Differential Equations (Undetermined Coefficients) For the non-homogeneous equation 
$$y" + y = 3\sin(2t) + t\cos(2t)$$
I tried to solve it by splitting it up, meaning I solved for $3\sin(2t)$ first, then I tried to solve for $t\cos(2t)$. 
The guess for $t$ would be $At + B$, and the guess for $\cos(2t)$ would be $A\sin(2t) + B\cos(2t)$, but how would I combine them together, since $t\cos(2t)$ is a product?
Also, in general, how do I combine guesses for products where the A/B/C/etc. overlap? For example, in my notes for a $g(t) = 4e^{-t}
\cos(2t))$ the guess was $e^{-t}(A\sin(2t) + B\cos(2t))$. What happened to the $A$ for the guess of $4e^{-t}$?
 A: You can try $(At+B)\sin{(2t)}+(Ct+D)\cos{(2t)}$ for both two terms. That will cover all cases. Remember you just want all possible terms that can give you $\sin{(2t)}$ and $t\cos{(2t)}$.
For the $e^{-t}$ one, the $A$ is absorbed into the two coefficients for $\sin$ and $\cos$.
A: Here is another viewpoint in case you find it strange what ansatz you should make for the particular solution. It uses the form of the functions on the right-hand-side.
You have the functions $u_1=3\sin 2t$ and $u_2=t\cos 2t$. The first one satisfies $(D^2+4)u_1=0$ (here, and below $D$ stands for differentiation) and the second one $(D^2+4)^2u_2=0$. Thus, if $y$ is supposed to be a solution to the original differential equation
$$
(D^2+1)y=u_1+u_2,
$$
it should also satisfy
$$
(D^2+4)^2(D^2+1)y=(D^2+4)^2(u_1+u_2)=0.
$$
Thus, $y$ should be a solution to a homoegeneous differential equation. Now, all solutions to this homogeneous differential equation are given by (I get the picture that you know of those)
$$
y=(a+bt)\cos 2t+(c+dt)\sin 2t+e\cos t+f\sin t.
$$
The part $e\cos t+f\sin t$ is already in your solution to the original homogeneous differential equation, so that can be omitted. We draw the conclusion that the ansatz for particular solution that you should do is
$$
y=(a+bt)\cos 2t+(c+dt)\sin 2t.
$$
If you understand what is written above, you will most likely be able to do the correct ansatz in all cases occuring in calculus courses. Also, when you get the hold of it, you don't have to do all the intermediate calculations, but just remember what types of function occur depending on the solution to the homogeneous differential equation and the form of the right-hand side.
