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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 elec...
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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 micros...
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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...
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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, 15...
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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,...
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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 elect...
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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.ed...
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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 Americ...
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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, c...
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TO BEVERLY
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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 Equa...
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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 S...
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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....
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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- ra...
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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 equ...
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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) a...
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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 differe...
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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 be...
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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 (birt...
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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...
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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 proportion...
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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 susce...
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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 th...
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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 differe...
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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 bo...
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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...
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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) ...
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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. Thes...
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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) =...
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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 f...
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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...
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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 ...
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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 direc...
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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 ...
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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,...
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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 complete...
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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
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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)
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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)
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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
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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)
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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)
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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;...
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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 = r...
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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...
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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 nonho...
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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 equatio...
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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 ...
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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 s...
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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 ...
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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 ...
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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...
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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′...
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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 ...
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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 r...
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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...
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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...
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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 c...
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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....
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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). (T...
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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 s...
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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 + ...
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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 g...
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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′ = ...
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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 ze...
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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....
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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 ...
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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/y...
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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′ = −2...
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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 ...
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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 equat...
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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 nex...
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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...
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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...
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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...
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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 (−∞, ∞) t...
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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 ...
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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...
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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...
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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 Sep...
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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...
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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 f...
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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 −...
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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)...
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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 =...
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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...
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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, ...
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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...
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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 =...
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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 +...
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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 add...
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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 (...
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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 cu...
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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 m...
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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 re...
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80 Chapter 2 First Order Equations 16.
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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....
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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 assumption...
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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 ope...
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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 integ...
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