Consider the Initial Value Problem $\frac{dy}{dx} = (y^2-1)\cos(x)$, $y=0$ when $x=0$. 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.
 A: 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 $(-1,1)$.
A: 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 $x\in [0,x_0)$.
