Exact Differential Equations | Test and Method of Solution

Sometimes the expression $M d x + N d y$ is the total differential of a single function $f (x, y)$ . When that happens the equation $M d x + N d y = 0$ says simply $d f = 0$ , and the solution is $f (x, y) = c$ . This section answers two questions: how to test whether an equation is of this type, and how to recover $f$ when it is.

Quick Reference

Item	Statement
Total differential	$d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y$
Exact equation	$M d x + N d y = 0$ where $M = \frac{\partial f}{\partial x}$ and $N = \frac{\partial f}{\partial y}$ for some $f$
Test for exactness	$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ , necessary and sufficient
General solution	$f (x, y) = c$
Step 1	$f (x, y) = \int M (x, y) d x + g (y)$
Step 2	Set $\frac{\partial f}{\partial y} = N$ and solve for g'(y)
Step 3	Integrate g'(y) and write $f (x, y) = c$
Alternative route	Start from $\frac{\partial f}{\partial y} = N$ and integrate in $y$ , with an "constant" $h (x)$

Where Exact Equations Come From

We remember from calculus that if a function $z = f (x, y)$ has continuous first partial derivatives, its total differential is defined by

dz = df = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy . \tag{1}

Therefore, if a family of curves is described by $f (x, y) = c$ , taking the differential of both sides gives

df = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy = 0 . \tag{2}

Recall that the differential of a constant is zero, which is what makes the right-hand side vanish.

This is the same elimination of arbitrary constants met in Section 1.3, now viewed through differentials. Every one-parameter family $f (x, y) = c$ produces a first-order differential equation, and equations produced this way are exactly the ones this section solves.

The family of curves $x^{3} y^{2} = c$ satisfies a differential equation. Find it.

Solution

Take the differential of both sides:

\frac{\partial}{\partial x} (x^{3} y^{2}) d x + \frac{\partial}{\partial y} (x^{3} y^{2}) d y = 0

3 x^{2} y^{2} d x + 2 x^{3} y d y = 0 .

Dividing both sides by $2 x^{3} y d x$ , the equation can be rewritten as

\frac{3 x^{2} y^{2}}{2 x^{3} y} + \frac{d y}{d x} = 0

or, simplifying the fraction,

\frac{dy}{dx} = -\frac{3y}{2x} \tag{3}

This is a differential equation, and we have arrived at it knowing its general solution in advance: the solution of (3) is $x^{3} y^{2} = c$ .

Read backwards, this is the whole idea of the section. Had we been handed $3 x^{2} y^{2} d x + 2 x^{3} y d y = 0$ cold, the task would be to recognize the left-hand side as $d (x^{3} y^{2})$ and read off the answer.

Graph of the family of curves x cubed times y squared equals c, for several values of c. — Graph of $x^{3} y^{2} = c$ for different values of $c$ . Each curve is one integral curve of the equation $\frac{d y}{d x} = - \frac{3 y}{2 x}$ .

Definition of an Exact Equation

Now consider a differential equation written in the form

M(x,y)\,dx + N(x,y)\,dy = 0 \tag{4}

where $M$ and $N$ are assumed to have continuous first partial derivatives.

If there exists a function $f (x, y)$ such that

\frac{\partial f}{\partial x} = M \quad\text{and}\quad \frac{\partial f}{\partial y} = N \tag{5}

then the differential form $M d x + N d y$ is called an exact differential, and Equation (4) is called an exact differential equation.

In that case (4) can be rewritten as

\frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y = 0 or d f = 0,

and the general solution of Equation (4) is

f (x, y) = c .

The solution $f (x, y) = c$ is an implicit solution, in contrast with an explicit solution $y = ϕ (x)$ .

The function $f$ is sometimes called a potential function, by analogy with physics, where a conservative force field is the gradient of a potential and the level curves of that potential are the curves of constant energy.

Two questions now need answering: when is a differential equation exact, and if it is exact, how do we find its solution?

Test for Exactness

The equations (5) lead to

\frac{\partial^{2} f}{\partial y \partial x} = \frac{\partial M}{\partial y}, and \frac{\partial^{2} f}{\partial x \partial y} = \frac{\partial N}{\partial x} .

