I have to solve the radial Schrodinger equation for a particle subjected to the potential:
$$V(r)= \dfrac{A}{r^2}-\dfrac{B}{r}$$
Where $r$ is the radial component (spherical coordinates) and $A,B>0$ While I was uncoupling the equations it was useful to define the following constants/function: $$k^2\dot{=}\dfrac{2mE}{\hslash^2} \ \ a\dot{=}\dfrac{2Am}{\hslash^2} \ \ b\dot{=}\dfrac{2Bm}{\hslash^2} \ \ \sigma\dot{=}[l(l+1)-a] \text{ and } u(r)\dot{=}R(r)$$ where $R(r)$ is the radial part of the variable separation and $l$ is the $L^2$ correspondent quantum number. So, I`ve found the following diffential equation (I'll denote u' as differentiation with respect to r):
$$u''(r)+\Big(\dfrac{\sigma}{r}+k^2r+b\Big)u(r)=0$$
As I want my result to be square integrable function I`d impose that $$\lim _{r\rightarrow\infty}r^2R^2(r)=0 \Rightarrow u(\infty)=0$$ But I'm not really sure if, for the $r=0$ condition, I should take $$|u(0)|<\infty \quad\text{ or }\quad u(0)=0~?$$
Which choice of those last are more suitable for physical means? Is there a single choice? Would 'em be the same of hydrogen atom?