Quiz5BmodifiedSolution

# Quiz5BmodifiedSolution - r 2 r-2 = 0 i.e roots are-2 and 1...

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MATH 415.01 QUIZ 5B Feb 10, 2010 SHOW ALL YOUR WORK! GOOD LUCK! NAME: 1. (6.5 pts.) Use the method of reduction of order to ﬁnd a second solution to the following diﬀerential equation. t 2 y 00 + 3 ty 0 + y = 0 , t > 0; y 1 ( t ) = t - 1 Let y = v ( t ) y 1 ( t ) = t - 1 v ( t ). Then, y 0 = t - 1 v 0 - t - 2 v and y 00 = - t - 2 v 0 + t - 1 v 00 + 2 t - 3 v - t - 2 v 0 = t - 1 v 00 - 2 t - 2 v 0 + 2 t - 3 v Plug these into the equation above. t 2 ( t - 1 v 00 - 2 t - 2 v 0 + 2 t - 3 v ) + 3 t ( t - 1 v 0 - t - 2 v ) + t - 1 v = 0 i.e. tv 00 - 2 v 0 + 2 t - 1 v + 3 v 0 - 3 t - 1 v + t - 1 v = 0 which is equal to tv 00 + v 0 = 0.If you let u = v 0 . equation will become t du dt = - u . This is separable and can be written as du u = - dt t . Solve for u here. ln u = - ln t i.e. u = t - 1 . Then ,solve for v : R v 0 = R t - 1 = ln t . Hence y = t - 1 ln t 2. (6 pts.) Find the solution of the given initial value problem below. y 00 + y 0 - 2 y = 2 t, y (0) = 0 , y 0 (0) = 1 Charactheristic equation:
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Unformatted text preview: r 2 + r-2 = 0 i.e. roots are -2 and 1. Then, solution of the homogenous part is y = c 1 e-2 t + c 2 e t Method of undetermined coeﬃcients: Y = At + B , Y = A , Y 00 = 0. Plug these in the equation above, A-2( At + B ) = 2 t i.e. A-2 B-2 At = 2 t . So, A =-1 and B =-1 / 2. General solution becomes y = c 1 e-2 t + c 2 e t-t-1 / 2 . Now using the initial conditions, y (0) = c 1 + c 2-1 / 2 = 0 and y (0) =-2 c 1 + c 2-1 = 1 we get c 1 =-1 / 2 and c 2 = 1. y =-1 / 2 e-2 t + e t-t-1 / 2 ....
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## This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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