Let $F:D\subset\Bbb{R}^n\rightarrow \Bbb{R}^n$ a vector field of class $C^1$ in an open subset $U$ of $\Bbb{R}^n$. Suppose that $p_0\in D$ is a point such that $F(p_0)=0$. I need to prove thar $F_t(p_0)=p_0$ and $dF_t=e^{tdF(p_0)}$.
Notation: $F_t(p)=\varphi(t,p)$ is the trajectory of $F$ passing through $p$.
I considered the Cauchy Problem
$$\begin{cases} x'=F(x) \\ x(0)=p_0 \end{cases}$$
By definition, the solution is $\varphi(t,p)=F_t(p)$. Therefore,
$$dF_t(p_0)=F(F_t(p_0)) \\ \implies \left.dF_t(p_0)\right\vert_{t=0}=F(F_0(p_0))=F(p_0)=0 $$
So, $F_t(p)$ is constant (with respect of $p$). If I could find any point such that $F_t(\tilde{p})=p_0$, I would conclude that $F_t(p_0)=p_0$, but I didn't. Is there something obvius that I'm not seeing?
And I tried to build some Cauchy Problems to get to the second affirmative, but I didn't got it.