Author Topic: First example from Thursday’s lecture  (Read 679 times)

Jiabei Bi

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First example from Thursday’s lecture
« on: September 14, 2018, 03:06:08 PM »
Hi,
I have a question for the first example from last night’s lecture. The condition is $y’=y²$ , $y(0) = 1$. I know how to get the solution. The answer is $y =1/(1-t)$. But I am confused about the restriction of variable $t$. Why the variable t has to be less than $1$?
Thanks!
« Last Edit: September 14, 2018, 04:58:20 PM by Victor Ivrii »

Victor Ivrii

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Re: First example from Thursday’s lecture
« Reply #1 on: September 14, 2018, 05:10:54 PM »
Solution "breaks" at $t=1$ and since initial condition is at $t=0$ we consider a maximal interval s.t.
1) solution does not break in it
2) it contains $t=0$