The Hamiltonian is

$\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}$

$\displaystyle \left( -\frac{\hbar^2}{2m}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - \frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} \right) R(r) = E R(r)$

rearrange, using Atomic unit

$\displaystyle \left( -\frac{1}{2}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - E -\frac{Z}{ r} + \frac{1}{2} \frac{l(l+1)}{r^2} \right) R(r) = 0$

Since the normalization condition of $R(r)$ is $\int_0^\infty R(r)R(r) r^2 dr = 1$, it is natural to define $u(r) = r R(r)$

$\displaystyle R = \frac{u}{r}, \frac{d}{dr}\left(r^2\frac{dR}{dr}\right) = r \frac{d^2u}{dr^2}$

$\displaystyle -\frac{1}{2}\frac{d^2u}{dr^2} - \left( E + \frac{Z}{ r} - \frac{l(l+1)}{2r^2} \right) u(r)= 0$

Substitute the eigen energy $E_n = - \frac{Z^2}{2n^2}$

$\displaystyle \frac{d^2u}{dr^2} + \left( - \frac{Z^2}{n^2} + \frac{2Z}{ r} - \frac{l(l+1)}{r^2} \right) u(r)= 0$

Set

$\displaystyle x = \frac{2Z}{n} r, \frac{d}{dr} = \frac{d}{dx} \frac{2Z}{n}$

$\displaystyle \left(\frac{2Z}{n}\right)^2\frac{d^2u(x)}{dx^2} - \left( - \frac{Z^2}{n^2} + \frac{4Z^2}{n x} - \frac{4Z^2}{n^2}\frac{l(l+1)}{x^2} \right) u(x)= 0$

Take out the $4Z^2/n^2$

$\displaystyle \frac{d^2u(x)}{dx^2} - \left(\frac{n}{ x} - \frac{1}{4} - \frac{l(l+1)}{x^2} \right) u(x)= 0$

Now, we know the short and long-range behaviour of $u(x)$, assume it to be

$u(x) = \exp\left(-\frac{x}{2} \right) x^{l+1} y(x)$

Then the equation of $y(x)$ is

$\displaystyle x \frac{d^2y}{dx^2} + \left( 2(l+1) - x\right) \frac{dy}{dx} +\left( n - l- 1 \right) y(x) = 0$

This is the Laguerre equation

$\displaystyle x \frac{d^2y}{dx^2} + \left( \alpha+1 - x\right) \frac{dy}{dx} + m y(x) = 0$

where $\alpha = 2l+1$ and $m = n-l -1$. Therefore, the solution of the radial equation is,

$R(r) = \frac{1}{r} A \exp\left(-\frac{x}{2} \right) x^{l+1} L_{n-l-1}^{2l+1}(x)$

where $A$ is normalization factor.

Notice that the Laguerre polynomial is only defined for $m \geq 0$, thus, $n > l$.