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Section 2.6 Exact Equations 85 Proof (a) If (My −Nx)/N is independent of y, then (2.6.8) holds with p = (My −Nx)/N and q ≡0. Therefore P (x) = ±e R p(x) dx and Q(y) = ±e R q(y) dy = ±e0 = ±1, so (2.6.10) is an integrating factor for (2.6.11) on R. (b) If (Nx −My)/M is independent of x then eqrefeq:2.6.8 holds with p ≡0... | Elementary Differential Equations with Boundary Value Problems_Page_95_Chunk2501 |
86 Chapter 2 Integrating Factors −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −4 −3 −2 −1 0 1 2 3 4 y x Figure 2.6.1 A direction field and integral curves for (2xy3 −2x3y3 −4xy2 + 2x) dx + (3x2y2 + 4y) dy = 0 Example 2.6.2 Find an integrating factor for 2xy3 dx + (3x2y2 + x2y3 + 1) dy = 0 (2.6.18) and solve the equation. Solution In (... | Elementary Differential Equations with Boundary Value Problems_Page_96_Chunk2502 |
Section 2.6 Exact Equations 87 µ(y) = ey is an integrating factor. Multiplying (2.6.18) by µ yields the exact equation 2xy3ey dx + (3x2y2 + x2y3 + 1)ey dy = 0. (2.6.19) To solve this equation, we must find a function F such that Fx(x, y) = 2xy3ey (2.6.20) and Fy(x, y) = (3x2y2 + x2y3 + 1)ey. (2.6.21) Integrating (2.6.20... | Elementary Differential Equations with Boundary Value Problems_Page_97_Chunk2503 |
88 Chapter 2 Integrating Factors Example 2.6.3 Find an integrating factor for (3xy + 6y2) dx + (2x2 + 9xy) dy = 0 (2.6.23) and solve the equation. Solution In (2.6.23) M = 3xy + 6y2, N = 2x2 + 9xy, and My −Nx = (3x + 12y) −(4x + 9y) = −x + 3y. Therefore My −Nx M = −x + 3y 3xy + 6y2 and Nx −My N = x −3y 2x2 + 9xy, so Th... | Elementary Differential Equations with Boundary Value Problems_Page_98_Chunk2504 |
Section 2.6 Exact Equations 89 −4 −3 −2 −1 0 1 2 3 4 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y x Figure 2.6.3 A direction field and integral curves for (3xy + 6y2) dx + (2x2 + 9xy) dy = 0 We leave it to you to use the method of Section 2.5 to show that this equation has the implicit solution x3y2 + 3x2y3 = c. (2.6.25) This is als... | Elementary Differential Equations with Boundary Value Problems_Page_99_Chunk2505 |
90 Chapter 2 Integrating Factors −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 2.6.4 A direction field and integral curves for −y dx + (x + x6) dy = 0 Solution In (2.6.26) M = −y, N = x + x6, and My −Nx = −1 −(1 + 6x5) = −2 −6x5. We look for functions p = p(x) and q = q... | Elementary Differential Equations with Boundary Value Problems_Page_100_Chunk2506 |
Section 2.6 Exact Equations 91 We leave it to you to use the method of the Section 2.5 to show that this equation has the implicit solution y x 5 + y5 = k. Solving for y yields y = k1/5x(1 + x5)−1/5, which we rewrite as y = cx(1 + x5)−1/5 by renaming the arbitrary constant. This is also a solution of (2.6.26). Figure... | Elementary Differential Equations with Boundary Value Problems_Page_101_Chunk2507 |
92 Chapter 2 Integrating Factors 7. (xy + x + 2y + 1) dx + (x + 1) dy = 0 8. (27xy2 + 8y3) dx + (18x2y + 12xy2) dy = 0 9. (6xy2 + 2y) dx + (12x2y + 6x + 3) dy = 0 10. y2 dx + xy2 + 3xy + 1 y dy = 0 11. (12x3y + 24x2y2) dx + (9x4 + 32x3y + 4y) dy = 0 12. (x2y + 4xy + 2y) dx + (x2 + x) dy = 0 13. −y dx + (x4 −x) dy =... | Elementary Differential Equations with Boundary Value Problems_Page_102_Chunk2508 |
Section 2.6 Exact Equations 93 30. According to Theorem 2.1.2, the general solution of the linear nonhomogeneous equation y′ + p(x)y = f(x) (A) is y = y1(x) c + Z f(x)/y1(x) dx , (B) where y1 is any nontrivial solution of the complementary equation y′ +p(x)y = 0. In this exercise we obtain this conclusion in a diff... | Elementary Differential Equations with Boundary Value Problems_Page_103_Chunk2509 |
CHAPTER 3 Numerical Methods In this chapter we study numerical methods for solving a first order differential equation y′ = f(x, y). SECTION 3.1 deals with Euler’s method, which is really too crude to be of much use in practical appli- cations. However, its simplicity allows for an introduction to the ideas required to ... | Elementary Differential Equations with Boundary Value Problems_Page_105_Chunk2510 |
96 Chapter 3 Numerical Methods 3.1 EULER’S METHOD If an initial value problem y′ = f(x, y), y(x0) = y0 (3.1.1) can’t be solved analytically, it’s necessary to resort to numerical methods to obtain useful approximations to a solution of (3.1.1). We’ll consider such methods in this chapter. We’re interested in computing ... | Elementary Differential Equations with Boundary Value Problems_Page_106_Chunk2511 |
Section 3.1 Euler’s Method 97 which isn’t useful, since we don’t know y(x1). Therefore we replace y(x1) by its approximate value y1 and redefine y2 = y1 + hf(x1, y1). Having computed y2, we can compute y3 = y2 + hf(x2, y2). In general, Euler’s method starts with the known value y(x0) = y0 and computes y1, y2, ..., yn su... | Elementary Differential Equations with Boundary Value Problems_Page_107_Chunk2512 |
