Given\(\ \xi =1,\eta =2x\) find Lie group \(\bar {x},\bar {y}\). Since\[ \xi \left ( x,y\right ) =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}\] Then \begin {align} \frac {d\bar {x}}{d\epsilon } & =\xi \left ( \bar {x},\bar {y}\right ) \nonumber \\ & =1 \tag {1} \end {align}
Similarly, since \[ \eta \left ( x,y\right ) =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}\] Then \begin {align} \frac {d\bar {y}}{d\epsilon } & =\eta \left ( \bar {x},\bar {y}\right ) \nonumber \\ & =2\bar {y} \tag {2} \end {align}
Where in both odes (1,2) we have the condition that at \(\epsilon =0\) then \(\bar {x}=x,\bar {y}=y\). Starting with (1), solving it gives\[ \bar {x}=\epsilon +c_{1}\left ( x,y\right ) \] Where \(c_{1}\left ( x,y\right ) \) is arbitrary function which acts like constant of integration since \(\bar {x}\left ( x,y\right ) \) is function of two variables. At \(\epsilon =0\) then \(c_{1}\left ( x,y\right ) =x\). Hence the above is\begin {equation} \bar {x}=\epsilon +x \tag {3} \end {equation} And from (2), solving give\[ \bar {y}=2\bar {x}\epsilon +c_{2}\left ( x,y\right ) \] But at \(\epsilon =0\) \(,\bar {y}=y,\bar {x}=x\) then the above gives \(c_{2}=y\). Hence the above becomes\[ \bar {y}=2\bar {x}\epsilon +y \] But \(\bar {x}=\epsilon +x\) from (3), hence the above becomes\begin {align*} \bar {y} & =2\left ( \epsilon +x\right ) \epsilon +y\\ & =2\epsilon ^{2}+2\epsilon x+y \end {align*}
Therefore Lie group is\begin {align*} \bar {x} & =\epsilon +x\\ \bar {y} & =2\epsilon ^{2}+2\epsilon x+y \end {align*}