Consider in $\mathbb{R}$ the Cauchy problem: $$(*) \quad x'(t)=\frac{1}{1+t^2+x^2(t)} \quad, \quad x(0)=0$$ Show that the solution of $(*)$ is bounded.
We know that the unique solution $\psi(t)$ of $(*)$ is defined in $\mathbb{R}$ and it's an odd function, but how to prove that $\psi(t)$ is bounded ?
Since $x'(t)$ is always positive, it follows that $x(t)$ is strictly increasing.
Hence, since $x(0)=0$, it follows that $x(t)$ is positive if $t > 0$ and $x(t)$ is negative if $t < 0$.
Let $y(t)=\tan^{-1}(t)-x(t)$.
Then $$ y'(t) = \frac{1}{1+t^2} - \frac{1}{1+t^2+x^2(t)} \ge 0 $$ hence $y$ is non-decreasing.
Thus, since $y(0)=0$, it follows that $y(t)\ge 0$ for $t\ge 0$ and $y(t)\le 0$ for $t\le 0$.
Then for $t\ge 0$ we get $$ 0\le x(t)\le \tan^{-1}(t) < \frac{\pi}{2} $$ and for $t\le 0$ we get $$ -\frac{\pi}{2} < \tan^{-1}(t) \le x(t)\le 0 $$ Therefore $x(t)$ is bounded.