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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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is w = 3 : *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 Al)g = pAl g so p0Ai = 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 i 2(Al + x)g and acts
upward. The resultant force must be equal to the mass of
3
the block times the acceleration, namely pi x". Hence
p , . p|
3 * 2_
pi 3g - Pol 2(Al + x)g
the block is Pi ' V' + poi gx harmonic motion with frequency (p0g/pl ) 1/2 period 2p(pl /p0g) 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 . V 27
u(t) = e
-t/8
(c1cos-
-t + c2sin-
-1. u(0) = 0 ^ c = 0
, V 27
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 ^ ^ + 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
0.25u" + 0.25u' + 16u = 4cos2t where u is measured in ft and t in sec. To determine the steady state response we need only compute a particular solution of the nonhomogeneous D.E. since the solutions of the homogeneous D.E. decay to zero as t ^ . We assume u(t) = Acos2t + Bsin2t, and substitute in the D.E.:
- Acos2t - Bsin2t + (1/2)(-Asin2t + Bcos2t) + 16(Acos2t +
Bsin2t) = 4cos2t. Hence 15A + (1/2)B = 4 and
-(1/2)A + 15B = 0, from which we obtain A = 240/901 and
B = 8/901. The steady state response is
u(t) = (24 0cos2t + 8sin2t)/901.
62
Section 3. 9
11b. In order to determine the value of m that maximizes the steady state response, we note that the present problem has exactly the form of the problem considered in the text. Referring to Eqs.(8) and (9), the response is a
maximum when A is a minimum since F0 is constant. A, as
given in Eq.(10), will be a minimum when
2/2 2\ 2 2 2 2-/
f(m) = m (w0 - w ) + g w , where w0 = k/m, is a
minimum. We calculate df/dm and set this quantity equal
to zero to obtain m = k/w . We verify that this value of
m gives a minimum of f(m) by the second derivative test.
For this problem k = 16 lb/ft and w = 2 rad/sec so the value of m that maximizes the response of the system is m = 4 slugs.
15. We must solve the three I.V.P.: (1)u1 + u1 = F0t,
0 < t < P, u1(0) = u1(0) = 0; (2) u2 + u2 = F0(2p-t),
p < t < 2p, u2(p) = u1(p), u 2(p) = u 1(P); and (3) u3 + u3 = 0, 2p < t, u3(2p) = u2(2p), u3(2p) = u2(2p).
The conditions at p and 2p insure the continuity of u and u' at those points. The general solutions of the D.E. are u1 = b1cost + b2sint + F0t, u2 = c1cost + c2sint + F0(2p-t), and u3 = d1cost + d2sint. The I.C. and matching conditions, in order, give b1 = 0, b2 + F0 = 0,
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