# exam1 - Exam I This is not the original form but the...

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Unformatted text preview: Exam I This is not the original form, but the questions are the same. 1. Solve the initial-value problem: y + y = 1 + e x , y (0) = 5 . We use the method of integrating factors. Recall μ = exp R p ( x ) dx , so here μ = e x . Multiplying both sides by μ we have e x y + e x y = e x + e 2 x . We recognize the left-hand side as d dx ( e x y ). Thus d dx ( e x y ) = e x + e 2 x . Integrating both sides, we find e x y = e x + 1 2 e 2 x + C. (1) We plug in our initial condition and find 5 = 1 + 1 2 + C, so C = 7 2 . Plugging this into (1) and dividing both sides by e x , we conclude y = 1 + 1 2 e x + 7 2 e- x . 2. Find the general solutions of dy dx + y x = g ( x ) where g ( x ) = , for x < 1 1 , for x ≥ 1 . We solve problems like this in two steps: For x < 1: Here dy/dx + y/x = 0. Using an integrating factor of μ = x , we see d dx ( xy ) = 0 and hence y = C/x here. For x ≥ 1: Here dy/dx + y/x = 1. Again, μ = x , so now d dx ( xy ) = x . Thus xy = x 2 2 + C 2 , so y = x 2 + C 2 x ....
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## This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.

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exam1 - Exam I This is not the original form but the...

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