Link to actual problem [11099] \[ \boxed {y^{\prime \prime }-y^{\prime }+\left (a \,{\mathrm e}^{2 \lambda x} \left (b \,{\mathrm e}^{\lambda x}+c \right )^{n}+\frac {1}{4}-\frac {\lambda ^{2}}{4}\right ) y=0} \]
type detected by program
{"unknown"}
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 &= {\mathrm e}^{-\frac {x \left (\lambda -1\right )}{2}} \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a \left (b \,{\mathrm e}^{\lambda x}+c \right )^{n +2}}{\lambda ^{2} b^{2} \left (n +2\right )^{2}}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x \left (\lambda -1\right )}{2}} y}{\operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a \left (b \,{\mathrm e}^{\lambda x}+c \right )^{n} \left (b \,{\mathrm e}^{\lambda x}+c \right )^{2}}{\lambda ^{2} b^{2} \left (n +2\right )^{2}}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a \left (b \,{\mathrm e}^{\lambda x}+c \right )^{n +2}}{\lambda ^{2} b^{2} \left (n +2\right )^{2}}\right ) \left (b \,{\mathrm e}^{\frac {x \left (\lambda +1\right )}{2}}+{\mathrm e}^{-\frac {x \left (\lambda -1\right )}{2}} c \right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x \left (\lambda -1\right )}{2}} y}{\operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a \left (b \,{\mathrm e}^{\lambda x}+c \right )^{n} \left (b \,{\mathrm e}^{\lambda x}+c \right )^{2}}{\lambda ^{2} b^{2} \left (n +2\right )^{2}}\right ) \left (b \,{\mathrm e}^{\frac {x \left (\lambda +1\right )}{2}} {\mathrm e}^{\frac {x \left (\lambda -1\right )}{2}}+c \right )}\right ] \\ \end{align*}