text stringlengths 1 7.76k | source stringlengths 17 81 |
|---|---|
Appendix C Physical Constants [m0141] The speed of light in free space (c), which is the phase velocity of any electromagnetic radiation in free space, is ∼= 2.9979 × 108 m/s. This is commonly rounded up to 3 × 108 m/s. This rounding incurs error of ∼= 0.07%, which is usually much less than other errors present in electrical engineering calculations. The charge of an electron is ∼= −1.602 × 10−19 C. The constant e ≜+1.602176634 × 10−19 C is known as the “elementary charge,” so the charge of the electron is said to be −e. The permittivity of free space (ǫ0) is ∼= 8.854 × 10−12 F/m. The permeability of free space (µ0) is 4π × 10−7 H/m. The wave impedance of free space (η0) is the ratio of the magnitude of the electric field intensity to that of the magnetic field intensity in free space and is p µ0/ǫ0 ∼= 376.7 Ω. This is also sometimes referred to as the intrinsic impedance of free space. Boltzmann’s constant is ∼= 1.381 × 10−23 J/K, the amount of energy associated with a change of one degree of temperature. This is typically assigned the symbol k (unfortunately, the same symbol often used to represent wavenumber). Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2 | Electromagnetics_Vol2_Page_225_Chunk2401 |
Index acceptance angle, 140 aluminum, 43, 46, 207 Ampere’s law general form, 7, 8, 26, 34, 105, 146, 154, 155, 159 magnetostatics, 6, 18, 124 antenna electrically-short dipole (ESD), 155–157, 159, 161, 168–172, 175–177, 198 folded half-wave dipole, 136 half-wave dipole, 136, 159, 161–162, 198 isotropic, 199, 201 microstrip patch, 174 quarter-wave monopole, 136 yagi, 181 antenna aperture, 196 antenna factor, 183 attenuation, 129–133, 135 attenuation rate, 40–41, 44 balun, 136 binomial series, 42 Biot-Savart law, 18–20, 154 Boltzmann’s constant, 198, 212 boundary conditions, 5, 6, 8, 187, 190 Brewster’s angle, 86 carbon, 207 CGS (system of units), 2 characteristic impedance, 122, 124 charge density line, 5 surface, 5 volume, 5 cladding, 138 conductivity of common materials, 207–208 conductor good, 43–45, 48, 169 perfect, 6, 43, 48, 50, 58 poor, 41–43 cone of acceptance, 140 constitutive relationships, 7 copper, 43, 207 Coulomb force, 21 Coulomb’s law, 18 coupling, 196 critical angle, 88 curl, 210 current, 5, 48 current density surface, 5, 163 volume, 5, 163 current moment, 20, 150, 152, 162 cutoff frequency, 100, 106, 117, 123 cyclotron motion, 11 dB, see decibel dBm, 40 decibel, 39–40 delay spread, 143 dielectric (material), 123 examples, 205 dielectric constant, 205 Dirac delta function, 145, 150, 162 directivity, 177–179, 199 dispersion, 54, 95, 102, 118, 141 displacement current, 7, 34 divergence, 210 divergence theorem, 26, 211 E-plane, 180 effective aperture, 196–200 effective height, 183 effective length, 183, 190 electric field intensity, 5, 11 electric flux density, 5 electrical length, 154, 168 electrically-short dipole, see antenna electrically-short dipole (ESD), 178 electron, 212 electrostatics (description), 5 213 | Electromagnetics_Vol2_Page_226_Chunk2402 |
214 INDEX English system of units, 2 evanescent waves, see waves far field, 153, 161, 167 Faraday’s law, 7, 23 ferrite, 207 fiber optics, 88 flux electric, 7 magnetic, 6, 7 force, 20 FR4, 41, 127, 128, 136, 205 Friis transmission equation, 201–202 gain power, 39 voltage, 40 Gauss’ law electric field, 5, 18 magnetic field, 6, 148 geranium, 205 glass, 205 gold, 43, 207 good conductor, 50 Goos-H¨anchen effect, 89 gradient, 209 group velocity, 95–97, 102, 118 H-plane, 180 half-power beamwidth (HPBW), 182 half-wave matching (slab), 65–66 Hertzian dipole, 147, 152–156, 159, 166, 169, 176, 191 homogeneity, 8 homogeneous (media), 8 impedance surface, 54 wire, 50–54 independence of path, 5 index of refraction, 59, 81, 138, 205 indium phosphide, 205 inductance, 53 equivalent, 53 inverse square law, 18 iron, 206, 207 isotropic, 182 isotropic (media), 9 isotropy, 9 Jefimenko’s equations, 18 Johnson-Nyquist noise, 198 joule heating, 27 Joule’s law, 27 Kirchoff’s voltage law electrostatics, 5 Laplace’s Equation, 133 Laplacian (operator), 210 lever arm, 15 linear (media), 9 linearity, 9 Lorentz force, 11 Lorentz reciprocity theorem, 187–188 Lorenz gauge condition, 149 loss resistance, 174 loss tangent, 30, 34–35, 41, 43 magnetic field intensity, 6 magnetic flux density, 6, 11, 148 magnetic vector potential, 147–152, 159 magnetostatics (description), 6 materials (properties), 8 materials, magnetic, 6 Maxwell’s equations, 7, 145, 147 differential phasor form, 7, 30, 108, 187 source-free lossless region, 8 static differential form, 6 static integral form, 6 Maxwell-Faraday equation, 7, 8, 26, 102 metric system, 2 microstrip, 108, 123–129, 136 mode, 100, 106, 112, 116 motional emf, 23 motor, 13, 17 mutual coupling, 196 notation, 3 numerical aperture, 140 Ohm’s law, 6, 25, 27, 30, 50 ohmic loss, 27, 34 omnidirectional, 182 optical fiber, 138–143 parallel-plate waveguide, see waveguide, see waveguide, see waveguide, see waveguide, see waveguide path gain, 201 path loss, 201 pattern (radiation), 179–182, 200 | Electromagnetics_Vol2_Page_227_Chunk2403 |
INDEX 215 normalized, 180 permeability, 6 of common materials, 206–207 relative, 6, 206 permittivity, 5 complex-valued, 30, 33–34 effective, 128 of common materials, 205–206 relative, 5, 205 phase velocity, 95–97, 118 in microstrip, 128 phasor, 7 plane of incidence, 70 plane wave relationships, 36, 73, 77, 107, 154, 157, 161 plasma, 147 platinum, 207 polarization, 179–182 polarizing angle, 86 polycarbonate, 205 polyethylene, 42, 205 polypropylene, 205 polystyrene, 205 polytetrafluoroethylene, 205 potential difference, 21 power handling, see transmission line Poynting vector, 27–29, 37 Poynting’s theorem, 25–28, 146 printed circuit board (PCB), 123, 128, 174 material, 205 prism, 82 propagation constant, 150 attenuation, 32, 36, 40, 46, 150 phase, 30, 31, 36, 150 quarter-wave matching (slab), 66–67 radiation condition, 151 radiation efficiency, 169, 174, 178 radiation resistance, 174, 175 radome, 60, 64 ray-fixed coordinates, 68, 70 Rayleigh-Jeans law, 198 reciprocity, 186–190 rectangular waveguide, see waveguide reference phase, 67 reference polarization, 67 reflection coefficient, 57 refraction, 81 resistivity, 207 RG-59, 52, 53, 131 right hand rule magnetostatics, 19, 124 Stokes’ theorem, 211 sampling function, 145, 150 semiconductor (material), 205 separation of variables, 111, 114 SI (system of units), 2 silica, 205 silicon, 205, 207 skin depth, 45–46, 50, 51, 129 skin effect, 49, 53 slab, 60–67 Snell’s law, 81, 83, 85, 88, 92, 140 soil, 206, 207 speed of light, 212 standing wave, 63, 161 steel, 206 Stokes’ theorem, 211 stripline, 123 styrofoam, 205 superposition, 9, 108, 111, 114, 146, 163, 166 surface wave, see waves Teflon, 205 thermal radiation, 198 thermodynamic equilibrium, 199 thermodynamics, 198 time-harmonic, 7, 147 time-invariance (media), 9 torque, 15 total internal reflection, 82, 84, 86, 88–90, 138 transducer, 166 transmission line analogy for wave propagation, 58, 63 coaxial, 52, 53, 103, 121, 129–135 differential, 121 lossy, 36 lumped-element model, 122, 124 microstrip, see microstrip parallel wire, 121–123 power handling, 133–136 single-ended, 121, 123, 136 stripline, 123 transverse electromagnetic (TEM), 32, 124 transverse electric, 70–75, 83–84, 110, 113–116 transverse electromagnetic (TEM), 71, 103, 110, 122 transverse magnetic, 70–71, 76–80, 85–87, 110–113 twin lead, 121 | Electromagnetics_Vol2_Page_228_Chunk2404 |
216 INDEX units, 1–2 vector arithmetic, 211 identity, 211 position-free, 15 vector effective length, 183, 190, 192, 195, 196 water, 205, 207 wave equation electromagnetic, 36, 147, 149, 150 magnetic vector potential, 149, 150 source-free lossless region, 8 source-free lossy region, 30–32 wave impedance, 8, 37, 42, 154, 212 waveguide parallel plate, 97–108 rectangular, 110–119 wavelength in microstrip, 128 waves evanescent, 90–92 plane, 8, 36–37 power density, 8, 37 surface, 92 transmission line analogy, 36 unidirectional, 108–110 work, 20 yagi, see antenna | Electromagnetics_Vol2_Page_229_Chunk2405 |
Electromagnetics, volume 2, by Steven W. Ellingson is a 216-page peer-reviewed open textbook designed especially for electrical engineering students in the third year of a bachelor of science degree program. It is intended as the primary textbook for the second semester of a two-semester undergraduate engineering electromagnetics sequence. The book addresses magnetic force and the Biot-Savart law; general and lossy media; parallel plate and rectangular waveguides; parallel wire, microstrip, and coaxial transmission lines; AC current flow and skin depth; reflection and transmission at planar boundaries; fields in parallel plate, parallel wire, and microstrip transmission lines; optical fiber; and radiation and antennas. This textbook is part of the Open Electromagnetics Project led by Steven W. Ellingson at Virginia Tech (https://www.faculty.ece.vt.edu/swe/oem). The project goal is to create publicly available, no-cost, openly licensed content for courses in engineering electromagnetics. The project is motivated by two things: lowering the cost of learning materials for students and giving faculty the freedom to adopt, modify, and improve their educational resources. For each book in the series, the following are freely available for all to use: openly licensed and accessible PDFs, problem sets and solutions, slides of figures, editable LaTeX source files, and information for the book’s collaborator portal and listserv. Books in this series Electromagnetics, volume 1, https://doi.org/10.21061/electromagnetics-vol-1 Electromagnetics, volume 2, https://doi.org/10.21061/electromagnetics-vol-2 Advance Praise for Electromagnetics “I commend the author for the hard work and generosity required to create a book like this and to make it available for free. The overall format is pleasing and the material is generally rigorous and complete. Well done.” — Karl Warnick, Brigham Young University “I really liked some of the ways the author related the math to physics and explained why the math made sense. … The figures are generally quite good. … The examples in the book are simple and straightforward.” — Randy Haupt, Colorado School of Mines Cover Design: Robert Browder This book is licensed with a Creative Commons Attribution-ShareAlike License 4.0. Virginia Tech Publishing ∙ Blacksburg, Virginia | Electromagnetics_Vol2_Page_230_Chunk2406 |
Trinity University Trinity University Digital Commons @ Trinity Digital Commons @ Trinity Faculty Authored and Edited Books & CDs 12-2013 Elementary Differential Equations with Boundary Value Problems Elementary Differential Equations with Boundary Value Problems William F. Trench Trinity University, wtrench@trinity.edu Follow this and additional works at: https://digitalcommons.trinity.edu/mono Part of the Mathematics Commons Recommended Citation Recommended Citation Trench, William F., "Elementary Differential Equations with Boundary Value Problems" (2013). Faculty Authored and Edited Books & CDs. 9. https://digitalcommons.trinity.edu/mono/9 This Book is brought to you for free and open access by Digital Commons @ Trinity. It has been accepted for inclusion in Faculty Authored and Edited Books & CDs by an authorized administrator of Digital Commons @ Trinity. For more information, please contact jcostanz@trinity.edu. | Elementary Differential Equations with Boundary Value Problems_Page_1_Chunk2407 |
ELEMENTARY DIFFERENTIAL EQUATIONS WITH BOUNDARY VALUE PROBLEMS William F. Trench Andrew G. Cowles Distinguished Professor Emeritus Department of Mathematics Trinity University San Antonio, Texas, USA wtrench@trinity.edu This book has been judged to meet the evaluation criteria set by the Edi- torial Board of the American Institute of Mathematics in connection with the Institute’s Open Textbook Initiative. It may be copied, modified, re- distributed, translated, and built upon subject to the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. FREE DOWNLOAD: STUDENT SOLUTIONS MANUAL | Elementary Differential Equations with Boundary Value Problems_Page_2_Chunk2408 |
Free Edition 1.01 (December 2013) This book was published previously by Brooks/Cole Thomson Learning, 2001. This free edition is made available in the hope that it will be useful as a textbook or reference. Reproduction is permitted for any valid noncommercial educational, mathematical, or scientific purpose. However, charges for profit beyond reasonable printing costs are prohibited. | Elementary Differential Equations with Boundary Value Problems_Page_3_Chunk2409 |
TO BEVERLY | Elementary Differential Equations with Boundary Value Problems_Page_4_Chunk2410 |
Contents Chapter 1 Introduction 1 1.1 Applications Leading to Differential Equations 1.2 First Order Equations 5 1.3 Direction Fields for First Order Equations 16 Chapter 2 First Order Equations 30 2.1 Linear First Order Equations 30 2.2 Separable Equations 45 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 55 2.4 Transformation of Nonlinear Equations into Separable Equations 62 2.5 Exact Equations 73 2.6 Integrating Factors 82 Chapter 3 Numerical Methods 3.1 Euler’s Method 96 3.2 The Improved Euler Method and Related Methods 109 3.3 The Runge-Kutta Method 119 Chapter 4 Applications of First Order Equations1em 130 4.1 Growth and Decay 130 4.2 Cooling and Mixing 140 4.3 Elementary Mechanics 151 4.4 Autonomous Second Order Equations 162 4.5 Applications to Curves 179 Chapter 5 Linear Second Order Equations 5.1 Homogeneous Linear Equations 194 5.2 Constant Coefficient Homogeneous Equations 210 5.3 Nonhomgeneous Linear Equations 221 5.4 The Method of Undetermined Coefficients I 229 iv | Elementary Differential Equations with Boundary Value Problems_Page_5_Chunk2411 |
5.5 The Method of Undetermined Coefficients II 238 5.6 Reduction of Order 248 5.7 Variation of Parameters 255 Chapter 6 Applcations of Linear Second Order Equations 268 6.1 Spring Problems I 268 6.2 Spring Problems II 279 6.3 The RLC Circuit 290 6.4 Motion Under a Central Force 296 Chapter 7 Series Solutions of Linear Second Order Equations 7.1 Review of Power Series 306 7.2 Series Solutions Near an Ordinary Point I 319 7.3 Series Solutions Near an Ordinary Point II 334 7.4 Regular Singular Points Euler Equations 342 7.5 The Method of Frobenius I 347 7.6 The Method of Frobenius II 364 7.7 The Method of Frobenius III 378 Chapter 8 Laplace Transforms 8.1 Introduction to the Laplace Transform 393 8.2 The Inverse Laplace Transform 405 8.3 Solution of Initial Value Problems 413 8.4 The Unit Step Function 419 8.5 Constant Coefficient Equations with Piecewise Continuous Forcing Functions 430 8.6 Convolution 440 8.7 Constant Cofficient Equations with Impulses 452 8.8 A Brief Table of Laplace Transforms Chapter 9 Linear Higher Order Equations 9.1 Introduction to Linear Higher Order Equations 465 9.2 Higher Order Constant Coefficient Homogeneous Equations 475 9.3 Undetermined Coefficients for Higher Order Equations 487 9.4 Variation of Parameters for Higher Order Equations 497 Chapter 10 Linear Systems of Differential Equations 10.1 Introduction to Systems of Differential Equations 507 10.2 Linear Systems of Differential Equations 515 10.3 Basic Theory of Homogeneous Linear Systems 521 10.4 Constant Coefficient Homogeneous Systems I 529 | Elementary Differential Equations with Boundary Value Problems_Page_6_Chunk2412 |
vi Contents 10.5 Constant Coefficient Homogeneous Systems II 542 10.6 Constant Coefficient Homogeneous Systems II 556 10.7 Variation of Parameters for Nonhomogeneous Linear Systems 568 Chapter 11 Boundary Value Problems and Fourier Expansions 580 11.1 Eigenvalue Problems for y′′ + λy = 0 580 11.2 Fourier Series I 586 11.3 Fourier Series II 603 Chapter 12 Fourier Solutions of Partial Differential Equations 12.1 The Heat Equation 618 12.2 The Wave Equation 630 12.3 Laplace’s Equation in Rectangular Coordinates 649 12.4 Laplace’s Equation in Polar Coordinates 666 Chapter 13 Boundary Value Problems for Second Order Linear Equations 13.1 Boundary Value Problems 676 13.2 Sturm–Liouville Problems 687 | Elementary Differential Equations with Boundary Value Problems_Page_7_Chunk2413 |
Preface Elementary Differential Equations with Boundary Value Problems is written for students in science, en- gineering, and mathematics who have completed calculus through partial differentiation. If your syllabus includes Chapter 10 (Linear Systems of Differential Equations), your students should have some prepa- ration in linear algebra. In writing this book I have been guided by the these principles: • An elementary text should be written so the student can read it with comprehension without too much pain. I have tried to put myself in the student’s place, and have chosen to err on the side of too much detail rather than not enough. • An elementary text can’t be better than its exercises. This text includes 2041 numbered exercises, many with several parts. They range in difficulty from routine to very challenging. • An elementary text should be written in an informal but mathematically accurate way, illustrated by appropriate graphics. I have tried to formulate mathematical concepts succinctly in language that students can understand. I have minimized the number of explicitly stated theorems and def- initions, preferring to deal with concepts in a more conversational way, copiously illustrated by 299 completely worked out examples. Where appropriate, concepts and results are depicted in 188 figures. Although I believe that the computer is an immensely valuable tool for learning, doing, and writing mathematics, the selection and treatment of topics in this text reflects my pedagogical orientation along traditional lines. However, I have incorporated what I believe to be the best use of modern technology, so you can select the level of technology that you want to include in your course. The text includes 414 exercises – identified by the symbols C and C/G – that call for graphics or computation and graphics. There are also 79 laboratory exercises – identified by L – that require extensive use of technology. In addition, several sections include informal advice on the use of technology. If you prefer not to emphasize technology, simply ignore these exercises and the advice. There are two schools of thought on whether techniques and applications should be treated together or separately. I have chosen to separate them; thus, Chapter 2 deals with techniques for solving first order equations, and Chapter 4 deals with applications. Similarly, Chapter 5 deals with techniques for solving second order equations, and Chapter 6 deals with applications. However, the exercise sets of the sections dealing with techniques include some applied problems. Traditionally oriented elementary differential equations texts are occasionally criticized as being col- lections of unrelated methods for solving miscellaneous problems. To some extent this is true; after all, no single method applies to all situations. Nevertheless, I believe that one idea can go a long way toward unifying some of the techniques for solving diverse problems: variation of parameters. I use variation of parameters at the earliest opportunity in Section 2.1, to solve the nonhomogeneous linear equation, given a nontrivial solution of the complementary equation. You may find this annoying, since most of us learned that one should use integrating factors for this task, while perhaps mentioning the variation of parameters option in an exercise. However, there’s little difference between the two approaches, since an integrating factor is nothing more than the reciprocal of a nontrivial solution of the complementary equation. The advantage of using variation of parameters here is that it introduces the concept in its simplest form and vii | Elementary Differential Equations with Boundary Value Problems_Page_8_Chunk2414 |
