which, by the chain rule, is the same as d 
x y
x = 0 dx But this means that 
x y
x = C with C constant. If we now read this argument from the last line back to the first, the conclusion is that the equation 
x y = C implicitly defines a function y
x that is the general solution of the differential equation. Thus, finding a function that satisfies equation (1.7) is equivalent to solving the differential equation. Before taking this further, consider an example.  EXAMPLE 1.17  The differential equation y
 = −  2xy3 + 2 3x2 y2 + 8e4y  is neither separable nor linear. Write it in the form   M + Ny
 = 2xy3 + 2 + 3x2 y2 + 8e4y y
 = 0  (1.8)  with M
x y = 2xy3 + 2  and N
x y = 3x2 y2 + 8e4y   Equation (1.8) can in turn be written M dx + N dy = 
2xy3 + 2 dx + 
3x2 y2 + 8e4y  dy = 0 Now let 
x y = x2 y3 + 2x + 2e4y  Soon we will see where this came from, but for now, observe that  = 2xy3 + 2 = M x  and   = 3x2 y2 + 8e4y = N y  With this choice of 
x y, equation (1.9) becomes   dx + dy = 0 x y or d 
x y = 0  (1.9)  28  CHAPTER 1  First-Order Differential Equations  The general solution of this equation is 
x y = C or, in this example, x2 y3 + 2x + 2e4y = C This implicitly defines the general solution of the differential equation (1.8). To verify this, differentiate the last equation implicitly with respect to x: 2xy3 + 3x2 y2 y
 + 2 + 8e4y y
 = 0 or 2xy3 + 2 + 
3x2 y2 + 8e4y y
 = 0 This is equivalent to the original differential equation y
 = −  2xy3 + 2  3x2 y2 + 8e4y  With this as background, we will make the following definitions.  DEFINITION 1.3  Potential Function  A function  is a potential function for the differential equation M
x y + N
x yy
 = 0 on a region R of the plane if, for each 
x y in R,  = M
x y and x  DEFINITION 1.4   = N
x y y  Exact Differential Equation  When a potential function exists on a region R for the differential equation M + Ny
 = 0, then this equation is said to be exact on R.  The differential equation of Example 1.17 is exact (over the entire plane), because we exhibited a potential function for it, defined for all 
x y. Once a potential function is found, we can write an equation implicitly defining the general solution. Sometimes we can explicitly solve for the general solution, and sometimes we cannot. Now go back to Example 1.17. We want to explore how the potential function that materialized there was found. Recall we required that  = 2xy3 + 2 = M x  and   = 3x2 y2 + 8e4y = N y  1.4 Exact Differential Equations  29  Pick either of these equations to begin and integrate it. Say we begin with the first. Then integrate with respect to x: 
x y =  
   dx = 2xy3 + 2 dx x  
  = x2 y3 + 2x + g
y In this integration with respect to x we held y fixed, hence we must allow that y appears in the "constant" of integration. If we calculate / x, we get 2xy2 + 2 for any function g
y. Now we know  to within this function g. Use the fact that we know / y to write  = 3x2 y2 + 8e4y y =  y  
x2 y3 + 2x + g
y = 3x2 y2 + g 
 
y  This equation holds if g 
 
y = 8e4y , hence we may choose g
y = 2e4y . This gives the potential function 
x y = x2 y3 + 2x + 2e4y  If we had chosen to integrate / y first, we would have gotten 
x y =  
   3x2 y2 + 8e4y dy  = x2 y3 + 2e4y + h
x Here h can be any function of one variable, because no matter how h
x is chosen,  x2 y3 + 2e4y + h
x = 3x2 y2 + 8e4y    y  as required. Now we have two expressions for / x:  = 2xy3 + 2 x =   x2 y3 + 2e4y + h
x = 2xy3 + h
 
x   x  This equation forces us to choose h so that h
 
x = 2, and we may therefore set h
x = 2x. This gives 
x y = x2 y3 + 2e4y + 2x as we got before. Not every first-order differential equation is exact. For example, consider y + y
 = 0  30  CHAPTER 1  First-Order Differential Equations  If there were a potential function , then we would have  = y x   = 1 y  Integrate / x = y with respect to x to get 
x y = xy + g
y. Substitute this into / y = 1 to get  y  
xy + g
y = x + g 
 