98 Chapter 3 Numerical Methods Solution Table 3.1.1 shows the values of the exact solution (3.1.6) at the specified points, and the ap- proximate values of the solution at these points obtained by Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025. In examining this table, keep in mind that the approximate ... | Elementary Differential Equations with Boundary Value Problems_Page_108_Chunk2513 |
Section 3.1 Euler’s Method 99 except in this case we can’t solve (3.1.7) exactly. The results in the “Exact” column were obtained by using a more accurate numerical method known as the Runge-Kutta method with a small step size. They are exact to eight decimal places. Since we think it’s important in evaluating the accu... | Elementary Differential Equations with Boundary Value Problems_Page_109_Chunk2514 |
100 Chapter 3 Numerical Methods where K is a constant independent of n. There are two sources of error (not counting roundoff) in Euler’s method: 1. The error committed in approximating the integral curve by the tangent line (3.1.2) over the interval [xi, xi+1]. 2. The error committed in replacing y(xi) by yi in (3.1.2... | Elementary Differential Equations with Boundary Value Problems_Page_110_Chunk2515 |
Section 3.1 Euler’s Method 101 Note that the magnitude of the local truncation error in Euler’s method is determined by the second derivative y′′ of the solution of the initial value problem. Therefore the local truncation error will be larger where |y′′| is large, or smaller where |y′′| is small. Since the local trunc... | Elementary Differential Equations with Boundary Value Problems_Page_111_Chunk2516 |
102 Chapter 3 Numerical Methods (since C = 1 + Rh). From this and (3.1.16), |y(b) −yn| = |en| ≤(1 + Rh)n −1 R Mh 2 . (3.1.17) Since Taylor’s theorem implies that 1 + Rh < eRh (verify), (1 + Rh)n < enRh = eR(b−x0) (since nh = b −x0). This and (3.1.17) imply that |y(b) −yn| ≤Kh, (3.1.18) with K = M eR(b−x0) −1 2R . Becau... | Elementary Differential Equations with Boundary Value Problems_Page_112_Chunk2517 |
Section 3.1 Euler’s Method 103 because it was impossible to evaluate this integral exactly in terms of elementary functions. Use step sizes h = 0.2, h = 0.1, and h = 0.05 to find approximate values of the solution of (3.1.22) at x = 0, 0.2, 0.4, 0.6, ..., 2.0 by (a) Euler’s method; (b) the Euler semilinear method. SOLUT... | Elementary Differential Equations with Boundary Value Problems_Page_113_Chunk2518 |
104 Chapter 3 Numerical Methods x h = 0.2 h = 0.1 h = 0.05 “Exact” 0.0 3.000000000 3.000000000 3.000000000 3.000000000 0.2 3.330594477 3.329558853 3.328788889 3.327851973 0.4 3.980734157 3.974067628 3.970230415 3.966059348 0.6 5.106360231 5.087705244 5.077622723 5.067039535 0.8 7.021003417 6.980190891 6.958779586 6.936... | Elementary Differential Equations with Boundary Value Problems_Page_114_Chunk2519 |
Section 3.1 Euler’s Method 105 Table 3.1.8. Numerical solution of y′ −2y = x/(1 + y2), y(1) = 7, by the Euler semilinear method. x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 7.000000000 7.000000000 7.000000000 7.000000000 1.1 8.552262113 8.551993978 8.551867007 8.551744786 1.2 10.447568674 10.447038547 10.446787646 10.4465... | Elementary Differential Equations with Boundary Value Problems_Page_115_Chunk2520 |
106 Chapter 3 Numerical Methods x h = 0.1 h = 0.05 h = 0.025 “Exact” x h = 0.1 h = 0.05 h = 0.025 h = .0125 2.0 2.000000000 2.000000000 2.000000000 2.000000000 2.1 0.708426286 0.702568171 0.701214274 0.701162906 2.2 0.214501852 0.222599468 0.228942240 0.236986800 2.3 0.069861436 0.083620494 0.092852806 0.103815729 2.4 ... | Elementary Differential Equations with Boundary Value Problems_Page_116_Chunk2521 |
Section 3.1 Euler’s Method 107 at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. Compare these approximate values with the values of the exact solution y = 1 3x2 (9 lnx + x3 + 2), which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.1.1. 8. C Use Euler’s method with step sizes h = 0.05, h... | Elementary Differential Equations with Boundary Value Problems_Page_117_Chunk2522 |
108 Chapter 3 Numerical Methods 12. C Use Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + (y + 1)(y −1)(y −2) x + 1 = 0, y(1) = 0 (Exercise 2.2.14) at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. 13. C Use Euler’s method and the Euler semi... | Elementary Differential Equations with Boundary Value Problems_Page_118_Chunk2523 |
Section 3.2 The Improved Euler Method and Related Methods 109 a useful way to evaluate the definite integral. In this case we must resort to approximate methods. There’s a class of such methods called numerical quadrature, where the approximation takes the form Z b a f(x) dx ≈ n X i=0 cif(xi), (B) where a = x0 < x1 < · ... | Elementary Differential Equations with Boundary Value Problems_Page_119_Chunk2524 |
110 Chapter 3 Numerical Methods were 12, 24, and 48, as in the three applications of Euler’s method; however, you can see from the third column of Table 3.2.1 that the approximation to e obtained by the improved Euler method with only 12 evaluations of f is better than the approximation obtained by Euler’s method with ... | Elementary Differential Equations with Boundary Value Problems_Page_120_Chunk2525 |
