3.32 ODE change of variable on both dependent and independent variable?

This verifies solution given in https://math.stackexchange.com/questions/3477732/can-t-see-that-an-ode-is-equivalent-to-a-bessel-equation Where a change of variables on

\[ \xi ^2 \frac {d^2\eta }{d\xi ^2} + \xi \frac {d\eta }{d\xi } - (\xi ^2+n^2)\eta =0 \]

Was made using

\[ \eta =\frac {y}{x^\alpha }, \quad \xi =\beta x^\gamma , \]

To produce the ode

\[ \frac {d^2y}{dx^2} - \frac {(2\alpha -1)}{x}\frac {dy}{dx} - (\beta ^2 \gamma ^2 x^{2\gamma -2}+\frac {n^2\gamma ^2-\alpha ^2}{x^2})y=0. \]

In Maple the above is done using

restart; 
ode := zeta^2*diff(eta(zeta),zeta$2) + zeta*diff(eta(zeta),zeta) - (zeta^2 + n^2)*eta(zeta) = 0; 
the_tr:={zeta=beta*x^gamma,eta(zeta)=y(x)/x^alpha}; 
PDEtools:-dchange(the_tr,ode,{y(x),x},'known'={eta(zeta)},'uknown'={y(x)},'params'={alpha,beta,gamma}); 
simplify(%); 
numer(lhs(%))=0; 
simplify(numer(lhs(%))/(x^(1-alpha)))=0; 
numer(lhs(%))=0; 
collect(%,[y(x),diff(y(x),x),diff(y(x),x$2)]);
 

Which gives

\[ \left (-\gamma ^{2} x^{-1+2 \gamma } \beta ^{2} x -\gamma ^{2} n^{2}+\alpha ^{2}\right ) y \left (x \right )+\left (-2 \alpha x +x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right ) = 0 \]

Here is another example. Here to make change of variables to polar coordinates by making \(x=r \cos \theta \) and \(y=r \sin \theta \) The ode is

\[ \frac {y-x y'}{\sqrt {1+(y')^2}}=x^2+y^2 \]

In Maple

restart; 
ode := (y(x)-x*diff(y(x),x))/sqrt(1+ diff(y(x),x)^2) = x^2+y(x)^2; 
the_tr:={x=r(t)*cos(t),y(x)=r(t)*sin(t)}; 
PDEtools:-dchange(the_tr,ode,{r(t),t},'known'={y(x)},'uknown'={r(t)});
 

Which gives

\[ \frac {r \left (t \right ) \sin \left (t \right )-\frac {r \left (t \right ) \cos \left (t \right ) \left (\left (\frac {d}{d t}r \left (t \right )\right ) \sin \left (t \right )+r \left (t \right ) \cos \left (t \right )\right )}{\left (\frac {d}{d t}r \left (t \right )\right ) \cos \left (t \right )-r \left (t \right ) \sin \left (t \right )}}{\sqrt {1+\frac {\left (\left (\frac {d}{d t}r \left (t \right )\right ) \sin \left (t \right )+r \left (t \right ) \cos \left (t \right )\right )^{2}}{\left (\left (\frac {d}{d t}r \left (t \right )\right ) \cos \left (t \right )-r \left (t \right ) \sin \left (t \right )\right )^{2}}}} = r \left (t \right )^{2} \left (\cos ^{2}\left (t \right )\right )+r \left (t \right )^{2} \left (\sin ^{2}\left (t \right )\right ) \]

Here is another example. Where we want to change \(R(r)\) to \(y(x)\) everywhere

\[ \frac {d^{2}}{d r^{2}}R \left (r \right )+\frac {d}{d r}R \left (r \right )+R \left (r \right ) = 0 \]

restart; 
ode:=diff(R(r),r$2)+diff(R(r),r)+R(r)=0; 
the_tr:={r=x,R(r)=y(x)}; 
PDEtools:-dchange(the_tr,ode,{y(x),x},'known'={R(r),r},'uknown'={y(x),x});
 

\[ \frac {d^{2}}{d x^{2}}y \left (x \right )+\frac {d}{d x}y \left (x \right )+y \! \left (x \right ) = 0 \]

The format of the transformation is old_independent_variable=new_independent_variable and old_dependent_variable=new_dependent_variable