viii Preface focuses the student’s attention on the idea of seeking a solution y of a differential equation by writing it as y = uy1, where y1 is a known solution of related equation and u is a function to be determined. I use this idea in nonstandard ways, as follows: • In Section 2.4 to solve nonlinear first order equations, such as Bernoulli equations and nonlinear homogeneous equations. • In Chapter 3 for numerical solution of semilinear first order equations. • In Section 5.2 to avoid the necessity of introducing complex exponentials in solving a second or- der constant coefficient homogeneous equation with characteristic polynomials that have complex zeros. • In Sections 5.4, 5.5, and 9.3 for the method of undetermined coefficients. (If the method of an- nihilators is your preferred approach to this problem, compare the labor involved in solving, for example, y′′ + y′ + y = x4ex by the method of annihilators and the method used in Section 5.4.) Introducing variation of parameters as early as possible (Section 2.1) prepares the student for the con- cept when it appears again in more complex forms in Section 5.6, where reduction of order is used not merely to find a second solution of the complementary equation, but also to find the general solution of the nonhomogeneous equation, and in Sections 5.7, 9.4, and 10.7, that treat the usual variation of parameters problem for second and higher order linear equations and for linear systems. Chapter 11 develops the theory of Fourier series. Section 11.1 discusses the five main eigenvalue prob- lems that arise in connection with the method of separation of variables for the heat and wave equations and for Laplace’s equation over a rectangular domain: Problem 1: y′′ + λy = 0, y(0) = 0, y(L) = 0 Problem 2: y′′ + λy = 0, y′(0) = 0, y′(L) = 0 Problem 3: y′′ + λy = 0, y(0) = 0, y′(L) = 0 Problem 4: y′′ + λy = 0, y′(0) = 0, y(L) = 0 Problem 5: y′′ + λy = 0, y(−L) = y(L), y′(−L) = y′(L) These problems are handled in a unified way for example, a single theorem shows that the eigenvalues of all five problems are nonnegative. Section 11.2 presents the Fourier series expansion of functions defined on on [−L, L], interpreting it as an expansion in terms of the eigenfunctions of Problem 5. Section 11.3 presents the Fourier sine and cosine expansions of functions defined on [0, L], interpreting them as expansions in terms of the eigenfunctions of Problems 1 and 2, respectively. In addition, Sec- tion 11.2 includes what I call the mixed Fourier sine and cosine expansions, in terms of the eigenfunctions of Problems 4 and 5, respectively. In all cases, the convergence properties of these series are deduced from the convergence properties of the Fourier series discussed in Section 11.1. Chapter 12 consists of four sections devoted to the heat equation, the wave equation, and Laplace’s equation in rectangular and polar coordinates. For all three, I consider homogeneous boundary conditions of the four types occurring in Problems 1-4. I present the method of separation of variables as a way of choosing the appropriate form for the series expansion of the solution of the given problem, stating— without belaboring the point—that the expansion may fall short of being an actual solution, and giving an indication of conditions under which the formal solution is an actual solution. In particular, I found it necessary to devote some detail to this question in connection with the wave equation in Section 12.2. In Sections 12.1 (The Heat Equation) and 12.2 (The Wave Equation) I devote considerable effort to devising examples and numerous exercises where the functions defining the initial conditions satisfy | Elementary Differential Equations with Boundary Value Problems_Page_9_Chunk2415 |
Preface ix the homogeneous boundary conditions. Similarly, in most of the examples and exercises Section 12.3 (Laplace’s Equation), the functions defining the boundary conditions on a given side of the rectangular domain satisfy homogeneous boundary conditions at the endpoints of the same type (Dirichlet or Neu- mann) as the boundary conditionsimposed on adjacent sides of the region. Therefore the formal solutions obtained in many of the examples and exercises are actual solutions. Section 13.1 deals with two-point value problems for a second order ordinary differential equation. Conditions for existence and uniqueness of solutions are given, and the construction of Green’s functions is included. Section 13.2 presents the elementary aspects of Sturm-Liouville theory. You may also find the following to be of interest: • Section 2.6 deals with integrating factors of the form µ = p(x)q(y), in addition to those of the form µ = p(x) and µ = q(y) discussed in most texts. • Section 4.4 makes phase plane analysis of nonlinear second order autonomous equations accessi- ble to students who have not taken linear algebra, since eigenvalues and eigenvectors do not enter into the treatment. Phase plane analysis of constant coefficient linear systems is included in Sec- tions 10.4-6. • Section 4.5 presents an extensive discussion of applications of differential equations to curves. • Section 6.4 studies motion under a central force, which may be useful to students interested in the mathematics of satellite orbits. • Sections 7.5-7 present the method of Frobenius in more detail than in most texts. The approach is to systematize the computations in a way that avoids the necessity of substituting the unknown Frobenius series into each equation. This leads to efficiency in the computation of the coefficients of the Frobenius solution. It also clarifies the case where the roots of the indicial equation differ by an integer (Section 7.7). • The free Student Solutions Manual contains solutions of most of the even-numbered exercises. • The free Instructor’s Solutions Manual is available by email to wtrench@trinity.edu, subject to verification of the requestor’s faculty status. The following observations may be helpful as you choose your syllabus: • Section 2.3 is the only specific prerequisite for Chapter 3. To accomodate institutions that offer a separate course in numerical analysis, Chapter 3 is not a prerequisite for any other section in the text. • The sections in Chapter 4 are independent of each other, and are not prerequisites for any of the later chapters. This is also true of the sections in Chapter 6, except that Section 6.1 is a prerequisite for Section 6.2. • Chapters 7, 8, and 9 can be covered in any order after the topics selected from Chapter 5. For example, you can proceed directly from Chapter 5 to Chapter 9. • The second order Euler equation is discussed in Section 7.4, where it sets the stage for the method of Frobenius. As noted at the beginning of Section 7.4, if you want to include Euler equations in your syllabus while omitting the method of Frobenius, you can skip the introductory paragraphs in Section 7.4 and begin with Definition 7.4.2. You can then cover Section 7.4 immediately after Section 5.2. • Chapters 11, 12, and 13 can be covered at any time after the completion of Chapter 5. William F. Trench | Elementary Differential Equations with Boundary Value Problems_Page_10_Chunk2416 |
CHAPTER 1 Introduction IN THIS CHAPTER we begin our study of differential equations. SECTION 1.1 presents examples of applications that lead to differential equations. SECTION 1.2 introduces basic concepts and definitions concerning differential equations. SECTION 1.3 presents a geometric method for dealing with differential equations that has been known for a very long time, but has become particularly useful and important with the proliferation of readily available differential equations software. 1 | Elementary Differential Equations with Boundary Value Problems_Page_11_Chunk2417 |
2 Chapter 1 Introduction 1.1 APPLICATIONS LEADING TO DIFFERENTIAL EQUATIONS In order to apply mathematical methods to a physical or “real life” problem, we must formulate the prob- lem in mathematical terms; that is, we must construct a mathematical model for the problem. Many physical problems concern relationships between changing quantities. Since rates of change are repre- sented mathematically by derivatives, mathematical models often involve equations relating an unknown function and one or more of its derivatives. Such equations are differential equations. They are the subject of this book. Much of calculus is devoted to learning mathematical techniques that are applied in later courses in mathematics and the sciences; you wouldn’t have time to learn much calculus if you insisted on seeing a specific application of every topic covered in the course. Similarly, much of this book is devoted to methods that can be applied in later courses. Only a relatively small part of the book is devoted to the derivation of specific differential equations from mathematical models, or relating the differential equations that we study to specific applications. In this section we mention a few such applications. The mathematical model for an applied problem is almost always simpler than the actual situation being studied, since simplifying assumptions are usually required to obtain a mathematical problem that can be solved. For example, in modeling the motion of a falling object, we might neglect air resistance and the gravitational pull of celestial bodies other than Earth, or in modeling population growth we might assume that the population grows continuously rather than in discrete steps. A good mathematical model has two important properties: • It’s sufficiently simple so that the mathematical problem can be solved. • It represents the actual situation sufficiently well so that the solution to the mathematical problem predicts the outcome of the real problem to within a useful degree of accuracy. If results predicted by the model don’t agree with physical observations, the underlying assumptions of the model must be revised until satisfactory agreement is obtained. We’ll now give examples of mathematical models involving differential equations. We’ll return to these problems at the appropriate times, as we learn how to solve the various types of differential equations that occur in the models. All the examples in this section deal with functions of time, which we denote by t. If y is a function of t, y′ denotes the derivative of y with respect to t; thus, y′ = dy dt . Population Growth and Decay Although the number of members of a population (people in a given country, bacteria in a laboratory cul- ture, wildflowers in a forest, etc.) at any given time t is necessarily an integer, models that use differential equations to describe the growth and decay of populations usually rest on the simplifying assumption that the number of members of the population can be regarded as a differentiable function P = P (t). In most models it is assumed that the differential equation takes the form P ′ = a(P )P, (1.1.1) where a is a continuous function of P that represents the rate of change of population per unit time per individual. In the Malthusian model, it is assumed that a(P ) is a constant, so (1.1.1) becomes P ′ = aP. (1.1.2) | Elementary Differential Equations with Boundary Value Problems_Page_12_Chunk2418 |
Section 1.1 Applications Leading to Differential Equations 3 (When you see a name in blue italics, just click on it for information about the person.) This model assumes that the numbers of births and deaths per unit time are both proportional to the population. The constants of proportionality are the birth rate (births per unit time per individual) and the death rate (deaths per unit time per individual); a is the birth rate minus the death rate. You learned in calculus that if c is any constant then P = ceat (1.1.3) satisfies (1.1.2), so (1.1.2) has infinitely many solutions. To select the solution of the specific problem that we’re considering, we must know the population P0 at an initial time, say t = 0. Setting t = 0 in (1.1.3) yields c = P (0) = P0, so the applicable solution is P (t) = P0eat. This implies that lim t→∞P (t) = ∞ if a > 0, 0 if a < 0; that is, the population approaches infinity if the birth rate exceeds the death rate, or zero if the death rate exceeds the birth rate. To see the limitations of the Malthusian model, suppose we’re modeling the population of a country, starting from a time t = 0 when the birth rate exceeds the death rate (so a > 0), and the country’s resources in terms of space, food supply, and other necessities of life can support the existing popula- tion. Then the prediction P = P0eat may be reasonably accurate as long as it remains within limits that the country’s resources can support. However, the model must inevitably lose validity when the pre- diction exceeds these limits. (If nothing else, eventually there won’t be enough space for the predicted population!) This flaw in the Malthusian model suggests the need for a model that accounts for limitations of space and resources that tend to oppose the rate of population growth as the population increases. Perhaps the most famous model of this kind is the Verhulst model, where (1.1.2) is replaced by P ′ = aP (1 −αP ), (1.1.4) where α is a positive constant. As long as P is small compared to 1/α, the ratio P ′/P is approximately equal to a. Therefore the growth is approximately exponential; however, as P increases, the ratio P ′/P decreases as opposing factors become significant. Equation (1.1.4) is the logistic equation. You will learn how to solve it in Section 1.2. (See Exer- cise 2.2.28.) The solution is P = P0 αP0 + (1 −αP0)e−at , where P0 = P (0) > 0. Therefore limt→∞P (t) = 1/α, independent of P0. Figure 1.1.1 shows typical graphs of P versus t for various values of P0. Newton’s Law of Cooling According to Newton’s law of cooling, the temperature of a body changes at a rate proportional to the difference between the temperature of the body and the temperature of the surrounding medium. Thus, if Tm is the temperature of the medium and T = T(t) is the temperature of the body at time t, then T ′ = −k(T −Tm), (1.1.5) where k is a positive constant and the minus sign indicates; that the temperature of the body increases with time if it’s less than the temperature of the medium, or decreases if it’s greater. We’ll see in Section 4.2 that if Tm is constant then the solution of (1.1.5) is T = Tm + (T0 −Tm)e−kt, (1.1.6) | Elementary Differential Equations with Boundary Value Problems_Page_13_Chunk2419 |
4 Chapter 1 Introduction P t 1/α Figure 1.1.1 Solutions of the logistic equation where T0 is the temperature of the body when t = 0. Therefore limt→∞T(t) = Tm, independent of T0. (Common sense suggests this. Why?) Figure 1.1.2 shows typical graphs of T versus t for various values of T0. Assuming that the medium remains at constant temperature seems reasonable if we’re considering a cup of coffee cooling in a room, but not if we’re cooling a huge cauldron of molten metal in the same room. The difference between the two situations is that the heat lost by the coffee isn’t likely to raise the temperature of the room appreciably, but the heat lost by the cooling metal is. In this second situation we must use a model that accounts for the heat exchanged between the object and the medium. Let T = T(t) and Tm = Tm(t) be the temperatures of the object and the medium respectively, and let T0 and Tm0 be their initial values. Again, we assume that T and Tm are related by (1.1.5). We also assume that the change in heat of the object as its temperature changes from T0 to T is a(T −T0) and the change in heat of the medium as its temperature changes from Tm0 to Tm is am(Tm−Tm0), where a and am are positive constants depending upon the masses and thermal properties of the object and medium respectively. If we assume that the total heat of the in the object and the medium remains constant (that is, energy is conserved), then a(T −T0) + am(Tm −Tm0) = 0. Solving this for Tm and substituting the result into (1.1.6) yields the differential equation T ′ = −k 1 + a am T + k Tm0 + a am T0 for the temperature of the object. After learning to solve linear first order equations, you’ll be able to show (Exercise 4.2.17) that T = aT0 + amTm0 a + am + am(T0 −Tm0) a + am e−k(1+a/am)t. | Elementary Differential Equations with Boundary Value Problems_Page_14_Chunk2420 |
Section 1.1 Applications Leading to Differential Equations 5 T t t Tm Figure 1.1.2 Temperature according to Newton’s Law of Cooling Glucose Absorption by the Body Glucose is absorbed by the body at a rate proportionalto the amount of glucose present in the bloodstream. Let λ denote the (positive) constant of proportionality. Suppose there are G0 units of glucose in the bloodstream when t = 0, and let G = G(t) be the number of units in the bloodstream at time t > 0. Then, since the glucose being absorbed by the body is leaving the bloodstream, G satisfies the equation G′ = −λG. (1.1.7) From calculus you know that if c is any constant then G = ce−λt (1.1.8) satisfies (1.1.7), so (1.1.7) has infinitely many solutions. Setting t = 0 in (1.1.8) and requiring that G(0) = G0 yields c = G0, so G(t) = G0e−λt. Now let’s complicate matters by injecting glucose intravenously at a constant rate of r units of glucose per unit of time. Then the rate of change of the amount of glucose in the bloodstream per unit time is G′ = −λG + r, (1.1.9) where the first term on the right is due to the absorption of the glucose by the body and the second term is due to the injection. After you’ve studied Section 2.1, you’ll be able to show (Exercise 2.1.43) that the solution of (1.1.9) that satisfies G(0) = G0 is G = r λ + G0 −r λ e−λt. | Elementary Differential Equations with Boundary Value Problems_Page_15_Chunk2421 |
6 Chapter 1 Introduction Graphs of this function are similar to those in Figure 1.1.2. (Why?) Spread of Epidemics One model for the spread of epidemics assumes that the number of people infected changes at a rate proportional to the product of the number of people already infected and the number of people who are susceptible, but not yet infected. Therefore, if S denotes the total population of susceptible people and I = I(t) denotes the number of infected people at time t, then S −I is the number of people who are susceptible, but not yet infected. Thus, I′ = rI(S −I), where r is a positive constant. Assuming that I(0) = I0, the solution of this equation is I = SI0 I0 + (S −I0)e−rSt (Exercise 2.2.29). Graphs of this function are similar to those in Figure 1.1.1. (Why?) Since limt→∞I(t) = S, this model predicts that all the susceptible people eventually become infected. Newton’s Second Law of Motion According to Newton’s second law of motion, the instantaneous acceleration a of an object with con- stant mass m is related to the force F acting on the object by the equation F = ma. For simplicity, let’s assume that m = 1 and the motion of the object is along a vertical line. Let y be the displacement of the object from some reference point on Earth’s surface, measured positive upward. In many applications, there are three kinds of forces that may act on the object: (a) A force such as gravity that depends only on the position y, which we write as −p(y), where p(y) > 0 if y ≥0. (b) A force such as atmospheric resistance that depends on the position and velocity of the object, which we write as −q(y, y′)y′, where q is a nonnegative function and we’ve put y′ “outside” to indicate that the resistive force is always in the direction opposite to the velocity. (c) A force f = f(t), exerted from an external source (such as a towline from a helicopter) that depends only on t. In this case, Newton’s second law implies that y′′ = −q(y, y′)y′ −p(y) + f(t), which is usually rewritten as y′′ + q(y, y′)y′ + p(y) = f(t). Since the second (and no higher) order derivative of y occurs in this equation, we say that it is a second order differential equation. Interacting Species: Competition Let P = P (t) and Q = Q(t) be the populations of two species at time t, and assume that each population would grow exponentially if the other didn’t exist; that is, in the absence of competition we would have P ′ = aP and Q′ = bQ, (1.1.10) where a and b are positive constants. One way to model the effect of competition is to assume that the growth rate per individual of each population is reduced by an amount proportional to the other population, so (1.1.10) is replaced by P ′ = aP −αQ Q′ = −βP + bQ, | Elementary Differential Equations with Boundary Value Problems_Page_16_Chunk2422 |
