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Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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− 1 · 3 · 5
2 · 4 · 6
1
s7
+ · · ·
]
.
Assuming that term-by-term inversion is valid,
y(t) = c
[
1− 1
2
t2
2!
+
1 · 3
2 · 4
t4
4!
− 1 · 3 · 5
2 · 4 · 6
t6
6!
+ · · ·
]
= c
[
1− 2!
22
t2
2!
+
4!
22 · 42
t4
4!
− 6!
22 · 42 · 62
t6
6!
+ · · ·
]
.
It follows that
y(t) = c
[
1− 1
22
t2 +
1
22 · 42 t
4 − 1
22 · 42 · 62 t
6 + · · ·
]
= c
∞∑
n=0
(−1)n
22n(n!)2
t2n.
The series is evidently the expansion, about x = 0 , of J0(t).
36.(b) Taking the Laplace transform of the given Legendre equation,
L [ y ′′]− L [ t2 y ′′]− 2L [ t y ′] + α(α+ 1)L [ y ] = 0 .
Using the differentiation property of the transform,
L [ y ′′]− d
2
ds2
L [ y ′′] + 2 d
ds
L [ y ′] + α(α+ 1)L [ y ] = 0 .
That is,[
s2Y (s)− s y(0)− y ′(0)]− d2
ds2
[
s2Y (s)− s y(0)− y ′(0)]+
+ 2
d
ds
[s Y (s)− y(0)] + α(α+ 1)Y (s) = 0 .
Invoking the initial conditions, we have
s2Y (s)− 1− d
2
ds2
[
s2Y (s)− 1]+ 2 d
ds
[s Y (s)] + α(α+ 1)Y (s) = 0 .
After carrying out the differentiation, the equation simplifies to
d2
ds2
[
s2Y (s)
]− 2 d
ds
[s Y (s)]− [s2 + α(α+ 1)]Y (s) = −1 .
That is,
s2
d2
ds2
Y (s) + 2s
d
ds
Y (s)− [s2 + α(α+ 1)]Y (s) = −1 .
37. By definition of the Laplace transform, given the appropriate conditions,
L [ g(t)] =
∫ ∞
0
e−st
[∫ t
0
f(τ)dτ
]
dt =
∫ ∞
0
∫ t
0
e−stf(τ)dτdt .
6.3 243
Assuming that the order of integration can be exchanged,
L [ g(t)] =
∫ ∞
0
f(τ)
[∫ ∞
τ
e−stdt
]
dτ =
∫ ∞
0
f(τ)
[
e−sτ
s
]
dτ.
(Note the region of integration is the area between the lines τ(t) = t and τ(t) = 0.)
Hence
L [ g(t)] = 1
s
∫ ∞
0
f(τ) e−sτdτ =
1
s
L [ f(t)] .
6.3
1.
3.
244 Chapter 6. The Laplace Transform
5.
6.
8.(a)
(b) f(t) = 1− 2u1(t) + 2u2(t)− 2u3(t) + u4(t).
6.3 245
9.(a)
(b) f(t) = 1 + (e−(t−2) − 1)u2(t).
11.(a)
(b) f(t) = t− u1(t)− u2(t) + (2− t)u3(t).
12.(a)
(b) f(t) = t+ (2− t)u2(t) + (5− t)u5(t) + (t− 7)u7(t).
246 Chapter 6. The Laplace Transform
13. Using the Heaviside function, we can write
f(t) = (t− 2)2 u2(t) .
The Laplace transform has the property that
L [uc(t)f(t− c)] = e−csL [f(t)] .
Hence
L
[
(t− 2)2 u2(t)
]
=
2 e−2s
s3
.
15. The function can be expressed as
f(t) = (t− pi) [upi(t)− u2pi(t)] .
Before invoking the translation property of the transform, write the function as
f(t) = (t− pi)upi(t)− (t− 2pi)u2pi(t)− pi u2pi(t) .
It follows that
L [ f(t)] = e
−pis
s2
− e
−2pis
s2
− pie
−2pis
s
.
16. It follows directly from the translation property of the transform that
L [ f(t)] = e
−s
s
+ 2
e−3s
s
− 6e
−4s
s
.
17. Before invoking the translation property of the transform, write the function
as
f(t) = (t− 2)u2(t)− u2(t)− (t− 3)u3(t)− u3(t) .
It follows that
L [ f(t)] = e
−2s
s2
− e
−2s
s
− e
−3s
s2
− e
−3s
s
.
18. It follows directly from the translation property of the transform that
L [ f(t)] = 1
s2
− e
−s
s2
.
19. Using the fact that L [eatf(t)] = L [f(t)]s→s−a ,
L−1
[
3!
(s− 2)4
]
= t3e2t .
21. First consider the function
G(s) =
2(s− 1)
s2 − 2s+ 2 .
Completing the square in the denominator,
G(s) =
2(s− 1)
(s− 1)2 + 1 .
6.3 247
It follows that
L−1 [G(s)] = 2 et cos t .
Hence
L−1
[
e−2sG(s)
]
= 2 et−2 cos (t− 2)u2(t) .
22. The inverse transform of the function 2/(s2 − 4) is f(t) = sinh 2t . Using the
translation property of the transform,
L−1
[
2 e−2s
s2 − 4
]
= sinh (2(t− 2)) · u2(t) .
23. First consider the function
G(s) =
(s− 2)
s2 − 4s+ 3 .
Completing the square in the denominator,
G(s) =
(s− 2)
(s− 2)2 − 1 .
It follows that
L−1 [G(s)] = e2t cosh t .
Hence
L−1
[
(s− 2)e−s
s2 − 4s+ 3
]
= e2(t−1) cosh (t− 1)u1(t) .
