Link to actual problem [1448] \[ \boxed {9 x^{2} y^{\prime \prime }-3 x \left (2 x^{2}+11\right ) y^{\prime }+\left (10 x^{2}+13\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Regular singular point. Difference is 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 &= {\mathrm e}^{\frac {x^{2}}{6}} x^{\frac {4}{3}} \operatorname {WhittakerM}\left (\frac {1}{6}, 1, \frac {x^{2}}{3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{-\frac {x^{2}}{6}} y}{x^{\frac {4}{3}} \operatorname {WhittakerM}\left (\frac {1}{6}, 1, \frac {x^{2}}{3}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= {\mathrm e}^{\frac {x^{2}}{6}} x^{\frac {4}{3}} \operatorname {WhittakerW}\left (\frac {1}{6}, 1, \frac {x^{2}}{3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{-\frac {x^{2}}{6}} y}{x^{\frac {4}{3}} \operatorname {WhittakerW}\left (\frac {1}{6}, 1, \frac {x^{2}}{3}\right )}\right ] \\ \end{align*}