Let $d\in\mathbb N$, $U\subseteq\mathbb R^d$ be open, $\tau>0$, $v:[0,\tau]\times U\to\mathbb R^d$ and $x_0\in U$.
Which assumption on $v$ do we need to impose in order to show that there is a (unique) $X:[0,\tau]\to U$ with$^1$ $$X(t)=x_0+\int_0^tv(s,X(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau]?\tag1$$
If necessary, replace $U$ with $\overline U$ and/or assume that $U$ is bounded. Moreover, it would be sufficient for me to find a $\delta>0$ such that the restriction of $X$ to $[0,\delta]$ maps into $U$.
Remark: Assuming $U=\mathbb R^d$, a classical result is the following:
If $v$ has at most linear growth in the second argument, i.e. $$\left\|v(t,x)\right\|\le c(1+\left\|x\right\|)\;\;\;\text{for all }t\in[0,\tau]\text{ and }x\in\mathbb R^d\tag2$$ for some $c\ge0$, and $v$ is locally Lipschitz continuous in the second argument uniformly with respect to the first argument, i.e. $$\left\|v(t,x)-v(t,y)\right\|\le c_n\left\|x-y\right\|\;\;\;\text{for all }t\in[0,\tau]\text{ and }x,y\in\overline B_n(0)\tag3$$ for all $n\in\mathbb N$ for some $(c_n)_{n\in\mathbb N}\subseteq[0,\infty)$, then there is a unique continuous$^2$ $X:[0,\tau]\to\mathbb R^d$ satisfying $(1)$.
So, the crucial thing is to find assumptions on $v$ which ensure that $X$ maps into $U$.
$^1$ Clearly, we need at least that $[0,\tau]\ni t\mapsto v(t,X(t))$ is integrable with respect to the Lebesgue measure on $[0,\tau]$.
$^2$ I'm only able to prove the uniqueness inside $C^0([0,\tau],\mathbb R^d)$. Are we able to drop the continuity, i.e. is the corresponding $X$ the unique plain function satisfying $(1)$?