Books
in black and white
Main menu
Share a book About us Home
Books
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
Ads

Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
Previous << 1 .. 298 299 300 301 302 303 < 304 > 305 306 307 308 309 310 .. 486 >> Next

1 - x
( x2 xn — (x3 x4 xn —
20. y — ^0 ( 1 + x + ——+ ••• +----------- + •••) + 2 I ——|—— + ••• +-------- + ••• I
7 0\ 2! n! / \3! 4! n! I
— ae + 2( ex — 1 — x — y) — ce* — 2 — 2x — x2
x2 x4 x6 (—1)nx2n
21. y — a0 I 1--t—I—2-----------3--+ ••• +----~n—: + '
0 \ 2 222! 233! 2nn!
x2 x3 x4 x5 -
+ ( x + — — — —--------------+------+ •
2 3 2 • 4 3 • 5
2/2 ( x2 x3 x4 x5
— a0 e—x /2 + x +----------------------1---------+ •
0 y 2 3 2 • 4 3 • 5
23. 1,1 — 3x2, 1 — 10x2 + fx4; x, x — §x3, x — 14x3 + 21 x5
24. (a) 1, x, (3x2 — 1)/2, (Sx3 — 3x)/2, (35x4 — 30x2 + 3)/8, (63x5 — 70x3 + 15x)/8
(c) P1, 0; P2, ±0.57735; P3, 0, ±0.77460; P4, ±0.33998, ±0.86114;
P5, 0, ±0.53847, ±0.90618
Answers to Problems
699
Section 5.4, page 259
17, 21, 21a, 22, 25, 31
See SSM for 1. x = 0, regular 2. x = 0, regular; x = 1, irregular
detailed solutions 3. x = 0, irregular; x = 1, regular 4. x = 0, irregular; x =±1, regular
to 1, 5 5. x = 1, regular; x = — 1, irregular 6. x = 0, regular
7. x = — 3, regular 8. x = 0, 1, regular; x = 1, irregular
9. x = 1, regular; x = — 2, irregular 10. x = 0, 3, regular
12, 17 11. x = 1, — 2, regular 12. x = 0, regular
13. x = 0, irregular 14. x = 0, regular
15. x = 0, regular 16. x = 0, ±nn, regular
19, 21, 23 17. 19. x = 0, ±nn, regular ( x2 x4 \ 7= ^ 2 • 5 + 2 • 4 • 5 • 9 • ) 18. x = 0, irregular; x = ±nn, regular
22. Irregular singular point 23. Regular singular point
25 24. Regular singular point 25. Irregular singular point
26. Irregular singular point Section 5.5, page 265 27. Irregular singular point
2, 4, 9, 13, 16 1. 3. 4. 12 y = c1 x + c2 x y = c1 x2 + c2 x2 ln | x| y = c1 x—1 cos(2 ln |x|) + c2x— 1 sin(2 ln 2. | x| ) y = cjx + 1|—1/2 + c2|x + 1|— 3/2
5. 7. y = c1 x + c2 x ln | x| y = d|x |(— 5+'f2§)/2 + c2|x |(—5—^9)/2 6. y = c1(x— 1)—3 + c2(x— 1)—4
9. y
y=
y=
2x
10. y=
11. y=
12. y= 14. y = 16. y = 18. ? >
20. a >
21. (a)
(b)
(c)
(d)
(e)
24. y = 26. y = c1
|x|3/2 cos( 1v',3ln |x|) + c2|x|3/2 sin(2\/,3ln |x|) x3 + c2x3 ln |x|
(x — 2)—2 cos(2 ln |x — 2|) + c2(x — 2)—2 sin(2ln |x — 2|)
^ 2
|x| 1/2cos(4vT5ln |x|) + c2|x| 1/2 sin( 1vT5ln |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
a < 1 and ? > 0
a < 1 and ? > 0, or a = 1 and ? > 0
a > 1 and ? > 0
a > 1 and ? > 0, or a = 1 and ? > 0
a = 1 and ? > 0
17.
