Question

can someone please explain Exactness and Uniqueness theorem to me for Differential Equations?

Answer 1

Suppose you have a DE like this: \( y \prime = f(y,t) \) with y(0) = 0. Then, if f(y,t) is continuous on some interval in the plane and its partial derivative with respect to y is continuous on the same interval, then the DE has an unique solution on the interval. What it is saying is that, as long as we have an initial condition passing through the origin (if it does not, we can rotate the plane) and the function and its y derivative is continuous, we can guarantee that there is exactly one solution in the interval that the function is continuous. That's, of course, for first order ODE. I am not up to generalize this, the proof is hard. If you want some intuition, look into Boyce's Elementary DE, chapter 2, section 2.8. It has a nice discussion on it for the simpler cases.

Answer 2

A typo: I meant that the function and its derivative are continuous on a region, then the solution is unique on a given interval.

Answer 3

Also, check: http://tutorial.math.lamar.edu/Classes/DE/IoV.aspx

Answer 4

basically, how I understand it is that the function will satisfy the existence and uniqueness theorem only if 1) the given function dy/dx is differentiable [i.e. continuous] and 2) the partial derivative with respect to y is continuous...meaning that it's just a real number when you plug the initial conditions in that they give you.
Is this correct?

Answer 5

Sorry mate, I am back now. Yeah, I think you can think it like that. What is the most important for this theorem, in my opinion, is to realize that you have to pick a region, so to speak. There may be multiple regions where we have an unique solution, but you want the one that satisfy (x0, y0) given in the exercise.