Exact Differentials

The Total Differential

For $f = f(x, y)$:

$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$

This describes an exact differential: the differential of a function.

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Definition

An expression $P(x,y)dx + Q(x,y)dy$ is exact if there exists $f(x,y)$ such that:

$df = P\,dx + Q\,dy$

That is:

• $\frac{\partial f}{\partial x} = P$
• $\frac{\partial f}{\partial y} = Q$

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Test for Exactness

A differential $P\,dx + Q\,dy$ is exact if and only if:

$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

Proof (if exact):

• If $df = P\,dx + Q\,dy$, then $P = f_x$ and $Q = f_y$
• $P_y = f_{xy}$ and $Q_x = f_{yx}$
• By equality of mixed partials: $P_y = Q_x$

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Finding the Potential Function

If $P\,dx + Q\,dy$ is exact, find $f$ by integration:

Step 1: Integrate $P$ with respect to $x$:

$f(x, y) = \int P(x, y)\,dx + g(y)$

where $g(y)$ is an arbitrary function (the “constant” of integration).

Step 2: Differentiate with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\int P\,dx + g'(y)$

Step 3: Set equal to $Q$ and solve for $g'(y)$:

$Q(x, y) = \frac{\partial}{\partial y}\int P\,dx + g'(y)$

Step 4: Integrate $g'(y)$ to find $g(y)$.

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Example

Is $(2xy + \cos y)\,dx + (x^2 - x\sin y)\,dy$ exact?

Check: $P_y = 2x - \sin y$, $Q_x = 2x - \sin y$. Yes, exact.

Find $f$:

Integrate $P = 2xy + \cos y$ wrt $x$:

$f = x^2y + x\cos y + g(y)$

Differentiate:

$f_y = x^2 - x\sin y + g'(y)$

Set equal to $Q = x^2 - x\sin y$:

$x^2 - x\sin y + g'(y) = x^2 - x\sin y$

So $g'(y) = 0$, hence $g(y) = C$.

$f(x, y) = x^2y + x\cos y + C$

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Integrating Factors

If $P\,dx + Q\,dy$ is not exact, sometimes we can find $\mu(x, y)$ such that:

$\mu P\,dx + \mu Q\,dy$

is exact. Then $\mu$ is an integrating factor.

For the ODE $P\,dx + Q\,dy = 0$, we then solve:

$d(\mu f) = 0 \Rightarrow \mu f = \text{constant}$

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Physical Interpretation

Thermodynamics: State functions (energy $U$, entropy $S$, etc.) have exact differentials.

• $dU = TdS - PdV$ is exact (energy is a state function)
• Heat $đQ$ and work $đW$ are not exact (path-dependent)

The notation $đ$ indicates inexact differentials.

State functions: The change depends only on initial and final states, not the path.

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Exact vs Inexact: Summary

Property	Exact	Inexact
Definition	$df$ for some $f$	Not $df$ for any $f$
Test	$\partial P/\partial y = \partial Q/\partial x$	$\partial P/\partial y \neq \partial Q/\partial x$
Integral	$\int P\,dx + Q\,dy$ path-independent	Path-dependent
Example in physics	State functions	Heat, work

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Why Exact Differentials Matter

• Identify when an ODE has a simple solution $f(x,y) = C$
• Connect to conservative fields in vector calculus
• Fundamental to thermodynamics (state vs path functions)

The exactness test quickly distinguishes solvable cases from ones needing integrating factors.