# If the IVP is x'=1+x^2, x(0)=0, why must the solution be tan(t), (-pi/2<t<pi/2) and not just tan(t)?

I understand that neither $$\tan(t)$$ nor its derivative $$\sec^2(t)$$ are defined at $$t = k\pi /2$$, where $$k$$ is any integer. If I plug in $$x(t) = \tan(t)$$, that leaves me with $$\sec^2(t) = 1 + \tan^2(t)$$ and I see no problem with that (trig identity). Both $$x'$$ and $$x$$ have the same domain, so I don't see why I'd have to limit it to $$(-\pi/2, \pi/2)$$. I read somewhere that the solution has to be smooth? I understand that $$x(t)$$ has to be differentiable, because we're using its derivative in the DE. My question is why does the solution have to be one continuous line? Why can't it be two or more continuous lines? What's so bad about discontinuities if they line up?

P.S. The problem I'm referring to is Example 7.7 found on page 78 in Differential Equations by Polking, Boggess, and Arnold 2nd Edition.

First lets try to extend the solution for $$x>\pi/2$$. We already see we can do that in infinitely many ways all of which are equally correct (in the sense of pointwise fulfilling the ODE): $$\begin{cases} \tan(x)&-\pi/2 For arbitrary $$c$$ and $$\epsilon$$ depending on $$c$$ sufficiently small. So we see that with a definition of solutions like that uniqueness of solutions does no longer exist. But it gets even worse: What is stopping us from cutting the piecewise function sooner? We could define another solution $$\begin{cases} \tan(x)&-\pi/2 Yes it is not continous (or even defined) at $$x=\pi/4$$, but thats the same scenarion with $$\tan(x)$$ at $$x=\pi/2$$. But why stop there? Let's cut the domain in infinitely many small pieces all with a different constant $$c$$...