Link to actual problem [1235] \[ \boxed {\left (-x^{3}+1\right ) y^{\prime \prime }+15 y^{\prime } x^{2}-36 y x=0} \] 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 &= \operatorname {hypergeom}\left (\left [\frac {10}{3}+\frac {2 \sqrt {7}}{3}, \frac {10}{3}-\frac {2 \sqrt {7}}{3}\right ], \left [\frac {2}{3}\right ], x^{3}\right ) \left (x^{2}+x +1\right )^{6} \left (-1+x \right )^{6}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\operatorname {hypergeom}\left (\left [\frac {10}{3}+\frac {2 \sqrt {7}}{3}, \frac {10}{3}-\frac {2 \sqrt {7}}{3}\right ], \left [\frac {2}{3}\right ], x^{3}\right ) \left (x^{2}+x +1\right )^{6} \left (-1+x \right )^{6}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x \operatorname {hypergeom}\left (\left [\frac {11}{3}+\frac {2 \sqrt {7}}{3}, \frac {11}{3}-\frac {2 \sqrt {7}}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right ) \left (x^{2}+x +1\right )^{6} \left (-1+x \right )^{6}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{x \operatorname {hypergeom}\left (\left [\frac {11}{3}+\frac {2 \sqrt {7}}{3}, \frac {11}{3}-\frac {2 \sqrt {7}}{3}\right ], \left [\frac {4}{3}\right ], x^{3}\right ) \left (x^{2}+x +1\right )^{6} \left (-1+x \right )^{6}}\right ] \\ \end{align*}