Link to actual problem [5022] \[ \boxed {\left (x^{2}-5 x +6\right ) y^{\prime \prime }-3 x y^{\prime }-y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Ordinary point", "second order series method. Taylor series method"}
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 &= \left (-2+x \right )^{-8+\sqrt {5}} \left (x -3\right )^{10} \operatorname {hypergeom}\left (\left [8-\sqrt {5}, 3-\sqrt {5}\right ], \left [1-2 \sqrt {5}\right ], \frac {1}{-2+x}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (-2+x \right )^{8} \left (-2+x \right )^{-\sqrt {5}} y}{\left (x -3\right )^{10} \operatorname {hypergeom}\left (\left [8-\sqrt {5}, 3-\sqrt {5}\right ], \left [1-2 \sqrt {5}\right ], \frac {1}{-2+x}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (-2+x \right )^{-8-\sqrt {5}} \left (x -3\right )^{10} \operatorname {hypergeom}\left (\left [\sqrt {5}+3, 8+\sqrt {5}\right ], \left [1+2 \sqrt {5}\right ], \frac {1}{-2+x}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (-2+x \right )^{8} \left (-2+x \right )^{\sqrt {5}} y}{\left (x -3\right )^{10} \operatorname {hypergeom}\left (\left [\sqrt {5}+3, 8+\sqrt {5}\right ], \left [1+2 \sqrt {5}\right ], \frac {1}{-2+x}\right )}\right ] \\ \end{align*}