= Solution
Use <separation of variables>, writing $\Psi=F(r)\Theta(\theta)$. After multiplication by $r^2/(F\Theta)$, the radial and angular terms are functions of independent variables, so set
$$
\frac{(r^2F')'}F=-\frac{(\sin\theta\,\Theta')'}{\sin\theta\,\Theta}=\lambda.
$$
With $x=\cos\theta$, the angular equation is
$$
\frac d{dx}\left[(1-x^2)\frac{d\Theta}{dx}\right]+\lambda\Theta=0.
$$
Regularity at both poles selects $\lambda=\ell(\ell+1)$, $\ell=0,1,\ldots$, and the <Legendre polynomials> $P_\ell(x)$; the other angular branch is singular at a pole. The radial <Euler-Cauchy equation> is $r^2F''+2rF'-\ell(\ell+1)F=0$. A power $r^s$ gives $(s-\ell)(s+\ell+1)=0$, hence $F=a_\ell r^\ell+b_\ell r^{-\ell-1}$.
Superposing these axisymmetric <spherical harmonics> gives
$$
\boxed{\Psi(r,\theta)=\sum_{\ell=0}^\infty\left(a_\ell r^\ell+\frac{b_\ell}{r^{\ell+1}}\right)P_\ell(\cos\theta).}
$$
The coefficients must give convergence appropriate to the domain. In a region containing the origin, regularity removes the negative powers; in an exterior region with zero potential at infinity, it removes the growing powers and the constant. No such radial restriction is needed for a general annular region.
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