Link to actual problem [1415] \[ \boxed {x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+x \left (10 x^{2}+3\right ) y^{\prime }-\left (-14 x^{2}+15\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
type detected by program
{"second order series method. Regular singular point. 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^{3}}{\left (x^{2}+1\right )^{\frac {5}{2}}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (x^{2}+1\right )^{\frac {5}{2}} y}{x^{3}}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {x^{3} \operatorname {arctanh}\left (\frac {1}{\sqrt {x^{2}+1}}\right )}{\left (x^{2}+1\right )^{\frac {5}{2}}}-\frac {x}{\left (x^{2}+1\right )^{2}}+\frac {2}{3 x \left (x^{2}+1\right )^{2}}+\frac {8}{x^{3} \left (x^{2}+1\right )^{2}}+\frac {16}{3 x^{5} \left (x^{2}+1\right )^{2}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {\left (x^{2}+1\right )^{\frac {9}{2}} x^{5} y}{x^{8} \left (x^{2}+1\right )^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {x^{2}+1}}\right )-\left (x^{2}+1\right )^{\frac {5}{2}} \left (x^{2}+2\right ) \left (x^{4}-\frac {8}{3} x^{2}-\frac {8}{3}\right )}\right ] \\ \end{align*}