Question

I have initial problem
y'=y^2
y(0)=1.
I solved it and I got that it's solution is
y=1/(1-t).
In book, there's written that the solution exists only in the interval −∞ < t < 1. Why? Why not in
1< t < ∞ too?

Answer 1

dy/dx = y^2

y^-2 dy = dx ; y!= 0

-y^-1 = x + C ; y!= 0

-y^-1 = x + C ; C = -1

y = 1/(-x+1)

since there is a split in the domain; and x=0 is only in the left side of the domain; then the initial value does not exist in the right side of the domain

Answer 2

something to do with continuity we have to split the function into 2 parts; each one being continuous in its own right i believe

Answer 3

i think its because the initial value y(0)=1is on the left side of the graph, since you are dealing only on the half left side of the graph,,,try to graph this one and you will see it

Answer 4

so, if it was, for example, y(2)=-1, solution would exist only in interval 1< t < ∞ ?

Answer 5

correct, if the initial value wanted a t value from the other part of the domain, then to maintain continuity, you would opt for the other side :)

Answer 6

thanks :)