Solving for Eigenvalues of Bessel like differential equations How to solve for $\lambda$ and $R(r)$ in this Bessel like differential equation when $R(r) = 0 \ \forall \ r\geq r_{0}$.
$$\frac{\partial^{2}R}{\partial r^{2}} + \frac{1}{r}\frac{\partial R}{\partial r}+ \left(\frac{\beta^2}{r^2}-\alpha^2 r^2\right)R = -\lambda R . $$
I am not sure how to transform this into a more standard form or how to solve for eigenvalues for this system.
 A: Hint:
$\dfrac{\partial^2R}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial R}{\partial r}+\left(\dfrac{\beta^2}{r^2}-\alpha^2r^2\right)R=-\lambda R$
$\dfrac{\partial^2R}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial R}{\partial r}+\left(\dfrac{\beta^2}{r^2}-\alpha^2r^2+\lambda\right)R=0$
Let $x=r^2$ ,
Then $\dfrac{\partial R}{\partial r}=\dfrac{\partial R}{\partial x}\dfrac{\partial x}{\partial r}=2r\dfrac{\partial R}{\partial x}$
$\dfrac{\partial^2R}{\partial r^2}=\dfrac{\partial}{\partial r}\left(2r\dfrac{\partial R}{\partial x}\right)=2r\dfrac{\partial}{\partial r}\left(\dfrac{\partial R}{\partial x}\right)+2\dfrac{\partial R}{\partial x}=2r\dfrac{\partial}{\partial x}\left(\dfrac{\partial R}{\partial x}\right)\dfrac{\partial x}{\partial r}+2\dfrac{\partial R}{\partial x}=2r\dfrac{\partial^2R}{\partial x^2}2r+2\dfrac{\partial R}{\partial x}=4r^2\dfrac{\partial^2R}{\partial x^2}+2\dfrac{\partial R}{\partial x}=4x\dfrac{\partial^2R}{\partial x^2}+2\dfrac{\partial R}{\partial x}$
$\therefore4x\dfrac{\partial^2R}{\partial x^2}+2\dfrac{\partial R}{\partial x}+2\dfrac{\partial R}{\partial x}+\left(\dfrac{\beta^2}{x}-\alpha^2x+\lambda\right)R=0$
$4x\dfrac{\partial^2R}{\partial x^2}+4\dfrac{\partial R}{\partial x}-\left(\alpha^2x-\lambda-\dfrac{\beta^2}{x}\right)R=0$
$x^2\dfrac{\partial^2R}{\partial x^2}+x\dfrac{\partial R}{\partial x}-\left(\dfrac{\alpha^2x^2}{4}-\dfrac{\lambda x}{4}-\dfrac{\beta^2}{4}\right)R=0$
Let $R=e^{-\frac{\lambda x}{4}}U$ ,
Then $\dfrac{\partial R}{\partial x}=e^{-\frac{\lambda x}{4}}\dfrac{\partial U}{\partial x}-\dfrac{\lambda e^{-\frac{\lambda x}{4}}U}{4}$
$\dfrac{\partial^2R}{\partial x^2}=e^{-\frac{\lambda x}{4}}\dfrac{\partial^2U}{\partial x^2}-\dfrac{\lambda e^{-\frac{\lambda x}{4}}}{4}\dfrac{\partial U}{\partial x}-\dfrac{\lambda e^{-\frac{\lambda x}{4}}}{4}\dfrac{\partial U}{\partial x}+\dfrac{\lambda^2e^{-\frac{\lambda x}{4}}U}{16}=e^{-\frac{\lambda x}{4}}\dfrac{\partial^2U}{\partial x^2}-\dfrac{\lambda e^{-\frac{\lambda x}{4}}}{2}\dfrac{\partial U}{\partial x}+\dfrac{\lambda^2e^{-\frac{\lambda x}{4}}U}{16}$
$\therefore x^2e^{-\frac{\lambda x}{4}}\dfrac{\partial^2U}{\partial x^2}-\dfrac{\lambda x^2e^{-\frac{\lambda x}{4}}}{2}\dfrac{\partial U}{\partial x}+\dfrac{\lambda^2x^2e^{-\frac{\lambda x}{4}}U}{16}+xe^{-\frac{\lambda x}{4}}\dfrac{\partial U}{\partial x}-\dfrac{\lambda xe^{-\frac{\lambda x}{4}}U}{4}-\left(\dfrac{\alpha^2x^2}{4}-\dfrac{\lambda x}{4}-\dfrac{\beta^2}{4}\right)e^{-\frac{\lambda x}{4}}U=0$
$x^2\dfrac{\partial^2U}{\partial x^2}-\dfrac{\lambda x^2-2x}{2}\dfrac{\partial U}{\partial x}-\left(\dfrac{(4\alpha^2-\lambda^2)x^2}{16}-\dfrac{\beta^2}{4}\right)U=0$
