Lista 10 Transformada de Laplace

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UFBA - Ca´lculo C - 2017/1
Lista de Exerc´ıcios 10 - Transformada de Laplace
(1) Encontre a transformada de laplace da func¸a˜o f(t) = cos(at), t ≤ 0, sendo a uma constante diferente
de zero.
(2) Encontre a transformada de Laplace da func¸a˜o dada, sendo a e b constantes ambas diferentes de
zero. Lembre-se que cosh(x) =
ex + e−x
2
e senh(x) =
ex − e−x
2
.
(a) f(t) = cosh(bt).
(b) f(t) = senh(bt).
(c) f(t) = eat cosh(bt).
(d) f(t) = eat senh(bt).
(e) f(t) = teat.
(f) f(t) = t sen(at).
(g) f(t) = t cosh(at).
(h) f(t) = tneat, n inteiro positivo.
(i) f(t) = t2 sen(at).
(j) f(t) = t2 senh(at).
(3) Em cada caso, determine se a func¸a˜o f = f(t) e´ cont´ınua, seccionalmente cont´ınua ou nenhuma das
duas opc¸o˜es, no intervalo [0, 3]. (Dica: fac¸a o gra´fico da func¸a˜o)
(a) f(t) =

t2, 0 ≤ t ≤ 1;
2 + t, 1 < t ≤ 2;
6− t, 2 < t ≤ 3.
(b) f(t) =

t2, 0 ≤ t ≤ 1;
(t− 1)−1, 1 < t ≤ 2;
1, 2 < t ≤ 3.
(c) f(t) =

t2, 0 ≤ t ≤ 1;
1, 1 < t ≤ 2;
3− t, 2 < t ≤ 3.
(d) f(t) =

t, 0 ≤ t ≤ 1;
3− t, 1 < t ≤ 2;
1, 2 < t ≤ 3.
(4) Encontre a transformada de Laplace inversa.
(a) F (s) =
3
s2 + 4
.
(b) F (s) =
4
(s− 1)3 .
(c) F (s) =
2
s2 + 3s− 4.
(d) F (s) =
3s
s2 − s− 6.
(e) F (s) =
2s+ 2
s2 + 2s+ 5
.
(f) F (s) =
2s− 3
s2 − 4 .
(g) F (s) =
2s+ 1
s2 − 2s+ 2.
(h) F (s) =
8s2 − 4s+ 12
s(s2 + 4)
.
(i) F (s) =
1− 2s
s2 + 4s+ 5
.
(j) F (s) =
2s− 3
s2 + 2s+ 10
.
(5) Encontre a soluc¸a˜o da EDO com condic¸o˜es iniciais.
(a) y′′′′ − 4y′′′ + 6y′′ − 4y′ + y = 0, y(0) = 0, y′(0) = 1, y′′(0) = 0, y′′′(0) = 1.
(b) y′′′′ − y = 0, y(0) = 1, y′(0) = 0, y′′(0) = 1, y′′′(0) = 0.
(c) y′′′′ − 4y = 0, y(0) = 1, y′(0) = 0, y′′(0) = −2, y′′′(0) = 0.
(6) Esboce o gra´fico da func¸a˜o dada no intervalo [0,∞].
(a) f(t) = u1(t) + 2u3(t)− 6u4(t).
(b) f(t) = (t− 3)u2(t)− (t− 2)u3(t).
(c) f(t) = (t− pi)2upi(t).
(d) f(t) = sen(t− 3)u3(t).
(e) f(t) = 2(t− 1)u2(t).
(f) f(t) = (t− 1)u1(t)− 2(t− 2)u2(t) + (t− 3)u3(t).
1
2
(7) Encontre a transformada de Laplace da func¸a˜o.
(a) f(t) =
{
0, t < 2
(t− 2)2, t ≥ 2.
(b) f(t) =
{
0, t < 1
t2 − 2t+ 2, t ≥ 1.
(c) f(t) =

