Differential equation $au''(x)+b\frac{u(x)}{x}+Eu=0$ $$au''(x)+b\frac{u(x)}{x}+Eu=0$$
The schrodinger equation of a particle in potential $1/|x|$, where some constants are not written out, e.g. $\bar{h}$.
Can anyone give solution for this? Or it can just be solved numericly?
 A: $$au''(x)+b\frac{u'(x)}{x}+E\,u(x)=0 \tag 1$$
This is an ODE of Bessel kind: http://mathworld.wolfram.com/BesselDifferentialEquation.html
See Eqs.(3-5) :
$$y''+(2p+1)\frac{y'}{x}+\left(\alpha^2x^{2r-2}+\frac{\beta^2}{x^2}\right)y(x)=0$$
which solution is :
$$y(x)=x^{-p}\big(c_1J_\nu (X) +c_2Y_\nu(X)\big) \quad\text{ with }
\begin{cases}
\nu=\frac{\sqrt{p^2-\beta^2}}{r}\\
X=\frac{\alpha}{r}x^r
\end{cases}$$
$J_\nu (X)$ is the Bessel function of first kind and $Y_\nu (X)$ is the Bessel function of second kind.
In case of Eq.$(1)$ :
$\quad u(x)=y(x) \quad;\quad 2p+1=\frac{b}{a}\quad;\quad 2r-2=0\quad;\quad \alpha^2=\frac{E}{a} \quad;\quad \beta=0$
$p=\frac{b-a}{2a}\quad;\quad r=1\quad;\quad \alpha=\sqrt{\frac{E}{a}}\quad;\quad \nu=\frac{b-a}{2a}\quad;\quad X=\sqrt{\frac{E}{a}}\:x$
The solution of Eq.$(1)$ is :
$$u(x)=x^{-p}\big(c_1J_\nu (X) +c_2Y_\nu(X)\big) \quad\text{ with }
\begin{cases}
p=\frac{b-a}{2a}\\
\nu=\frac{b-a}{2a}\\
X=\sqrt{\frac{E}{a}}\:x
\end{cases}$$
SECOND EQUATION : 
$$au''(x)+b\frac{u'(x)}{|x|}+E\,u(x)=0 $$
$$u(x)=x^{-p}\big(c_1J_\nu (X) +c_2Y_\nu(X)\big) \quad\text{ with }
\begin{cases}
B=\text{sgn}(x)\:b\\
p=\frac{B-a}{2a}\\
\nu=\frac{B-a}{2a}\\
X=\sqrt{\frac{E}{a}}\:x
\end{cases}$$
