Link to actual problem [1377] \[ \boxed {x^{2} \left (4 x +1\right ) y^{\prime \prime }-x \left (-4 x +1\right ) y^{\prime }+\left (x +1\right ) y=0} \] 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 &= x \left (1+4 x \right )^{-1+\frac {i}{2}} \operatorname {hypergeom}\left (\left [-\frac {i}{2}, 1-\frac {i}{2}\right ], \left [1-i\right ], \frac {1}{1+4 x}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (1+4 x \right )^{1-\frac {i}{2}} y}{x \operatorname {hypergeom}\left (\left [-\frac {i}{2}, 1-\frac {i}{2}\right ], \left [1-i\right ], \frac {1}{1+4 x}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x \left (1+4 x \right )^{-1-\frac {i}{2}} \operatorname {hypergeom}\left (\left [\frac {i}{2}, 1+\frac {i}{2}\right ], \left [1+i\right ], \frac {1}{1+4 x}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (1+4 x \right )^{1+\frac {i}{2}} y}{x \operatorname {hypergeom}\left (\left [\frac {i}{2}, 1+\frac {i}{2}\right ], \left [1+i\right ], \frac {1}{1+4 x}\right )}\right ] \\ \end{align*}