First example from Thursday’s lecture

[Jiabei Bi]

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!

[Victor Ivrii]

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 $t=0$