Section 1.2 Basic Concepts 7 where α and β are positive constants. (Since negative population doesn’t make sense, this system works only while P and Q are both positive.) Now suppose P (0) = P0 > 0 and Q(0) = Q0 > 0. It can be shown (Exercise 10.4.42) that there’s a positive constant ρ such that if (P0, Q0) is above the line L through the origin with slope ρ, then the species with population P becomes extinct in finite time, but if (P0, Q0) is below L, the species with population Q becomes extinct in finite time. Figure 1.1.3 illustrates this. The curves shown there are given parametrically by P = P (t), Q = Q(t), t > 0. The arrows indicate direction along the curves with increasing t. P Q L Figure 1.1.3 Populations of competing species 1.2 BASIC CONCEPTS A differential equation is an equation that contains one or more derivatives of an unknown function. The order of a differential equation is the order of the highest derivative that it contains. A differential equation is an ordinary differential equation if it involves an unknown function of only one variable, or a partial differential equation if it involves partial derivatives of a function of more than one variable. For now we’ll consider only ordinary differential equations, and we’ll just call them differential equations. Throughout this text, all variables and constants are real unless it’s stated otherwise. We’ll usually use x for the independent variable unless the independent variable is time; then we’ll use t. The simplest differential equations are first order equations of the form dy dx = f(x) or, equivalently, y′ = f(x), where f is a known function of x. We already know from calculus how to find functions that satisfy this kind of equation. For example, if y′ = x3, | Elementary Differential Equations with Boundary Value Problems_Page_17_Chunk2423 |
8 Chapter 1 Introduction then y = Z x3 dx = x4 4 + c, where c is an arbitrary constant. If n > 1 we can find functions y that satisfy equations of the form y(n) = f(x) (1.2.1) by repeated integration. Again, this is a calculus problem. Except for illustrative purposes in this section, there’s no need to consider differential equations like (1.2.1).We’ll usually consider differential equations that can be written as y(n) = f(x, y, y′, . . ., y(n−1)), (1.2.2) where at least one of the functions y, y′, ..., y(n−1) actually appears on the right. Here are some exam- ples: dy dx −x2 = 0 (first order), dy dx + 2xy2 = −2 (first order), d2y dx2 + 2 dy dx + y = 2x (second order), xy′′′ + y2 = sin x (third order), y(n) + xy′ + 3y = x (n-th order). Although none of these equations is written as in (1.2.2), all of them can be written in this form: y′ = x2, y′ = −2 −2xy2, y′′ = 2x −2y′ −y, y′′′ = sin x −y2 x , y(n) = x −xy′ −3y. Solutions of Differential Equations A solution of a differential equation is a function that satisfies the differential equation on some open interval; thus, y is a solution of (1.2.2) if y is n times differentiable and y(n)(x) = f(x, y(x), y′(x), . . ., y(n−1)(x)) for all x in some open interval (a, b). In this case, we also say that y is a solution of (1.2.2) on (a, b). Functions that satisfy a differential equation at isolated points are not interesting. For example, y = x2 satisfies xy′ + x2 = 3x if and only if x = 0 or x = 1, but it’s not a solution of this differential equation because it does not satisfy the equation on an open interval. The graph of a solution of a differential equation is a solution curve. More generally, a curve C is said to be an integral curve of a differential equation if every function y = y(x) whose graph is a segment of C is a solution of the differential equation. Thus, any solution curve of a differential equation is an integral curve, but an integral curve need not be a solution curve. | Elementary Differential Equations with Boundary Value Problems_Page_18_Chunk2424 |
Section 1.2 Basic Concepts 9 Example 1.2.1 If a is any positive constant, the circle x2 + y2 = a2 (1.2.3) is an integral curve of y′ = −x y . (1.2.4) To see this, note that the only functions whose graphs are segments of (1.2.3) are y1 = p a2 −x2 and y2 = − p a2 −x2. We leave it to you to verify that these functions both satisfy (1.2.4) on the open interval (−a, a). However, (1.2.3) is not a solution curve of (1.2.4), since it’s not the graph of a function. Example 1.2.2 Verify that y = x2 3 + 1 x (1.2.5) is a solution of xy′ + y = x2 (1.2.6) on (0, ∞) and on (−∞, 0). Solution Substituting (1.2.5) and y′ = 2x 3 −1 x2 into (1.2.6) yields xy′(x) + y(x) = x 2x 3 −1 x2 + x2 3 + 1 x = x2 for all x ̸= 0. Therefore y is a solution of (1.2.6) on (−∞, 0) and (0, ∞). However, y isn’t a solution of the differential equation on any open interval that contains x = 0, since y is not defined at x = 0. Figure 1.2.1 shows the graph of (1.2.5). The part of the graph of (1.2.5) on (0, ∞) is a solution curve of (1.2.6), as is the part of the graph on (−∞, 0). Example 1.2.3 Show that if c1 and c2 are constants then y = (c1 + c2x)e−x + 2x −4 (1.2.7) is a solution of y′′ + 2y′ + y = 2x (1.2.8) on (−∞, ∞). Solution Differentiating (1.2.7) twice yields y′ = −(c1 + c2x)e−x + c2e−x + 2 and y′′ = (c1 + c2x)e−x −2c2e−x, | Elementary Differential Equations with Boundary Value Problems_Page_19_Chunk2425 |
10 Chapter 1 Introduction x y 0.5 1.0 1.5 2.0 −0.5 −1.0 −1.5 −2.0 2 4 6 8 −2 −4 −6 −8 Figure 1.2.1 y = x2 3 + 1 x so y′′ + 2y′ + y = (c1 + c2x)e−x −2c2e−x +2 −(c1 + c2x)e−x + c2e−x + 2 +(c1 + c2x)e−x + 2x −4 = (1 −2 + 1)(c1 + c2x)e−x + (−2 + 2)c2e−x +4 + 2x −4 = 2x for all values of x. Therefore y is a solution of (1.2.8) on (−∞, ∞). Example 1.2.4 Find all solutions of y(n) = e2x. (1.2.9) Solution Integrating (1.2.9) yields y(n−1) = e2x 2 + k1, where k1 is a constant. If n ≥2, integrating again yields y(n−2) = e2x 4 + k1x + k2. If n ≥3, repeatedly integrating yields y = e2x 2n + k1 xn−1 (n −1)! + k2 xn−2 (n −2)! + · · · + kn, (1.2.10) | Elementary Differential Equations with Boundary Value Problems_Page_20_Chunk2426 |
Section 1.2 Basic Concepts 11 where k1, k2, ..., kn are constants. This shows that every solution of (1.2.9) has the form (1.2.10) for some choice of the constants k1, k2, ..., kn. On the other hand, differentiating (1.2.10) n times shows that if k1, k2, ..., kn are arbitrary constants, then the function y in (1.2.10) satisfies (1.2.9). Since the constants k1, k2, ..., kn in (1.2.10) are arbitrary, so are the constants k1 (n −1)!, k2 (n −2)!, · · · , kn. Therefore Example 1.2.4 actually shows that all solutions of (1.2.9) can be written as y = e2x 2n + c1 + c2x + · · · + cnxn−1, where we renamed the arbitrary constants in (1.2.10) to obtain a simpler formula. As a general rule, arbitrary constants appearing in solutions of differential equations should be simplified if possible. You’ll see examples of this throughout the text. Initial Value Problems In Example 1.2.4 we saw that the differential equation y(n) = e2x has an infinite family of solutions that depend upon the n arbitrary constants c1, c2, ..., cn. In the absence of additional conditions, there’s no reason to prefer one solution of a differential equation over another. However, we’ll often be interested in finding a solution of a differential equation that satisfies one or more specific conditions. The next example illustrates this. Example 1.2.5 Find a solution of y′ = x3 such that y(1) = 2. Solution At the beginning of this section we saw that the solutions of y′ = x3 are y = x4 4 + c. To determine a value of c such that y(1) = 2, we set x = 1 and y = 2 here to obtain 2 = y(1) = 1 4 + c, so c = 7 4. Therefore the required solution is y = x4 + 7 4 . Figure 1.2.2 shows the graph of this solution. Note that imposing the condition y(1) = 2 is equivalent to requiring the graph of y to pass through the point (1, 2). We can rewrite the problem considered in Example 1.2.5 more briefly as y′ = x3, y(1) = 2. We call this an initial value problem. The requirement y(1) = 2 is an initial condition. Initial value problems can also be posed for higher order differential equations. For example, y′′ −2y′ + 3y = ex, y(0) = 1, y′(0) = 2 (1.2.11) | Elementary Differential Equations with Boundary Value Problems_Page_21_Chunk2427 |
12 Chapter 1 Introduction is an initial value problem for a second order differential equation where y and y′ are required to have specified values at x = 0. In general, an initial value problem for an n-th order differential equation requires y and its first n−1 derivatives to have specified values at some point x0. These requirements are the initial conditions. 1 2 3 4 5 0 1 2 −1 −2 (1,2) x y Figure 1.2.2 y = x2 + 7 4 We’ll denote an initial value problem for a differential equation by writing the initial conditions after the equation, as in (1.2.11). For example, we would write an initial value problem for (1.2.2) as y(n) = f(x, y, y′, . . ., y(n−1)), y(x0) = k0, y′(x0) = k1, . . ., y(n−1) = kn−1. (1.2.12) Consistent with our earlier definition of a solution of the differential equation in (1.2.12), we say that y is a solution of the initial value problem (1.2.12) if y is n times differentiable and y(n)(x) = f(x, y(x), y′(x), . . ., y(n−1)(x)) for all x in some open interval (a, b) that contains x0, and y satisfies the initial conditions in (1.2.12). The largest open interval that contains x0 on which y is defined and satisfies the differential equation is the interval of validity of y. Example 1.2.6 In Example 1.2.5 we saw that y = x4 + 7 4 (1.2.13) is a solution of the initial value problem y′ = x3, y(1) = 2. Since the function in (1.2.13) is defined for all x, the interval of validity of this solution is (−∞, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_22_Chunk2428 |
Section 1.2 Basic Concepts 13 Example 1.2.7 In Example 1.2.2 we verified that y = x2 3 + 1 x (1.2.14) is a solution of xy′ + y = x2 on (0, ∞) and on (−∞, 0). By evaluating (1.2.14) at x = ±1, you can see that (1.2.14) is a solution of the initial value problems xy′ + y = x2, y(1) = 4 3 (1.2.15) and xy′ + y = x2, y(−1) = −2 3. (1.2.16) The interval of validity of (1.2.14) as a solution of (1.2.15) is (0, ∞), since this is the largest interval that contains x0 = 1 on which (1.2.14) is defined. Similarly, the interval of validity of (1.2.14) as a solution of (1.2.16) is (−∞, 0), since this is the largest interval that contains x0 = −1 on which (1.2.14) is defined. Free Fall Under Constant Gravity The term initial value problem originated in problems of motion where the independent variable is t (representing elapsed time), and the initial conditions are the position and velocity of an object at the initial (starting) time of an experiment. Example 1.2.8 An object falls under the influence of gravity near Earth’s surface, where it can be as- sumed that the magnitude of the acceleration due to gravity is a constant g. (a) Construct a mathematical model for the motion of the object in the form of an initial value problem for a second order differential equation, assuming that the altitude and velocity of the object at time t = 0 are known. Assume that gravity is the only force acting on the object. (b) Solve the initial value problem derived in (a) to obtain the altitude as a function of time. SOLUTION(a) Let y(t) be the altitude of the object at time t. Since the acceleration of the object has constant magnitude g and is in the downward (negative) direction, y satisfies the second order equation y′′ = −g, where the prime now indicates differentiation with respect to t. If y0 and v0 denote the altitude and velocity when t = 0, then y is a solution of the initial value problem y′′ = −g, y(0) = y0, y′(0) = v0. (1.2.17) SOLUTION(b) Integrating (1.2.17) twice yields y′ = −gt + c1, y = −gt2 2 + c1t + c2. Imposing the initial conditions y(0) = y0 and y′(0) = v0 in these two equations shows that c1 = v0 and c2 = y0. Therefore the solution of the initial value problem (1.2.17) is y = −gt2 2 + v0t + y0. | Elementary Differential Equations with Boundary Value Problems_Page_23_Chunk2429 |
14 Chapter 1 Introduction 1.2 Exercises 1. Find the order of the equation. (a) d2y dx2 + 2 dy dx d3y dx3 + x = 0 (b) y′′ −3y′ + 2y = x7 (c) y′ −y7 = 0 (d) y′′y −(y′)2 = 2 2. Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function. (a) y = ce2x; y′ = 2y (b) y = x2 3 + c x; xy′ + y = x2 (c) y = 1 2 + ce−x2; y′ + 2xy = x (d) y = (1 + ce−x2/2); (1 −ce−x2/2)−1 2y′ + x(y2 −1) = 0 (e) y = tan x3 3 + c ; y′ = x2(1 + y2) (f) y = (c1 + c2x)ex + sin x + x2; y′′ −2y′ + y = −2 cos x + x2 −4x + 2 (g) y = c1ex + c2x + 2 x; (1 −x)y′′ + xy′ −y = 4(1 −x −x2)x−3 (h) y = x−1/2(c1 sinx + c2 cos x) + 4x + 8; x2y′′ + xy′ + x2 −1 4 y = 4x3 + 8x2 + 3x −2 3. Find all solutions of the equation. (a) y′ = −x (b) y′ = −x sin x (c) y′ = x lnx (d) y′′ = x cos x (e) y′′ = 2xex (f) y′′ = 2x + sinx + ex (g) y′′′ = −cos x (h) y′′′ = −x2 + ex (i) y′′′ = 7e4x 4. Solve the initial value problem. (a) y′ = −xex, y(0) = 1 (b) y′ = x sinx2, y rπ 2 = 1 (c) y′ = tan x, y(π/4) = 3 (d) y′′ = x4, y(2) = −1, y′(2) = −1 (e) y′′ = xe2x, y(0) = 7, y′(0) = 1 (f) y′′ = −x sin x, y(0) = 1, y′(0) = −3 (g) y′′′ = x2ex, y(0) = 1, y′(0) = −2, y′′(0) = 3 (h) y′′′ = 2 + sin 2x, y(0) = 1, y′(0) = −6, y′′(0) = 3 (i) y′′′ = 2x + 1, y(2) = 1, y′(2) = −4, y′′(2) = 7 5. Verify that the function is a solution of the initial value problem. (a) y = x cos x; y′ = cos x −y tan x, y(π/4) = π 4 √ 2 (b) y = 1 + 2 lnx x2 + 1 2; y′ = x2 −2x2y + 2 x3 , y(1) = 3 2 | Elementary Differential Equations with Boundary Value Problems_Page_24_Chunk2430 |
Section 1.2 Basic Concepts 15 (c) y = tan x2 2 ; y′ = x(1 + y2), y(0) = 0 (d) y = 2 x −2; y′ = −y(y + 1) x , y(1) = −2 6. Verify that the function is a solution of the initial value problem. (a) y = x2(1 + ln x); y′′ = 3xy′ −4y x2 , y(e) = 2e2, y′(e) = 5e (b) y = x2 3 + x −1; y′′ = x2 −xy′ + y + 1 x2 , y(1) = 1 3, y′(1) = 5 3 (c) y = (1 + x2)−1/2; y′′ = (x2 −1)y −x(x2 + 1)y′ (x2 + 1)2 , y(0) = 1, y′(0) = 0 (d) y = x2 1 −x; y′′ = 2(x + y)(xy′ −y) x3 , y(1/2) = 1/2, y′(1/2) = 3 7. Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only force acting on it thereafter is gravity. Take g = 32 ft/sec2. (a) Find the highest altitude attained by the object. (b) Determine how long it takes for the object to fall to the ground. 8. Let a be a nonzero real number. (a) Verify that if c is an arbitrary constant then y = (x −c)a (A) is a solution of y′ = ay(a−1)/a (B) on (c, ∞). (b) Suppose a < 0 or a > 1. Can you think of a solution of (B) that isn’t of the form (A)? 9. Verify that y = ( ex −1, x ≥0, 1 −e−x, x < 0, is a solution of y′ = |y| + 1 on (−∞, ∞). HINT: Use the definition of derivative at x = 0. 10. (a) Verify that if c is any real number then y = c2 + cx + 2c + 1 (A) satisfies y′ = −(x + 2) + p x2 + 4x + 4y 2 (B) on some open interval. Identify the open interval. (b) Verify that y1 = −x(x + 4) 4 also satisfies (B) on some open interval, and identify the open interval. (Note that y1 can’t be obtained by selecting a value of c in (A).) | Elementary Differential Equations with Boundary Value Problems_Page_25_Chunk2431 |
16 Chapter 1 Introduction 1.3 DIRECTION FIELDS FOR FIRST ORDER EQUATIONS It’s impossible to find explicit formulas for solutions of some differential equations. Even if there are such formulas, they may be so complicated that they’re useless. In this case we may resort to graphical or numerical methods to get some idea of how the solutions of the given equation behave. In Section 2.3 we’ll take up the question of existence of solutions of a first order equation y′ = f(x, y). (1.3.1) In this section we’ll simply assume that (1.3.1) has solutions and discuss a graphical method for ap- proximating them. In Chapter 3 we discuss numerical methods for obtaining approximate solutions of (1.3.1). Recall that a solution of (1.3.1) is a function y = y(x) such that y′(x) = f(x, y(x)) for all values of x in some interval, and an integral curve is either the graph of a solution or is made up of segments that are graphs of solutions. Therefore, not being able to solve (1.3.1) is equivalent to not knowing the equations of integral curves of (1.3.1). However, it’s easy to calculate the slopes of these curves. To be specific, the slope of an integral curve of (1.3.1) through a given point (x0, y0) is given by the number f(x0, y0). This is the basis of the method of direction fields. If f is defined on a set R, we can construct a direction field for (1.3.1) in R by drawing a short line segment through each point (x, y) in R with slope f(x, y). Of course, as a practical matter, we can’t actually draw line segments through every point in R; rather, we must select a finite set of points in R. For example, suppose f is defined on the closed rectangular region R : {a ≤x ≤b, c ≤y ≤d}. Let a = x0 < x1 < · · · < xm = b be equally spaced points in [a, b] and c = y0 < y1 < · · · < yn = d be equally spaced points in [c, d]. We say that the points (xi, yj), 0 ≤i ≤m, 0 ≤j ≤n, form a rectangular grid (Figure 1.3.1). Through each point in the grid we draw a short line segment with slope f(xi, yj). The result is an approximation to a direction field for (1.3.1) in R. If the grid points are sufficiently numerous and close together, we can draw approximate integral curves of (1.3.1) by drawing curves through points in the grid tangent to the line segments associated with the points in the grid. | Elementary Differential Equations with Boundary Value Problems_Page_26_Chunk2432 |
Section 1.3 Direction Fields for First Order Equations 17 y x a b c d Figure 1.3.1 A rectangular grid Unfortunately, approximating a direction field and graphing integral curves in this way is too tedious to be done effectively by hand. However, there is software for doing this. As you’ll see, the combina- tion of direction fields and integral curves gives useful insights into the behavior of the solutions of the differential equation even if we can’t obtain exact solutions. We’ll study numerical methods for solving a single first order equation (1.3.1) in Chapter 3. These methods can be used to plot solution curves of (1.3.1) in a rectangular region R if f is continuous on R. Figures 1.3.2, 1.3.3, and 1.3.4 show direction fields and solution curves for the differential equations y′ = x2 −y2 1 + x2 + y2 , y′ = 1 + xy2, and y′ = x −y 1 + x2 , which are all of the form (1.3.1) with f continuous for all (x, y). −4 −3 −2 −1 0 1 2 3 4 −4 −3 −2 −1 0 1 2 3 4 y x Figure 1.3.2 A direction field and integral curves for y = x2 −y2 1 + x2 + y2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y x Figure 1.3.3 A direction field and integral curves for y′ = 1 + xy2 | Elementary Differential Equations with Boundary Value Problems_Page_27_Chunk2433 |
18 Chapter 1 Introduction −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 y x Figure 1.3.4 A direction and integral curves for y′ = x −y 1 + x2 The methods of Chapter 3 won’t work for the equation y′ = −x/y (1.3.2) if R contains part of the x-axis, since f(x, y) = −x/y is undefined when y = 0. Similarly, they won’t work for the equation y′ = x2 1 −x2 −y2 (1.3.3) if R contains any part of the unit circle x2 + y2 = 1, because the right side of (1.3.3) is undefined if x2 + y2 = 1. However, (1.3.2) and (1.3.3) can written as y′ = A(x, y) B(x, y) (1.3.4) where A and B are continuous on any rectangle R. Because of this, some differential equation software is based on numerically solving pairs of equations of the form dx dt = B(x, y), dy dt = A(x, y) (1.3.5) where x and y are regarded as functions of a parameter t. If x = x(t) and y = y(t) satisfy these equations, then y′ = dy dx = dy dt dx dt = A(x, y) B(x, y), so y = y(x) satisfies (1.3.4). | Elementary Differential Equations with Boundary Value Problems_Page_28_Chunk2434 |
