Find the flow of a vector field Question: Let $\mathbb{X}$ be the vector field given by $\mathbb{X}(x,y)=(x,y)$
Compute its flow $\Phi(x,y)$
Attempt: We have $\dot{x}(t)=x\therefore$$$\int_{x_0}^{x(t)}dx'=\int_{0}^{t}x(t')dt'$$
$$\left.x'\right|_{x_0}^{x(t)}=\left.\pm e^{t'}\right|_0^t$$
$$x(t)-x_0=\pm e^t-1$$
$$x(t)=\pm e^t+x_0-1$$
And something similar for $y$
Giving flow:
$\Phi(x,y)=(\pm e^t+x-1,\pm e^t+y-1)$
Is this right? Any help or comments would be greatly appreciated
 A: Hint: vector field generates the system of differential equations:
$$ \dot{{\rm x}} = {\mathbb X}({\rm x}), $$
where ${\rm x} = (x,y)$. This system could be rewritten as
$$ \left \lbrace 
\begin{array}{ccc}
\dot{x} &= x \\
\dot{y} &= y 
\end{array}
 \right . $$
Note on solving:
If you solve $\dot{x} = x $ by separation of variables, then you should obtain this
$$ \frac{d x}{dt} = x $$
$$ \frac{d x}{x} = dt $$
$$ \int\limits_{x_0}^{x(t^\ast)} \frac{d x}{x} = \int\limits_{0}^{t^\ast} dt $$
$$ \ln{\frac{x(t^\ast)}{x_0}} = t^\ast  $$
$$ x(t^\ast) = x_0 \cdot e^{t^\ast} $$
A: The vector field $X(x,y):=(x,y)$ encodes the system of ODEs
$$\dot x=x,\quad \dot y=y\ ,$$
which is separated as it stands. Given an initial point $(x_0,y_0)\in{\mathbb R}^2$ the solution is obviously given by
$$x(t)=e^t x_0,\quad y(t)=e^t y_0\qquad(-\infty< t<\infty)\ .$$
In terms of "flow" this means that
$$\Phi_t(x_0,y_0)=e^t(x_0,y_0)\ ,$$
where at the end we may drop the index ${}_0\,$:
$$\Phi_t(x,y)=e^t(x,y)\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ .$$
As expected, one has
$$\Phi_{t+s}(x,y)=\Phi_s\circ \Phi_t(x,y)\ .$$
