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 .. 311 312 313 314 315 316 < 317 > 318 319 320 321 322 323 .. 486 >> Next

7. $>n(x) = sin x, where satisfies Vk = — tan Vk n;
k = 0.6204, k2 = 2.7943, kn = (2n — 1)2/4 for large n
8. $n(x) = cos^kx, where ^Jk~n satisfies v! = cot Vi;
k1 = 0.7402, k2 = 11.7349, kn = (n — 1)2n2 for large n
9. 0n(x) = sin YYx + ^fk~ncos x, where YY satisfies
(k — 1) sin Vk — 2 Vk cos VI = 0;
Y = 1.7071, k2 = 13.4924, kn = (n — 1)2n2 for large n
10. k0 = 0; ^0(x) = 1 — x
For n = 1, 2, 3,...,$> (x) = sinii x — i cos ii x and k = — i2n, where i satisfies i = tan i.
k1 = —20.1907, k2 = —59.6795, kn = — (2n + 1)2n2/4 for large n
2
12. i(x) = e x 13. i(x) = 1 /x
14. i(x) = e—x 15. i(x) = (1 — x2)—1/2
16. a2X" + (k — k)X = 0, T" + cT' + kT = 0
17. (a) s (x) = ex (b) kn = n2n2, (x) = ex sin nn x; n = 1, 2, 3,...
18. Positive eigenvaluesk = kn, where YV satisfies Vk = § tan3 VkL; corresponding eigenfunctions are </>n(x) = e—2xsm3^>. If L = 2,k0 = 0 is eigenvalue, 00(x) = xe—2x
is eigenfunction; if L = 2 ,k = 0 is not eigenvalue. If L < 2, there are no negative eigenvalues; if L > 2, there is one negative eigenvalue k = — i2, where i is a root of i = 3 tanh 3iL; corresponding eigenfunction is <Y 1 (x) = e—2x sinh 3ix.
19. No real eigenvalues.
20. Only eigenvalue is k = 0; eigenfunction is $(x) = x — 1.
21. (a) 2 sin Vk — Vk cosVk = 0; k1 = 18.2738, k2 = 57.7075
(b) 2 sin^Y! — \fl cosh Vi = 0, i = —k; k_ 1 = —3.6673
732
Answers to Problems
See SSM for 24. (a) Xn = ?4n, where ?n is a root of sin ?L sinh ?L = 0, hence Xn = (nn/L)4;
X1 = 97.409/L4, 0n (x) = sin(nn x/L)
to 24b (b) Xn = ?4n, where ?n is a root of sin ?L cosh ?L — cos ?L sinh ?L = 0;
, sin ? x sinh ? L — sin ? L sinh ? x
X.. = 237.72/L4, ^^
1 sinh ?nL
(x) =
(c) Xn = ?^, where ?n is a root of 1 + cosh ?L cos ?L = 0; X. = 12.362/L4,
[(sin ?nx — sinh ?nx)(cos ?nL + cosh ?nL) + (sin ?nL + sinh ?nL )(cosh ?nx — cos ?nx)]
cos ?nL + cosh ?nL
25. (c) 0n (x) = sin ^JX, x, where Xn satisfies cos ^/X, L — Y^fX, L sin ^/X, L = 0;
X. = 1.1597/L2, X2 = 13.276/L2
Section 11.2, page 639
1. 0n(x) = V2sin(n — 2)nx; n = 1, 2,...
1,3,5 2. 0n(x) = -J2cos(n — 1 )nx; n = 1, 2,...
3. 00(x) = 1, 0n(x) = \/2cosnnx; n = 1, 2,...
-?/2 cos VX, x — — —
4. 0n(x) = (1 + sin2 yX^) 1/2 , where Xn satisfies cos VXn — VXn sinVXn = 0
2^2
5. 0 (x) = \fl ex sinnnx; n = 1, 2,... 6. a = -; n = 1, 2,...
n ^ „ i n (2n — 1)n
„ ^V2(—1)n—1 , „ V '
7,10,14,17,21a 7. an = - rrrr; n = 1, 2—.
(2n — 1) n
2 /2
8. a =--------------{1 — cos[(2n — 1)n/4]}; n = 1, 2,...
n (2n — 1)n
2^2 sin(n — 1 )(n/2)
9. a = 1 2 2-; n = 1, 2,...
n (n — 1 )2n 2
In Problems 10 through 13, an = (1 + sin2 )1/2 and cos yX — ^JX~n sin= 0.
