Link to actual problem [1439] \[ \boxed {x^{2} \left (x^{2}+1\right ) y^{\prime \prime }-x \left (-2 x^{2}+7\right ) y^{\prime }+12 y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Regular singular point. Difference is integer"}
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 &= \frac {x^{6}}{\left (x^{2}+1\right )^{\frac {7}{2}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (x^{2}+1\right )^{\frac {7}{2}} y}{x^{6}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {8 x^{6}}{\left (x^{2}+1\right )^{3}}-\frac {15 x^{6} \operatorname {arctanh}\left (\frac {1}{\sqrt {x^{2}+1}}\right )}{\left (x^{2}+1\right )^{\frac {7}{2}}}-\frac {9 x^{4}}{\left (x^{2}+1\right )^{3}}-\frac {2 x^{2}}{\left (x^{2}+1\right )^{3}}\right ] \\ \left [R &= x, S \left (R \right ) &= -\frac {\left (x^{2}+1\right )^{\frac {13}{2}} y}{15 \left (x^{4} \left (x^{2}+1\right )^{3} \operatorname {arctanh}\left (\frac {1}{\sqrt {x^{2}+1}}\right )-\frac {8 \left (x^{4}-\frac {9}{8} x^{2}-\frac {1}{4}\right ) \left (x^{2}+1\right )^{\frac {7}{2}}}{15}\right ) x^{2}}\right ] \\ \end{align*}