Solve With Variation of Parameters

[TranscendArcu]

Homework Statement

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

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?

 

[mighty2000]

Thank you for sharing your solution attempt with us. I believe the mistake in your solution lies in the calculation of the Wronskian. The correct Wronskian for the given solutions is W(t,t*e^t) = t^3e^t, which will change the rest of your calculations accordingly. After correcting this, I believe your particular solution will match the answer in the back of the book. Keep up the good work!