Link to actual problem [10908] \[ \boxed {x y^{\prime \prime }+\left (a \,x^{2}+b x +2\right ) y^{\prime }+\left (c \,x^{2}+d x +b \right ) y=0} \]
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 &= \frac {\operatorname {hypergeom}\left (\left [1-\frac {a^{2} d -a b c +c^{2}}{2 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {\left (x \,a^{2}+a b -2 c \right )^{2}}{2 a^{3}}\right ) {\mathrm e}^{-\frac {x \left (x \,a^{2}+2 a b -2 c \right )}{2 a}}}{x}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x \,{\mathrm e}^{\frac {x \left (x \,a^{2}+2 a b -2 c \right )}{2 a}} y}{\operatorname {hypergeom}\left (\left [\frac {2 a^{3}-a^{2} d +a b c -c^{2}}{2 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {\left (x \,a^{2}+a b -2 c \right )^{2}}{2 a^{3}}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {\operatorname {hypergeom}\left (\left [\frac {3}{2}-\frac {a^{2} d -a b c +c^{2}}{2 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {\left (x \,a^{2}+a b -2 c \right )^{2}}{2 a^{3}}\right ) {\mathrm e}^{-\frac {x \left (x \,a^{2}+2 a b -2 c \right )}{2 a}} \left (x \,a^{2}+a b -2 c \right )}{x}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x \left (x \,a^{2}+2 a b -2 c \right )}{2 a}} x y}{\operatorname {hypergeom}\left (\left [\frac {3 a^{3}-a^{2} d +a b c -c^{2}}{2 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {\left (x \,a^{2}+a b -2 c \right )^{2}}{2 a^{3}}\right ) \left (x \,a^{2}+a b -2 c \right )}\right ] \\ \end{align*}