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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.
Download (direct link): elementarydifferentialequations2001.pdf
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(x — 2)—2 cos(2 ln |x — 2|) + c2(x — 2)—2 sin(2ln |x — 2|)
1e 2
|x| 1/2cos(tv/t5ln |x|) + c2|x| 1/2 sin(tv/t5ln |x|)
x + c2 x
13.
1/2
y
= 2x3/2 - x—1
(2lnx) — x 1/2sin(2lnx) 15. y — 2x2 — 7x2ln |x|
x 1 cos(2 ln x)
0
1
เ < 1 and โ > 0
เ < 1 and โ > 0, or เ — 1 and โ > 0
เ > 1 and โ > 0
เ > 1 and โ > 0, or เ — 1 and โ > 0
เ — 1 and โ > 0
17.
19.
เ < 1 ส — 2
c,x 1 + c2 x2
27. y —
28. y —
29. y — 31. x >
+ c2 x 5 + x
25. y — c1 x2 + c2x2 lnx + 4 lnx + 4
x + c2x2 + 3x2 ln x + ln x + 2 cos(2lnx) + c2 sin(2lnx) + 1 sin(ln x) x— 3/2 cos(3 lnx) + c2x— 3/2 sin(2 lnx)
c1 — k1, c2 — k2; x < 0 : c1( — 1)r1 — k1, c2 — k2
700
Answers to Problems
See SSM for detailed solutions to 2
Section 5.6, page 271
1. r(2r — 1) = 0; an = —
(n + r )[2(n + r) — 1]’
r, = 2, r2 = 0
x1/2
x2
1 — — +
2-5 2 • 4 • 5 • 9 2-4-6-5-9-13
+
+
(—1)nx2 n
2nn!5 • 9 • 13 • • • (4n + 1)
+
x2
K2(x) = 1 — 7-ร +
2-3 2-4-3-7 2-4-6-3-7- 11
+
+
(— 1)nx2 n
2nn!3 • 7 • 11 • • • (4n — 1)
+
2. r2 — 9 = 0;
n—2
(n + r)2 —
rl = 3 , r2 = 3
ำ,(x) = x
x1/3
1 I x\2
1------------+D (ฟ) +
1
าฟ( 2)'
+
1!(1 + f) V2/ ' 2!(1 + f)(2 + 3) V2
(—1)“ / x\2m
+
m!(1 + 3)(2 + 3) • • • (m + 1)
(2)
+
ำ2(ี) = X
_ x—1/3
1 / x \ 2
1--------IT) ( 2) +
1
d( D4
+
1! (1 — f) V2! ' 2! (1 — -)(2 — ณ) V2,
(—1)“ / x \2m
+
“! (1 — 3)(2 — 3) • • • (“ — f)
(2)
+
Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....
a
3. r(r — 1) = 0; an = —
n—1
ำ,(x) = x
(n + r)(n + r — 1) ’ 1
(—1)n
x x2
1 — -------- + --------- + • • • +
1!2! 2!3! n! (n + 1)!
r1 = 1 , r2 = 0
xn +
a
n—1
4. r2 = 0; an = 9,
(n + r)2
r1 = r2 = 0
y, (x) = 1 + —2 +
x2
(l!)2 (2!)2
+
xn
‘ -----๎
(n!)2
an—2
+
5. r(3r — 1) = 0; a =- ,
n (n + r)[3(n + r) — 1]
r = 1 r = 0
i1 3, 2 U
1/3
y,(x) = x
- — _L(?\+ _^ x:\ +.
1!7 \ 2 2!7 • 13 \ 2 /
(—1)m
m!7 • 13 • • • (6m + 1) \ 2
+
y2(x) 1 1!5 \ 2 ) + 2!5 • 11 \ 2
xr +
1 x2
x2
-ณ +• • • +
(—1)m
m!5 • 11 • • • (6m — 1)
Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....
a
n2
4
6
4
6
9
Answers to Problems
701
See SSM for detailed solutions to 11
14
15
16ab
6. r2 - 2 = 0; a =------------a"-า---; r = 42, r = —2
n (n + r)2 - 2 1 2
71 (x) = x^
72(x) = x-^
x x2
1(1 + 2V2) 2! (1 + 2v/2')(2 + 2^2)
I (-1)n M I '
n!(1 + 2\/2)(2 + 2^2) •••(n + 2V2) .
x x2
1(1 - 2V2) 2! (1 - 2\/2)(2 - 2^2)
I (-1)n n I "
n! (1 - 2\f2)(2 - 2^2) •••(n - 2^2) .
7. r2 = 0; (n + r)an = an-1; r1 = r2 = 0
x2 x3 xn
71(x) = 1 + x + - + - + ••• + - + ••• = ex
8. 2r2 + r — 1 = 0; (2n + 2r — 1)(n + r + 1)an + 2an-2 = 0; r1 = 1, r2 = —1
_ 1/2 I , x2 : x4 : (-1)mx2m
y1 (x) = x1/2 1-------1-------------••• I-------------------------1 •
^ ‘ 7 2!7 • 11 m!7 • 11 ••• (4m + 3)
(x4 (— 1)mx2 m
1 - x2 + ---------------+--------------(-^x----+ ฆ
2!5 m!5 • 9 ••• (4m - 3)
9. r2 — 4r + 3 = 0; (n + r — 3)(n + r — 1)an — (n -— r — 2)an-1 = 0; r1 = 3, r2 = 1
3 2 x2 2 xn
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