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# 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
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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
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
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
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)
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