# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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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 (ã ln x). 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 ã. 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 df2'

(21)

d2u du

% 2^2 + à%^ + â u = 0, 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(%) =

c1% r1 + c2% r2

+ c2ln %)%r1 c1%A cos(l ln %) + c2%x sin(^ ln %),

(23)

depending on whether the zeros of F(r) = r (r — 1) + ar + â 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),

5.5 Euler Equations

265

(11), and (18) to obtain real-valued solutions valid in any interval not containing the origin (also see Problems 30 and 31). These results are summarized in the following theorem.

Theorem 5.5.1 The general solution of the Euler equation (1),

x2 y" + a xy' + â y = 0,

in any interval not containing the origin is determined by the roots r1 and r2 of the

equation

F (r) = r (r --- 1) + ar + â = 0.

If the roots are real and different, then

y (24)

II

î

r

+

î

tsj

If the roots are real and equal, then

y = (Cj + c2ln |x|)|xÃ1. (25)

If the roots are complex, then

y = |x|A[î1 cos(iln |x|) + C2sin(^ ln |x|)], (26)

where rp r2 = X ± i[i.

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