$r^2 R_{rr}+rR_r+(r^2 \lambda - s^2) R = 0$ - Bessel's equation

Question

Let the equation $r^2 R_{rr}+rR_r+(r^2 \lambda - s^2) R = 0$, where $R=R(r)$. Is anyone could explain to me how convert this D.E into Bessel's equation?

Answer 1

If $\lambda=0$, there is no hope (but it can be easily proven that $R(r)=A\,r^{+s}+B\,r^{-s}$ for some $A,B$ if $s\neq 0$, and $R(r)=A+B\ln(r)$ for some $A,B$ if $s=0$). If $\lambda\neq 0$, let $\lambda=\nu^2$. Define $\rho:=\nu\,r$. Show that $R(r)=u(\nu\,r)$ where $u$ is a solution to the Bessel equation of order $s$.