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

This Question is Closed

amistre64 Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.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
 2 years ago

mark_o. Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

gorica Group TitleBest ResponseYou've already chosen the best response.0
so, if it was, for example, y(2)=1, solution would exist only in interval 1< t < ∞ ?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.2
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 :)
 2 years ago
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