# Differential equation defined only on $(-1,1)$

We are going over differential equations, and I'm trying to think up of an example of a differential equation where the solution satisfying $$x(0) = 0$$ is defined for $$t\in(-1,1)$$, but I'm a bit stuck.

Any help appreciated!

• What about \begin{align} x'(t) &= 0 \\ x(0) &= 0 \end{align} ? – Mattos Oct 10 '18 at 1:21

An example differential equation satisfying the given conditions would be $$(1-t^2)x'(t)^2=1$$ whose solutions with $$x(0)=0$$ are $$x(t)=\pm\sin^{-1}t$$. The inverse sine, of course, is only defined on $$(-1,1)$$.
A piece of the tangent function comes to mind (compressed if you want it only defined on $$(-1,1)$$.)
So consider $$y=\tan(\pi t/2)$$. Then $$y'=\frac{\pi}{2}\sec^2(\pi t/2)$$ is already a differential equation, technically. You could make it autonomous since $$\tan$$ and $$\sec$$ have a Pythagorean relationship: $$y'=\frac{\pi}{2}\left(1+\tan^2(\pi t/2)\right)$$ $$y'=\frac{\pi}{2}\left(1+y^2\right)$$