Link to actual problem [1322] \[ \boxed {8 x^{2} y^{\prime \prime }+x \left (x^{2}+2\right ) y^{\prime }+y=0} \] 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 {{\mathrm e}^{-\frac {x^{2}}{32}} \operatorname {WhittakerM}\left (-\frac {5}{16}, \frac {1}{16}, \frac {x^{2}}{16}\right )}{x^{\frac {5}{8}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x^{2}}{32}} x^{\frac {5}{8}} y}{\operatorname {WhittakerM}\left (-\frac {5}{16}, \frac {1}{16}, \frac {x^{2}}{16}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {{\mathrm e}^{-\frac {x^{2}}{32}} \operatorname {WhittakerW}\left (-\frac {5}{16}, \frac {1}{16}, \frac {x^{2}}{16}\right )}{x^{\frac {5}{8}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x^{2}}{32}} x^{\frac {5}{8}} y}{\operatorname {WhittakerW}\left (-\frac {5}{16}, \frac {1}{16}, \frac {x^{2}}{16}\right )}\right ] \\ \end{align*}