Section 1.3 Direction Fields for First Order Equations 19 Eqns. (1.3.2) and (1.3.3) can be reformulated as in (1.3.4) with dx dt = −y, dy dt = x and dx dt = 1 −x2 −y2, dy dt = x2, respectively. Even if f is continuous and otherwise “nice” throughout R, your software may require you to reformulate the equation y′ = f(x, y) as dx dt = 1, dy dt = f(x, y), which is of the form (1.3.5) with A(x, y) = f(x, y) and B(x, y) = 1. Figure 1.3.5 shows a direction field and some integral curves for (1.3.2). As we saw in Example 1.2.1 and will verify again in Section 2.2, the integral curves of (1.3.2) are circles centered at the origin. x y Figure 1.3.5 A direction field and integral curves for y′ = −x y Figure 1.3.6 shows a direction field and some integral curves for (1.3.3). The integral curves near the top and bottom are solution curves. However, the integral curves near the middle are more complicated. For example, Figure 1.3.7 shows the integral curve through the origin. The vertices of the dashed rectangle are on the circle x2 + y2 = 1 (a ≈.846, b ≈.533), where all integral curves of (1.3.3) have infinite slope. There are three solution curves of (1.3.3) on the integral curve in the figure: the segment above the level y = b is the graph of a solution on (−∞, a), the segment below the level y = −b is the graph of a solution on (−a, ∞), and the segment between these two levels is the graph of a solution on (−a, a). USING TECHNOLOGY | Elementary Differential Equations with Boundary Value Problems_Page_29_Chunk2435 |
20 Chapter 1 Introduction As you study from this book, you’ll often be asked to use computer software and graphics. Exercises with this intent are marked as C (computer or calculator required), C/G (computer and/or graphics required), or L (laboratory work requiring software and/or graphics). Often you may not completely understand how the software does what it does. This is similar to the situation most people are in when they drive automobiles or watch television, and it doesn’t decrease the value of using modern technology as an aid to learning. Just be careful that you use the technology as a supplement to thought rather than a substitute for it. y x Figure 1.3.6 A direction field and integral curves for y′ = x2 1 −x2 −y2 x y (a,−b) (a,b) (−a,b) (−a,−b) 1 2 −1 −2 1 2 −1 −2 Figure 1.3.7 1.3 Exercises In Exercises 1–11 a direction field is drawn for the given equation. Sketch some integral curves. | Elementary Differential Equations with Boundary Value Problems_Page_30_Chunk2436 |
Section 1.3 Direction Fields for First Order Equations 21 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x y 1 A direction field for y′ = x y | Elementary Differential Equations with Boundary Value Problems_Page_31_Chunk2437 |
22 Chapter 1 Introduction 0 0.5 1 1.5 2 2.5 3 3.5 4 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 x y 2 A direction field for y′ = 2xy2 1 + x2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x y 3 A direction field for y′ = x2(1 + y2) | Elementary Differential Equations with Boundary Value Problems_Page_32_Chunk2438 |
Section 1.3 Direction Fields for First Order Equations 23 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x y 4 A direction field for y′ = 1 1 + x2 + y2 0 0.5 1 1.5 2 2.5 3 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x y 5 A direction field for y′ = −(2xy2 + y3) | Elementary Differential Equations with Boundary Value Problems_Page_33_Chunk2439 |
24 Chapter 1 Introduction −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x y 6 A direction field for y′ = (x2 + y2)1/2 0 1 2 3 4 5 6 7 −3 −2 −1 0 1 2 3 x y 7 A direction field for y′ = sin xy | Elementary Differential Equations with Boundary Value Problems_Page_34_Chunk2440 |
Section 1.3 Direction Fields for First Order Equations 25 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x y 8 A direction field for y′ = exy 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x y 9 A direction field for y′ = (x −y2)(x2 −y) | Elementary Differential Equations with Boundary Value Problems_Page_35_Chunk2441 |
26 Chapter 1 Introduction 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 x y 10 A direction field for y′ = x3y2 + xy3 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 2.5 3 3.5 4 x y 11 A direction field for y′ = sin(x −2y) | Elementary Differential Equations with Boundary Value Problems_Page_36_Chunk2442 |
Section 1.3 Direction Fields for First Order Equations 27 In Exercises 12-22 construct a direction field and plot some integral curves in the indicated rectangular region. 12. C/G y′ = y(y −1); {−1 ≤x ≤2, −2 ≤y ≤2} 13. C/G y′ = 2 −3xy; {−1 ≤x ≤4, −4 ≤y ≤4} 14. C/G y′ = xy(y −1); {−2 ≤x ≤2, −4 ≤y ≤4} 15. C/G y′ = 3x + y; {−2 ≤x ≤2, 0 ≤y ≤4} 16. C/G y′ = y −x3; {−2 ≤x ≤2, −2 ≤y ≤2} 17. C/G y′ = 1 −x2 −y2; {−2 ≤x ≤2, −2 ≤y ≤2} 18. C/G y′ = x(y2 −1); {−3 ≤x ≤3, −3 ≤y ≤2} 19. C/G y′ = x y(y2 −1); {−2 ≤x ≤2, −2 ≤y ≤2} 20. C/G y′ = xy2 y −1; {−2 ≤x ≤2, −1 ≤y ≤4} 21. C/G y′ = x(y2 −1) y ; {−1 ≤x ≤1, −2 ≤y ≤2} 22. C/G y′ = − x2 + y2 1 −x2 −y2 ; {−2 ≤x ≤2, −2 ≤y ≤2} 23. L By suitably renaming the constants and dependent variables in the equations T ′ = −k(T −Tm) (A) and G′ = −λG + r (B) discussed in Section 1.2 in connection with Newton’s law of cooling and absorption of glucose in the body, we can write both as y′ = −ay + b, (C) where a is a positive constant and b is an arbitrary constant. Thus, (A) is of the form (C) with y = T, a = k, and b = kTm, and (B) is of the form (C) with y = G, a = λ, and b = r. We’ll encounter equations of the form (C) in many other applications in Chapter 2. Choose a positive a and an arbitrary b. Construct a direction field and plot some integral curves for (C) in a rectangular region of the form {0 ≤t ≤T, c ≤y ≤d} of the ty-plane. Vary T, c, and d until you discover a common property of all the solutions of (C). Repeat this experiment with various choices of a and b until you can state this property precisely in terms of a and b. 24. L By suitably renaming the constants and dependent variables in the equations P ′ = aP (1 −αP ) (A) and I′ = rI(S −I) (B) | Elementary Differential Equations with Boundary Value Problems_Page_37_Chunk2443 |
28 Chapter 1 Introduction discussed in Section 1.1 in connection with Verhulst’s population model and the spread of an epidemic, we can write both in the form y′ = ay −by2, (C) where a and b are positive constants. Thus, (A) is of the form (C) with y = P , a = a, and b = aα, and (B) is of the form (C) with y = I, a = rS, and b = r. In Chapter 2 we’ll encounter equations of the form (C) in other applications.. (a) Choose positive numbers a and b. Construct a direction field and plot some integral curves for (C) in a rectangular region of the form {0 ≤t ≤T, 0 ≤y ≤d} of the ty-plane. Vary T and d until you discover a common property of all solutions of (C) with y(0) > 0. Repeat this experiment with various choices of a and b until you can state this property precisely in terms of a and b. (b) Choose positive numbers a and b. Construct a direction field and plot some integral curves for (C) in a rectangular region of the form {0 ≤t ≤T, c ≤y ≤0} of the ty-plane. Vary a, b, T and c until you discover a common property of all solutions of (C) with y(0) < 0. You can verify your results later by doing Exercise 2.2.27. | Elementary Differential Equations with Boundary Value Problems_Page_38_Chunk2444 |
CHAPTER 2 First Order Equations IN THIS CHAPTER we study first order equations for which there are general methods of solution. SECTION 2.1 deals with linear equations, the simplest kind of first order equations. In this section we introduce the method of variation of parameters. The idea underlying this method will be a unifying theme for our approach to solving many different kinds of differential equations throughout the book. SECTION 2.2 deals with separable equations, the simplest nonlinear equations. In this section we intro- duce the idea of implicit and constant solutions of differential equations, and we point out some differ- ences between the properties of linear and nonlinear equations. SECTION 2.3 discusses existence and uniqueness of solutions of nonlinear equations. Although it may seem logical to place this section before Section 2.2, we presented Section 2.2 first so we could have illustrative examples in Section 2.3. SECTION 2.4 deals with nonlinear equations that are not separable, but can be transformed into separable equations by a procedure similar to variation of parameters. SECTION 2.5 covers exact differential equations, which are given this name because the method for solving them uses the idea of an exact differential from calculus. SECTION 2.6 deals with equations that are not exact, but can made exact by multiplying them by a function known called integrating factor. 29 | Elementary Differential Equations with Boundary Value Problems_Page_39_Chunk2445 |
30 Chapter 2 First Order Equations 2.1 LINEAR FIRST ORDER EQUATIONS A first order differential equation is said to be linear if it can be written as y′ + p(x)y = f(x). (2.1.1) A first order differential equation that can’t be written like this is nonlinear. We say that (2.1.1) is homogeneous if f ≡0; otherwise it’s nonhomogeneous. Since y ≡0 is obviously a solution of the homgeneous equation y′ + p(x)y = 0, we call it the trivial solution. Any other solution is nontrivial. Example 2.1.1 The first order equations x2y′ + 3y = x2, xy′ −8x2y = sin x, xy′ + (ln x)y = 0, y′ = x2y −2, are not in the form (2.1.1), but they are linear, since they can be rewritten as y′ + 3 x2 y = 1, y′ −8xy = sin x x , y′ + ln x x y = 0, y′ −x2y = −2. Example 2.1.2 Here are some nonlinear first order equations: xy′ + 3y2 = 2x (because y is squared), yy′ = 3 (because of the product yy′), y′ + xey = 12 (because of ey). General Solution of a Linear First Order Equation To motivate a definition that we’ll need, consider the simple linear first order equation y′ = 1 x2 . (2.1.2) From calculus we know that y satisfies this equation if and only if y = −1 x + c, (2.1.3) where c is an arbitrary constant. We call c a parameter and say that (2.1.3) defines a one–parameter family of functions. For each real number c, the function defined by (2.1.3) is a solution of (2.1.2) on | Elementary Differential Equations with Boundary Value Problems_Page_40_Chunk2446 |
Section 2.1 Linear First Order Equations 31 (−∞, 0) and (0, ∞); moreover, every solution of (2.1.2) on either of these intervals is of the form (2.1.3) for some choice of c. We say that (2.1.3) is the general solution of (2.1.2). We’ll see that a similar situation occurs in connection with any first order linear equation y′ + p(x)y = f(x); (2.1.4) that is, if p and f are continuous on some open interval (a, b) then there’s a unique formula y = y(x, c) analogous to (2.1.3) that involves x and a parameter c and has the these properties: • For each fixed value of c, the resulting function of x is a solution of (2.1.4) on (a, b). • If y is a solution of (2.1.4) on (a, b), then y can be obtained from the formula by choosing c appropriately. We’ll call y = y(x, c) the general solution of (2.1.4). When this has been established, it will follow that an equation of the form P0(x)y′ + P1(x)y = F (x) (2.1.5) has a general solution on any open interval (a, b) on which P0, P1, and F are all continuous and P0 has no zeros, since in this case we can rewrite (2.1.5) in the form (2.1.4) with p = P1/P0 and f = F/P0, which are both continuous on (a, b). To avoid awkward wording in examples and exercises, we won’t specify the interval (a, b) when we ask for the general solution of a specific linear first order equation. Let’s agree that this always means that we want the general solution on every open interval on which p and f are continuous if the equation is of the form (2.1.4), or on which P0, P1, and F are continuous and P0 has no zeros, if the equation is of the form (2.1.5). We leave it to you to identify these intervals in specific examples and exercises. For completeness, we point out that if P0, P1, and F are all continuous on an open interval (a, b), but P0 does have a zero in (a, b), then (2.1.5) may fail to have a general solution on (a, b) in the sense just defined. Since this isn’t a major point that needs to be developed in depth, we won’t discuss it further; however, see Exercise 44 for an example. Homogeneous Linear First Order Equations We begin with the problem of finding the general solution of a homogeneous linear first order equation. The next example recalls a familiar result from calculus. Example 2.1.3 Let a be a constant. (a) Find the general solution of y′ −ay = 0. (2.1.6) (b) Solve the initial value problem y′ −ay = 0, y(x0) = y0. SOLUTION(a) You already know from calculus that if c is any constant, then y = ceax satisfies (2.1.6). However, let’s pretend you’ve forgotten this, and use this problem to illustrate a general method for solving a homogeneous linear first order equation. We know that (2.1.6) has the trivial solution y ≡0. Now suppose y is a nontrivial solution of (2.1.6). Then, since a differentiable function must be continuous, there must be some open interval I on which y has no zeros. We rewrite (2.1.6) as y′ y = a | Elementary Differential Equations with Boundary Value Problems_Page_41_Chunk2447 |
32 Chapter 2 First Order Equations x 0.2 0.4 0.6 0.8 1.0 y 0.5 1.0 1.5 2.0 2.5 3.0 a = 2 a = 1.5 a = 1 a = −1 a = −2.5 a = −4 Figure 2.1.1 Solutions of y′ −ay = 0, y(0) = 1 for x in I. Integrating this shows that ln|y| = ax + k, so |y| = ekeax, where k is an arbitrary constant. Since eax can never equal zero, y has no zeros, so y is either always positive or always negative. Therefore we can rewrite y as y = ceax (2.1.7) where c = ek if y > 0, −ek if y < 0. This shows that every nontrivial solution of (2.1.6) is of the form y = ceax for some nonzero constant c. Since setting c = 0 yields the trivial solution, all solutions of (2.1.6) are of the form (2.1.7). Conversely, (2.1.7) is a solution of (2.1.6) for every choice of c, since differentiating (2.1.7) yields y′ = aceax = ay. SOLUTION(b) Imposing the initial condition y(x0) = y0 yields y0 = ceax0, so c = y0e−ax0 and y = y0e−ax0eax = y0ea(x−x0). Figure 2.1.1 show the graphs of this function with x0 = 0, y0 = 1, and various values of a. Example 2.1.4 (a) Find the general solution of xy′ + y = 0. (2.1.8) (b) Solve the initial value problem xy′ + y = 0, y(1) = 3. (2.1.9) | Elementary Differential Equations with Boundary Value Problems_Page_42_Chunk2448 |
Section 2.1 Linear First Order Equations 33 SOLUTION(a) We rewrite (2.1.8) as y′ + 1 xy = 0, (2.1.10) where x is restricted to either (−∞, 0) or (0, ∞). If y is a nontrivial solution of (2.1.10), there must be some open interval I on which y has no zeros. We can rewrite (2.1.10) as y′ y = −1 x for x in I. Integrating shows that ln|y| = −ln |x| + k, so |y| = ek |x|. Since a function that satisfies the last equation can’t change sign on either (−∞, 0) or (0, ∞), we can rewrite this result more simply as y = c x (2.1.11) where c = ek if y > 0, −ek if y < 0. We’ve now shown that every solution of (2.1.10) is given by (2.1.11) for some choice of c. (Even though we assumed that y was nontrivial to derive (2.1.11), we can get the trivial solution by setting c = 0 in (2.1.11).) Conversely, any function of the form (2.1.11) is a solution of (2.1.10), since differentiating (2.1.11) yields y′ = −c x2 , and substituting this and (2.1.11) into (2.1.10) yields y′ + 1 xy = −c x2 + 1 x c x = −c x2 + c x2 = 0. Figure 2.1.2 shows the graphs of some solutions corresponding to various values of c SOLUTION(b) Imposing the initial condition y(1) = 3 in (2.1.11) yields c = 3. Therefore the solution of (2.1.9) is y = 3 x. The interval of validity of this solution is (0, ∞). The results in Examples 2.1.3(a) and 2.1.4(b) are special cases of the next theorem. Theorem 2.1.1 If p is continuous on (a, b), then the general solution of the homogeneous equation y′ + p(x)y = 0 (2.1.12) on (a, b) is y = ce−P(x), where P (x) = Z p(x) dx (2.1.13) is any antiderivative of p on (a, b); that is, P ′(x) = p(x), a < x < b. (2.1.14) | Elementary Differential Equations with Boundary Value Problems_Page_43_Chunk2449 |
34 Chapter 2 First Order Equations x y c > 0 c < 0 c > 0 c < 0 Figure 2.1.2 Solutions of xy′ + y = 0 on (0, ∞) and (−∞, 0) Proof If y = ce−P(x), differentiating y and using (2.1.14) shows that y′ = −P ′(x)ce−P(x) = −p(x)ce−P(x) = −p(x)y, so y′ + p(x)y = 0; that is, y is a solution of (2.1.12), for any choice of c. Now we’ll show that any solution of (2.1.12) can be written as y = ce−P(x) for some constant c. The trivial solution can be written this way, with c = 0. Now suppose y is a nontrivial solution. Then there’s an open subinterval I of (a, b) on which y has no zeros. We can rewrite (2.1.12) as y′ y = −p(x) (2.1.15) for x in I. Integrating (2.1.15) and recalling (2.1.13) yields ln|y| = −P (x) + k, where k is a constant. This implies that |y| = eke−P(x). Since P is defined for all x in (a, b) and an exponential can never equal zero, we can take I = (a, b), so y has zeros on (a, b) (a, b), so we can rewrite the last equation as y = ce−P(x), where c = ek if y > 0 on (a, b), −ek if y < 0 on (a, b). REMARK: Rewriting a first order differential equation so that one side depends only on y and y′ and the other depends only on x is called separation of variables. We did this in Examples 2.1.3 and 2.1.4, and in rewriting (2.1.12) as (2.1.15).We’llapply this method to nonlinear equations in Section 2.2. | Elementary Differential Equations with Boundary Value Problems_Page_44_Chunk2450 |
Section 2.1 Linear First Order Equations 35 Linear Nonhomogeneous First Order Equations We’ll now solve the nonhomogeneous equation y′ + p(x)y = f(x). (2.1.16) When considering this equation we call y′ + p(x)y = 0 the complementary equation. We’ll find solutions of (2.1.16) in the form y = uy1, where y1 is a nontrivial solution of the com- plementary equation and u is to be determined. This method of using a solution of the complementary equation to obtain solutions of a nonhomogeneous equation is a special case of a method called variation of parameters, which you’ll encounter several times in this book. (Obviously, u can’t be constant, since if it were, the left side of (2.1.16) would be zero. Recognizing this, the early users of this method viewed u as a “parameter” that varies; hence, the name “variation of parameters.”) If y = uy1, then y′ = u′y1 + uy′ 1. Substituting these expressions for y and y′ into (2.1.16) yields u′y1 + u(y′ 1 + p(x)y1) = f(x), which reduces to u′y1 = f(x), (2.1.17) since y1 is a solution of the complementary equation; that is, y′ 1 + p(x)y1 = 0. In the proof of Theorem 2.2.1 we saw that y1 has no zeros on an interval where p is continuous. Therefore we can divide (2.1.17) through by y1 to obtain u′ = f(x)/y1(x). We can integrate this (introducing a constant of integration), and multiply the result by y1 to get the gen- eral solution of (2.1.16). Before turning to the formal proof of this claim, let’s consider some examples. Example 2.1.5 Find the general solution of y′ + 2y = x3e−2x. (2.1.18) By applying (a) of Example 2.1.3 with a = −2, we see that y1 = e−2x is a solution of the com- plementary equation y′ + 2y = 0. Therefore we seek solutions of (2.1.18) in the form y = ue−2x, so that y′ = u′e−2x −2ue−2x and y′ + 2y = u′e−2x −2ue−2x + 2ue−2x = u′e−2x. (2.1.19) Therefore y is a solution of (2.1.18) if and only if u′e−2x = x3e−2x or, equivalently, u′ = x3. Therefore u = x4 4 + c, | Elementary Differential Equations with Boundary Value Problems_Page_45_Chunk2451 |
