Existence of solutions to first order ODE The fundamental theorem of autonomous ODE states that if $V:\Bbb R^n\to\Bbb R^n$ is a smooth map, then the initial value problem
$$
\begin{aligned}
\dot{y}^i(t) &= V^i(y^1(t),\ldots,y^n(t)),&i=1,\ldots,n \\
y^i(t_0) &= c^i, &i=1,\ldots,n
\end{aligned}\tag{1}
$$
for $t_0\in\Bbb R$ and $c=(c^1,\ldots,c^n)\in\Bbb R^n$ has the following existence property:

Existence: For any $t_0\in\Bbb R$ and $x_0\in\Bbb R^n$, there exist an open interval $J$ containing $t_0$ and an open subset $U$ containing $x_0$ such that for each $c\in U$, there is a smooth map $y:J\to\Bbb R^n$ that solves $(1)$.

Now here is my question:

Question: Suppose we already know that a solution exists with initial value $y(t_0)=x_0$ on an interval $J_0$ containing $t_0$. Does the interval $J$ above can be assumed to contain $J_0$?

A priori, there is noting telling us that in the statement of the theorem. My question can be rephrased as follows.

Reformulation of the Question: Let $y:J\to\Bbb R^n$ be a smooth solution to $(1)$ with initial value $y(t_0)=x_0$. Is there an open set $U$ containing $x_0$ such that for all $c\in U$ there is a smooth solution $z:J\to\Bbb R^n$ to $(1)$ with initial value $z(t_0)=c$?

Edit: And what about the case where $J$ is a compact interval?

The answer to the reformulation is negative. Consider the problem 
\begin{equation}
\begin{cases}
y'=y^2 \\
y(0)=c
\end{cases}
\end{equation}
Its solution is $y(t)=\frac{1}{c^{-1}-t}$ and it is defined on $J_c=(-\infty, c^{-1})$. So, for example, the solution with initial datum $c=1$ is defined on $(-\infty, 1)$ while the solution with initial datum $1+\epsilon$ is defined on a strictly smaller interval, no matter how small $\epsilon$ is. 
