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

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15. For X < 0 there are no eigenvalues, as shown in Problem 11. For X = 0 we have y(x) = c1 + c2x, so y'(0) = c2 = 0 and

204

Section 10.2

y' (n) = c2 = 0, and thus X = 0 is an eigenvalue, with y0(x) = 1 as the eigenfunction. For X > 0 we again have

y(x) = c^in^/Xx + c2co^\fXx, so y'(0) = \fX c1 = 0 and y'(L) = -c2\fX sin^/X L = 0. We know X > 0, in this case, so the eigenvalues are given by sin^X^L = 0 or \[X L = nn. Thus X n = (nn /L) 2 and yn(x) = cos(nnx/L) for n = 1,2,3... .

Section 10.2, Page 555

We look for values of T for which sinh2(x+T) = sinh2x for all x. Expanding the left side of this equation gives sinh2xcosh2T + cosh2xsinh2T = sinh2x, which will be satisfied for all x if we can choose T so that cosh2T = 1 and sinh2T = 0. The only value of T satisfying these two constraints is T = 0. Since T is not positive we conclude that the function sinh2x is not periodic.

We look for values of T for which tann(x+T) = tannx.

Expanding the left side gives

tann(x+T) = (tannx + tannT)/(1-tannxtannT) which is equal

to tannx only for tannT = 0. The only positive solutions of this last equation are T = 1,2,3... and hence tannx is periodic with fundamental period T = 1.

7.

To start, let n = 0, then f(x)

for n

I 0 -1 < x < 0

I 1 0 < x < 1

0 1 < x < 2 I 0 3 < x < 4

; and for n = 2, f(x) = \ .

| 1 2 < x < 3 | 1 4 < x < 5

continuing in this fashion, and drawing a graph, it can be

f(x) =

seen that T

2.

10.

The graph of f(x) is:

We note that f(x) is a straight line with a slope of one in any interval. Thus f(x) has the form x+b in any interval for the correct value of b. Since f(x+2) = f(x), we may set x = -1/2 to obtain f(3/2) = f(-1/2). Noting that 3/2 is

on the interval 1 < x < 2[f(3/2) = 3/2 + b] and that -1/2

is on the interval -1 < x < 0[f(-1/2) = -1/2 + 1], we

conclude that 3/2 + b = -1/2 + 1, or b = -1 for the

interval 1 < x < 2. For the interval 8 < x < 9 we have

f(x+8) = f(x+6) = ... = f(x) by successive applications of the periodicity condition. Thus for x = 1/2 we have f(17/2) = f(1/2) or 17/2 + b = 1/2 so b = -8 on the

interval 8 < x < 9.

1,

By

Section 10.2

205

In Problems 13 through 18 it is often necessary to use integration by parts to evaluate the coefficients, although all the details will not be shown here.

13a. The function represents a sawtooth wave. It is periodic with period 2L.

13b. The Fourier series is of the form

f(x) = a0/2 + (amcosmnx/L + bmsinmnx/L ), where the

m=1

coefficients are computed from Eqs. (13) and (14). Substituting for f(x) in these equations yields

a0 = (l/L)IL (-x)dx = 0 and am = (l/L)IL (-x)cos(mnx/L)dx = 0,

-L -L

m = 1,2... (these can be shown by direct integration, or using the fact that Jag(x)dx = 0 when g(x) is an odd

function). Finally, bm = (l/L)JL (-x)sin(mnx/L)dx

L

L

(x/mn)cos(mnx/L)

- (l/mn) I L cos(mnx/L)dx

-L -L

L

= 2L(-1) m/mn

= (2Lcosmn)/mn - (L/m2n 2)sin(mnx/L)

-L

Substituting these terms in the above Fourier series for f(x) yields the desired answer.

15a. See the next page.

15b. In this case f(x) is periodic of period 2ï and thus

L = ï in Eqs. (9), (13,) and (14). The constant a0 is

found to be a0 = (1/ï)I°xdx = -ï/2 since f(x) is zero on

-ï

the interval [0,ï]. Likewise

an = (1/ï)I°xcosnxdx = [1 - (-1)ï]/ï2ï, using integration

-ï

by parts and recalling that cos^ = (-1) n. Thus an = 0 for n even and an = 2/ï2ï for n odd, which may be written as a2n-1 = 2/(2n-1) 2ï since 2n-1 is always an odd number.

In a similar fashion bn = (1/ï) I °xsinnxdx = (-1) n+1/n and

thus the desired solution is obtained. Notice that in this case both cosine and sine terms appear in the Fourier series for the given f(x).

206

Section 10.2

15a.

21a.

21b. a

1 J2—dx = —

2 J-2 2 12

n = 1 J-2-

12 nnx

4 a0 2

—, so — = — and

3 2 3

2x

-cos-------dx

2 -2 2 2

2

1 2x nnx 8x nnx

— [------sin-------- + --------cos--------

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