I think I know what a convex set is, which is that given two points from the set the line formed between them is also in the set. And then there is another type of sets, which I don't know the correct term in english, so I'll just name them "star" sets, which are sets that have a center, and given a random point of a star set, the line between that point and its center is also contained in a the star set. Thus, all convex sets are star sets but not all star sets are convex.
But say I have the folowing differential equation: $$ A(x,y)+B(x,y)\frac{dy}{dx}=0 $$ Let U be a star set: $$ A:U\to I\!R $$ $$B:U\to I\!R$$
and $$ \begin{equation} \frac{\partial A}{\partial y}=\frac{\partial B}{\partial x} \end{equation} $$
Then the differential equation is exact. Now my question is, I thought that if the last condition was met, it automaticaly implied that the differential equation was exact, what is the need of stating that U is a star set?
Please explain in layman terms, because I have looked for a reason for this but the mathematical language is just so overwhelming I get lost in it.
In a convex set, or a star set, then if you have smooth functions $A$, $B$ with $$\newcommand{\p}{\partial}\frac{\p A}{\p y}=\frac{\p B}{\p x}\tag{1}$$ then there is a smooth $C$ such that $$A=\frac{\p C}{\p x}\qquad\text{and}\qquad B=\frac{\p C}{\p y}.\tag{2}$$ Therefore both these notions of "exactness" coincide.
This is not the case for general open sets. If $U=\Bbb R^2-\{(0,0)\}$ take $$A=\frac{-y}{x^2+y^2}\qquad\text{and}\qquad B=\frac{x}{x^2+y^2}$$ then $(1)$ holds, but there is no $C$ such that $(2)$ holds.