Section 3.2 The Improved Euler Method and Related Methods 111 The improved Euler method requires two evaluations of f(x, y) per step, while Euler’s method requires only one. However, we’ll see at the end of this section that if f satisfies appropriate assumptions, the local truncation error with the improved Euler metho... | Elementary Differential Equations with Boundary Value Problems_Page_121_Chunk2526 |
112 Chapter 3 Numerical Methods at x = 0, 0.1, 0.2, 0.3, ..., 1.0. For comparison, it also shows the corresponding approximate values obtained with Euler’s method in 3.1.2, and the values of the exact solution y = e−2x 4 (x4 + 4). The results obtained by the improved Euler method with h = 0.1 are better than those obta... | Elementary Differential Equations with Boundary Value Problems_Page_122_Chunk2527 |
Section 3.2 The Improved Euler Method and Related Methods 113 at x = 0, 0.2, 0.4, 0.6, ..., 2.0 by (a) the improved Euler method; (b) the improved Euler semilinear method. (We used Euler’s method and the Euler semilinear method on this problem in 3.1.4.) SOLUTION(a) Rewriting (3.2.6) as y′ = 1 + 2xy, y(0) = 3 and apply... | Elementary Differential Equations with Boundary Value Problems_Page_123_Chunk2528 |
114 Chapter 3 Numerical Methods A Family of Methods with O(h3) Local Truncation Error We’ll now derive a class of methods with O(h3) local truncation error for solving (3.2.1). For simplicity, we assume that f, fx, fy, fxx, fyy, and fxy are continuous and bounded for all (x, y). This implies that if y is the solution o... | Elementary Differential Equations with Boundary Value Problems_Page_124_Chunk2529 |
Section 3.2 The Improved Euler Method and Related Methods 115 which is true if σ + ρ = 1 and ρθ = 1 2. (3.2.10) Since y′ = f(x, y), we can now conclude from (3.2.8) that y(xi+1) = y(xi) + h [σf(xi, yi) + ρf(xi + θh, y(xi + θh))] + O(h3) (3.2.11) if σ, ρ, and θ satisfy (3.2.10). However, this formula would not be useful... | Elementary Differential Equations with Boundary Value Problems_Page_125_Chunk2530 |
116 Chapter 3 Numerical Methods which can be organized as k1i = f(xi, yi), k2i = f xi + 2h 3 , yi + 2h 3 k1i , yi+1 = yi + h 4(k1i + 3k2i). Letting ρ = 1 yields the midpoint method, yi+1 = yi + hf xi + h 2 , yi + h 2 f(xi, yi) , which can be organized as k1i = f(xi, yi), k2i = f xi + h 2 , yi + h 2 k1i , yi... | Elementary Differential Equations with Boundary Value Problems_Page_126_Chunk2531 |
Section 3.2 The Improved Euler Method and Related Methods 117 at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. Compare these approximate values with the values of the exact solution y = 1 3x2 (9 lnx + x3 + 2) which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.2.2. 8. C Use the improved... | Elementary Differential Equations with Boundary Value Problems_Page_127_Chunk2532 |
118 Chapter 3 Numerical Methods 12. C Use the improved Euler method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + (y + 1)(y −1)(y −2) x + 1 = 0, y(1) = 0 (Exercise 2.2.14) at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. 13. C Use the improved Euler met... | Elementary Differential Equations with Boundary Value Problems_Page_128_Chunk2533 |
Section 3.3 The Runge-Kutta Method 119 28. C Do Exercise 11 with “improved Euler method” replaced by “Heun’s method.” 29. C Do Exercise 12 with “improved Euler method” replaced by “midpoint method.” 30. C Do Exercise 12 with “improved Euler method” replaced by “Heun’s method.” 31. Show that if f, fx, fy, fxx, fyy, and ... | Elementary Differential Equations with Boundary Value Problems_Page_129_Chunk2534 |
120 Chapter 3 Numerical Methods Example 3.3.1 Use the Runge-Kutta method with h = 0.1 to find approximate values for the solution of the initial value problem y′ + 2y = x3e−2x, y(0) = 1, (3.3.2) at x = 0.1, 0.2. Solution Again we rewrite (3.3.2) as y′ = −2y + x3e−2x, y(0) = 1, which is of the form (3.3.1), with f(x, y) ... | Elementary Differential Equations with Boundary Value Problems_Page_130_Chunk2535 |
Section 3.3 The Runge-Kutta Method 121 at x = 0, 0.1, 0.2, 0.3, ..., 1.0. For comparison, it also shows the corresponding approximate values obtained with the improved Euler method in Example 3.2.2, and the values of the exact solution y = e−2x 4 (x4 + 4). The results obtained by the Runge-Kutta method are clearly bett... | Elementary Differential Equations with Boundary Value Problems_Page_131_Chunk2536 |
122 Chapter 3 Numerical Methods Example 3.3.4 Tables 3.3.3 and 3.3.4 show results obtained by applying the Runge-Kutta and Runge- Kutta semilinear methods to to the initial value problem y′ −2xy = 1, y(0) = 3, which we considered in Examples 3.1.4 and 3.2.4. Table 3.3.3. Numerical solution of y′ −2xy = 1, y(0) = 3, by ... | Elementary Differential Equations with Boundary Value Problems_Page_132_Chunk2537 |
