Chapter 2 · Section 2.1

2.1 Exact Equations and Integrating Factors

Suppose you walk on a hilly landscape whose height is $F (x, y)$ , and you always stay at the same altitude. Your path is a level curve $F (x, y) = c$ . Along that path the height does not change, so its differential is zero: $d F = 0$ . Exact equations are exactly this situation, seen backwards. We are handed $d F = 0$ in disguise, and our job is to find the landscape $F$ .

Two ways to write a first-order equation

Every first-order equation in this chapter can be written in one of two forms. The derivative form is

y^{'} = f (x, y),

and the differential form is

M (x, y) d x + N (x, y) d y = 0.

Moving between them is just algebra. Take, for instance,

y^{'} = \frac{4 x - y}{x + 2 y ^{3}} .

Write $y^{'} = d y / d x$ and cross-multiply:

(x + 2 y^{3}) d y (4 x - y) d x - (x + 2 y^{3}) d y = (4 x - y) d x = 0.

So $M = 4 x - y$ and $N = - (x + 2 y^{3})$ . Going the other way, the equation $(y e^{x} + 1) d x + (x - 2 y) d y = 0$ becomes

y^{'} = - \frac{y e ^{x} + 1}{x - 2 y} .

In derivative form the roles are fixed: $x$ is the independent variable and $y$ the unknown function. In differential form $x$ and $y$ stand on an equal footing, and we are free to pick either one as the unknown. Our default is $y$ as a function of $x$ ; whenever we switch, we will say so. This section works mostly with the differential form.

Total differentials and exact equations

Recall from calculus how a small change in $x$ and $y$ changes a function of two variables.

Definition 2.1.1 (Total differential)

Let $F (x, y)$ have continuous first partial derivatives in a region $D$ of the plane. Its total differential is

d F = \frac{\partial F}{\partial x} d x + \frac{\partial F}{\partial y} d y .

For example, if $F (x, y) = x^{3} y^{2} - 4 x y$ , then $F_{x} = 3 x^{2} y^{2} - 4 y$ and $F_{y} = 2 x^{3} y - 4 x$ , so

d F = (3 x^{2} y^{2} - 4 y) d x + (2 x^{3} y - 4 x) d y .

Now reverse the question. Someone gives you an expression $M d x + N d y$ . Is it the total differential of some function?

Definition 2.1.2 (Exact differential, exact equation)

The expression $M (x, y) d x + N (x, y) d y$ is an exact differential in a region $D$ if there is a function $F (x, y)$ with

\frac{\partial F}{\partial x} = M (x, y), \frac{\partial F}{\partial y} = N (x, y) for all (x, y) \in D .

If $M d x + N d y$ is an exact differential, the equation $M d x + N d y = 0$ is called an exact equation, and $F$ is called a potential function for it.

Why do we care? Because an exact equation says $d F = 0$ , and that is easy to solve.

Why the level curves are the solutions

Let $y = φ (x)$ be a differentiable function. Along its graph, the height is $F (x, φ (x))$ . By the chain rule,

\frac{d}{d x} F (x, φ (x)) = F_{x} (x, φ (x)) + F_{y} (x, φ (x)) φ^{'} (x) = M (x, φ (x)) + N (x, φ (x)) φ^{'} (x) .

The last line is zero exactly when $φ$ solves $M + N y^{'} = 0$ . So:

every solution keeps $F$ constant, that is, it lies on a level curve $F (x, y) = c$ ;
every differentiable function defined implicitly by $F (x, y) = c$ is a solution.

Near any point where $N = F_{y} \neq = 0$ , the implicit function theorem guarantees that $F (x, y) = c$ really does define $y$ as a differentiable function of $x$ . Hence $F (x, y) = c$ is a one-parameter family of solutions, written in implicit form.

Two questions remain. How can we tell whether $M d x + N d y$ is exact without guessing $F$ ? And if it is, how do we find $F$ ? One theorem answers both.

The exactness test

Theorem 2.1.1 (Test for exactness)

Let $M (x, y)$ and $N (x, y)$ be continuous, with continuous partial derivatives $\partial M / \partial y$ and $\partial N / \partial x$ , on an open rectangle $R = {(x, y) : a < x < b, c < y < d}$ . Then $M d x + N d y$ is an exact differential in $R$ if and only if

\frac{\partial M}{\partial y} (x, y) = \frac{\partial N}{\partial x} (x, y) for all (x, y) \in R .

