Link to actual problem [11013] \[ \boxed {x \left (x^{2}+a \right ) y^{\prime \prime }+\left (b \,x^{2}+c \right ) y^{\prime }+s x y=0} \]
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 &= x^{1-\frac {c}{a}} \left (x^{2}+a \right )^{-\frac {b}{2}+1+\frac {c}{2 a}} \operatorname {hypergeom}\left (\left [-\frac {b}{4}+\frac {5}{4}-\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}, -\frac {b}{4}+\frac {5}{4}+\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}\right ], \left [\frac {3}{2}-\frac {c}{2 a}\right ], -\frac {x^{2}}{a}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{\frac {c}{a}} \left (x^{2}+a \right )^{\frac {b}{2}} \left (x^{2}+a \right )^{-\frac {c}{2 a}} y}{x \left (x^{2}+a \right ) \operatorname {hypergeom}\left (\left [-\frac {b}{4}+\frac {5}{4}-\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}, -\frac {b}{4}+\frac {5}{4}+\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}\right ], \left [\frac {-c +3 a}{2 a}\right ], -\frac {x^{2}}{a}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (x^{2}+a \right )^{-\frac {b}{2}+1+\frac {c}{2 a}} \operatorname {hypergeom}\left (\left [\frac {3}{4}-\frac {\sqrt {b^{2}-2 b -4 s +1}\, a +a b -2 c}{4 a}, \frac {3}{4}+\frac {\sqrt {b^{2}-2 b -4 s +1}\, a -a b +2 c}{4 a}\right ], \left [\frac {1}{2}+\frac {c}{2 a}\right ], -\frac {x^{2}}{a}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (x^{2}+a \right )^{\frac {b}{2}} \left (x^{2}+a \right )^{-\frac {c}{2 a}} y}{\left (x^{2}+a \right ) \operatorname {hypergeom}\left (\left [-\frac {\sqrt {b^{2}-2 b -4 s +1}\, a +a b -3 a -2 c}{4 a}, \frac {\sqrt {b^{2}-2 b -4 s +1}\, a +\left (-b +3\right ) a +2 c}{4 a}\right ], \left [\frac {a +c}{2 a}\right ], -\frac {x^{2}}{a}\right )}\right ] \\ \end{align*}