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> Also, it’s not obvious from looking at the equation that there should be a problem at t = 1. I found this curious, it seemed obvious from the equation there
by conistonwater 9y ago
> Also, it’s not obvious from looking at the equation that there should be a problem at t = 1.
I found this curious, it seemed obvious from the equation there is going to be a problem. If you think about it, u'=u^2+1 has the solution tan (up to some constants), which is singular. So a function that satisfies y'=y^2+t^2 will satisfy y'>u' for t>1, and so should also have a singularity, you just don't quite know where it's going to be.
- monochromatic 9y agoEverything is obvious once you know about it. But be honest--if you looked at that equation in the course of your daily work (rather than in a blog post about solutions that don't exist), would you immediately think "well that's gonna blow up real quick"? I sure wouldn't.
- conistonwater 9y agoYes, I would, that was the point I was trying to make, it looks like it has a singularity. In my comment I was trying to explain how one might be able to spot it in advance.
- monochromatic 9y agoI guess. It doesn't jump out at me the way that 1/x or tan(x) would though. But I haven't had a lot of reason to mess with differential equations since undergrad, so maybe I'm just rusty.