5.1.10.1 Example 1 on how to find Lie group $(x,y)$ given Lie infinitesimal xi and eta

Given$\xi =1,\eta =2x$ find Lie group $\overline{x},\overline{y}$.  Since

$$\xi (x,y)={\frac{\partial \overline{x}}{\partial ϵ}|}_{ϵ=0}$$

Then

$$\begin{array}{rl}\frac{d\overline{x}}{dϵ}& =\xi (\overline{x},\overline{y})\\ \text{(1)}& & =1\end{array}$$

Similarly, since

$$\eta (x,y)={\frac{\partial \overline{y}}{\partial ϵ}|}_{ϵ=0}$$

Then

$$\begin{array}{rl}\frac{d\overline{y}}{dϵ}& =\eta (\overline{x},\overline{y})\\ \text{(2)}& & =2\overline{y}\end{array}$$

Where in both odes (1,2) we have the condition that at $ϵ=0$ then $\overline{x}=x,\overline{y}=y$. Starting with (1), solving it gives

$$\overline{x}=ϵ+c_1(x,y)$$

Where $c_1(x,y)$ is arbitrary function which acts like constant of integration since $\overline{x}(x,y)$ is function of two variables. At $ϵ=0$ then $c_1(x,y)=x$. Hence the above is

$$\begin{array}{}\text{(3)}& \overline{x}=ϵ+x\end{array}$$

And from (2), solving give

$$\overline{y}=2\overline{x}ϵ+c_2(x,y)$$

But at $ϵ=0$ $,\overline{y}=y,\overline{x}=x$ then the above gives $c_2=y$. Hence the above becomes

$$\overline{y}=2\overline{x}ϵ+y$$

But $\overline{x}=ϵ+x$ from (3), hence  the above becomes

$$\begin{array}{rl}\overline{y}& =2(ϵ+x)ϵ+y\\ & =2{ϵ}^2+2ϵx+y\end{array}$$

Therefore Lie group is

$$\begin{array}{rl}\overline{x}& =ϵ+x\\ \overline{y}& =2{ϵ}^2+2ϵx+y\end{array}$$