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 :)
 A: Hint: $y'=t-y^2=(\sqrt{t}-y)(\sqrt{t}+y)$ and that's ok for $\sqrt{t}$ since the hypothesis are asking for a solution in $[0; +\infty)$.
Looking for a lower bound: let's assume a positive solution $y(t)$
Looking ofr an upper bound: assuming a positive function $y$, with the initial condition $y(0)=0$, then it has to be an increasing function, then, $y' \geq 0$. That is satisfied if:

*

*$\sqrt{t}-y\geq0$ and $\sqrt{t}+y \geq0$
or

*

*$\sqrt{t}-y\leq0$ and $\sqrt{t}+y\leq0$
First set of conditions reveal that $\sqrt{t}\geq y$ and $\sqrt{t}\geq-y$. Since $\sqrt{t}$ is strictly positive, if $y\leq\sqrt{t}$ then obviously $-y\leq\sqrt{t}$. Second set of conditions reveal that $\sqrt{t}\leq y$ and $\sqrt{t}\leq-y$, but they contradict each other, because $-y$ assumes negative values, and the square root assumes positive values.
Then, an upper bound for the solution is indeed, $y\leq\sqrt{t}$, and we found that $0\leq y(t) \leq \sqrt{t}$, for $t \in [0,+\infty)$
You could work also out what would happen assuming a negative valued solution, but that's not what the exercise is asking.
A: 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 $t>0$.
A: 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.
