F

But since N = dF/dt, we can equate N = y2 and dF/dt = y2 + >p'(t), to get *'(') = 0

Step iii Integration of the last result gives us

^(t) = jy(t) dt = jo dt = k and now we have a specific form of 4>{t). It happens in the present case that \p(t) is simply a constant; more generally, it can be a nonconstant function of t.

Step iv The results of Steps i and iii can be combined to yield

The solution of the exact differential equation should then be F(y, t) = c. But since the constant k can be merged into c, we may write the solution simply as y2t = c or y(t) = ct 1/2

where c is arbitrary.

Example 2 Solve the equation (t + 2y) dy + (y + 3r2) dt = 0. First let us check whether this is an exact differential equation. Setting M = t + 2y and N = y + 3f2, we find that dM/dt = 1 = dN/dy. Thus the equation passes the exactness test. To find its solution, we again follow the procedure outlined in Example 1.

Step i Apply (14.19) and write

[constant merged into »/>(/)] Sthp ii Differentiate this result with respect to t, to get

Step iii Integrate this last result to get xp(f) = Jit2 dt = t3 [constant may be omitted]

Step iv Combine the results of Steps i and iii to get the complete form of the function F(y, t):

which implies that the solution of the given differential equation is yt + v2 + t3 = c

You should verify that setting the total differential of this equation equal to zero will indeed produce the given differential equation.

This four-step procedure can be used to solve any exact differential equation. Interestingly, it may even be applicable when the given equation is not exact. To see this, however, we must first introduce the concept of integrating factor.

Integrating Factor

Sometimes an inexact differential equation can be made exact by multiplying every term of the equation by a particular common factor. Such a factor is called an integrating factor.

Example 3 The differential equation

is not exact, because it does not satisfy (14.18):

dM d , „ J dN d . , , — = —(2 i) = 2 + — = = 1 dt dty ' dy dy

However, if we multiply each term by y, the given equation will turn into (14.16), which has been established to be exact. Thus y is an integrating factor for the differential equation in the present example.

When an integrating factor can be found for an inexact differential equation, it is always possible to render it exact, and then the four-step solution procedure can be readily put to use.

Solution of First-Order Linear Differential Equations

The general first-order linear differential equation dy

dt which, in the format of (14.17), can be expressed as (14.20) dy+{uy~w)dt = 0 has the integrating factor e/uc" = expj ju dt\

This integrating factor, whose form is by no means intuitively obvious, can be "discovered" as follows. Let 7 be the (yet unknown) integrating factor. Multiplication of (14.20) through by 7 should convert it into an exact differential equation

The exactness test dictates that dM/dt = dN/dy. Visual inspection of the M and N expressions suggests that, since M consists of 7 only, and since u and w are functions of t alone, the exactness test will reduce to a very simple condition if 7 is also a function of t alone. For then the test becomes

Thus the special form I = l(t) can indeed work, provided it has a rate of growth equal to u, or more explicitly, u(t). Accordingly, I(t) should take the specific form

As can be easily verified, however, the constant A can be set equal to 1 without affecting the ability of I(t) to meet the exactness test. Thus we can use the simpler form elud' as the integrating factor.

Substitution of this integrating factor into (14.20') yields the exact differential equation

(14.20") eiudt dy + e'uc"{uy - w) dt = 0 which can then be solved by the four-step procedure.

dl/dt 1

Step i First, we apply (14.19) to obtain

The result of integration emerges in this simple form because the integrand is independent of the variable y.

Step ii Next, we differentiate the above result with respect to t, to get (9 F

And, since this can be equated to N = e^"d'(uy — w), we have i'(t) = -wefud'

Step iii Straight integration now yields

Inasmuch as the functions u = u(t) and w = w(t) have not been given specific forms, nothing further can be done about this integral, and we must be contented with this rather general expression for ipi*)-

Step iv Substituting this ip(t) expression into the result of Step i, we find that

So the general solution of the exact differential equation (14.20")—and of the equivalent, though inexact, first-order linear differential equation (14.20)—is

0 0

Post a comment