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Unformatted text preview: Numerical Analysis in Engineering ME 140A, Fall 2007 Homework #6 Solution Dec 6, 2007 By: Mohamad NasrAzadani: mmnasr@engr.ucsb.edu 1. a) One can integrate dy dx = (1 + 2 x ) y, y (0) = 1 (1) using separation of variables technique. Toward this goal, we may write dy y = (1 + 2 x ) dx Z dy y = Z (1 + 2 x ) dx 2 y = x + x 2 + C, y (0) = 1 C = 2 y = 1 2 ( x + x 2 ) + 1 2 (2) b) Euler scheme is applied as following to solve equation (1), numeri cally y n +1 = y n + f ( x n , y n ) x (3) where f ( x, y ) is given by f ( x, y ) = (1 + 2 x ) y (4) One can see the results obtained with Euler scheme in figure 1 for vaious x s. 1 c) Heun (predictorcorrector) scheme (with no iteration) is implemented as following to solve equation (1), numerically y n +1 = y n + f ( x n , y n ) x y n +1 = y n + f ( x n , y n ) + f ( x n +1 , y n +1 ) 2 x. (5) One can see the results obtained with Heun scheme in figure 2 for vaious x s. d) Fourth order RungeKutta is implemented as following to solve equa tion (1), numerically k 1 = f ( x n , y n ) k 2 = f ( x n + 0 . 5 x, y n + 0 . 5 xk 1 ) k 3 = f ( x n + 0 . 5 x, y n + 0 . 5 xk 2 ) k 4 = f ( x n + x, y n + xk 3 ) y n +1 = y n + x 6 ( k 1 + 2 k 2 + 2 k 3 + k 4 ) (6) One can see the results obtained with 4th order RungeKutta method in figure 3 for vaious x s. In figure 4), the results for x = 0 . 025 obtained using the mentioned methods are plotted and comnpared with the analytical solutions. Also, in figure 5, relative error at y (1) is plotted vs x in logarithmic scale. The slope of each line would give us the order of accuracy of each numerical technique.each numerical technique....
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This note was uploaded on 11/05/2009 for the course ME 140A taught by Professor Meiburg during the Fall '08 term at UCSB.
 Fall '08
 Meiburg

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