0, t < pi
t− pi, pi ≤ t < 2pi
0, t ≥ 2pi.
(d) f(t) = u1(t) + 2u3(t)− 6u4(t).
(e) f(t) = (t− 3)u2(t)− (t− 2)u3(t).
(f) f(t) = t− u1(t)(t− 1).
(8) Calcule a transformada de Laplace inversa da func¸a˜o.
(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) =
2e−2s
s2 − 4.
(e) F (s) =
(s− 2)e−s
s2 − 4s+ 3.
(f) F (s) =
e−s + e−2s − e−3s − e−4s
s
.
(9) Seja f = f(t), t ≥ 0, uma func¸a˜o e suponha que F (s) = L {f(t)} existe para s > a ≥ 0.
(a) Mostre que, se c > 0 e´ uma constante, enta˜o
L {f(ct)} = 1
c
F
(s
c
)
, s > ca.
(b) Mostre que, se k > 0 e´ uma constante, enta˜o
L −1{F (ks)} = 1
k
f
(
t
k
)
.
(c) Mostre que, se a e b sa˜o constantes com a > 0, enta˜o
L −1{F (as+ b)} = 1
a
e−bt/af
(
t
a
)
.
(10) Use os resultados do exerc´ıcio anterior para calcular a transformada de Laplace inversa da func¸a˜o.
(a) F (s) =
2n+1
n!
.
(b) F (s) =
2s+ 1
4s2 + 4s+ 5
.
(c) F (s) =
1
9s2 − 12s+ 3.
(d) F (s) =
e2e−4s
2s− 1 .
(11) Encontre a soluc¸a˜o da EDO com condic¸o˜es iniciais.
(a) y′′ + y = f(t); y(0) = 0, y′(0) = 1; f(t) =
{
1, 0 ≤ t < pi/2
0, t ≥ pi/2.
(b) y′′ + 2y′ + 2y = h(t); y(0) = 0, y′(0) = 1; h(t) =

