# Thread: Exact Differential Form Question

1. ## Exact Differential Form Question

If an equation is an exact equation in the form M(x,y)dx + N(x,y)dy = 0, and M(x,y)dx = N(x,y)dy, how does that satisfy it equaling zero (e.g. dM/dy = 4xy and dN/dx = 4xy , 4xy + 4xy ≠ 0)? Am I missing some fundamental understanding here?

2. ## Re: Exact Differential Form Question

I think you have some confusion. For me, the easiest way to think about these equations is to work backwards from the answer. Suppose we have a function $$F(x,y)=c \tag{1}$$ where $c$ is some constant. Then of $x$ and $y$ only one is independent. That is, if we are given a value of $x$, we can find out which values of $y$ satisfy the equation (1). Now, we may differentiate equation (1) with respect to $x$ to get $${\partial F \over \partial x} + {\partial F \over \partial y}{\mathrm d y \over \mathrm d x} = 0$$ and we can multiply through by $\mathrm d x$ (which isn't really a valid operation, but it looks OK so we'll do it anyway) to get $${\partial F \over \partial x} \mathrm d x + {\partial F \over \partial y}\mathrm d y = 0 \tag{2}$$

Now if we write $M(x,y) = {\partial F \over \partial x}$ and $N(x,y) = {\partial F \over \partial y}$, we have your original equation.

Next we differentiate $M(x,y)$ with respect to $y$, noting that $x$ and $y$ here are independent because we haven't set $M$ equal to anything. This gets us $${\partial M \over \partial y} = {\partial \over \partial y} {\partial F \over \partial x} = { \partial F \over \partial x \partial y}$$
And we differentiate $N(x,y)$ with respect to $x$, to get $${\partial N \over \partial y} = {\partial \over \partial x} {\partial F \over \partial y} = { \partial F \over \partial y \partial x}$$
But $${ \partial F \over \partial x \partial y} = { \partial F \over \partial y \partial x}$$ so we conclude that if equation (2) is exact then $${\partial M \over \partial y} = {\partial N \over \partial x}$$