General solution of $\ddot{y} + 4\omega^2y = 2\omega gt\sin{\lambda} \equiv ct$ In solving a problem involving differential equations, I come across the following:

$$\ddot{y} + 4\omega^2y = 2\omega gt\sin{\lambda} \equiv ct$$
The general solution is the general solution of the homogeneous equation and one particular solution of the inhomogeneous equation, i.e.,
$$y = \frac{c}{4\omega^2}t + A\sin{2\omega t} + B\cos{2\omega t}$$

I'm at a loss as to how it got to $y$. I can only think of the following:
$$\dot y + 4\omega^2yt = \omega gt^2\sin{\lambda} + C$$
which of course is nowhere near what I read. I'd appreciate if someone can point me in the right direction.
 A: When they write $\omega gt \sin \lambda\equiv ct$ they mean to quickly introduce the substitution $c:=\omega g \sin\lambda$. Note furthermore that differentiating any polynomial twice will reduce its degree by two, so twice differentiating a linear map will send it to $0$ (i.e. annihilate, or kill, it). Hence if we introduce for our particular solution a linear map (as the RHS is just a linear map), we can simply choose $y=ct / (4\omega^2)$. This way
$$(0)+4\omega^2\left(\frac{ct}{4\omega^2}\right)=ct.$$
The homogeneous part to the solution will solve $\ddot{y}+4\omega^2y=0$. The characteristic equation has solutions $r=\pm 2\omega i$, so the general solution will be of the form
$$\frac{ct}{4\omega^2}+\alpha e^{2\omega i}+\beta e^{-2\omega i}$$
or, after a change of variables,
$$\frac{ct}{4\omega^2}+A\sin2\omega t+B\cos2\omega t.$$
Edit: Apparently your scratchwork indicates you believed one could simply integrate $y$ to get $yt+C$. I  sternly agree with joriki's comment: this is false and it's going to be difficult to work with differential equations if you don't have the more elementary calculus fully understood yet.
