Link to actual problem [9387] \[ \boxed {y^{\prime \prime }+\left (x a +b \right ) y^{\prime }+\left (\operatorname {a1} \,x^{2}+\operatorname {b1} x +\operatorname {c1} \right ) y=0} \]
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 &= \operatorname {hypergeom}\left (\left [\frac {1}{4}+\frac {a^{3}-2 a^{2} \operatorname {c1} +2 a b \operatorname {b1} -2 \operatorname {a1} \,b^{2}-4 a \operatorname {a1} +8 \operatorname {a1} \operatorname {c1} -2 \operatorname {b1}^{2}}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {\left (x \,a^{2}+a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )^{2}}{2 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}\right ) {\mathrm e}^{-\frac {x \left (\left (x a +2 b \right ) \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}+\left (a^{2}-4 \operatorname {a1} \right ) \left (x \,a^{2}+2 a b -4 \operatorname {a1} x -4 \operatorname {b1} \right )\right )}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x \left (\left (x a +2 b \right ) \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}+\left (a^{2}-4 \operatorname {a1} \right ) \left (x \,a^{2}+2 a b -4 \operatorname {a1} x -4 \operatorname {b1} \right )\right )}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}} y}{\operatorname {hypergeom}\left (\left [\frac {\left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}+a^{3}-2 a^{2} \operatorname {c1} +\left (2 b \operatorname {b1} -4 \operatorname {a1} \right ) a +\left (-2 b^{2}+8 \operatorname {c1} \right ) \operatorname {a1} -2 \operatorname {b1}^{2}}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {\left (x \,a^{2}+a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )^{2}}{2 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{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 [\frac {3}{4}+\frac {a^{3}-2 a^{2} \operatorname {c1} +2 a b \operatorname {b1} -2 \operatorname {a1} \,b^{2}-4 a \operatorname {a1} +8 \operatorname {a1} \operatorname {c1} -2 \operatorname {b1}^{2}}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {\left (x \,a^{2}+a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )^{2}}{2 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}\right ) {\mathrm e}^{-\frac {x \left (\left (x a +2 b \right ) \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}+\left (a^{2}-4 \operatorname {a1} \right ) \left (x \,a^{2}+2 a b -4 \operatorname {a1} x -4 \operatorname {b1} \right )\right )}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}} \left (x \,a^{2}+a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {x \left (\left (x a +2 b \right ) \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}+\left (a^{2}-4 \operatorname {a1} \right ) \left (x \,a^{2}+2 a b -4 \operatorname {a1} x -4 \operatorname {b1} \right )\right )}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}} y}{\operatorname {hypergeom}\left (\left [\frac {3 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}+a^{3}-2 a^{2} \operatorname {c1} +\left (2 b \operatorname {b1} -4 \operatorname {a1} \right ) a +\left (-2 b^{2}+8 \operatorname {c1} \right ) \operatorname {a1} -2 \operatorname {b1}^{2}}{4 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {\left (x \,a^{2}+a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )^{2}}{2 \left (a^{2}-4 \operatorname {a1} \right )^{{3}/{2}}}\right ) \left (x \,a^{2}+a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )}\right ] \\ \end{align*}