The condition is easy to remember: differentiate $M$ with respect to the other variable $y$ , differentiate $N$ with respect to the other variable $x$ , and compare. Let us see why it works. There are two directions to prove.

Part 1: exact implies $M_{y} = N_{x}$

Assume a potential $F$ exists, so $F_{x} = M$ and $F_{y} = N$ in $R$ .

Differentiate the first relation with respect to $y$ : $F_{x y} = M_{y}$ .
Differentiate the second relation with respect to $x$ : $F_{y x} = N_{x}$ .
By hypothesis $M_{y}$ and $N_{x}$ are continuous, so both mixed partial derivatives of $F$ are continuous. A standard theorem of calculus (Schwarz's, or Clairaut's, theorem) says that continuous mixed partials are equal: $F_{x y} = F_{y x}$ .
Therefore $M_{y} = N_{x}$ everywhere in $R$ .

Part 2: $M_{y} = N_{x}$ implies exact

This is the useful direction, because the proof builds $F$ . Pick any point $(x_{0}, y_{0})$ in $R$ and define

F (x, y) = \int_{x_{0}}^{x} M (s, y) d s + \int_{y_{0}}^{y} N (x_{0}, t) d t .

In words: walk from $(x_{0}, y_{0})$ vertically up to $(x_{0}, y)$ , collecting $N d y$ , then horizontally to $(x, y)$ , collecting $M d x$ . Because $R$ is a rectangle, both legs of the walk stay inside $R$ . We now check the two conditions of Definition 2.1.2.

Figure 2.1.1. The walk used to define

F

in the proof: first vertically from

(x_{0}, y_{0})

(x_{0}, y)

, then horizontally to

(x, y)

. Both legs are straight segments parallel to the axes, so they never leave the rectangle

R

The $x$ -derivative. The second integral does not depend on $x$ . By the fundamental theorem of calculus, $F_{x} (x, y) = M (x, y) .$
The $y$ -derivative, first integral. Here $y$ sits inside the integrand. Since $M_{y}$ is continuous on the rectangle, we may differentiate under the integral sign: $\frac{\partial}{\partial y} \int_{x_{0}}^{x} M (s, y) d s = \int_{x_{0}}^{x} M_{y} (s, y) d s .$
Use the hypothesis. Replace $M_{y}$ by $N_{x}$ and integrate with respect to $s$ : $\int_{x_{0}}^{x} N_{x} (s, y) d s = N (x, y) - N (x_{0}, y) .$
The $y$ -derivative, second integral. By the fundamental theorem of calculus, $\frac{\partial}{\partial y} \int_{y_{0}}^{y} N (x_{0}, t) d t = N (x_{0}, y) .$
Add the pieces. $F_{y} (x, y) = [N (x, y) - N (x_{0}, y)] + N (x_{0}, y) = N (x, y) .$

Both conditions hold, so $M d x + N d y = d F$ is exact. ∎

Why the region matters

The proof needed every horizontal and vertical walk to stay inside the region. A rectangle (or a disc, or any region "without holes") is fine. On a region with a hole, $M_{y} = N_{x}$ is not enough. The classic example is

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

defined everywhere except the origin. A direct computation gives $M_{y} = N_{x} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ . Yet going once around the unit circle, $\oint M d x + N d y = 2 π \neq = 0$ . If a potential $F$ existed on the punctured plane, this loop integral would be $F (end) - F (start) = 0$ . So there is no potential on the whole punctured plane, although on any rectangle that avoids the origin there is one (the polar angle).

Finding the potential function

The proof suggests a practical method. We do not need the definite integrals; indefinite integrals with a careful "constant" are enough.

We want $F$ with $F_{x} = M$ . Integrating with respect to $x$ , while holding $y$ fixed, gives $F$ up to anything that does not depend on $x$ , that is, up to a function of $y$ alone:

F (x, y) = \int M (x, y) d x + h (y) .

Next we force the second condition, $F_{y} = N$ :

\frac{\partial}{\partial y} \int M (x, y) d x + h^{'} (y) = N (x, y) .

Solve this for $h^{'} (y)$ . When the equation is exact, all the $x$ 's cancel and $h^{'}$ depends on $y$ only. Integrate to get $h (y)$ , and you have $F$ .

