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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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25
?-1{H(s)} = (4/25)(5t - 4 + 4e-t/2cost - 3e-t/2sint),
which yields the desired solution. The graph of the forcing function is a ramp (f(t) = t) for 0 < t < p/2 and a constant (f(t) = p/2) for t > p/2. The solution will be a damped sinusoid oscillating about the "ramp" (20t-16)/25 for 0 < t < p/2 and oscillating about 2p/5 for t > p/2.
10. Note that g(t) = sint - up(t)sint = sint + up(t)sin(t-p). Proceeding as in Problem 8 we find
- ps 1
Y(s) = (1+e )------------------. The correct partial
22 (s +1)(s +s+5/4)
as+b cs+d
fraction expansion of the quotient is -------------- + ,
s2+1 s2+s+5/4
where
a+c = 0, a+b+d = 0, (5/4)a+b+c = 0 and (5/4)b+d = 1 by
equating coefficients. Solving for the constants yields the desired solution.
16b. Taking the Laplace transform of the D.E. we obtain
U(s2+s/4+1) = k(e 3s/2-e 5s/2)/s, since the I.C. are zero. Solving for U and using partial fractions yields
-3s/2 -5s/2 1 s +1/4
U(s) = k(e -e )(-). Thus, if
s s2 +s/4 +1
1 s +1/4
H(s) = (---------------), then
s s2+s/4+1
, , -t/8, 3\fl _ /7 . 3^7^, n
h(t) = 1 - e (cos 1+ sin 1) and
8 21 8
u(t) = ku3/2(t)h(t-3/2) - ku5/2(t)h(t-5/2).
16c. In all cases the plot will be zero for 0 < t < 3/2. For 3/2 < t < 5/2 the plot will be the system response
114
Section 6.4
(damped sinusoid) to a step input of magnitude k. For t > 5/2, the plot will be the system response to the
I.C. u(5-/2), u'(5_/2) with no forcing function. The graph shown is for k = 2.
Varying k will just affect the amplitude. Note that the amplitude never reaches 2, which would be the steady state response for the step input 2u3/2(t).
Note also that the solution and its derivative are continuous at t = 5/2.
19a. The graph on 0 ? t < 6p will depend on how large n is.
For instance, if n = 2 then
1 0 ? t < p, 2p ? t < 6p
f(t) =
For
-1 -p ? t < 2p
n > 6, f(t) =
1 0 ? t < P, 2p ? t < 3P, 4P ? t < 5P
-1 p ? t < 2 p, 3p ? t <4p, 5p t < 6 p
19b. Taking the Laplace transform of the D.E. and using the I.C.
have Y(s) =
1 [1 + 2 ? (-1) VPks],
/ 2 s(s +1)
-pks
since
?{upk(t)} =
e
Since
k=1
1
1s
s
2
-, we then obtain
s(s +1)
s +1
y(t) = 1 - cost + 2^^ (-1) kupk(t)[1 - cos(t-pk)], using line
k=1
13 in Table 6.2.1.
k
19d. Since cos(t-pk) = (-1) cost, the solution in part b can be written as
y(t) = 1 - cost + 2 ? (-1)
Upk(t) - 2
? (-1)
2k
cost
k=1
k=1
s
k
we
= 1 - cost - 2ncost + 2 (-1) kupk(t) which diverges for n?w.
k=1
Section 6.4
115
20. In this case
1
Y(s) = ----------------
2
s(s +.1S+1) fractions we have 1
H(s) = ----------------
2
[1 + 2 (-1) ke pks]. Using partial
n=1
s(s +.1s+1)
1 s+.1
s 2
s s +.1s+1 1 s+.05
s
.05
2 2 2 2 (s+.05) +b (s+.05) +b
, where
2
b = [1-(.05) ] = .9975. Now let
h(t) = ?-1{H(s)} = 1 - e .°5tcosbt - e .°5tsinbt. Hence,
b
y(t) = h(t) + 2 (-1) kupk(t) h(t-pk), and thus, for t > n the
n=1
solution will be approximated by
±1 - Ae .°5(t np) cos[b(t-np) + 8], and therefore converges as t.
20a. y(t) for n = 30
y(t) for n = 31
20b. From the graph of part a, A @ 12.5 and the frequency is 2p.
20c. From the graph
(or analytically) A = 10 and the frequency is 2 p.
116
Section 6.5
Section 6.5, Page 328
1. Proceeding as in Example 1, we take the Laplace transform of the D.E. and apply the I.C.:
(s2 + 2 s + 2 ) Y ( s ) = s + 2 + e-ps. Thus,
Y(s) = (s+2)/[(s + 1) 2 + 1] + e ps/[(s + 1) 2 + 1]. We write
22 the first term as (s + 1)/[(s + 1) + 1] + 1/[(s + 1) + 1].
Applying Theorem 6.3.1 and using Table 6.2.1, we obtain the solution, y = e-tcost + e-tsint - up(t)e (t p)sint.
Note that sin(t-p) = -sint.
3. Taking the Laplace transform and using the I.C. we have
-10s
2 1 - 5s e
(s + 3s+2)Y(s) =  + e + ------------. Thus
2s
1/2 e-5s -10s 1/2 1/2 1
Y(s) = ---------- + + e (----------------1---------------) and hence
s2+3s+2 s2+3s+2 s s+2 s+1
y(t) = h(t) + u5(t)h(t-5) + u10(t)[ + ie-2(t-10)-e-(t-10)]
2 5 10 2 2
where h(t) = e t - e 2t.
5. The Laplace transform of the D.E. is
(s2+2s+3)Y(s) = ------- + e 3ps, so
s2+1
1 -3 ps 1
Y(s) = ------------------ + e [------]. Using partial
(s2+1)(s2+2s+3) s2+2s+3
fractions or a computer algebra system we obtain
1 . 1 1 -t r- 1
4 4 . 4    _ _ .
where h(t) = e tsi^V"2t.
y(t) = sint - cost + e cosy 2 t + ; u3p(t)h(t-3p),
7. Taking the Laplace transform of the D.E. yields
(s2+1)Y(s) - y'(0) = J0 e"st5 (t-2p)costdt. Since
S(t-2p) = 0 for t n 2p the integral on the right is equal
f -st 2 ps
to I e o(t-2p) costdt which equals e cos2p from
Eq.(16). Substituting for y'(0) and solving for Y(s)
-2 ps
1e
gives Y(s) = ---- + ------ and hence
22 s2 +1 s2 +1
[sint 0 ? t < 2p y(t) = sint + u2p (t)sin(t-2p) = <!
[ 2sint 2p ? t
10. See the solution for Problem 7.
Section 6.5
117
13a. From Eq. (22) y(t) will complete one cycle when \J15 (t-5)/4 = 2p or T = t - 5 = 8p/V 15 , which is consistent with the plot in Fig. 6.5.3. Since an impulse causes a discontinuity in the first derivative, we need
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