How to solve this second order DE? The equation is:
$$y'' -2y' -3y = -3t*e^{-t}$$
So I understand that the complementary solution is:
$$y_c = C_1e^{-t}+ C_2e^{3t}$$
And I'm pretty sure the general form of the particular solution is:
$$y_p= (A t + B)e^{-t}$$
It's either that or:
$$y_p = (A  t^2+B  t)  e^{-t}$$
I tried finding A and B with both options for $y_p$. With the second, I couldn't find a solution at all. With the first however, I did get $A = (3/2)  t$, which I found is wrong, according to the back of the book.
Can anyone help me find out what I've done wrong? I spent over an hour on this and I just can't find anything else to try.
Thanks.
 A: For problems like this, there's a way to avoid guesswork and solve the equation directly.  Unfortunately, that means integration by parts in this case.  I assume you've used integrating factors before?
$$y''-2y'-3y=(y'-3y)'+(y'-3y)=-3te^{-t}$$
Assuming that "*" was multiplication and not convolution or something.  In any case, that wouldn't change the method, only the final result.
$$e^t(y'-3y)'+e^t(y'-3y)=[e^t(y'-3y)]'=-3t$$
$$e^t(y'-3y)=-\frac32t^2+k_1$$
$$e^{-3t}(y'-3y)=(e^{-3t}y)'=-\frac32t^2e^{-4t}+k_1e^{-4t}$$
$$e^{-3t}y=\int-\frac32t^2e^{-4t}dt-\frac{k_1}4e^{-4t}+k_2$$
Use integration by parts twice to eliminate the $t^2$.  After that, the remaining integral will yield a constant times $e^{-4t}$ which can be combined with the other like term to produce a new constant times $e^{-4t}$.  Then multiply both sides by $e^{3t}$ and you're done.
A: Hint. A special solution would be of the form
$$
(at^2+bt)\mathrm{e}^{-t}.
$$
You need to add a power of $t$, since $\mathrm{e}^{-t}$ is a solution of the corresponding homogeneous equation.
General solution
$$
c_1\mathrm{e}^{-t}+c_2\mathrm{e}^{3t}+(at^2+bt)\mathrm{e}^{-t}.
$$