Section 3.3 The Runge-Kutta Method 123 y−1 = z1, y−2 = z2, . . ., y−n = zn are approximate values of the solution of (3.3.3) at x0 −h, x0 −2h, ..., x0 −nh = a. The justification for this answer is sketched in Exercise 23. Note how easy it is to make the change the given problem (3.3.3) to the modified problem (3.3.4): fir... | Elementary Differential Equations with Boundary Value Problems_Page_133_Chunk2538 |
124 Chapter 3 Numerical Methods x y Exact 0.00 3.466212070 3.466212074 0.10 3.516161955 3.516161958 0.20 3.567241862 3.567241864 0.30 3.619284615 3.619284617 0.40 3.672141529 3.672141530 0.50 3.725680888 3.725680889 0.60 3.779786399 3.779786399 0.70 3.834355648 3.834355648 0.80 3.889298649 3.889298649 0.90 3.944536474 ... | Elementary Differential Equations with Boundary Value Problems_Page_134_Chunk2539 |
Section 3.3 The Runge-Kutta Method 125 8. C Use the Runge-Kutta method with step sizes h = 0.05, h = 0.025, and h = 0.0125 to find approximate values of the solution of the initial value problem y′ = y2 + xy −x2 x2 , y(1) = 2 at x = 1.0, 1.05, 1.10, 1.15 ..., 1.5. Compare these approximate values with the values of the ... | Elementary Differential Equations with Boundary Value Problems_Page_135_Chunk2540 |
126 Chapter 3 Numerical Methods 12. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + (y + 1)(y −1)(y −2) x + 1 = 0, y(1) = 0 (Exercise 2.2.14), at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. 13. C Use the Runge-Kutta method a... | Elementary Differential Equations with Boundary Value Problems_Page_136_Chunk2541 |
Section 3.3 The Runge-Kutta Method 127 24. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of y′ = y2 + xy −x2 x2 , y(2) = −1 at x = 1.1, 1.2, 1.3, ...2.0. Compare these approximate values with the values of the exact solution y = x(4 −3x2) 4 + 3x2... | Elementary Differential Equations with Boundary Value Problems_Page_137_Chunk2542 |
128 Chapter 3 Numerical Methods (b) L For several choices of a, b, A, B, C, and D apply (A) to f(x) = A + Bx + Cx + Dx3, with n = 10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain what you find. (c) L For several choices of a, b, A, B, C, D, and E apply (A) to f(x) = A + Bx + Cx2 + Dx3 + ... | Elementary Differential Equations with Boundary Value Problems_Page_138_Chunk2543 |
CHAPTER 4 Applications of First Order Equations IN THIS CHAPTER we consider applications of first order differential equations. SECTION 4.1 begins with a discussion of exponential growth and decay, which you have probably al- ready seen in calculus. We consider applications to radioactive decay, carbon dating, and compo... | Elementary Differential Equations with Boundary Value Problems_Page_139_Chunk2544 |
130 Chapter 4 Applications of First Order Equations 4.1 GROWTH AND DECAY Since the applications in this section deal with functions of time, we’ll denote the independent variable by t. If Q is a function of t, Q′ will denote the derivative of Q with respect to t; thus, Q′ = dQ dt . Exponential Growth and Decay One of t... | Elementary Differential Equations with Boundary Value Problems_Page_140_Chunk2545 |
Section 4.1 Growth and Decay 131 Q t a > 0 a < 0 Q0 Figure 4.1.1 Exponential growth and decay so e−kτ = 1 2. Taking logarithms yields −kτ = ln 1 2 = −ln 2, so the half-life is τ = 1 k ln2. (4.1.5) (Figure 4.1.2). The half-life is independent of t0 and Q0, since it’s determined by the properties of material, not by the ... | Elementary Differential Equations with Boundary Value Problems_Page_141_Chunk2546 |
132 Chapter 4 Applications of First Order Equations Q t τ Q0 .5Q0 Figure 4.1.2 Half-life of a radioactive substance Therefore the mass left after 810 years will be Q(810) = 4e−(810 ln 2)/1620 = 4e−(ln 2)/2 = 2 √ 2 g. SOLUTION(b) Setting t = t1 in (4.1.7) and requiring that Q(t1) = 1.5 yields 3 2 = 4e(−t1 ln 2)/1620. Di... | Elementary Differential Equations with Boundary Value Problems_Page_142_Chunk2547 |
Section 4.1 Growth and Decay 133 If interest is compounded semiannually, the value of the account is multiplied by (1 + r/2) every 6 months. Since this occurs twice annually, the value of the account after t years is Q(t) = Q0 1 + r 2 2t . In general, if interest is compounded n times per year, the value of the acco... | Elementary Differential Equations with Boundary Value Problems_Page_143_Chunk2548 |
134 Chapter 4 Applications of First Order Equations Example 4.1.2 If $150 is deposited in a bank that pays 5 1 2% annual interest compounded continuously, the value of the account after t years is Q(t) = 150e.055t dollars. (Note that it’s necessary to write the interest rate as a decimal; thus, r = .055.) Therefore, af... | Elementary Differential Equations with Boundary Value Problems_Page_144_Chunk2549 |
Section 4.1 Growth and Decay 135 Q t a/k Figure 4.1.3 Q(t) approaches the steady state value a k as t →∞ and Q = ue−kt = a k + ce−kt. Since Q(0) = Q0, setting t = 0 here yields Q0 = a k + c or c = Q0 −a k . Therefore Q = a k + Q0 −a k e−kt. (4.1.12) SOLUTION(b) Since k > 0, limt→∞e−kt = 0, so from (4.1.12) lim t→∞Q... | Elementary Differential Equations with Boundary Value Problems_Page_145_Chunk2550 |
