Upper and lower bounds for solution to $ y' = t - y^2, y(0)=0.$ I am working on the initial value problem
$$ y' = t - y^2,\qquad y(0)=0.$$
Assuming that the solution exists on $[0,\infty)$, I am supposed to show that
$$ 0\leq y(t) \leq \sqrt{t}\qquad \forall t\geq 0.$$
I've managed to prove an upper bound of $t^2/2$ by using the fundamental theorem of calculus, but I don't seem to be getting anywhere near the $\sqrt{t}$ that the question is asking for. Any help would be appreciated :)

Consider a point $(t_0,y_0)$ such that $ y_0 > \sqrt{t_0}$ and a solution to the differential equation $y(t)$ with boundary condition $y(t_0) = y_0$. We clearly have $y'(t_0) < 0$ and that if $0 < t < t_0$ and $y(t) > y(t_0)$, then $y'(t) < 0$. So $y$ must be a decreasing function on this interval. Therefore $y(0) > y(t_0) > \sqrt{t_0} > 0$, and so the solution with $y(0) = 0$ does not pass through this point.
