Math 528
Homework 4 Solutions
Assignment:
Section 6.1:
Section 6.2:
Section 6.3:
Section 6.5:
Graded:
Section 6.1:
Section 6.2:
Section 6.3:
Section 6.5:
#
#
#
#
1-16
1-15
1-11
1-13
all, 25-32 all
odd
odd, 13-17 all
odd, 17-25 odd, 26
#
#
#
#
2,
1,
3,
3,
8, 12, 14, 16, 22, 26, 28, 30, 32
11, 15
14, 16
9, 13, 19, 23
6,
7,
7,
5,
Section 6.1:
Note
∞
Z
Z
T
g(t) dt = lim
g(t) dt
T →∞
0
0
and
g(t)|∞
t=0 = lim g(T ) − g(0)
T →∞
1. If f (t) = 3t + 12 then
L (f ) =
Z
∞
e−st (3t + 12)dt
0
Z ∞
Z
−st
=3
te dt + 12
0
∞
e−st dt
0
3
12
= 2+
s
s
2. If f (t) = (a − bt)2 then
Z ∞
L (f ) =
e−st (a − bt)2 dt
0
Z ∞
Z
2
−st
e dt − 2ab
=a
0
2
=
0
2
a
2ab 2b
− 2 + 3
s
s
s
∞
te
−st
dt + b
2
Z
0
∞
t2 e−st dt
3. If f (t) = cos πt then
L (f ) =
Z
∞
e−st cos πt dt
∞ 0
Z ∞
1 −st
s
=
e sin πt
+
e−st sin πt dt
π
π 0
t=0
∞
Z
s
1 −st
s ∞ −st
=
− e cos πt −
e cos πt dt
π
π
π 0
0
Z
s
s2 ∞ −st
= 2− 2
e cos πt dt
π
π 0
Thus, we see
L (f ) =
s
s2
−
L (f )
π2 π2
so
L (f ) =
s2
s
.
+ π2
4. If f (t) = cos2 ωt then
L (f ) =
Z
∞
e−st cos2 ωt dt
Z0 ∞
1 −st
e (1 + cos 2ωt)dt
2
0
Z
1
1 ∞ −st
=
+
e cos 2ωt dt.
2s 2 0
=
Let
Z
I=
∞
e−st cos 2ωt dt
0
Then
∞
Z ∞
1 −st
2ω
I = − e cos 2ωt −
e−st sin 2ωt dt
s
s 0
t=0
∞
1
4ω 2
2ω
= + 2 e−st sin 2ωt − 2 I
s
s
s
t=0
2
1 4ω
= − 2 I.
s
s
Thus,
Z
∞
e−st cos 2ωt dt =
0
so
s2
s
+ 4ω 2
s
1
s2 + 2ω 2
+
L (f ) =
= 3
.
2s 2s2 + 8ω 2
s + 4sω 2
6. If f (t) = e−t sinh 4t then
Z
L (f ) =
=
=
=
=
=
∞
e−(s+1)t sinh 4t dt
0
Z
1 ∞ −(s+1)t 4t
e − e−4t dt
e
2 0
Z
1 ∞ −(s−3)t
e
− e−(s+5)t dt
2 0
∞ 1
1 −(s+5)t 1 −(s−3)t
e
+
e
−
2
s−3
s+5
t=0
1
1
−
2s − 6 2s + 10
16
4
= 2
(2s − 6)(2s + 10)
s + 2s − 15
8. If f (t) = 1.5 sin(3t − π/2) = −1.5 cos 3t then
Z ∞
L (f ) =
−1.5e−st cos 3t dt
0
∞ Z
1 −st
4.5 ∞ −st
= 1.5
e cos 3t
e sin 3t dt
+
s
s 0
0
∞ 1.5 4.5
1 −st
9
=−
+
− e sin 3t
− 2 L (f )
s
s
s
s
0
1.5
9
=−
− 2 L (f )
s
s
Thus,
L (f ) =
10.
f (t) =
−1.5s
s2 + 9
k, 0 ≤ t ≤ c
0, t > c
c
Z
L (f ) =
ke−st dt
0
k
=
1 − e−cs
s
12.

