Link to actual problem [10045] \[ \boxed {y^{\prime \prime } \left (x +y\right )+{y^{\prime }}^{2}-y^{\prime }=0} \]
type detected by program
{"second_order_integrable_as_is"}
type detected by Maple
[[_2nd_order, _exact, _nonlinear], [_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_y_y1], [_2nd_order, _reducible, _mu_xy]]
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 &= \frac {y}{x}, S \left (R \right ) &= \ln \left (x \right )\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= -\frac {\left (3 x +y \right ) \left (-y +x \right )}{3}, \underline {\hspace {1.25 ex}}\eta &= -\frac {\left (x +3 y \right ) \left (-y +x \right )}{3}\right ] \\ \left [R &= \frac {x +y}{x^{2}-2 x y+y^{2}}, S \left (R \right ) &= \frac {3 \left (x +y\right ) \left (4 x^{2} \ln \left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}+1\right )+4 \ln \left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}+1\right ) x y-4 x^{2} \ln \left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}-1\right )-4 \ln \left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}-1\right ) x y-8 x^{2} \operatorname {arctanh}\left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}\right )-8 \,\operatorname {arctanh}\left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}\right ) x y-x^{2} \sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}+2 \sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}\, x y-\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}\, y^{2}+x^{2}-2 x y+y^{2}\right )}{\left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}+1\right ) \left (\sqrt {\frac {8 x \left (x +y\right )}{x^{2}-2 x y+y^{2}}+1}-1\right ) \left (x -y\right )^{4}}\right ] \\ \end{align*}
\begin{align*} \\ \left [R &= x, S \left (R \right ) &= \frac {\ln \left (x +y\right )}{\textit {\_y1} \left (\textit {\_y1} -1\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {1}{2} x^{2}+x y -\frac {3}{2} x^{2} \textit {\_y1} -\frac {1}{4} y^{2} \textit {\_y1} -\frac {5}{4} \textit {\_y1} x y +\frac {3}{4} \textit {\_y1}^{2} x^{2}+\frac {3}{4} y x \,\textit {\_y1}^{2}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {8 \,\operatorname {arctanh}\left (\frac {-3 \textit {\_y1}^{2} x +5 \textit {\_y1} x +2 y \textit {\_y1} -4 x}{\sqrt {9 \textit {\_y1}^{4} x^{2}-18 \textit {\_y1}^{3} x^{2}+25 \textit {\_y1}^{2} x^{2}-32 x^{2} \textit {\_y1} +16 x^{2}}}\right )}{\sqrt {9 \textit {\_y1}^{4} x^{2}-18 \textit {\_y1}^{3} x^{2}+25 \textit {\_y1}^{2} x^{2}-32 x^{2} \textit {\_y1} +16 x^{2}}}\right ] \\ \end{align*}