Link to actual problem [1284] \[ \boxed {\left (x^{2}+1\right ) y^{\prime \prime }+\left (x^{2}+2\right ) y^{\prime }+y x=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -3, 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 &= {\mathrm e}^{-\frac {\arctan \left (x \right )}{2}} \left (x +i\right )^{\frac {i}{4}} \left (-x +i\right )^{1+\frac {i}{4}} \operatorname {HeunC}\left (2 i, -1+\frac {i}{2}, 1+\frac {i}{2}, 2 i, -\frac {1}{8}-i, -\frac {i x}{2}+\frac {1}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {\arctan \left (x \right )}{2}} \left (x +i\right )^{-\frac {i}{4}} \left (-x +i\right )^{-1-\frac {i}{4}} y}{\operatorname {HeunC}\left (2 i, -1+\frac {i}{2}, 1+\frac {i}{2}, 2 i, -\frac {1}{8}-i, -\frac {i x}{2}+\frac {1}{2}\right )}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= {\mathrm e}^{-\frac {\arctan \left (x \right )}{2}} \left (x +i\right )^{1-\frac {i}{4}} \left (-x +i\right )^{1+\frac {i}{4}} \operatorname {HeunC}\left (2 i, 1-\frac {i}{2}, 1+\frac {i}{2}, 2 i, -\frac {1}{8}-i, -\frac {i x}{2}+\frac {1}{2}\right )\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{\frac {\arctan \left (x \right )}{2}} \left (x +i\right )^{-1+\frac {i}{4}} \left (-x +i\right )^{-1-\frac {i}{4}} y}{\operatorname {HeunC}\left (2 i, 1-\frac {i}{2}, 1+\frac {i}{2}, 2 i, -\frac {1}{8}-i, -\frac {i x}{2}+\frac {1}{2}\right )}\right ] \\ \end{align*}