y = 1  But this can hold only if g 
 
y = 1 − x, an impossibility if g is to be independent of x. Therefore, y + y
 = 0 has no potential function. This differential equation is not exact (even though it is easily solved either as a separable or as a linear equation). This example suggests the need for a convenient test for exactness. This is provided by the following theorem, in which a "rectangle in the plane" refers to the set of points on or inside any rectangle having sides parallel to the axes.  THEOREM 1.1  Test for Exactness  Suppose M
x y, N
x y, M/ y, and N/ x are continuous for all 
x y within a rectangle R in the plane. Then, M
x y + N
x yy
 = 0 is exact on R if and only if, for each 
x y in R, M N =  y x Proof  If M + Ny
 = 0 is exact, then there is a potential function  and  = M
x y and x   = N
x y y  Then, for 
x y in R, M = y y     x    2  2   = = = y x x y x     y   =  N  x  Conversely, suppose M/ y and N/ x are continuous on R. Choose any 
x0  y0  in R and define, for 
x y in R, 
x y =  
  x  x0  M
 y0  d +  
  y  N
x  d  y0  Immediately we have, from the fundamental theorem of calculus,  = N
x y y  (1.10)  1.4 Exact Differential Equations  31  since the first integral in equation (1.10) is independent of y. Next, compute 
 x 
 y  = M
 y0  d + N
x  d x x x0 x y0 
 y N = M
x y0  + 
x  d x y0 
 y M 
x  d = M
x y0  + y y0 = M
x y0  + M
x y − M
x y0  = M
x y and the proof is complete. For example, consider again y + y
 = 0. Here M
x y = y and N
x y = 1, so N =0 x  and  M =1 y  throughout the entire plane. Thus, y + y
 = 0 cannot be exact on any rectangle in the plane. We saw this previously by showing that this differential equation can have no potential function.  EXAMPLE 1.18  Consider x2 + 3xy + 
4xy + 2xy
 = 0 Here M
x y = x2 + 3xy and N
x y = 4xy + 2x. Now N = 4y + 2 x  M = 3x y  and  and 3x = 4y + 2 is satisfied by all 
x y on a straight line. However, N/ x = M/ y cannot hold for all 
x y in an entire rectangle in the plane. Hence this differential equation is not exact on any rectangle.  EXAMPLE 1.19  Consider ex sin
y − 2x + 
ex cos
y + 1 y
 = 0 With M
x y = ex sin
y − 2x and N
x y = ex cos
y + 1, we have M N = ex cos
y = x y for all 
x y. Therefore this differential equation is exact. To find a potential function, set  = ex sin
y − 2x x  and   = ex cos
y + 1 y  32  CHAPTER 1  First-Order Differential Equations  Choose one of these equations and integrate it. Integrate the second equation with respect to y: 
 
x y = 
ex cos
y + 1 dy = ex sin
y + y + h
x Then we must have  = ex sin
y − 2x x =  x  
ex sin
y + y + h
x = ex sin
y + h
 
