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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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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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