Presenting the Cauchy-Euler Differential Equation as Successive First Order Derivatives Consider the following ordinary differential equation (ODE)
$$\frac{d^2w}{dr^2}(r)+\frac{1}{r}\frac{dw}{dr}(r)-\frac{\alpha^2}{r^2}w(r)=p(r)\tag{1}$$
where $\alpha$ is a non-negative real number. This is a Cauchy-Euler second order ODE. Clearly, its homogeneous solution $w_h(r)$ can be found easily by employing a solution of the form $r^m$ and inserting it into the ODE. Carrying out this procedure one finds
$$
w_h(r)=
\begin{cases}
A_0+B_0r^2+C_0\ln(r)+D_0r^2\ln r, & \alpha=0 \\
A_1r+B_1r^{-1}+C_1r^3+D_1r\ln r, & \alpha=\pm 1 \\
A_{\alpha}r^{\alpha}+B_{\alpha}r^{-\alpha}+C_{\alpha}r^{\alpha+2}+D_{\alpha}r^{-\alpha+2}, & \text{Otherwise}
\end{cases}
\tag{2}$$
Now, I am inclined to obtain a particular solution $w_p(r)$ for an arbitrary given function $p(r)$. I saw somewhere that the ODE is re-written in the following form
$$r^{\alpha-1}\frac{d}{dr}\Big(r^{1-2\alpha}\frac{d}{dr}\Big(r^{\alpha}w(r)\Big)\Big)=p(r)\tag{3}$$
so one can easily found the particular solution by successive integration.

I want to know that how one transforms $(1)$ to $(3)$? What is the main Idea behind this transformation?

Of course, I don't want to do the reverse procedure which is carrying out the derivatives in $(3)$ and seeing that $(1)$ is obtained. In fact, I am interested in a solution supposing that we were not aware of $(3)$.
 A: One motivation is from observing that $\frac{d^p}{dr^p}\left(r^q w(r)\right)=a_0r^{q-p}w(r)+a_1r^{q-p+1}w'(r)+...$ which has all the terms $r^{q-p+t}w^t(r)$ in a Cauchy-Euler equation. It is not too far of a stretch, then, to intersperse $r^{s}$ terms with each successive derivative.
So, working from generality,
\begin{align*}
&r^{k} \frac{d}{dr} \left( r^{m}\frac{d}{dr}\left[r^n w(r)\right]\right)\\
&=r^{k} \frac{d}{dr} \left( r^{m}[n r^{n-1} w(r)+r^n w'(r)]\right)\\
&=r^{k} \frac{d}{dr} \left(n r^{m+n-1} w(r)+r^{m+n}w'(r)\right)\\
&=r^{k} \left(n(m+n-1) r^{m+n-2} w(r) + nr^{m+n-1}w'(r) + (m+n)r^{m+n-1}w'(r)+r^{m+n}w''(r)\right)\\
&=r^{k} \left(n(m+n-1) r^{m+n-2} w(r) + (m+2n)r^{m+n-1}w'(r)+r^{m+n}w''(r)\right)\\
&=r^{m+n+k} \left(n(m+n-1) r^{-2} w(r) + (m+2n)r^{-1}w'(r)+w''(r)\right)
\end{align*}
Now, you can equate terms and solve a system of 3 variables $m, n, k$:
\begin{align*}
n(m+n-1)&=-\alpha^2 \\
m+2n&=1 \\ 
k+m+n &= 0
\end{align*}
By inspection, $$\begin{align}n&=\pm\alpha\\m&=1\mp2\alpha\\k&=\pm\alpha-1\end{align}$$
But, there are other ways of solving Cauchy-Euler equations. Most textbooks would suggest a transformation with $r=e^t$ to obtain a constant coefficient nonhomogeneous differential equation. 