36 Chapter 2 First Order Equations −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 x y Figure 2.1.3 A direction field and integral curves for y′ + 2y = x2e−2x and y = ue−2x = e−2x x4 4 + c is the general solution of (2.1.18). Figure 2.1.3 shows a direction field and some integral curves for (2.1.18). Example 2.1.6 (a) Find the general solution y′ + (cot x)y = x csc x. (2.1.20) (b) Solve the initial value problem y′ + (cot x)y = x csc x, y(π/2) = 1. (2.1.21) SOLUTION(a) Here p(x) = cot x and f(x) = x csc x are both continuous except at the points x = rπ, where r is an integer. Therefore we seek solutions of (2.1.20) on the intervals (rπ, (r + 1)π). We need a nontrival solution y1 of the complementary equation; thus, y1 must satisfy y′ 1 + (cot x)y1 = 0, which we rewrite as y′ 1 y1 = −cot x = −cos x sin x . (2.1.22) Integrating this yields ln |y1| = −ln | sinx|, where we take the constant of integration to be zero since we need only one function that satisfies (2.1.22). Clearly y1 = 1/ sin x is a suitable choice. Therefore we seek solutions of (2.1.20) in the form y = u sin x, | Elementary Differential Equations with Boundary Value Problems_Page_46_Chunk2452 |
Section 2.1 Linear First Order Equations 37 so that y′ = u′ sin x −u cos x sin2 x (2.1.23) and y′ + (cot x)y = u′ sin x −u cos x sin2 x + u cot x sin x = u′ sin x −u cos x sin2 x + u cos x sin2 x = u′ sin x. (2.1.24) Therefore y is a solution of (2.1.20) if and only if u′/ sin x = x csc x = x/ sinx or, equivalently, u′ = x. Integrating this yields u = x2 2 + c, and y = u sin x = x2 2 sinx + c sin x. (2.1.25) is the general solution of (2.1.20) on every interval (rπ, (r + 1)π) (r =integer). SOLUTION(b) Imposing the initial condition y(π/2) = 1 in (2.1.25) yields 1 = π2 8 + c or c = 1 −π2 8 . Thus, y = x2 2 sinx + (1 −π2/8) sin x is a solution of (2.1.21). The interval of validity of this solution is (0, π); Figure 2.1.4 shows its graph. 1 2 3 − 15 − 10 − 5 5 10 15 x y Figure 2.1.4 Solution of y′ + (cot x)y = x csc x, y(π/2) = 1 | Elementary Differential Equations with Boundary Value Problems_Page_47_Chunk2453 |
38 Chapter 2 First Order Equations REMARK: It wasn’t necessary to do the computations (2.1.23) and (2.1.24) in Example 2.1.6, since we showed in the discussion preceding Example 2.1.5 that if y = uy1 where y′ 1 + p(x)y1 = 0, then y′ + p(x)y = u′y1. We did these computations so you would see this happen in this specific example. We recommend that you include these “unnecesary” computations in doing exercises, until you’re confident that you really understand the method. After that, omit them. We summarize the method of variation of parameters for solving y′ + p(x)y = f(x) (2.1.26) as follows: (a) Find a function y1 such that y′ 1 y1 = −p(x). For convenience, take the constant of integration to be zero. (b) Write y = uy1 (2.1.27) to remind yourself of what you’re doing. (c) Write u′y1 = f and solve for u′; thus, u′ = f/y1. (d) Integrate u′ to obtain u, with an arbitrary constant of integration. (e) Substitute u into (2.1.27) to obtain y. To solve an equation written as P0(x)y′ + P1(x)y = F (x), we recommend that you divide through by P0(x) to obtain an equation of the form (2.1.26) and then follow this procedure. Solutions in Integral Form Sometimes the integrals that arise in solving a linear first order equation can’t be evaluated in terms of elementary functions. In this case the solution must be left in terms of an integral. Example 2.1.7 (a) Find the general solution of y′ −2xy = 1. (b) Solve the initial value problem y′ −2xy = 1, y(0) = y0. (2.1.28) SOLUTION(a) To apply variation of parameters, we need a nontrivial solution y1 of the complementary equation; thus, y′ 1 −2xy1 = 0, which we rewrite as y′ 1 y1 = 2x. | Elementary Differential Equations with Boundary Value Problems_Page_48_Chunk2454 |
Section 2.1 Linear First Order Equations 39 Integrating this and taking the constant of integration to be zero yields ln|y1| = x2, so |y1| = ex2. We choose y1 = ex2 and seek solutions of (2.1.28) in the form y = uex2, where u′ex2 = 1, so u′ = e−x2. Therefore u = c + Z e−x2dx, but we can’t simplify the integral on the right because there’s no elementary function with derivative equal to e−x2. Therefore the best available form for the general solution of (2.1.28) is y = uex2 = ex2 c + Z e−x2dx . (2.1.29) SOLUTION(b) Since the initial condition in (2.1.28) is imposed at x0 = 0, it is convenient to rewrite (2.1.29) as y = ex2 c + Z x 0 e−t2dt , since Z 0 0 e−t2 dt = 0. Setting x = 0 and y = y0 here shows that c = y0. Therefore the solution of the initial value problem is y = ex2 y0 + Z x 0 e−t2dt . (2.1.30) For a given value of y0 and each fixed x, the integral on the right can be evaluated by numerical methods. An alternate procedure is to apply the numerical integration procedures discussed in Chapter 3 directly to the initial value problem (2.1.28). Figure 2.1.5 shows graphs of of (2.1.30) for several values of y0. x y Figure 2.1.5 Solutions of y′ −2xy = 1, y(0) = y0 | Elementary Differential Equations with Boundary Value Problems_Page_49_Chunk2455 |
40 Chapter 2 First Order Equations An Existence and Uniqueness Theorem The method of variation of parameters leads to this theorem. Theorem 2.1.2 Suppose p and f are continuous on an open interval (a, b), and let y1 be any nontrivial solution of the complementary equation y′ + p(x)y = 0 on (a, b). Then: (a) The general solution of the nonhomogeneous equation y′ + p(x)y = f(x) (2.1.31) on (a, b) is y = y1(x) c + Z f(x)/y1(x) dx . (2.1.32) (b) If x0 is an arbitrary point in (a, b) and y0 is an arbitrary real number, then the initial value problem y′ + p(x)y = f(x), y(x0) = y0 has the unique solution y = y1(x) y0 y1(x0) + Z x x0 f(t) y1(t) dt on (a, b). Proof (a) To show that (2.1.32) is the general solution of (2.1.31) on (a, b), we must prove that: (i) If c is any constant, the function y in (2.1.32) is a solution of (2.1.31) on (a, b). (ii) If y is a solution of (2.1.31) on (a, b) then y is of the form (2.1.32) for some constant c. To prove (i), we first observe that any function of the form (2.1.32) is defined on (a, b), since p and f are continuous on (a, b). Differentiating (2.1.32) yields y′ = y′ 1(x) c + Z f(x)/y1(x) dx + f(x). Since y′ 1 = −p(x)y1, this and (2.1.32) imply that y′ = −p(x)y1(x) c + Z f(x)/y1(x) dx + f(x) = −p(x)y(x) + f(x), which implies that y is a solution of (2.1.31). To prove (ii), suppose y is a solution of (2.1.31) on (a, b). From the proof of Theorem 2.1.1, we know that y1 has no zeros on (a, b), so the function u = y/y1 is defined on (a, b). Moreover, since y′ = −py + f and y′ 1 = −py1, u′ = y1y′ −y′ 1y y2 1 = y1(−py + f) −(−py1)y y2 1 = f y1 . | Elementary Differential Equations with Boundary Value Problems_Page_50_Chunk2456 |
Section 2.1 Linear First Order Equations 41 Integrating u′ = f/y1 yields u = c + Z f(x)/y1(x) dx , which implies (2.1.32), since y = uy1. (b) We’ve proved (a), where R f(x)/y1(x) dx in (2.1.32) is an arbitrary antiderivative of f/y1. Now it’s convenient to choose the antiderivative that equals zero when x = x0, and write the general solution of (2.1.31) as y = y1(x) c + Z x x0 f(t) y1(t) dt . Since y(x0) = y1(x0) c + Z x0 x0 f(t) y1(t) dt = cy1(x0), we see that y(x0) = y0 if and only if c = y0/y1(x0). 2.1 Exercises In Exercises 1–5 find the general solution. 1. y′ + ay = 0 (a=constant) 2. y′ + 3x2y = 0 3. xy′ + (lnx)y = 0 4. xy′ + 3y = 0 5. x2y′ + y = 0 In Exercises 6–11 solve the initial value problem. 6. y′ + 1 + x x y = 0, y(1) = 1 7. xy′ + 1 + 1 lnx y = 0, y(e) = 1 8. xy′ + (1 + x cot x)y = 0, y π 2 = 2 9. y′ − 2x 1 + x2 y = 0, y(0) = 2 10. y′ + k xy = 0, y(1) = 3 (k= constant) 11. y′ + (tan kx)y = 0, y(0) = 2 (k = constant) In Exercises 12 –15 find the general solution. Also, plot a direction field and some integral curves on the rectangular region {−2 ≤x ≤2, −2 ≤y ≤2}. 12. C/G y′ + 3y = 1 13. C/G y′ + 1 x −1 y = −2 x 14. C/G y′ + 2xy = xe−x2 15. C/G y′ + 2x 1 + x2 y = e−x 1 + x2 | Elementary Differential Equations with Boundary Value Problems_Page_51_Chunk2457 |
42 Chapter 2 First Order Equations In Exercises 16 –24 find the general solution. 16. y′ + 1 xy = 7 x2 + 3 17. y′ + 4 x −1y = 1 (x −1)5 + sinx (x −1)4 18. xy′ + (1 + 2x2)y = x3e−x2 19. xy′ + 2y = 2 x2 + 1 20. y′ + (tan x)y = cos x 21. (1 + x)y′ + 2y = sin x 1 + x 22. (x −2)(x −1)y′ −(4x −3)y = (x −2)3 23. y′ + (2 sinx cos x)y = e−sin2 x 24. x2y′ + 3xy = ex In Exercises 25–29 solve the initial value problem and sketch the graph of the solution. 25. C/G y′ + 7y = e3x, y(0) = 0 26. C/G (1 + x2)y′ + 4xy = 2 1 + x2 , y(0) = 1 27. C/G xy′ + 3y = 2 x(1 + x2), y(−1) = 0 28. C/G y′ + (cot x)y = cos x, y π 2 = 1 29. C/G y′ + 1 xy = 2 x2 + 1, y(−1) = 0 In Exercises 30–37 solve the initial value problem. 30. (x −1)y′ + 3y = 1 (x −1)3 + sin x (x −1)2 , y(0) = 1 31. xy′ + 2y = 8x2, y(1) = 3 32. xy′ −2y = −x2, y(1) = 1 33. y′ + 2xy = x, y(0) = 3 34. (x −1)y′ + 3y = 1 + (x −1) sec2 x (x −1)3 , y(0) = −1 35. (x + 2)y′ + 4y = 1 + 2x2 x(x + 2)3 , y(−1) = 2 36. (x2 −1)y′ −2xy = x(x2 −1), y(0) = 4 37. (x2 −5)y′ −2xy = −2x(x2 −5), y(2) = 7 In Exercises 38–42 solve the initial value problem and leave the answer in a form involving a definite integral. (You can solve these problems numerically by methods discussed in Chapter 3.) 38. y′ + 2xy = x2, y(0) = 3 39. y′ + 1 xy = sin x x2 , y(1) = 2 40. y′ + y = e−x tan x x , y(1) = 0 | Elementary Differential Equations with Boundary Value Problems_Page_52_Chunk2458 |
Section 2.1 Linear First Order Equations 43 41. y′ + 2x 1 + x2 y = ex (1 + x2)2 , y(0) = 1 42. xy′ + (x + 1)y = ex2, y(1) = 2 43. Experiments indicate that glucose is absorbed by the body at a rate proportional to the amount of glucose present in the bloodstream. Let λ denote the (positive) constant of proportionality. Now suppose glucose is injected into a patient’s bloodstream at a constant rate of r units per unit of time. Let G = G(t) be the number of units in the patient’s bloodstream at time t > 0. Then G′ = −λG + r, where the first term on the right is due to the absorption of the glucose by the patient’s body and the second term is due to the injection. Determine G for t > 0, given that G(0) = G0. Also, find limt→∞G(t). 44. (a) L Plot a direction field and some integral curves for xy′ −2y = −1 (A) on the rectangular region {−1 ≤x ≤1, −.5 ≤y ≤1.5}. What do all the integral curves have in common? (b) Show that the general solution of (A) on (−∞, 0) and (0, ∞) is y = 1 2 + cx2. (c) Show that y is a solution of (A) on (−∞, ∞) if and only if y = 1 2 + c1x2, x ≥0, 1 2 + c2x2, x < 0, where c1 and c2 are arbitrary constants. (d) Conclude from (c) that all solutions of (A) on (−∞, ∞) are solutions of the initial value problem xy′ −2y = −1, y(0) = 1 2. (e) Use (b) to show that if x0 ̸= 0 and y0 is arbitrary, then the initial value problem xy′ −2y = −1, y(x0) = y0 has infinitely many solutionson (−∞, ∞). Explain why this does’nt contradict Theorem 2.1.1(b). 45. Suppose f is continuous on an open interval (a, b) and α is a constant. (a) Derive a formula for the solution of the initial value problem y′ + αy = f(x), y(x0) = y0, (A) where x0 is in (a, b) and y0 is an arbitrary real number. (b) Suppose (a, b) = (a, ∞), α > 0 and lim x→∞f(x) = L. Show that if y is the solution of (A), then lim x→∞y(x) = L/α. | Elementary Differential Equations with Boundary Value Problems_Page_53_Chunk2459 |
44 Chapter 2 First Order Equations 46. Assume that all functions in this exercise are defined on a common interval (a, b). (a) Prove: If y1 and y2 are solutions of y′ + p(x)y = f1(x) and y′ + p(x)y = f2(x) respectively, and c1 and c2 are constants, then y = c1y1 + c2y2 is a solution of y′ + p(x)y = c1f1(x) + c2f2(x). (This is theprinciple of superposition.) (b) Use (a) to show that if y1 and y2 are solutions of the nonhomogeneous equation y′ + p(x)y = f(x), (A) then y1 −y2 is a solution of the homogeneous equation y′ + p(x)y = 0. (B) (c) Use (a) to show that if y1 is a solution of (A) and y2 is a solution of (B), then y1 + y2 is a solution of (A). 47. Some nonlinear equations can be transformed into linear equations by changing the dependent variable. Show that if g′(y)y′ + p(x)g(y) = f(x) where y is a function of x and g is a function of y, then the new dependent variable z = g(y) satisfies the linear equation z′ + p(x)z = f(x). 48. Solve by the method discussed in Exercise 47. (a) (sec2 y)y′ −3 tany = −1 (b) ey2 2yy′ + 2 x = 1 x2 (c) xy′ y + 2 lny = 4x2 (d) y′ (1 + y)2 − 1 x(1 + y) = −3 x2 49. We’ve shown that if p and f are continuous on (a, b) then every solution of y′ + p(x)y = f(x) (A) on (a, b) can be written as y = uy1, where y1 is a nontrivial solution of the complementary equa- tion for (A) and u′ = f/y1. Now suppose f, f′, ..., f(m) and p, p′, ..., p(m−1) are continuous on (a, b), where m is a positive integer, and define f0 = f, fj = f′ j−1 + pfj−1, 1 ≤j ≤m. Show that u(j+1) = fj y1 , 0 ≤j ≤m. | Elementary Differential Equations with Boundary Value Problems_Page_54_Chunk2460 |
Section 2.2 Separable Equations 45 2.2 SEPARABLE EQUATIONS A first order differential equation is separable if it can be written as h(y)y′ = g(x), (2.2.1) where the left side is a product of y′ and a function of y and the right side is a function of x. Rewriting a separable differential equation in this form is called separation of variables. In Section 2.1 we used separation of variables to solve homogeneous linear equations. In this section we’ll apply this method to nonlinear equations. To see how to solve (2.2.1), let’s first assume that y is a solution. Let G(x) and H(y) be antiderivatives of g(x) and h(y); that is, H′(y) = h(y) and G′(x) = g(x). (2.2.2) Then, from the chain rule, d dxH(y(x)) = H′(y(x))y′(x) = h(y)y′(x). Therefore (2.2.1) is equivalent to d dxH(y(x)) = d dxG(x). Integrating both sides of this equation and combining the constants of integration yields H(y(x)) = G(x) + c. (2.2.3) Although we derived this equation on the assumption that y is a solution of (2.2.1), we can now view it differently: Any differentiable function y that satisfies (2.2.3) for some constant c is a solution of (2.2.1). To see this, we differentiate both sides of (2.2.3), using the chain rule on the left, to obtain H′(y(x))y′(x) = G′(x), which is equivalent to h(y(x))y′(x) = g(x) because of (2.2.2). In conclusion, to solve (2.2.1) it suffices to find functions G = G(x) and H = H(y) that satisfy (2.2.2). Then any differentiable function y = y(x) that satisfies (2.2.3) is a solution of (2.2.1). Example 2.2.1 Solve the equation y′ = x(1 + y2). Solution Separating variables yields y′ 1 + y2 = x. Integrating yields tan−1 y = x2 2 + c Therefore y = tan x2 2 + c . | Elementary Differential Equations with Boundary Value Problems_Page_55_Chunk2461 |
46 Chapter 2 First Order Equations Example 2.2.2 (a) Solve the equation y′ = −x y . (2.2.4) (b) Solve the initial value problem y′ = −x y , y(1) = 1. (2.2.5) (c) Solve the initial value problem y′ = −x y , y(1) = −2. (2.2.6) SOLUTION(a) Separating variables in (2.2.4) yields yy′ = −x. Integrating yields y2 2 = −x2 2 + c, or, equivalently, x2 + y2 = 2c. The last equation shows that c must be positive if y is to be a solution of (2.2.4) on an open interval. Therefore we let 2c = a2 (with a > 0) and rewrite the last equation as x2 + y2 = a2. (2.2.7) This equation has two differentiable solutions for y in terms of x: y = p a2 −x2, −a < x < a, (2.2.8) and y = − p a2 −x2, −a < x < a. (2.2.9) The solution curves defined by (2.2.8) are semicircles above the x-axis and those defined by (2.2.9) are semicircles below the x-axis (Figure 2.2.1). SOLUTION(b) The solutionof (2.2.5) is positive when x = 1; hence, it is of the form (2.2.8). Substituting x = 1 and y = 1 into (2.2.7) to satisfy the initial condition yields a2 = 2; hence, the solution of (2.2.5) is y = p 2 −x2, − √ 2 < x < √ 2. SOLUTION(c) The solution of (2.2.6) is negative when x = 1 and is therefore of the form (2.2.9). Substituting x = 1 and y = −2 into (2.2.7) to satisfy the initial condition yields a2 = 5. Hence, the solution of (2.2.6) is y = − p 5 −x2, − √ 5 < x < √ 5. | Elementary Differential Equations with Boundary Value Problems_Page_56_Chunk2462 |
Section 2.2 Separable Equations 47 x y 1 2 −1 −2 1 2 −1 −2 (a) (b) Figure 2.2.1 (a) y = √ 2 −x2, − √ 2 < x < √ 2; (b) y = − √ 5 −x2, − √ 5 < x < √ 5 Implicit Solutions of Separable Equations In Examples 2.2.1 and 2.2.2 we were able to solve the equation H(y) = G(x) + c to obtain explicit formulas for solutions of the given separable differential equations. As we’ll see in the next example, this isn’t always possible. In this situation we must broaden our definition of a solution of a separable equation. The next theorem provides the basis for this modification. We omit the proof, which requires a result from advanced calculus called as the implicit function theorem. Theorem 2.2.1 Suppose g = g(x) is continous on (a, b) and h = h(y) are continuous on (c, d). Let G be an antiderivative of g on (a, b) and let H be an antiderivative of h on (c, d). Let x0 be an arbitrary point in (a, b), let y0 be a point in (c, d) such that h(y0) ̸= 0, and define c = H(y0) −G(x0). (2.2.10) Then there’s a function y = y(x) defined on some open interval (a1, b1), where a ≤a1 < x0 < b1 ≤b, such that y(x0) = y0 and H(y) = G(x) + c (2.2.11) for a1 < x < b1. Therefore y is a solution of the initial value problem h(y)y′ = g(x), y(x0) = x0. (2.2.12) It’s convenient to say that (2.2.11) with c arbitrary is an implicit solution of h(y)y′ = g(x). Curves defined by (2.2.11) are integral curves of h(y)y′ = g(x). If c satisfies (2.2.10), we’ll say that (2.2.11) is an implicit solution of the initial value problem (2.2.12). However, keep these points in mind: • For some choices of c there may not be any differentiable functions y that satisfy (2.2.11). | Elementary Differential Equations with Boundary Value Problems_Page_57_Chunk2463 |
48 Chapter 2 First Order Equations • The function y in (2.2.11) (not (2.2.11) itself) is a solution of h(y)y′ = g(x). Example 2.2.3 (a) Find implicit solutions of y′ = 2x + 1 5y4 + 1. (2.2.13) (b) Find an implicit solution of y′ = 2x + 1 5y4 + 1, y(2) = 1. (2.2.14) SOLUTION(a) Separating variables yields (5y4 + 1)y′ = 2x + 1. Integrating yields the implicit solution y5 + y = x2 + x + c. (2.2.15) of (2.2.13). SOLUTION(b) Imposing the initial condition y(2) = 1 in (2.2.15) yields 1 + 1 = 4 + 2 + c, so c = −4. Therefore y5 + y = x2 + x −4 is an implicit solution of the initial value problem (2.2.14). Although more than one differentiable func- tion y = y(x) satisfies 2.2.13) near x = 1, it can be shown that there’s only one such function that satisfies the initial condition y(1) = 2. Figure 2.2.2 shows a direction field and some integral curves for (2.2.13). Constant Solutions of Separable Equations An equation of the form y′ = g(x)p(y) is separable, since it can be rewritten as 1 p(y)y′ = g(x). However, the division by p(y) is not legitimate if p(y) = 0 for some values of y. The next two examples show how to deal with this problem. Example 2.2.4 Find all solutions of y′ = 2xy2. (2.2.16) Solution Here we must divide by p(y) = y2 to separate variables. This isn’t legitimate if y is a solution of (2.2.16) that equals zero for some value of x. One such solution can be found by inspection: y ≡0. Now suppose y is a solution of (2.2.16) that isn’t identically zero. Since y is continuous there must be an interval on which y is never zero. Since division by y2 is legitimate for x in this interval, we can separate variables in (2.2.16) to obtain y′ y2 = 2x. | Elementary Differential Equations with Boundary Value Problems_Page_58_Chunk2464 |
