Why aren't exact differential equations considered PDE?

Exact differential equations come from finding the total differential from some multivariable function.

In the exact differential equation $M\mathrm{d}x+N\mathrm{d}y=0$

M and N are considered to be partial derivatives of some potential function... So why aren't exact differential equations considered PDEs? After all, you're finding the potential function given it's partial derivatives...

Thanks.

• Because it is essentially a differential equation $dF(x(t),y(t))/dt=0$. – Troy Woo Sep 13 '14 at 18:25
• But it's not always the case that both x and y are functions of t. Or is this irrelevant? – DLV Sep 13 '14 at 18:26
• It has to be, $x,y$ are functionally dependent. If you don't like $t$, how about $x$? – Troy Woo Sep 13 '14 at 18:28
• No no, I'm not against the letter t. I'm just saying that x and y are not always functions of the same variable. Or do they have to be? How do you know they have to be? – DLV Sep 13 '14 at 18:30
• Since $Mdx+Ndy$ is exact, it has a primitive $F(x,y)$. So $Mdx+Ndy=0$ means $F(x,y)=c$ a constant. This is a level set or simply a curve in $\mathbb R^2$. So a curve can always be parametrized. You could just check up implicit function theorem. – Troy Woo Sep 13 '14 at 18:42

Because the partial differentials part is just a method of solving them, it's in the intermediate steps of a solution, not in the DE itself from the start. A bad example(can't think of a better one right now) would be considering $x-2=0$ a second degree polynomial because you can introduce parameters and make it $x^2=4, x>0$.
A more specialized example would be $$y'=y \iff y'-y=0 \stackrel{\cdot e^{-x}}{\iff}\frac{y'}{e^x}-\frac{y}{e^x}=0$$ Now, would you consider $y'=y$ a PDE?