Use Method of Undetermined Coefficients to find general solution The problem is $u''-w_{o}^{2}u=\cos(w_{o}t)$.
The only thing I am having trouble with is what to have $u(t)$ to be. I tried a linear combination of sin and cos but that didn't work so I'm having a hard time especially since you have to have this to find the particular solution. 
 A: Because you didn't have trouble with the rest, I'm assuming you know that the general solution will be of the form
$$ u = c_1 e^{w_0 t} + c_2 e^{-w_0 t} + U(t). $$
So, we'll use the method of undetermined coefficients. Let $U(t) = A\cos(w_0 t) + B\sin(w_0 t)$.  Taking the second derivative yields $U''(t) = -w_0^2 A \cos(w_0 t) - w_0^2 B\sin(w_0 t)$.
Plugging $U(t)$ in as $u$ and $U''(t)$ as $u''$, we get $\left(-w_0^2 A \cos(w_0 t) - w_0^2 B\sin(w_0 t)\right) - w_0^2 \left(A\cos(w_0 t) + B\sin(w_0 t)\right) = \cos(w_0 t)$.
This simplifies to $-2w_0^2A\cos(w_0 t)-2w_0^2B\sin(w_0 t) = \cos(w_0 t)$.
From this we get our system of equations that will give us the coefficients $A$ and $B$:
$$
\begin{align}
-2w_0^2 A - 0 B &= 1 \\
0A -2w_0^2 B &= 0. \\
\end{align}
$$
You can take it from here to get the values of $A$ and $B$, and then plug the new $U(t)$ into your general solution.
A: Outline Solve $f'' - w_0^2 f = 0$ first. 
Then find out a function $g$ which satisfies $g'' - w_0^2 g = \cos(w_0 t)$. The answer is of the form $f + g$. 
To find out such a function $g$, assume $g(t) = a \sin(w_0 t) + b \cos(w_0 t)$ and solve for $a$ and $b$ for example.
