This section shows how to obtain eq. (8) in paper "Computer Algebra Solving of First Order ODEs Using Symmetry Methods" 1996 by Durate, Terrab, Mota. Which is an alternative equation to solve instead of the main Lie condition for symmetry we were looking at above.
Starting with the main linearized symmetry pde\begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {14} \end {equation} Assuming anstaz \begin {equation} \eta =\xi \omega +\chi \tag {A} \end {equation} Hence \begin {align*} \eta _{x} & =\xi _{x}\omega +\xi \omega _{x}+\chi _{x}\\ \eta _{y} & =\xi _{y}\omega +\xi \omega _{y}+\chi _{y} \end {align*}
Then (14) becomes\begin {align} \left ( \xi _{x}\omega +\xi \omega _{x}+\chi _{x}\right ) +\omega \left ( \left ( \xi _{y}\omega +\xi \omega _{y}+\chi _{y}\right ) -\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\left ( \xi \omega +\chi \right ) & =0\nonumber \\ \xi _{x}\omega +\xi \omega _{x}+\chi _{x}+\xi _{y}\omega ^{2}+\xi \omega _{y}\omega +\chi _{y}\omega -\omega \xi _{x}-\omega ^{2}\xi _{y}-\omega _{x}\xi -\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \\ \xi _{x}\omega +\chi _{x}+\xi _{y}\omega ^{2}+\xi \omega _{y}\omega +\chi _{y}\omega -\omega \xi _{x}-\omega ^{2}\xi _{y}-\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \\ \chi _{x}+\xi _{y}\omega ^{2}+\xi \omega _{y}\omega +\chi _{y}\omega -\omega ^{2}\xi _{y}-\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \\ \chi _{x}+\xi \omega _{y}\omega +\chi _{y}\omega -\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \end {align}
Or\begin {equation} \chi _{x}+\chi _{y}\omega -\omega _{y}\chi =0 \tag {1} \end {equation} And hence (1) is now solved for \(\chi \left ( x,y\right ) \). If we are able to find \(\chi \) then we can use the anstaz \(\eta =\xi \omega +\chi \). This leaves only one unknown \(\xi \). The paper does not explain how to solve for this, \(\xi \), which I assume is by using (14) again. The paper only said
The knowledge of \(\chi \), in turn, allows one to set \(\xi \) and \(\eta ~\)as desired using (A)
Which is not too clear how in practice this is done. I need to work an example showing this. The paper says that (1) is solved for \(\chi \left ( x,y\right ) \) by using bivariate polynomial anstaz. The degree can be set by a user, or Maple internally determines this.