Link to actual problem [6589] \[ \boxed {x^{2} \left (x -5\right )^{2} y^{\prime \prime }+4 y^{\prime } x +\left (x^{2}-25\right ) y=0} \] 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 &= x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \left (x -5\right )^{\frac {2}{25}-\frac {\sqrt {2941}}{50}-\frac {i \sqrt {3}}{2}} \operatorname {HeunC}\left (\frac {4}{25}, i \sqrt {3}, \frac {\sqrt {2941}}{25}, -\frac {108}{625}, \frac {1}{2}, -\frac {5}{x -5}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{-\frac {\sqrt {2941}}{50}} \left (x -5\right )^{\frac {\sqrt {2941}}{50}} \left (x -5\right )^{\frac {i \sqrt {3}}{2}} y}{x^{\frac {21}{50}} \left (x -5\right )^{\frac {2}{25}} \operatorname {HeunC}\left (\frac {4}{25}, i \sqrt {3}, \frac {\sqrt {2941}}{25}, -\frac {108}{625}, \frac {1}{2}, -\frac {5}{x -5}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x^{\frac {21}{50}+\frac {\sqrt {2941}}{50}} \left (x -5\right )^{\frac {2}{25}+\frac {i \sqrt {3}}{2}-\frac {\sqrt {2941}}{50}} \operatorname {HeunC}\left (\frac {4}{25}, -i \sqrt {3}, \frac {\sqrt {2941}}{25}, -\frac {108}{625}, \frac {1}{2}, -\frac {5}{x -5}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{-\frac {\sqrt {2941}}{50}} \left (x -5\right )^{-\frac {i \sqrt {3}}{2}} \left (x -5\right )^{\frac {\sqrt {2941}}{50}} y}{x^{\frac {21}{50}} \left (x -5\right )^{\frac {2}{25}} \operatorname {HeunC}\left (\frac {4}{25}, -i \sqrt {3}, \frac {\sqrt {2941}}{25}, -\frac {108}{625}, \frac {1}{2}, -\frac {5}{x -5}\right )}\right ] \\ \end{align*}