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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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22 s (s +s+5/4)
= e-«s/)i/5 16/25 + (16/25)s-4/25 1
^ s2 s (s + 1/2) 2+1 ^
-ps/2
= (1-e 'lH(s), where we have used partial fractions and completed the square in the denominator of the last term. Since the numerator of the last term of H
16
can be written as —[(s+1/2) - 3/4], we see that
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"
(2 0t—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
, , —1/8, 3\[¿ _ ó[¿ . 3^/7 ^ ï
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 < 2 p
n > 6, f(t) =
1 0 < t < p, 2p < t < 3p, 4p < t < 5p
-1 p < t < 2p, 3p < t <4p, 5p t < 6p
19b. Taking the Laplace transform of the D.E. and using the I.C.
have Y(s) =
1---- [1 + 2 X (-1) Vpks],
2
s(s +1) -pks
since
?{upk(t)} =
e
Since
k=1
1
1 s
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 X (-1)
Upk(t) - 2
X (-1)
2k
cost
k=1
k=1
s
k
we
= 1 - cost - 2ncost + 2 (-1) kupk(t) which diverges for n^ro.
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 .05tsinbt. 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 .05(t np) cos[b(t-np) + 8], and therefore converges as t——•.
20a. y(t) for n = 30
y(t) for n = 31
2 0b. From the graph of part a, A @ 12.5 and the frequency is 2p.
2 0c. From the graph
(or analytically) A = 10 and the frequency is 2p.
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].
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