136 Chapter 4 Applications of First Order Equations achieved its steady state value long ago as a result of its creation and decomposition over millions of years. These assumptions led Libby to conclude that the ratio of carbon-14 to carbon-12 has been nearly constant for a long time. This constant, which we denote by ... | Elementary Differential Equations with Boundary Value Problems_Page_146_Chunk2551 |
Section 4.1 Growth and Decay 137 and therefore Q satisfies the differential equation Q′ −.06Q = 2600. (4.1.13) (Of course, we must recognize that the solution of this equation is an approximation to the true value of Q at any given time. We’ll discuss this further below.) Since e.06t is a solution of the complementary e... | Elementary Differential Equations with Boundary Value Problems_Page_147_Chunk2552 |
138 Chapter 4 Applications of First Order Equations 4.1 Exercises 1. The half-life of a radioactive substance is 3200 years. Find the quantity Q(t) of the substance left at time t > 0 if Q(0) = 20 g. 2. The half-life of a radioactive substance is 2 days. Find the time required for a given amount of the material to deca... | Elementary Differential Equations with Boundary Value Problems_Page_148_Chunk2553 |
Section 4.1 Growth and Decay 139 17. A tank is empty at t = 0. Water is added to the tank at the rate of 10 gal/min, but it leaks out at a rate (in gallons per minute) equal to the number of gallons in the tank. What is the smallest capacity the tank can have if this process is to continue forever? 18. A person deposit... | Elementary Differential Equations with Boundary Value Problems_Page_149_Chunk2554 |
140 Chapter 4 Applications of First Order Equations (a) Determine the time T(α) when the loan will be paid off and the amount S(α) that the home- buyer will save. (b) Suppose P0 = $50, 000, r = 8%, and N = 15. Compute the savings realized by accelerated payments with α = 1.05, 1.10, and 1.15. 24. A benefactor wishes to... | Elementary Differential Equations with Boundary Value Problems_Page_150_Chunk2555 |
Section 4.2 Cooling and Mixing 141 so T = ue−kt = Tm + ce−kt. If T(0) = T0, setting t = 0 here yields c = T0 −Tm, so T = Tm + (T0 −Tm)e−kt. (4.2.2) Note that T −Tm decays exponentially, with decay constant k. Example 4.2.1 A ceramic insulator is baked at 400◦C and cooled in a room in which the temperature is 25◦C. Afte... | Elementary Differential Equations with Boundary Value Problems_Page_151_Chunk2556 |
142 Chapter 4 Applications of First Order Equations t T 5 10 15 20 25 30 100 150 200 250 300 350 400 50 Figure 4.2.1 T = 25 + 375e−(t/4)ln 15/7 We obtain k from the stated condition that the temperature of the object is 50◦F at 11:07. Since 11:07 is t = 2 on our time scale, we can determine k by substituting T = 50 and... | Elementary Differential Equations with Boundary Value Problems_Page_152_Chunk2557 |
Section 4.2 Cooling and Mixing 143 T t −5 5 10 15 20 25 30 35 40 20 40 60 80 −20 100 T=72 Figure 4.2.2 T = −20 + 80e−t 2 ln 8 7 Therefore the object was placed outside about 2 minutes and 5 seconds before 11:05; that is, at 11:02:55. Mixing Problems In the next two examples a saltwater solution with a given concentrati... | Elementary Differential Equations with Boundary Value Problems_Page_153_Chunk2558 |
144 Chapter 4 Applications of First Order Equations 600 gal 4 gal/min; .5 lb/gal 4 gal/min Figure 4.2.3 A mixing problem The rate in is 1 2 lb/gal × (4 gal/min) = 2 lb/min. Determining the rate out requires a little more thought. We’re removing 4 gallons of the mixture per minute, and there are always 600 gallons in... | Elementary Differential Equations with Boundary Value Problems_Page_154_Chunk2559 |
Section 4.2 Cooling and Mixing 145 100 200 300 400 500 600 700 800 900 50 100 150 200 250 300 t Q Figure 4.2.4 Q = 300 −260e−t/150 so Q = ue−t/150 = 300 + ce−t/150 (4.2.6) (Figure 4.2.4). Since Q(0) = 40, c = −260; therefore, Q = 300 −260e−t/150. SOLUTION(b) From (4.2.6), we see that that limt→∞Q(t) = 300 for any value... | Elementary Differential Equations with Boundary Value Problems_Page_155_Chunk2560 |
146 Chapter 4 Applications of First Order Equations 2t+200 liters 4 liters/min; .25 g/liter Figure 4.2.5 Another mixing problem The rate in is 1 4 g/liter × (4 liters/min) = 1 g/min. (4.2.8) To determine the rate out, we observe that since the mixture is being removed from the tank at the constant rate of 2 liters/m... | Elementary Differential Equations with Boundary Value Problems_Page_156_Chunk2561 |
Section 4.2 Cooling and Mixing 147 Hence, u = (t + 100)2 2 + c. (4.2.11) Since Q(0) = 10 and u = (t + 100)Q, (4.2.11) implies that (100)(10) = (100)2 2 + c, so c = 100(10) −(100)2 2 = −4000 and therefore u = (t + 100)2 2 −4000. Hence, Q = u t + 200 = t + 100 2 − 4000 t + 100. Now let K(t) be the concentration of salt a... | Elementary Differential Equations with Boundary Value Problems_Page_157_Chunk2562 |
148 Chapter 4 Applications of First Order Equations 4.2 Exercises 1. A thermometer is moved from a room where the temperature is 70◦F to a freezer where the tem- perature is 12◦F . After 30 seconds the thermometer reads 40◦F. What does it read after 2 minutes? 2. A fluid initially at 100◦C is placed outside on a day whe... | Elementary Differential Equations with Boundary Value Problems_Page_158_Chunk2563 |
