A simple second order ODE This might be a very naive question, but is there a solution for a second order ODE of the form 
$$y''(x) = f(x)y(x)$$
where $f(x)$ is a general function? Any information is appreciated.
Thanks.
 A: $$y''(x)=f(x)y(x)$$
Citation from Fabian : "This problem does not have an explicit solution for generic $f(x)$ " . I fully agree.
Are there special cases of $f(x)$ where this has a solution? 
Of course, yes. Some examples below (far to be exhaustive) :
Case: $f(x)=c^2 \quad\to\quad y(x)=c_1e^{cx}+c_2e^{-cx}=c_3\cosh(cx)+c_4\sinh(cx)$
Case: $f(x)=-c^2 \quad\to\quad y(x)=c_1\cos(cx)+c_2\sin(cx)$
Case: $f(x)=x \quad\to\quad y(x)=c_1Ai(x)+c_2Bi(x) \quad $ Ary functions.
Case: $f(x)=x^2 \quad\to\quad y(x)=c_1 D_{-1/2}(\sqrt{2}x) +c_2 D_{-1/2}(i\sqrt{2}x) \quad $ Parabolic cylinder function.
Case: $f(x)=-\lambda^2 x^{\frac{1}{\nu}-2} \quad\to\quad y(x)=c_1 \sqrt{x}J_{\nu}(\lambda x) +c_2 \sqrt{x}Y_{\nu}(\lambda x) \quad $ Bessel functions.
Case: $f(x)=\lambda^2 x^{\frac{1}{\nu}-2} \quad\to\quad y(x)=c_1 \sqrt{x}I_{\nu}(\lambda x) +c_2 \sqrt{x}K_{\nu}(\lambda x) \quad $ Modified Bessel functions.
Case: $f(x)=-a+2b\cos(2x) \quad\to\quad y(x)= c_1C(a\:,\:b\:;\:x)+c_1S(a\:,\:b\:;\:x)$ Mathieu functions. 
Case: $f(x)=\frac{A}{x^2}+\frac{B}{x}+C  \quad\to\quad y(x)=e^{-\frac{\gamma}{2}x}x^{\frac{\beta}{2}}\left( c_1 M(\alpha\:,\:\beta\:;\:\gamma x)+c_2 U(\alpha\:,\:\beta\:;\:\gamma x) \right) \quad$ with $\begin{cases}
\gamma=\pm2\sqrt{C}\\
\beta=1\pm 2\sqrt{A+\frac{1}{4}}\\
\alpha=\frac{\beta}{2}+\frac{B}{\gamma}
\end{cases}\quad$
Kummer and Tricomi functions (confluent hypergeometric functions).
Etc.
A: We have a 2nd order ODE of the form
$$\ddot y = f (t) y$$
Let $x := (y, \dot y)$. We then have a linear time-varying (LTV) system of the form
$$\dot x = \begin{bmatrix} 0 & 1\\ f (t) & 0\end{bmatrix} x$$
LTV systems have been studied thoroughly in control theory. Take a look at Antsaklis & Michel.
