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?

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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?

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

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

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