Link to actual problem [9690] \[ \boxed {y^{\prime \prime }+\frac {\left (x^{2} a +a -2\right ) y^{\prime }}{x \left (x^{2}-1\right )}+\frac {b y}{x^{2}}=0} \]
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 &= x^{\frac {a}{2}-\frac {1}{2}+\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}} \left (x^{2}-1\right )^{-a +2} \operatorname {hypergeom}\left (\left [-\frac {a}{2}+\frac {3}{2}, -\frac {a}{2}+\frac {3}{2}+\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], \left [1+\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], x^{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{-\frac {a}{2}} \sqrt {x}\, x^{-\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}} \left (x^{2}-1\right )^{a} y}{\left (x^{2}-1\right )^{2} \operatorname {hypergeom}\left (\left [-\frac {a}{2}+\frac {3}{2}, -\frac {a}{2}+\frac {3}{2}+\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], \left [1+\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], x^{2}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x^{\frac {a}{2}-\frac {1}{2}-\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}} \left (x^{2}-1\right )^{-a +2} \operatorname {hypergeom}\left (\left [-\frac {a}{2}+\frac {3}{2}, -\frac {a}{2}+\frac {3}{2}-\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], \left [1-\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], x^{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{-\frac {a}{2}} \sqrt {x}\, x^{\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}} \left (x^{2}-1\right )^{a} y}{\left (x^{2}-1\right )^{2} \operatorname {hypergeom}\left (\left [-\frac {a}{2}+\frac {3}{2}, -\frac {a}{2}+\frac {3}{2}-\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], \left [1-\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right ], x^{2}\right )}\right ] \\ \end{align*}