x  Then h
 
x = −2x and we may choose h
x = −x2 . A potential function is 
x y = ex sin
y + y − x2  The general solution of the differential equation is defined implicitly by ex sin
y + y − x2 = C Note of Caution: If  is a potential function for M + Ny
 = 0,  itself is not the solution. The general solution is defined implicitly by the equation 
x y = C.  SECTION 1.4  PROBLEMS  In each of Problems 1 through 8, determine where (if anywhere) in the plane the differential equation is exact. If it is exact, find a potential function and the general solution, perhaps implicitly defined. If the equation is not exact, do not attempt a solution at this time. 1. 2y2 + yexy + 
4xy + xexy + 2yy
 = 0 2. 4xy + 2x + 
2x2 + 3y2 y
 = 0 3. 4xy + 2x2 y + 
2x2 + 3y2 y
 = 0 4. 2 cos
x + y − 2x sin
x + y − 2x sin
x + yy
 = 0 1 5. + y + 
3y2 + xy
 = 0 x 6. ex sin
y2  + xex sin
y2  + 
2xyex sin
y2  + ey y
 = 0 7. sinh
x sinh
y + cosh
x cosh
yy
 = 0 8. 4y4 + 3 cos
x + 
16y3 x − 3 cos
yy
 = 0 In each of Problems 9 through 14, determine if the differential equation is exact in some rectangle containing in its interior the point where the initial condition is given. If so, solve the initial value problem. This solution may be implicitly defined. If the differential equation is not exact, do not attempt a solution.  9. 3y4 − 1 + 12xy3 y
 = 0 y
1 = 2 10. 2y − y2 sec2 
xy2  + 
2x − 2xy sec2 
xy2 y
 = 0 y
1 = 2 11. x cos
2y − x − sin
2y − x − 2x cos
2y − xy
 = 0 y
/12 = /8 y 12. 1 + ey/x − ey/x + ey/x y
 = 0 y
1 = −5 x 13. y sinh
y − x − cosh
y − x + y sinh
y − xy
 = 0 y
4 = 4 14. ey + 
xey − 1y
 = 0 y
5 = 0 In Problems 15 and 16, choose a constant  so that the differential equation is exact, then produce a potential function and obtain the general solution. 15. 2xy3 − 3y − 
3x + x2 y2 − 2yy
 = 0 16. 3x2 + xy − x2 y−1 y
 = 0 17. Let  be a potential function for M + Ny
 = 0 in some region R of the plane. Show that for any constant c,  + c is also a potential function. How does the general solution of M + Ny
 = 0 obtained by using  differ from that obtained using  + c?  1.5 Integrating Factors  1.5  33  Integrating Factors "Most" differential equations are not exact on any rectangle. But sometimes we can multiply the differential equation by a nonzero function 
x y to obtain an exact equation. Here is an example that suggests why this might be useful.  EXAMPLE 1.20  The equation y2 − 6xy + 
3xy − 6x2 y
 = 0  (1.11)  is not exact on any rectangle. Multiply it by 
x y = y to get y3 − 6xy2 + 
3xy2 − 6x2 yy
 = 0  (1.12)  Wherever y = 0, equations (1.11) and (1.12) have the same solution. The reason for this is that equation (1.12) is just y y2 − 6xy + 
3xy − 6x2 y
 = 0 and if y = 0, then necessarily y2 − 6xy + 
3xy − 6x2 y
 = 0. Now notice that equation (1.12) is exact (over the entire plane), having potential function 
x y = xy3 − 3x2 y2  Thus the general solution of equation (1.12) is defined implicitly by xy3 − 3x2 y2 = C and, wherever y = 0, this defines the general solution of equation (1.11) as well. To review what has just occurred, we began with a nonexact differential equation. We multiplied it by a function  chosen so that the new equation was exact. We solved this exact equation, then found that this solution also worked for the original, nonexact equation. The function  therefore enabled us to solve a nonexact equation by solving an exact one. This idea is worth pursuing, and we begin by giving a name to .  DEFINITION 1.5  Let M
x y and N
x y be defined on a region R of the plane. Then 
x y is an integrating factor for M + Ny
 = 0 if 
x y = 0 for all 
x y in R, and M + Ny
 = 0 is exact on R.  How do we find an integrating factor for M + Ny
 = 0? For  to be an integrating factor, M + Ny
 = 0 must be exact (in some region of the plane), hence x  
N =  y  
M  (1.13)  34  CHAPTER 1  First-Order Differential Equations  in this region. This is a starting point. Depending on M and N , we may be able to determine  from this equation. Sometimes equation (1.13) becomes simple enough to solve if we try  as a function of just x or just y.  EXAMPLE 1.21  The differential equation x − xy − y
 = 0 is not exact. Here M = x − xy and N = −1 and equation (1.13) is x  
− =  y  

