general solution to a Non homogeneous differential equation Determine the general solution to the following differential equation:
$$y'' + 9y = t^2e^{3t}$$
The homogeneous solution is 
$$y(t)=C_1\cos(3t) +C_2\sin(3t)$$
Solving the non homogeneous solution I let 
$$y(t)=(At^2+Bt+C)e^{3t}$$
I found $A=1/36$, $B=-5/216$, $C=7/864$
Does anybody know if this is correct?

$y" + 9y = t^2e^{3t}$
$(D^2 + 9)y = t^2e^{3t}$
Auxiliary equation: $m^2 + 9 = 0$ or $m = \pm3i$
$CF = C_1\cos(3t) + C_2\sin(3t)$
$PI = \frac{1}{D^2 + 9}t^2e^{3t}$
Isolating $e^{3t}$ and replacing $D$ by $D+3 $ (coeff. of t)
$PI = e^{3t}\frac{1}{(D+3)^2 + 9}t^2$
$PI = e^{3t}\frac{1}{D^2+6D+18}t^2$
$PI = e^{3t}\frac{1}{18}\frac{1}{1 + \frac{D^2+6D}{18}}t^2$
$PI = \frac{e^{3t}}{18} \bigg[1-\frac{D^2+6D}{18}+ (\frac{D^2+6D}{18})^2 \cdots\bigg]t^2$
$\bigg[D(t^2) = 2t \ , \ D^2(t^2) = 2 \ , \ D^3(t^2) = D^4(t^2) = \cdots = 0\bigg]$
$PI = \frac{e^{3t}}{18} \bigg[1-\frac{D^2+6D}{18}+ \frac{D^4+12D^3 + 36D^2}{18^2} \cdots\bigg]t^2$
$PI = \frac{e^{3t}}{18} \bigg[t^2 - \frac{2+6(2t)}{18} + \frac{0 + 0 + 36(2)}{18^2}\bigg]$
$PI = \frac{e^{3t}}{18}[t^2 - \frac{2t}{3} + \frac{1}{9}]$
$$PI = e^{3t}\bigg[\frac{t^2}{18} - \frac{t}{27} + \frac{1}{162}\bigg]$$
Thus, $$A = \frac{1}{18} \ , \ B = -\frac{1}{27} \ , \ C = \frac{1}{162}$$ 