Section 4.2 Cooling and Mixing 149 Assuming that the gas is always uniformly distributed in the room and its initial concentration c0 is at a safe level, find the smallest value of q2 required to maintain safe conditions in the laboratory for all time. 14. A 1200-gallon tank initially contains 40 pounds of salt dissolve... | Elementary Differential Equations with Boundary Value Problems_Page_159_Chunk2564 |
150 Chapter 4 Applications of First Order Equations (b) Find T(t) and Tm(t) for t > 0. (c) Find limt→∞T(t) and limt→∞Tm(t). 18. Control mechanisms allow fluid to flow into a tank at a rate proportional to the volume V of fluid in the tank, and to flow out at a rate proportional to V 2. Suppose V (0) = V0 and the constants ... | Elementary Differential Equations with Boundary Value Problems_Page_160_Chunk2565 |
Section 4.3 Elementary Mechanics 151 4.3 ELEMENTARY MECHANICS Newton’s Second Law of Motion In this section we consider an object with constant mass m moving along a line under a force F . Let y = y(t) be the displacement of the object from a reference point on the line at time t, and let v = v(t) and a = a(t) be the v... | Elementary Differential Equations with Boundary Value Problems_Page_161_Chunk2566 |
152 Chapter 4 Applications of First Order Equations Solving this equation yields v as a function of t. If we know y(t0) for some time t0, we can integrate v to obtain y as a function of t. Equations of the form (4.3.3) occur in problems involving motion through a resisting medium. Motion Through a Resisting Medium Unde... | Elementary Differential Equations with Boundary Value Problems_Page_162_Chunk2567 |
Section 4.3 Elementary Mechanics 153 v v F1 = − kv F1 = − kv (a) (b) Figure 4.3.1 Resistive forces so c = v0 + mg k and (4.3.7) becomes v = −mg k + v0 + mg k e−kt/m. Letting t →∞here shows that the terminal velocity is lim t→∞v(t) = −mg k , which is independent of the initial velocity v0 (Figure 4.3.2). Example 4.3... | Elementary Differential Equations with Boundary Value Problems_Page_163_Chunk2568 |
154 Chapter 4 Applications of First Order Equations − mg/k t v Figure 4.3.2 Solutions of mv′ = −mg −kv Since e−t/10 is a solution of the complementary equation, the solutions of this equation are of the form v = ue−t/10, where u′e−t/10 = −32, so u′ = −32et/10. Hence, u = −320et/10 + c, so v = ue−t/10 = −320 + ce−t/10. ... | Elementary Differential Equations with Boundary Value Problems_Page_164_Chunk2569 |
Section 4.3 Elementary Mechanics 155 where k is a constant. Since the magnitude of the resistance is 8 N when v = 2 m/s, k(22) = 8, so k = 2 N-s2/m2. Since m = 10 and g = 9.8, (4.3.9) becomes 10v′ = −98 + 2v2 = 2(v2 −49). (4.3.10) If v0 = −7, then v ≡−7 for all t ≥0. If v0 ̸= −7, we separate variables to obtain 1 v2 −4... | Elementary Differential Equations with Boundary Value Problems_Page_165_Chunk2570 |
156 Chapter 4 Applications of First Order Equations t v v = − (mg/k)1/2 Figure 4.3.3 Solutions of mv′ = −mg + kv2, v(0) = v0 ≤0 Example 4.3.4 A 10-kg mass is launched vertically upward from Earth’s surface with an initial velocity of v0 m/s. The only forces acting on the mass are gravity and atmospheric resistance prop... | Elementary Differential Equations with Boundary Value Problems_Page_166_Chunk2571 |
Section 4.3 Elementary Mechanics 157 0.2 0.4 0.6 0.8 1 10 20 30 40 50 t v Figure 4.3.4 Solutions of (4.3.15) for various v0 > 0 Solving this for v yields v = 7 tan −7t 5 + tan−1 v0 7 . (4.3.17) Using the identity tan(A −B) = tan A −tan B 1 + tan A tan B with A = tan−1(v0/7) and B = 7t/5, and noting that tan(tan−1 θ... | Elementary Differential Equations with Boundary Value Problems_Page_167_Chunk2572 |
158 Chapter 4 Applications of First Order Equations y = − R y = 0 y = h y Figure 4.3.5 Escape velocity Escape Velocity Suppose a space vehicle is launched vertically and its fuel is exhausted when the vehicle reaches an altitude h above Earth, where h is sufficiently large so that resistance due to Earth’s atmosphere ca... | Elementary Differential Equations with Boundary Value Problems_Page_168_Chunk2573 |
Section 4.3 Elementary Mechanics 159 so (4.3.19) implies that d2y dt2 = − gR2 (y + R)2 . (4.3.20) We’ll show that there’s a number ve, called the escape velocity, with these properties: 1. If v0 ≥ve then v(t) > 0 for all t > 0, and the vehicle continues to climb for all t > 0; that is, it “escapes” Earth. (Is it really... | Elementary Differential Equations with Boundary Value Problems_Page_169_Chunk2574 |
160 Chapter 4 Applications of First Order Equations If (4.3.24) holds then the parenthetical expression in (4.3.23) is negative and the vehicle will attain a maximum altitude ym > h that satisfies the equation 0 = gR2 ym + R + v2 0 2 −gR2 h + R . The velocity will be zero at the maximum altitude, and the object will ... | Elementary Differential Equations with Boundary Value Problems_Page_170_Chunk2575 |
