Link to actual problem [1258] \[ \boxed {\left (1+2 x \right ) y^{\prime \prime }-\left (1-2 x \right ) y^{\prime }-\left (3-2 x \right ) y=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 1, y^{\prime }\left (1\right ) = -2] \end {align*}
With the expansion point for the power series method at \(x = 1\).
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 &= {\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}} \left (2 x +1\right )^{2} \operatorname {KummerM}\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}, 3, \frac {i \sqrt {3}\, \left (2 x +1\right )}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {\left (1+i \sqrt {3}\right ) x}{2}} y}{\left (2 x +1\right )^{2} \operatorname {KummerM}\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}, 3, \frac {i \sqrt {3}\, \left (2 x +1\right )}{2}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= {\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}} \left (2 x +1\right )^{2} \operatorname {KummerU}\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}, 3, \frac {i \sqrt {3}\, \left (2 x +1\right )}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {\left (1+i \sqrt {3}\right ) x}{2}} y}{\left (2 x +1\right )^{2} \operatorname {KummerU}\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}, 3, \frac {i \sqrt {3}\, \left (2 x +1\right )}{2}\right )}\right ] \\ \end{align*}