There are no constant solutions to this DE. Can you please explain why the range of the solution of this IVP is a subset of $(-1,1)$.
I understand that this is a fairly straightforward IVP to solve, but I believe the answer to the question is intended to be inferred from information obvious without finding the solution.
Thanks in advance for your help.
On
Putting aside the given initial condition, the constant functions $y=1$ and $y=-1$ are solutions.
You could not have another continuous solution that takes a value inside $(-1,1)$ and then also takes a value outside of that interval. If you did, by the intermediate value theorem, there would be an $x_1$ where $y(x_1)=1$ (or equals $-1$) and now your proposed other solution crosses one of those constant solutions.
There cannot be a crossing like that, since the two curves would satisfy the same first order differential equation with the same "initial" condition $y(x_1)=1$.
So since it's given that $y$ takes a value inside $(-1,1)$ (namely $0$) then the solution $y$ can never take a value outside $(-1,1)$.
First, you can easily solve this equation and see what happen. But you can also see it without solving this equation. Let $g(x,y)=(y^2-1)\cos(x)$. The function $g(x,\cdot )$ is locally-Lipschitz, and thus your IVP has a unique local solution. Let $x_0>0$ and $y:[0,x_0)\to \mathbb R$ a solution of your IVP. Since $y(0)=0$, there is $u<x_0$ s.t. $y(x)\in (-1,1)$ for all $x\in [0,u)$. Suppose that there is $\bar x\in[u,x_0)$ s.t. $y(\bar x)=\pm 1$ (suppose WLOG that $y(\bar x)=1$). By unicity of the solution, $y\equiv 1$ on $[0,x_0)$ which contradict $y(0)=0$. Therefore $u=x_0$, i.e. $y(x)\in (-1,1)$ for all $x\in [0,x_0)$.