Transformation of PDE derived Bessel's eqn into standard form

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Often we solve a PDE and arrive at the equation \begin{align} r^2 R'' + r R' + (\lambda^2 r^2-\mu^2)R =0\\ \end{align} which has to be transform into the familiar standard form, then we can solve the equation like demonstrated in the article Bessel's Differential Equation.

In order to transfrom we substitue $x=\lambda r$ \begin{align} r^2 \frac{R''}{\lambda^2} + r \frac{R'}{\lambda} + (x^2-\mu^2)R =0\\ \end{align} Now we need to know how $R(r)$ depends on $x$ \begin{align} \frac{\partial R}{\partial r}=\frac{\partial R}{\partial x}\frac{\partial x}{\partial r}=\frac{\partial R}{\partial x} \lambda \end{align} and denote the derivative $dR/dx=dy/dx$ in order to distiguish $R'=dR/dr$ from $y'=dR/dx$. \begin{align} R'=\frac{\partial R}{\partial x} \lambda=y' \lambda \end{align} and \begin{align} \frac{\partial^2 R}{\partial r^2}&=\frac{\partial }{\partial r} \left( \frac{\partial R}{\partial x}\frac{\partial x}{\partial r} \right)=\frac{\partial }{\partial x} \left( \frac{\partial R}{\partial x}\frac{\partial x}{\partial r} \right) \frac{\partial x}{\partial r}=\frac{\partial^2 R}{\partial x^2} \lambda^2\\ R''&=y'' \lambda^2\\ \end{align} We use this partial derivatives to get Bessel's Differential Equation in terms of $y(x)$ \begin{align} x^2 y'' +x y' + (x^2-\mu^2)y =0\\ \end{align}