Link to actual problem [1394] \[ \boxed {x^{2} \left (x^{2}+2\right ) y^{\prime \prime }+x \left (-x^{2}+14\right ) y^{\prime }+2 \left (x^{2}+9\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Regular singular point. Repeated root"}
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 (x^{2}+2\right )^{5} x^{-9+i} \operatorname {hypergeom}\left (\left [3-\frac {i}{2}, 3-\frac {i}{2}\right ], \left [1-i\right ], -\frac {2}{x^{2}}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{9-i} y}{\left (x^{2}+2\right )^{5} \operatorname {hypergeom}\left (\left [3-\frac {i}{2}, 3-\frac {i}{2}\right ], \left [1-i\right ], -\frac {2}{x^{2}}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (x^{2}+2\right )^{5} x^{-9-i} \operatorname {hypergeom}\left (\left [3+\frac {i}{2}, 3+\frac {i}{2}\right ], \left [1+i\right ], -\frac {2}{x^{2}}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {x^{9+i} y}{\left (x^{2}+2\right )^{5} \operatorname {hypergeom}\left (\left [3+\frac {i}{2}, 3+\frac {i}{2}\right ], \left [1+i\right ], -\frac {2}{x^{2}}\right )}\right ] \\ \end{align*}