# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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M(x, y) + N(x, y) y = 0 (6)

be given. Suppose that we can identify a function — such that

d— d —

XX (x, y) = M(x, y), XX (x, y) = N(x, y), (7)

d x d y

and such that —(x, y) = c defines y = fi(x) implicitly as a differentiable function of x. Then

d— d— dy d

M(x, y) + N(x, y)y = — + — — = ——[x,$(x)] d x d y dx dx

and the differential equation (6) becomes

d—[x ,<p(x)] = 0. (8)

dx

In this case Eq. (6) is said to be an exact differential equation. Solutions of Eq. (6), or the equivalent Eq. (8), are given implicitly by

— (x, y) = c, (9)

where c is an arbitrary constant.

2.6 Exact Equations and Integrating Factors

91

Theorem 2.6.1

In Example 1 it was relatively easy to see that the differential equation was exact and, in fact, easy to find its solution, by recognizing the required function ty. For more complicated equations it may not be possible to do this so easily. A systematic way of determining whether a given differential equation is exact is provided by the following theorem.

Let the functions M, N, My, and Nx, where subscripts denote partial derivatives, be continuous in the rectangular11 region R: a < x < f3,y < y <8. Then Eq. (6),

M(x, y) + N(x, y)/ = 0, is an exact differential equation in R if and only if

My(x, y) = Nx(x, y) (10)

at each point of R. That is, there exists a function ty satisfying Eqs. (7),

tyx (x, y) = M(x, y), tyy(x, y) = N(x, y),

if and only if M and N satisfy Eq. (10).

The proof of this theorem has two parts. First, we show that if there is a function ty ich that Eqs. (7) are true, thi N from Eqs. (7), we obtain

such that Eqs. (7) are true, then it follows that Eq. (10) is satisfied. Computing My and

My(x, y) = tyxy(x, y), Nx(x, y) = tyyx(x, y). (11)

Since My and Nx are continuous, it follows that tyxy and tyyx are also continuous. This

guarantees their equality, and Eq. (10) follows.

We now show that if M and N satisfy Eq. (10), then Eq. (6) is exact. The proof involves the construction of a function ty satisfying Eqs. (7),

tyx(x, y) = M(x, y), tyy(x, y) = N(x, y).

Integrating the first of Eqs. (7) with respect to x, holding y constant, we find that

ty(x, y) = y M(x, y) dx + h(y). (12)

The function h is an arbitrary function of y, playing the role of the arbitrary constant. Now we must show that it is always possible to choose h(y) so that tyy = N. From

Eq. (12)

d C

fy(x, y) = djJM(x, y) dx + h (y) = j My(x, y) dx + h'(y).

11It is not essential that the region be rectangular, only that it be simply connected. In two dimensions this means that the region has no holes in its interior. Thus, for example, rectangular or circular regions are simply connected, but an annular region is not. More details can be found in most books on advanced calculus.

92

Chapter 2. First Order Differential Equations

EXAMPLE

2

Setting —y = N and solving for h'(y), we obtain

h'(y) = N(x, y) — j My(x, y) dx. (13)

To determine h( y) from Eq. (13), it is essential that, despite its appearance, the right side of Eq. (13) be a function of y only. To establish this fact, we can differentiate the quantity in question with respect to x, obtaining

Nx (x, y) — My(x, y),

which is zero on account of Eq. (10). Hence, despite its apparent form, the right side of Eq. (13) does not, in fact, depend on x, and a single integration then gives h(y). Substituting for h(y) in Eq. (12), we obtain as the solution of Eqs. (7)

f(x, y) = / M(x, y) dx + /[ N(x, y) - j My(x, y) dx

dy. (14)

It should be noted that this proof contains a method for the computation of —(x, y) and thus for solving the original differential equation (6). It is usually better to go through this process each time it is needed rather than to try to remember the result given in Eq. (14). Note also that the solution is obtained in implicit form; it may or may not be feasible to find the solution explicitly.

Solve the differential equation

(y cos x + 2xey) + (sin x + x2ey — 1) y = 0. (15)

It is easy to see that

My(x, y) = cos x + 2xey = Nx(x, y),

so the given equation is exact. Thus there is a — (x, y) such that

—x(x, y) = y cos x + 2xey,

—y(x, y) = sin x + x2 ey — 1.

Integrating the first of these equations, we obtain

— (x, y) = y sin x + x2ey + h(y). (16)

Setting —y = N gives

—y(x, y) = sin x + x2 ey + h'(y) = sin x + x2 ey — 1.

Thus h'(y) = — 1 and h(y) = —y. The constant of integration can be omitted since

any solution of the preceding differential equation is satisfactory; we do not require the most general one. Substituting for h(y) in Eq. (16) gives

— (x, y) = y sin x + x2 ey — y.

Hence solutions of Eq. (15) are given implicitly by

y sin x + x2ey — y = c. (17)

2.6 Exact Equations and Integrating Factors

93

Solve the differential equation

(3xy + y2) + (x2 + xy) y/ = 0. (18)

Here,

My(x, y) = 3x + 2y, Nx(x, y) = 2x + y;

since My = Nx, the given equation is not exact. To see that it cannot be solved by the procedure described previously, let us seek a function — such that

—x(x, y) = 3xy + y2, —y(x, y) = x2 + xy. (19)

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