Link to actual problem [11896] \[ \boxed {\left (2 x^{2}-3\right ) y^{\prime \prime }-2 y^{\prime } x +y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, y^{\prime }\left (0\right ) = 5] \end {align*}
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^{2}-3\right )^{\frac {3}{4}} \operatorname {LegendreP}\left (\frac {\sqrt {2}}{2}-\frac {1}{2}, \frac {3}{2}, \frac {\sqrt {6}\, x}{3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\left (2 x^{2}-3\right )^{\frac {3}{4}} \operatorname {LegendreP}\left (\frac {\sqrt {2}}{2}-\frac {1}{2}, \frac {3}{2}, \frac {\sqrt {6}\, x}{3}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \left (2 x^{2}-3\right )^{\frac {3}{4}} \operatorname {LegendreQ}\left (\frac {\sqrt {2}}{2}-\frac {1}{2}, \frac {3}{2}, \frac {\sqrt {6}\, x}{3}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {y}{\left (2 x^{2}-3\right )^{\frac {3}{4}} \operatorname {LegendreQ}\left (\frac {\sqrt {2}}{2}-\frac {1}{2}, \frac {3}{2}, \frac {\sqrt {6}\, x}{3}\right )}\right ] \\ \end{align*}