Link to actual problem [1378] \[ \boxed {x^{2} \left (x +1\right ) y^{\prime \prime }+x \left (1+2 x \right ) y^{\prime }+y x=0} \] 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 &= \operatorname {hypergeom}\left (\left [\frac {1}{2}-\frac {i \sqrt {3}}{2}, \frac {1}{2}-\frac {i \sqrt {3}}{2}\right ], \left [1-i \sqrt {3}\right ], -\frac {1}{x}\right ) x^{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\sqrt {x}\, x^{-\frac {i \sqrt {3}}{2}} y}{\operatorname {hypergeom}\left (\left [\frac {1}{2}-\frac {i \sqrt {3}}{2}, \frac {1}{2}-\frac {i \sqrt {3}}{2}\right ], \left [1-i \sqrt {3}\right ], -\frac {1}{x}\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 {1}{2}+\frac {i \sqrt {3}}{2}, \frac {1}{2}+\frac {i \sqrt {3}}{2}\right ], \left [1+i \sqrt {3}\right ], -\frac {1}{x}\right ) x^{-\frac {1}{2}-\frac {i \sqrt {3}}{2}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\sqrt {x}\, x^{\frac {i \sqrt {3}}{2}} y}{\operatorname {hypergeom}\left (\left [\frac {1}{2}+\frac {i \sqrt {3}}{2}, \frac {1}{2}+\frac {i \sqrt {3}}{2}\right ], \left [1+i \sqrt {3}\right ], -\frac {1}{x}\right )}\right ] \\ \end{align*}