Prove that the solution of an initial value problem is asymptotic to a function

Question

I'm studying qualitatively the following nonlinear initial value problem:
\begin{cases} y'=t-y^2 \\ y(0)=0 \end{cases}
I' ve found that the solution is increasing for $t>0$ and remains below the graph of the function $f(t)=\sqrt{t}$ . Is there an analytical method to prove or disprove that the solution $y(t) \sim_{+\infty} f(t)$ ?

Answer 1

The curve $y=\sqrt t+a$, has a slope of $\frac1{2\sqrt t}$ which is positive and smaller than $1$ for $t>\frac14$. On this curve the ODE has slope $$ \dot y = t-(\sqrt t+a)^2=-2a\sqrt t-a^2 $$ This is negative for $a>0$ so that no solution can cross such an upper bound from below.

Now for $a<0$ and $t>a^2$ one gets $$ \dot y > a^2 $$ Sharpen that to $a\le -1$ and you get that $\dot y>1$ so that any curve that has $y(1)\in[0,1]$ will stay between these bounds, $$ \sqrt{t}-1\le y(t)\le\sqrt{t}. $$ Similarly one can take $a=-\frac12$ for $t\ge 4$ etc., narrowing the distance of the boundary by delaying the onset of the bound.