Link to actual problem [10981] \[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }+\left (2 a +1\right ) y^{\prime }-b \left (2 a +b \right ) y=0} \]
type detected by program
{"unknown"}
type detected by Maple
[[_2nd_order, _with_linear_symmetries]]
Maple symgen result This shows Maple’s found \(\xi ,\eta \) and the corresponding canonical coordinates \(R,S\)\begin{align*} \\ \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \operatorname {hypergeom}\left (\left [\frac {\sqrt {1+4 b \left (2 a +b \right )}}{2}-\frac {1}{2}, -\frac {1}{2}-\frac {\sqrt {1+4 b \left (2 a +b \right )}}{2}\right ], \left [-a -\frac {1}{2}\right ], \frac {x}{2}+\frac {1}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\operatorname {hypergeom}\left (\left [\frac {\sqrt {8 a b +4 b^{2}+1}}{2}-\frac {1}{2}, -\frac {1}{2}-\frac {\sqrt {8 a b +4 b^{2}+1}}{2}\right ], \left [-a -\frac {1}{2}\right ], \frac {x}{2}+\frac {1}{2}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (\frac {x}{2}+\frac {1}{2}\right )^{a +\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {\sqrt {1+4 b \left (2 a +b \right )}}{2}+1+a , 1-\frac {\sqrt {1+4 b \left (2 a +b \right )}}{2}+a \right ], \left [\frac {5}{2}+a \right ], \frac {x}{2}+\frac {1}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {8 \left (\frac {x}{2}+\frac {1}{2}\right )^{-a} y}{\left (2 x +2\right )^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {\sqrt {8 a b +4 b^{2}+1}}{2}+1+a , 1-\frac {\sqrt {8 a b +4 b^{2}+1}}{2}+a \right ], \left [\frac {5}{2}+a \right ], \frac {x}{2}+\frac {1}{2}\right )}\right ] \\ \end{align*}