0, 0 ≤ t < pi
1, pi ≤ t < 2pi
0, t ≥ 2pi.
(c) y′′ + 4y = sen(t)− u2pi(t) sen(t− 2pi); y(0) = 0; y′(0) = 0.
(d) y′′ + 4y = sen(t)− upi(t) sen(t− pi); y(0) = 0; y′(0) = 0.
(e) y′′ + 3y′ + 2y = f(t); y(0) = 0, y′(0) = 0; f(t) =
{
1, 0 ≤ t < 10
0, t ≥ 10.
(f) y′′ + 3y′ + 2y = u2(t); y(0) = 0, y′(0) = 1.
(g) y′′ + y = u3pi(t); y(0) = 1, y′(0) = 0.
(h) y′′ + y′ + 54y = t− upi/2(t)(t− pi/2); y(0) = 0, y′(0) = 0.
3
(i) y′′ + y = g(t); y(0) = 0, y′(0) = 1; g(t) =
{
t/2, 0 ≤ t < 6
3, t ≥ 6.
(j) y′′ + y′ + 54y = g(t); y(0) = 0, y
′(0) = 0; g(t) =
{
sen(t), 0 ≤ t < pi
0, t ≥ pi.
(k) y′′ + 4y = upi(t)− u3pi(t); y(0) = 0, y′(0) = 0.
(l) yiv − y = u1(t)− u2(t); y(0) = 0, y′(0) = 0, y′′(0) = 0, y′′′(0) = 0.
(m) yiv + 5y′′ + 4y = 1− upi(t); y(0) = 0, y′(0) = 0, y′′(0) = 0, y′′′(0) = 0.
(12) Encontre a soluc¸a˜o da EDO com condic¸o˜es iniciais.
(a) y′′ + 2y′ + 2y = δ(t− pi); y(0) = 1, y′(0) = 0.
(b) y′′ + 4y = δ(t− pi)− δ(t− 2pi); y(0) = 0, y′(0) = 0.
(c) y′′ + 3y′ + 2y = δ(t− 5) + u10(t); y(0) = 0, y′(0) = 1/2.
(d) y′′ − y = −20 δ(t− 3); y(0) = 1, y′(0) = 0.
(13) Encontre a transformada de Laplace da func¸a˜o.
(a) f(t) =
∫ t
0
(t− u)2 cos(2u) du.
(b) f(t) =
∫ t
0
e−(t−u) sen(u) du.
(c) f(t) =
∫ t
0
(t− u)eu du.
(d) f(t) =
∫ t
0
sen(t− u) cos(u) du.
(14) Encontre a transformada de Laplace inversa da func¸a˜o.
(a) F (s) =
1
s4(s2 + 1)
.
(b) F (s) =
s
(s+ 1)(s2 + 4)
.
(c) F (s) =
1
(s+ 1)2(s2 + 4)
.
(d) F (s) =
G(s)
s2 + 1
.
(15) Encontre a soluc¸a˜o da EDO com condic¸o˜es iniciais.
(a) y′′ + ω2y = g(t); y(0) = 0, y′(0) = 1, ω constante na˜o nula.
(b) 4y′′ + 4y′ + 17y = g(t); y(0) = 0, y′(0) = 0.
(c) y′′ + 4y′ + 4y = g(t); y(0) = 2, y′(0) = −3.
(d) yiv + 5y′′ + 4y = g(t); y(0) = 1, y′(0) = 0, y′′(0) = 0, y′′′(0) = 0.
Respostas
(1) L {cos(at)} = s
s2 + a2
, s > 0
(2) (a) L {cosh(bt)} = s
s2 − b2 , s > |b|
(b) L { senh(bt)} = b
s2 − b2 , s > |b|
(c) L {eat cosh(bt)} = s− a
(s− a)2 − b2 , s− a > |b|
(d) L {eat senh(bt)} = b
(s− a)2 − b2 , s− a > |b|
(e) L {teat} = 1
s− a , s > a
(f) L {t sen(at)} = 2as
(s2 + a2)2
, s > 0
(g) L {t cosh(at)} = s
2 + a2
(s− a)2(s+ a)2 , s > |a|
(h) L {tneat} = n!
(s− a)n+1 , s > a
(i) L {t2 sen(at)} = 2a(3s
3 − a2)
(s2 + a2)3
, s > 0
(j) L {t2 senh(at)} = 2a(3s
2 + a2)
(s2 − a2)3 , s > |a|
4
(3) (a) seccionalmente cont´ınua
(b) seccionalmente cont´ınua
(c) nenhuma das duas
(d) cont´ınua
(4) (a) f(t) = 32 sen(t)
(b) f(t) = 2t2et
(c) f(t) = 25e
t − 25e−4t
(d) f(t) = 95e
3t + 65e