 t, 0 ≤ t ≤ 1
1, 1 ≤ t ≤ 2
f (t) =

0, t > 2
L (f ) =
Z
1
Z
2
e−st dt
0
1
2
1
t −st
1 −st 1 −st = − e − 2e
− se s
s
0
1
1
1 −s 1 −2s
= 2 − 2e − e
s
s
s
13.
te
−st
dt +

 1, 0 ≤ t ≤ 1
−1, 1 ≤ t ≤ 2
f (t) =

0, t > 2
L (f ) =
1
Z
e
−st
Z
dt −
0
2
e−st dt
1
1
=
1 − 2e−s + e−2s
s
1
= (1 − e−s )2
s
14.
f (t) =
k, a ≤ t ≤ b
0, 0 < t < a, t > b
L (f ) =
Z
b
ke−st
a
k −as
=
e
− e−bs
s
16.


t,
0≤t≤1
2 − t, 1 ≤ t ≤ 2
f (t) =

0,
t>2
L (f ) =
Z
1
Z
2
dt +
(2 − t)e−st dt
0
1
1 2
t−2
t
1
1
−st −st = − − 2 e +
+ 2 e s s
s
s
0
1
−s
−2s
−s 2
1 − 2e + e
1−e
=
=
s2
s
26.
te
−st
5s + 1
s2 − 25
5s + 1
−1
−1
L (F ) = L
s2 − 25
1 −1
s
5
−1
= 5L
+ L
s2 − 25
5
s2 − 25
1
= 5 cosh 5t + sinh 5t.
5
F (s) =
28.
F (s) =
L
−1
1
√
2)(s − 3)
1
1
1
−1
√ L
√ −
√
(F ) = √
2+ 3
s − 3 s − (− 2)
1
1
1
−1
−1
√
√
√
= √
L
−L
2+ 3
s− 3
s − (− 2)
√
=
30.
(s +
√
e
√
− e− 2t
√
√
2+ 3
3t
4s + 32
s
4
=4 2
+8 2
2
s − 16
s − 16
s − 16
4t
−4t
8e4t − 8e−4t
4e + 4e
L −1 (F ) = 4 cosh 4t + 8 sinh 4t =
+
= 6e4t − 2e−4t
2
2
F (s) =
32.
1
1
F (s) =
=
(s + a)(s + b)
b−a
L
−1
1
1
−
s+a s+b
1
1
1
−1
(F ) =
L
−
b−a
s − (−a) s − (−b)
e−at − e−bt
=
b−a
Section 6.2:
1.
y 0 + 5.2y = 19.4 sin 2t,
y(0) = 0
Let L (y) = Y then the ODE becomes
sY − y(0) + 5.2Y =
38.8
s2 + 4
38.8
sY + 5.2Y = 2
s +4
38.8
1
s
5.2
Y = 2
= 1.25
−
+
(s + 4)(s + 5.2)
s + 5.2 s2 + 4 s2 + 4
Taking the inverse Laplace transform of Y we find
y = 1.25e−5.2t − 1.25 cos 2t + 3.25 sin 2t.
3.
y 00 − y 0 − 6y = 0,
y(0) = 11, y 0 (0) = 28
Let L (y) = Y then the ODE becomes
s2 Y − sy(0) − y 0 (0) − sY + y(0) − 6Y = 0
s2 Y − 11s − 28 − sY + 11 − 6Y = 0
10
1
11s + 17
=
+
Y = 2
s −s−6
s−3 s+2
Taking the inverse Laplace transform of Y we find
y = 10e3t + e−2t .
4.
y 00 + 9y = 10e−t ,
y(0) = 0, y 0 (0) = 0
Let L (y) = Y then the ODE becomes
s2 Y − sy(0) − y 0 (0) + 9Y =
s2 Y + 9Y =
10
s+1
1
s+1
1
(s + 1)(s2 + 9)
1
1
s−1
=
−
10 s + 1 s2 + 9
Y =
Taking the inverse Laplace transform of Y we find
y=
1 −t
1
1
e −
cos 3t +
sin 3t.
10
10
30
7.
y 00 + 7y 0 + 12y = 21e3t ,
y(0) = 3.5, y 0 (0) = −10
Let L (y) = Y then the ODE becomes
s2 Y − sy(0) − y 0 (0) + 7sY − 7y(0) + 12Y =
s2 Y − 3.5s + 10 + 7sY − 24.5 + 12Y =
7s2 + 8s − 45
2(s − 3)(s + 3)(s + 4)
1
1
1
5
=
+
+
2 s−3 s+3 s+4
Y =
Taking the inverse Laplace transform of Y we find
1
1
5
y = e3t + e−3t + e−4t .