x − xy  Write this as −    = 
x − xy − x x y  Now observe that this equation is simplified if we try to find  as just a function of x, because in this event / y = 0 and we are left with just  = x x This is separable. Write 1 d = x dx  and integrate to obtain ln = 21 x2  Here we let the constant of integration be zero because we need only one integrating factor. From the last equation, choose 2  
x = ex /2  a nonzero function. Multiply the original differential equation by ex 
x − xye  x2 /2  −e  x2 /2  2 /2  to obtain  y
 = 0  This equation is exact over the entire plane, and we find the potential function 
x y = 2 
1 − yex /2 . The general solution of this exact equation is implicitly defined by 
1 − yex  2 /2  = C  In this case, we can explicitly solve for y to get y
x = 1 − Ce−x /2  2  and this is also the general solution of the original equation x − xy − y
 = 0. If we cannot find an integrating factor that is a function of just x or just y, then we must try something else. There is no template to follow, and often we must start with equation (1.13) and be observant.  1.5 Integrating Factors  35  EXAMPLE 1.22  Consider 2y2 − 9xy + 
3xy − 6x2 y
 = 0. This is not exact. With M = 2y2 − 9xy and N = 3xy − 6x2 , begin looking for an integrating factor by writing equation (1.13): x  
3xy − 6x2  =  y  
2y2 − 9xy   This is 
3xy − 6x2     + 
3y − 12x = 
2y2 − 9xy + 
4y − 9x x y  (1.14)  If we attempt  = 
x, then / y = 0 and we obtain 
3xy − 6x2    + 
3y − 12x = 
4y − 9x x  which cannot be solved for  as just a function of x. Similarly, if we try  = 
y, so / x = 0, we obtain an equation we cannot solve. We must try something else. Notice that equation (1.14) involves only integer powers of x and y. This suggests that we try 
x y = xa yb . Substitute this into equation (1.14) and attempt to choose a and b. The substitution gives us 3axa yb+1 − 6axa+1 yb + 3xa yb+1 − 12xa+1 yb = 2bxa yb+1 − 9bxa+1 yb + 4xa yb+1 − 9xa+1 yb  Assume that x = 0 and y = 0. Then we can divide by xa yb to get 3ay − 6ax + 3y − 12x = 2by − 9bx + 4y − 9x Rearrange terms to write 
1 + 2b − 3ay = 
−3 + 9b − 6ax Since x and y are independent, this equation can hold for all x and y only if 1 + 2b − 3a = 0  and  − 3 + 9b − 6a = 0  Solve these equations to obtain a = b = 1. An integrating factor is 
x y = xy. Multiply the differential equation by xy to get 2xy3 − 9x2 y2 + 
3x2 y2 − 6x3 yy
 = 0 This is exact with potential function 
x y = x2 y3 − 3x3 y2 . For x = 0 and y = 0, the solution of the original differential equation is given implicitly by x2 y3 − 3x3 y2 = C The manipulations used to find an integrating factor may fail to find some solutions, as we saw with singular solutions of separable equations. Here are two examples in which this occurs.  EXAMPLE 1.23  Consider 2xy − y
 = 0 y−1  (1.15)  36  CHAPTER 1  First-Order Differential Equations  We can solve this as a separable equation, but here we want to make a point about integrating factors. Equation (1.15) is not exact, but 
x y = 
y − 1/y is an integrating factor for y = 0, a condition not required by the differential equation itself. Multiplying the differential equation by 
x y yields the exact equation 2x −  y−1 
 y = 0 y  with potential function 
x y = x2 − y + lny and general solution defined by x2 − y + lny = C  for y = 0  This is also the general solution of equation (1.15), but the method used has required that y = 0. However, we see immediately that y = 0 is also a solution of equation (1.15). This singular solution is not contained in the expression for the general solution for any choice of C.  EXAMPLE 1.24  The equation y − 3 − xy
 = 0  (1.16)  is not exact, but 
x y = 1/x
y − 3 is an integrating factor for x = 0 and y = 3, conditions not required by the differential equation itself. Multiplying equation (1.16) by 
x y yields the exact equation 1 
 1 − y = 0 x y−3 with general solution defined by lnx + C = lny − 3 This is also the general solution of equation (1.16) in any region of the plane not containing the lines x = 0 or y = 3. This general solution can be solved for y explicitly in terms of x. First, any real number is the natural logarithm of some positive number, so write the arbitrary constant as C = ln
k, in which k can be any positive number. The equation for the general solution becomes lnx + ln
k = lny − 3 or lnkx = lny − 3 But then y − 3 = ±kx. Replacing ±k with K, which can now be any nonzero real number, we obtain y = 3 + Kx as the general solution of equation (1.16). Now observe that y = 3 is a solution of equation (1.16). This solution was "lost", or at least not found, in using the integrating factor as a method of solution. However, y = 3 is not a singular solution because we can include it in the expression y = 3 + Kx by allowing K = 0. Thus the general solution of equation (1.16) is y = 3 + Kx, with K any real number.  1.5 Integrating Factors  1.5.1  37  Separable Equations and Integrating Factors  We will point out a connection between separable equations and integrating factors. The separable equation y
 = A
xB
y is in general not exact. To see this, write it as A
xB
y − y
 = 0 so in the present context we have M
x y = A
xB
y and N
x y = −1. Now x  
−1 = 0  and  y  A
xB
y = A
xB
 