Method in a nutshell

Identify $M$ and $N$ and check $M_{y} = N_{x}$ . If it fails, the equation is not exact (try Section 2.4 or another method).
Integrate $M$ with respect to $x$ , treating $y$ as a constant: $F = \int M d x + h (y)$ .
Differentiate this $F$ with respect to $y$ and set it equal to $N$ . Solve for $h^{'} (y)$ . (Check: it must not contain $x$ .)
Integrate to find $h (y)$ . No constant of integration is needed here.
Write the solution as $F (x, y) = c$ . For an initial condition $y (x_{0}) = y_{0}$ , take $c = F (x_{0}, y_{0})$ .

Common mistake

Do not "integrate $M$ in $x$ , integrate $N$ in $y$ , and add". Terms that contain both variables would be counted twice. Either use the $h (y)$ step, or, if you add, keep each term only once.

A built-in alarm

Step 3 doubles as a check on your work. If, after simplifying, $h^{'} (y)$ still contains $x$ , something is wrong: either the equation is not exact after all (go back to step 1), or there is an algebra slip in step 2. The proof of Theorem 2.1.1 guarantees that for an exact equation the $x$ 's always cancel.

Do not be disappointed if the answer stays implicit. A relation $F (x, y) = c$ is a perfectly good solution. Solving it for $y$ is a bonus that is possible only for some $F$ , and even then you must choose the right branch, as Example 2.1.3 below shows.

It does not matter which variable you start with. If $\int N d y$ looks easier, compute $F = \int N d y + g (x)$ and then use $F_{x} = M$ to find $g (x)$ .

Worked examples

Example 2.1.1

Solve $(6 x^{2} y + y^{3} + 2) d x + (2 x^{3} + 3 x y^{2} - 4 y) d y = 0$ .

Solution

Test. $M = 6 x^{2} y + y^{3} + 2$ and $N = 2 x^{3} + 3 x y^{2} - 4 y$ . $M_{y} = 6 x^{2} + 3 y^{2}, N_{x} = 6 x^{2} + 3 y^{2} .$ They agree everywhere, so the equation is exact in the whole plane.
Integrate $M$ in $x$ . $F = \int (6 x^{2} y + y^{3} + 2) d x = 2 x^{3} y + x y^{3} + 2 x + h (y) .$
Match $N$ . $F_{y} = 2 x^{3} + 3 x y^{2} + h^{'} (y) =! 2 x^{3} + 3 x y^{2} - 4 y .$ The $x$ -terms cancel, leaving $h^{'} (y) = - 4 y$ .
Find $h$ . $h (y) = - 2 y^{2}$ .
Answer. $2 x^{3} y + x y^{3} + 2 x - 2 y^{2} = c .$ Check: $F_{x} = 6 x^{2} y + y^{3} + 2 = M$ and $F_{y} = 2 x^{3} + 3 x y^{2} - 4 y = N$ .

Figure 2.1.2. Level curves

F (x, y) = c

of the potential

F = 2 x^{3} y + x y^{3} + 2 x - 2 y^{2}

from Example 2.1.1. Each level curve is made of solution curves of the exact equation. Where a level curve crosses the dashed curve

N = 0

, its tangent is vertical, and near such a point

y

is not a differentiable function of

x

Try the exactness test on your own equations. The solver below accepts polynomial $M$ and $N$ . It checks $M_{y} = N_{x}$ , builds $F$ step by step when the equation is exact, and, when it is not, looks for an integrating factor that depends on $x$ only or on $y$ only (we explain those in Section 2.4).

Example 2.1.2

(a) Show that $(3 y + x) d x + (x - 2 y) d y = 0$ is not exact.
(b) Solve $(y cos x + 2 x e^{y}) d x + (sin x + x^{2} e^{y} + 3) d y = 0$ .

Solution

(a) Test. $M_{y} = 3$ and $N_{x} = 1$ . Since $3 \neq = 1$ , the expression is not exact in any rectangle. No potential function exists, and the method of this section does not apply directly.
(b) Test. $M = y cos x + 2 x e^{y}$ and $N = sin x + x^{2} e^{y} + 3$ . $M_{y} = cos x + 2 x e^{y}, N_{x} = cos x + 2 x e^{y} .$ Exact in the whole plane.
Integrate $M$ in $x$ . Remember that $y$ is a constant here: $F = \int (y cos x + 2 x e^{y}) d x = y sin x + x^{2} e^{y} + h (y) .$
Match $N$ . $F_{y} = sin x + x^{2} e^{y} + h^{'} (y) =! sin x + x^{2} e^{y} + 3,$ so $h^{'} (y) = 3$ and $h (y) = 3 y$ .
Answer. $y sin x + x^{2} e^{y} + 3 y = c .$

Example 2.1.3

Solve the initial-value problem $(y^{2} + 2 x) d x + (2 x y - 3) d y = 0, y (1) = 1,$ and find an explicit formula for the solution.

