Link to actual problem [6588] \[ \boxed {\left (x^{3}+4 x \right ) y^{\prime \prime }-2 y^{\prime } x +6 y=0} \] 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 &= \left (x +2 i\right )^{-\frac {i}{4}} \left (2 i-x \right )^{\frac {i}{4}} {\mathrm e}^{\frac {\arctan \left (\frac {x}{2}\right )}{2}} \operatorname {HeunG}\left (2, 3 i, -1, 0, -\frac {i}{2}, 0, 1-\frac {i x}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (x +2 i\right )^{\frac {i}{4}} \left (2 i-x \right )^{-\frac {i}{4}} {\mathrm e}^{-\frac {\arctan \left (\frac {x}{2}\right )}{2}} y}{\operatorname {HeunG}\left (2, 3 i, -1, 0, -\frac {i}{2}, 0, 1-\frac {i x}{2}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (2 i-x \right )^{\frac {i}{4}} \left (x +2 i\right )^{1+\frac {i}{4}} {\mathrm e}^{\frac {\arctan \left (\frac {x}{2}\right )}{2}} \operatorname {HeunG}\left (2, -\frac {1}{4}+\frac {7 i}{2}, \frac {i}{2}, 1+\frac {i}{2}, 2+\frac {i}{2}, 0, 1-\frac {i x}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (2 i-x \right )^{-\frac {i}{4}} \left (x +2 i\right )^{-1-\frac {i}{4}} {\mathrm e}^{-\frac {\arctan \left (\frac {x}{2}\right )}{2}} y}{\operatorname {HeunG}\left (2, -\frac {1}{4}+\frac {7 i}{2}, \frac {i}{2}, 1+\frac {i}{2}, 2+\frac {i}{2}, 0, 1-\frac {i x}{2}\right )}\right ] \\ \end{align*}