From calculus we know that mixed partial derivatives are equal,

\frac{\partial^{2} f}{\partial y \partial x} = \frac{\partial^{2} f}{\partial x \partial y}

provided these partial derivatives are continuous. This is Clairaut's theorem. Hence

M d x + N d y = 0 is exact ⟹ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} .

So the condition $\partial M / \partial y = \partial N / \partial x$ is necessary for exactness. Conversely, it can be shown that this condition is also sufficient:

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ⟹ M d x + N d y = 0 is exact.

Let $M (x, y)$ and $N (x, y)$ have continuous first partial derivatives on a rectangular region of the $x y$ -plane. Then the equation $M d x + N d y = 0$ is exact if and only if

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

Proof of sufficiency

To prove that $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ is sufficient for exactness, let

P (x, y) = \int M (x, y) d x,

the partial integral of $M$ with respect to $x$ , holding $y$ fixed. By construction,

\frac{\partial P}{\partial x} = M (x, y) .

Differentiating both sides with respect to $y$ ,

\frac{\partial^{2} P}{\partial y \partial x} = \frac{\partial M}{\partial y} .

By hypothesis the right side equals $\frac{\partial N}{\partial x}$ , and by equality of mixed partials the left side equals $\frac{\partial^{2} P}{\partial x \partial y}$ . Therefore

\frac{\partial N}{\partial x} = \frac{\partial^{2} P}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{\partial P}{\partial y}) .

Both $N$ and $\frac{\partial P}{\partial y}$ have the same partial derivative with respect to $x$ , so they differ by at most a function of $y$ alone:

N = \frac{\partial P}{\partial y} + g (y)

where $g (y)$ is an arbitrary function of $y$ . This function is our integration "constant", which must be allowed to depend on $y$ because $y$ was held fixed during the partial integration.

Consequently,

M d x + N d y = \frac{\partial P}{\partial x} d x + \frac{\partial P}{\partial y} d y + g (y) d y = d {P (x, y) + G (y)},

where $G (y)$ is any antiderivative of $g$ , so that $d G (y) = g (y) d y$ .

We have shown that if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ , then $M d x + N d y$ is the differential of a function, namely $P (x, y) + G (y)$ . That is precisely the statement that the equation is exact.

The proof is not merely an existence argument. It is a recipe: partially integrate $M$ in $x$ , then correct by a function of $y$ alone. The method of solution below is exactly this proof carried out on a concrete equation.

Method of Solution

Now we find the general solution of $M d x + N d y = 0$ when the equation is exact. Let $f (x, y)$ be a function whose differential is $M d x + N d y$ .

Step 1. Because $\frac{\partial f}{\partial x} = M$ ,

f(x,y) = \int M(x,y)\,dx + g(y) \tag{6}

where $g (y)$ is the "constant" of integration. Remember that if we differentiate (6) with respect to $x$ , both $g (y)$ and $y$ behave as constants, so the term $g (y)$ disappears and the identity $\partial f / \partial x = M$ is restored regardless of what $g$ is.

Step 2. Because $\frac{\partial f}{\partial y} = N$ , differentiate (6) with respect to $y$ and set the result equal to $N$ :

N = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \int M (x, y) d x + \frac{d g (y)}{d y} .

Therefore

\frac{d g (y)}{d y} = N - \frac{\partial}{\partial y} \int M (x, y) d x

Step 3. Integrate both sides with respect to $y$ :

g (y) = \int (N - \frac{\partial}{\partial y} \int M (x, y) d x) d y

Putting the pieces together,

f (x, y) = \int M (x, y) d x + \overset{g (y)}{\overset{⏞}{\int (N - \frac{\partial}{\partial y} \int M (x, y) d x) d y}}

and the general solution of the differential equation is $f (x, y) = c$ , where $c$ is an arbitrary constant.

Remark. The two variables play symmetric roles, so you may instead start from $\frac{\partial f}{\partial y} = N$ and integrate with respect to $y$ while holding $x$ fixed, introducing a "constant" $h (x)$ . Choose whichever of $M$ and $N$ is easier to integrate.

Built-in check. In Step 2, the expression that remains for g'(y) must be free of $x$ . If an $x$ survives, either the equation was not exact or an algebra slip occurred. This makes Step 2 a free verification of Step 1.

Worked Examples

Find the general solution of the equation

(x^{3} + y^{2} + y) d x + (y^{3} + 2 x y + x) d y = 0 .

