Solving a second order nonhomogeneous differential equation $ x'' +4x' +13x = 360te^{-2t}\sin(3t) $ 
I need to find the solution to the following equation:
  $$x'' +4x' +13x = 360te^{-2t}\sin(3t).$$

So far I've found the roots of the characteristic equation which are a conjugate pair $\lambda_1 = -2-3i$ and $\lambda_2=-2+3i$.
Next I've concluded that $e^{-2t}\sin(3t) = \text{Im}(e^{(-2+3i)t})$ and I tried to find a solution in a form $(At^2+Bt+C)e^{(-2+3i)t}$ by plugging it into the initial equation and finding the coefficients but I don't know whether I'm on the right track or not.
I would greatly appreciate some help with solving this equation.
 A: You are on the right track. You may also let $C=0$, because $e^{(-2+3i)t} $ is a solution of the homogeneous ode, and try a particular solution of the form
$$(At^2+Bt)e^{(-2+3i)t}$$
where $A$ and $B$ are complex constants to be found. If you prefer to work with real constants try
$$e^{-2t}((at^2+bt)\cos(3t)+(ct^2+dt)\sin(3t)).$$
A: The complex version of the equation is
$$ z'' + 4z' + 13z = 360te^{(-2+3i)t} $$
in which $x(t)$ is the imaginary part of $z(t)$
The particular solution should have the form
$$z_p(t) = (At+B)te^{(-2+3i)t} = (At^2+Bt)e^{(-2+3i)t}$$
where all coefficients are complex. The factors break down as follows:


*

*$e^{(-2+3i)t}$ from the exponential factor

*$At+B$ from the linear factor

*An additional factor of $t$ due to the repeated root


Taking derivatives
\begin{align} 
{z_p}' &= \big[(-2+3i)At^2 + \big(2A + (-2+3i)B\big)t +B\big]e^{(-2+3i)t}
\\ 
{z_p}'' &= \big[(-2+3i)^2At^2 + \big(4(-2+3i)A + (-2+3i)^2B\big)t + \big(2A + 2(-2+3i)B\big)\big]e^{(-2+3i)t}
\end{align}
Plugging in
$$ {z_p}'' + 4{z_p}' + 13{z_p} = \big[12iAt + \big(2A + 6iB\big)\big]e^{(-2+3i)t} $$
This gives
\begin{align} 12iA &= 360 \\ 2A + 6iB &= 0 \end{align}
or $A = -30i$, $B = 10$. The full (particular) solution is
$$ z_p(t) = (-30it^2 +10t)e^{(-2+3i)t} $$
Taking the imaginary part, we obtain
$$ x_p(t) = e^{-2t}\big[30t^2\sin(3t) + 10t\cos(3t)\big] $$
