MATH 215/255
Fall 2014
Assignment 6
§6.1, §6.2
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 6.1.8: Find the Laplace transform of cos2 (ωt).
Answer. As follows from trigonometric identities, cos2 (ωt) = 21 (1+cos(2ωt)). Using
the linearity of Laplace transform and by looking up the table,
L{ 21 (1 + cos(2ωt))} = 21 L{1} + 12 L{cos(2ωt)} =
• 6.1.12: Find the Laplace transform of f (t) =
s
1
+ 2
2s 2s + 8ω 2
t if t ≥ 1,
0 if t < 1.
Answer. We rewrite f (t) using Heaviside function, u(t), as f (t) = tu(t − 1). Using
the second shifting property and by looking up the table, we get
L{(t − 1 + 1)u(t − 1)} = e−s L{t + 1} = e−s (s−1 + s−2 )
Alternative solution. Using the definition of Laplace transform and the fact that
f (t) = 0 for t < 1, we get
Z ∞
Z ∞
−ts
L{f (t)} =
e f (t) dt =
e−ts t dt.
0
1
The latter integral can be calculated by parts as
Z ∞
Z
t −ts ∞ 1 ∞ −ts
e−s e−s
−ts
e t dt = − e
e dt =
+ 2 .
+
s
s 1
s
s
1
1
• 6.1.13: Find the inverse Laplace transform of
Answer.
s
.
(s2 + s + 2)(s + 4)
First, we rewrite the fraction as
s
As + B
C
(As + B)(s + 4) + C(s2 + s + 2)
=
+
=
.
(s2 + s + 2)(s + 4)
s2 + s + 2 s + 4
(s2 + s + 2)(s + 4)
Since s = (As + B)(s + 4) + C(s2 + s + 2), we have
(A + C)s2 + (4A + B + C − 1)s + 4B + 2C = 0.
So that A = −C, B = − 12 C, and 4(−C) + (− 21 C) + C = 1, i.e. C = − 27 . It allows to
find A = 72 and B = 17 , so finally
s + 12
s
1
2s + 1
2
2
1
=
−
=
−
.
(s2 + s + 2)(s + 4)
7 s2 + s + 2 s + 4
7 (s + 21 )2 + 74
s+4
Then, by looking up the table, we have
( )
1
√
s
+
2
1
2 −1t
2
2
2 cos( 7 t) − e−4t .
L−1
e
−
=
2
1
7
2
7 (s + 2 ) + 4
s+4
7
7
• 6.1.102: Find the inverse Laplace transform of
Answer.
8
.
+ 2)
s3 (s
First, we rewrite the fraction as
8
A
B
C
D
A(s + 2) + Bs(s + 2) + Cs2 (s + 2) + Ds3
=
+
+
+
=
,
s3 (s + 2)
s3 s2
s
s+2
s3 (s + 2)
so that 8 = (C + D)s3 + (B + 2C)s2 + (A + 2B)s + 2A. The latter gives A = 4,
B = −2, C = 1, and D = −1. The linearity of inverse Laplace transform can be used
to find
n1o
n1o
n4
n2o
n 1 o
2
1
1 o
L−1 3 − 2 + −
= 2L−1 3 − 2L−1 2 + L−1
− L−1
s
s
s s+2
s
s
s
s+2
= 2t2 − 2t + 1 − e−2t .
• 6.1.104: Find the Laplace transform of sin(t)e−t .
Answer.
To deal with the exponent, we can use the shifting property to write
L{e−t sin(t)} = F (s + 1),
where F (s) = 1/(s2 + 1) is the Laplace transform of sin(t). Finally,
L{e−t sin(t)} =
1
1
= 2
.
(s + 1)2 + 1
s + 2s + 2
• 6.2.2: Using the Heaviside function write down the piecewise function that is 0 for
t < 0, t2 for t in [0, 1], and t for t > 1.
Answer. To make any function f (t) zero outside of any interval a < t < b we need to
multiply it by u(t − a) and u(b − t). If either a or b is absent, then only one Heaviside
function should be used. Using this property, we can write
f (t) = 0u(−t) + t2 u(t − 0)u(1 − t) + tu(t − 1) = t2 u(t)u(1 − t) + tu(t − 1).
Alternative answer.