y  and in general A
xB
 
y = 0. However, 
y = 1/B
y is an integrating factor for the separable equation. If we multiply the differential equation by 1/B
y, we get A
x −  1 
 y = 0 B
y  an exact equation because x  −  1 = A
x = 0 B
y y  The act of separating the variables is the same as multiplying by the integrating factor 1/B
y.  1.5.2  Linear Equations and Integrating Factors  Consider the linear equation y
 + p
xy = q
x. We can write this as p
xy − q
x + y
 = 0, so in the present context, M
x y = p
xy − q
x and N
x y = 1. Now x  1 =0  and  y  p
xy − q
x = p
x  so the linear equation is not exact unless p
x is identically zero. However, 
x y = e is an integrating factor. Upon multiplying the linear equation by , we get p
xy − q
x e    p
x dx  and this is exact because   x  SECTION 1.5  e  p
x dx  = p
xe    p
x dx  =  +e   y      p
x dx  p
x dx 
  y = 0  p
xy − q
x e    p
x dx     PROBLEMS  1. Determine a test involving M and N to tell when M + Ny
 = 0 has an integrating factor that is a function of y only. 2. Determine a test to determine when M + Ny
 = 0 has an integrating factor of the form 
x y = xa yb for some constants a and b. 3. Consider y − xy
 = 0. (a) Show that this equation is not exact on any rectangle.  (b) Find an integrating factor 
x that is a function of x alone. (c) Find an integrating factor 
y that is a function of y alone. (d) Show that there is also an integrating factor 
x y = xa yb for some constants a and b. Find all such integrating factors.  CHAPTER 1  38  First-Order Differential Equations  In each of Problems 4 through 12, (a) show that the differential equation is not exact, (b) find an integrating factor, (c) find the general solution (perhaps implicitly defined), and (d) determine any singular solutions the differential equation might have. 
  4. xy − 3y = 2x  13. 1 + xy
 = 0 y
e4  = 0 14. 3y + 4xy
 = 0 y
1 = 6 15. 2
y3 − 2 + 3xy2 y
 = 0 y
3 = 1  3  16. y
1 + x + 2xy
 = 0 y
4 = 6  5. 1 + 
3x − e−2y y
 = 0  2  6. 6x2 y + 12xy + y2 + 
6x2 + 2yy
 = 0  17. 2xy + 3y
 = 0 y
0 = 4 (Hint: try  = ya ebx )  7. 4xy + 6y2 + 
2x2 + 6xyy
 = 0  18. 2y
1 + x2  + xy
 = 0 y
2 = 3 (Hint: try  = xa ebx )  8. y2 + y − xy
 = 0  19. sin
x − y + cos
x − y − cos
x − yy
 = 0  2  y
0 = 7/6  9. 2xy2 + 2xy + 
x2 y + x2 y
 = 0 10. 2y2 − 9xy + 
3xy − 6x2 y
 = 0 (Hint: try 
x y = xa y b ) 
  11. y + y = y (Hint: try 
x y = e y ) 4  ax b  12. x2 y
 + xy = −y−3/2 (Hint: try 
x y = xa yb ) In each of Problems 13 through 20, find an integrating factor, use it to find the general solution of the differential equation, and then obtain the solution of the initial value problem.  1.6  20. 3x2 y + y3 + 2xy2 y
 = 0 y
2 = 1  21. Show that any nonzero constant multiple of an integrating factor for M + Ny
 = 0 is also an integrating factor. 22. Let 
x y be an integrating factor for M + Ny
 = 0 and suppose that the general solution is defined by 
x y = C. Show that 
x yG

x y is also an integrating factor, for any differentiable function G of one variable.  Homogeneous, Bernoulli, and Riccati Equations In this section we will consider three additional kinds of first-order differential equations for which techniques for finding solutions are available.  1.6.1  Homogeneous Differential Equations  DEFINITION 1.6  Homogeneous Equation  A first-order differential equation is homogeneous if it has the form y y
 = f  x  1.6 Homogeneous, Bernoulli, and Riccati Equations  39  In a homogeneous equation, y
 is isolated on one side, and the other side is some expression in which y and x must always appear in the combination y/x. For example, y x y
 = sin y x is homogeneous, while y
 = x2 y is not. Sometimes algebraic manipulation will put a first order equation into the form of the homogeneous equation. For example, y y
 = (1.17) x+y is not homogeneous. However, if x = 0, we can write this as y
 =  y/x  1 + y/x  (1.18)  a homogeneous equation. Any technique we develop for homogeneous equations can therefore be used on equation (1.18). However, this solution assumes that x = 0, which is not required in equation (1.17). Thus, as we have seen before, when we perform manipulations on a differential equation, we must be careful that solutions have not been overlooked. A solution of equation (1.18) will also satisfy (1.17), but equation (1.17) may have other solutions as well. Now to the point. A homogeneous equation is always transformed into a separable one by the transformation y = ux. To see this, compute y
 = u
 x + x
 u = u
 x + u and write u = y/x. Then y
 = f
y/x becomes u
 x + u = f
u We can write this as 1 du 1 =  f
u − u dx x or, in differential form, 1 1 du = dx f
u − u x and the variables (now x and u) have been separated. Upon integrating this equation, we obtain the general solution of the transformed equation. Substituting u = y/x then gives the general solution of the original homogeneous equation.  EXAMPLE 1.25  Consider xy
 = Write this as y
 =  y2 + y x  y x  2  y +  x  Let y = ux. Then u
 x + u = u2 + u  40  CHAPTER 1  First-Order Differential Equations  or u
 x = u 2  Write this as 1 1 du = dx 2 u x and integrate to obtain −  1 = lnx + C u  Then u
x =  −1  lnx + C  the general solution of the transformed equation. The general solution of the original equation is y=  −x  lnx + C  EXAMPLE 1.26 A Pursuit Problem  A pursuit problem is one of determining a trajectory so that one object intercepts another. Examples involving pursuit problems are missiles fired at airplanes and a rendezvous of a shuttle with a space station. These are complex problems that require numerical approximation techniques. We will consider a simple pursuit problem that can be solved explicitly. Suppose a person jumps into a canal of constant width w and swims toward a fixed point directly opposite the point of entry into the canal. The person's speed is v and the water current's speed is s. Assume that, as the swimmer makes his way across, he always orients to point toward the target. We want to determine the swimmer's trajectory. Figure 1.13 shows a coordinate system drawn so that the swimmer's destination is the origin and the point of entry into the water is 
w 0. At time t the swimmer is at the point 
x
t y
t. The horizontal and vertical components of his velocity are, respectively, x
 
