The two dimensional system is given below
$\dot{x}=x$,
$\dot{y}=-y+x^2$.
Hence the vector field is given by $f(x,y)=(x,-y+x^2)$. In the texbook it is given that the flow generated by this vector field is $\phi_t(x,y)=(e^tx,e^{-t}y+\frac{1}{3}(e^{2t}-e^{-t})x^2)$. How they deduced this flow of that given vector field $f$?
My approach was I solved the system of differential equations that are
$\dot{x}=x$ and $\dot{y}=-y+x^2$. But I didn’t get exactly what they did. Any help/hint/solution will be really helpful for me. Thank you in advance.
Given a differential equation (possibly a system)
\begin{align} X'(t) &= F(X(t)), & X(0) &= X_{0} \end{align}
the flow of the equation set is defined by $\varphi(X_{0}, t) = X(t)$. This implies that
\begin{align} \varphi(X_{0}, 0) &= X_{0}, & \frac{d}{dt} \varphi(X_{0}, t) &= F(\varphi(X_{0}, t)) \end{align}
Solving the equation in $x$ first, then substituting this result into the equation for $y$ and solving using an integrating factor yields
\begin{align} x &= x_{0} e^{t}, & y &= c e^{-t} + \frac{1}{3} x_{0}^{2} e^{2t} \end{align}
where $x(0) = x_{0}$ and $c$ are the integration constants. If we set $y(0) = y_{0}$ then we have
$$y_{0} = c + \frac{1}{3} x_{0}^{2} \implies c = y_{0} - \frac{1}{3} x_{0}^{2}$$
and hence
$$\varphi(x_{0}, y_{0}, t) = (x, y) = \left( x_{0} e^{t}, y_{0} e^{-t} + \frac{1}{3} x_{0}^{2} (e^{2t} - e^{-t}) \right)$$