Solution to $x''(t)+a(t)x(t)=0$? Consider everything one dimensional, i.e. I want to find a solution $x:\mathbb{R}\to\mathbb{R}$ to the ODE above with $a:\mathbb{R}\to\mathbb{R}$ at least continuous (you can assume it bounded in case, anyway I'm interested mainly in the solution on compact intervals $[0,T]$ rather than the long time behaviour). Is there a nice way to write the solution explicitly, at least under reasonable assumptions on $a$?
The only case in which I know this can be done it's when $a$ is constant, and in that case the behaviour is radically different depending on the sign of $a$; in general I would prefer not to impose a condition like $a\geq 0$ or $a\leq 0$ for all $t$, but if that helps also a formula in that particular case would be very appreciated!
 A: Your equation is essentially not simpler than the general second-order linear ODE
$$y''(t)+p(t) y'(t)+q(t) y(t)=0 \tag{1} .$$
In order to see this, try the substitution
$$y(t)=\mathrm{e}^{-\int p(t)/2 \mathrm{d} t}  x(t).$$
It is well known that equation $(1)$ cannot be solved in the general case.
A: I want to add another remark to why this is difficult in general. One might be tempted to write
\begin{equation}
 \frac{d}{dx}\begin{bmatrix} x \\ \dot x\end{bmatrix} 
 = \begin{bmatrix}0&1\\-a(t)  &0\end{bmatrix} 
\cdot \begin{bmatrix} x \\ \dot x\end{bmatrix} 
\end{equation}
And then say that the solution to the time dependent linear system $\dot x = A(t) x$ is simply $x(t)=e^{\int_0^t A(\tau)d\tau}x_0$. However this formula almost always fails, since $\frac{d}{dt} e^{B(t)}=\dot Be^{B(t)} $ is only true when $B$ and $\dot B$ commute. 
So let $b(t) = -\int a(t)$. Then in the case at hand,  $B = \int A = B_0 + \begin{bmatrix} 0 & t \\ b(t) &0\end{bmatrix}$. Hence the formula is only applicable if $[B,\dot B] =0$, which is a very strong restriction on what $b(t)$ can possibly be.