The function that is equal to 1 on an interval [a, b] and zero
outside may be written as u(t − a) − u(t − b), and the function that is equal to 1 on
the interval [a, ∞) and zero outside is u(t − a). Multiplying t2 or t by cutoff functions
of this kind, we can match the piecewise definition of the desired function:
f (t) = t2 u(t) − u(t − 1) + tu(t − 1).
2
• 6.2.6: Solve x00 + x = u(t − 1) for initial conditions x(0) = 0 and x0 (0) = 0.
Answer.
Application of Laplace transform to this differential equation gives
s2 X(s) − sx(0) − x0 (0) + X(s) =
e−s
.
s
Substituting the initial conditions yields to
X=
e−s
s −s 1
=
e
−
.
s(s2 + 1)
s 1 + s2
The second shifting property can be used to find
n 1
s o
x = L−1 e−s
−
= u(t−1)(1 − cos(t−1)).
s 1 + s2
• simplified 6.2.8: Solve x000 + x = 0 for initial conditions x(0) = 1, x0 (0) = 0, and
x00 (0) = 0.
Answer.
Application of Laplace transform to this differential equation gives
s3 X(s) − s2 x(0) − sx0 (0) − x00 (0) + X(s) = 0.
Substituting the initial conditions, we get
X(s) =
s2
.
1 + s3
We rewrite the fraction as
s2
A
Bs + C
(A + B)s2 + (B + C − A)s + A + C
=
+
=
.
1 + s3
s + 1 s2 − s + 1
1 + s3
So that C = −A, B = 1 − A, 1 − 3A = 0, and finally
s − 21
s2
1 1
X(s) =
=
+2
.
1 + s3
3 s+1
(s − 12 )2 + 43
The application of inverse Laplace transform gives
√3 1
1 −t
t
x =
e + 2e 2 cos
t .
3
2
• 6.2.9: Show the second shifting property: L{f (t − a)u(t − a)} = e−as L{f (t)}.
Answer.
By definition of Laplace transform,
Z ∞
Z
−ts
L{f (t − a)u(t − a)} =
e f (t − a)u(t − a) dt =
0
∞
e−ts f (t − a) dt.
a
By introducing τ = t − a, the last integral can be simplified as
Z ∞
Z ∞
e−ts f (t − a) dt =
e−(a+τ )s f (τ ) dτ = e−as L{f (τ )}.
a
Note that
0
Z
L{f (τ )} =
∞
−τ s
e
Z
f (τ )dτ =
0
0
3
∞
e−ts f (t)dt = L{f (t)}.
• 6.2.102: Solve x00 − x = (t2 − 1)u(t − 1) for initial conditions x(0) = 1, x0 (0) = 2
using the Laplace transform.
Answer.
Application of Laplace transform to differential equation gives
n
h
io
2
2
s2 X(s) − sx(0) − x0 (0) − X(s) = L u(t − 1) (t − 1)2 + 2(t − 1) = e−s 3 + 2 .
s
s
Substitution of the initial conditions leads to
e−s 2
2
s+2
X(s) = 2
+
+ 2
.
3
2
s −1 s
s
s −1
The first fraction can be written as
2
A
B
C
D E
1 2
+
=
+
+ 3+ 2+
2
3
2
s −1 s
s
s−1 s+1 s
s
s
3
3
A(s + 1)s + B(s − 1)s + C(s2 − 1) + Ds(s2 − 1) + Es2 (s2 − 1)
,
=
(s2 − 1)s3
so that
A+B+E = 0
A−B+D = 0
C −E = 0
−D = 2
−C = 2.
The solution of the system is A = 2, B = 0, C = −2, D = −2, and E = −2. The
second fraction can be simplified as
s+2
a1
a2
(a1 + a2 )s + a1 − a2
=
+
=
,
2
s −1
s−1 s+1
s2 − 1
so that a1 = 3/2, and a2 = −1/2. Note that
e−s
2
s−1
= e−s L{2et }
= L{2et−1 u(t − 1)}
by the second shifting property. Similarly,
e−s
2
s3
= e−s L{t2 }
= L{(t − 1)2 u(t − 1)}.
The solution is
)
2
2
2
2
3/2
1/2
−
−
−
+
−
x = L
e
s − 1 s3 s2
s
s−1 s+1
3
1
= u(t − 1) 2et−1 − (t − 1)2 − 2(t − 1) − 2 + et − e−t
2
2
3
1
= u(t − 1) 2et−1 − t2 − 1 + et − e−t .
2
2
(
−1
−s
4