Link to actual problem [7654] \[ \boxed {2 x^{2} \left (2+x \right ) y^{\prime \prime }-x \left (4-7 x \right ) y^{\prime }-\left (5-3 x \right ) y=0} \]
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 {x^{\frac {5}{2}}}{\left (2+x \right )^{\frac {7}{2}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (2+x \right )^{\frac {7}{2}} y}{x^{\frac {5}{2}}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {33 x^{\frac {3}{2}} \sqrt {2}}{\left (2+x \right )^{3}}+\frac {15 x^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {2}\, \sqrt {2+x}}{2}\right )}{\left (2+x \right )^{\frac {7}{2}}}+\frac {52 \sqrt {x}\, \sqrt {2}}{\left (2+x \right )^{3}}+\frac {32 \sqrt {2}}{\sqrt {x}\, \left (2+x \right )^{3}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\sqrt {x}\, \left (2+x \right )^{\frac {13}{2}} y}{15 x^{3} \left (2+x \right )^{3} \operatorname {arctanh}\left (\frac {\sqrt {2}\, \sqrt {2+x}}{2}\right )+33 \left (2+x \right )^{\frac {7}{2}} \left (x^{2}+\frac {52}{33} x +\frac {32}{33}\right ) \sqrt {2}}\right ] \\ \end{align*}