Elementary Differential Equations and Boundary Value Problems  Boyce W.E.
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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 = c1erit + c2er2t + 3x106, 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  4e1000t+ 3) coulombs. Substituting t = .001 sec we obtain
Q(.001) = 106(e4  4e1 + 3) = 1.5468 x 106 coulombs. Since exponentials are to a negative power Q(t) ^ 3x10 coulombs as t ^ •, which is the steady state charge.
Section 3.9
63
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.
u*
w  0.5
W ¦ 1.0
II  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.
64
CHAPTER 4
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.
3 2t2+1 3t2+t

t
2
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
22 c1(2t3) + c2(2t +1) + c3(3t +t) =
2
(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
2t
and e are solutions can be verified by direct substitution. Computing the Wronskian we obtain,
nt t t 2t,
W(e ,e ,e )
t t 2t
e e e 1 1 1
t t t 2t 1 1 2
e e 2 = e
e
2

t t 4 1 1 4
e e 
2
rt
= 6e
2t
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 sint = (1  cos2t)/2 = —(5) + (1/2)cos2t and
10
2
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
rt
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
solutions e , e , e and e . Since
Section 4.2
65
W(et, e t, e2t, e 2t) Ï 0, the four functions form a fundamental set of solutions.
21. Comparing this D.E. to that of Problem 20 we see that
p1(t) = 2 and thus from the results of Problem 20 we have
 f2dt  2t
W = ce = ce .
27. As in Problem 26, let y = v(t)et. Differentiating three times and substituting into the D.E. yields (2t)etv"/ + (3t)etv" = 0. Dividing by (2t)et and letting w = v" we obtain the first order separable
t3 1
equation w = w = (1 + )w. Separating t and w,
t2 t2
integrating, and then solving for w yields w = v" = c1(t2)e t. Integrating this twice then gives
v = c1tet + c2t + c3 so that y = vet = c1t + c2tet + c3et,
which is the complete solution, since it contains the given y^t) and three constants.