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Unformatted text preview: + te t = N t , the equation is exact. We choose to integrate N over y instead of M over t (Why?). Then F = R Ndy + h ( t ) = ( te t + 2) y + h ( t ). Since F t = M , we obtain e t + te t + h ( t ) = e t + te t . Thus, we can set h = 0. Therefore, F = ( te t +2) y and ( te t +2) y = C is the solution. For t = 0 we have 2 y (0) = C or 2(1) = C . Thus, C =2. The ANSWER: y =2 te t + 2 . 1...
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This note was uploaded on 12/15/2011 for the course MAP 2302 taught by Professor Tuncer during the Spring '08 term at University of Florida.
 Spring '08
 TUNCER

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