t = −v cos
 and  y
 
t = s − v sin
  with  the angle between the positive x axis and 
x
t y
t at time t. From these equations, dy y
 
t s − v sin
 s = 
 = = tan
 − sec
 dx x 
t −v cos
 v  y (x, y) v  α  v sin(α) v cos(α)  (0, 0) FIGURE 1.13  (w, 0) The swimmer's path.  x  1.6 Homogeneous, Bernoulli, and Riccati Equations  41  From Figure 1.13, tan
 =  y x  and  1  2 x + y2  x  sec
 =  Therefore dy y s1  2 = − x + y2  dx x v x Write this as the homogeneous equation dy y s = − dx x v   1+  y  2  x  and put y = uv to obtain 1 s1 du = − dx √ 2 v x 1+u Integrate to get     s   lnu + 1 + u2  = − lnx + C v  Take the exponential of both sides of this equation:      u + 1 + u2  = eC e−
s lnx/v  We can write this as   u + 1 + u2 = Kx−s/v   This equation can be solved for u. First write  1 + u2 = Kx−s/v − u and square both sides to get 1 + u2 = K 2 e−2s/v − 2Kue−s/v + u2  Now u2 cancels and we can solve for u: 1 1 1 s/v u
x = Kx−s/v − x  2 2K Finally, put u = y/x to get 1 1 1+s/v 1  x y
x = Kx1−s/v − 2 2K To determine K, notice that y
w = 0, since we put the origin at the point of destination. Thus, 1 1 1 1+s/v Kw1−s/v − w =0 2 2K and we obtain K = ws/v  Therefore, y
x =  w x 2 w  1−s/v  −  x w  1+s/v    42  CHAPTER 1  First-Order Differential Equations y  s/v  34  0.30 0.25 1  0.20  s/v  2  0.15  s/v  3  0.10  s/v  5  1  1  0.05 0.0  0.2  0.4  0.6  0.8  1.0  x  Graphs of w  x 1−s/v  x 1+s/v y= − 2 w w 1 1 1 3 for s/v equal to 5 , 3 , 2 and 4 , and w chosen as 1.  FIGURE 1.14  As might be expected, the path the swimmer takes depends on the width of the canal, the speed of the swimmer, and the speed of the current. Figure 1.14 shows trajectories corresponding to s/v equal to 15 , 13 , 21 and 43 , with w = 1.  1.6.2  The Bernoulli Equation  DEFINITION 1.7  A Bernoulli equation is a first order equation, y
 + P
xy = R
xy  in which  is a real number.  A Bernoulli equation is separable if  = 0 and linear if  = 1. About 1696, Leibniz showed that a Bernoulli equation with  = 1 transforms to a linear equation under the change of variables: v = y1−  This is routine to verify. Here is an example.  EXAMPLE 1.27  Consider the equation 1 y
 + y = 3x2 y3  x which is Bernoulli with P
x = 1/x, R
x = 3x2 , and  = 3. Make the change of variables v = y−2   1.6 Homogeneous, Bernoulli, and Riccati Equations  43  Then y = v−1/2 and 1 y
 
