prelim1_2 - Prelim 1 Problem 3 dy 1 = t y t2 dx t y(1 = 1...

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Unformatted text preview: Prelim 1: Problem 3 dy 1 = - t y + t2 dx t y(1) = 1 Solution dy 1 - - t y = t2 dx t dy 1 + t- y = t2 dx t The integrating factor is (x) = e =e = (t-1/t)dx t2 2 -ln t t2 2 e t Multiplying the given differential equation by the integrating factor, e t2 2 t2 t dy 1 e 2 + t- y dx t t e = t2 t2 2 t2 2 t d e y dx t e t2 2 t2 2 = te y t t2 2 = te t2 2 dx y e t =e t2 2 +C t2 2 y = t + Cte Using the initial condition y(1) = 1, - 1 = 1 + Ce- 2 C=0 Therefore, the solution to the given initial value problem is y(t) = t 1 1 ...
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This note was uploaded on 04/01/2008 for the course MATH 2930 taught by Professor Terrell,r during the Fall '07 term at Cornell.

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