Link to actual problem [9706] \[ \boxed {y^{\prime \prime }-\frac {12 y}{\left (x +1\right )^{2} \left (x^{2}+2 x +3\right )}=0} \]
type detected by program
{"kovacic", "second_order_bessel_ode"}
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^{2}+2 x +3}{\left (1+x \right )^{2}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (1+x \right )^{2} y}{x^{2}+2 x +3}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {\left (3 x^{2}+6 x +9\right ) \arctan \left (\frac {\sqrt {2}\, \left (1+x \right )}{2}\right )-\sqrt {2}\, \left (x^{3}+2 x^{2}+4 x +1\right )}{\left (1+x \right )^{2}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (1+x \right )^{2} y}{\left (3 x^{2}+6 x +9\right ) \arctan \left (\frac {\sqrt {2}\, \left (1+x \right )}{2}\right )-\sqrt {2}\, \left (x^{3}+2 x^{2}+4 x +1\right )}\right ] \\ \end{align*}