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# m238sp06t1s - Dierential Equations test one solutions...

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Differential Equations test one solutions Tuesday, April 25, 2006 1 . Given the function y = 4+ x 2 sin s 2 ds ( a ) Calculate y (2) = 4+ 2 2 sin s 2 ds = 4+0 = 4 ( b ) Calculate dy dx = sin x 2 2 . Is the function y = x + 1 x a solution of dy dx + y 2 = 3+ x 2 ? Yes because: dy dx + y 2 = d dx x + 1 x + x + 1 x 2 = 1 - 1 x 2 + x 2 + 2 + 1 x 2 = 3 + x 2 3 . Determine whether each of the following equations is linear or non-linear. ( a ) dy dt = t - y 2 non-linear ( b ) t dy dt = e t y - cos t 2 linear ( c ) y dy dt = 4 t - y non-linear 4 . Which of the following equations is exact? ( a ) 3 x 2 - 4 y dx +(2 y - 4 x ) dy = 0 3 x 2 - 4 y y ? = (2 y - 4 x ) x - 4 = - 4 exact ( b ) dy dx + x 2 y = x or ( x 2 y - x ) dx +(1) dy = 0 x 2 y - x y ? = (1) x x 2 = 0 not exact

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page two 5 . Given the initial value problem t dy dt + 3 t - 2 y = 4 t + 1 and y (3) = 5 determine the largest t -interval over which a unique solution is guaranteed to exist. Rewrite the differential equation as dy dt + 3 t ( t - 2) y = 4 t ( t + 1) The coefficients are not continuous at t = 0 and t = 2. The solution to the initial value
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