2
2
2
21
s−3
21
s−3
11.
y 00 + 3y 0 + 2.25y = 9t3 + 64,
y(0) = 1, y 0 (0) = 31.5
Let L (y) = Y then the ODE becomes
54 64
+
s4
s
54 64
s2 Y − s − 31.5 + 3sY − 3 + 2.25Y = 4 +
s
s
s2 Y − sy(0) − y 0 (0) + 3sY − 3y(0) + 2.25Y =
Y =
s5 + 34.5s4 + 64s3 + 54
1
32 32 24
1
+
= 2 − 3 + 4 +
4
2
s (s + 1.5)
s
s
s
s + 1.5 (s + 1.5)2
Taking the inverse Laplace transform of Y we find
y = 4t3 − 16t2 + 32t + (1 + t)e−1.5t .
15.
y 00 + 3y 0 − 4y = 6e2t−3 ,
y(1.5) = 4, y 0 (1.5) = 5
Set t = t̃ + 1.5. Then the problem is
ỹ 00 + 3ỹ 0 − 4ỹ = 6e2t̃ ,
ỹ(0) = 4, ỹ 0 (0) = 5
Let L (ỹ) = Ỹ then the ODE becomes
s2 Ỹ − sỹ(0) − ỹ 0 (0) + 3sỸ − 3ỹ(0) − 4Ỹ =
s2 Ỹ − 4s − 5 + 3sỸ − 12 − 4Ỹ =
4s2 + 9s − 28
(s + 4)(s − 1)(s − 2)
3
1
=
+
s−1 s−2
Ỹ =
Taking the inverse Laplace transform of Ỹ we find
ỹ = 3et̃ + e2t̃
Now replacing t̃ with t − 1.5 we get the solution
y = 3et−1.5 + e2t−3
6
s−2
6
s−2
Section 6.3:
3.
t − 2,
t>2
This function can be represented as (t − 2)u(t − 2). Its transform is given by
e−2s
L ((t − 2)u(t − 2)) = e L (t) = 2 .
s
2s
7.
e−πt ,
2<t<4
This function can be represented as f (t) = e−πt (u(t − 2) − u(t − 4)). Its trans-
form is given by
L (f ) = L (e−πt (u(t − 2) − u(t − 4)))
= L (e−πt u(t − 2)) − L (e−πt u(t − 4))
= e−2s L (e−π(t+2) ) − e−4s L (e−π(t+4) )
e−2π−2s − e−4π−4s
=
s+π
14.
L (f ) =
Since
4 (e−2s − 2e−5s )
s
8
4
has inverse 4 and − has inverse −8, by the second shifting theorem
s
s
f (t) = 4u(t − 2) − 8u(t − 5).
16.
2 (e−s − e−3s )
L (f ) =
s2 − 4
Since
s2
2
has inverse sinh 2t, by the second shifting theorem
−4
f (t) = sinh 2(t − 1)u(t − 1) − sinh 2(t − 3)u(t − 3)
Section 6.5:
1.
Z
1∗1=
t
dτ = t.
0
3.
t
e ∗e
−t
Z
t
eτ e−t+τ dτ
0
Z t
−t
e2τ dτ
=e
0
t !
1 2τ −t
=e
e
2 0
=
=
et − e−t
= sinh t
2
5.
Z
t
(sin ωτ )(cos ω(t − τ ))dτ
sin ωt ∗ cos ωt =
0
Z
t
sin ωτ (cos ωt cos ωτ + sin ωt sin ωτ )dτ
Z t
Z t
= cos ωt
sin ωτ cos ωτ dτ + sin ωt
sin2 ωτ dτ
0
Z t
Z t 0
1
1
2
= sin t + sin ωt
dτ −
cos 2ωτ dτ
ω
2
0
0
1
1
1
= cos ωt sin2 ωt + t sin ωt −
sin ωt sin 2ωt
ω
2
2ω
1
= t sin ωt
2
=
0
7.
t
Z
t
τ et−τ dτ
0
Z t
t
=e
τ e−τ dτ
0
t
−τ
−τ t
= e −τ e − e 0
t∗e =
= et − t − 1
9. We see the given equation can be written as a convolution, y − 1 ∗ y = 1.
Writing Y = L (y) and applying the convolution theorem we obtain,
Y (s) −
1
Y (s)
= .
s
s
The solution is
Y (s) =
1
s−1
and gives the answer
y(t) = et .
