It's known that, if I have positive initial condition and $f(t,0)>0$, so $y(t)$, solution of ODE $$y'(t)=f(t,y),$$ is nonnegative, because if $T$ is the first time s.t. $y(T)=0$ so $y'(t)>0$ for $t\in (T-\epsilon,T+\epsilon)$, contradiction.
I'd like to know more about the case $f(t,0)=0$. If the initial condition is positive and $f(t,0)=0$, what can I conclude about positiveness?
Thank you so much.
$y(t)=0$ is a solution, and if you have the uniqueness property, then no other solution can cross or branch off or in to this solution. Meaning every other solution has one definite sign.