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mt2_solution

# mt2_solution - Numerical Analysis in Engineering ME 140A...

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Numerical Analysis in Engineering ME 140A, Fall 2007 Midterm #2 solutions Problem 1. Solve through separation of variables: dy dx = x 2 1 + y 2 (1) Separate the variables and integrate (1 + y 2 ) dy = x 2 dx y + y 3 3 = x 3 3 + c where c is a constant of integration that has to be determined from initial conditions. Note that the above equation is a cubic equation in y . Problem 2. Solve by finding an integrating factor: xy + 2 y = sin x, x > 0 (2) Hint: (sin x - x cos x ) = x sin x . Solution : One can show that (see class notes) a first order ODE in the form of y + p ( x ) y = g ( x ) has an integrating factor defined by μ ( x ) = e p ( x ) dx . (3) For the given ODE, dividing both sides by x , gives us y + 2 x y = sin x x (4) 1

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which is in the form that allows us to use the integrating factor method. Therefore, μ ( x ) is found as μ ( x ) = e 2 x dx = e 2 ln x = e ln ( x 2 ) = x 2 . (5) Multiplying μ ( x ) to both sides of equation (4) will give us x 2 y + 2 xy = x sin x d dx x 2 y = x sin x d x 2 y = x sin xdx (6) Now, one can easily integrate both sides of equation (6) to find d x 2 y = x sin xdx x 2 y = C +
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mt2_solution - Numerical Analysis in Engineering ME 140A...

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