Link to actual problem [7559] \[ \boxed {\left (3 x^{2}+x +1\right ) y^{\prime \prime }+\left (2+15 x \right ) y^{\prime }+12 y=0} \]
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 {\left (i \sqrt {11}-6 x -1\right )^{\frac {3}{2}} \left (-36 x^{2}-12 x -12\right )^{-\frac {1}{4}+\frac {i \sqrt {11}}{44}} {\mathrm e}^{\frac {\sqrt {11}\, \arctan \left (\frac {\left (6 x +1\right ) \sqrt {11}}{11}\right )}{22}} \operatorname {hypergeom}\left (\left [\frac {\sqrt {1078+66 i \sqrt {11}}}{44}+\frac {1}{2}-\frac {\sqrt {1078-66 i \sqrt {11}}}{44}, \frac {\sqrt {1078+66 i \sqrt {11}}}{44}+\frac {1}{2}-\frac {\sqrt {1078-66 i \sqrt {11}}}{44}\right ], \left [1-\frac {\sqrt {1078-66 i \sqrt {11}}}{22}\right ], \frac {1}{2}+\frac {i \left (-6 x -1\right ) \sqrt {11}}{22}\right )}{\left (3 x^{2}+x +1\right )^{\frac {5}{4}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (3 x^{2}+x +1\right )^{\frac {5}{4}} \left (-36 x^{2}-12 x -12\right )^{\frac {1}{4}} \left (-36 x^{2}-12 x -12\right )^{-\frac {i \sqrt {11}}{44}} {\mathrm e}^{-\frac {\sqrt {11}\, \arctan \left (\frac {\left (6 x +1\right ) \sqrt {11}}{11}\right )}{22}} y}{\left (i \sqrt {11}-6 x -1\right )^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {1}{2}+\frac {i \sqrt {11}}{22}, \frac {1}{2}+\frac {i \sqrt {11}}{22}\right ], \left [-\frac {1}{2}+\frac {i \sqrt {11}}{22}\right ], \frac {1}{2}+\frac {i \left (-6 x -1\right ) \sqrt {11}}{22}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {\left (i \sqrt {11}+6 x +1\right )^{\frac {5}{4}-\frac {i \sqrt {11}}{44}} \left (i \sqrt {11}-6 x -1\right )^{\frac {5}{4}+\frac {i \sqrt {11}}{44}} {\mathrm e}^{\frac {\sqrt {11}\, \arctan \left (\frac {\left (6 x +1\right ) \sqrt {11}}{11}\right )}{22}} \operatorname {hypergeom}\left (\left [\frac {\sqrt {1078+66 i \sqrt {11}}}{44}+\frac {1}{2}+\frac {\sqrt {1078-66 i \sqrt {11}}}{44}, \frac {\sqrt {1078+66 i \sqrt {11}}}{44}+\frac {1}{2}+\frac {\sqrt {1078-66 i \sqrt {11}}}{44}\right ], \left [1+\frac {\sqrt {1078-66 i \sqrt {11}}}{22}\right ], \frac {1}{2}+\frac {i \left (-6 x -1\right ) \sqrt {11}}{22}\right )}{\left (3 x^{2}+x +1\right )^{\frac {5}{4}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (3 x^{2}+x +1\right )^{\frac {5}{4}} \left (i \sqrt {11}+6 x +1\right )^{\frac {i \sqrt {11}}{44}} \left (i \sqrt {11}-6 x -1\right )^{-\frac {i \sqrt {11}}{44}} {\mathrm e}^{-\frac {\sqrt {11}\, \arctan \left (\frac {\left (6 x +1\right ) \sqrt {11}}{11}\right )}{22}} y}{\left (i \sqrt {11}+6 x +1\right )^{\frac {5}{4}} \left (i \sqrt {11}-6 x -1\right )^{\frac {5}{4}} \operatorname {hypergeom}\left (\left [2, 2\right ], \left [\frac {5}{2}-\frac {i \sqrt {11}}{22}\right ], \frac {1}{2}+\frac {i \left (-6 x -1\right ) \sqrt {11}}{22}\right )}\right ] \\ \end{align*}