24. Write the function as
F (s) =
e−s
s
+
e−2s
s
− e
−3s
s
− e
−4s
s
.
It follows from the translation property of the transform, that
L−1
[
e−s + e−2s − e−3s − e−4s
s
]
= u1(t) + u2(t)− u3(t)− u4(t) .
25.(a) By definition of the Laplace transform,
L [ f(ct)] =
∫ ∞
0
e−stf(ct)dt .
Making a change of variable, τ = ct , we have
L [ f(ct)] = 1
c
∫ ∞
0
e−s(τ/c)f(τ)dτ =
1
c
∫ ∞
0
e−(s/c)τf(τ)dτ .
Hence L [ f(ct)] = 1c F ( sc ) , where s/c > a .
(b) Using the result in part (a),
L
[
f
(
t
k
)]
= k F (ks).
248 Chapter 6. The Laplace Transform
Hence
L−1 [F (ks)] = 1
k
f
(
t
k
)
.
(c) From part (b), L−1 [F (as)] = 1af( ta ) Note that as+ b = a(s+ b/a). Using the
fact that L [ectf(t)] = L [f(t)]s→s−c ,
L−1 [F (as+ b)] = e−bt/a 1
a
f
(
t
a
)
.
26. First write
F (s) =
n!
( s2 )
n+1
.
Let G(s) = n!/sn+1. Based on the results in Problem 25,
1
2
L−1
[
G
(s
2
)]
= g(2t),
in which g(t) = tn. Hence
L−1 [F (s)] = 2 (2t)n = 2n+1tn.
29. First write
F (s) =
e−4(s−1/2)
2(s− 1/2) .
Now consider
G(s) =
e−2s
s
.
Using the result in Problem 25(b),
L−1 [G(2s)] = 1
2
g
(
t
2
)
,
in which g(t) = u2(t). Hence L−1 [G(2s)] = 12 u2(t/2) = 12 u4(t). It follows that
L−1 [F (s)] = 1
2
et/2 u4(t).
30. By definition of the Laplace transform,
L [ f(t)] =
∫ ∞
0
e−stu1(t)dt .
That is,
L [ f(t)] =
∫ 1
0
e−st dt =
1− e−s
s
.
31. First write the function as f(t) = u0(t)− u1(t) + u2(t)− u3(t) . It follows that
L [ f(t)] =
∫ 1
0
e−stdt +
∫ 3
2
e−stdt .
6.3 249
That is,
L [ f(t)] = 1− e
−s
s
+
e−2s − e−3s
s
=
1− e−s + e−2s − e−3s
s
.
32. The transform may be computed directly. On the other hand, using the trans-
lation property of the transform,
L [f(t)] = 1
s
+
2n+1∑
k=1
(−1)k e
−ks
s
=
1
s
[
2n+1∑
k=0
(−e−s)k
]
=
1
s
1− (−e−s)2n+2
1 + e−s
.
That is,
L [f(t)] = 1− (e
−2s)n+1
s(1 + e−s)
.
35. The given function is periodic, with T = 2 . Using the result of Problem 34,
L [f(t)] = 1
1− e−2s
∫ 2
0
e−stf(t)dt =
1
1− e−2s
∫ 1
0
e−stdt .
That is,
L [f(t)] = 1− e
−s
s(1− e−2s) =
1
s(1 + e−s)
.
37. The function is periodic, with T = 1 . Using the result of Problem 34,
L [f(t)] = 1
1− e−s
∫ 1
0
t e−stdt .
It follows that
L [f(t)] = 1− e
−s(1 + s)
s2(1− e−s) .
38. The function is periodic, with T = pi . Using the result of Problem 34,
L [f(t)] = 1
1− e−pis
∫ pi
0
sin t · e−stdt .
We first calculate ∫ pi
0
sin t · e−stdt = 1 + e
−pis
1 + s2
.
Hence
L [f(t)] = 1 + e
−pis
(1− e−pis)(1 + s2) .
250 Chapter 6. The Laplace Transform
39.(a)
L [f(t)] = L [1]− L [u1(t)] = 1
s
− e
−s
s
.
(b)
Let F (s) = L [1− u1(t)]. Then
L
[∫ t
0
[1− u1(τ)] dτ
]
=
1
s
F (s) =
1− e−s
s2
.
(c)
Let G(s) = L [g(t)]. Then
L [h(t)] = G(s)− e−sG(s) = 1− e
−s
s2
− e−s 1− e
−s
s2
=
(1− e−s)2
s2
.
6.4 251
40.(a)
(b) The given function is periodic, with T = 2 . Using the result of Problem 34,
L [f(t)] = 1
1− e−2s
∫ 2
0
e−stp(t)dt .
Based on the piecewise definition of p(t),∫ 2
0
e−stp(t)dt =
∫ 1
0
t e−stdt+
∫ 2
1
(2− t)e−stdt = 1
s2
(1− e−s)2.
Hence
L [p(t)] = (1− e
−s)
s2(1 + e−s)
.
(c) Since p(t) satisfies the hypotheses of Theorem 6.2.1 ,
L [p ′(t)] = sL [p(t)]− p(0) .
Using the result of Problem 36 from the Student Solutions Manual,
L [p ′(t)] = (1− e
−s)
s(1 + e−s)
.
We note the p(0) = 0 , hence
L [p(t)] = 1
s
[
(1− e−s)
s(1 + e−s)
]
.
6.4
2.(a) Let h(t) be the forcing function on the right-hand-side. Taking the Laplace
transform of both sides of the ODE, we obtain
s2 Y (s)− s y(0)− y ′(0)

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