Differential Equations Variation of Params Problems

1. Mar 19, 2010

lubuntu

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

$$t^{2}$$ * y'' - 2y = $$3t^{2}$$ - 1

$$y_{1}$$ = t$$^{2}$$

$$y_{2}$$ = 1/t

2. Relevant equations

Variation of Params forumla
wronskian det

3. The attempt at a solution

W = t^2 * -t^-2 -[1/t * 2t] = -3

Y = -t^2 * Integral[ (-1/3)(1/t)(3t^2 -1)] + 1/t * Integral[(-1/3)(t^2)(3t^2 -1)]

I get an answer that is not correct so I guess I set this part up wrong somehow anyone see the error?

My answer: Y = 3/10*t^4 - t^2 * ln|t|/3

Correct Answer: Y = t^2*ln|t| + 1/2

Last edited: Mar 19, 2010
2. Mar 19, 2010

lubuntu

So I figured I ended up getting the correct answer but going the long way and solving the system using

u*t^2 + v * 1/t = 3t^2 -1

but I still have no idea why the integral gave me the wrong answer apparently I missed something somewhere, if you feel like going through please let me know what I messed up on.