Statement: The equation $M(x,y)dx + N(x,y)dy = 0$ is exact if and only if:

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Proof:

Necessity ($\Rightarrow$): Assume the equation is exact. Then there exists $F(x,y)$ such that:

$\frac{\partial F}{\partial x} = M(x,y) \quad \text{and} \quad \frac{\partial F}{\partial y} = N(x,y)$

Taking the partial derivative of the first equation with respect to $y$:

$\frac{\partial^2 F}{\partial y \partial x} = \frac{\partial M}{\partial y}$

Taking the partial derivative of the second equation with respect to $x$:

$\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial N}{\partial x}$

By Clairaut's theorem (equality of mixed partial derivatives), if $F$ has continuous second partial derivatives, then:

$\frac{\partial^2 F}{\partial y \partial x} = \frac{\partial^2 F}{\partial x \partial y}$

Therefore:

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Sufficiency ($\Leftarrow$): Assume $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. We must show there exists $F(x,y)$ such that $\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$.

Define $F(x,y)$ by:

$F(x,y) = \int_{x_0}^x M(s,y)ds + \int_{y_0}^y N(x_0,t)dt$

where $(x_0, y_0)$ is any fixed point in the domain.

Step 1: Show $\frac{\partial F}{\partial x} = M(x,y)$.

By Leibniz's rule for differentiation under the integral sign:

$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}\int_{x_0}^x M(s,y)ds + \frac{\partial}{\partial x}\int_{y_0}^y N(x_0,t)dt$

The first term gives $M(x,y)$ by the fundamental theorem of calculus.

The second term is zero because $N(x_0,t)$ does not depend on $x$. Therefore:

$\frac{\partial F}{\partial x} = M(x,y) \quad \checkmark$

Step 2: Show $\frac{\partial F}{\partial y} = N(x,y)$.

$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}\int_{x_0}^x M(s,y)ds + \frac{\partial}{\partial y}\int_{y_0}^y N(x_0,t)dt$

The second term gives $N(x_0,y)$ by the fundamental theorem.

For the first term, differentiate under the integral:

$\frac{\partial}{\partial y}\int_{x_0}^x M(s,y)ds = \int_{x_0}^x \frac{\partial M}{\partial y}(s,y)ds$

Using the exactness condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$:

$= \int_{x_0}^x \frac{\partial N}{\partial x}(s,y)ds = N(x,y) - N(x_0,y)$

by the fundamental theorem of calculus.

Therefore:

$\frac{\partial F}{\partial y} = N(x,y) - N(x_0,y) + N(x_0,y) = N(x,y) \quad \checkmark$

We have constructed $F(x,y)$ satisfying both required conditions, proving the equation is exact. ∎