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Unformatted text preview: Chapter 3
k3 = hf k2 h x + ,y + 2 2 = 0.005 k2 y+ 2
2 − 2e(x+h/2) y + k2 2 + e2(x+h/2) + e(x+h/2) , k4 = hf (x + h, y + k3 ) = 0.005 (y + k3 )2 − 2e(x+h) (y + k3 ) + e2(x+h) + e(x+h) . Hence in Step 4 we have x = x + 0.005 , 1 y = y + (k 1 + 2 k 2 + 2 k 3 + k 4 ) . 6 Using the initial conditions x0 = 0, y0 = 3, c = 2, and N = 400 on Step 3 we obtain k1 = 0.005 32 − 2e0 (3) + e2(0) + e0 = 0.025 , k2 = 0.005 (3 + 0.0125)2 − 2e(0+0.0025) (3 + 0.0125) + e2(0+0.0025) + e(0+0.0025) ≈ 0.02522 , k3 = 0.005 (3 + 0.01261)2 − 2e(0+0.0025) (3 + 0.01261) + e2(0+0.0025) + e(0+0.0025) ≈ 0.02522 , k4 = 0.005 (3 + 0.02522)2 − 2e(0+0.0025) (3 + 0.02522) + e2(0+0.0025) + e(0+0.0025) ≈ 0.02543 . Thus, Step 4 yields x = 0 + 0.005 = 0.005 and y ≈3+ 1 (0.025 + 2(0.02522) + 2(0.02522) + 0.02543) ≈ 3.02522 . 6 Thus the approximate value at x = 0.005 is 3.02522. By repeating Steps 3 and 4 of the subroutine we ﬁnd that, at x = 0.505, y = 2.0201 · 1013 . The next iteration gives a ﬂoating point overﬂow. This would lead one to think the asymptote occurs at x = 0.51 . As a check lets apply the 4th order RungeKutta subroutine with the initial conditions x0 = 0, y0 = 3, c = 1, and N = 400. This gives a ﬁner step size of h = 0.0025. With these inputs, we ﬁnd y (0.5025) ≈ 4.0402 · 1013 . Repeating the subroutine one more time with a step size of 0.00125, we obtain the value y (0.50125) ≈ 8.0804 · 1013 . Therefore we conclude that the vertical asymptote occurs at x = 0.50 and not at 0.51 as was earlier thought. 160 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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