Link to actual problem [11891] \[ \boxed {\left (x^{3}-1\right ) y^{\prime \prime }+y^{\prime } x^{2}+x y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Ordinary point", "second order series method. Taylor series method"}
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 &= \left (x^{3}-1\right )^{\frac {2}{3}} x \operatorname {hypergeom}\left (\left [1+\frac {i}{3}, 1-\frac {i}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\left (x^{3}-1\right )^{\frac {2}{3}} x \operatorname {hypergeom}\left (\left [1+\frac {i}{3}, 1-\frac {i}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (x^{3}-1\right )^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {2}{3}+\frac {i}{3}, \frac {2}{3}-\frac {i}{3}\right ], \left [\frac {2}{3}\right ], x^{3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\left (x^{3}-1\right )^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {2}{3}+\frac {i}{3}, \frac {2}{3}-\frac {i}{3}\right ], \left [\frac {2}{3}\right ], x^{3}\right )}\right ] \\ \end{align*}