Solution

Test. $M_{y} = 2 y$ and $N_{x} = 2 y$ . Exact.
Potential. $F = \int (y^{2} + 2 x) d x = x y^{2} + x^{2} + h (y)$ . Then $F_{y} = 2 x y + h^{'} (y) =! 2 x y - 3,$ so $h (y) = - 3 y$ and $F = x y^{2} + x^{2} - 3 y$ .
Use the initial condition. $c = F (1, 1) = 1 + 1 - 3 = - 1$ . The implicit solution is $x y^{2} + x^{2} - 3 y = - 1.$
Solve for $y$ . This is a quadratic in $y$ : $x y^{2} - 3 y + (x^{2} + 1) = 0$ . For $x \neq = 0$ , $y = \frac{3 \pm 9 - 4 x ( x ^{2} + 1 )}{2 x} = \frac{3 \pm 9 - 4 x - 4 x ^{3}}{2 x} .$
Pick the branch. At $x = 1$ the square root equals $1$ , so the two branches give $y = 2$ and $y = 1$ . The initial condition picks the minus sign.
Tidy up. Multiply top and bottom by $3 + 9 - 4 x - 4 x^{3}$ . The numerator becomes $9 - (9 - 4 x - 4 x^{3}) = 4 x (x^{2} + 1)$ , and the factor $x$ cancels: $y = \frac{2 ( x ^{2} + 1 )}{3 + 9 - 4 x - 4 x ^{3}} .$ This form also works at $x = 0$ , where it gives $y = 1/3$ (check: $0 + 0 - 1 = - 1$ ).
Interval. We need $9 - 4 x - 4 x^{3} > 0$ . The left side decreases strictly (its derivative is $- 4 - 12 x^{2} < 0$ ), so it has exactly one real zero, $x^{*} \approx 1.060$ . The solution lives on $- \infty < x < x^{*}$ . At $x^{*}$ we have $N = 2 x y - 3 = 0$ : the slope becomes infinite, just as the remark on $N \neq = 0$ predicted.

Figure 2.1.3. The level curve

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

of Example 2.1.3 (dashed) has several branches. The initial condition

y (1) = 1

selects the solid one, given by the minus sign. It exists for

- \infty < x < x^{*} \approx 1.060

; at

x^{*}

it meets the other branch, through

(1, 2)

, with a vertical tangent.

Integrating factors

Most equations are not exact. Sometimes, though, multiplying by a well-chosen function makes them exact. Consider $(3 x y + 2 y^{3}) d x + (x^{2} + 3 x y^{2}) d y = 0$ . Here $M_{y} = 3 x + 6 y^{2}$ but $N_{x} = 2 x + 3 y^{2}$ , so it is not exact. Multiply every term by $x$ :

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

Now the new coefficients satisfy $\partial_{y} (3 x^{2} y + 2 x y^{3}) = 3 x^{2} + 6 x y^{2} = \partial_{x} (x^{3} + 3 x^{2} y^{2})$ . The equation has become exact.

Definition 2.1.3 (Integrating factor)

A function $μ (x, y)$ , not identically zero, is an integrating factor of $M d x + N d y = 0$ if

μ (x, y) M (x, y) d x + μ (x, y) N (x, y) d y = 0

is exact. With continuous partial derivatives, this means $\frac{\partial ( μ M )}{\partial y} = \frac{\partial ( μ N )}{\partial x}$ .

Wherever $μ \neq = 0$ , the new equation has the same solutions as the old one, because we can divide $μ$ back out. So we solve the exact equation and keep its solutions.

Watch the zeros of

μ

Multiplying by $μ$ can introduce extra curves where $μ = 0$ (they solve the new equation trivially but may not solve the old one). Dividing by an expression can remove solutions. After using an integrating factor, check the curves where $μ$ is zero or undefined against the original equation.