19.
a < 1
X = 2
c,x 1 + c2 x2
27. y =
28. y =
29. y = 31. x >
+ c2 x 5 + 12 x
25. y = c1 x2 + c2x2 lnx + 4 lnx + 4
x + c2x2 + Sx2 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 = *1, c2 = k2; x < 0 : c1( —1)6 = k1, c2 = k2
700
Answers to Problems
See SSM for detailed solutions to 2
Section 5.6, page 271
1. r(2r - 1) = o; an = -
(n + r)[2(n + r) - 1]’
r1 = 2, r2 = o
x1/2
y1 (x) = x
x2
1 - — +
2-5 2-4-5-9 2-4-6-5-9-13
+
+
(-1)nx2 n
2nn!5 • 9 • 13 • • • (4n + 1)
+
x2
y2(x) = 1 - 7-T +
2-3 2-4-3-7 2-4-6-3-7- 11
+
+
(- 1)nx2 n
2nn!3 • 7 • 11 • • • (4n - 1)
+ •
2. r2 - 9 = o;
an-2
(n + r)2 -
r1 = 3 ’ r2 = 3
y1(x) = x
x1/3
1 I x\2
1------------+T) (?) +
1
Tt( 2)'
+
1!(1 + 1 ) V2Z ' 2!(1 + 1 )(2 + 3) V2
(-1)“ / x \2m
+ •
m! (1 + 3)(2 + 3) ••• (m + 1 )
(2)
+ •
y2(x) = x
_ x-1/3
1
tA 2)4
+
1! (1 - 1 ) V2! ' 2! (1 - 1 )(2 - 3) V2,
(-1)m / x\ 2m
+ •
m! (1 - 3)(2 - 3) •••(m -1 )
(2)
+ •
Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....
a
3. r(r - 1) = o; an = -
n-1
y1(x) = x
(n + r)(n + r - 1) ’ 1
(-1)n
x x2
1 — + + ••• +
1!2! 2!3! n! (n + 1)!
r1 = 1 , r2 = o
xn + •••
a
n-1
4. r2 = o; an = 2’
(n + r)2
r1 = r2 = o
y1 (x) = 1 + —2 +
x2
(1!)2 (2!)2
+ •
xn ?
(n!)2
an-2
+
5. r(3r - 1) = o; a =--------------------------- ,
n (n + r)[3(n + r) - 1]
r = 1 r = o
i1 3 ’ 2 U
1/3
y(x) = x
1 - _L (?\ + _^ xS +.
1!7 \ 2 2I7.1M2
(- 1)n
m!7 • 13 • • • (6m + 1) \ 2
+
y2(x) 1 1!5 \ 2 ) + 2!5 • 11 \ 2
x2 +
1 x2
x2
-I +••• +
(-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 =-------------^-----; r1 = 42, r2 = -42
n (n + r)2 - 2 1 2
y1 (x) = x^
72(x) = x-'2
x x2
1(1 + 242) 2! (1 + 242)(2 + 242)
I (-1)n n I '
n!(1 + 242)(2 + 242) •• • (n + 2V2) .
x x2
1(1 - 242) 2! (1 - 242)(2 - 2^2)
I (-1)n f I "
n! (1 - 242)(2 - 2^2) • • • (n - 242) .
7. r2 = 0; (n + r)an = an-1; r1 = r2 = 0
x2 x3 xn
y1 (x) = 1 + x + 2! + 37 + • • • + n! + • • • = ^
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 I 1-----------\------------• • • +------------------------------+ •
1 7 2!7 11 m!7 11 (4m \ 3)
x4 ( 1 )mx2m
y2(x) = x—1 1 - x2 + -------------------------+---------------(_^L---+ .
J2 y 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
Previous << 1 .. 298 299 300 301 302 303 < 304 > 305 306 307 308 309 310 .. 486 >> Next