Link to actual problem [742] \[ \boxed {\left (x^{3}+1\right ) y^{\prime \prime }+4 y^{\prime } x +y=0} \] 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 &= \frac {{\mathrm e}^{-\frac {2 \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, x}{3}-\frac {\sqrt {3}}{3}\right )}{3}} \operatorname {HeunG}\left (\frac {-\sqrt {3}+3 i}{\sqrt {3}+3 i}, -\frac {5}{2}-\frac {5 i \sqrt {3}}{6}, -\frac {1}{3}, \frac {2}{3}, -\frac {4}{3}, \frac {4}{3}-\frac {2 i \sqrt {3}}{3}, \frac {-2 x -2}{i \sqrt {3}-3}\right ) \left (2 x -1+i \sqrt {3}\right )^{\frac {2}{3}-\frac {i \sqrt {3}}{3}} \left (-2 x +1+i \sqrt {3}\right )^{\frac {\sqrt {3}+5 i}{3 \sqrt {3}+3 i}}}{\left (x^{2}-x +1\right )^{\frac {1}{3}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}} \left (x^{2}-x +1\right )^{\frac {1}{3}} \left (2 x -1+i \sqrt {3}\right )^{\frac {i \sqrt {3}}{3}} \left (-2 x +1+i \sqrt {3}\right )^{-\frac {2}{3}-\frac {i \sqrt {3}}{3}} y}{\operatorname {HeunG}\left (\frac {-\sqrt {3}+3 i}{\sqrt {3}+3 i}, -\frac {5}{2}-\frac {5 i \sqrt {3}}{6}, -\frac {1}{3}, \frac {2}{3}, -\frac {4}{3}, \frac {4}{3}-\frac {2 i \sqrt {3}}{3}, \frac {-2 x -2}{i \sqrt {3}-3}\right ) \left (2 x -1+i \sqrt {3}\right )^{\frac {2}{3}}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {{\mathrm e}^{-\frac {2 \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, x}{3}-\frac {\sqrt {3}}{3}\right )}{3}} \operatorname {HeunG}\left (\frac {-\sqrt {3}+3 i}{\sqrt {3}+3 i}, \frac {3 i \sqrt {3}}{2}+\frac {9}{2}, 2, 3, \frac {10}{3}, \frac {4}{3}-\frac {2 i \sqrt {3}}{3}, \frac {-2 x -2}{i \sqrt {3}-3}\right ) \left (1+x \right )^{\frac {7}{3}} \left (2 x -1+i \sqrt {3}\right )^{\frac {2}{3}-\frac {i \sqrt {3}}{3}} \left (-2 x +1+i \sqrt {3}\right )^{\frac {\sqrt {3}+5 i}{3 \sqrt {3}+3 i}}}{\left (x^{2}-x +1\right )^{\frac {1}{3}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}} \left (x^{2}-x +1\right )^{\frac {1}{3}} \left (2 x -1+i \sqrt {3}\right )^{\frac {i \sqrt {3}}{3}} \left (-2 x +1+i \sqrt {3}\right )^{-\frac {2}{3}-\frac {i \sqrt {3}}{3}} y}{\operatorname {HeunG}\left (\frac {-\sqrt {3}+3 i}{\sqrt {3}+3 i}, \frac {3 i \sqrt {3}}{2}+\frac {9}{2}, 2, 3, \frac {10}{3}, \frac {4}{3}-\frac {2 i \sqrt {3}}{3}, \frac {-2 x -2}{i \sqrt {3}-3}\right ) \left (1+x \right )^{\frac {7}{3}} \left (2 x -1+i \sqrt {3}\right )^{\frac {2}{3}}}\right ] \\ \end{align*}