Toronto Math Forum

\begin{equation}
t^{2}y" - 2y = t^{4}e^{t},    t>0
\end{equation}

Solve Euler for homogeneous, and find the two solutions to the system:

\begin{equation}
\begin{split}
x = \ln(t)\\
y"- y'- 2y = 0\\
r^{2}-r-2 = 0\\
r_1 = -1,  r_2 = 2\\
\end{split}
\end{equation}
\begin{equation}
\begin{split}
y_1 = e^{-x} = t^{-1}\\
y_2 = e^{2x} = t^2
\end{split}
\end{equation}

Then, using the formula for variation of parameters, you get:
\begin{equation}
y_p(t) = -\frac {t^{2}}{3} \int{te^{t}dt} + \frac{t^{-1}}{3} \int{t^{4}e^{t}dt}
\end{equation}
\begin{equation}
y_p(t) = -\frac{t^{2}e^t}{3}(t-1) + \frac{t^{-1}e^{t}}{3}(t^{4} - 4t^{3} + 12t^{2} - 24t + 24)
\end{equation}
Thus, the general solution is given by:
\begin{equation}
Y(t) = c_{1}t^{-1} + c_2t^{2} + y_{p}(t)
\end{equation}