Exact Equations

Exact equations arise naturally from the total differential of a function of two variables. They provide an elegant connection between differential equations and multivariable calculus, particularly the theory of conservative vector fields.

DefinitionExact Differential Equation

A first-order differential equation of the form:

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

is called exact if there exists a function $F(x,y)$ such that:

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

In this case, the solution is given implicitly by $F(x,y) = C$ .

The criterion for exactness comes from the equality of mixed partial derivatives.

Exactness Criterion

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

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

This follows from the equality of mixed partials: $\frac{\partial^2 F}{\partial y \partial x} = \frac{\partial^2 F}{\partial x \partial y}$ .

ExampleTesting for Exactness

Consider $(2xy + 3)dx + (x^2 + 4y)dy = 0$ .

Here $M(x,y) = 2xy + 3$ and $N(x,y) = x^2 + 4y$ .

$\frac{\partial M}{\partial y} = 2x, \quad \frac{\partial N}{\partial x} = 2x$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ , the equation is exact.

Solution Method

To solve an exact equation:

Verify exactness: Check that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
Find F: Integrate $M$ with respect to $x$ : $F(x,y) = \int M(x,y)dx + g(y)$ where $g(y)$ is an unknown function of $y$ only
Determine g(y): Differentiate $F$ with respect to $y$ and set equal to $N$ : $\frac{\partial F}{\partial y} = N(x,y)$
Write solution: The implicit solution is $F(x,y) = C$

ExampleSolving an Exact Equation

Solve $(2xy + 3)dx + (x^2 + 4y)dy = 0$ .

We already verified exactness. Now integrate $M$ with respect to $x$ :

$F(x,y) = \int (2xy + 3)dx = x^2y + 3x + g(y)$

Differentiate with respect to $y$ :

$\frac{\partial F}{\partial y} = x^2 + g'(y)$

Set equal to $N$ :

$x^2 + g'(y) = x^2 + 4y$

Therefore $g'(y) = 4y$ , giving $g(y) = 2y^2$ .

The solution is:

$F(x,y) = x^2y + 3x + 2y^2 = C$

Remark

Not all first-order equations can be written in exact form, but some non-exact equations can be made exact by multiplying by an appropriate integrating factor $\mu(x,y)$ . Finding such integrating factors is generally difficult, but in special cases (when $\mu$ depends only on $x$ or only on $y$ ), systematic methods exist.

ExampleNon-Exact Equation with Integrating Factor

Consider $y\,dx - x\,dy = 0$ .

Here $M = y$ and $N = -x$ .

$\frac{\partial M}{\partial y} = 1, \quad \frac{\partial N}{\partial x} = -1$

Not exact! However, multiplying by $\mu = 1/y^2$ gives:

$\frac{1}{y}dx - \frac{x}{y^2}dy = 0$

Now $\frac{\partial M}{\partial y} = -1/y^2 = \frac{\partial N}{\partial x}$ , so it's exact.

The solution is $F = x/y = C$ , or $x = Cy$ .

Exact equations provide deep insight into the geometric structure of differential equations and their relationship to gradient fields and path independence in vector calculus.