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Given $\frac{dy}{dx}=\sqrt{x-y}$; $\ \ $ $y(2)=2$

Why is existence not guaranteed using the Existence and Uniqueness Theorem for Differential Equations?

I thought that if $f(x,y)$ was continuous "near" the initial value, it guarantees existence of a solution of the given initial value problem.

In this case, wouldn't it exist? $f(2, 2)=\sqrt{2-2}=\sqrt{0}=0$

I understand, however, that it is not unique: $\frac{\partial f}{\partial y}=-\frac{1}{2\sqrt{x-y}}$, where it would not be continuous at $(2,2)$: $-\frac{1}{2\sqrt{2-2}}=-\frac{1}{0}$

Where am I going wrong?

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    $\begingroup$ Your $f$ is not continuous in a neighborhood of $(2,2)$ even though it is in fact continuous at $(2,2)$. $\endgroup$ – Ian Sep 7 '15 at 21:20
  • $\begingroup$ $\sqrt{x-y}$ is not even defined in a neighborhood of $(2, 2)$. $\endgroup$ – user99914 Sep 7 '15 at 21:24
  • $\begingroup$ I guess I don't fully understand what "in the neighborhood of" means. Could someone explain how I would mathematically show that $f$ is not defined in a neighborhood of $(2,2)$ or what this exactly means? $\endgroup$ – bobtran12 Sep 7 '15 at 21:31
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    $\begingroup$ One way of saying it is that there should be a rectangle centered at $(2,2)$ such that $f$ is continuous on the entire rectangle. But there is not. $\endgroup$ – Ian Sep 7 '15 at 21:36
  • $\begingroup$ Oh, I see. Is there a way of showing this mathematically or would I have to picture the graph and/or plug in nearby points to prove this? $\endgroup$ – bobtran12 Sep 7 '15 at 21:41

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