Link to actual problem [14009] \[ \boxed {x^{2} \left (2 x +1\right ) y^{\prime \prime }+x y^{\prime }+\left (4 x^{3}-4\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Regular singular point. 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 &= {\mathrm e}^{-i \sqrt {2}\, x} \operatorname {HeunC}\left (i \sqrt {2}, 4, -2, \frac {1}{2}, 5, -2 x \right ) x^{2}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{i \sqrt {2}\, x} y}{\operatorname {HeunC}\left (i \sqrt {2}, 4, -2, \frac {1}{2}, 5, -2 x \right ) x^{2}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= {\mathrm e}^{-i \sqrt {2}\, x} \operatorname {HeunC}\left (i \sqrt {2}, 4, -2, \frac {1}{2}, 5, -2 x \right ) x^{2} \left (\int \frac {\left (1+2 x \right ) {\mathrm e}^{2 i \sqrt {2}\, x}}{x^{5} \operatorname {HeunC}\left (i \sqrt {2}, 4, -2, \frac {1}{2}, 5, -2 x \right )^{2}}d x \right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{i \sqrt {2}\, x} y}{\operatorname {HeunC}\left (i \sqrt {2}, 4, -2, \frac {1}{2}, 5, -2 x \right ) x^{2} \left (\int \frac {\left (1+2 x \right ) {\mathrm e}^{2 i \sqrt {2}\, x}}{x^{5} \operatorname {HeunC}\left (i \sqrt {2}, 4, -2, \frac {1}{2}, 5, -2 x \right )^{2}}d x \right )}\right ] \\ \end{align*}