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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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1. 1 II u2(t) - (t - 2)u3(t) 12. f (t) = t - u (t)(t - 1), t > 0
In each of Problems 13 through 18 find the inverse Laplace transform of the given function.
e-2s
13. F (s) = 3! 14. F (s)
(s - 2)4
15. F (s) = 2(s - 1)e-2s s2 2s + 2 16. F (s)
17. F (s) = (s - 2)e-s s2 - 4s + 3 18. F(s)
s2 + s 2
2e
-2s
s2 4
e-s + e-2s - e-3s - e-4s
6.3 Step Functions
315
19. Suppose that F(s) = L{ f (f)} exists for s > a > 0.
(a) Show that if c is a positive constant, then
c \c
(b) Show that if k is a positive constant, then
L{ f(ct)} = C^(C) , s > constant, then
Fi f(t).
kk
(c) Show that if a and b are constants with a > 0, then
C~l{F(as + b)} = 1 e-bt/af(^j .
In each of Problems 20 through 23 use the results of Problem 19 to find the inverse Laplace transform of the given function.
20. F(s) =
2n+1n\
r.n+1
22. F(s) =
9s2 - 12s + 3
2s + 1
21. F (s) = 2--------------------
4s2 + 4s + 5
23. F (s) =
2s 1
In each of Problems 24 through 27 find the Laplace transform of the given function. In Problem 27 assume that term-by-term integration of the infinite series is permissible.
24. f (t) =
0 < t < 1 t1
25. f(t) =
0 < t < 1
1 < t < 2 2<t<3 t3
2n+1
26. f (t) = 1 U1(t) + + U2n (t) U2n+1(t) = 1 + "Y2 (-1)kuk (t)
k=1
TO
27. f (t) = 1 + (1)kuk(t)- See Figure 6.3.6.
FIGURE 6.3.6 A square wave.
28. Let f satisfy f (t + T) = f (t) for all t > 0 and for some fixed positive number T; f is said to be periodic with period T on 0 < t < to. Show that
L{ f (t)} =
f
J0
e-st f(t) dt
1- e-
1
e
316
Chapter 6. The Laplace Transform
In each of Problems 29 through 32, use the result of Problem 28 to find the Laplace transform of the given function.
1, 0 < t < 1,
29. f(t) =
0, 1 < t < 2;
f (t + 2) = f (t).
Compare with Problem 27.
30. f(t) =
1, -1,
f (t + 2) = f (t). See Figure 6.3.7.
0 < t < 1,
1 < t < 2;
31. f(t) = t, 0 < t < 1;
f (t + 1) = f (t).
See Figure 6.3.8.
32. f (t) = sin t, 0 < t < n;
f (t + n) = f (t).
See Figure 6.3.9.
2n 3n t
FIGURE 6.3.9 A rectified sine wave.
33. (a) If f (t) = 1 u1(t), find L{ f (t)}; compare with Problem 24. Sketch the graph of 7 = f (t). t
(b) Let g(t) = f (%) d%, where the function f is defined in part (a). Sketch the graph
*^0
of 7 = g(t) and find L{g(t)}.
(c) Let h(t) = g(t) u1 (t)g(t 1), where g is defined in part (b). Sketch the graph of 7 = h(t) and find L{h(t)}.
34. Consider the function p defined by
p(t) =
t,
2 t,
0 < t < 1,
1 < t < 2;
P(t + 2) = p(t).
(a) Sketch the graph of 7 = p(t).
(b) Find L{p(t)} by noting that p is the periodic extension of the function h in Problem 33(c) and then using the result of Problem 28.
6.4 Differential Equations with Discontinuous Forcing Functions
317
(c) Find L{p(t)} by noting that
p(t) = f(t) dt,
Jo
where f is the function in Problem 30, and then using Theorem 6.2.1.
6.4 Differential Equations with Discontinuous Forcing Functions
In this section we turn our attention to some examples in which the nonhomogeneous term, or forcing function, is discontinuous.
1
Find the solution of the differential equation
2 / + / + 2y = g(t), (1)
where
g(t) = u5(t) - u20(t) = J 0 0 5 and t > 20. (2)
Assume that the initial conditions are
y(0) = 0, y (0) = 0. (3)
This problem governs the charge on the capacitor in a simple electric circuit with a unit voltage pulse for 0 t < 20. Alternatively, y may represent the response of a damped oscillator subject to the applied force g(t).
The Laplace transform of Eq. (1) is
2s2Y(s) - 2sy(0) - 2/(0) + sY(s) - y(0) + 2Y(s) = L{uo(t)} - L^t)}
= (e-0s - e-20s)/s.
Introducing the initial values (3) and solving for Y(s), we obtain
e-5s _ -20s
Y(s) = e 2 e . (4)
s (2s + s + 2)
To find y = $(t) it is convenient to write Y(s) as
Y(s) = (e-0s - e-20s) H(s), (0)
where
H (s) = 1/s (2s2 + s + 2). (6)
Then, if h(t) = L-1{H(s)}, we have
y = m = uo(t)h(t - 0) - u20(t)h(t - 20). (7)
318
Chapter 6. The Laplace Transform
Observe that we have used Theorem 6.3.1 to write the inverse transforms of e5s H(s) and e~20sH(s), respectively. Finally, to determine h(t) we use the partial fraction expansion of H(s):
a bs + c
H (s) = + TT2 T. (8)
s 2s + s + 2
Upon determining the coefficients we find that a = 2, b = 1, and c = 1. Thus
H s ) =
s 2s + s + 2
1/2 /1 \ (s + 4 ) + 4
V (s + 4)2 +
K2 , 15 (9)
4> ' 16
s
so that, by referring to lines 9 and 10 of Table 6.2.1, we obtain
h(t) = 1 2[e1/4cos(VT51/4) + (VT5/15)e1/4 sin(VT5 t/4)]. (10)
In Figure 6.4.1 the graph of y = fi(t) from Eqs. (7) and (10) shows that the solution consists of three distinct parts. For 0 < t < 5 the differential equation is
2/ + y + 2 y = 0 (11)
and the initial conditions are given by Eq. (3). Since the initial conditions impart no energy to the system, and since there is no external forcing, the system remains at rest; that is, y = 0 for 0 < t < 5. This can be confirmed by solving Eq. (11) subject to the initial conditions (3). In particular, evaluating the solution and its derivative at t = 5, or more precisely, as t approaches 5 from below, we have
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