Section 2.2 Separable Equations 49 1 1.5 2 2.5 3 3.5 4 −1 −0.5 0 0.5 1 1.5 2 x y Figure 2.2.2 A direction field and integral curves for y′ = 2x + 1 5y4 + 1 Integrating this yields −1 y = x2 + c, which is equivalent to y = − 1 x2 + c. (2.2.17) We’ve now shown that if y is a solution of (2.2.16) that is not identically zero, then y must be of the form (2.2.17). By substituting (2.2.17) into (2.2.16), you can verify that (2.2.17) is a solution of (2.2.16). Thus, solutions of (2.2.16) are y ≡0 and the functions of the form (2.2.17). Note that the solution y ≡0 isn’t of the form (2.2.17) for any value of c. Figure 2.2.3 shows a direction field and some integral curves for (2.2.16) Example 2.2.5 Find all solutions of y′ = 1 2x(1 −y2). (2.2.18) Solution Here we must divide by p(y) = 1 −y2 to separate variables. This isn’t legitimate if y is a solution of (2.2.18) that equals ±1 for some value of x. Two such solutions can be found by inspection: y ≡1 and y ≡−1. Now suppose y is a solution of (2.2.18) such that 1 −y2 isn’t identically zero. Since 1 −y2 is continuous there must be an interval on which 1 −y2 is never zero. Since division by 1 −y2 is legitimate for x in this interval, we can separate variables in (2.2.18) to obtain 2y′ y2 −1 = −x. | Elementary Differential Equations with Boundary Value Problems_Page_59_Chunk2465 |
50 Chapter 2 First Order Equations −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y x Figure 2.2.3 A direction field and integral curves for y′ = 2xy2 A partial fraction expansion on the left yields 1 y −1 − 1 y + 1 y′ = −x, and integrating yields ln y −1 y + 1 = −x2 2 + k; hence, y −1 y + 1 = eke−x2/2. Since y(x) ̸= ±1 for x on the interval under discussion, the quantity (y −1)/(y + 1) can’t change sign in this interval. Therefore we can rewrite the last equation as y −1 y + 1 = ce−x2/2, where c = ±ek, depending upon the sign of (y −1)/(y + 1) on the interval. Solving for y yields y = 1 + ce−x2/2 1 −ce−x2/2 . (2.2.19) We’ve now shown that if y is a solution of (2.2.18) that is not identically equal to ±1, then y must be as in (2.2.19). By substituting (2.2.19) into (2.2.18) you can verify that (2.2.19) is a solution of (2.2.18). Thus, the solutions of (2.2.18) are y ≡1, y ≡−1 and the functions of the form (2.2.19). Note that the | Elementary Differential Equations with Boundary Value Problems_Page_60_Chunk2466 |
Section 2.2 Separable Equations 51 constant solution y ≡1 can be obtained from this formula by taking c = 0; however, the other constant solution, y ≡−1, can’t be obtained in this way. Figure 2.2.4 shows a direction field and some integrals for (2.2.18). −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −3 −2 −1 0 1 2 3 x y Figure 2.2.4 A direction field and integral curves for y′ = x(1 −y2) 2 Differences Between Linear and Nonlinear Equations Theorem 2.1.2 states that if p and f are continuous on (a, b) then every solution of y′ + p(x)y = f(x) on (a, b) can be obtained by choosing a value for the constant c in the general solution, and if x0 is any point in (a, b) and y0 is arbitrary, then the initial value problem y′ + p(x)y = f(x), y(x0) = y0 has a solution on (a, b). The not true for nonlinear equations. First, we saw in Examples 2.2.4 and 2.2.5 that a nonlinear equation may have solutions that can’t be obtained by choosing a specific value of a constant appearing in a one-parameter family of solutions. Second, it is in general impossible to determine the interval of validity of a solution to an initial value problem for a nonlinear equation by simply examining the equation, since the interval of validity may depend on the initial condition. For instance, in Example 2.2.2 we saw that the solution of dy dx = −x y , y(x0) = y0 is valid on (−a, a), where a = p x2 0 + y2 0. | Elementary Differential Equations with Boundary Value Problems_Page_61_Chunk2467 |
52 Chapter 2 First Order Equations Example 2.2.6 Solve the initial value problem y′ = 2xy2, y(0) = y0 and determine the interval of validity of the solution. Solution First suppose y0 ̸= 0. From Example 2.2.4, we know that y must be of the form y = − 1 x2 + c. (2.2.20) Imposing the initial condition shows that c = −1/y0. Substituting this into (2.2.20) and rearranging terms yields the solution y = y0 1 −y0x2 . This is also the solution if y0 = 0. If y0 < 0, the denominator isn’t zero for any value of x, so the the solution is valid on (−∞, ∞). If y0 > 0, the solution is valid only on (−1/√y0, 1/√y0). 2.2 Exercises In Exercises 1–6 find all solutions. 1. y′ = 3x2 + 2x + 1 y −2 2. (sin x)(sin y) + (cos y)y′ = 0 3. xy′ + y2 + y = 0 4. y′ ln |y| + x2y = 0 5. (3y3 + 3y cos y + 1)y′ + (2x + 1)y 1 + x2 = 0 6. x2yy′ = (y2 −1)3/2 In Exercises 7–10 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region. 7. C/G y′ = x2(1 + y2); {−1 ≤x ≤1, −1 ≤y ≤1} 8. C/G y′(1 + x2) + xy = 0; {−2 ≤x ≤2, −1 ≤y ≤1} 9. C/G y′ = (x −1)(y −1)(y −2); {−2 ≤x ≤2, −3 ≤y ≤3} 10. C/G (y −1)2y′ = 2x + 3; {−2 ≤x ≤2, −2 ≤y ≤5} In Exercises 11 and 12 solve the initial value problem. 11. y′ = x2 + 3x + 2 y −2 , y(1) = 4 12. y′ + x(y2 + y) = 0, y(2) = 1 In Exercises 13-16 solve the initial value problem and graph the solution. 13. C/G (3y2 + 4y)y′ + 2x + cos x = 0, y(0) = 1 | Elementary Differential Equations with Boundary Value Problems_Page_62_Chunk2468 |
Section 2.2 Separable Equations 53 14. C/G y′ + (y + 1)(y −1)(y −2) x + 1 = 0, y(1) = 0 15. C/G y′ + 2x(y + 1) = 0, y(0) = 2 16. C/G y′ = 2xy(1 + y2), y(0) = 1 In Exercises 17–23 solve the initial value problem and find the interval of validity of the solution. 17. y′(x2 + 2) + 4x(y2 + 2y + 1) = 0, y(1) = −1 18. y′ = −2x(y2 −3y + 2), y(0) = 3 19. y′ = 2x 1 + 2y, y(2) = 0 20. y′ = 2y −y2, y(0) = 1 21. x + yy′ = 0, y(3) = −4 22. y′ + x2(y + 1)(y −2)2 = 0, y(4) = 2 23. (x + 1)(x −2)y′ + y = 0, y(1) = −3 24. Solve y′ = (1 + y2) (1 + x2) explicitly. HINT: Use the identity tan(A + B) = tan A + tan B 1 −tan A tan B . 25. Solve y′p 1 −x2 + p 1 −y2 = 0 explicitly. HINT: Use the identity sin(A−B) = sin A cos B − cos A sin B. 26. Solve y′ = cos x sin y , y(π) = π 2 explicitly. HINT: Use the identity cos(x + π/2) = −sin x and the periodicity of the cosine. 27. Solve the initial value problem y′ = ay −by2, y(0) = y0. Discuss the behavior of the solution if (a) y0 ≥0; (b) y0 < 0. 28. The population P = P (t) of a species satisfies the logistic equation P ′ = aP (1 −αP ) and P (0) = P0 > 0. Find P for t > 0, and find limt→∞P (t). 29. An epidemic spreads through a population at a rate proportional to the product of the number of people already infected and the number of people susceptible, but not yet infected. Therefore, if S denotes the total population of susceptible people and I = I(t) denotes the number of infected people at time t, then I′ = rI(S −I), where r is a positive constant. Assuming that I(0) = I0, find I(t) for t > 0, and show that limt→∞I(t) = S. 30. L The result of Exercise 29 is discouraging: if any susceptible member of the group is initially infected, then in the long run all susceptible members are infected! On a more hopeful note, suppose the disease spreads according to the model of Exercise 29, but there’s a medication that cures the infected population at a rate proportional to the number of infected individuals. Now the equation for the number of infected individuals becomes I′ = rI(S −I) −qI (A) where q is a positive constant. | Elementary Differential Equations with Boundary Value Problems_Page_63_Chunk2469 |
54 Chapter 2 First Order Equations (a) Choose r and S positive. By plotting direction fields and solutions of (A) on suitable rectan- gular grids R = {0 ≤t ≤T, 0 ≤I ≤d} in the (t, I)-plane, verify that if I is any solution of (A) such that I(0) > 0, then limt→∞I(t) = S −q/r if q < rS and limt→∞I(t) = 0 if q ≥rS. (b) To verify the experimental results of (a), use separation of variables to solve (A) with initial condition I(0) = I0 > 0, and find limt→∞I(t). HINT: There are three cases to consider: (i) q < rS; (ii) q > rS; (iii) q = rS. 31. L Consider the differential equation y′ = ay −by2 −q, (A) where a, b are positive constants, and q is an arbitrary constant. Suppose y denotes a solution of this equation that satisfies the initial condition y(0) = y0. (a) Choose a and b positive and q < a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids R = {0 ≤t ≤T, c ≤y ≤d} (B) in the (t, y)-plane, discover that there are numbers y1 and y2 with y1 < y2 such that if y0 > y1 then limt→∞y(t) = y2, and if y0 < y1 then y(t) = −∞for some finite value of t. (What happens if y0 = y1?) (b) Choose a and b positive and q = a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that there’s a number y1 such that if y0 ≥y1 then limt→∞y(t) = y1, while if y0 < y1 then y(t) = −∞for some finite value of t. (c) Choose positive a, b and q > a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that no matter what y0 is, y(t) = −∞for some finite value of t. (d) Verify your results experiments analytically. Start by separating variables in (A) to obtain y′ ay −by2 −q = 1. To decide what to do next you’ll have to use the quadratic formula. This should lead you to see why there are three cases. Take it from there! Because of its role in the transition between these three cases, q0 = a2/4b is called a bifur- cation value of q. In general, if q is a parameter in any differential equation, q0 is said to be a bifurcation value of q if the nature of the solutions of the equation with q < q0 is qualitatively different from the nature of the solutions with q > q0. 32. L By plotting direction fields and solutions of y′ = qy −y3, convince yourself that q0 = 0 is a bifurcation value of q for this equation. Explain what makes you draw this conclusion. 33. Suppose a disease spreads according to the model of Exercise 29, but there’s a medication that cures the infected population at a constant rate of q individuals per unit time, where q > 0. Then the equation for the number of infected individuals becomes I′ = rI(S −I) −q. Assuming that I(0) = I0 > 0, use the results of Exercise 31 to describe what happens as t →∞. | Elementary Differential Equations with Boundary Value Problems_Page_64_Chunk2470 |
Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 55 34. Assuming that p ̸≡0, state conditions under which the linear equation y′ + p(x)y = f(x) is separable. If the equation satisfies these conditions, solve it by separation of variables and by the method developed in Section 2.1. Solve the equations in Exercises 35–38 using variation of parameters followed by separation of variables. 35. y′ + y = 2xe−x 1 + yex 36. xy′ −2y = x6 y + x2 37. y′ −y = (x + 1)e4x (y + ex)2 38. y′ −2y = xe2x 1 −ye−2x 39. Use variation of parameters to show that the solutions of the following equations are of the form y = uy1, where u satisfies a separable equation u′ = g(x)p(u). Find y1 and g for each equation. (a) xy′ + y = h(x)p(xy) (b) xy′ −y = h(x)p y x (c) y′ + y = h(x)p(exy) (d) xy′ + ry = h(x)p(xry) (e) y′ + v′(x) v(x) y = h(x)p (v(x)y) 2.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUATIONS Although there are methods for solving some nonlinear equations, it’s impossible to find useful formulas for the solutions of most. Whether we’re looking for exact solutions or numerical approximations, it’s useful to know conditions that imply the existence and uniqueness of solutions of initial value problems for nonlinear equations. In this section we state such a condition and illustrate it with examples. y x a b c d Figure 2.3.1 An open rectangle | Elementary Differential Equations with Boundary Value Problems_Page_65_Chunk2471 |
56 Chapter 2 First Order Equations Some terminology: an open rectangle R is a set of points (x, y) such that a < x < b and c < y < d (Figure 2.3.1). We’ll denote this set by R : {a < x < b, c < y < d}. “Open” means that the boundary rectangle (indicated by the dashed lines in Figure 2.3.1) isn’t included in R . The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for first order nonlinear differential equations. We omit the proof, which is beyond the scope of this book. Theorem 2.3.1 (a) If f is continuous on an open rectangle R : {a < x < b, c < y < d} that contains (x0, y0) then the initial value problem y′ = f(x, y), y(x0) = y0 (2.3.1) has at least one solution on some open subinterval of (a, b) that contains x0. (b) If both f and fy are continuous on R then (2.3.1) has a unique solution on some open subinterval of (a, b) that contains x0. It’s important to understand exactly what Theorem 2.3.1 says. • (a) is an existence theorem. It guarantees that a solution exists on some open interval that contains x0, but provides no information on how to find the solution, or to determine the open interval on which it exists. Moreover, (a) provides no information on the number of solutions that (2.3.1) may have. It leaves open the possibility that (2.3.1) may have two or more solutionsthat differ for values of x arbitrarily close to x0. We will see in Example 2.3.6 that this can happen. • (b) is a uniqueness theorem. It guarantees that (2.3.1) has a unique solution on some open interval (a,b) that contains x0. However, if (a, b) ̸= (−∞, ∞), (2.3.1) may have more than one solution on a larger interval that contains (a, b). For example, it may happen that b < ∞and all solutions have the same values on (a, b), but two solutions y1 and y2 are defined on some interval (a, b1) with b1 > b, and have different values for b < x < b1; thus, the graphs of the y1 and y2 “branch off” in different directions at x = b. (See Example 2.3.7 and Figure 2.3.3). In this case, continuity implies that y1(b) = y2(b) (call their common value y), and y1 and y2 are both solutions of the initial value problem y′ = f(x, y), y(b) = y (2.3.2) that differ on every open interval that contains b. Therefore f or fy must have a discontinuity at some point in each open rectangle that contains (b, y), since if this were not so, (2.3.2) would have a unique solution on some open interval that contains b. We leave it to you to give a similar analysis of the case where a > −∞. Example 2.3.1 Consider the initial value problem y′ = x2 −y2 1 + x2 + y2 , y(x0) = y0. (2.3.3) Since f(x, y) = x2 −y2 1 + x2 + y2 and fy(x, y) = −2y(1 + 2x2) (1 + x2 + y2)2 | Elementary Differential Equations with Boundary Value Problems_Page_66_Chunk2472 |
Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 57 are continuous for all (x, y), Theorem 2.3.1 implies that if (x0, y0) is arbitrary, then (2.3.3) has a unique solution on some open interval that contains x0. Example 2.3.2 Consider the initial value problem y′ = x2 −y2 x2 + y2 , y(x0) = y0. (2.3.4) Here f(x, y) = x2 −y2 x2 + y2 and fy(x, y) = − 4x2y (x2 + y2)2 are continuous everywhere except at (0, 0). If (x0, y0) ̸= (0, 0), there’s an open rectangle R that contains (x0, y0) that does not contain (0, 0). Since f and fy are continuous on R, Theorem 2.3.1 implies that if (x0, y0) ̸= (0, 0) then (2.3.4) has a unique solution on some open interval that contains x0. Example 2.3.3 Consider the initial value problem y′ = x + y x −y , y(x0) = y0. (2.3.5) Here f(x, y) = x + y x −y and fy(x, y) = 2x (x −y)2 are continuous everywhere except on the line y = x. If y0 ̸= x0, there’s an open rectangle R that contains (x0, y0) that does not intersect the line y = x. Since f and fy are continuous on R, Theorem 2.3.1 implies that if y0 ̸= x0, (2.3.5) has a unique solution on some open interval that contains x0. Example 2.3.4 In Example 2.2.4 we saw that the solutions of y′ = 2xy2 (2.3.6) are y ≡0 and y = − 1 x2 + c, where c is an arbitrary constant. In particular, this implies that no solution of (2.3.6) other than y ≡0 can equal zero for any value of x. Show that Theorem 2.3.1(b) implies this. Solution We’ll obtain a contradiction by assuming that (2.3.6) has a solution y1 that equals zero for some value of x, but isn’t identically zero. If y1 has this property, there’s a point x0 such that y1(x0) = 0, but y1(x) ̸= 0 for some value of x in every open interval that contains x0. This means that the initial value problem y′ = 2xy2, y(x0) = 0 (2.3.7) has two solutions y ≡0 and y = y1 that differ for some value of x on every open interval that contains x0. This contradicts Theorem 2.3.1(b), since in (2.3.6) the functions f(x, y) = 2xy2 and fy(x, y) = 4xy. are both continuous for all (x, y), which implies that (2.3.7) has a unique solution on some open interval that contains x0. | Elementary Differential Equations with Boundary Value Problems_Page_67_Chunk2473 |
58 Chapter 2 First Order Equations Example 2.3.5 Consider the initial value problem y′ = 10 3 xy2/5, y(x0) = y0. (2.3.8) (a) For what points (x0, y0) does Theorem 2.3.1(a) imply that (2.3.8) has a solution? (b) For what points (x0, y0) does Theorem 2.3.1(b) imply that (2.3.8) has a unique solution on some open interval that contains x0? SOLUTION(a) Since f(x, y) = 10 3 xy2/5 is continuous for all (x, y), Theorem 2.3.1 implies that (2.3.8) has a solution for every (x0, y0). SOLUTION(b) Here fy(x, y) = 4 3xy−3/5 is continuous for all (x, y) with y ̸= 0. Therefore, if y0 ̸= 0 there’s an open rectangle on which both f and fy are continuous, and Theorem 2.3.1 implies that (2.3.8) has a unique solution on some open interval that contains x0. If y = 0 then fy(x, y) is undefined, and therefore discontinuous; hence, Theorem 2.3.1 does not apply to (2.3.8) if y0 = 0. Example 2.3.6 Example 2.3.5 leaves open the possibility that the initial value problem y′ = 10 3 xy2/5, y(0) = 0 (2.3.9) has more than one solution on every open interval that contains x0 = 0. Show that this is true. Solution By inspection, y ≡0 is a solution of the differential equation y′ = 10 3 xy2/5. (2.3.10) Since y ≡0 satisfies the initial condition y(0) = 0, it’s a solution of (2.3.9). Now suppose y is a solution of (2.3.10) that isn’t identically zero. Separating variables in (2.3.10) yields y−2/5y′ = 10 3 x on any open interval where y has no zeros. Integrating this and rewriting the arbitrary constant as 5c/3 yields 5 3y3/5 = 5 3(x2 + c). Therefore y = (x2 + c)5/3. (2.3.11) Since we divided by y to separate variables in (2.3.10), our derivation of (2.3.11) is legitimate only on open intervals where y has no zeros. However, (2.3.11) actually defines y for all x, and differentiating (2.3.11) shows that y′ = 10 3 x(x2 + c)2/3 = 10 3 xy2/5, −∞< x < ∞. | Elementary Differential Equations with Boundary Value Problems_Page_68_Chunk2474 |
Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 59 x y Figure 2.3.2 Two solutions (y = 0 and y = x1/2) of (2.3.9) that differ on every interval containing x0 = 0 Therefore (2.3.11) satisfies (2.3.10) on (−∞, ∞) even if c ≤0, so that y( p |c|) = y(− p |c|) = 0. In particular, taking c = 0 in (2.3.11) yields y = x10/3 as a second solution of (2.3.9). Both solutions are defined on (−∞, ∞), and they differ on every open interval that contains x0 = 0 (see Figure 2.3.2.) In fact, there are four distinct solutions of (2.3.9) defined on (−∞, ∞) that differ from each other on every open interval that contains x0 = 0. Can you identify the other two? Example 2.3.7 From Example 2.3.5, the initial value problem y′ = 10 3 xy2/5, y(0) = −1 (2.3.12) has a unique solution on some open interval that contains x0 = 0. Find a solution and determine the largest open interval (a, b) on which it’s unique. Solution Let y be any solution of (2.3.12). Because of the initial condition y(0) = −1 and the continuity of y, there’s an open interval I that contains x0 = 0 on which y has no zeros, and is consequently of the form (2.3.11). Setting x = 0 and y = −1 in (2.3.11) yields c = −1, so y = (x2 −1)5/3 (2.3.13) for x in I. Therefore every solution of (2.3.12) differs from zero and is given by (2.3.13) on (−1, 1); that is, (2.3.13) is the unique solution of (2.3.12) on (−1, 1). This is the largest open interval on which (2.3.12) has a unique solution. To see this, note that (2.3.13) is a solution of (2.3.12) on (−∞, ∞). From | Elementary Differential Equations with Boundary Value Problems_Page_69_Chunk2475 |
