Mid2Solution

Mid2Solution - Numerical Analysis in Engineering ME 140A,...

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Numerical Analysis in Engineering ME 140A, Fall 2008 Midterm #2 Solution By: Mohamad M. Nasr-Azadani mmnasr@engr.ucsb.edu December 3, 2008 Problem 1 Using separation of variables, the given ODE can be re-arranged as y 2 dy = 2 dx x . (1) Integrating both sides of equation (1) will give us Z y 2 dy = Z 2 dx x y 3 3 = 2 ln | x | + C (2) Here, C is a constant which can be found using the initial condition. Equation (2) gives y as a function of x implicitly. Finally, y is found as y = (6 ln | x | + 3 C ) 1 3 . (3) 1
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For the first order ODE with the form given by y 0 + p ( x ) = g ( x ) (4) the integrating factor μ ( x ) is found by μ ( x ) = e R p ( x ) dx . (5) Hence, we have μ ( x ) = e R 2 xdx = e x 2 (6) Multiplying μ ( x ) to the both sides of equation (4), we have d dx [ μy ] = g ( x ) μ ( x ) (7) Integrating both sides of equation (7) results in μ ( x ) y = Z g ( x ) μ ( x ) dx (8) Using equation (6), we have e x 2 y = Z xe x 2 dx = Z 1 2 (2 x ) e x 2 dx = 1 2 e x 2 + C
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This note was uploaded on 01/04/2011 for the course ME 140A taught by Professor Meiburg during the Spring '08 term at UCSB.

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Mid2Solution - Numerical Analysis in Engineering ME 140A,...

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