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

ISBN 0-471-31999-6

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d d

—L [xr ] = — [xrF (r)]. d r d r

Substituting for F(r), interchanging differentiation with respect to x and with respect to r, and noting that d(xr )/dr = xr ln x, we obtain

L [xr ln x] = (r — r)xr ln x + 2(r — r1)xr. (9)

The right side of Eq. (9) is zero for r = rx; consequently,

y2(x) = xrl lnx,

x > 0

(10)

262

Chapter 5. Series Solutions of Second Order Linear Equations

EXAMPLE

2

is a second solution of Eq. (1). It is easy to show that W(xri, xri ln x) = x2ri 1. Hence xri and xri ln x are linearly independent for x > 0, and the general solution of Eq. (1) is

y = (c1 + c2ln x )xri, x > 0. (11)

Solve

x2y" + 5xy; + 4y = 0, x > 0. (12)

Substituting y = xr in Eq. (12) gives

xr [r (r — 1) + 5r + 4] = xr (r2 + 4r + 4) = 0.

Hence r1 = r2 = —2, and

y = x—2(c1 + c2lnx), x > 0. (13)

Complex Roots. Finally, suppose that the roots r1 and r2 are complex conjugates, say, r1 = X + ip and r2 = X — ip, with p = 0. We must now explain what is meant by xr when r is complex. Remembering that

xr = er ln x (14)

when x > 0 and r is real, we can use this equation to define xr when r is complex. Then

xX+ip e(X+iP) lnx eX lnxeiP lnx xXeiP lnx

x — e — e e — x e

= xX[cos(p lnx) + i sin(p lnx)], x > 0. (15)

With this definition of xr for complex values of r, it can be verified that the usual laws

of algebra and the differential calculus hold, and hence xr1 and xr2 are indeed solutions of Eq. (1). The general solution of Eq. (1) is

y = c1 xX+I p + c2xX—Ip. (16)

The disadvantage of this expression is that the functions xX+'p and xX 1 p are complex-

2

X—i p

valued. Recall that we had a similar situation for the second order differential equation with constant coefficients when the roots of the characteristic equation were complex. In the same way as we did then we observe that the real and imaginary parts of xX+ip, namely,

xX cos(plnx) and xX sin(plnx), (17)

are also solutions of Eq. (1). A straightforward calculation shows that

W [xX cos(p ln x ), xX sin(p ln x )] = px2X—1.

Hence these solutions are also linearly independent for x > 0, and the general solution of Eq. (1) is

y = c1 xX cos(p lnx) + c2xX sin(p lnx), x > 0. (18)

5.5 Euler Equations

263

Solve

3 x y + x/+ y = 0. (19)

Substituting y = xr in Eq. (19) gives

xr [r (r — 1) + r + 1] = xr (r2 + 1) = 0.

Hence r = ±i, and the general solution is

y = c1 cos(lnx) + c2sin(lnx), x > 0. (20)

Now let us consider the qualitative behavior of the solutions of Eq. (1) near the singular point x = 0. This depends entirely on the nature of the exponents r1 and r2. First, if r is real and positive, then xr ^ 0 as x tends to zero through positive values. On the other hand, if r is real and negative, then xr becomes unbounded, while if r = 0, then xr = 1. These possibilities are shown in Figure 5.5.1 for various values of r .If r is complex, then atypical solution is xX cos(r lnx). This function becomes unbounded or approaches zero if X is negative or positive, respectively, and also oscillates more and more rapidly as x ^ 0. This behavior is shown in Figures 5.5.2 and 5.5.3 for selected values of X and r. If X = 0, the oscillation is of constant amplitude. Finally, if there are repeated roots, then one solution is of the form xr ln x, which tends to zero if r > 0 and becomes unbounded if r < 0. An example of each case is shown in Figure 5.5.4.

The extension of the solutions of Eq. (1) into the interval x < 0 can be carried out in a relatively straightforward manner. The difficulty lies in understanding what is meant by xr when x is negative and r is not an integer; similarly, ln x has not been defined for x < 0. The solutions of the Euler equation that we have given for x > 0 can be shown to be valid for x < 0, but in general they are complex-valued. Thus in Example 1 the solution x1/2 is imaginary for x < 0.

FIGURE 5.5.1 Solutions of an Euler equation; real roots.

264

Chapter 5. Series Solutions of Second Order Linear Equations

FIGURE 5.5.2 Solution of an Euler equation; FIGURE 5.5.3 Solution of an Euler equa-complex roots with negative real part. tion; complex roots with positive real part.

FIGURE 5.5.4 Solutions of an Euler equation; repeated roots.

It is always possible to obtain real-valued solutions of the Euler equation (1) in the interval x < 0 by making the following change of variable. Let x = —?, where ? > 0, and let y = u(?). Then we have

d2y d

dx2 d?

dy

dx

du d? du

d ? dx d ? ’

Thus Eq. (1), for x < 0, takes the form

du\ d? d ? / dx

d 2u d?2 '

(21)

-, d2u du

? ^2 + a?^? + ?u = o.

d ? 2 d ?

? > 0.

(22)

But this is exactly the problem that we have just solved; from Eqs. (6), (11), and (18) we have

u(?) =

Ci? L + C2? r2 (Ci + C2ln ?)?ri

c1?? cos(^ ln?) + c2?x sin(^ ln?),

(23)

depending on whether the zeros of F(r) = r (r — 1) + ar + /3 are real and different, real and equal, or complex conjugates. To obtain u in terms of x we replace ? by —x in Eqs. (23).

We can combine the results for x > 0 and x < 0 by recalling that |x |= x when x > 0 and that |x| = — x when x < 0. Thus we need only replace x by |x | in Eqs. (6),

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