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

ISBN 0-471-31999-6

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order to assert that x = 0 is a regular singular point we

2

must demonstrate that xp(x) and xq(x), with xp(x) = 1 at 2

x = 0 and x q(x) = -1 at x = 0, have convergent power series (are analytic) about x = 0. We know that cosx is analytic so we need only consider (sinx)/x. Now

sinx = (-1) nx2n+1/(2n+1)! for -

n=0

(sinx)/x — (-1) nx2n/(2n+1)! and hence is analytic.

n—0

Thus we may conclude that x — 0 is a regular singular point.

17b. From part a) it follows that the indicial equation is

2

r(r-1) + r - 1 = r - 1 = 0 and the roots are r-, = 1,

r2 — -1.

17c. To find the first few terms of the solution corresponding to r1 — 1, assume that

, 2 . 2 3

y — x(a0 + a^x + a2x + ...) — a0x + a^x + a2x + ... . Substituting this series for y in the D.E. and expanding sinx and cosx about x — 0 yields

2 2 3

x (2a-, + 6a2x + 12a3x + 20a4x + ...) +

(x - x3/3! + x5/5! - ...)(a0 + 2a1x + 3a2x2 + 4a3x3 + 5a4x4+

oo

Section 5.7

95

\ /-i 2/~i 4 / . . x / 2 3 4

...) - (1 - x/2! + x/4! - ...)(aoX + a^x + a2x + a3x +

5 2

a4x + ...) = 0. Collecting terms, (2a1 + 2a1 - a1)x + (6a2 + 3a2 - ao/6 - a2 + ao/2)x + (12a3 + 4a3 - 2ai/6 - a3+

a1/2)x4 + (20a4 + 5a4 - 3a2/6 + a0/120 - a4 + a2/2 -

5 2 3

a0/24)x + ... = 0. Simplifying, 3aj_x + (8a2 + a0/3)x + (15a3 + a1/6)x4 + (24a4 - a0/30)x5 + ... = 0. Thus, a1 = 0, a2 = -a0/4!, a3 = 0, a4 = a0/6!,... . Hence

y1(x) = x - x3/4! + x5/6! + ... where we have set a0 = 1.

From Eq. (24) the second solution has the form

y2 (x) = ayx (x)lnx + x-1(1+^cnxn)

n=1

1 2 3

= ayx (x)lnx + — + cx + c2x + c3x + c4x + ..., so

x

-1 -2 2

y2 = ay^nx + ay^ — x + c2 + 2c3x + 3c4x + ..., and

-1 -2 -3

y2 = ay! lnx + 2ay!x — ay!x + 2x + 2c3 + 3c4x + ....

When these are substituted in the given D.E. the terms

including lnx will appear as

a[x2y1 + (sinx)y1 — (cosx)y^, which is zero since y! is a solution. For the remainder of the terms, use y1 = x — x3/24 + x5/720 ? shown earlier to obtain

23

—c1 + (2/3+2a)x + (3c3+c1/2)x + (4/45+c2/3+8c4)x +...= 0.

These yield c1 = 0, a = —1/3, c3 = 0, and

c4 = —c2/24 — 1/90. We may take c2 = 0, since this term

will simply generate y1(x) over again. Thus

1 — 1 1 3

y2(x) = y-, (x)lnx + x — —x . If a computer algebra

2 3 1 90

system is used, then additional terms in each series may be obtained without much additional effort. The next terms, in each case, are shown here:

3 5 . - 7

x x 4 3x

y-, (x) = x — — + — ------- + ... and

1 24 720 1451520

1 1 x4 41x6

y2 (x) = y1 (x)lnx + — [1— + ---------- — ...].

2 3 1 x 90 120960

y1 = x — x3/24 + x5/720 and the cosx and sinx series as

18. We first write the D.E. in the standard form as given for Theorem 5.7.1 except that we are expanding in powers of (x-1) rather than powers of x:

96

Section 5.7

9 .. »9

(x-1) y + (x-1)[(x-1)/2lnx]y + [(x-1) /lnx]y = 0. since ln1 = 0, x = 1 is a singular point. To show it is a regular singular point of this D.E. we must show that (x-1)/lnx is analytic at x = 1; it will then follow that 9

(x-1) /lnx = (x-1)[(x-1)/lnx] is also analytic at x = 1. If we expand lnx in a Taylor series about x = 1

1 9 1 3

we find that lnx = (x-1) - —(x-1) + —(x-1) - ... .

9 3

Thus

1 1 9 -l 1

(x-1)/lnx = [1 - - (x-1) + - (x-1) 9 -...] 1 = 1 + — (x-1)+...

9 3 9

has a power series expansion about x = 1, and hence is

analytic. We can use the above result to obtain the

indicial equation at x = 1. We have

9 „ 11 (x-1) y + (x-1)[ + -(x-1) + ...]y' + [(x-1) +

9 4

19

— (x-1) + ...]y = 0. Thus p0 = 1/9, q0 = 0 and the

indicial equation is r(r-1) + r/9 = 0. Hence r = 1/9 and

r = 0. In order to find the first three non-zero terms

in a series solution corresponding to r = 1/9, it is

better to keep the differential equation in its original

form and to substitute the above power series for lnx:

1 9 1 3 1 4 1 ,

[(x-1) - - (x-1) 9 + - (x-1) 3 - - (x-1) 4 + ...]y" + -y' + y = 0.

9 3 4 9

Next we substitute y = a0(x-1) 1/9 + a1(x-1) 3/9 + a9(x-1) 5/9

+ ... and collect coefficients of like powers of (x-1) which are then set equal to zero. This requires some algebra before we find that 6a1/4 + 9a0/8 = 0 and

5a9 + 5a1/8 - a0/19 = 0. These equations yield

a1 = -3a0/4 and a9 = 53a0/480. With a0 = 1 we obtain the

solution

1/9 3 3/9 53 5/9

y-,(x) = (x-1) - — (x-1) +-----------------(x-1) + ... . Since

1 4 480

the radius of convergence of the series for ln x is 1, we

would expect p = 1.

90a. If we write the D.E. in the standard form as given in

Theorem 5.7.1 we obtain x9y" + x[a/x]y' + [p/x]y = 0 where

xp(x) = a/x and x9q(x) = P/x. Neither of these terms are

analytic at x = 0 so x = 0 is an irregular singular point.

Section 5.7

97

2 0b. Substituting y = xr^i anxn in x3y" + axy' + Py = 0 gives

n=0

? ? ?

X ' , x x n+r+1 ^ X ' , x n+r n X’ n+r ~

> (n+r)(n+r-1)anx + a > (n+^a^ + p > a^ = 0.

n=0 n=0 n=0

Shifting the index in the first series and collecting coefficients of common powers of x we obtain (ar + P)a0xr

?

+ (n+r-1)(n+r-2)an-1 + [a(n+r) + p]anxn+r = 0. Thus

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