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

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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 + a1x + a2x + ...) = a0x + a^ + 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

\ /-³ 2/~i 4 / . . x / 2 3 4

...) - (1 - x/2! + x/4! - ...)(aox + a1x + a2x + a3x +

5 2

a4x + ...) = 0. Collecting terms, (2a1 + 2a1 - a1)x +

(6a2 + 3a2 - ao/6 - a2 + ao/2)x + (12a3 + 4a3 - 2aj/6 - a3+

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

5 2 3

a0/24)x + ... = 0. Simplifying, 3a1x + (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

to

Ó2 (x) = ay1 (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 + ay1x — x + c2 + 2c3x + 3c4x + ..., and

-1 -2 -3

y2 = ay1 lnx + 2ay1x — ay1x + 2x + 2c3 + 3c4x + ....

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

including lnx will appear as

a[x2y1 + (sinx)y'1 — (cosx)y1], which is zero since y1 is a solution. For the remainder of the terms, use y1 = x — x3/24 + x5/720 a 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) =---y1 (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) = y, (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

22 (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

(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 2 1 3 we find that lnx = (x-1) - —(x-1) + —(x-1) - ... .

23

Thus

1 1 2 -1 1

(x-1)/lnx = [1 - - (x-1) + - (x-1) 2 -...] 1 = 1 + — (x-1)+...

2 3 2

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

2 „ 11 (x-1) 2y" + (x-1)[ + -(x-1) + ...]y' + [(x-1) +

24

12

— (x-1) + ...]y = 0. Thus p0 = 1/2, q0 = 0 and the

indicial equation is r(r-1) + r/2 = 0. Hence r = 1/2 and

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

in a series solution corresponding to r = 1/2, it is

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