# Solve With Variation of Parameters

1. Apr 16, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

Find the particular solution to $t^2 y'' - t(t + 2)y' + (t+2)y = 2t^3$ given that y1 = t and y2 = tet are solutions. Also, require that t > 0

3. The attempt at a solution

Rewrite the original equation as $y'' - ((t + 2)/t)y' + ((t+2)/t^2)y = 2t$

So first I calculate the Wronskian: $W(t,t*e^t) = t^2e^t$. Thus, I have that
$Y = -t \int \frac{t*e^t * 2t}{t^2e^t} dt + t*e^t \int \frac{2t^2}{t^2*e^t}dt = -t \int 2 dt + t*e^t \int 2*e^{-t} dt = -2t^2 - 2t$, which I think is the particular solution.

However, the answer in the back of the book has no -2t term, so where have I gone wrong?