Section 4.3 Elementary Mechanics 161 10. An object weighing 256 lb is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the square of the speed. The magnitude of the resisting force is 1 lb when |v| = 4 ft/s. Find v for t > 0, and find its terminal velocity. 11. An object with ma... | Elementary Differential Equations with Boundary Value Problems_Page_171_Chunk2576 |
162 Chapter 4 Applications of First Order Equations 17. A space probe is to be launched from a space station 200 miles above Earth. Determine its escape velocity in miles/s. Take Earth’s radius to be 3960 miles. 18. A space vehicle is to be launched from the moon, which has a radius of about 1080 miles. The acceleratio... | Elementary Differential Equations with Boundary Value Problems_Page_172_Chunk2577 |
Section 4.4 Autonomous Second Order Equations 163 curve of (4.4.4). We’ll call these integral curves trajectories of (4.4.1), and we’ll call (4.4.4) the phase plane equivalent of (4.4.1). In this section we’ll consider autonomous equations that can be written as y′′ + q(y, y′)y′ + p(y) = 0. (4.4.5) Equations of this fo... | Elementary Differential Equations with Boundary Value Problems_Page_173_Chunk2578 |
164 Chapter 4 Applications of First Order Equations for all t > 0. Figure 4.4.1 illustrates the geometrical interpretation of this definition in the Poincaré phase plane: if (y0, v0) is in the smaller shaded circle (with radius δ), then (y(t), v(t)) must be in in the larger circle (with radius ϵ) for all t > 0. y y v ε ... | Elementary Differential Equations with Boundary Value Problems_Page_174_Chunk2579 |
Section 4.4 Autonomous Second Order Equations 165 constant, or is conserved. In particular, if a trajectory passes through a given point (y0, v0) then c = v2 0 2 + P (y0). Example 4.4.1 [The Undamped Spring - Mass System] Consider an object with mass m suspended from a spring and moving vertically. Let y be the displac... | Elementary Differential Equations with Boundary Value Problems_Page_175_Chunk2580 |
166 Chapter 4 Applications of First Order Equations y v Figure 4.4.3 Trajectories of my′′ + ky = 0 which implies that mv2 + ky2 = ρ (4.4.10) (ρ = 2c). This defines an ellipse in the Poincaré phase plane (Figure 4.4.3). We can identify ρ by setting t = 0 in (4.4.10); thus, ρ = mv2 0 + ky2 0, where y0 = y(0) and v0 = v(0)... | Elementary Differential Equations with Boundary Value Problems_Page_176_Chunk2581 |
Section 4.4 Autonomous Second Order Equations 167 y = R y = − R t y Figure 4.4.4 y = R sin(ω0t + φ) (see (4.4.11)), (4.4.12) implies that that there’s a constant φ such that sin−1 y R = ω0t + φ or y = R sin(ω0t + φ) for all t in I. Although we obtained this function by assuming that v > 0, you can easily verify tha... | Elementary Differential Equations with Boundary Value Problems_Page_177_Chunk2582 |
168 Chapter 4 Applications of First Order Equations m y L Figure 4.4.5 The undamped pendulum (a) Stable equilibrium (b) Unstable equilibrium Figure 4.4.6 (a) Stable equilibrium (b) Unstable equilibrium However, if n is odd, the mass is directly above the axle (Figure 4.4.6 (b)) and gravity increases any deviation from ... | Elementary Differential Equations with Boundary Value Problems_Page_178_Chunk2583 |
Section 4.4 Autonomous Second Order Equations 169 The curves defined by (4.4.15) are the trajectories of (4.4.13). They are periodic with period 2π in y, which isn’t surprising, since if y = y(t) is a solution of (4.4.13) then so is yn = y(t) + 2nπ for any integer n. Figure 4.4.7 shows trajectories over the interval [−π... | Elementary Differential Equations with Boundary Value Problems_Page_179_Chunk2584 |
170 Chapter 4 Applications of First Order Equations π − π x y Figure 4.4.7 Trajectories of the undamped pendulum Figure 4.4.8 The whirling undamped pendulum ymax − ymax Figure 4.4.9 The oscillating undamped pendulum Stability and Instability Conditions for y′′ + p(y) = 0 It can be shown (Exercise 23) that an equilibriu... | Elementary Differential Equations with Boundary Value Problems_Page_180_Chunk2585 |
Section 4.4 Autonomous Second Order Equations 171 It can also be shown (Exercise 24) that y is unstable if there’s a b > y such that p(y) < 0 if y < y < b (4.4.20) or an a < y such that p(y) > 0 if a < y < y. (4.4.21) If we regard p(y) as a force acting on a unit mass, (4.4.20) means that the force tends to increase al... | Elementary Differential Equations with Boundary Value Problems_Page_181_Chunk2586 |
172 Chapter 4 Applications of First Order Equations 1 y v (a) (b) (c) (b) Figure 4.4.10 Trajectories of y′′ + y(y −1) = 0 The separatrix contains four trajectories of (4.4.22). One is the point (0, 0), the trajectory of the equi- librium y = 0. Since distinct trajectories can’t intersect, the segments of the separatrix... | Elementary Differential Equations with Boundary Value Problems_Page_182_Chunk2587 |
Section 4.4 Autonomous Second Order Equations 173 and considering the direction fields for this equation. In the following examples we’ll also be showing computer generated trajectories of this equation, obtained by numerical methods. The exercises call for similar computations. The methods discussed in Chapter 3 are no... | Elementary Differential Equations with Boundary Value Problems_Page_183_Chunk2588 |
