Poisson's PDE practical example An infinite tube of radius $a$ is kept at an electrical potential. The Poisson PDE gives us the potential distribution:
$$\nabla^{2} V=0=\frac{\partial^{2} V}{\partial r^{2}}+\frac{1}{r} \frac{\partial V}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} V}{\partial \theta^{2}}$$
(There's no $z$-based term because due to infinity of the tube $\frac{\partial V}{\partial z}=0$)
Regardless for a minute of BCs, this is the solution I found on a reputable site:
$$V=\frac{\left(V_{1}+V_{2}\right)}{2}+2 \frac{\left(V_{1}-V_{2}\right)}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2 n-1)}\left(\frac{r}{a}\right)^{2 n-1} \cos (2 n-1) \theta$$
What struck me is that if we assume:
$$V(r,\theta)=R(r)\Theta(\theta)$$
then according this result:
$$R(r)=\left(\frac{r}{a}\right)^{2 n-1}$$
and:
$$\Theta(\theta)=B_n\cos (2 n-1) \theta$$

Intuitively I didn't expect these results, especially for $R(r)$. So I kind of routinely checked the separation of variables, as follws.
$$\nabla^{2} V=0=\frac{\partial^{2} V}{\partial r^{2}}+\frac{1}{r} \frac{\partial V}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} V}{\partial \theta^{2}}$$
$$V(r,\theta)=R(r)\Theta(\theta)$$
$$\Theta R''+\frac{1}{r}\Theta R'+\frac{1}{r^2}R\Theta''=0$$
$$\frac{R''}{R}+\frac{R'}{rR}+\frac{\Theta''}{r^2\Theta}=0$$
$$r^2\frac{R''}{R}+r\frac{R'}{R}+\frac{\Theta''}{\Theta}=0$$
$$r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Theta''}{\Theta}=-m^2$$
$$\Theta''-m^2\Theta=0\tag{1}$$
$$r^2\frac{R''}{R}+r\frac{R'}{R}+m^2=0$$
$$r^2R''+rR'+m^2R=0\tag{2}$$
The resp. solutions of the ODEs are:
$$\Theta(\theta)=c_1e^{m\theta}+c_2e^{-m\theta}$$
$$R(r)=c_3\sin(m\ln r)+c_4\cos(m\ln r)$$
Needless to say, $(1)$ and $(2)$ DO NOT yield the solutions outlined above. So what's causing this discrepancy?
 A: Use $r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{\Theta''}{\Theta}=m^2$ instead.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Use superposition of a constant
$\ds{\,\,\,{V_{1} + V_{2} \over 2}\,\,\,}$ potential and another one with opposites values
$\ds{\,\,\,\pm\,{V_{1} - V_{2} \over 2}}$.
\begin{align}
&\mbox{Namely,}\quad\left\{\begin{array}{lcrcl}
\ds{V_{1}} & \ds{=} &
\ds{V_{1} + V{2} \over 2} &  \ds{\color{red}{+}}  & \ds{V_{1} - V{2} \over 2}
\\[1mm]
\ds{V_{2}} & \ds{=} &
\ds{V_{1} + V{2} \over 2} &  \ds{\color{red}{-}}  & \ds{V_{1} - V{2} \over 2}
\end{array}\right.
\end{align}
The first one has the trivial solution $\ds{V_{1} + V_{2} \over 2}$ while for the $\ds{\underline{second\ one}}$ the solution has the general form
\begin{align}
&\pars{A + B\theta}\bracks{C\ln\pars{r} + D}
\\ &\
+ \sum_{n = 0}^{\infty}\bracks{A_{n}r^{n}\sin\pars{n\theta + \delta_{n}} + B_{n}r^{-n}\sin\pars{n\theta + \gamma_{n}}}
\end{align}
However, the solution must vanish out as $\ds{r \to \infty}$ which reduces the solution to
$$
A + B\theta +
\sum_{n = 0}^{\infty}\bracks{a_{n}\sin\pars{n\theta} + b_{n}\cos\pars{n\theta}}r^{-n}
$$
Since the solution changes its sign under
$\ds{\theta \mapsto 2\pi - \theta}$ and it's unchanged under
$\ds{\theta \mapsto \pi - \theta}$, the solution becomes
\begin{align}
&
\sum_{n = 0}^{\infty}a_{2n + 1}
\sin\pars{\bracks{2n + 1}\theta}\pars{1 \over r}^{2n + 1}\quad
\mbox{and, in addition,}
\\[5mm] & 
\sum_{n = 0}^{\infty}a_{2n + 1}
\sin\pars{\bracks{2n + 1}\theta}\pars{1 \over a}^{2n + 1}
\\ = &\
\left\{\begin{array}{lcl}
\ds{\phantom{-}{V_{1} - V_{2} \over 2}} & \mbox{if} &
\ds{0 \leq \theta < \pi}
\\[1mm]
\ds{-\,{V_{1} - V_{2} \over 2}} & \mbox{if} &
\ds{\pi \leq \theta < 2\pi}
\end{array}\right. 
\end{align}
Multiply both members by $\ds{{2 \over \pi}\,\sin\pars{\bracks{2n + 1}\theta}}$ and integrate over $\ds{\theta \in \pars{0,\pi}}$:
$$
a_{2n + 1}{1 \over a^{2n + 1}} =
{V_{1} - V_{2} \over \pi}\ \underbrace{\int_{0}^{\pi}
\sin\pars{\bracks{2n + 1}\theta}\dd\theta}
_{\ds{2 \over 2n + 1}}
$$
The final solution becomes
$$
{V_{1} + V_{2} \over 2} +
{2 \over \pi}\pars{V_{1} - V_{2}}
\sum_{n = 0}^{\infty}{\sin\pars{\bracks{2n + 1}\theta} \over 2n + 1}
\pars{a \over r}^{2n + 1}
$$
which can be written in a closed form:
$$
{V_{1} + V_{2} \over 2} +
{V_{1} - V_{2} \over \pi}
\arctan\pars{2\xi\sin\pars{\theta} \over
\xi^{2} - 1}\,,
\quad \xi \equiv {r \over a} > 1
$$
