Getting particular solution for harmonic oscillator I have $\ddot u + \omega^2\cdot u = \frac{K(t)}{m}$ whereas $u$ is the "distance" of a mass $m$.
Now I need to find a particular solution to it for the case that $K(t)=K_0$, so it is a constant. I somehow can't solve it.
My first confusion is about the exersice itself. Do they actually mean the general solution for the non-homogenic harm. osc. or "just" the particular solution?
Anyway, that's what I'd do:
First let's find the homogenic solution, I won't include much details since it's a really good known solution.
Ansatz: $u(t)=e^{\lambda t}$ so we get $\lambda_{1,2}=\pm i \omega$
Now: $u_h(t)=Ae^{\omega i t} + Be^{-\omega i t}$ 
or by using the addition theorem of cos/sin we can also write it down as:
$u_h(t)=C\cdot \cos(\omega t - \varphi)$
For getting the particular solution I'd use the homogenic solution as a ansatz but consider the integration constant as a function of t. 
Set $k:=\omega t -\varphi$
Ansatz: $u_p(t)=C(t)\cos(k)$
Now: $\dot u = C'\cos(k) - C\omega\sin(k)$
$\ddot u = C''\cos(k) - C'\omega\sin(k) - C'\omega\sin(k) - C\omega^2\cos(k)$
We put it inot the DGL and get:
$C''\omega \cos(k) - 2C'\omega\sin(k) - 2C\omega^2\cos(k)=\frac{K_0}{m}$
I see tha I can either get rid of cosinus or sinus by just choosing k accordingly. (we can do that, since $K_0=const$) but in both cases, I don't really get the desired result. 
Can someone please give me a hint on how to solve this? :)
The solution should be: $u_{sp}(t)=\frac{K_0}{m\omega^2}$ (no idea what sp stands for)

When the right hand side is a constant, try $u(t)=u_0$, then $\ddot{u}(t)=0$, so
$$\omega^2 u_0=\frac{K_0}{m}$$
implies
$$u_0=\frac{K_0}{m\omega^2}$$
That means $u(t)=\frac{K_0}{m\omega^2}$ solves the DE.