174 Chapter 4 Applications of First Order Equations −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 v y Figure 4.4.11 A typical direction field for my′′ + cy′ + ky = 0 with 0 < c < c1 −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 v y Figure 4.... | Elementary Differential Equations with Boundary Value Problems_Page_184_Chunk2589 |
Section 4.4 Autonomous Second Order Equations 175 c/mL in this equation.) To plot a direction field for (4.4.29) we write its phase plane equivalent as dv dy = −c mL −g Lv sin y. Figure 4.4.15 shows trajectories of four solutions of (4.4.29), all satisfying y(0) = 0. For each m = 0, 1, 2, 3, imparting the initial veloci... | Elementary Differential Equations with Boundary Value Problems_Page_185_Chunk2590 |
176 Chapter 4 Applications of First Order Equations 4.4 Exercises In Exercises 1–4 find the equations of the trajectories of the given undamped equation. Identify the equilibrium solutions, determine whether they are stable or unstable, and plot some trajectories. HINT: Use Eqn. (4.4.8) to obtain the equations of the tr... | Elementary Differential Equations with Boundary Value Problems_Page_186_Chunk2591 |
Section 4.4 Autonomous Second Order Equations 177 inside the limit cycle spiral outward to it, while trajectories outside the limit cycle spiral inward to it (Figure 4.4.16). Use your favorite differential equations software to verify this for µ = .5, 1.1.5, 2. Use a grid with −4 < y < 4 and −4 < v < 4. y v Figure 4.4.... | Elementary Differential Equations with Boundary Value Problems_Page_187_Chunk2592 |
178 Chapter 4 Applications of First Order Equations (d) Conclude from symmetry that the time required for (y(t), v(t)) to traverse the trajectory v2 = v2 0 −v2 c sin2 y/2 is T = 4T1, and that consequently y(t + T) = y(t) and v(t + T) = v(t); that is, the oscillation is periodic with period T. (e) Show that if v0 = vc, ... | Elementary Differential Equations with Boundary Value Problems_Page_188_Chunk2593 |
Section 4.5 Applications to Curves 179 4.5 APPLICATIONS TO CURVES One-Parameter Families of Curves We begin with two examples of families of curves generated by varying a parameter over a set of real numbers. Example 4.5.1 For each value of the parameter c, the equation y −cx2 = 0 (4.5.1) defines a curve in the xy-plane... | Elementary Differential Equations with Boundary Value Problems_Page_189_Chunk2594 |
180 Chapter 4 Applications of First Order Equations x y Figure 4.5.2 A family of lines defined by y = x + c x y Figure 4.5.3 A family of circles defined by x2 + y2 −c2 = 0 Equations (4.5.1) and (4.5.2) define one–parameter families of curves. (Although (4.5.2) isn’t in the form (4.5.3), it can be written in this form as y... | Elementary Differential Equations with Boundary Value Problems_Page_190_Chunk2595 |
Section 4.5 Applications to Curves 181 To find a differential equation for a one–parameter family we differentiate its defining equation (4.5.5) implicitly with respect to x, to obtain Hx(x, y, c) + Hy(x, y, c)y′ = 0. (4.5.7) If this equation doesn’t, then it’s a differential equation for the family. If it does contain c... | Elementary Differential Equations with Boundary Value Problems_Page_191_Chunk2596 |
182 Chapter 4 Applications of First Order Equations We must choose the plus sign in (4.5.11) if x < x0 and the minus sign if x > x0; thus, x0 = x + p x2 −y if x < x0 and x0 = x − p x2 −y if x > x0. Since y′ = 2x0, this implies that y′ = 2 x + p x2 −y , if x < x0 (4.5.12) and y′ = 2 x − p x2 −y , if x > ... | Elementary Differential Equations with Boundary Value Problems_Page_192_Chunk2597 |
Section 4.5 Applications to Curves 183 Letting x0 = 3 in (4.5.10) shows that (2, 3) is on the line y = −9 + 6x, which is tangent to the parabola at (x0, x2 0) = (3, 9), as shown in Figure 4.5.5 Letting x0 = 1 in (4.5.10) shows that (2, 3) is on the line y = −1 + 2x, which is tangent to the parabola at (x0, x2 0) = (1, ... | Elementary Differential Equations with Boundary Value Problems_Page_193_Chunk2598 |
184 Chapter 4 Applications of First Order Equations From Figure 4.5.6, P is the midpoint of the line segment connecting (xI, 0) and (0, yI) if and only if xI = 2x0 and yI = 2y(x0). Substituting the first of these conditions into (4.5.14) or the second into (4.5.15) yields y(x0) + y′(x0)x0 = 0. Since x0 is arbitrary we d... | Elementary Differential Equations with Boundary Value Problems_Page_194_Chunk2599 |
Section 4.5 Applications to Curves 185 x0 x0 2 x y Figure 4.5.8 x = 1 x y Figure 4.5.9 Therefor y′ y = − 1 x2 −x = − 1 x(x −1) = 1 x − 1 x −1, so ln|y| = ln|x| −ln|x −1| + k = ln x x −1 + k, and y = cx x −1. If c = 0, the graph of this function is the x-axis. If c ̸= 0, it’s a hyperbola with vertical asymptote x = 1 an... | Elementary Differential Equations with Boundary Value Problems_Page_195_Chunk2600 |
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