In this problem you will use variation of parameters to solve the nonhomogeneous equation $$t^{2}y''+2ty'-2y = 2t^{2}-2t^{3}$$

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A. Plug $y=t^n$ into the associated homogeneous equation (with "$0$" instead of "$2t^{2}-2t^{3}$") to get an equation with only $t$ and $n$.

$\quad$

(Note: Do not cancel out the $t$, or webwork won't accept your answer!)

B. Solve the equation above for $n$ (use $t\neq 0$ to cancel out the $t$).
    You should get two values for $n$, which give two fundamental solutions of the form $y = t^n$.

$\quad \phantom{\mathrm{W}(y_1,\,)} y_1 =$

$\qquad y_2 =$

$\quad \mathrm{W}\bigl(y_1, y_2\bigr) =$

C. To use variation of parameters, the linear differential equation must be written in standard form $y'' + py' + qy = g$.
    What is the function $g$?

$\qquad g(t) =$

D. Compute the following integrals.

$\quad \displaystyle \int \frac{y_1 g}{W} dt =$

$\quad \displaystyle \int \frac{y_2 g}{W} dt =$

E. Write the general solution. (Use c1 and c2 for $c_1$ and $c_2$).

$\quad y =$

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If you don't get this in 3 tries, you can get a hint to help you find the fundamental solutions.

Hint: