Link to actual problem [1368] \[ \boxed {25 x^{2} y^{\prime \prime }+x \left (15+x \right ) y^{\prime }+\left (x +1\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Regular singular point. Repeated root"}
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}{50}} \operatorname {WhittakerM}\left (\frac {7}{10}, 0, \frac {x}{25}\right )}{x^{\frac {3}{10}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x}{50}} x^{\frac {3}{10}} y}{\operatorname {WhittakerM}\left (\frac {7}{10}, 0, \frac {x}{25}\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}{50}} \operatorname {WhittakerW}\left (\frac {7}{10}, 0, \frac {x}{25}\right )}{x^{\frac {3}{10}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x}{50}} x^{\frac {3}{10}} y}{\operatorname {WhittakerW}\left (\frac {7}{10}, 0, \frac {x}{25}\right )}\right ] \\ \end{align*}