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mathlilly
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Show by verifying the hypothesis of the Existence and Uniqueness Theorem that the initial value problem x dot = 1+x^2, x(0) = 0 has a unique solution. Find the solution. what is the maximal interval of definition of the solution?
is this right?
Because f: R> R is continuous, then for any x knot which is an element of R, there is an interval (alpha,beta) containing 0 and there is a solution x(t) of x dot = f(x). The limit exist, therefore a solution exists. Because f is differentiable and f' is continuous, then x(t) is unique.
I don't know how to do the 2nd part.
 one year ago
 one year ago
mathlilly Group Title
Show by verifying the hypothesis of the Existence and Uniqueness Theorem that the initial value problem x dot = 1+x^2, x(0) = 0 has a unique solution. Find the solution. what is the maximal interval of definition of the solution? is this right? Because f: R> R is continuous, then for any x knot which is an element of R, there is an interval (alpha,beta) containing 0 and there is a solution x(t) of x dot = f(x). The limit exist, therefore a solution exists. Because f is differentiable and f' is continuous, then x(t) is unique. I don't know how to do the 2nd part.
 one year ago
 one year ago

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corey1234 Group TitleBest ResponseYou've already chosen the best response.0
For this differential equation, it is non linear and first order. This means you need to use the existence and uniqueness theorem that goes like this. Call x dot f(x,t). If f and partial f/partial x is continuous on an interval containing 0 (the x knot), then there exists and unique solution to the IVP on an interval tdelta<t<t+delta where delta is a some positive number. You do not know delta, but it is not zero. I will draw the work. So f= 1+x^2 which is a polynomial and partial f/partial x is 2x which is a polynomial so they are continuous everywhere so there exists a solution within (0 delta, 0 + delta).
 one year ago
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