Link to actual problem [9823] \[ \boxed {x^{2} y^{\prime \prime \prime }-3 \left (p +q \right ) x y^{\prime \prime }+3 p \left (3 q +1\right ) y^{\prime }-x^{2} y=0} \]
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 [\right ], \left [-p +\frac {2}{3}, -q +\frac {1}{3}\right ], \frac {x^{3}}{27}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\operatorname {hypergeom}\left (\left [\right ], \left [-p +\frac {2}{3}, -q +\frac {1}{3}\right ], \frac {x^{3}}{27}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x^{2+3 q} \operatorname {hypergeom}\left (\left [\right ], \left [q +\frac {5}{3}, \frac {4}{3}+q -p \right ], \frac {x^{3}}{27}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{-3 q} y}{x^{2} \operatorname {hypergeom}\left (\left [\right ], \left [q +\frac {5}{3}, \frac {4}{3}+q -p \right ], \frac {x^{3}}{27}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x^{3 p +1} \operatorname {hypergeom}\left (\left [\right ], \left [p +\frac {4}{3}, \frac {2}{3}-q +p \right ], \frac {x^{3}}{27}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{-3 p} y}{x \operatorname {hypergeom}\left (\left [\right ], \left [p +\frac {4}{3}, \frac {2}{3}-q +p \right ], \frac {x^{3}}{27}\right )}\right ] \\ \end{align*}