128ahw11sum10 - MATH 128A, SUMMER 2010, HOMEWORK 11...

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Unformatted text preview: MATH 128A, SUMMER 2010, HOMEWORK 11 SOLUTION BENJAMIN JOHNSON Homework 11: Due Monday, August 2 5.9; 4a 5.10; 1, 2a 5.11; 1a section 5.9 4. Use the Runge-Kutta for Systems Algorithm to approximate the solutions of the following higher-order differential equations, and compare the results to the actual solutions. a. y 00- 3 y + 2 y = 6 e- t , 0 t 1, y (0) = y (0) = 2, with h = 0 . 1 Solution: Let u 1 ( t ) = y ( t ) and u 2 ( t ) = y ( t ). Then u 1 ( t ) = y ( t ) = u 2 ( t ) and u 2 ( t ) = y 00 ( t ) = 6 e- t + 3 y- 2 y = 6 e- t + 3 u 2- 2 u 1 . We also have u 1 (0) = y (0) = 2 and u 2 (0) = y (0) = 2. The system of first-order IVPs is thus: u 1 = u 2 , u 2 = 6 e- t + 3 u 2- 2 u 1 , t 1 , u 1 (0) = u 2 (0) = 2 . Using the online Java Interface for the Runge-Kutta for Systems Algorithm at http://www.as.ysu.edu/~faires/Numerical-Analysis/DiskMaterial/... programs/Java/Algo57.htm with N = 10, I got the following: Runge-Kutta Method for Systems with m = 2. T W1 W2 0.000000000 2.000000000 2.000000000 0.100000000 2.242468099 2.875595247 0.200000000 2.580967377 3.927146008 0.300000000 3.035177986 5.197761232 0.400000000 3.629545279 6.739956583 0.500000000 4.394323086 8.617777272 0.600000000 5.366851934 10.909389903 0.700000000 6.593124154 13.710247574 0.800000000 8.129699354 17.1369555360....
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128ahw11sum10 - MATH 128A, SUMMER 2010, HOMEWORK 11...

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