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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) = (240cos2t + 8sin2t)/901.
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,
-b1 + pF0 = -c1 + pF0, -b2 + F0 = -c2 - F0, c1 = d1, and c2 - F 0 = d2. Solving these equations we obtain t - sint , 0 < t < p
(2p - t) - 3sint , p < t < 2p
u = F0
4sint , 2p < t.
16. The I.V.P. is Q" + 5x103 Q' + 4x106 Q = 12, Q(0) = 0, and Q'(0) = 0. The particular solution is of the form Q = A, so that upon substitution into the D.E. we obtain 4x106A = 12 or A = 3x10 6. The general solution of the D.E. is Q = c1er1t + c2er2t + 3x10-6, where r1 and r2 satisfy r2 + 5x103r + 4x106 = 0 and thus are r1 = -1000 and r2 = -4000. The I.C. yield c1 = -4x10 6 and c2 = 10 6 and
thus Q = 10-6(e-4000t - 4e-1000t+ 3) coulombs. Substituting t = .001 sec we obtain
Q(.001) = 10-6(e-4 - 4e-1 + 3) = 1.5468 x 10-6 coulombs. Since exponentials are to a negative power Q(t) —— 3x10 coulombs as t — •, which is the steady state charge.
22. The amplitude of the steady state response is seven or
eight times the amplitude (3) of the forcing term. This large an increase is due to the fact that the forcing function has the same frequency as the natural frequency, w0, of the system.
There also appears to be a phase lag of approximately 1/4 of a period. That is, the maximum of the response occurs 1/4 of a period after the maximum of the forcing function. Both these results are substantially different than those of either Problems 21 or 23.
w - 0.5
» « 1.0
« - 2.0
• - 1.5
From viewing the above graphs, it appears that the system exhibits a beat near w = 1.5, while the pattern for w = 1.0 is more irregular. However, the system exhibits the resonance characteristic of the linear system for w near 1, as the amplitude of the response is the largest here.
Section 4.1, Page 212
2. Writing the equation in standard form, we obtain
y"' + [(sint)/t]y" + (3/t)y = cost/t. The functions p1(t) = sint/t, p3(t) = 3/t and g(t) = cost/t have discontinuities at t = 0. Hence Theorem 4.1.1 guarantees
that a solution exists for t < 0 anc for t > 0.
2t-3 2t2+1 3t2+t
8. We have W(f1, f2, f3) = 2 4t 6t+1 = 0 for all t
0 4 6
Thus by the extension of Theorem 3.3.1 the given functions are linearly dependent. Thus
2 2 c1(2t-3) + c2(2t +1) + c3(3t +t) =
(2c2+3c3)t + (2c1+c3)t + (-3c1+c2) = 0 when
(2c2 + 3c3) = 0, 2c1 + c3 = 0 and -3c1 + c2 = 0. Thus c1 = 1 c2 = 3 and c3 = -2.
13. That et, e t
and e are solutions can be verified by direct substitution. Computing the Wronskian we obtain,
, t -t -2t,
W(e ,e ,e )
t -t -2t
e e e 1 1 1
t e -t -e -2e-2t -2t = e 1 -1 -2
t e -t e 4e-2t 1 1 4
17. To show that the given Wronskian is zero, it is helpful,
in evaluating the Wronskian, to note that (sin2t)' =
2sintcost = sin2t. This result can be obtained directly
21 since sin t = (1 - cos2t)/2 = —(5) + (-1/2)cos2t and
hence sin t is a linear combination of 5 and cos2t. Thus the functions are linearly dependent and their Wronskian is zero.
19c. If we let L[y] = ylv - 5y" + 4y and if we use the result
rt 4 2 rt
of Problem 19b, we have L[e ] = (r - 5r + 4)e . Thus
e will be a solution of the D.E. provided 22
(r -4)(r -1) = 0. Solving for r, we obtain the four
t -t 2t , -2t 0.
solutions e , e , e and e . Since
W(et, e t, e2t, e 2t) n 0, the four functions form a fundamental set of solutions.