solution of second order ODEs I would like to know the exact solution $u(t,x)$ of the following ODEs
$\frac{\partial^2 u}{\partial ^2 t} + 2\alpha \frac{\partial u}{\partial t} + (\alpha^2 - c(t)^2)u = 0 $ where $\alpha$ be a constant and $c(t) = 1+ \epsilon \sin(\omega t)$ 
thanks to all !
 A: This is what maple gives as a result.
$$u \left( x,t \right) =\int \!{\frac {(-4\,\cos \left( 2\,\omega\,t
 \right) {\alpha}^{4}{\epsilon}^{2}u-\cos \left( 2\,\omega\,t \right) 
{\alpha}^{2}{\epsilon}^{2}{\omega}^{2}u-4\,\sin \left( 2\,\omega\,t
 \right) {\alpha}^{3}{\epsilon}^{2}\omega\,u-\sin \left( 2\,\omega\,t
 \right) \alpha\,{\epsilon}^{2}{\omega}^{3}u+16\,\sin \left( \omega\,t
 \right) {\alpha}^{4}\epsilon\,u+16\,\sin \left( \omega\,t \right) {
\alpha}^{2}\epsilon\,{\omega}^{2}u+16\,{{\rm e}^{-2\,\alpha\,t}}{\it 
\_C1}\,{\alpha}^{5}+20\,{{\rm e}^{-2\,\alpha\,t}}{\it \_C1}\,{\alpha}^
{3}{\omega}^{2}+4\,{{\rm e}^{-2\,\alpha\,t}}{\it \_C1}\,\alpha\,{
\omega}^{4}-8\,\cos \left( \omega\,t \right) {\alpha}^{3}\epsilon\,
\omega\,u-8\,\cos \left( \omega\,t \right) \alpha\,\epsilon\,{\omega}^
{3}u-8\,{\alpha}^{6}u+4\,{\alpha}^{4}{\epsilon}^{2}u-10\,{\alpha}^{4}{
\omega}^{2}u+5\,{\alpha}^{2}{\epsilon}^{2}{\omega}^{2}u-2\,{\alpha}^{2
}{\omega}^{4}u+{\epsilon}^{2}{\omega}^{4}u+8\,{\alpha}^{4}u+10\,{
\alpha}^{2}{\omega}^{2}u+2\,{\omega}^{4}u)}{16\,{\alpha}^{5}+20\,{
\alpha}^{3}{\omega}^{2}+4\,\alpha\,{\omega}^{4}}}\,{\rm d}t+{\it \_C2}
$$
A: Can you help me to find the exactly formular solution of $T"(t) = (1+\epsilon \sin(\omega t)^2) T(t)$ where $T(t)$ be one variable function depending only on $t$.