60 Chapter 2 First Order Equations Exercise 2.2.15, there are infinitely many other solutions of (2.3.12) that differ from (2.3.13) on every open interval larger than (−1, 1). One such solution is y = ( (x2 −1)5/3, −1 ≤x ≤1, 0, |x| > 1. (Figure 2.3.3). 1 −1 x y (0, −1) Figure 2.3.3 Two solutions of (2.3.12) on (−∞, ∞) that coincide on (−1, 1), but on no larger open interval x y (0,1) Figure 2.3.4 The unique solution of (2.3.14) Example 2.3.8 From Example 2.3.5, the initial value problem y′ = 10 3 xy2/5, y(0) = 1 (2.3.14) has a unique solution on some open interval that contains x0 = 0. Find the solution and determine the largest open interval on which it’s unique. Solution Let y be any solution of (2.3.14). Because of the initial condition y(0) = 1 and the continuity of y, there’s an open interval I that contains x0 = 0 on which y has no zeros, and is consequently of the form (2.3.11). Setting x = 0 and y = 1 in (2.3.11) yields c = 1, so y = (x2 + 1)5/3 (2.3.15) for x in I. Therefore every solution of (2.3.14) differs from zero and is given by (2.3.15) on (−∞, ∞); that is, (2.3.15) is the unique solution of (2.3.14) on (−∞, ∞). Figure 2.3.4 shows the graph of this solution. 2.3 Exercises In Exercises 1-13 find all (x0, y0) for which Theorem 2.3.1 implies that the initial value problem y′ = f(x, y), y(x0) = y0 has (a) a solution (b) a unique solution on some open interval that contains x0. | Elementary Differential Equations with Boundary Value Problems_Page_70_Chunk2476 |
Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 61 1. y′ = x2 + y2 sin x 2. y′ = ex + y x2 + y2 3. y′ = tan xy 4. y′ = x2 + y2 ln xy 5. y′ = (x2 + y2)y1/3 6. y′ = 2xy 7. y′ = ln(1 + x2 + y2) 8. y′ = 2x + 3y x −4y 9. y′ = (x2 + y2)1/2 10. y′ = x(y2 −1)2/3 11. y′ = (x2 + y2)2 12. y′ = (x + y)1/2 13. y′ = tan y x −1 14. Apply Theorem 2.3.1 to the initial value problem y′ + p(x)y = q(x), y(x0) = y0 for a linear equation, and compare the conclusions that can be drawn from it to those that follow from Theorem 2.1.2. 15. (a) Verify that the function y = ( (x2 −1)5/3, −1 < x < 1, 0, |x| ≥1, is a solution of the initial value problem y′ = 10 3 xy2/5, y(0) = −1 on (−∞, ∞). HINT: You’ll need the definition y′(x) = lim x→x y(x) −y(x) x −x to verify that y satisfies the differential equation at x = ±1. (b) Verify that if ϵi = 0 or 1 for i = 1, 2 and a, b > 1, then the function y = ϵ1(x2 −a2)5/3, −∞< x < −a, 0, −a ≤x ≤−1, (x2 −1)5/3, −1 < x < 1, 0, 1 ≤x ≤b, ϵ2(x2 −b2)5/3, b < x < ∞, is a solution of the initial value problem of (a) on (−∞, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_71_Chunk2477 |
62 Chapter 2 First Order Equations 16. Use the ideas developed in Exercise 15 to find infinitely many solutionsof the initial value problem y′ = y2/5, y(0) = 1 on (−∞, ∞). 17. Consider the initial value problem y′ = 3x(y −1)1/3, y(x0) = y0. (A) (a) For what points (x0, y0) does Theorem 2.3.1 imply that (A) has a solution? (b) For what points (x0, y0) does Theorem 2.3.1 imply that (A) has a unique solution on some open interval that contains x0? 18. Find nine solutions of the initial value problem y′ = 3x(y −1)1/3, y(0) = 1 that are all defined on (−∞, ∞) and differ from each other for values of x in every open interval that contains x0 = 0. 19. From Theorem 2.3.1, the initial value problem y′ = 3x(y −1)1/3, y(0) = 9 has a unique solution on an open interval that contains x0 = 0. Find the solution and determine the largest open interval on which it’s unique. 20. (a) From Theorem 2.3.1, the initial value problem y′ = 3x(y −1)1/3, y(3) = −7 (A) has a unique solution on some open interval that contains x0 = 3. Determine the largest such open interval, and find the solution on this interval. (b) Find infinitely many solutions of (A), all defined on (−∞, ∞). 21. Prove: (a) If f(x, y0) = 0, a < x < b, (A) and x0 is in (a, b), then y ≡y0 is a solution of y′ = f(x, y), y(x0) = y0 on (a, b). (b) If f and fy are continuous on an open rectangle that contains (x0, y0) and (A) holds, no solution of y′ = f(x, y) other than y ≡y0 can equal y0 at any point in (a, b). 2.4 TRANSFORMATION OF NONLINEAR EQUATIONS INTO SEPARABLE EQUATIONS In Section 2.1 we found that the solutions of a linear nonhomogeneous equation y′ + p(x)y = f(x) | Elementary Differential Equations with Boundary Value Problems_Page_72_Chunk2478 |
Section 2.4 Transformation of Nonlinear Equations into Separable Equations 63 are of the form y = uy1, where y1 is a nontrivial solution of the complementary equation y′ + p(x)y = 0 (2.4.1) and u is a solution of u′y1(x) = f(x). Note that this last equation is separable, since it can be rewritten as u′ = f(x) y1(x). In this section we’ll consider nonlinear differential equations that are not separable to begin with, but can be solved in a similar fashion by writing their solutions in the form y = uy1, where y1 is a suitably chosen known function and u satisfies a separable equation. We’llsay in this case that we transformed the given equation into a separable equation. Bernoulli Equations A Bernoulli equation is an equation of the form y′ + p(x)y = f(x)yr, (2.4.2) where r can be any real number other than 0 or 1. (Note that (2.4.2) is linear if and only if r = 0 or r = 1.) We can transform (2.4.2) into a separable equation by variation of parameters: if y1 is a nontrivial solution of (2.4.1), substituting y = uy1 into (2.4.2) yields u′y1 + u(y′ 1 + p(x)y1) = f(x)(uy1)r, which is equivalent to the separable equation u′y1(x) = f(x) (y1(x))r ur or u′ ur = f(x) (y1(x))r−1 , since y′ 1 + p(x)y1 = 0. Example 2.4.1 Solve the Bernoulli equation y′ −y = xy2. (2.4.3) Solution Since y1 = ex is a solution of y′ −y = 0, we look for solutions of (2.4.3) in the form y = uex, where u′ex = xu2e2x or, equivalently, u′ = xu2ex. Separating variables yields u′ u2 = xex, and integrating yields −1 u = (x −1)ex + c. Hence, u = − 1 (x −1)ex + c | Elementary Differential Equations with Boundary Value Problems_Page_73_Chunk2479 |
64 Chapter 2 First Order Equations −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 x y Figure 2.4.1 A direction field and integral curves for y′ −y = xy2 and y = − 1 x −1 + ce−x . Figure 2.4.1 shows direction field and some integral curves of (2.4.3). Other Nonlinear Equations That Can be Transformed Into Separable Equations We’ve seen that the nonlinear Bernoulli equation can be transformed into a separable equation by the substitution y = uy1 if y1 is suitably chosen. Now let’s discover a sufficient condition for a nonlinear first order differential equation y′ = f(x, y) (2.4.4) to be transformable into a separable equation in the same way. Substituting y = uy1 into (2.4.4) yields u′y1(x) + uy′ 1(x) = f(x, uy1(x)), which is equivalent to u′y1(x) = f(x, uy1(x)) −uy′ 1(x). (2.4.5) If f(x, uy1(x)) = q(u)y′ 1(x) for some function q, then (2.4.5) becomes u′y1(x) = (q(u) −u)y′ 1(x), (2.4.6) which is separable. After checking for constant solutions u ≡u0 such that q(u0) = u0, we can separate variables to obtain u′ q(u) −u = y′ 1(x) y1(x). | Elementary Differential Equations with Boundary Value Problems_Page_74_Chunk2480 |
Section 2.4 Transformation of Nonlinear Equations into Separable Equations 65 Homogeneous Nonlinear Equations In the text we’ll consider only the most widely studied class of equations for which the method of the preceding paragraph works. Other types of equations appear in Exercises 44–51. The differential equation (2.4.4) is said to be homogeneous if x and y occur in f in such a way that f(x, y) depends only on the ratio y/x; that is, (2.4.4) can be written as y′ = q(y/x), (2.4.7) where q = q(u) is a function of a single variable. For example, y′ = y + xe−y/x x = y x + e−y/x and y′ = y2 + xy −x2 x2 = y x 2 + y x −1 are of the form (2.4.7), with q(u) = u + e−u and q(u) = u2 + u −1, respectively. The general method discussed above can be applied to (2.4.7) with y1 = x (and therefore y′ 1 = 1). Thus, substituting y = ux in (2.4.7) yields u′x + u = q(u), and separation of variables (after checking for constant solutions u ≡u0 such that q(u0) = u0) yields u′ q(u) −u = 1 x. Before turning to examples, we point out something that you may’ve have already noticed: the defini- tion of homogeneous equation given here isn’t the same as the definition given in Section 2.1, where we said that a linear equation of the form y′ + p(x)y = 0 is homogeneous. We make no apology for this inconsistency, since we didn’t create it historically, homo- geneous has been used in these two inconsistent ways. The one having to do with linear equations is the most important. This is the only section of the book where the meaning defined here will apply. Since y/x is in general undefined if x = 0, we’ll consider solutions of nonhomogeneous equations only on open intervals that do not contain the point x = 0. Example 2.4.2 Solve y′ = y + xe−y/x x . (2.4.8) Solution Substituting y = ux into (2.4.8) yields u′x + u = ux + xe−ux/x x = u + e−u. Simplifying and separating variables yields euu′ = 1 x. | Elementary Differential Equations with Boundary Value Problems_Page_75_Chunk2481 |
66 Chapter 2 First Order Equations Integrating yields eu = ln |x| + c. Therefore u = ln(ln |x| + c) and y = ux = x ln(ln |x| + c). Figure 2.4.2 shows a direction field and integral curves for (2.4.8). 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 x y Figure 2.4.2 A direction field and some integral curves for y′ = y + xe−y/x x Example 2.4.3 (a) Solve x2y′ = y2 + xy −x2. (2.4.9) (b) Solve the initial value problem x2y′ = y2 + xy −x2, y(1) = 2. (2.4.10) SOLUTION(a) We first find solutions of (2.4.9) on open intervals that don’t contain x = 0. We can rewrite (2.4.9) as y′ = y2 + xy −x2 x2 for x in any such interval. Substituting y = ux yields u′x + u = (ux)2 + x(ux) −x2 x2 = u2 + u −1, so u′x = u2 −1. (2.4.11) | Elementary Differential Equations with Boundary Value Problems_Page_76_Chunk2482 |
Section 2.4 Transformation of Nonlinear Equations into Separable Equations 67 By inspection this equation has the constant solutions u ≡1 and u ≡−1. Therefore y = x and y = −x are solutions of (2.4.9). If u is a solution of (2.4.11) that doesn’t assume the values ±1 on some interval, separating variables yields u′ u2 −1 = 1 x, or, after a partial fraction expansion, 1 2 1 u −1 − 1 u + 1 u′ = 1 x. Multiplying by 2 and integrating yields ln u −1 u + 1 = 2 ln|x| + k, or u −1 u + 1 = ekx2, which holds if u −1 u + 1 = cx2 (2.4.12) where c is an arbitrary constant. Solving for u yields u = 1 + cx2 1 −cx2 . −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y x Figure 2.4.3 A direction field and integral curves for x2y′ = y2 + xy −x2 x y 1 2 Figure 2.4.4 Solutions of x2y′ = y2 + xy −x2, y(1) = 2 Therefore y = ux = x(1 + cx2) 1 −cx2 (2.4.13) is a solution of (2.4.10) for any choice of the constant c. Setting c = 0 in (2.4.13) yields the solution y = x. However, the solution y = −x can’t be obtained from (2.4.13). Thus, the solutions of (2.4.9) on intervals that don’t contain x = 0 are y = −x and functions of the form (2.4.13). | Elementary Differential Equations with Boundary Value Problems_Page_77_Chunk2483 |
68 Chapter 2 First Order Equations The situation is more complicated if x = 0 is the open interval. First, note that y = −x satisfies (2.4.9) on (−∞, ∞). If c1 and c2 are arbitrary constants, the function y = x(1 + c1x2) 1 −c1x2 , a < x < 0, x(1 + c2x2) 1 −c2x2 , 0 ≤x < b, (2.4.14) is a solution of (2.4.9) on (a, b), where a = −1 √c1 if c1 > 0, −∞ if c1 ≤0, and b = 1 √c2 if c2 > 0, ∞ if c2 ≤0. We leave it to you to verify this. To do so, note that if y is any function of the form (2.4.13) then y(0) = 0 and y′(0) = 1. Figure 2.4.3 shows a direction field and some integral curves for (2.4.9). SOLUTION(b) We could obtain c by imposing the initial condition y(1) = 2 in (2.4.13), and then solving for c. However, it’s easier to use (2.4.12). Since u = y/x, the initial condition y(1) = 2 implies that u(1) = 2. Substituting this into (2.4.12) yields c = 1/3. Hence, the solution of (2.4.10) is y = x(1 + x2/3) 1 −x2/3 . The interval of validity of this solution is (− √ 3, √ 3). However, the largest interval on which (2.4.10) has a unique solution is (0, √ 3). To see this, note from (2.4.14) that any function of the form y = x(1 + cx2) 1 −cx2 , a < x ≤0, x(1 + x2/3) 1 −x2/3 , 0 ≤x < √ 3, (2.4.15) is a solution of (2.4.10) on (a, √ 3), where a = −1/√c if c > 0 or a = −∞if c ≤0. (Why doesn’t this contradict Theorem 2.3.1?) Figure 2.4.4 shows several solutions of the initial value problem (2.4.10). Note that these solutions coincide on (0, √ 3). In the last two examples we were able to solve the given equations explicitly. However, this isn’t always possible, as you’ll see in the exercises. 2.4 Exercises In Exercises 1–4 solve the given Bernoulli equation. 1. y′ + y = y2 2. 7xy′ −2y = −x2 y6 3. x2y′ + 2y = 2e1/xy1/2 4. (1 + x2)y′ + 2xy = 1 (1 + x2)y In Exercises 5 and 6 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region. | Elementary Differential Equations with Boundary Value Problems_Page_78_Chunk2484 |
Section 2.4 Transformation of Nonlinear Equations into Separable Equations 69 5. C/G y′ −xy = x3y3; {−3 ≤x ≤3, 2 ≤y ≥2} 6. C/G y′ −1 + x 3x y = y4; {−2 ≤x ≤2, −2 ≤y ≤2} In Exercises 7–11 solve the initial value problem. 7. y′ −2y = xy3, y(0) = 2 √ 2 8. y′ −xy = xy3/2, y(1) = 4 9. xy′ + y = x4y4, y(1) = 1/2 10. y′ −2y = 2y1/2, y(0) = 1 11. y′ −4y = 48x y2 , y(0) = 1 In Exercises 12 and 13 solve the initial value problem and graph the solution. 12. C/G x2y′ + 2xy = y3, y(1) = 1/ √ 2 13. C/G y′ −y = xy1/2, y(0) = 4 14. You may have noticed that the logistic equation P ′ = aP (1 −αP ) from Verhulst’s model for population growth can be written in Bernoulli form as P ′ −aP = −aαP 2. This isn’t particularly interesting, since the logistic equation is separable, and therefore solvable by the method studied in Section 2.2. So let’s consider a more complicated model, where a is a positive constant and α is a positive continuous function of t on [0, ∞). The equation for this model is P ′ −aP = −aα(t)P 2, a non-separable Bernoulli equation. (a) Assuming that P (0) = P0 > 0, find P for t > 0. HINT: Express your result in terms of the integral R t 0 α(τ)eaτ dτ. (b) Verify that your result reduces to the known results for the Malthusian model where α = 0, and the Verhulst model where α is a nonzero constant. (c) Assuming that lim t→∞e−at Z t 0 α(τ)eaτ dτ = L exists (finite or infinite), find limt→∞P (t). In Exercises 15–18 solve the equation explicitly. 15. y′ = y + x x 16. y′ = y2 + 2xy x2 17. xy3y′ = y4 + x4 18. y′ = y x + sec y x | Elementary Differential Equations with Boundary Value Problems_Page_79_Chunk2485 |
70 Chapter 2 First Order Equations In Exercises 19-21 solve the equation explicitly. Also, plot a direction field and some integral curves on the indicated rectangular region. 19. C/G x2y′ = xy + x2 + y2; {−8 ≤x ≤8, −8 ≤y ≤8} 20. C/G xyy′ = x2 + 2y2; {−4 ≤x ≤4, −4 ≤y ≤4} 21. C/G y′ = 2y2 + x2e−(y/x)2 2xy ; {−8 ≤x ≤8, −8 ≤y ≤8} In Exercises 22–27 solve the initial value problem. 22. y′ = xy + y2 x2 , y(−1) = 2 23. y′ = x3 + y3 xy2 , y(1) = 3 24. xyy′ + x2 + y2 = 0, y(1) = 2 25. y′ = y2 −3xy −5x2 x2 , y(1) = −1 26. x2y′ = 2x2 + y2 + 4xy, y(1) = 1 27. xyy′ = 3x2 + 4y2, y(1) = √ 3 In Exercises 28–34 solve the given homogeneous equation implicitly. 28. y′ = x + y x −y 29. (y′x −y)(ln |y| −ln |x|) = x 30. y′ = y3 + 2xy2 + x2y + x3 x(y + x)2 31. y′ = x + 2y 2x + y 32. y′ = y y −2x 33. y′ = xy2 + 2y3 x3 + x2y + xy2 34. y′ = x3 + x2y + 3y3 x3 + 3xy2 35. L (a) Find a solution of the initial value problem x2y′ = y2 + xy −4x2, y(−1) = 0 (A) on the interval (−∞, 0). Verify that this solution is actually valid on (−∞, ∞). (b) Use Theorem 2.3.1 to show that (A) has a unique solution on (−∞, 0). (c) Plot a direction field for the differential equation in (A) on a square {−r ≤x ≤r, −r ≤y ≤r}, where r is any positive number. Graph the solution you obtained in (a) on this field. (d) Graph other solutions of (A) that are defined on (−∞, ∞). | Elementary Differential Equations with Boundary Value Problems_Page_80_Chunk2486 |
Section 2.4 Transformation of Nonlinear Equations into Separable Equations 71 (e) Graph other solutions of (A) that are defined only on intervals of the form (−∞, a), where is a finite positive number. 36. L (a) Solve the equation xyy′ = x2 −xy + y2 (A) implicitly. (b) Plot a direction field for (A) on a square {0 ≤x ≤r, 0 ≤y ≤r} where r is any positive number. (c) Let K be a positive integer. (You may have to try several choices for K.) Graph solutions of the initial value problems xyy′ = x2 −xy + y2, y(r/2) = kr K , for k = 1, 2, ..., K. Based on your observations, find conditions on the positive numbers x0 and y0 such that the initial value problem xyy′ = x2 −xy + y2, y(x0) = y0, (B) has a unique solution (i) on (0, ∞) or (ii) only on an interval (a, ∞), where a > 0? (d) What can you say about the graph of the solution of (B) as x →∞? (Again, assume that x0 > 0 and y0 > 0.) 37. L (a) Solve the equation y′ = 2y2 −xy + 2x2 xy + 2x2 (A) implicitly. (b) Plot a direction field for (A) on a square {−r ≤x ≤r, −r ≤y ≤r} where r is any positive number. By graphing solutions of (A), determine necessary and sufficient conditions on (x0, y0) such that (A) has a solution on (i) (−∞, 0) or (ii) (0, ∞) such that y(x0) = y0. 38. L Follow the instructions of Exercise 37 for the equation y′ = xy + x2 + y2 xy . 39. L Pick any nonlinear homogeneous equation y′ = q(y/x) you like, and plot direction fields on the square {−r ≤x ≤r, −r ≤y ≤r}, where r > 0. What happens to the direction field as you vary r? Why? | Elementary Differential Equations with Boundary Value Problems_Page_81_Chunk2487 |
72 Chapter 2 First Order Equations 40. Prove: If ad −bc ̸= 0, the equation y′ = ax + by + α cx + dy + β can be transformed into the homogeneous nonlinear equation dY dX = aX + bY cX + dY by the substitution x = X −X0, y = Y −Y0, where X0 and Y0 are suitably chosen constants. In Exercises 41-43 use a method suggested by Exercise 40 to solve the given equation implicitly. 41. y′ = −6x + y −3 2x −y −1 42. y′ = 2x + y + 1 x + 2y −4 43. y′ = −x + 3y −14 x + y −2 In Exercises 44–51 find a function y1 such that the substitution y = uy1 transforms the given equation into a separable equation of the form (2.4.6). Then solve the given equation explicitly. 44. 3xy2y′ = y3 + x 45. xyy′ = 3x6 + 6y2 46. x3y′ = 2(y2 + x2y −x4) 47. y′ = y2e−x + 4y + 2ex 48. y′ = y2 + y tan x + tan2 x sin2 x 49. x(lnx)2y′ = −4(lnx)2 + y ln x + y2 50. 2x(y + 2√x)y′ = (y + √x)2 51. (y + ex2)y′ = 2x(y2 + yex2 + e2x2) 52. Solve the initial value problem y′ + 2 xy = 3x2y2 + 6xy + 2 x2(2xy + 3) , y(2) = 2. 53. Solve the initial value problem y′ + 3 xy = 3x4y2 + 10x2y + 6 x3(2x2y + 5) , y(1) = 1. 54. Prove: If y is a solution of a homogeneous nonlinear equation y′ = q(y/x), so is y1 = y(ax)/a, where a is any nonzero constant. 55. A generalized Riccati equation is of the form y′ = P (x) + Q(x)y + R(x)y2. (A) (If R ≡−1, (A) is a Riccati equation.) Let y1 be a known solution and y an arbitrary solution of (A). Let z = y −y1. Show that z is a solution of a Bernoulli equation with n = 2. | Elementary Differential Equations with Boundary Value Problems_Page_82_Chunk2488 |
