Page 116 - Advanced engineering mathematics

96 CHAPTER 3 The Laplace Transform

1
24. e −s is charged to a potential of 5 volts and the switch is
s − 5 opened at time zero and closed 5 seconds later. Graph
1 this output.
25. e −21s
s(s + 16) 34. Determine the output voltage in the RL circuit of
2
2w
26. Determine L[e −2t
t e cos(3w)dw]. Hint: Use the Figure 3.20 if the current is initially zero and
0

first shifting theorem. 0 for 0 ≤ t < 5
E(t) =
2 for t ≥ 5.
In each of Problems 27 through 32, solve the initial value
problem. Graph this output function.
R

27. y + 4y = f (t); y(0) = 1, y (0) = 0, with

0for 0 ≤ t < 4
f (t) =
3for t ≥ 4
28. y − 2y − 3y = f (t); y(0) = 1, y (0) = 0, with E(t)

0 for 0 ≤ t < 4 L
f (t) =
12 for t ≥ 4

29. y − 8y = g(t); y(0) = y (0) = y (0) = 0, with

0for 0 ≤ t < 6
g(t) =
2for t ≥ 6

30. y + 5y + 6y = f (t); y(0) = y (0) = 0, with FIGURE 3.20 The RL circuit of

Problem 34, Section 3.3.

−2 for 0 ≤ t < 3
f (t) =
0 for t ≥ 3 35. Solve for the current in the RL circuit of Problem 34
if the current is initially zero and
31. y − y + 4y − 4y = 0; y(0) = y (0) = 0, y (0) = 1,

with k for 0 ≤ t < 5
E(t) =

1for 0 ≤ t < 5 0 for t ≥ 5.
f (t) =
2for t ≥ 5 36. Show that Heaviside’s formula can be written
n

32. y − 4y + 4y = f (t); y(0) =−2, y (0) = 1, with p(a j )
−1
a j t
L [F](t) = e .

q (a j )
t for 0 ≤ t < 3 j=1
f (t) =
t + 2for t ≥ 3 Hint: Write
p(s) p(s)
33. Determine the output voltage in the circuit of (s − a j ) = .
Figure 3.18, assuming that at time zero the capacitor q(s) (q(s) − q(a j ))/(s − a j )
3.4 Convolution
If f (t) and g(t) are defined for t ≥0, then the convolution f ∗ g of f with g is the function
defined by
t
( f ∗ g)(t) = f (t − τ)g(τ)dτ
0
for t ≥ 0 such that this integral converges.
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October 14, 2010 14:14 THM/NEIL Page-96 27410_03_ch03_p77-120