x = − v−3/2 v
 
x 2 so the differential equation becomes 1 1 − v−3/2 v
 
x + v−1/2 = 3x2 v−3/2  2 x or, upon multiplying by −2v3/2 , 2 v
 − v = −6x2  x a linear equation. An integrating factor is e− factor to get    
2/x dx  = x−2 . Multiply the last equation by this  x−2 v
 − 2x−3 v = −6 which is 
x−2 v
 = −6 Integrate to get x−2 v = −6x + C so v = −6x3 + Cx2  The general solution of the Bernoulli equation is y
x =   1.6.3  1 v
x  =√  1 Cx2 − 6x3    The Riccati Equation  DEFINITION 1.8  A differential equation of the form y
 = P
xy2 + Q
xy + R
x is called a Riccati equation.  A Riccati equation is linear exactly when P
x is identically zero. If we can somehow obtain one solution S
x of a Riccati equation, then the change of variables y = S
x +  1 z  44  CHAPTER 1  First-Order Differential Equations  transforms the Riccati equation to a linear equation. The strategy is to find the general solution of this linear equation and from it produce the general solution of the original Riccati equation.  EXAMPLE 1.28  Consider the Riccati equation 1 1 2 y
 = y2 + y −  x x x By inspection, y = S
x = 1 is one solution. Define a new variable z by putting 1 y = 1+  z Then y
 = −  1 
 z z2  Substitute these into the Riccati equation to get     1 1 1 2 1 1 2 − 2 z
 = + 1+ 1+ −  z x z x z x or 3 1 z
 + z = −  x x This is linear. An integrating factor is e    
3/x dx  = x3 . Multiply by x3 to get  x3 z
 + 3x2 z = 
x3 z
 = −x2  Integrate to get 1 x3 z = − x3 + C 3 so 1 C z
x = − + 3  3 x The general solution of the Riccati equation is y
x = 1 +  1 1 = 1+  z
x −1/3 + C/x3  This solution can also be written y
x = in which K = 3C is an arbitrary constant.  K + 2x3  K − x3  1.6 Homogeneous, Bernoulli, and Riccati Equations  PROBLEMS  SECTION 1.6  In each of Problems 1 through 14, find the general solution. These problems include all types considered in this section. 1 1 1. y = 2 y2 − y + 1 x x 
  2 1 2. y + y = 3 y−4/3 x x 
  3. y
 + xy = xy2 4. y
 =  x y + y x  5. y
 =  y x+y  6. y
 =  4 1 2 1 y − y− 2x x x  In each of Problems 16 through 19, use the idea of Problem 15 to find the general solution. 16. y
 =  y−3 x+y−1  17. y
 =  3x − y − 9 x+y+1  18. y
 =  x + 2y + 7 −2x + y − 9  19. y
 =  2x − 5y − 9 −4x + y + 9  20. Continuing from Problem 15, consider the case that ae − bd = 0. Now let u = 
ax + by/a, assuming that a = 0. Show that this transforms the differential equation of Problem 15 into the separable equation   du b au + c = 1+ F  dx a du + r  7. 
x − 2yy
 = 2x − y 8. xy
 = x cos
y/x + y 1 1 9. y
 + y = 4 y−3/4 x x  In each of Problems 21 through 24, use the method of Problem 20 to find the general solution.  10. x2 y
 = x2 + y2 1 2 11. y
 = − y2 + y x x 12. x3 y
 = x2 y − y3 13. y
 = −e−x y2 + y + ex 2 3 14. y
 + y = y2 x x 15. Consider the differential equation  y
 = F  45   ax + by + c  dx + ey + r  in which a, b, c, d, e, and r are constants and F is a differentiable function of one variable. (a) Show that this equation is homogeneous if and only if c = r = 0. (b) If c and/or r is not zero, this equation is called nearly homogeneous. Assuming that ae − bd = 0, show that it is possible to choose constants h and k so that the transformation X = x + h Y = y + k converts this nearly homogeneous equation into a homogeneous one. Hint: Put x = X − h y = Y − k into the differential equation and obtain a differential equation in X and Y . Use the conclusion of (a) to choose h and k so that this equation is homogeneous.  21. y
 =  x−y+2 x−y+3  22. y
 =  3x + y − 1 6x + 2y − 3  23. y
 =  x − 2y 3x − 6y + 4  24. y
 =  x−y+6 3x − 3y + 4  25. (The Pursuing Dog) A man stands at the junction of two perpendicular roads and his dog is watching him from one of the roads at a distance A feet away. At a given instant the man starts to walk with constant speed v along the other road, and at the same time the dog begins to run toward the man with speed 2v. Determine the path the dog will take, assuming that it always moves so that it is facing the man. Also determine when the dog will eventually catch the man. (This is American Mathematical Monthly problem 3942, 1941). 26. (Pursuing Bugs) One bug is located at each corner of a square table of side length a. At a given time they begin moving at constant speed v, each pursuing its neighbor to the right. (a) Determine the curve of pursuit of each bug. Hint: Use polar coordinates with the origin at the  CHAPTER 1  46  First-Order Differential Equations  center of the table and the polar axis containing one of the corners. When a bug is at 
f
 , its target is at 
f
  + /2. Use the chain rule to write  speed . The bug moves toward the center of the disk at constant speed v. (a) Derive a differential equation for the path of the bug, using polar coordinates.  dy dy/d =  dx dx/d  (b) How many revolutions will the disk make before the bug reaches the center? (The solution will be in terms of the angular speed and radius of the disk.)  where y
