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 :)

Since $y'(0)=0$, there is some $\delta>0$ such that for $t \in (0,\delta)$,
we have $y(t)^2<t$, so $y'(t)>0$. By the intermediate value theorem, $y(t)>0$ in $(0,\delta)$. Suppose that
$$\tau:=\inf \Bigl\{t\ge \delta: \, y(t) \in \{0,\sqrt{t}\} \Bigr\}<\infty \,. \tag{*} $$
(Recall the convention that the infimum of an empty set is infinity). If $y(\tau)=0$ then $y'(\tau)=\tau$, a contradiction since $y(t)>0$ on $(0,\tau)$. If $y(\tau)=\sqrt{\tau}$, then $y'(\tau)=0$, so $g(t)=t-y^2$ satisfies $g(\tau)=0$ and $g'(\tau)=1$, a contradiction since $g(t)>0$ for
$t \in (0,\tau)$. Thus $\tau$ defined in $(*)$ must be infinite, so
$0<y(t)<\sqrt{t}$ for all $t>0$.
