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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/(1t).
In book, there's written that the solution exists only in the interval −∞ < t < 1. Why? Why not in
1< t < ∞ too?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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amistre64Best 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
 one year ago

amistre64Best 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
 one year ago

mark_o.Best 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
 one year ago

goricaBest 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 < ∞ ?
 one year ago

amistre64Best 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 :)
 one year ago
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