I'm having trouble on this problem. I did it and received an incorrect answer, so I tried pluggin it into Wolframalpha.
Wolframalpha:
$y = c_1 e^{-3t} + c_2 + 72t^2-48t + 5sin(3t) - 5cos(3t)$
I successful got the homogenous part
$c_1 e^{-3t} + c_2$
When I did the particular side I received the sin and cos, but I'm not sure where the $72t^2 $and $-48t$ came from.
Any help would be appreciated. I have a final in 2 weeks and I'm trying to understand this.
Are you sure you plugged it in correctly? Wolfram alpha gave me this: $$y=c_1e^{-3t}+c_2-3\sin(3t)-13\cos(3t)$$
Either way, if you proceed by the method of undetermined coefficients, the particular solution you seek will have the form $$y_p = A\sin(3t)+B\cos(3t)$$ Plugging this into the ODE (verification is left to you) leaves you with $$(-9A-9B)\sin(3t) + (9A-9B)\cos(3t) = 144\sin(3t) + 90\cos(3t)$$ and hence $$\left\{\begin{aligned} -9A-9B &= 144\\ 9A-9B &= 90 \end{aligned}\right.$$ Solving this system should give you the coefficients for the desired particular solution.