Given $\epsilon>0$, I need to create a first order ODE that is defined everywhere on $\mathbb{R}$, such that all solutions to this ODE are only defined on an interval of length $\leq \epsilon$.
My attempt: $$x'(t) = \frac{1}{\log(10)}\left(\frac{1}{\epsilon/t -1}\right)\frac{\epsilon}{t}.$$ This has solutions $$ x(t) = \log_{10}\left( \frac{\epsilon}{t}-1 \right) + c $$ for $c\in\mathbb{R}$.
The problem is that my ODE and solutions are only defined on the interval $(0,\epsilon)$, not all of $\mathbb{R}$.
My professor said that I was close, but my ODE should be autonomous so it is defined everywhere. There is no way that I can see how to extend my example to be what is desired. I do not think I am close at all. Any hints or help would be appreciated.
We know that the trigonometric tangent function has poles at periodical positions. The differential equation for the tangent is $$ y'=1+y^2 $$ Now change the time scale so that the distance between two poles is reduced from $\pi$ to $ϵ$.