2

First-Order Differential Equations

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2.4 Exact Equations

Copyright © Cengage Learning. All rights reserved.

Exact Equations (1 of 18)
Differential of a Function of Two Variables
If z = f(x, y) is a function of two variables with continuous
first partial derivatives in a region R of the xy-plane, then
recall from calculus that its differential is defined to be

In the special case when f(x, y) = c, where c is a constant,

then (1) gives

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Exact Equations (2 of 18)
In other words, given a one-parameter family of functions
f(x, y) = c, we can generate a first-order differential equation
by computing the differential of both sides of the equality.
For example, if then (2) gives the first-
order DE

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Exact Equations (3 of 18)
A Definition
Of course, not every first-order DE written in differential form
M(x, y) dx + N(x, y) dy = 0 corresponds to a differential of
f(x, y) = c.

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Exact Equations (4 of 18)
Definition 2.4.1 Exact Equation
A differential expression M(x, y) dx + N(x, y) dy is an exact
differential in a region R of the xy-plane if it corresponds
to the differential of some function f(x, y) defined in R. A
first-order differential equation of the form

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

is said to be an exact equation if the expression on the

left-hand side is an exact differential.

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Exact Equations (5 of 18)
For example, is an exact equation,
because its left-hand side is an exact differential:

Notice that if we make the identifications and

then

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Exact Equations (6 of 18)
Theorem 2.4.1 Criterion for an Exact Differential
Let M(x, y) and N(x, y) be continuous and have continuous
first partial derivatives in a rectangular region R defined by
a < x < b, c < y < d. Then a necessary and sufficient
condition that M(x, y) dx + N(x, y) dy be an exact
differential is

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Exact Equations (7 of 18)
Method of Solution
Given an equation in the differential form M(x, y) dx + N(x, y)
dy = 0, determine whether the equality in (4) holds.

If it does, then there exists a function f for which

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Exact Equations (8 of 18)
We can find f by integrating M(x, y) with respect to x while
holding y constant:

where the arbitrary function g(y) is the “constant” of

integration.

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Exact Equations (9 of 18)
Now differentiate (5) with respect to y and assume that
∂f ⁄ ∂y = N(x, y):

This gives

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Exact Equations (10 of 18)
Finally, integrate (6) with respect to y and substitute the result
in (5). The implicit solution of the equation is f(x, y) = c.

Some observations are in order. First, it is important to realize

that the expression N(x, y) − (∂ ⁄ ∂y ) ∫ M(x, y) dx in (6) is
independent of x, because

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Exact Equations (11 of 18)
Second, we could just as well start the foregoing procedure
with the assumption that ∂f ⁄ ∂y = N(x, y). After integrating N
with respect to y and then differentiating that result, we would
find the analogues of (5) and (6) to be, respectively,

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Example 1 – Solving an Exact DE
Solve

Solution:
With M(x, y) = 2xy and we have

Thus the equation is exact, and so by Theorem 2.4.1 there

exists a function f(x, y) such that

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Example 1 – Solution (1 of 2)
From the first of these equations we obtain, after integrating,

Taking the partial derivative of the last expression with

respect to y and setting the result equal to N(x, y) gives

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Example 1 – Solution (2 of 2)
It follows that g′(y) = −1 and g(y) = −y. Hence
so the solution of the equation in implicit form is

The explicit form of the solution is easily seen to be

and is defined on any interval not containing
either x = 1 or x = −1.

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Exact Equations (12 of 18)
Note
The solution of the DE in Example 1 is not
Rather, it is f(x, y) = c; if a constant is used in the integration
of g′(y), we can then write the solution as f(x, y) = 0.

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Example 3 – An Initial-Value Problem
Solve

Solution:
By writing the differential equation in the form

we recognize that the equation is exact because

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Example 3 – Solution (1 of 3)
Now

The last equation implies that h′(x) = cos x sin x. Integrating

gives

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Example 3 – Solution (2 of 3)
Thus

where 2c1 has been replaced by c. The initial condition y = 2

when x = 0 demands that and so c = 3.

An implicit solution of the problem is then

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Example 3 – Solution (3 of 3)
The solution curve of the IVP is the curve drawn in blue in
figure; it is part of an interesting family of curves.

Figure 2.4.1

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Exact Equations (13 of 18)
Integrating Factors
The left-hand side of the linear equation y′ + P(x)y = f(x) can
be transformed into a derivative when we multiply the
equation by an integrating factor.

The same basic idea sometimes works for a nonexact

differential equation M(x, y) dx + N(x, y) dy = 0.

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Exact Equations (14 of 18)
That is, it is sometimes possible to find an integrating factor
μ(x, y) so that after multiplying, the left-hand side of

μ(x, y)M(x, y) dx + μ(x, y)N(x, y) dy = 0 (8)

is an exact differential.

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Exact Equations (15 of 18)
In an attempt to find μ, we turn to the criterion (4) for
exactness. Equation (8) is exact if and only if (μM)y = (μN)x,
where the subscripts denote partial derivatives.

By the Product Rule of differentiation the last equation is the

same as μMy + μyM = μNx + μxN or

μxN − μyM = (My − Nx)μ. (9)

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Exact Equations (16 of 18)
If μ is only dependent on x, μx = dμ ⁄ dx and μy = 0, so (9) can
be written as

We are still at an impasse if the quotient (My − Nx) ⁄ N

depends on both x and y.

Now, we assume (My − Nx) ⁄ N only depends on x.

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Exact Equations (17 of 18)
We can finally determine μ because (10) is separable as
well as linear.

In like manner, it follows from (9) that if μ depends only on

the variable y, then

In this case, if (Nx − My) ⁄ M is a function of y only, then we

can solve (11) for μ.

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Exact Equations (18 of 18)
We summarize the results for the differential equation

M(x, y) dx + N(x, y) dy = 0. (12)

• If (My − Nx) ⁄ N is a function of x alone, then an integrating

factor for (12) is

• If (Nx − My) ⁄ M is a function of y alone, then an integrating

factor for (12) is

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Example 4 – A Nonexact DE Made Exact (1 of 3)

The nonlinear first-order differential equation

is not exact.

With the identifications we find

the partial derivatives My = x and Nx = 4x.

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Example 4 – A Nonexact DE Made Exact (2 of 3)

The first quotient from (13) gets us nowhere, since

depends on x and y.

However, (14) yields a quotient that depends only on y:

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Example 4 – A Nonexact DE Made Exact (3 of 3)

The integrating factor is then After

we multiply the given DE by the resulting equation is

You should verify that the last equation is now exact as well
as show, using the method of this section, that a family of
solutions is

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