Link to actual problem [11082] \[ \boxed {x^{2} \left (a \,x^{n}+b \right )^{2} y^{\prime \prime }+\left (n +1\right ) x \left (a^{2} x^{2 n}-b^{2}\right ) y^{\prime }+y c=0} \]
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 &= x \left (a \,x^{n}+b \right )^{-1-\frac {1}{n}} \sqrt {a \,x^{2 n}+b \,x^{n}}\, \left (\frac {x^{n}}{a \,x^{n}+b}\right )^{\frac {\sqrt {\frac {\left (n +2\right )^{2} b^{2}-4 c}{n^{2} a^{2}}}\, a}{2 b}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (a \,x^{n}+b \right ) \left (a \,x^{n}+b \right )^{\frac {1}{n}} \left (\frac {x^{n}}{a \,x^{n}+b}\right )^{-\frac {\sqrt {\frac {\left (n +2\right )^{2} b^{2}-4 c}{n^{2} a^{2}}}\, a}{2 b}} y}{x \sqrt {a \,x^{2 n}+b \,x^{n}}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= x \left (a \,x^{n}+b \right )^{-1-\frac {1}{n}} \sqrt {a \,x^{2 n}+b \,x^{n}}\, \left (\frac {x^{n}}{a \,x^{n}+b}\right )^{-\frac {\sqrt {\frac {\left (n +2\right )^{2} b^{2}-4 c}{n^{2} a^{2}}}\, a}{2 b}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (a \,x^{n}+b \right ) \left (a \,x^{n}+b \right )^{\frac {1}{n}} \left (\frac {x^{n}}{a \,x^{n}+b}\right )^{\frac {\sqrt {\frac {\left (n +2\right )^{2} b^{2}-4 c}{n^{2} a^{2}}}\, a}{2 b}} y}{x \sqrt {a \,x^{2 n}+b \,x^{n}}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= \left (x^{n}\right )^{\frac {1}{n}} \left (a \,x^{n}+b \right ), \underline {\hspace {1.25 ex}}\eta &= 0\right ] \\ \left [R &= y, S \left (R \right ) &= \frac {x \left (x^{n}\right )^{-\frac {1}{n}} \left (\frac {\ln \left (x^{n}\right )}{b}-\frac {\ln \left (a \,x^{n}+b \right )}{b}\right )}{n}\right ] \\ \end{align*}