4 ms·
If you get rid of the t^2 term it still goes to infinity at t=1. dy/dt = y(t)^2; y(0)=1 has the solution y = 1/(1-x)
by tlb 9y ago
If you get rid of the t^2 term it still goes to infinity at t=1.
dy/dt = y(t)^2; y(0)=1
has the solution y = 1/(1-x)
- nokcha 9y ago>has the solution y = 1/(1-x) 1/(1-t), not 1/(1-x), right?
- tlb 9y agoRight
- aprescott 9y agoIsn't the purpose of the example to make it hard to just trivially deduce y? If you know y = 1/(1 - t) then yeah it's undefined at t = 1. The example given doesn't seem to lend itself to that though, and a lesson is to know when your tooling isn't good enough.
- JustFinishedBSG 9y agoBut if v' > u' with v(0) = u(0) and u blows up then we know that v blows up. Because u blows up at 1 then we know that v will blow up at or before 1