Consider the following ODE $$y'(t) = -\sqrt{y(t)},$$ with initial condition $y(0) = 0$. Does this initial value problem have a unique solution ($y \equiv 0$ on $\mathbb{R}$) if we further assume that $y(t) \geq 0$ for every $t$?
We can see that if $y(t) \geq 0$ for every $t$, then $y' \leq 0$, so $y$ is decreasing. As $y(0) = 0$, we get that $y \equiv 0$ on $[0, \infty)$. However, can we conclude that $y \equiv 0$ on $\mathbb{R}$?
As suggested in the other answer, the claim is false since
$$ y(t) = \begin{cases} \frac{t^2}{4} & \text{ if } t\le 0, \\0 & \text{ if } t>0\end{cases}$$
is also a solution. Thus $y(t) \ge 0$ alone is not sufficient.
This is the case of having non-unique solutions. Indeed we can construct infinitely many solutions: for any $c<0$
$$ y(t) = \begin{cases} \frac{(t-c)^2}{4} & \text{ if } t\le c,\\ 0 & \text{ if } t>c\end{cases}$$
are all solutions (That non-uniqueness implies the existence of infinitely many solutions is a general phenomenon: see this).
This is a first order non-linear ODE which is also separable. Clearly $y(t)=0$ is a solution which satisfies $y(t)\geq0$ for every t.
If $y\neq0$ then sepating the ODE we obtain $\frac{dy}{\sqrt{y}}=-1dt$ and integrating yields $2y^{\frac{1}{2}}=-t+c$, then from the initial condition we obtain $c=0$. Thus $y(t)=\frac{t^{2}}{4}$ is another solution valid for $t\in (-\infty,0]$ since $\sqrt{\frac{t^{2}}{4}}=|\frac{t}{2}|.$
