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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

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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 $\ds{A,B \in \mathbb{R}}$.

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