Differential equation $au''(x)+b\frac{u(x)}{x}+Eu=0$

Question

$$au''(x)+b\frac{u(x)}{x}+Eu=0$$ 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?

Answer 1

$$au''(x)+b\frac{u'(x)}{x}+E\,u(x)=0 \tag 1$$ 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 : $$au''(x)+b\frac{u'(x)}{|x|}+E\,u(x)=0 $$ $$u(x)=x^{-p}\big(c_1J_\nu (X) +c_2Y_\nu(X)\big) \quad\text{ with } \begin{cases} B=\text{sgn}(x)\:b\\ p=\frac{B-a}{2a}\\ \nu=\frac{B-a}{2a}\\ X=\sqrt{\frac{E}{a}}\:x \end{cases}$$