−2t
(e) f(t) = 2e−t cos(2t)
(f) f(t) = 2 cosh(2t)− 32 senh(2t)
(g) f(t) = 2et cos(t) + 3et sen(t)
(h) f(t) = 3− 2 sen(2t) + 5 cos(2t)
(i) f(t) = −2e−2t cos(t) + 5e−2t sen(t)
(j) f(t) = 2e−t cos(3t)− 53e−t sen(3t)
(5) (a) y = tet − t2et + 23 t3et (b) y = cosh(t) (c) y = cos(
√
2t)
(6)
(7) (a) F (s) = e−s/s3
(b) F (s) = e−s(s2 + 2)/s3
(c) F (s) =
e−pis
s2
− e
−2pis
s2
(1 + pis)
(d) F (s) =
1
s
(e−s + 2e−3s − 6e−4s)
(e) F (s) = s−2
[
(1− s) e−2s − (1 + s)e−3s]
(f) F (s) = (1− e−s)/s2
(8) (a) f(t) = t3e2t
(b) f(t) = 13u2(t)
[
et−2 − e−2(t−2)]
(c) f(t) = 2u2(t)e
t−2 cos(t− 2)
(d) f(t) = u2(t) senh[2(t− 2)]
(e) f(t) = u1(t)e
2(t−1) cosh(t− 1)
(f) f(t) = u1(t) + u2(t)− u3(t)− u4(t)
(9)
(10) (a) f(t) = 2(2t)n
(b) f(t) = 12e
−t/2 cos(t)
(c) f(t) = 16e
t/3(e2t/3 − 1)
(d) f(t) = 12e
t/2u2(t/2)
(11) (a) y = 1− cos(t) + sen(t)− upi/2(t)(1− sen(t))
(b) y = e−t sen(t) + 12upi(t)
(
1 + e−(t−pi) cos(t) + e−(t−pi) sen(t)
)
− 12u2pi(t)
(
1 + e−(t−2pi) cos(t)− e−(t−2pi) sen(t))
(c) y = 16(1− u2pi(t))(2 sen(t)− sen(2t))
(d) y = 16(2 sen(t)− sen(2t))− 16upi(t)(2 sen(t)− sen(2t))
(e) y = 12 +
1
2e
−t − e−t − u10(t)
(
1
2 +
1
2e
−2(t−10) − e−(t−10))
(f) y = e−t − e−2t + u2(t)
(
1
2 − e−(t−2) + 12e−2(t−2)
)
(g) y = cos(t) + u3pi(t) (1− cos(t− 3pi))
(h) y = h(t)− upi/2(t)h(t− pi/2), h(t) = 425
(−4 + 5t+ 4e−t/2 cos(t)− 3e−t/2 sen(t))
(i) y = 12 sen(t) +
1
2 t− 16u6(t) (t− 6− sen(t− 6))
(j) y = h(t)− upi(t)h(t− pi), h(t) = 417
(−4 cos(t) + sen(t) + 4e−t/2 cos(t) + e−t/2 sen(t))
(k) y = upi(t)
(
1
4 − 14 cos(2t− 2pi)
)− u3pi(t) (14 − 14 cos(2t− 6pi))
(l) y = u1(t)h(t− 1)− u2(t)h(t− 2), h(t) = −1 + (cos(t) + cosh(t))/2
(m) y = h(t)− upi(t)h(t−pi), h(t) = (3− 4 cos(t) + cos(2t))/12
(12) (a) y = e−t cos(t) + e−t sen(t) + upi(t)e−(t−pi) sen(t− pi)
(b) y = 12upi(t) sen(2(t− pi))− 12u2pi(t) sen(2(t− pi))
(c) y = −12e−2t + 12e−t + u5(t)
(−e−2(t−5) + e−(t−5))+ u10(t) (12 + 12e−2(t−10) + e−(t−10))
(d) y = cosh(t)− 20u3(t)senh(t− 3)
5
(13) (a) F (s) =
2
s2(s2 + 4)
(b) F (s) =
1
(s+ 1)(s2 + 1)
(c) F (s) =
1
s2(s− 1)
(d) F (s) =
s
(s2 + 1)2
(14) (a) f(t) = 16
∫ t
0
(t− u)3 sen(u) du
(b) f(t) =
∫ t
0
e−(t−u) cos(u) du
(c) f(t) = 12
∫ t
0
(t− u)e−(t−u) sen(2u) du
(d) f(t) =
∫ t
0
sen(t− u)g(u) du
(15) (a) y =
1
ω
sen(ωt) +
1
ω
∫ t
0
sen (ω(t− u)) g(u)du
(b) y =
1
8
∫ t
0
e−(t−u)/2 sen (2(t− u)) g(u)du
(c) y = 2e−2t + te−2t +
∫ t
0
(t− u)e−2(t−u)g(u)du
(d) y =
4
3
cos(t)− 1
3
cos(2t) +
1
6
∫ t
0
[2 sen(t− u)− sen (2(t− u))] g(u)du