Test for exactness

Statement

Let $R = (\alpha, \beta) \times (\gamma, \delta)$ be an open rectangle in $\mathbb{R}^2$. Let $M, N : R \to \mathbb{R}$ be continuous functions such that the partial derivatives $\partial M / \partial y$ and $\partial N / \partial x$ exist and are continuous on $R$. Then the equation $M\, dx + N\, dy = 0$ is exact on $R$ if and only if $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \quad \text{on } R.$$

Remarks

The condition is necessary on each open set: if $\psi_x = M$ and $\psi_y = N$, the mixed partial derivatives of $\psi$ are equal, so $M_y = N_x$. For the converse, fix $(x_0, y_0) \in R$ and set $\psi(x, y) = \int_{x_0}^{x} M(s, y_0)\, ds + \int_{y_0}^{y} N(x, t)\, dt$. The converse holds on each simply connected open set. It can fail on other sets: $(-y\, dx + x\, dy)/(x^2 + y^2) = 0$ satisfies the condition on $\mathbb{R}^2 \setminus \{0\}$, but it is not exact there.

Explanations

L1 human

If $\psi$ exists, then $M$ and $N$ are its two partial derivatives. Their cross derivatives must agree, because the order of differentiation does not matter for nice functions. On a rectangle, this easy test is also enough.