Lista de Exercícios 8

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Universidade de Bras´ılia
Departamento de Matema´tica
Ca´lculo 2 – 2o/2017 – Turma F
Lista de Exerc´ıcios no 8
1. Encontre a transformada de Laplace de cada uma das func¸o˜es a seguir:
(a) tneat, n e´ um inteiro positivo e a e´ uma constante real
(b) f(t) =

0, t < pi
t− pi, pi ≤ t < 2pi
0, t ≥ 2pi
2. Lembre-se que cosh(bt) = (ebt + e−bt)/2 e que senh(bt =)(ebt − e−bt)/2. Encontre a
transformada de Laplace das func¸o˜es dadas, sendo a e b constantes reais:
(a) cosh(bt) (c)eat cosh(bt)
(b) senh(bt) (d) eatsenh(bt)
3. Encontre a transformada de Laplace inversa da func¸a˜o dada:
(a) F (s) =
3!
(s− 2)4
(b) F (s) =
e−2s
s2 + s− 2
(c) F (s) =
2(s− 1)e−2s
s2 − 2s+ 2
(d) F (s) =
e−s + e−2s − e−3s − e−4s
s
4. Use a transformada de Laplace para resolver o PVI dado:
(a) y′′ − 4y′ + 4y = 0, y(0) = 1, y′(0) = 1
(b) y′′ − 2y′ − 2y = 0, y(0) = 2, y′(0) = 0
(c) y(iv) − 4y′′′ + 6y′′ − 4y′ + y = 0, y(0) = 0, y′(0) = 1, y′′(0) = 0, y′′′(0) = 1
(d) y′′ − 2y′ + 2y = e−t, y(0) = 0, y′(0) = 1
(e) y′′ + y = f(t), y(0) = 0, y′(0) = 1, f(t) =
{
1, t < pi/2
0, pi/2 ≤ t <∞
(f) y′′ + 4y = sent− u2pi(t)sen(t− 2pi), y(0) = 0, y′(0) = 0
(g) y(iv) − y = u1(t)− u2(t), y(0) = 0, y′(0) = 0, y′′(0) = 0, y′′′(0) = 0
(h) y′′ + 2y′ + 2y = δ(t− pi), y(0) = 1, y′(0) = 0
(i) y′′ + 4y = δ(t− pi)− δ(t− 2pi), y(0) = 0, y′(0) = 0
(j) y′′ + 3y′ + 2y = δ(t− 5) + u10(t), y(0) = 0, y′(0) = 1/2
(k) y′′ + 2y′ + 3y = sent+ δ(t− 3pi), y(0) = 0, y′(0) = 0
(l) y′′ + y = δ(t− 2pi) cos t, y(0) = 0, y′(0) = 1
(m) y′′ + y = upi/2(t) + 3δ(t− 3pi/2)− u2pi(t), y(0) = 0, y′(0) = 0
RESPOSTAS
(1)
(a)
n!
(s− a)n+1 , s > a
(b) F (s) =
e−pis
s2
− e
−2pis
s2
(1 + pis)
(2)
(a)
s
s2 − b2 , s > |b|
(b)
b
s2 − b2 , s > |b|
(c)
s− a
(s− a)2 − b2 , s− a > |b|
(d)
b
(s− a)2 − b2 , s− a > |b|
(3)
(a) f(t) = t3e2t
(b) f(t) = 1
3
u2(t)[e
t−2 − e−2(t−2)]
(c) f(t) = 2u2(t)e
t−2 cos(t− 2)
(d) f(t) = u1(t) + u2(t)− u3(t)− u4(t)
(4)
(a) y = e2t − te2t
(b) y = 2et cosh(
√
3t)− (2/√3)etsenh(√3t)
(c) y = tet − t2et + 2
3
t3et
(d) y = 1
5
(e−t − et cos t+ 7etsent)
(e) y = 1− cost+ sent− upi/2(t)(1− sent)
(f) y = 1
6
[1− u2pi(t)](2sent− sen(2t))
(g) y = u1(t)h(t− 1)− u2(t)h(t− 2), h(t) = −1 + (cos t+ cosh t)/2
(h) y = e−tcost+ e−tsent+ upi(t)e−(t−pi)sen(t− pi)
(i) y = 1
2
upi(t)sen2(t− pi)− 12u2pi(t)sen2(t− 2pi)
(j) y = −1
2
e−2t + 1
2
e−t + u5(t)[−e−2(t−5) + e−(t−5)] + u10(t)[12 + 12e−2(t−10) − e−(t−10)]
(k) y = 1
4
sent− 1
4
cost+ 1
4
e−tcos(
√
2t) + (1/
√
2)u3pi(t)e
−(t−3pi)sen
√
2(t− 3pi)
(l) y = sent+ u2pi(t)sen(t− 2pi)
(m) y = upi/2(t)[1− cos(t− pi/2)] + 3u3pi/2(t)sen(t− 3pi/2)− u2pi(t)[1− cos(t− 2pi)]