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Answer Guide 2 Math 427K: Unique Number 55160 Wednesday, September 7, 2011 In each of the °rst four exercises, use an integrating factor to °nd the general solution x ( t ) of the di/erential equation. 1. dx dt + 2 x t = t ° 1 ( t > 0) Solution: The general solution is: x ( t ) = Ct ° 2 + 1 4 t 2 ° 1 3 t: ( F ) 2. x 0 ° x tan t ° 2 sin t = 0 ( ° ° 2 < t < ° 2 ) Solution: The general solution is: x ( t ) = C ° cos(2 t ) 2 cos t : ( F ) 3. (1 + t 2 ) x 0 ° 2 tx = 3 + 3 t 2 Solution: The general solution is: x ( t ) = ( C + 3 arctan t )(1 + t 2 ) : ( F ) 4. x 0 ° x t ln t + 3 t = 0 ( t > 0) Solution: The general solution is: x ( t ) = j ln t j ( C ° 3 j ln j ln t jj ) : ( F ) Four the next four exercises, °nd the general solution and then solve the initial value problem. 5. dx dt = x t ( x ° 1) ; x (1) = 2 (An implicit solution is OK here.) Solution: The general solution in implicit form is t = C e x ( t ) x ( t ) : ( F ) The initial condition forces C = 2 e ° 2 .

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6. dx dt = t 3 ° x ; x (0) = ° 4 Solution: The general solution is x ( t ) = 3 ± p 9 ° A ° t 2 = 3 ± p B ° t 2 : ( F ) The initial condition makes us choose the negative branch and set B = 49 , yielding x ( t ) = 3 ° p 49 ° t 2 : ( FF ) 7. dx dt = t 2 (1 + x 2 ) ; x (0) = 0 (What is the interval of existence?) Solution: The general solution is x ( t ) = tan ° C + t 3 = 3 ± : ( F )
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