Link to actual problem [9781] \[ \boxed {y^{\prime \prime \prime }-y^{\prime \prime } x^{2}+\left (a +b -1\right ) x y^{\prime }-b y a=0} \]
type detected by program
{"unknown"}
type detected by Maple
[[_3rd_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 {a}{3}, -\frac {b}{3}\right ], \left [\frac {1}{3}, \frac {2}{3}\right ], \frac {x^{3}}{3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\operatorname {hypergeom}\left (\left [-\frac {a}{3}, -\frac {b}{3}\right ], \left [\frac {1}{3}, \frac {2}{3}\right ], \frac {x^{3}}{3}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \operatorname {hypergeom}\left (\left [\frac {1}{3}-\frac {a}{3}, \frac {1}{3}-\frac {b}{3}\right ], \left [\frac {2}{3}, \frac {4}{3}\right ], \frac {x^{3}}{3}\right ) x\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\operatorname {hypergeom}\left (\left [\frac {1}{3}-\frac {a}{3}, \frac {1}{3}-\frac {b}{3}\right ], \left [\frac {2}{3}, \frac {4}{3}\right ], \frac {x^{3}}{3}\right ) x}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \operatorname {hypergeom}\left (\left [-\frac {a}{3}+\frac {2}{3}, -\frac {b}{3}+\frac {2}{3}\right ], \left [\frac {4}{3}, \frac {5}{3}\right ], \frac {x^{3}}{3}\right ) x^{2}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\operatorname {hypergeom}\left (\left [-\frac {a}{3}+\frac {2}{3}, -\frac {b}{3}+\frac {2}{3}\right ], \left [\frac {4}{3}, \frac {5}{3}\right ], \frac {x^{3}}{3}\right ) x^{2}}\right ] \\ \end{align*}