# Math Help - Exact equation

1. ## Exact equation

Prove that if the equation M(x,y)dx +N(x,y)dy = 0 (*) has a solution then it has an integrating factor.
My solution
If (*) is exact then easily, it has a solution, and the factor is 1.
If (*) is not exact suppose it does not have the factor. Then the solution has the form g(x,y)=C = Mdx + Ndy => it is exact (contradict).
Is that right?

2. ## Re: Exact equation

No, it is not right, since you cannot consider g(x,y) when (*) is not exact.

You could argue as follows.
When the equation has a solution g(x,y)=c, the integrating factor is 1.
Conversely, suppose that m is an integrating factor.
Then mMdx+mNdy is exact, and so there exists G(x,y) such that...

3. ## Re: Exact equation

I think that we should say that any differentiable function of two variables can be written in the form $F(x,y) = 0$ and so has a total differential $$F_x(x,y) \mathrm d x + F_y(x,y) \mathrm d y = 0$$ and for any non-zero function $u(x,y)$ we can write
$${F_x(x,y) \over u(x,y)} \mathrm d x + {F_y(x,y) \over u(x,y)} \mathrm d y = 0$$

Now, if there is no non-zero $u(x,y)$ such that $M(x,y) = {F_x(x,y) \over u(x,y)}$ and $N(x,y) = {F_y(x,y) \over u(x,y)}$, then we cannot transform $M(x,y) \mathrm d x + N(x,y) \mathrm d y = 0$ into the exact differential of $F(x,y)$ and so $F(x,y) = 0$ is not a solution of the given equation.

More generally, if there is no function $u(x,y)$ such that $u(x,y)M(x,y) \mathrm d x + u(x,y)N(x,y) \mathrm d y = 0$ is a total differential of any function $G(x,y)$, then there is no function $G(x,y) = 0$ that is a solution of the given equation.

I suspect that we require $M(x,y)$ and $N(x,y)$ to be non-zero for that to be a valid conclusion, but if either (or both) are zero, then we have only trivial solutions in two variables (i.e. the solution is independent of one (or both) variable).

4. ## Re: Exact equation

Lol, only 4 years late.

5. ## Re: Exact equation

So it is.

Shame I wasn't a member when it was posted eh?

6. ## Re: Exact equation

Nonetheless, a very formidable answer for our ever growing community, thanks!