Solution

Here $M = x^{3} + y^{2} + y$ and $N = y^{3} + 2 x y + x$ . Testing for exactness,

\frac{\partial M}{\partial y} = 2 y + 1, \frac{\partial N}{\partial x} = 2 y + 1,

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

Let $d f = M d x + N d y$ . Because $\frac{\partial f}{\partial x} = M$ ,

\begin{aligned} f(x,y) &= \int M\,dx + g(y) \\ &= \int \left(x^3 + y^2 + y\right)dx + g(y) \\ &= \frac{x^4}{4} + y^2 x + yx + g(y) . \end{aligned}

Here $y$ was treated as a constant throughout, so $y^{2}$ and $y$ integrated to $y^{2} x$ and $y x$ .

Now impose $\frac{\partial f}{\partial y} = N$ :

\begin{aligned} \frac{\partial}{\partial y}\left[\frac{x^4}{4} + y^2 x + yx + g(y)\right] &= y^3 + 2xy + x \\ 2yx + x + g'(y) &= y^3 + 2xy + x \\ g'(y) &= y^3 \end{aligned}

All the $x$ terms cancelled, exactly as they must. Integrating,

g (y) = \frac{y^{4}}{4} + c,

and we may drop the constant $c$ here, since a single arbitrary constant will appear at the end.

Thus

f (x, y) = \frac{x^{4}}{4} + y^{2} x + y x + \frac{y^{4}}{4}

and the general solution is

\frac{x^{4}}{4} + y^{2} x + y x + \frac{y^{4}}{4} = C

where $C$ is a constant determined from an initial condition. Multiplying by $4$ gives the tidier form $x^{4} + 4 y^{2} x + 4 y x + y^{4} = 4 C$ .

Find the general solution of the equation $e^{y} d x + (x e^{y} + 2 y) d y = 0$ .

Solution

Here

M (x, y) = e^{y}, N (x, y) = x e^{y} + 2 y .

Because

\frac{\partial M}{\partial y} = e^{y} = \frac{\partial N}{\partial x},

this equation is exact.

We are looking for $f (x, y)$ such that

\left\{ \begin{aligned} \frac{\partial f}{\partial x} &= M(x,y) = e^{y}, \\[6pt] \frac{\partial f}{\partial y} &= N(x,y) = x e^{y} + 2y \end{aligned} \right.

Integrating the first equation with respect to $x$ , treating $y$ as constant,

f (x, y) = \int e^{y} d x + g (y) = x e^{y} + g (y) .

Therefore

\frac{\partial f}{\partial y} = x e^{y} + g'(y) .

This partial derivative must equal $N (x, y) = x e^{y} + 2 y$ :

x e^{y} + g'(y) = x e^{y} + 2y \quad \Longrightarrow \quad g'(y) = 2y \quad \Longrightarrow \quad g(y) = y^2 + c .

Finally

f (x, y) = x e^{y} + y^{2} + c = c_{1} .

Combining the two constants, the solution is simply

x e^{y} + y^{2} = c_{2} .

Solve the initial value problem $(2 x + y) d x + (x - 6 y) d y = 0$ , $y (1) = 0$ .

Solution

Here $M = 2 x + y$ and $N = x - 6 y$ , with

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

so the equation is exact.

Integrating $M$ with respect to $x$ ,

f (x, y) = x^{2} + x y + g (y) .

Differentiating with respect to $y$ and matching $N$ ,

x + g'(y) = x - 6y \quad \Longrightarrow \quad g'(y) = -6y \quad \Longrightarrow \quad g(y) = -3y^2 .

So the general solution is

x^{2} + x y - 3 y^{2} = C .

Applying $y (1) = 0$ gives $1 + 0 - 0 = C$ , so $C = 1$ and the solution is

x^{2} + x y - 3 y^{2} = 1 .

Exercises

Determine whether $(3 x^{2} y + 2) d x + (x^{3} + 4 y) d y = 0$ is exact, and solve it if it is.

Solution

With $M = 3 x^{2} y + 2$ and $N = x^{3} + 4 y$ ,

\frac{\partial M}{\partial y} = 3 x^{2} = \frac{\partial N}{\partial x},

so the equation is exact.

