# Prove a solution of a differential equation is bounded

I am interested in the differential equation

$$(\mathcal{S})\left\{ \begin{array}{l} y'(t) = \displaystyle \frac{y^{2}(t)}{1+t^{2}+y^{2}(t)} \\[2mm] y(0) = \displaystyle \frac{3}{4} \end{array} \right.$$

Let $y$ be a maximal solution of $(\mathcal{S})$. I already proved that $y$ is defined on $\mathbb{R}$. However, I do not see how to prove that $\displaystyle \lim \limits_{t \to +\infty} y(t)$ is lower than $14$ and $\displaystyle \lim \limits_{t \to -\infty} y(t)$ is greater than $\frac{1}{3}$. A hint would be appreciated !

• Where do the constants $14$ and $1/3$ come from? The text of the exercise? – TZakrevskiy Mar 13 '15 at 22:48
• @TZakrevskiy : Yes, from the text of the exercise! – Odile Mar 13 '15 at 23:14
• $\frac13$ is odd. – Did Mar 14 '15 at 15:45

From the equation $0<y'\le1$, so that $y$ is increasing and $$\frac34\le y(t)\le\frac34+t,\quad t\ge0.$$ Substituting into the equation we get $$y'\le\frac{y^2}{1+t^2+(3/4)^2}=\frac{y^2}{t^2+(5/4)^2}.$$ Integrate to obtain $$-\frac{1}{y}+\frac43\le\frac45\arctan\frac{4\,t}{5},\quad t\ge0,$$ and from here $$y(t)\le\frac{1}{4/3-(4/5)\arctan t}<\frac{1}{4/3-(4/5)(\pi/2)}=13.038\dots$$