Link to actual problem [2903] \[ \boxed {y^{\prime \prime }-y^{\prime } x^{2}-3 y x=0} \] 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 {x^{3}}{3}} x\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{-\frac {x^{3}}{3}} y}{x}\right ] \\ \end{align*}
\begin{align*} \left [\underline {\hspace {1.25 ex}}\xi &= 0, \underline {\hspace {1.25 ex}}\eta &= \frac {\left (x^{6} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, \frac {x^{3}}{3}\right )+5 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {x^{3}}{3}\right ) x^{3}+10 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {x^{3}}{3}\right )\right ) {\mathrm e}^{\frac {x^{3}}{6}}}{x^{4}}\right ] \\ \left [R &= x, S \left (R \right ) &= \frac {{\mathrm e}^{-\frac {x^{3}}{6}} x^{4} y}{x^{6} \operatorname {WhittakerM}\left (\frac {1}{3}, \frac {5}{6}, \frac {x^{3}}{3}\right )+5 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {x^{3}}{3}\right ) x^{3}+10 \operatorname {WhittakerM}\left (\frac {4}{3}, \frac {5}{6}, \frac {x^{3}}{3}\right )}\right ] \\ \end{align*}