11. We see the given equation can be written as a convolution, y+y∗t = 1. Writing
Y = L (y) and applying the convolution theorem we obtain,
Y (s) −
1
Y (s)
= .
2
s
s
The solution is
Y (s) =
s2
s
+1
and gives the answer
y(t) = cos t.
13. We see the given equation can be written as a convolution, y + y ∗ 2et = tet .
Writing Y = L (y) and applying the convolution theorem we obtain,
Y (s) +
1
2Y
=
s−1
(s − 1)2
The solution is
Y (s) =
17.
L (f ) =
s2
1
−1
and gives the answer
y(t) = sinh t.
5.5
= 5.5L (e−1.5t )L (e4t ) = L (5.5e−1.5t ∗ e4t )
(s + 1.5)(s − 4)
f (t) = 5.5e−1.5t ∗ e4t
Z t
= 5.5
e−1.5τ e4(t−τ ) dτ
0
Z t
4t
= 5.5e
e−5.5τ dτ
0
= e (1 − e−5.5t
= e4t − e−1.5t
4t
19.
L (f ) =
2πs
= 2L (cos πt)L (sin πt) = L (2 cos πt ∗ sin πt)
(s2 + π 2 )2
f (t) = 2 cos πt ∗ sin πt
Z t
= 2 (cos πτ )(sin(π(t − τ )))dτ
0
Z t
Z t
2
= 2 sin πt (cos πτ ) dτ − 2 cos πt (sin πτ )(cos πτ ) dτ
0
0
Z t
1
(1 + cos 2πτ ) dτ − (sin2 πt)(cos πt)
= sin πt
π
0
1
1
= sin πt t +
sin 2πt − (sin2 πt)(cos πt)
2π
π
1
1
= sin πt t + (sin πt)(cos πt) − (sin2 πt)(cos πt)
π
π
= t sin πt
21.
L (f ) =
ω
= L (t)L (sin ωt) = L (t ∗ sin ωt)
+ ω2)
s2 (s2
f (t) = t ∗ sin ωt = sin ωt ∗ t
Z t
(sin ωτ )(t − τ )dτ
=
0
Z t
Z t
τ sin ωτ dτ
sin ωτ dτ −
=t
0
0
t !
t !
t
τ
1
= − cos ωτ +
cos ωτ − 2 sin ωτ ω
ω
ω
0
0
t
t
1
t
cos ωt + + cos ωt − 2 sin ωt
ω
ω ω
ω
t
1
= − 2 sin ωt
ω ω
ωt − sin ωt
=
ω2
=−
23.
L (f ) =
40.5
= 13.5L (1)L (sinh 3t) = L (13.5 ∗ sinh 3t)
s(s2 − 9)
f (t) = 13.5 ∗ sinh 3t = sinh t ∗ 13.5
Z t
sinh 3τ dτ
= 13.5
0
= 4.5 cosh 3τ |t0
= 4.5(cosh 3t − 1)
25.
L (f ) =
18s
= 3L (cos 6t)L (sin 6t) = L (3 cos 6t ∗ sin 6t)
(s2 + 36)2
f (t) = 3 cos 6t ∗ sin 6t
Z t
= 3 (cos 6τ )(sin(6(t − τ )))dτ
0
Z t
Z t
2
sin 6τ cos 6τ dτ
cos 6τ dτ − 3 cos 6t
= 3 sin 6t
0
0
Z t
t 3
1
2
= sin 6t
(1 + cos 12τ ) dτ − cos 6t sin 6τ 0
2
4
0
1
1
3
sin 12t − sin2 6t cos 6t
= sin 6t t +
2
12
4
3
1 2
1
= t sin 6t + sin 6t cos 6t − sin2 6t cos 6t
2
4
4
3
= t sin 6t
2
26. 17:
L (f ) =
5.5
1
1
=
−
(s + 1.5)(s − 4)
s − 4 s + 1.5
f (t) = e4t − e−1.5t
21:
23:
ω
1 1
1
L (f ) = 2 2
=
−
s (s + ω 2 )
ω s2 s2 + ω 2
1
1
ωt − sin ωt
f (t) =
t − sin ωt =
ω
ω
ω2
40.5
L (f ) =
= 2.25
s(s2 − 9)
1
1
2
+
−
s+3 s−3 s
f (t) = 2.25 e−3t + e3t − 2 = 4.5(cosh 3t − 1)