Pendulum with changing length over time. What's wrong? I tried to find the equation of this pendulum, but I think I did something wrong. I know I have to get the Bessel's equation but I can't see it. 
It's a simple 2-D pendulum, without any dissipation.
The length is $l$.
\begin{gather}
x(t)= l(t)  \sin\theta(t), \\
y(t)=l(t)  \cos\theta(t), \\
\dot x(t)= \dot l(t) \sin\theta(t) + l(t) \dot \theta(t) \cos\theta(t) \\
\dot y(t)= \dot l(t) \cos\theta(t) - l(t) \dot \theta(t) \sin\theta 
\end{gather}
and:
\begin{equation}
\dot x^2 + \dot y^2 = \dot l^2 + l^2 \dot \theta^2
\end{equation}
The potential and kinetic energy:
\begin{gather}
V=-mgy= -mgl(t) \cos\theta(t), \\
T=\frac{m}{2} (\dot l^2(t)+ l^2(t) \dot \theta^2(t))
\end{gather}
So the Langrangian is:
\begin{equation}
L=T-V= \frac{m}{2}(\dot l^2 + l^2 \dot \theta^2) + mgl \cos\theta
\end{equation}
After this, we get:
\begin{gather}
\frac{ \partial L}{\partial \theta} = -mgl \sin\theta, \\
\frac{ d}{dt} \frac{\partial L}{\partial \dot \theta}= m2l \dot l \dot \theta +ml^2 \ddot \theta.
\end{gather}
From thism we get the equation:
\begin{equation}
\ddot \theta + 2 \dot l \dot \theta = g \sin\theta.
\end{equation}
And for the radial part:
\begin{gather}
\frac{\partial L}{\partial l} = ml \dot \theta^2 + mg \cos\theta , \\
\frac{d}{dt} \frac{\partial L}{\partial \dot l}= m \ddot l.
\end{gather}
The equation is:
\begin{equation}
\dot l = l \dot \theta^2 + g \cos\theta
\end{equation}
My question is how can I get to the Bessel's equation from here? 
 A: The correct equation of motion for $\theta$ is
\begin{equation}
l\ddot \theta + 2 \dot l \dot \theta = -g \sin\theta.
\end{equation}
Now if you assume $l(t)=l_{0}+vt$ then you will get a Bessel Differential Equations.
\begin{equation}
\ddot \theta+\frac{2}{l}\dot \theta+\frac{g}{v^{2}l}\theta=0
\end{equation}
The solution for which is the Bessel Function. ($\sin\theta$ approximated by $\theta$ for small angle)
A: First of all, two tiny mistakes: your $\theta$ equation is missing an $\ell$, and the sign of $g  \sin \theta$ is flipped. That is, it should be
$$ 2 \dot{\ell} \dot{\theta} + \ell \ddot{\theta} = -g \sin \theta
$$
Now, I have not seen Bessel equation emerge for pendulum dynamics for arbitrary length variability. It only arises, as far as I have seen, for the linearly-changing length, that is, for $\ell=\ell_0+v t$, and even in that case, under the small-$\theta$ approximation, to the first non-zero order or $\theta$. 
In this special case, the equation reduces to $(\ell_0+v t) \ddot{\theta}+2 v \dot{\theta}+g \theta = 0$. There is a change-of-variables that transforms this into Bessel's ode, which is $\psi= \theta \sqrt{\ell}$ and $\phi=\frac{2}{v}\sqrt{\ell g}$. 
The calculations can be googled, and for example can be found here. This eventuates in the Bessel equation for the $\psi$ and $\phi$ pair. 
Another pendulum-related case that I know where Bessel's equation emerges is when the deformation of the rope from a straight line is taken into account, and the first corrections invoke Bessel equations. 
Other than these two special cases, I don't think one can evoke Bessel's equation in pendulum dynamics.
