Consider a first order equation in $\mathbb{R}$ with $f(t,x)$ defined on $\mathbb{R}\times \mathbb{R}$. Assume the equation $x'=f(t,x)$. Suppose $xf(t,x)<0$ for $|x|>R$ where $R$ is a fixed positive constant number. I have to show that all solutions exist for all $t>0$.
The way I look at it is that $x x'=xf(t,x)<0$ thus $x'<0$ for $x> R$ and $x'>0$ for $x<-R$, therefore for solutions starting at $x>R$ the solution cannot pass $x=-R$ line since it would need to go back up and for solutions starting at $x<-R$ the solution starts growing but it cannot go beyond $x=R$ since it needs to start decreasing again.
I need help understanding the answer below.Why is $|x(\tau)|=2R$ below?
Thank you,Klara
