I'm interested in the following equation
$$\ddot{f} - \cos^2(\omega t) \dot{f} - f = 0$$
with an initial condition $f(0) = f_0$. I don't think there is a closed-form solution, but do you know if this equation has been studied somewhere, or where I should look about it?
Hint:
Let $r=\tan \omega t$ ,
Then $\dfrac{df}{dt}=\dfrac{df}{dr}\dfrac{dr}{dt}=\omega(\sec^2\omega t)\dfrac{df}{dr}$
$\dfrac{d^2f}{dt^2}=\dfrac{d}{dt}\left(\omega(\sec^2\omega t)\dfrac{df}{dr}\right)=\omega(\sec^2\omega t)\dfrac{d}{dt}\left(\dfrac{df}{dr}\right)+2\omega^2(\sec^2\omega t\tan \omega t)\dfrac{df}{dr}=\omega(\sec^2\omega t)\dfrac{d}{dr}\left(\dfrac{df}{dr}\right)\dfrac{dr}{dt}+2\omega^2(\sec^2\omega t\tan \omega t)\dfrac{df}{dr}=\omega(\sec^2\omega t)\dfrac{d^2f}{dr^2}\omega(\sec^2\omega t)+2\omega^2(\sec^2\omega t\tan \omega t)\dfrac{df}{dr}=\omega^2(\sec^4\omega t)\dfrac{d^2f}{dr^2}+2\omega^2(\sec^2\omega t\tan \omega t)\dfrac{df}{dr}$
$\therefore\omega^2(\sec^4\omega t)\dfrac{d^2f}{dr^2}+2\omega^2(\sec^2\omega t\tan \omega t)\dfrac{df}{dr}-\omega\dfrac{df}{dr}-f=0$
$\omega^2(r^2+1)^2\dfrac{d^2f}{dr^2}+(2\omega^2r(r^2+1)-\omega)\dfrac{df}{dr}-f=0$
$\dfrac{d^2f}{dr^2}+\left(\dfrac{2r}{r^2+1}-\dfrac{1}{\omega(r^2+1)^2}\right)\dfrac{df}{dr}-\dfrac{f}{\omega^2(r^2+1)^2}=0$