# Definition of unique solution of initial value problem

I am confused about the definition of unique solution of ODE. I wonder which of following definitions ((1) or (2)) is correct?

Given an ODE $y'=f(x,y)$ and the initial condition $y(0) = 1, y'(0) = 2$ .
If $y=f(x)$ (defined on (-3,3) for example) satisfies the ODE and the initial condition then $y$ is said to be the unique solution on (-3,3) if

1. For any function z(x) that satisfies the ODE and the initial condition on $(-3,3)$, we have $z(x) = y(x) \quad \forall x \in (-3,3)$ .
2. For any function z(x) that satisfies the ODE and the initial condition on $(a,b) \subset (-3,3)$ and $0 \in (a,b)$, we have $z(x) = y(x) \quad \forall x \in (a,b)$ .

I mean if a solution $y_1$ if said to be unique on an interval $I$, is there any chance that there exists another solution $y_2$ defined on a subset $B$ of $(a,b)$ but does not coincide with $y_1$ on $B$? I think they are equivalent and tried to prove it by extending $y_2$ on $I$ but I am not sure if one can do that.

So, even if we decided to take your number $2$ as the definition of uniqueness, the truth is that we could always extend the solution from $(a,b)$ to the larger interval, thus making $1$ and $2$ equivalent. Summing up, one can take your number $1$ as the definition in case $(-3,3)$ is not necessarily the maximal interval:
For some initial condition $x(t_0)=x_0$, the solution is said to be unique on the interval $(a,b)$ containing $t_0$ (and $(a,b)$ may not be the maximal interval!) if any other solution on the same interval coincides.