Conversion to bessel equation I came across an equation(HERE,pg 2 at the end) which was as follows:
$$ xF''(x) +F'(x) + \frac{\omega^2}{g}F  = 0  \hspace{1.5 cm} (A)$$
I was trying to convert it to the form 
$$x^2 y'' +xy' +(x^2 - n^2)y = 0  \hspace{1.5 cm} (1)$$
I understood that it was a zero order Bessel equation.,i.e $ n  = 0 $
So,
$$x^2 y'' +xy' +(x^2)y = 0$$
Dividing by $x$
$$x y'' +y' +(x)y = 0$$
But I don't know what  to do next, to bring it to form of (1).
I feel that there is this x missing in ,i.e:  $\frac{\omega^2}{g}F(x) $.
Can you please guide me how do i convert (A) to (1).
Thank You very much
Anupam Bisht
 A: Let $x=aX^2\quad\to\quad dx=2aXdX\quad\to\quad \frac{dX}{dx}=\frac{1}{2aX}$
$\frac{dF}{dx}=\frac{dF}{dX}\frac{dX}{dx}=\frac{1}{2aX}\frac{dF}{dX}$
$\frac{d^2F}{dx^2}=\frac{d}{dX}\left( \frac{1}{2aX}\frac{dF}{dX}\right)\frac{dX}{dx}=\frac{1}{2aX}\left( \frac{1}{2aX}\frac{d^2F}{dX^2}-\frac{1}{2aX^2}\frac{dF}{dX}\right)$
$$x\frac{d^2F}{dx^2}+\frac{dF}{dx}+\frac{\omega^2}{g}F=0$$
$$(aX^2)\frac{1}{2aX}\left( \frac{1}{2aX}\frac{d^2F}{dX^2}-\frac{1}{2aX^2}\frac{dF}{dX}\right) +\frac{1}{2aX}\frac{dF}{dX}+\frac{\omega^2}{g}F=0$$
$$\frac{1}{4a}\frac{d^2F}{dX^2}+\frac{1}{4aX}\frac{dF}{dX}+\frac{\omega^2}{g}F=0$$
$$X^2\frac{d^2F}{dX^2}+X\frac{dF}{dX}+4a\frac{\omega^2}{g}X^2F=0$$
Let  $a=\frac{g}{4\omega^2}$
$$X^2\frac{d^2F}{dX^2}+X\frac{dF}{dX}+X^2F=0$$
$$F(X)=c_1J_0(X)+c_2Y_0(X)\qquad\text{where}\qquad
X=\pm\sqrt{\frac{4\omega^2}{g}x}$$
Note :
When one is faced with a non-standard form of Bessel equation, it is of use to try the changes of variable and function :
$$\begin{cases}
x=aX^p \\
F(x)=bX^qG(X)
\end{cases}$$
In the general case, it is an arduous task. So, usally one start with the change of variable only and see if it is sufficient. Fortunately, that was the case and identifying the coefficients of the transformed equation to whose of the standared Bessel equation leads to $p=2$.