V2(2cos yi; — 1) _
Xnan ’
V2 sin^/2) _
21 bc, 23abc, 24 25bc
10. a \fl sin JT —7=^—; n = : 1,2,... 11.
n VXn an
12. a V/2(1 — cos XX n = 1, 2,... 13.
n X a nn
14. Not self-adjoint 15.
16. Not self-adjoint 17.
18. Self-adjoint
21. (a) If a2 = 0 or b2 = 0, then the corresponding b
25. (a) X1 = n /L ; 01 (x) = sin (n x /L)
(b) X , = (4.4934)2/L2; 01(x) = sm^X x—
(c) X1 = (2n )2/L2; 01 (x) = 1 — cos(2n x / L)
26. X1 = n 2/4L2; 01(x) = : 1 — cos(n x /2 L)
, n = 1, 2,..
Xn an
n = 1, 2,...
Section 11.3, page 651
X (—1)n+1 sinnnx X (—1)n+1 sin(n — 1 )nx
X (n2n2 — 2)nn X [(n — 2 )2n2 — 2] (n — 2 )2n2
n
Answers to Problems
733
See SSM for detailed solutions
to 3, 5, 8
10, 11, 12, 14, 18
19, 22
24, 28a
28bcd
30bcd
30e
cos(2n - 1)n x
i TO
3. y =--------------4 V
4 — \(2n - 1)2n2 - 2](2n - 1)2n2
(2cos Jk~ - 1) cos Jk~x sin(nn/2) sinnnx
4. y - 2> -----------------12 5. y - 8 > ?
i—* \ t\ _ owi dr.2 /F7
=1 Xn&n - 2)(1 + sin2 JK) n=1 (n2^2 - 2)n2^2
6-9. For each problem the solution is
“ c f1
7 =J2 YJL^ ^n (x), cn = f (x)$n (x) dx, * = Xn,
n=1 xn - * Jo
where 0n(x) is given in Problems 1, 2, 3, and 4, respectively, in Section 11.2, and kn is the corresponding eigenvalue. In Problem 8 summation starts at n = 0.
1 1 1/1'
10. a =—, y =—cosnx4--------^ x + csinnx
2 2n2 n2 V 2 /
11. No solution 12. a is arbitrary, y = ccos nx + a/n2
13. a = 0, y = csinnx — (x/2n) sinnx 17. v(x) = a + (b — a)x
2 x
18. v(x) - 1 - 2x
19. u(x, t) - V2
n \n
n x
sin-----
2
4c 77
- ^Y-(n 1)2Z2 _ e-(n-1/2) n t] sin(n - 2)nx,
n-2 (2n - !) n
4V2(-1)n+1
n - 1, 2,...
n (2n - 1)2n
r c
20. u(x, t) -72^ '
n-1
4n - 1
(e-t - e-v) + ane-
(1 + sm2^k)1/2’
42 sin 42.(1 - cos 4kn)
n JTB(1 + sin2 yxy2’ n kn(1 + sin^yxn)1/2’
and kn satisfies cos 44 — 4^^ sin 4^ = 0.
sin (nn /2) _n2n 2t
21. u(x, t) = 8 y — (1 — e n 1) sin nn x
n=1 nn
~ c (e— 1 — e— (n— 1/2)2n21) sin(n — 1 )n x
22. u(x, t) —-----------------rW ---------------------------------- ,
n=1 (n— 1 )2n2 — 1
2V2(2n — 1)n + 4V2(— 1)n
c = ---------------------------------
n (2n — 1)2n2
23. (a)r(x)wt = [p(x)wx]x — q(x)w, w(0, t) = 0, w(1, t) = 0, w(x, 0) = f(x) — v(x)
2 4 ^ e—(2n—1)2n2t sin(2n — 1)nx
24. u(x, t) = x2 — 2x + 1 + -> -----------------------------------------2-------------
n 2n — 1
n=1
25. u(x, t) = — cos nx + e—9n t/4cos(3nx/2)
Previous << 1 .. 311 312 313 314 315 316 < 317 > 318 319 320 321 322 323 .. 486 >> Next