Link to actual problem [7870] \[ \boxed {2 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+x \left (11 x^{2}+11 x +9\right ) y^{\prime }+\left (7 x^{2}+10 x +6\right ) y=0} \]
type detected by program
{"kovacic"}
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 {\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, x}{3}+\frac {\sqrt {3}}{3}\right )}{6}} \operatorname {HeunG}\left (\frac {\sqrt {3}+i}{i-\sqrt {3}}, 0, 0, \frac {5}{2}, \frac {1}{2}, \frac {5 \sqrt {3}+3 i}{3 \sqrt {3}+3 i}, -\frac {2 x}{1+i \sqrt {3}}\right ) \left (i \sqrt {3}+2 x +1\right )^{\frac {5 \sqrt {3}+3 i}{6 \sqrt {3}+6 i}} \left (i \sqrt {3}-2 x -1\right )^{\frac {64 i \sqrt {3}+2368}{\left (\sqrt {3}+i\right )^{3} \left (i-\sqrt {3}\right )^{4} \left (13 \sqrt {3}+9 i\right )}}}{\left (x^{2}+x +1\right )^{\frac {1}{4}} x^{2}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 x +1\right )}{3}\right )}{6}} \left (x^{2}+x +1\right )^{\frac {1}{4}} x^{2} \left (i \sqrt {3}+2 x +1\right )^{-\frac {3}{4}+\frac {i \sqrt {3}}{12}} \left (i \sqrt {3}-2 x -1\right )^{-\frac {3}{4}-\frac {i \sqrt {3}}{12}} y}{\operatorname {HeunG}\left (\frac {\sqrt {3}+i}{i-\sqrt {3}}, 0, 0, \frac {5}{2}, \frac {1}{2}, \frac {5 \sqrt {3}+3 i}{3 \sqrt {3}+3 i}, -\frac {2 x}{1+i \sqrt {3}}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {{\mathrm e}^{-\frac {\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, x}{3}+\frac {\sqrt {3}}{3}\right )}{6}} \operatorname {HeunG}\left (\frac {\sqrt {3}+i}{i-\sqrt {3}}, -\frac {64}{\left (i \sqrt {3}-1\right )^{3} \left (i-\sqrt {3}\right )^{4}}, \frac {1}{2}, 3, \frac {3}{2}, \frac {5 \sqrt {3}+3 i}{3 \sqrt {3}+3 i}, -\frac {2 x}{1+i \sqrt {3}}\right ) \left (i \sqrt {3}+2 x +1\right )^{\frac {5 \sqrt {3}+3 i}{6 \sqrt {3}+6 i}} \left (i \sqrt {3}-2 x -1\right )^{\frac {64 i \sqrt {3}+2368}{\left (\sqrt {3}+i\right )^{3} \left (i-\sqrt {3}\right )^{4} \left (13 \sqrt {3}+9 i\right )}}}{\left (x^{2}+x +1\right )^{\frac {1}{4}} x^{\frac {3}{2}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2 x +1\right )}{3}\right )}{6}} \left (x^{2}+x +1\right )^{\frac {1}{4}} x^{\frac {3}{2}} \left (i \sqrt {3}+2 x +1\right )^{-\frac {3}{4}+\frac {i \sqrt {3}}{12}} \left (i \sqrt {3}-2 x -1\right )^{-\frac {3}{4}-\frac {i \sqrt {3}}{12}} y}{\operatorname {HeunG}\left (\frac {\sqrt {3}+i}{i-\sqrt {3}}, -\frac {8}{\left (i-\sqrt {3}\right )^{4}}, \frac {1}{2}, 3, \frac {3}{2}, \frac {5 \sqrt {3}+3 i}{3 \sqrt {3}+3 i}, -\frac {2 x}{1+i \sqrt {3}}\right )}\right ] \\ \end{align*}