# Non-exact differential equation

1. Dec 29, 2012

### Settembrini

I try to show, that equation
$\frac{-y}{ x^{2}+y ^{2} } + \frac{x}{ x^{2}+y ^{2}}y'=0$
is not exact in $\mathbb{R^{2}} \setminus \{(0,0)\}$.
It's obvious that I have to use the fact, that the set is not simply connected, but I don't know how to do it.

2. Dec 29, 2012

### Staff: Mentor

I would try this:
Assume that it is exact and that there is a potential function F. Integrate its derivative (as given by your problem) in a circle around (0,0). If such a function F exists, the result has to be 0.

3. Dec 30, 2012

### Settembrini

I'm not sure, if we can use the line integral here. We are trying to show, that there doesn't exist function F such that F is exact differential, that is
$$\frac{ \partial F}{ \partial x}=\frac{-y}{ x^{2}+y ^{2} }$$ and $$\frac{ \partial F}{ \partial y}=\frac{x}{ x^{2}+y ^{2}}$$
Existence of such function is equivalent to $$\int_{L}\frac{-y}{ x^{2}+y ^{2} } \mbox{d}x +\frac{x}{ x^{2}+y ^{2}} \mbox{d}y=0$$ for every closed curve L. Although, this statement is true only if the domain is simply connected; and our domain isn't of this kind.

4. Dec 30, 2012

### lurflurf

consider
(y/x)'
or
(Arctan(y/x))'

5. Dec 30, 2012

### Settembrini

I know how to solve this equation and how to find F in any "regular" domain, for example in real plane $$\mathbb{R^{2}}$$. Problems appear in the neighbourhood of point (0,0) in our domain, because all methods of solving this kind of equation, I know are valid only in simply connected domain.

6. Dec 30, 2012

### lurflurf

I do not know why you could not use integrals.

http://en.wikipedia.org/wiki/Atan2

Suppose
F=C+Arctan2(y,x)
F'=0
but consider the nonexistence of the limit (by inequality of directional limits)

$\lim_{(x,y) \rightarrow (0,0)} F(x,y)$

F cannot be continuous and the equation is not exact

Note that if we had excluded a path to infinity along with 0 we would have a simply connected region and an exact equation