hw2-1 - Homework #2 Section 1.4 t 1. y = , y (0) = 1 y (a)...

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Homework #2 Section 1.4 1. y 0 = t y , y (0) = 1 (a) h = 0 . 1 n t n = t n - 1 + h y n = y n - 1 + hf ( t n - 1 ,y n - 1 ) f ( t n ,y n ) 0 0 1 0 1 0.1 1 0.1 2 0.2 1.01 0.1980 3 0.3 1.0298 * The approximations at t = 0 . 1, t = 0 . 2,and t = 0 . 3 are y 1 (0 . 1) 1, y 2 (0 . 2) 1 . 01, and y 3 (0 . 3) 1 . 0298. (b) h = 0 . 05 n t n = t n - 1 + h y n = y n - 1 + hf ( t n - 1 ,y n - 1 ) f ( t n ,y n ) 0 0 1 0 1 0.05 1 0.05 2 0.1 1.0025 0.0998 3 0.15 1.0075 0.1489 4 0.2 1.0149 0.1971 5 0.25 1.0248 0.2440 6 0.3 1.03698 * The approximations at t = 0 . 1, t = 0 . 2,and t = 0 . 3 are y 2 (0 . 1) 1 . 0025, y 4 (0 . 2) 1 . 0149, and y 6 (0 . 3) 1 . 03698. (c) The analytic solution is determined by separation of variables. dy dt = t y Z ydy = Z tdt 1 2 y 2 = 1 2 t 2 + c y 2 = t 2 + 2 c y = ± t 2 + 2 c Using the initial condition, y (0) = 1 1 = ± 2 c c = 1 2 1
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y = t 2 + 1 and to four decimal place accuracy, y (0 . 2) = 1 . 0198. The errors in Euler’s ap- proximation are h = 0 . 1 error = y (0 . 2) - y 2 (0 . 2) = 1 . 0198 - 1 . 01 = 0 . 0098 h = 0 . 05 error = y (0 . 2) - y 4 (0 . 2) = 1 . 0198 - 1 . 0149 = 0 . 0050 . The smaller stepsize gives smaller error. 3. y 0 = 3 t 2 - y , y (0) = 1, [0 , 1] h = 0 . 1 t n y n t n y n 0 1 0.6 0.6822 0.1 0.9 0.7 0.7220 0.2 0.813 0.8 0.7968 0.3 0.7437 0.9 0.9091 0.4 0.6963 1 1.0612 0.5 0.6747 h = 0 . 2 t n y n 0 1 0.2 0.8 0.4 0.664 0.6 0.6272 0.8 0.71776 1 0.95821 The differential equation is not separable, so we have no exact solution for comparison. 8. y 0 = - t y , y (0) = 1 h = 0 . 1 t n y n t n y n 0 1 0.6 0.8405 0.1 1 0.7 0.7691 0.2 0.99 0.8 0.6781 0.3 0.9698 0.9 0.5601 0.4 0.9389 1 0.3994 0.5 0.8963 The analytical solution is similar to problem 1, and is given by y
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This note was uploaded on 03/23/2010 for the course APPM 2360 taught by Professor Williamheuett during the Spring '07 term at Colorado.

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hw2-1 - Homework #2 Section 1.4 t 1. y = , y (0) = 1 y (a)...

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