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.

Question

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.

anonymous · Answer

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 t-delta<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).