In general, finding $μ$ means solving a partial differential equation, which is harder than the original problem. But in many useful cases $μ$ depends on one variable only, and then it can be computed with a formula. That is the topic of Section 2.4. For now, we check a given factor.

Example 2.1.4

Show that $μ = x$ is an integrating factor of $(3 x y + 2 y^{3}) d x + (x^{2} + 3 x y^{2}) d y = 0$ , and solve the equation.

Solution

Original equation is not exact. $M_{y} = 3 x + 6 y^{2}$ , $N_{x} = 2 x + 3 y^{2}$ .
Multiply by $x$ . $μ M = 3 x^{2} y + 2 x y^{3}$ and $μ N = x^{3} + 3 x^{2} y^{2}$ . $(μ M)_{y} = 3 x^{2} + 6 x y^{2}, (μ N)_{x} = 3 x^{2} + 6 x y^{2} .$ Equal, so $μ = x$ is an integrating factor.
Potential. $F = \int (3 x^{2} y + 2 x y^{3}) d x = x^{3} y + x^{2} y^{3} + h (y) .$ Then $F_{y} = x^{3} + 3 x^{2} y^{2} + h^{'} (y)$ must equal $x^{3} + 3 x^{2} y^{2}$ , so $h^{'} = 0$ .
Solution. $x^{3} y + x^{2} y^{3} = c$ .
Check the zero of $μ$ . $μ = 0$ on the line $x = 0$ . Treating $x$ as a function of $y$ , the constant $x \equiv 0$ gives $d x = 0$ and $N (0, y) = 0$ , so it does solve the original equation. It is already contained in the family (take $c = 0$ ). Nothing is lost or gained.

Exercises

In Exercises 1–7, decide whether the equation is exact. If it is, solve it.

$(5 x - 2 y) d x + (6 y - 2 x) d y = 0$
Answer
Exact; $5 x^{2} - 4 x y + 6 y^{2} = c$ .
$(y^{3} - 6 x) d x + (3 x y^{2} + 5) d y = 0$
Answer
Exact; $x y^{3} - 3 x^{2} + 5 y = c$ .
$(4 x^{3} y - y^{2}) d x + (x^{4} - 2 x y + 5) d y = 0$
Answer
Exact; $x^{4} y - x y^{2} + 5 y = c$ .
$(x^{2} + y) d x + (3 x + y^{2}) d y = 0$
Answer
Not exact: $M_{y} = 1$ but $N_{x} = 3$ .
$(2 x cos y - 3) d x - (x^{2} sin y + 1) d y = 0$
Answer
Exact; $x^{2} cos y - 3 x - y = c$ .
$(3 y e^{3 x} - x) d x + (e^{3 x} + 2 y) d y = 0$
Answer
Exact; $y e^{3 x} - \frac{1}{2} x^{2} + y^{2} = c$ .
$(\frac{1}{x} + 2 y) d x + (2 x - \frac{1}{y}) d y = 0$ , for $x > 0, y > 0$ .
Answer
Exact; $ln x + 2 x y - ln y = c$ .
Solve $(4 x y - 3 x^{2}) d x + (2 x^{2} + 3 y^{2}) d y = 0$ , $y (1) = - 1$ .
Answer
$2 x^{2} y - x^{3} + y^{3} = - 4$ .
Solve $(y e^{x y} + 4 x) d x + (x e^{x y} - 2) d y = 0$ , $y (0) = 1$ .
Answer
$e^{x y} + 2 x^{2} - 2 y = - 1$ .
Find the constant $k$ that makes $(k x y^{2} + 2 y) d x + (3 x^{2} y + 2 x) d y = 0$ exact, and solve the resulting equation.
Answer
$k = 3$ ; $3 x^{2} y^{2} + 4 x y = c$ .
Show that $μ = y$ is an integrating factor of $(y^{2} + 2 x y) d x + (3 x y + 2 x^{2}) d y = 0$ and solve the equation.
Answer
After multiplying by $y$ : $(μ M)_{y} = (μ N)_{x} = 3 y^{2} + 4 x y$ ; solution $x y^{3} + x^{2} y^{2} = c$ .

Two ways to write a first-order equation

Total differentials and exact equations

The exactness test

Part 1: exact implies My​=Nx​

Part 2: My​=Nx​ implies exact

Finding the potential function

Worked examples

Integrating factors

Exercises

Part 1: exact implies $M_{y} = N_{x}$

Part 2: $M_{y} = N_{x}$ implies exact