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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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17. The mass is 8/32 lb-sec2/ft, and the spring constant is 8/(1/8) = 64 lb/ft. Hence (1/4)u" + gu' + 64u = 0 or u" + 4gu' + 256u = 0, where u is measured in ft, t in sec and the units of g are lb-sec/ft. We look for solutions of the D.E. of the form u = ert and find r2 + 4gr + 256 = 0, so r1,r2 = [-4g ± ‘sj 16g2 - 1024 ]/2.
The system will be overdamped, critically damped or
Section 3.8
59
underdamped as (16g2 - 1024) is > 0, =0, or < 0, respectively. Thus the system is critically damped when g = 8 lb-sec/ft.
r t r t
19. The general solution of the D.E. is u = Ae 1 + Be 2 where r^ ,r2 = [-g ± (g2 - 4km) 1/2]/2m provided
2
g - 4km n 0, and where A and B are determined by the
2
I.C. When the motion is overdamped, g - 4km > 0 and
r t r t
r! > r2. Setting u = 0, we obtain Ae 1 = - Be 2 or
(r -r )t
e 1 2 = - B/A. Since the exponential function is a
monotone function, there is at most one value of t
(when B/A < 0) for which this equation can be satisfied.
Hence u can vanish at most once. If the system is critically damped, the general solution is u(t) = (A + Bt)e-gt/2m. The exponential function is never zero; hence u can vanish only if A + Bt = 0. If B = 0
then u never vanishes; if B n 0 then u vanishes once at
t = - A/B provided A/B < 0.
20. The general solution of Eq.(21) for the case of critical damping is u = (A + Bt)e-gt/2m. The I.C. u(0) = u0 ^
A = u0 and u'(0) = v0 ^ A(-g/2m) + B = v0. Hence
u = [u0 + (vo + gu0/2m)t]e-gt/2m. If v0 = 0, then
u = u0(1 + gt/2m)e gt/2m, which is never zero since g and
m are postive. By L'Hopital's Rule u^0 as t^rc>.
Finally for u0 > 0, we want the condition which will
insure that v = 0 at least once. Since the exponential function is never zero we require
u0 + (v0 + gu0/2m)t = 0 at a positive value of t. This
requires that v0 + gu0/2m n 0 and that
t = -u0(v0 + u0g/2m) -1 > 0. We know that u0 > 0 so we must have v0 + gu0/2m < 0 or v0 < -gu0/2m.
2pg
23. From Problem 21: A = ------------ = Tdg/2m. Substituting the
m (2m)
(1/2) (3)
known values we find g = = 5 lb sec/ft.
.3
24. From Eq.(13) w 0 = ------ so P = 2p/V 2k/3 = p ^ k = 6.
3
Thus u(t) = cicos2t + c2sin2t and u(0) = 2 ^ ci = 2 and
v
u'(0) = v ^ c2 = v/2. Hence u(t) = 2cos2t + —sin2t =
2
60
Section 3 . 8
v2 v2
4 +— cos(2t-g). Thus ] 4 + — 4 V 4
3 and v
±2V5 .
I”
27. First, consider the static case. Let Dl denote the length of the block below the surface of the water. The weight of the block, which is a downward force, «
is w = pi 3i : *g. This is balanced by an equal and opposite buoyancy force B, which is
equal to the weight of the displaced
water. Thus B = (p0l Dl)g = pDl g so p0Dl = pi. Now let x be the displacement of the block from its equilibrium position. We take downward as the positive direction.
In a displaced position the forces acting on the block are its weight, which acts downward and is unchanged, and
2
the buoyancy force which is now p0 l 2(Dl + x)g and acts
upward. The resultant force must be equal to the mass of
3
the block times the acceleration, namely pl x". Hence
3 p l 2, Dl . pl
32
pl 3g - pol 2(Dl + x)g
the block is pl ' V' + pol gx harmonic motion with frequency (p0g/pl ) 1/2 period 2p(pl /pog) 1/2.
The D.E. for the motion of = 0. This gives a simple and natural
29a. The characteristic equation is 4r + r + 8 = 0, so r = (-1+^/127 )/8 and hence
V127 . V127
u(t) = e
-t/8
(c1cos-
-t + c2sin-
-1. u(0) = 0 ^ c1 = 0
, V127
and u (0) = 2 ^ -----------c2 = 2. Thus
u(t)
16
V 127
V 127
e-t/8sin --------1.
8
29c. The phase plot is the spiral shown and the direction of motion is clockwise since the graph starts at (0,2) and u increases initially.
8
8
30c. Using u(t) as found in part(b), show that ku2/2 + m(u') 2/2 = (ka2 + mb2)/2 for all t.
Section 3.9
61
Section 3.9, Page 205
1. We use the trigonometric identities
cos(A ± B) = cosA cosB + sinA sinB to obtain
cos(A + B) - cos(A - B) = -2sinA sinB. If we choose
A + B = 9t and A - B = 7t, then A = 8t and B = t.
Substituting in the formula just derived, we obtain cos9t - cos7t = -2sin8tsint.
5. The mass m = 4/32 = 1/8 lb-sec /ft and the spring
constant k = 4/(1/8) = 32 lb/ft. Since there is no
damping, the I.V.P. is (1/8)u" + 32u = 2cos3t,
u(0) = 1/6, u'(0) = 0 where u is measured in ft and t in
sec.
7a. From the solution to Problem 5, we have m = 1/8, F0 = 2,
2 2 w0 = 256, and w = 9, so Eq.(3) becomes
16
u = c-,cos16t + c2sin16t + ------ cos3t. The I.C.
1 2 247
u(0) = 1/6 ^ c1 + 16/247 = 1/6 and u'(0) = 0 ^ 16c2 = 0,
so the solution is u = (151/1482)cos16t + (16/247)cos3t ft.
7c. Resonance occurs when the frequency w of the forcing
function 4sinwt is the same as the natural frequency w 0 of the system. Since w 0 = 16, the system will resonate when w = 16 rad/sec.
10. The I.V.P. is .25u" + 16u = 8sin8t, u(0) = 3 and
u'(0) = 0. Thus, the particular solution has the form t(Acos8t + Bsin8t) and resonance occurs.
11a. For this problem the mass m = 8/32 lb-sec /ft and the spring constant k = 8/(1/2) = 16 lb/ft, so the D.E. is
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