Find a solution to
$[g(x)u'(x)]'=\delta(x)$
$-\infty<x<\infty$.
What i have so far:
$[g(x)u'(x)]'=g(x)u''(x)+g'(x)u'(x)=\delta(x)$
$\Rightarrow u''(x)+\frac{g'(x)}{g(x)}u'(x)=\frac{\delta(x)}{g(x)}$.
This is where i'm stuck, since what i know form my course of differential equations i only know how to solve when dealing with constant coefficients. Any hints or guidence would be greatly aprreciated.
I forgot to say that $g^{-1}$ is integrable.
On
Note that for $x\ne 0$, we can write
$$(g(x)u'(x))'=0$$
Therefore, we find that
$$ u(x)=\begin{cases}C^+\int_0^x \frac{1}{g(t)}\,dt+K^+&,x>0\\\\C^-\int_0^x \frac{1}{g(t)}\,dt+K^-&,x<0 \end{cases}$$
Since $u$ is continuous at $x=0$, then $K^+=K^-=u(0)$. Furthermore, $u'$ is discontinuous at $x=0$ with $u'(0^+)-u'(0^-)=1/g(0)$. Thus, $C^-=C^+-1$. Therefore, we can write
$$ u(x)=\begin{cases}C^+\int_0^x \frac{1}{g(t)}\,dt+u(0)&,x>0\\\\(C^+-1)\int_0^x \frac{1}{g(t)}\,dt+u(0)&,x<0 \end{cases}$$
You can integrate once directly, $$ g(x)u'(x)=H(x)+C $$ with $H$ the Heaviside function of a jump from zero to one at zero. After that you just have to integrate $$ g(x)=D+\int\frac{H(x)+C}{g(x)}dx $$ and fix constants to fit initial conditions.