Section 2.5 Exact Equations 73 In Exercises 56–59, given that y1 is a solution of the given equation, use the method suggested by Exercise 55 to find other solutions. 56. y′ = 1 + x −(1 + 2x)y + xy2; y1 = 1 57. y′ = e2x + (1 −2ex)y + y2; y1 = ex 58. xy′ = 2 −x + (2x −2)y −xy2; y1 = 1 59. xy′ = x3 + (1 −2x2)y + xy2; y1 = x 2.5 EXACT EQUATIONS In this section it’s convenient to write first order differential equations in the form M(x, y) dx + N(x, y) dy = 0. (2.5.1) This equation can be interpreted as M(x, y) + N(x, y) dy dx = 0, (2.5.2) where x is the independent variable and y is the dependent variable, or as M(x, y) dx dy + N(x, y) = 0, (2.5.3) where y is the independent variable and x is the dependent variable. Since the solutions of (2.5.2) and (2.5.3) will often have to be left in implicit, form we’ll say that F (x, y) = c is an implicit solution of (2.5.1) if every differentiable function y = y(x) that satisfies F (x, y) = c is a solution of (2.5.2) and every differentiable function x = x(y) that satisfies F (x, y) = c is a solution of (2.5.3). Here are some examples: Equation (2.5.1) Equation (2.5.2) Equation (2.5.3) 3x2y2 dx + 2x3y dy = 0 3x2y2 + 2x3y dy dx = 0 3x2y2 dx dy + 2x3y = 0 (x2 + y2) dx + 2xy dy = 0 (x2 + y2) + 2xy dy dx = 0 (x2 + y2) dx dy + 2xy = 0 3y sin x dx −2xy cos x dy = 0 3y sin x −2xy cos x dy dx = 0 3y sin x dx dy −2xy cos x = 0 Note that a separable equation can be written as (2.5.1) as M(x) dx + N(y) dy = 0. We’ll develop a method for solving (2.5.1) under appropriate assumptions on M and N. This method is an extension of the method of separation of variables (Exercise 41). Before stating it we consider an example. | Elementary Differential Equations with Boundary Value Problems_Page_83_Chunk2489 |
74 Chapter 2 First Order Equations Example 2.5.1 Show that x4y3 + x2y5 + 2xy = c (2.5.4) is an implicit solution of (4x3y3 + 2xy5 + 2y) dx + (3x4y2 + 5x2y4 + 2x) dy = 0. (2.5.5) Solution Regarding y as a function of x and differentiating (2.5.4) implicitly with respect to x yields (4x3y3 + 2xy5 + 2y) + (3x4y2 + 5x2y4 + 2x) dy dx = 0. Similarly, regarding x as a function of y and differentiating (2.5.4) implicitly with respect to y yields (4x3y3 + 2xy5 + 2y)dx dy + (3x4y2 + 5x2y4 + 2x) = 0. Therefore (2.5.4) is an implicit solution of (2.5.5) in either of its two possible interpretations. You may think this example is pointless, since concocting a differential equation that has a given implicit solution isn’t particularly interesting. However, it illustrates the next important theorem, which we’ll prove by using implicit differentiation, as in Example 2.5.1. Theorem 2.5.1 If F = F (x, y) has continuous partial derivatives Fx and Fy, then F (x, y) = c (c=constant), (2.5.6) is an implicit solution of the differential equation Fx(x, y) dx + Fy(x, y) dy = 0. (2.5.7) Proof Regarding y as a function of x and differentiating (2.5.6) implicitly with respect to x yields Fx(x, y) + Fy(x, y) dy dx = 0. On the other hand, regarding x as a function of y and differentiating (2.5.6) implicitly with respect to y yields Fx(x, y) dx dy + Fy(x, y) = 0. Thus, (2.5.6) is an implicit solution of (2.5.7) in either of its two possible interpretations. We’ll say that the equation M(x, y) dx + N(x, y) dy = 0 (2.5.8) is exact on an an open rectangle R if there’s a function F = F (x, y) such Fx and Fy are continuous, and Fx(x, y) = M(x, y) and Fy(x, y) = N(x, y) (2.5.9) for all (x, y) in R. This usage of “exact” is related to its usage in calculus, where the expression Fx(x, y) dx + Fy(x, y) dy (obtained by substituting (2.5.9) into the left side of (2.5.8)) is the exact differential of F . Example 2.5.1 shows that it’s easy to solve (2.5.8) if it’s exact and we know a function F that satisfies (2.5.9). The important questions are: | Elementary Differential Equations with Boundary Value Problems_Page_84_Chunk2490 |
Section 2.5 Exact Equations 75 QUESTION 1. Given an equation (2.5.8), how can we determine whether it’s exact? QUESTION 2. If (2.5.8) is exact, how do we find a function F satisfying (2.5.9)? To discover the answer to Question 1, assume that there’s a function F that satisfies (2.5.9) on some open rectangle R, and in addition that F has continuous mixed partial derivatives Fxy and Fyx. Then a theorem from calculus implies that Fxy = Fyx. (2.5.10) If Fx = M and Fy = N, differentiating the first of these equations with respect to y and the second with respect to x yields Fxy = My and Fyx = Nx. (2.5.11) From (2.5.10) and (2.5.11), we conclude that a necessary condition for exactness is that My = Nx. This motivates the next theorem, which we state without proof. Theorem 2.5.2 [The Exactness Condition] Suppose M and N are continuous and have continuous par- tial derivatives My and Nx on an open rectangle R. Then M(x, y) dx + N(x, y) dy = 0 is exact on R if and only if My(x, y) = Nx(x, y) (2.5.12) for all (x, y) in R.. To help you remember the exactness condition, observe that the coefficients of dx and dy are differ- entiated in (2.5.12) with respect to the “opposite” variables; that is, the coefficient of dx is differentiated with respect to y, while the coefficient of dy is differentiated with respect to x. Example 2.5.2 Show that the equation 3x2y dx + 4x3 dy = 0 is not exact on any open rectangle. Solution Here M(x, y) = 3x2y and N(x, y) = 4x3 so My(x, y) = 3x2 and Nx(x, y) = 12x2. Therefore My = Nx on the line x = 0, but not on any open rectangle, so there’s no function F such that Fx(x, y) = M(x, y) and Fy(x, y) = N(x, y) for all (x, y) on any open rectangle. The next example illustrates two possible methods for finding a function F that satisfies the condition Fx = M and Fy = N if M dx + N dy = 0 is exact. Example 2.5.3 Solve (4x3y3 + 3x2) dx + (3x4y2 + 6y2) dy = 0. (2.5.13) Solution (Method 1) Here M(x, y) = 4x3y3 + 3x2, N(x, y) = 3x4y2 + 6y2, | Elementary Differential Equations with Boundary Value Problems_Page_85_Chunk2491 |
76 Chapter 2 First Order Equations and My(x, y) = Nx(x, y) = 12x3y2 for all (x, y). Therefore Theorem 2.5.2 implies that there’s a function F such that Fx(x, y) = M(x, y) = 4x3y3 + 3x2 (2.5.14) and Fy(x, y) = N(x, y) = 3x4y2 + 6y2 (2.5.15) for all (x, y). To find F , we integrate (2.5.14) with respect to x to obtain F (x, y) = x4y3 + x3 + φ(y), (2.5.16) where φ(y) is the “constant” of integration. (Here φ is “constant” in that it’s independent of x, the variable of integration.) If φ is any differentiable function of y then F satisfies (2.5.14). To determine φ so that F also satisfies (2.5.15), assume that φ is differentiable and differentiate F with respect to y. This yields Fy(x, y) = 3x4y2 + φ′(y). Comparing this with (2.5.15) shows that φ′(y) = 6y2. We integrate this with respect to y and take the constant of integration to be zero because we’re interested only in finding some F that satisfies (2.5.14) and (2.5.15). This yields φ(y) = 2y3. Substituting this into (2.5.16) yields F (x, y) = x4y3 + x3 + 2y3. (2.5.17) Now Theorem 2.5.1 implies that x4y3 + x3 + 2y3 = c is an implicit solution of (2.5.13). Solving this for y yields the explicit solution y = c −x3 2 + x4 1/3 . Solution (Method 2) Instead of first integrating (2.5.14) with respect to x, we could begin by integrating (2.5.15) with respect to y to obtain F (x, y) = x4y3 + 2y3 + ψ(x), (2.5.18) where ψ is an arbitrary function of x. To determine ψ, we assume that ψ is differentiable and differentiate F with respect to x, which yields Fx(x, y) = 4x3y3 + ψ′(x). Comparing this with (2.5.14) shows that ψ′(x) = 3x2. Integrating this and again taking the constant of integration to be zero yields ψ(x) = x3. | Elementary Differential Equations with Boundary Value Problems_Page_86_Chunk2492 |
Section 2.5 Exact Equations 77 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 y x Figure 2.5.1 A direction field and integral curves for (4x3y3 + 3x2) dx + (3x4y2 + 6y2) dy = 0 Substituting this into (2.5.18) yields (2.5.17). Figure 2.5.1 shows a direction field and some integral curves of (2.5.13), Here’s a summary of the procedure used in Method 1 of this example. You should summarize procedure used in Method 2. Procedure For Solving An Exact Equation Step 1. Check that the equation M(x, y) dx + N(x, y) dy = 0 (2.5.19) satisfies the exactness condition My = Nx. If not, don’t go further with this procedure. Step 2. Integrate ∂F (x, y) ∂x = M(x, y) with respect to x to obtain F (x, y) = G(x, y) + φ(y), (2.5.20) where G is an antiderivative of M with respect to x, and φ is an unknown function of y. Step 3. Differentiate (2.5.20) with respect to y to obtain ∂F (x, y) ∂y = ∂G(x, y) ∂y + φ′(y). Step 4. Equate the right side of this equation to N and solve for φ′; thus, ∂G(x, y) ∂y + φ′(y) = N(x, y), so φ′(y) = N(x, y) −∂G(x, y) ∂y . | Elementary Differential Equations with Boundary Value Problems_Page_87_Chunk2493 |
78 Chapter 2 First Order Equations Step 5. Integrate φ′ with respect to y, taking the constant of integration to be zero, and substitute the result in (2.5.20) to obtain F (x, y). Step 6. Set F (x, y) = c to obtain an implicit solution of (2.5.19). If possible, solve for y explicitly as a function of x. It’s a common mistake to omit Step 6. However, it’s important to include this step, since F isn’t itself a solution of (2.5.19). Many equations can be conveniently solved by either of the two methods used in Example 2.5.3. How- ever, sometimes the integration required in one approach is more difficult than in the other. In such cases we choose the approach that requires the easier integration. Example 2.5.4 Solve the equation (yexy tan x + exy sec2 x) dx + xexy tan x dy = 0. (2.5.21) Solution We leave it to you to check that My = Nx on any open rectangle where tan x and sec x are defined. Here we must find a function F such that Fx(x, y) = yexy tan x + exy sec2 x (2.5.22) and Fy(x, y) = xexy tan x. (2.5.23) It’s difficult to integrate (2.5.22) with respect to x, but easy to integrate (2.5.23) with respect to y. This yields F (x, y) = exy tan x + ψ(x). (2.5.24) Differentiating this with respect to x yields Fx(x, y) = yexy tan x + exy sec2 x + ψ′(x). Comparing this with (2.5.22) shows that ψ′(x) = 0. Hence, ψ is a constant, which we can take to be zero in (2.5.24), and exy tan x = c is an implicit solution of (2.5.21). Attempting to apply our procedure to an equation that isn’t exact will lead to failure in Step 4, since the function N −∂G ∂y won’t be independent of x if My ̸= Nx (Exercise 31), and therefore can’t be the derivative of a function of y alone. Here’s an example that illustrates this. Example 2.5.5 Verify that the equation 3x2y2 dx + 6x3y dy = 0 (2.5.25) is not exact, and show that the procedure for solving exact equations fails when applied to (2.5.25). | Elementary Differential Equations with Boundary Value Problems_Page_88_Chunk2494 |
Section 2.5 Exact Equations 79 Solution Here My(x, y) = 6x2y and Nx(x, y) = 18x2y, so (2.5.25) isn’t exact. Nevertheless, let’s try to find a function F such that Fx(x, y) = 3x2y2 (2.5.26) and Fy(x, y) = 6x3y. (2.5.27) Integrating (2.5.26) with respect to x yields F (x, y) = x3y2 + φ(y), and differentiating this with respect to y yields Fy(x, y) = 2x3y + φ′(y). For this equation to be consistent with (2.5.27), 6x3y = 2x3y + φ′(y), or φ′(y) = 4x3y. This is a contradiction, since φ′ must be independent of x. Therefore the procedure fails. 2.5 Exercises In Exercises 1–17 determine which equations are exact and solve them. 1. 6x2y2 dx + 4x3y dy = 0 2. (3y cos x + 4xex + 2x2ex) dx + (3 sin x + 3) dy = 0 3. 14x2y3 dx + 21x2y2 dy = 0 4. (2x −2y2) dx + (12y2 −4xy) dy = 0 5. (x + y)2 dx + (x + y)2 dy = 0 6. (4x + 7y) dx + (3x + 4y) dy = 0 7. (−2y2 sin x + 3y3 −2x) dx + (4y cos x + 9xy2) dy = 0 8. (2x + y) dx + (2y + 2x) dy = 0 9. (3x2 + 2xy + 4y2) dx + (x2 + 8xy + 18y) dy = 0 10. (2x2 + 8xy + y2) dx + (2x2 + xy3/3) dy = 0 11. 1 x + 2x dx + 1 y + 2y dy = 0 12. (y sin xy + xy2 cos xy) dx + (x sin xy + xy2 cos xy) dy = 0 13. x dx (x2 + y2)3/2 + y dy (x2 + y2)3/2 = 0 14. | Elementary Differential Equations with Boundary Value Problems_Page_89_Chunk2495 |
80 Chapter 2 First Order Equations 16. | Elementary Differential Equations with Boundary Value Problems_Page_90_Chunk2496 |
Section 2.5 Exact Equations 81 (c) Plot a direction field and some integral curves for (A) on a rectangular region centered at the origin. What is the interval of validity of the solution of (B)? 28. L (a) Solve the exact equation (x2 + y2) dx + 2xy dy = 0 (A) implicitly. (b) For what choices of (x0, y0) does Theorem 2.3.1 imply that the initial value problem (x2 + y2) dx + 2xy dy = 0, y(x0) = y0, (B) has a unique solution y = y(x) on some open interval (a, b) that contains x0? (c) Plot a direction field and some integral curves for (A). From the plot determine, the interval (a, b) of (b), the monotonicity properties (if any) of the solution of (B), and limx→a+ y(x) and limx→b−y(x). HINT: Your answers will depend upon which quadrant contains (x0, y0). 29. Find all functions M such that the equation is exact. (a) M(x, y) dx + (x2 −y2) dy = 0 (b) M(x, y) dx + 2xy sin x cos y dy = 0 (c) M(x, y) dx + (ex −ey sin x) dy = 0 30. Find all functions N such that the equation is exact. (a) (x3y2 + 2xy + 3y2) dx + N(x, y) dy = 0 (b) (ln xy + 2y sin x) dx + N(x, y) dy = 0 (c) (x sin x + y sin y) dx + N(x, y) dy = 0 31. Suppose M, N, and their partial derivatives are continuous on an open rectangle R, and G is an antiderivative of M with respect to x; that is, ∂G ∂x = M. Show that if My ̸= Nx in R then the function N −∂G ∂y is not independent of x. 32. Prove: If the equations M1 dx+N1 dy = 0 and M2 dx+N2 dy = 0 are exact on an open rectangle R, so is the equation (M1 + M2) dx + (N1 + N2) dy = 0. 33. Find conditions on the constants A, B, C, and D such that the equation (Ax + By) dx + (Cx + Dy) dy = 0 is exact. 34. Find conditions on the constants A, B, C, D, E, and F such that the equation (Ax2 + Bxy + Cy2) dx + (Dx2 + Exy + F y2) dy = 0 is exact. | Elementary Differential Equations with Boundary Value Problems_Page_91_Chunk2497 |
82 Chapter 2 First Order Equations 35. Suppose M and N are continuous and have continuous partial derivatives My and Nx that satisfy the exactness condition My = Nx on an open rectangle R. Show that if (x, y) is in R and F (x, y) = Z x x0 M(s, y0) ds + Z y y0 N(x, t) dt, then Fx = M and Fy = N. 36. Under the assumptions of Exercise 35, show that F (x, y) = Z y y0 N(x0, s) ds + Z x x0 M(t, y) dt. 37. Use the method suggested by Exercise 35, with (x0, y0) = (0, 0), to solve the these exact equa- tions: (a) (x3y4 + x) dx + (x4y3 + y) dy = 0 (b) (x2 + y2) dx + 2xy dy = 0 (c) (3x2 + 2y) dx + (2y + 2x) dy = 0 38. Solve the initial value problem y′ + 2 xy = − 2xy x2 + 2x2y + 1, y(1) = −2. 39. Solve the initial value problem y′ −3 xy = 2x4(4x3 −3y) 3x5 + 3x3 + 2y , y(1) = 1. 40. Solve the initial value problem y′ + 2xy = −e−x2 3x + 2yex2 2x + 3yex2 ! , y(0) = −1. 41. Rewrite the separable equation h(y)y′ = g(x) (A) as an exact equation M(x, y) dx + N(x, y) dy = 0. (B) Show that applying the method of this section to (B) yields the same solutions that would be obtained by applying the method of separation of variables to (A) 42. Suppose all second partial derivatives of M = M(x, y) and N = N(x, y) are continuous and M dx + N dy = 0 and −N dx + M dy = 0 are exact on an open rectangle R. Show that Mxx + Myy = Nxx + Nyy = 0 on R. 43. Suppose all second partial derivatives of F = F (x, y) are continuous and Fxx + Fyy = 0 on an open rectangle R. (A function with these properties is said to be harmonic; see also Exercise 42.) Show that −Fy dx + Fx dy = 0 is exact on R, and therefore there’s a function G such that Gx = −Fy and Gy = Fx in R. (A function G with this property is said to be a harmonic conjugate of F .) | Elementary Differential Equations with Boundary Value Problems_Page_92_Chunk2498 |
Section 2.6 Exact Equations 83 44. Verify that the following functions are harmonic, and find all their harmonic conjugates. (See Exercise 43.) (a) x2 −y2 (b) ex cos y (c) x3 −3xy2 (d) cos x cosh y (e) sin x cosh y 2.6 INTEGRATING FACTORS In Section 2.5 we saw that if M, N, My and Nx are continuous and My = Nx on an open rectangle R then M(x, y) dx + N(x, y) dy = 0 (2.6.1) is exact on R. Sometimes an equation that isn’t exact can be made exact by multiplying it by an appro- priate function. For example, (3x + 2y2) dx + 2xy dy = 0 (2.6.2) is not exact, since My(x, y) = 4y ̸= Nx(x, y) = 2y in (2.6.2). However, multiplying (2.6.2) by x yields (3x2 + 2xy2) dx + 2x2y dy = 0, (2.6.3) which is exact, since My(x, y) = Nx(x, y) = 4xy in (2.6.3). Solving (2.6.3) by the procedure given in Section 2.5 yields the implicit solution x3 + x2y2 = c. A function µ = µ(x, y) is an integrating factor for (2.6.1) if µ(x, y)M(x, y) dx + µ(x, y)N(x, y) dy = 0 (2.6.4) is exact. If we know an integrating factor µ for (2.6.1), we can solve the exact equation (2.6.4) by the method of Section 2.5. It would be nice if we could say that (2.6.1) and (2.6.4) always have the same solutions, but this isn’t so. For example, a solution y = y(x) of (2.6.4) such that µ(x, y(x)) = 0 on some interval a < x < b could fail to be a solution of (2.6.1) (Exercise 1), while (2.6.1) may have a solution y = y(x) such that µ(x, y(x)) isn’t even defined (Exercise 2). Similar comments apply if y is the independent variable and x is the dependent variable in (2.6.1) and (2.6.4). However, if µ(x, y) is defined and nonzero for all (x, y), (2.6.1) and (2.6.4) are equivalent; that is, they have the same solutions. Finding Integrating Factors By applying Theorem 2.5.2 (with M and N replaced by µM and µN), we see that (2.6.4) is exact on an open rectangle R if µM, µN, (µM)y, and (µN)x are continuous and ∂ ∂y (µM) = ∂ ∂x(µN) or, equivalently, µyM + µMy = µxN + µNx on R. It’s better to rewrite the last equation as µ(My −Nx) = µxN −µyM, (2.6.5) which reduces to the known result for exact equations; that is, if My = Nx then (2.6.5) holds with µ = 1, so (2.6.1) is exact. You may think (2.6.5) is of little value, since it involves partial derivatives of the unknown integrating factor µ, and we haven’t studied methods for solving such equations. However, we’ll now show that (2.6.5) is useful if we restrict our search to integrating factors that are products of a function of x and a | Elementary Differential Equations with Boundary Value Problems_Page_93_Chunk2499 |
84 Chapter 2 Integrating Factors function of y; that is, µ(x, y) = P (x)Q(y). We’re not saying that every equation M dx + N dy = 0 has an integrating factor of this form; rather, we’re saying that some equations have such integrating factors.We’llnow develop a way to determine whether a given equation has such an integrating factor, and a method for finding the integrating factor in this case. If µ(x, y) = P (x)Q(y), then µx(x, y) = P ′(x)Q(y) and µy(x, y) = P (x)Q′(y), so (2.6.5) becomes P (x)Q(y)(My −Nx) = P ′(x)Q(y)N −P (x)Q′(y)M, (2.6.6) or, after dividing through by P (x)Q(y), My −Nx = P ′(x) P (x) N −Q′(y) Q(y) M. (2.6.7) Now let p(x) = P ′(x) P (x) and q(y) = Q′(y) Q(y) , so (2.6.7) becomes My −Nx = p(x)N −q(y)M. (2.6.8) We obtained (2.6.8) by assuming that M dx+N dy = 0 has an integrating factor µ(x, y) = P (x)Q(y). However, we can now view (2.6.7) differently: If there are functions p = p(x) and q = q(y) that satisfy (2.6.8) and we define P (x) = ±e R p(x) dx and Q(y) = ±e R q(y) dy, (2.6.9) then reversing the steps that led from (2.6.6) to (2.6.8) shows that µ(x, y) = P (x)Q(y) is an integrating factor for M dx +N dy = 0. In using this result, we take the constants of integration in (2.6.9) to be zero and choose the signs conveniently so the integrating factor has the simplest form. There’s no simple general method for ascertaining whether functions p = p(x) and q = q(y) satisfying (2.6.8) exist. However, the next theorem gives simple sufficient conditions for the given equation to have an integrating factor that depends on only one of the independent variables x and y, and for finding an integrating factor in this case. Theorem 2.6.1 Let M, N, My, and Nx be continuous on an open rectangle R. Then: (a) If (My −Nx)/N is independent of y on R and we define p(x) = My −Nx N then µ(x) = ±e R p(x) dx (2.6.10) is an integrating factor for M(x, y) dx + N(x, y) dy = 0 (2.6.11) on R. (b) If (Nx −My)/M is independent of x on R and we define q(y) = Nx −My M , then µ(y) = ±e R q(y) dy (2.6.12) is an integrating factor for (2.6.11) on R. | Elementary Differential Equations with Boundary Value Problems_Page_94_Chunk2500 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.