I'm solving Laplace's equation in a region with cylindrical symmetry (i.e. no polar angle dependence). Thus, from the outset, Laplace's equation becomes,
$$ \nabla^2\,f(\rho,z)=\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial f}{\partial\rho}\right)+\frac{\partial^2 f}{\partial z^2}=0 $$
Doing the standard method (for this particular linear DFQ), we try to look for solutions of the form $f(\rho,z)=P(\rho)Z(z)$. Substituting this into Laplace's equation and dividing by $P(\rho)Z(z)$, we get
$$\frac{1}{\rho P}\frac{d}{d\rho}\left(\rho P'\right)+\frac{Z''}{Z}=0$$
Since this must hold for every $\rho$ and $z$, the two must be equal to a constant (which are inverses of each other of course). For convenience (i.e. the standard next step), I'll set them equal to the following:
$$\frac{Z''}{Z}=k^2$$
$$\frac{1}{\rho P}\frac{d}{d\rho}\left(\rho P'\right)=-k^2$$
The equation for $Z$ is trivial. The equation for $P$, however, becomes
$$\rho P''+P'+k^2\rho P=0$$
Here is my question: The form of the equation for $P$ that I've gotten seems, after looking online, unusual. When I look for linearly independent solutions through the Frobenius method (power-series solution search), I get coefficients that have squares of double factorials, and I've never seen those arising before in Bessel functions. Moreover, I can't seem to match my differential equation with a particular form of the Bessel differential equation at all. What has gone wrong? Should I expect non-Bessel function solutions? \
I've solved my problem. I was obtaining the series representation of the Bessel function of the second kind (zero order), which I had never seen before. However, it can easily be put into the form of the standard Bessel equation by doing the following.
Define $x=k\rho$. The equation then takes the form,
\begin{align*} \rho P''+P'+k^2\rho P&=\frac{x}{k}\left(\frac{dx}{d \rho}\right)^2\frac{\partial^2 P}{\partial x^2}+\frac{dx}{d \rho}\frac{\partial P}{\partial x}+k^2\frac{x}{k}P\\ &=k\left(x\frac{\partial^2 P}{\partial x^2}+\frac{\partial P}{\partial x}+xP\right)\\ &=\frac{k}{x}\left(x^2\frac{\partial^2 P}{\partial x^2}+x\frac{\partial P}{\partial x}+(x^2-0^2)P\right)\\ &=0 \end{align*}
The first term $k/x$ is never zero, and this equality has to hold for all $x$, so the second term must vanish. We've now found a new (identical) differential equation that $P(x)$ satisfies, rather than $P(\rho)$.
$$x^2\frac{\partial^2 P}{\partial x^2}+x\frac{\partial P}{\partial x}+(x^2-0^2)P=0$$
This is the standard Bessel equation of order zero. The most general solution to this is
$$P(k\rho)=c_1J_0(k\rho)+c_2 Y_0(k\rho)$$