Let $$y(t):[0, \infty)\to \mathbb{R}$$ be the solution of the problem $$y'(t)=(1-y^2(t))\cosh y(t)\ \text{for}\ t>0, \ y(0)=y_0.$$
Find the set of real values of $y_0$ such that $y(t)$ mentioned above is bounded.
I don't know how to proceed for solving such equations.
I have written $$\frac{dy}{(1-y^2(t))\cosh y(t)}=dt.$$
But unable to proceed further. Please help.
The ODE $$ y'=(1-y^2)\cosh y=:f(y) \tag{1} $$ has two fixed points, $y=-1$ and $y=1$. Therefore, if $y_0=\pm 1$, then $y(t)=\pm 1$ for all $t\in[0,\infty)$, so $y(t)$ is obviously bounded in these cases. Now notice that
$$ \begin{cases} f(y)>0&\text{if $|y|<1,$} \\ f(y)<0&\text{if $|y|>1,$} \end{cases} \tag{2} $$ and that, by uniqueness$^{(*)}$, the solution of the IVP $y'(t)=f(y(t))$, $y(t_0)=y_0$ cannot cross the lines $y=\pm 1$ if $y_0\neq \pm 1$. Therefore,
If $y_0>1$, then $y(t)$ is in a region where $y'=f(y)<0$, hence $y_0\geq y(t)>1$ for all $t\in[0,\infty)$, so $y(t)$ is bounded;
If $-1<y_0<1$, then $y(t)$ is in a region where $y'=f(y)>0$, hence $y_0\leq y(t)<1$ for all $t\in[0,\infty)$, so $y(t)$ is bounded;
If $y_0<-1$, then $y(t)$ is in a region where $y'=f(y)<0$, hence $y(t)\leq y_0$ for all $t$ in the domain of $y(t)$, so the latter is bounded above; however, it is not bounded below, since the integral $$ t=\int_{y_0}^{y}\frac{du}{(1-u^2)\cosh u} \tag{3} $$ is well-defined for all $y\in(-\infty,y_0]$.
In summary: $y(t)$ is bounded if $y_0\geq -1$, and unbounded below if $y_0<-1$.
$^{(*)}$ It's not difficult to show that the ODE $(1)$ satisfies the hypotheses of the Picard–Lindelöf theorem.