= Solution
Introduce the reduced <radial wavefunction> $u(r)=r\psi(r)$. Multiplying the <radial Schrödinger equation for a central potential> by $-2mr/\hbar^2$ gives
$$
u''-\frac{\ell(\ell+1)}{r^2}u+\frac{2mKZ}{\hbar^2r}u+\frac{2mE}{\hbar^2}u=0.
$$
For the decaying <bound state>, take $\alpha>0$ and put $u=f e^{-\alpha r}$. Differentiation yields
$$
u''=e^{-\alpha r}\left(f''-2\alpha f'+\alpha^2f\right).
$$
Since $2mE/\hbar^2=-\alpha^2$, the two constant multiples of $f$ cancel. Dividing by the nonzero exponential gives
$$
\boxed{f''-2\alpha f'-\frac{\ell(\ell+1)}{r^2}f+\frac{2mKZ}{\hbar^2r}f=0.}
$$
This separates the large-radius exponential decay from the remaining <Coulomb bound state> equation.
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