Uniqueness of first order differential equation? We have a theorem that says: Let $h: I\rightarrow \mathbb{R}$ and $g:J \rightarrow \mathbb{R}$ be continuous functions, $t_0 \in I $ and $y_0 \in \text{int(J)}$, then the differential equation y'(t)=g(y(t))h(t) has a unique solution in some surrounding of $t_0$ if $g(y_0)\neq 0$
Now I wanted to ask you, how I determine the magnitude of this surrounding? Especially, if I integrate the differential equation and solve it for y(t), am I only able to say: My solution is only unique in the surround of $y_0$ for which g(y(t)) is not zero, is this the condition that determines my surrounding? I mean, I could have determined a solution that gives me zero for some values of t, but is the only solution for my given problem? Am I correct, that in this case, my theorem is not able to say something about the uniqueness of this solution?
 A: Short answer: local uniqueness + global existence $\implies$ global uniqueness.
The solution is unique as long as it exists and stays out of the zeros of $g$; that is, as long as your theorem applies. Indeed, given two solutions $y,z$ with the same initial value, consider the set $\{t \in I :y(t)=z(t)\}$ where $I$ is some interval in which $y,z$ exist. This set is closed by continuity, and open by local uniqueness. Therefore, it coincides with $I$.  
The size of the interval of existence comes from the Peano existence theorem. Draw a rectangle $$R=\{(t,y): |t-t_0|\le c, |y-y_0|\le b\}$$ around your initial point $(t_0,y_0)$. If $g(y)h(t)$ is continuous in this rectangle, the solution is safe there. Then you estimate for how long the solution is guaranteed to remain in the safe zone. Let $M$ be the supremum of $g(y)h(t)$ in $R$. The solution cannot exit through the horizontal side in time less than $b/M$. Therefore, it exists for $|t-t_0|\le \min(c,b/M)$. Of course, nothing here is specific to the particular structure of the right hand side that you have.
