Link to actual problem [1331] \[ \boxed {3 x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+5 x \left (x^{2}+1\right ) y^{\prime }-\left (-5 x^{2}+1\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Regular singular point. Difference not integer"}
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 {\left (x^{2}+1\right ) \operatorname {hypergeom}\left (\left [\frac {2}{3}+\frac {i \sqrt {14}}{6}, \frac {2}{3}-\frac {i \sqrt {14}}{6}\right ], \left [\frac {1}{3}\right ], -x^{2}\right )}{x}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x y}{\left (x^{2}+1\right ) \operatorname {hypergeom}\left (\left [\frac {2}{3}+\frac {i \sqrt {14}}{6}, \frac {2}{3}-\frac {i \sqrt {14}}{6}\right ], \left [\frac {1}{3}\right ], -x^{2}\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 {4}{3}+\frac {i \sqrt {14}}{6}, \frac {4}{3}-\frac {i \sqrt {14}}{6}\right ], \left [\frac {5}{3}\right ], -x^{2}\right ) x^{\frac {1}{3}} \left (x^{\frac {2}{3}}+1\right ) \left (x^{\frac {4}{3}}-x^{\frac {2}{3}}+1\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\operatorname {hypergeom}\left (\left [\frac {4}{3}+\frac {i \sqrt {14}}{6}, \frac {4}{3}-\frac {i \sqrt {14}}{6}\right ], \left [\frac {5}{3}\right ], -x^{2}\right ) x^{\frac {1}{3}} \left (x^{\frac {2}{3}}+1\right ) \left (x^{\frac {4}{3}}-x^{\frac {2}{3}}+1\right )}\right ] \\ \end{align*}