Link to actual problem [14392] \[ \boxed {y^{\prime } t -y-{y^{\prime }}^{2}+y^{\prime }=1} \]
type detected by program
{"clairaut"}
type detected by Maple
[[_1st_order, _with_linear_symmetries], _Clairaut]
Maple symgen result This shows Maple’s found \(\xi ,\eta \) and the corresponding canonical coordinates \(R,S\)\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 1, \underline {\hspace {1.25 ex}}\eta &= \frac {t}{2}+\frac {1}{2}\right ] \\ \left [R &= y-\frac {t^{2}}{4}-\frac {t}{2}, S \left (R \right ) &= t\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= \frac {t}{2}+\frac {1}{2}, \underline {\hspace {1.25 ex}}\eta &= y +1\right ] \\ \left [R &= \frac {y+1}{\left (t +1\right )^{2}}, S \left (R \right ) &= 2 \ln \left (t +1\right )\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= -\frac {7}{2}+t^{2}+\frac {3}{2} t -2 y, \underline {\hspace {1.25 ex}}\eta &= t y -1\right ] \\ \left [R &= \frac {t^{2}-4 y+2 t -3}{t^{2}-8 t y+16 y^{2}+2 t -8 y+1}, S \left (R \right ) &= \int _{}^{y}\frac {1}{\frac {\left (\frac {4 \left (t^{2}-4 y+2 t -3\right ) \textit {\_a}}{t^{2}-8 t y+16 y^{2}+2 t -8 y+1}+2 \sqrt {\frac {4 \left (t^{2}-4 y+2 t -3\right ) \textit {\_a}^{2}}{t^{2}-8 t y+16 y^{2}+2 t -8 y+1}-\frac {\left (t^{2}-4 y+2 t -3\right ) \textit {\_a}}{t^{2}-8 t y+16 y^{2}+2 t -8 y+1}-\frac {t^{2}-4 y+2 t -3}{t^{2}-8 t y+16 y^{2}+2 t -8 y+1}+\textit {\_a} +1}-\frac {t^{2}-4 y+2 t -3}{t^{2}-8 t y+16 y^{2}+2 t -8 y+1}+1\right ) \textit {\_a}}{\frac {t^{2}-4 y+2 t -3}{t^{2}-8 t y+16 y^{2}+2 t -8 y+1}-1}-1}d \textit {\_a}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= t^{3}+\frac {5}{2}+t^{2}-3 t y -\frac {3}{2} t +y, \underline {\hspace {1.25 ex}}\eta &= t^{2} y +t^{2}-2 y^{2}+2\right ] \\ \operatorname {FAIL} \\ \end{align*}