Limit for $t\to+\infty$ of a solution to an ODE Consider the Cauchy problem:
\begin{cases}
y'(t)=\frac{t^2}{2+\mathrm{sin}(t^2)}cos(y(t)^2) \\
y(0)=0
\end{cases}
I have to prove that $\mathrm{lim}_{t\to+\infty} y(t)=\sqrt{\pi/2}$.
I have tried to solve it by separating the variables $t$ and $y$:  dividing by $\cos(y(t)^2)$ both sides of the equation and integrating, one obtains:
$$
\int^{y(t)} _0 \frac{1}{\cos(u^2)}\mathrm{d}u=\int^{t} _0 \frac{s^2}{2+\sin(s^2)}\mathrm{d}s, 
$$
but it does not look very useful.
Does anyone know how to solve it?
 A: First observe that the RHS
$$
f:\mathbb{R}^2\rightarrow \mathbb{R}, ~f(t, y) \mapsto \frac{t^2}{2+\sin(t^2)} \cos(y(t)^2)
$$
is continuously differentiable in $y$. Therefore, for any pair $(s, x) \in \mathbb{R}^2$ there is exactly one solution $\tilde{y}$ such that $\tilde{y}(s)=x$. Note that because of this uniqueness property, graphs of solutions do never cross.
It is obvious that $y(t) = \pm\sqrt{\frac{\pi}{2}}$ is a solution. So a (unique!) solution with $y(0)=0$ is bounded in between $\left(-\sqrt{\frac{\pi}{2}}, \sqrt{\frac{\pi}{2}} \right)$. (Just graph it and you will see.) Also note that since $f>0$ on $(0, \mathbb{R}) \times \left(-\sqrt{\frac{\pi}{2}}, \sqrt{\frac{\pi}{2}} \right)$, we have that $y'> 0$ which means that $y$ increases strictly. Now, since $y$ is bounded and strictly increasing, it has a limit at $t \rightarrow \infty$.
Assume that $\ell := \displaystyle\lim_{t \rightarrow \infty} y(t) \in \left(0,  \sqrt{\frac{\pi}{2}}\right)$. Then , on $t\in[1, \infty)$, $y'(t) = f(t, y(t))$ has some interesting properties: Since $y(0)=0$ and due to $y$ being strictly increasing, we must have that $y(t)\in [y(1), \ell]$ for all $t \geq1$ where $y(1) \in (0, \ell]$. In fact, one can see:
$$
L:=\inf_{t\geq 1} f(t, y(t)) > 0
$$
It follows:
$$
\lim_{t \rightarrow \infty} y(t) =  \lim_{t \rightarrow \infty} \int^t_0 f(r, y(r))~\mathrm{d}r \geq \lim_{t \rightarrow \infty} \int^t_1 f(r, y(r))~\mathrm{d}r \geq \lim_{t \rightarrow \infty} \int^t_1 L~\mathrm{d}r =  \lim_{t \rightarrow \infty} L(t-1) = \infty 
$$
So somehow $\displaystyle \lim_{t \rightarrow \infty} y(t) = \infty$, which clearly is a contradiction to the assumption. So we must have that $\ell = \sqrt{\frac{\pi}{2}}$.


This argument is far from being trivial, but it is one of the most important ones in ODE theory.
