How to convert the following ODE to a Bessel differential equation?

Question

I've problems finding the steps of the solution of the following differential equation.

$$y''+\frac{1-2\alpha}{x}y'+\left(\beta^2+\frac{\alpha^2-v^2}{x^2}\right)y=0$$

I know that the general solution is:

$$ y = x^\alpha Z_v(\beta x)$$

Answer 1

Using the change of variables $t=\beta x$ and $w=yx^{-\alpha}$ it can be converted to a Bessel differential equation for $w(t)$.

For reference see: B. G. Korenev: Bessel Functions and Their Applications, page 32

The first steps:

$$\frac{\mathrm{d}w}{\mathrm{d}x}=x^{-\alpha}\frac{\mathrm{d}y}{\mathrm{d}x}+(-\alpha)x^{-\alpha-1}y$$

$$\frac{\mathrm{d}^2w}{\mathrm{d}x^2}=x^{-\alpha}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2(-\alpha)x^{-\alpha-1}\frac{\mathrm{d}y}{\mathrm{d}x}+(-\alpha)(-\alpha-1)x^{-\alpha-2}y$$

$$y=x^{\alpha}w$$ $$\frac{\mathrm{d}y}{\mathrm{d}x}=x^{\alpha}\frac{\mathrm{d}w}{\mathrm{d}x}+\alpha x^{\alpha-1}w$$ $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=...$$