3.1.6 The linearized symmetry condition

This was derived in the introduction

\begin{equation} \fbox {$\eta _x+\omega \left ( \eta _y-\xi _x\right ) -\omega ^2\xi _y-\omega _x\xi -\omega _y\eta =0$} \tag {14}\end{equation}

The above equation (14) is what is used to determine \(\xi ,\eta \). It is the linearized symmetry condition. There is an additional constraint not mentioned above which is

\[ \bar {x}_{x}\bar {y}_{y}\neq \bar {x}_{y}\bar {y}_{x}\]

The restricted form of (14) is

\[ \chi _{x}+\chi _{y}\omega -\chi \omega _{y}=0 \]

An important property is the following. Given any

\[ \xi =A,\eta =B \]

Then we can always write the above as

\[ \xi =0,\eta =B-\omega A \]

So that \(\xi =0\) can always be used if needed to simplify some things.

After finding \(\xi ,\eta \) from (14), the question now becomes is how to use them to solve the original ODE?