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?