The differential equation is $\frac{d^2y}{dt^2} - \frac{dy}{dt} -6y = e^{3t} - 3t^2$. I first found the homogeneous solution which I got as $$y(t) = c_1e^{-2t} + c_2e^{3t}$$ I am trying to figure out what the non-homogeneous part would look like. Looking online I assumed the form would look something like $$h(t) = Ate^{3t} + Bt^2 + Ct$$ But, upon working this part out I end up with the equation $$e^{3t} - 3t^2 = 5Ae^{3t} - 6Bt^2 - 2Bt - 6Ct + 2B - C$$ which leads to an inconsistent system
What would the form look like? Also, is the any way to quickly determine what the non-homogeneous part would look like?
As you have determined, the solution to the homogeneous equation is
$$y_h(t) = c_1 e^{−2t} + c_2 e^{3t}$$
Then we need to find the particular solution $y_p(t)$. We can guess that the particular solution looks like:
$$y_p(t) = Ate^{3t} + at^2 + bt +c$$
Plug this guess expression in the DE, we get:
$$y'' - y' - 6y$$
$$= (A(9t+6)e^{3t} + 2a) - (A(3t+1)e^{3t}+2at+b) - 6(Ate^{3t} + at^2 + bt +c)$$
$$= 5Ae^{3t} - 6at^2 - (6b + 2a)t + (2a - b -6c) \cdots (1)$$
Let $(1) = e^{3t} - 3t^2$, we get:
$$5A = 1$$
$$−6a = -3$$
$$−(6b+2a) = 0$$
$$2a−b−6c$$
These equations are easy to solve. So I will leave you with this.