Link to actual problem [1191] \[ \boxed {\left (3 x^{2}+1\right ) y^{\prime \prime }+3 y^{\prime } x^{2}-2 y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Ordinary point", "second order series method. Taylor series method"}
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 &= {\mathrm e}^{\frac {i \sqrt {3}}{6}+\frac {\sqrt {3}\, \arctan \left (\sqrt {3}\, x \right )}{6}} \operatorname {HeunC}\left (\frac {2 i \sqrt {3}}{3}, -1-\frac {i \sqrt {3}}{6}, 1-\frac {i \sqrt {3}}{6}, 0, -\frac {1}{24}, -\frac {i \sqrt {3}\, x}{2}+\frac {1}{2}\right ) \left (-9 x^{2}-3\right )^{-\frac {i \sqrt {3}}{12}} \left (-3 x +i \sqrt {3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{-\frac {i \sqrt {3}}{6}} {\mathrm e}^{-\frac {\sqrt {3}\, \arctan \left (\sqrt {3}\, x \right )}{6}} \left (-9 x^{2}-3\right )^{\frac {i \sqrt {3}}{12}} y}{\operatorname {HeunC}\left (\frac {2 i \sqrt {3}}{3}, -1-\frac {i \sqrt {3}}{6}, 1-\frac {i \sqrt {3}}{6}, 0, -\frac {1}{24}, -\frac {i \sqrt {3}\, x}{2}+\frac {1}{2}\right ) \left (-3 x +i \sqrt {3}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (-3 x +i \sqrt {3}\right )^{1-\frac {i \sqrt {3}}{12}} \left (3 x +i \sqrt {3}\right )^{1+\frac {i \sqrt {3}}{12}} {\mathrm e}^{\frac {i \sqrt {3}}{6}+\frac {\sqrt {3}\, \arctan \left (\sqrt {3}\, x \right )}{6}} \operatorname {HeunC}\left (\frac {2 i \sqrt {3}}{3}, \frac {i \sqrt {3}}{6}+1, 1-\frac {i \sqrt {3}}{6}, 0, -\frac {1}{24}, -\frac {i \sqrt {3}\, x}{2}+\frac {1}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (-3 x +i \sqrt {3}\right )^{\frac {i \sqrt {3}}{12}} \left (3 x +i \sqrt {3}\right )^{-\frac {i \sqrt {3}}{12}} {\mathrm e}^{-\frac {i \sqrt {3}}{6}} {\mathrm e}^{-\frac {\sqrt {3}\, \arctan \left (\sqrt {3}\, x \right )}{6}} y}{\left (-3 x +i \sqrt {3}\right ) \left (3 x +i \sqrt {3}\right ) \operatorname {HeunC}\left (\frac {2 i \sqrt {3}}{3}, \frac {i \sqrt {3}}{6}+1, 1-\frac {i \sqrt {3}}{6}, 0, -\frac {1}{24}, -\frac {i \sqrt {3}\, x}{2}+\frac {1}{2}\right )}\right ] \\ \end{align*}