<s>[INST] How can I use Bessel's equation to solve the Lengthening Pendulum differential equation? Taking a small extract of this previous bounty question of mine: 

It can be shown that the differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{1}$$ has the solution $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are
  constants.



To see how to use this, let us “solve” the differential equation: $$y^{\prime\prime}+9xy=0\tag{3}$$ If $(3)$ is of the type $(1)$, then we must have $$1-2a=0$$ $$2(c-1)=1$$ $$(bc)^2=9$$  $$a^2-p^2c^2=0$$ from these $4$ equations we find
  $$a=\dfrac12$$ $$c=\dfrac32$$ $$b=2$$ $$p=\dfrac{a}{c}=\dfrac13$$
Then the solution of $(3)$ is $$y=x^{1/2}Z_{1/3}\left(2x^{3/2}\right)$$ This means that the general solution of $(3)$ is $$y=x^{1/2}\left[AJ_{1/3}\left(2x^{3/2}\right)+BN_{1/3}\left(2x^{3/2}\right)\right]$$ where $A$ and $B$ are arbitrary constants.


The differential equation of a lengthening pendulum is $$l\dfrac{d^2\theta}{dl^2}+2\dfrac{d\theta}{dl}+\dfrac{g}{v^2}\theta=0\qquad\quad\tag{4}\longleftarrow\text{proved in this former question}$$
I have to solve this differential equation for $\theta$ by comparing $(4)$ with $(1)$ in the same manner as used to find the solution to $(3)$

So here is my attempt:
First I begin by writing $(1)$ in terms of the new variables $\theta$ and $l$ $$\frac{\mathrm{d}^2\theta}{\mathrm{d}l^2}+\left(\frac{1-2a}{l}\right)\frac{\mathrm{d}\theta}{\mathrm{d}l}+\left[\left(b\,c\,l^{c-1}\right)^2+\frac{a^2-p^2c^2}{l^2}\right]\theta=0\tag{5}$$
Now I compare $(4)$ with $(5)$ to obtain 
$$l=1$$
$$\frac{1-2a}{l}=2\implies a=-\frac12\qquad\text{since $l=1$}$$
$$\left(b\,c\,l^{c-1}\right)^2+\frac{a^2-p^2c^2}{l^2}=\frac{g}{v^2}\tag{6}$$
Substituting $a=\frac12$ and $l=1$ into $(6)$
$$\implies b^2c^2+\frac14 -p^2c^2=\frac{g}{v^2}\tag{7}$$
Now if $l=1$ we must have $$c=1$$
Substituting $c=1$ into $(7)$
$$\implies b^2+\frac14 -p^2=\frac{g}{v^2}\tag{8}$$
But I am unable to proceed from here and I have made a mistake anyway as I can tell you that the correct answer is $$\theta=l^{-1/2}Z_1\left(\frac{2{g}^{1/2}}{v}l^{1/2}\right)$$ Comparison with 
$$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$
shows that $$a=-\frac12, \quad c=\frac12, \quad b=\frac{2{g}^{1/2}}{v},\quad p=1$$
Could someone please help me find the correct values of $a,b,c,p$?
Any hints or advice is well appreciated.
 [/INST] Hint: In the pendulum example the variable $l$ plays the role of $x$ and $\theta$ the role of $y$.

Keeping this in mind and comparing
  \begin{align*}
\frac{d^2\theta}{dl^2}+\frac{2}{l}\frac{d\theta}{dl}+\frac{g}{v^2}\frac{\theta}{l}=0
\end{align*}
  with
  \begin{align*}
\frac{d^2\theta}{dl^2}+\left(\frac{1-2a}{l}\right)\frac{d\theta}{dl}+\left[(bcl^{c-1})^2+\frac{a^2-p^2c^2}{l^2}\right]\theta=0
\end{align*}
  we get the following conditions:
  \begin{align*}
1-2a&=2\\
2(c-1)&=-1\\
b^2c^2&=\frac{g}{v^2}\\
a^2-p^2c^2&=0
\end{align*}
  and conclude
  \begin{align*}
a=-\frac{1}{2},\quad c=\frac{1}{2},\quad b=\sqrt{\frac{4g}{v^2}}=\frac{2}{v}\sqrt{g},\quad   p=\sqrt{\frac{a^2}{c^2}}=1
\end{align*}

We  finally obtain 
\begin{align*}
\theta=l^aZ_p\left(bl^c\right)=\frac{1}{\sqrt{l}}Z_1\left(\frac{2}{v}\sqrt{gl}\right)
\end{align*}
</s>