Given any first order ODE \begin {equation} \frac {dy}{dx}=\omega \left ( x,y\right ) \tag {A} \end {equation} \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) are called the infinitesimals of the transformation. Maple has function called symgen in the DEtools package to determine these using 16 different algorithms. Starting with the Lie point transformation group \begin {align*} \bar {x} & \equiv \bar {x}\left ( x,y;\epsilon \right ) \\ \bar {y} & \equiv \bar {y}\left ( x,y;\epsilon \right ) \end {align*}
Expanding using Taylor series near\(\ \epsilon =0\) gives\begin {align*} \bar {x} & =x+\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}\epsilon +O\left ( \epsilon ^{2}\right ) \\ & =x+\epsilon \xi \left ( x,y\right ) +O\left ( \epsilon ^{2}\right ) \\ \bar {y} & =y+\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}\epsilon +O\left ( \epsilon ^{2}\right ) \\ & =y+\epsilon \eta \left ( x,y\right ) +O\left ( \epsilon ^{2}\right ) \end {align*}
Ignoring higher order terms gives\begin {align} \bar {x}\left ( x,y\right ) & =x+\epsilon \xi \left ( x,y\right ) \tag {1}\\ \bar {y}\left ( x,y\right ) & =y+\epsilon \eta \left ( x,y\right ) \tag {2} \end {align}
In the above \(\epsilon \) is the one parameter in the Lie symmetry group. The symmetry condition for (A) is that \[ \frac {d\bar {y}}{d\bar {x}}=\omega \left ( \bar {x},\bar {y}\right ) \] Whenever\[ \frac {dy}{dx}=\omega \left ( x,y\right ) \] Symmetry of an ODE means the ODE in \(\left ( x,y\right ) \) remain the same form (but using new variables \(\left ( \bar {x},\bar {y}\right ) \)) after applying the (non-trivial) transformation (1,2).
Nontrivial transformation means \(\epsilon \neq 0\). The first goal is to find the functions \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) which satisfy the symmetry condition above.
The symmetry condition is written as\begin {equation} \frac {d\bar {y}}{d\bar {x}}=\frac {\frac {d\bar {y}}{dx}}{\frac {d\bar {x}}{dx}}=\omega \left ( \bar {x},\bar {y}\right ) \tag {3} \end {equation} Where \(\frac {d\bar {y}}{dx}\) is the total derivative with respect to the \(x\) variable. Similarly for \(\frac {d\bar {x}}{dx}\). But\begin {align} \frac {d\bar {y}}{dx} & =\bar {y}_{x}+\bar {y}_{y}\frac {dy}{dx}\nonumber \\ & =\bar {y}_{x}+\bar {y}_{y}\omega \left ( x,y\right ) \tag {4} \end {align}
And\begin {align} \frac {d\bar {x}}{dx} & =\bar {x}_{x}+\bar {x}_{y}\frac {dy}{dx}\nonumber \\ & =\bar {x}_{x}+\bar {x}_{y}\omega \left ( x,y\right ) \tag {5} \end {align}
Substituting (4,5) into (3) gives the symmetry condition as\begin {equation} \frac {\bar {y}_{x}+\omega \left ( x,y\right ) \bar {y}_{y}}{\bar {x}_{x}+\omega \left ( x,y\right ) \bar {x}_{y}}=\omega \left ( \bar {x},\bar {y}\right ) \tag {6} \end {equation} But\begin {equation} \bar {x}_{x}=1+\epsilon \xi _{x} \tag {7} \end {equation} And similarly \begin {equation} \bar {x}_{y}=\epsilon \xi _{y} \tag {8} \end {equation} And\begin {equation} \bar {y}_{x}=\epsilon \eta _{x} \tag {9} \end {equation} And\begin {equation} \bar {y}_{y}=1+\epsilon \eta _{y} \tag {10} \end {equation} Substituting (7,8,9,10) back into the symmetry condition (6) gives \begin {align} \frac {\epsilon \eta _{x}+\omega \left ( 1+\epsilon \eta _{y}\right ) }{\left ( 1+\epsilon \xi _{x}\right ) +\omega \epsilon \xi _{y}} & =\omega \left ( x+\epsilon \xi ,y+\epsilon \eta \right ) \nonumber \\ \frac {\epsilon \eta _{x}+\omega +\omega s\eta _{y}}{1+\epsilon \xi _{x}+\omega \epsilon \xi _{y}} & =\omega \left ( x+\epsilon \xi ,y+\epsilon \eta \right ) \nonumber \\ \frac {\omega +s\left ( \eta _{x}+\omega \eta _{y}\right ) }{1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) } & =\omega \left ( x+\epsilon \xi ,y+\epsilon \eta \right ) \tag {11} \end {align}
The above is used to determine \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \). The above PDE is too complicated to use as is. It is linearized, and the linearized version is used to solve for \(\xi ,\eta \) near small \(\epsilon \).
