Bessel's equation and second order DE

Hi all!

I've hit a wall on the question I'm working on. Can someone look over my work and see where I may have possible gone wrong?

The question is:

Show that if $\displaystyle y(x)$ is a solution of Bessel's equation of order p, then, $\displaystyle w(z)=z^{-c}y(az^b)$ is a solution of $\displaystyle z^2w''+(2c+1)zw'+[a^2b^2z^{2b}+(c^2-p^2b^2)]w=0$.

Here is my working.

Considering that $\displaystyle w(z)=z^{-c}y(az^b)$ is a solution, then:

$\displaystyle w'(z)=-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c}$ and

$\displaystyle w''(z)=(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)$

Subing into $\displaystyle L[w]$,

$\displaystyle L[w]=z^2(c^2+c)z^{-c-2}y(az^b)-2cz^{-c-1}abz^{-b-1}y'(az^b)+z^{-c}a^2b^2z^{b-2}y''(az^b)$+$\displaystyle (2c+1)z(-cz^{-c-1}y(az^b)+abz^{b-1}y'(az^b)z^{-c})+[a^2b^2z^{2b}+(c^2-p^2b^2)]z^{-c}y(az^b)$.

I simplify to find:

$\displaystyle bz^{-c}(a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)).$

As$\displaystyle L[w]=0$, that means $\displaystyle a^2z^{2b}y''(az^b)+abz^by'(az^b)+b(a^2z^2b-p^2)y(az^b)=0$.

Making the substitution of $\displaystyle x=az^b$

I get the following:

$\displaystyle x^2y''(x)+bxy'(x)+b(x^2-p^2)y(x)=0$.

That equation is soo close to Bessel's equation. I go through my working and I can't find why I have those b's are there.

Thank you very much for your time.