TITLE: about uniqueness of IVP with separable ODE
QUESTION [2 upvotes]: if I have an IVP with a separable ODE, which theorem states uniqueness for it? Please help. I need to prove it without solving.

REPLY [1 votes]: There ain't no one. The standard example for non-uniqueness of solutions
$$
y'(t)=\sqrt{|y(t)|}
$$
is separable.
However, Cauchy's theorem states that if $f$ in $y'=f(t,y)$ is continuously differentiable (in $y$, continuous in $t$), then every IVP has a unique solution (which may reach infinity before reaching the end of the integration interval, see $y'=y^2$).
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