Eq. (11) is linearized by expanding the LHS and the RHS using Taylor series around \(\epsilon =0\) . Starting with the LHS first, let \(\frac {\omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) }{1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) }=\Delta _{LHS}\). Expanding this using Taylor series around\(\ \epsilon =0\) gives\begin {equation} \Delta _{LHS}=\Delta _{\epsilon =0}+\epsilon \frac {d}{d\epsilon }\ \left ( \Delta \right ) _{\epsilon =0}+h.o.t. \tag {11A} \end {equation} But \(\Delta _{\epsilon =0}=\omega \) and \begin {align*} \frac {d}{d\epsilon }\ \left ( \Delta _{LHS}\right ) & =\frac {\frac {d}{d\epsilon }\ \left [ \omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ] \left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) -\left ( \omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ) \frac {d}{d\epsilon }\ \left [ 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ] }{\left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) ^{2}}\\ & =\frac {\left ( \eta _{x}+\omega \eta _{y}\right ) \left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) -\left ( \omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ) \left ( \xi _{x}+\omega \xi _{y}\right ) }{\left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) ^{2}} \end {align*}
At \(\epsilon =0\) the above reduces to\begin {align} \frac {d}{d\epsilon }\ \left ( \Delta _{LHS}\right ) _{\epsilon =0} & =\left ( \eta _{x}+\omega \eta _{y}\right ) -\omega \left ( \xi _{x}+\omega \xi _{y}\right ) \nonumber \\ & =\eta _{x}+\omega \eta _{y}-\omega \xi _{x}-\omega ^{2}\xi _{y}\nonumber \\ & =\eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y} \tag {12} \end {align}
Therefore the LHS of Eq. (11A) becomes\begin {equation} \Delta _{LHS}=\omega +\epsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) \tag {11B} \end {equation} Now the RHS of Eq. (11) is linearized. Let \(\omega \left ( x+s\xi ,y+s\eta \right ) =\Delta _{RHS}\). Expansion around \(\epsilon =0\) gives\[ \Delta _{RHS}=\Delta _{\epsilon =0}+\epsilon \left ( \frac {d}{d\epsilon }\ \Delta \right ) _{\epsilon =0}+h.o.t. \] But \(\Delta _{\epsilon =0}=\omega \left ( x,y\right ) \) and \[ \frac {d}{d\epsilon }\ \Delta _{RHS}=\omega _{x}\xi +\omega _{y}\eta \] Hence the linearized RHS of (11) becomes \begin {equation} \Delta _{RHS}=\omega \left ( x,y\right ) +\epsilon \left ( \omega _{x}\xi +\omega _{y}\eta \right ) \tag {13} \end {equation} Substituting (11B,13) back into (11), gives the linearized version of (11) as\begin {align} \Delta _{LHS} & =\Delta _{RHS}\nonumber \\ \omega +\epsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) & =\omega +\epsilon \left ( \omega _{x}\xi +\omega _{y}\eta \right ) \nonumber \\ \epsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) & =\epsilon \left ( \omega _{x}\xi +\omega _{y}\eta \right ) \nonumber \\ \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y} & =\omega _{x}\xi +\omega _{y}\eta \nonumber \end {align}
Hence\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?