Find the general solution of the nonhomogeneous differential equation I have the differential equation 
\begin{align}
y'' + 9y = t^2e^{3t} + 6
\end{align}
I found the complementary equation which is
\begin{align}
c_1\cos(3t) + c_2\sin(3t).
\end{align}
I have no idea how to go about getting the particular solution.  Do I split the equations up?  Can someone point me in the right direction?  Thanks in advance
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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Define $\ds{z \equiv y' + 3y\ic \implies z' \equiv y'' + 3y'\ic =
y'' + 3\ic\pars{z - 3y\ic} = y'' + 3\ic z + 9y}$. So, I'll have

\begin{align}
y'' + 9y = z' - 3\ic z = t^{2}\expo{3t} + 6 \quad\mbox{and}\quad
y = {1 \over 3}\,\Im\pars{z}
\end{align}

\begin{align}
\totald{\pars{\expo{-3\ic t}z}}{t} & =
\expo{-3\ic t}\pars{t^{2}\expo{3t} + 6}
\\[5mm]
\implies 
\expo{-3\ic t}z & =
\int\expo{-3\ic t}\pars{t^{2}\expo{3t} + 6}\,\dd t + C\quad\pars{~C:\ Complex\ Constant~}
\\[5mm]
\implies
z & =
{1 \over 2}\,\ic +
\expo{3t}\,{-1 + \ic - 6\ic t + \pars{9 + 9\ic}t^{2} \over 54} +
C\expo{3\ic t}
\end{align}

$$\bbx{\ds{%
y\pars{t} =
{1 \over 6} +
\expo{3t}\,{1 - 6t + 9t^{2} \over 162} +
{1 \over 3}\Im\pars{C\expo{3\ic t}}}}
$$


Note that
  $\ds{{1 \over 3}\Im\pars{C\expo{3\ic t}} = A\sin\pars{3t} + B\cos\pars{3t}}$ with $\ds{A,B \in \mathbb{R}}$.

