Link to actual problem [1370] \[ \boxed {x^{2} \left (4 x +9\right ) y^{\prime \prime }+3 y^{\prime } x +\left (x +1\right ) y=0} \] 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 &= x^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{6}, -\frac {1}{6}\right ], \left [-\frac {1}{3}\right ], 1+\frac {4 x}{9}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{x^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{6}, -\frac {1}{6}\right ], \left [-\frac {1}{3}\right ], 1+\frac {4 x}{9}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x^{\frac {1}{3}} \left (9+4 x \right )^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [\frac {7}{6}, \frac {7}{6}\right ], \left [\frac {7}{3}\right ], 1+\frac {4 x}{9}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{x^{\frac {1}{3}} \left (9+4 x \right )^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [\frac {7}{6}, \frac {7}{6}\right ], \left [\frac {7}{3}\right ], 1+\frac {4 x}{9}\right )}\right ] \\ \end{align*}