Existence of solutions to y'' = f(x)y I was reading some results earlier concerning properties of solutions to equations of the form
$$y'' = f \left(x \right) y$$
paired with initial conditions $y (x_0) = y_0$, $y' (x_0) = y_0'$, where $f$ is continuous. I was wondering -- is there an easy proof that solutions exist in general? My work in progress is to allow $z$ to solve the first-order nonlinear nonhomogeneous system 
$$z' + z^2 = -f (x)$$
Then if we set
$$y = e^{\int z dx}$$
we obtain that 
$$y'' = (z' + z^2)y = f (x) y$$
I'm not happy with this solution however -- there are definitely some pieces missing (the integral of $z$ existing being just one of them) and I don't know if this method can be fixed. 
I'm aware that there are (complicated) existence and uniqueness results for more general systems, but I want to avoid this. I'm fine with using Picard-Lindelof, however. Any help is appreciated.

You can use Picard-Lindelof. Let $u = y'$, Then $$u' = f(x) y$$ so the vector $v = \binom u y$ satisfies the first order equation $$v' = \begin{pmatrix} 0 & f(x) \\1&0 \end{pmatrix}v =  A(x)v$$ which locally has a unique solution due to Picard-Lindelof (in fact, if $f$ is bounded, there is a unique global solution).
