# Is This Differential Equation Non-Exact in Non-Simply Connected Regions?

• Settembrini
In summary, The conversation is about showing that the 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)\}. The speaker suggests using the fact that the set is not simply connected and trying to find a potential function F. However, they are unsure if the line integral can be used. They mention that the existence of such a function is equivalent to a certain integral being 0 for every closed curve, but this is only true in simply connected domains. They also mention trying (y/x)' or (Arctan(y
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.

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.

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.

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

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.

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

## What is a non-exact differential equation?

A non-exact differential equation is a type of differential equation where the total differential of the solution cannot be expressed as a combination of the variables and their differentials. This means that there is no function that can be found to satisfy the equation.

## What makes a differential equation non-exact?

A differential equation is considered non-exact when the coefficients of the variables and their differentials are not equal. This can make it difficult to find a solution using traditional methods such as separation of variables or integrating factors.

## How do you solve a non-exact differential equation?

There are several methods for solving non-exact differential equations, including the method of integrating factors, the method of undetermined coefficients, and using power series solutions. However, these methods can be complex and may require advanced mathematical knowledge.

## What are some real-world applications of non-exact differential equations?

Non-exact differential equations are commonly used in physics, engineering, and other fields to model complex systems and phenomena. They can be used to study heat transfer, fluid dynamics, population dynamics, and many other systems.

## What is the difference between an exact and non-exact differential equation?

The main difference between an exact and non-exact differential equation is that an exact differential equation can be solved using traditional methods, while a non-exact differential equation requires more advanced techniques. In an exact differential equation, the coefficients of the variables and their differentials are equal, making it possible to find a function that satisfies the equation.

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