 = f
 sin
 and x
 = f
 cos
. (b) Determine the distance traveled by each bug. (c) Does any bug actually catch its quarry?  (c) Referring to (b), what is the total distance the bug will travel, taking into account the motion of the disk?  27. (The Spinning Bug) A bug steps onto the edge of a disk of radius a that is spinning at a constant angular  1.7  Applications to Mechanics, Electrical Circuits, and Orthogonal Trajectories 1.7.1  Mechanics  Before applying first-order differential equations to problems in mechanics, we will review some background. Newton's second law of motion states that the rate of change of momentum (mass times velocity) of a body is proportional to the resultant force acting on the body. This is a vector equation, but we will for now consider only motion along a straight line. In this case Newton's law is d F =k 
mv dt We will take k = 1, consistent with certain units of measurement, such as the English, MKS, or gcs systems. The mass of a moving object need not be constant. For example, an airplane consumes fuel as it moves. If m is constant, then Newton's law is dv = ma dt in which a is the acceleration of the object along the line of motion. If m is not constant, then F =m  dv dm +v  dt dt Newton's law of gravitational attraction states that if two objects have masses m1 and m2 , and they (or their center of masses) are at distance r from each other, then each attracts the other with a gravitational force of magnitude mm F = G 12 2  r This force is directed along the line between the centers of mass. G is the universal gravitational constant. If one of the objects is the earth, then F =m  F =G  mM  
R + x2  where M is the mass of the earth, R is its radius (about 3,960 miles), m is the mass of the second object, and x is its distance from the surface of the earth. This assumes that the earth is  1.7 Applications to Mechanics, Electrical Circuits, and Orthogonal Trajectories  47  spherical and that its center of mass is at the center of this sphere, a good enough approximation for some purposes. If x is small compared to R, then R + x is approximately R and the force on the object is approximately GM m R2 which is often written as mg. Here g = GM/R2 is approximately 32 feet per second per second or 98 meters per second per second. We are now ready to analyze some problems in mechanics. Terminal Velocity Consider an object that is falling under the influence of gravity, in a medium such as water, air or oil. This medium retards the downward motion of the object. Think, for example, of a brick dropped in a swimming pool or a ball bearing dropped in a tank of oil. We, want to analyze the object's motion. Let v
t be the velocity at time t. The force of gravity pulls the object down and has magnitude mg. The medium retards the motion. The magnitude of this retarding force is not obvious, but experiment has shown that its magnitude is proportional to the square of the velocity. If we choose downward as the positive direction and upward as negative, then Newton's law tells us that, for some constant , dv  dt If we assume that the object begins its motion from rest (dropped, not thrown) and if we start the clock at this instant, then v
0 = 0. We now have an initial value problem for the velocity:  v
 = g − v2  v
0 = 0 m This differential equation is separable. In differential form, F = mg − v2 = m  1 dv = dt g − 
/mv2 Integrate to get    m tanh−1 g  Solve for the velocity, obtaining v
t =      mg tanh     v = t + C mg     g 
t + C  m  Now use the initial condition to solve for the integration constant:     mg g tanh C v
0 = = 0  m Since tanh
 = 0 only if  = 0, this requires that C = 0 and the solution for the velocity is    mg g tanh t  v
t =  m Even in this √ generality, we can draw an important conclusion about the motion. As t increases, tanh
 g/mt approaches 1. This means that  mg  lim v
t = t→   CHAPTER 1  48  First-Order Differential Equations  This means that an object falling under the influence of gravity, through a retarding medium (with force proportional to the square of the velocity), will not√ increase in velocity indefinitely. Instead, the object's velocity approaches the limiting value mg/. If the medium is deep enough, the object will settle into a descent of approximately constant velocity. This number √ mg/ is called the terminal velocity of the object. Skydivers experience this phenomenon.  7 ft 9 ft  Motion of a Chain on a Pulley A 16 foot long chain weighing  pounds per foot hangs over a small pulley, which is 20 feet abov

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