Integrating $M$ in $x$ : $f = x^{3} y + 2 x + g (y)$ . Then \dfrac{\partial f}{\partial y} = x^3 + g'(y), and matching $N$ gives g'(y) = 4y, so $g (y) = 2 y^{2}$ .

The general solution is $x^{3} y + 2 x + 2 y^{2} = C$ .

Show that $y d x - x d y = 0$ is not exact, then verify that dividing by $y^{2}$ makes it exact and solve.

Solution

With $M = y$ and $N = - x$ ,

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

which differ, so the equation is not exact.

Dividing by $y^{2}$ gives $\frac{1}{y} d x - \frac{x}{y^{2}} d y = 0$ . Now $M = \frac{1}{y}$ and $N = - \frac{x}{y^{2}}$ , and

\frac{\partial M}{\partial y} = - \frac{1}{y^{2}} = \frac{\partial N}{\partial x},

so the new equation is exact. Integrating $M$ in $x$ gives $f = \frac{x}{y} + g (y)$ , and \dfrac{\partial f}{\partial y} = -\dfrac{x}{y^2} + g'(y) matches $N$ with g'(y) = 0.

The solution is $\frac{x}{y} = C$ , that is, $y = k x$ with $k = 1 / C$ .

The factor $1 / y^{2}$ used here is an integrating factor, the subject of the next section.

Solve $(\cos y + 2 x) d x - (x \sin y) d y = 0$ .

Solution

With $M = \cos y + 2 x$ and $N = - x \sin y$ ,

\frac{\partial M}{\partial y} = - \sin y = \frac{\partial N}{\partial x},

so the equation is exact.

Integrating $M$ in $x$ : $f = x \cos y + x^{2} + g (y)$ . Then \dfrac{\partial f}{\partial y} = -x\sin y + g'(y), and matching $N$ gives g'(y) = 0.

The general solution is $x \cos y + x^{2} = C$ .

Frequently Asked Questions

What does "exact" actually mean?

It means the expression $M d x + N d y$ is the total differential of some single function $f (x, y)$ . The equation then reads $d f = 0$ , whose content is that $f$ stays constant along solutions. Finding the solution amounts to identifying $f$ , and the answer is the level-curve family $f (x, y) = c$ .

Why does the test

\partial M / \partial y = \partial N / \partial x

work?

If the equation is exact then $M = f_{x}$ and $N = f_{y}$ , so $\partial M / \partial y$ and $\partial N / \partial x$ are the two mixed second partials of $f$ . Clairaut's theorem says these are equal when continuous, which gives the necessity. Sufficiency requires the constructive argument in the proof above, and it needs the region to be a rectangle or otherwise simply connected.

Why is the constant of integration a function

g (y)

rather than a number?

Because the integration in Step 1 is partial, performed with respect to $x$ while $y$ is held fixed. Anything not depending on $x$ differentiates to zero under $\partial / \partial x$ , and that includes every function of $y$ alone. So the most general "constant" available at that stage is $g (y)$ .

What if an

x

is left over when I solve for g'(y)?

Then something is wrong. Either the equation is not exact, so recheck the test, or there is an algebra error in Step 1 or Step 2. Since $g$ depends on $y$ alone, g'(y) cannot contain $x$ , which makes this a reliable self-check.

What do I do when the exactness test fails?

Look for an integrating factor, a function $μ (x, y)$ such that multiplying through by it makes the equation exact. That is the subject of Section 2.5, and the second exercise above is a small preview of the idea.

Which variable should I integrate first?

Whichever of $M$ and $N$ is simpler to integrate. The two routes give the same answer. In the second worked example, integrating $M = e^{y}$ with respect to $x$ was trivial, whereas starting from $N = x e^{y} + 2 y$ would have been more work for the same result.

Should I present the answer explicitly as

y = ϕ (x)

Usually not. The natural form of the answer is the implicit relation $f (x, y) = c$ , and solving for $y$ is often impossible, as in $x^{4} + 4 y^{2} x + 4 y x + y^{4} = C$ . Leave it implicit unless the algebra untangles cleanly.

Exact First-Order Equations

Quick Reference

Where Exact Equations Come From

Definition of an Exact Equation

Test for Exactness

Method of Solution

Worked Examples

Exercises

Frequently Asked Questions

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