= Solution
Put $R=f e^{-\gamma r}$. Then
$$
R'=e^{-\gamma r}(f'-\gamma f),
\qquad
R''=e^{-\gamma r}(f''-2\gamma f'+\gamma^2f).
$$
After substitution and cancellation of the $\gamma^2f$ terms, the equation becomes
$$
rf''+(2-2\gamma r)f'+(\beta-2\gamma)f=0.
$$
For the <power series> $f=\sum_{n=0}^{\infty}a_nr^n$, the coefficient of $r^{n-1}$ is
$$
n(n+1)a_n+(\beta-2\gamma n)a_{n-1}=0.
$$
Thus
$$
\boxed{a_n=\frac{2\gamma n-\beta}{n(n+1)}a_{n-1}}.
$$
If the series does not terminate, then $a_n/a_{n-1}\sim2\gamma/n$, so $f$ acquires the large-$r$ behaviour $e^{2\gamma r}$ and $R$ grows like $e^{\gamma r}$. Normalizability therefore requires termination, which occurs when
$$
2\gamma N-\beta=0
$$
for some positive integer $N$. Hence
$$
\gamma_N=\frac{\beta}{2N},
\qquad
E_N=-\frac{\hbar^2\gamma_N^2}{2m}
=-\frac{mq^4}{2\hbar^2N^2}.
$$
The ground state has $N=1$, so the <hydrogen ground-state energy> is
$$
\boxed{E_1=-\frac{mq^4}{2\hbar^2}}.
$$
This is the <series-termination quantization of the Coulomb radial equation>.
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