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# sol2 - te t = ∂N ∂t the equation is exact We choose to...

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MAP 2302 Fall-2011 Section 0100 Quiz 2 1. Solve the equation 2 txdx + ( t 2 - x 2 ) dt = 0 SOLUTION: This is a homogeneous equation. The substitution v = x t turns it into t dv dt + v = v 2 - 1 2 v which is a separable equation dv dt = - v 2 +1 2 v 1 t . Separate the variables to obtain - 2 v v 2 +1 dv = dt t . Integrate (using substitution u = v 2 + 1) to obtain - ln( v 2 + 1) = ln | t | + C . This means that 1 v 2 +1 = At or t 2 x 2 + t 2 = At . Solving it for x we obtain the ANSWER x = ± Kt - t 2 where K is a constant. 2. Solve the initial value problem ( e t y + te t y ) dt + ( te t + 2) dy = 0 , y (0) = - 1 SOLUTION: Let M = e t y + te t y and N = te t + 2. Since ∂M ∂y
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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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