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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
Download (direct link): elementarydifferentialequat2001.pdf
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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 (2-t)etv"/ + (3-t)etv" = 0. Dividing by (2-t)et and letting w = v" we obtain the first order separable
t-3 1
equation w = w = (-1 + -------------)w. Separating t and w,
t-2 t-2
integrating, and then solving for w yields w = v" = c1(t-2)e t. Integrating this twice then gives
v = c1te-t + c2t + c3 so that y = vet = c1t + c2tet + c3et,
which is the complete solution, since it contains the given y1(t) and three constants.
Section 4.2, Page 219
2. If -1 + iy^3 = Re10, then R = [(-1) 2 + (\/~3) 2] 1/2 = 2. The
angle 0 is given by Rcos0 = 2cos0 = -1 and
Rsin0 = 2sin0 = \[3 . Hence cos0 = -1/2 and
sin0 = \J~3 /2 which has the solution 0 = 2p/3. The angle
0 is only determined up to an additive integer multiple of ± 2p.
i0
8. Writing (1-i) in the form Re , we obtain
(1-i) = ,\/~2ei("p/4+2mp)where m is any integer. Hence,
(1-i) 1/2 = [21/2ei(-p/4+2mp)] 1/2 = 21/4ei(-p/8+mP) We obtain
the two square roots by setting m = 0,1. They are 21/4e-ip/8 and 21/4ei7p/8. Note that any other integer value of m gives one of these two values. Also note that 1-i
-in t ... ... „nr i(7p/4 + 2mp)
could be written as 1-i = y 2e .
rt
12. We look for solutions of the form y = e . Substituting in the D.E., we obtain the characteristic equation
32
r - 3r + 3r - 1 = 0 which has roots r = 1,1,1. Since the roots are repeated, the general solution is
t . t .2 t
y = c1e + c2te + c3t e .
66
Section 4.2
15.
23.
27.
29.
30.
31.
rt
We look for solutions of the form y = e . Substituting in the D.E. we obtain the characteristic equation
r6 + 1 = 0. The six roots of -1 are obtained by setting
m = 0,1,2,3,4,5 in (-1) 1/6 = ei(p+2mp)/6. They are
eip/6 = (^T + i)/2, eip/2 = i, ei5p/6 = (-y[T + i)/2,
ei7p/6 = (-j^3 - i)/2, ei3p/2 = -i, and
ei11p/6 = (V"3 - i)/2. Note that there are three pairs of conjugate roots. The general solution is
y = ^^3t/2[c1cos(t/2) + c2sin(t/2)]
+ e^^3t/2[c3cos(t/2) + c4sin(t/2)] + c5cost + c6sint.
32
The characteristic equation is r -5r + 3r + 1 = 0.
Using the procedure suggested following Eq. (12) we try r = 1 as a root and find that indeed it is. Factoring
2
out (r—1) we are then left with r - 4r - 1 = 0, which
has the roots 2 ± .
4 3
The characteristic equation in this case is 12r + 31r +
2
75r + 37r + 5 = 0. Using an equation solver we find
11
r =----,----, -1 ± 2i. Thus
4 3
y = c1e-t/4 + c2e-t/3 + e-t (c3cos2t +c4sin2t). As in
Problem 23, it is possible to find the first two of these roots without using an equation solver.
The characteristic equation is r3 + r = 0 and hence r = 0, +i, -i are the roots and the general solution is y(t) = c1 + c2cost + c3sint. y(0) = 0 implies
c1 + c2 = 0, y'(0) = 1 implies c3 = 1 and y"(0) = 2
implies -c2 = 2. Use this last equation in the first to
find c1 = 2 and thus y(t) = 2 - 2cost + sint, which
continues to oscillate as t —— •.
The general solution is given by Eq. (21).
The general solution would normally be written y(t) = c1 + c2t + c3e2t + c4te2t. However, in order to evaluate the c's when the initial conditions are given at t = 1, it is advantageous to rewrite
/4- \ /4- \ 4- 2(t-1) . 2(t-1)
y(t) as y(t) = c1 + c2t + c5e + c6(t-1)e .
Section 4.3 67
34. The characteristic equation is 4r3 + r + 5 = 0, which has 1
roots -1, — ± i. Thus 2
-1 t/2
y(t) = c1e + e (c2cost + c3sint),
y'(t) = -c1e t + et/2[(c2/2 + c3)cost + (-c2 + c3/2)sint]
and
y''(t) = c1e t + et/2[(-3c2/4 + c3)cost + (-c2 - 3c3/4)sint]. The I.C. then yield c1 + c2 = 2, -c1 + c2/2 + c3 = 1 and c1 - 3c2/4 + c3 = -1. Solving these last three equations give c1 = 2/13, c2 = 24/13 and c3 = 3/13.
37. The approach developed in this section for solving the
D.E. would normally yield y(t) = c1cost + c2sint + c5et +
c6e-t as the solution. Now use the definition of coshx
and sinht to yield the desired result. It is convenient
to use cosht and sinht rather than et and e-t because the I.C. are given at t = 0. Since cosht and sinht and all
of their derivatives are either 0 or 1 at t = 0, the
algebra in satisfying the I.C. is simplified.
-lodt
38a. Since p^t) = 0, W = ce J = c.
39a. As in Section 3.8, the force that the spring designated by k1 exerts on mass m1 is -3u1. By an analysis similar to that shown in Section 3.8, the middle spring exerts a force of -2(u1-u2) on mass m1 and a force of -2(u2-u1) on mass m2. In all cases the positive direction is taken in the direction shown in Figure 4.2.4.
39c. From Eq.(i) we have u1(0) = 2u2(0) - 5u1(0) = -1 and u/1"(0) = 2u"2(0) - 5u1(0) = 0. From Prob.3 9b we have ui = c1cost + c2sint + c3cos/61 + c4sin/6t. Thus c1+c3 = 1, c2+\/"6c4 = 0, -c1-6c3 = -1 and -c2-6V"6c4 = 0, which yield c1 = 1 and c2 = c3 = c4 = 0, so that u1 = cost. The first of Eqs.(i) then gives u2.
Section 4.3, Page 224
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