Q5B_230

# Q5B_230 - Math 415 Quiz 5B The dierential equation t2 y 5ty...

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Math 415 Quiz 5B The differential equation t 2 y 00 + 5 ty 0 + 3 y = 0 has y 1 ( t ) = t - 1 as one solution. Use the method of reduction of order to find a second solution. Since y 1 ( t ) = t - 1 is a solution, we assume that y 2 is of the form v ( t ) y 1 ( t ) for some unknown function v ( t ). Plugging this into the differential equation gives us 0 = t 2 ( v 00 y 1 + 2 v 0 y 0 1 + vy 00 1 ) + 5 t ( v 0 y 1 + vy 0 1 ) + 3 vy 1 = t 2 ( 1 t v 00 - 2 t 2 v 0 + 2 t 3 v ) + 5 t ( 1 t v 0 - 1 t 2 v ) + 3 t v = tv 00 + 3 v 0 . So we have a first-order linear differential equation in v 0 . For clarity, set w ( t ) = v 0 ( t ). Then we have the differential equation tw 0 + 3 w = 0, or in the normalized form w 0 + 3 t w = 0. This has integrating factor μ ( t ) = t 3 , i.e. t 3 w 0 + 3 t 2 w = ( t 3 w ) 0 = 0, and so w ( t ) = D t 3 . Which means that v ( t ) = R w ( t ) dt = C 1 t 4 + C 2 . Going back to our original assumption, we see that y 2 ( t ) = v ( t ) y 1 ( t ) = C 1 t 3 + C 2 t . Since we only need one solution, which together with y 1 forms a linearly independent pair, we can say that y 2 ( t ) = t - 3 . Find the solution of the initial value problem: y 00 + y 0 - 6 y = - 9 te - t , y (0) = 0 , y 0 (0) = 1 . The homogeneous equation y 00 + y 0 - 6

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