Link to actual problem [10858] \[ \boxed {y^{\prime \prime }+\left (a \,x^{2}+2 b \right ) y^{\prime }+\left (a b \,x^{2}-a x +b^{2}\right ) y=0} \]
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 [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{b x} y}{x}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {\left (x^{6} a^{2} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, \frac {a \,x^{3}}{3}\right )+5 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {a \,x^{3}}{3}\right ) a \,x^{3}+10 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {a \,x^{3}}{3}\right )\right ) {\mathrm e}^{-\frac {x \left (x^{2} a +6 b \right )}{6}}}{x^{4}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x \left (x^{2} a +6 b \right )}{6}} x^{4} y}{x^{6} a^{2} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, \frac {a \,x^{3}}{3}\right )+5 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {a \,x^{3}}{3}\right ) a \,x^{3}+10 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {a \,x^{3}}{3}\right )}\right ] \\ \end{align*}