# Show that the separable differential equation M(x) + N(y)y' = 0 is exact.

Show that the separable equation

$M(x) + N(y)\frac{dy}{dx} = 0$

is exact.

The homework sheet that the teacher gave us said it should be a one-liner. I'm not sure how to prove it that easy (as to put it in one line).

I have done this so far:

$M(x) dx + N(y) dy = 0$

I know that the derivative of M(x) with respect to y should equal the derivative of N(y) with respect to x. ($M_x = N_y$)

But if they don't than it's not exact, right?

Even if I bring the M(x)dx to the other side it would be

$N(y) dy = - M(x) dx$

Then I still have the issue that M_x might be different than N_y

Edit with @Evgeny comment in mind

Ok so, then I need to look at it as the derivative of $\frac {d M}{dy} = \frac {dN}{dx}$?

Couldn't they still be different?

• You should compare $M'_y$ and $N'_x$. This is a criteria for exact equation. And you're comparing $M'_x$ with $N'_y$ which is wrong. Sep 26, 2015 at 7:46
• @kingcobra1986 They are both zero !!! Sep 26, 2015 at 8:09
• @kingcobra1986 The expression M(x) means that M is a function of x, right? It does not depend on y, it is constant with respect to y, so the derivative with respect to y is... Sep 26, 2015 at 8:14
• @kingcobra1986 I suggest reviewing the theory: definition and meaning of derivative and partial derivative Sep 26, 2015 at 8:19
• I just understood your question better. The derivative of M'(x) with respect to y should be 0, as well as the derivative of N'(x) with respect to y. Therefore you will get 0=0 which is true. Another way to right that is $\psi _{xy} = \psi_{yx}$. Correct? Sep 26, 2015 at 8:27

$$M(x) + N(y)\frac{dy}{dx} = 0 \tag 1$$ is exact because an integral of $(1)$ is $$\int M(x) \, dx+\int N(y) \, dy = \text{ constant }$$ and is a solution of $(1).$