F18XD2
Solutions 1: The Laplace Transform
1.1.
(i)
Z
L{c} =
∞
−st
ce
dt = c
0
∞
Z
−st
e
0
1
dt = lim − e−st
b→∞
s
b
=
0
c
,
s
(ii) Use integration by parts with u = t, v = e−st :
(
b Z b −st )
Z ∞
e
1
1
dt = 2 ,
L{t} =
−t e−st +
te−st dt = lim
b→∞
s
s
s
0
0
0
(iii)
kt
L{e } =
Z
∞
kt −st
e e
dt =
0
Z
∞
Z
∞
−(s−k)t
e
0
e−(s−k)t
dt = lim −
b→∞
s−k
b
0
=
1
,
s−k
(iv)
Z
iat
L{e } =
∞
iat −st
e e
dt =
0
e
0
(v)
ct
−(s−ia)t
L{at + be } =
Z
−(s−ia)t b
1
e
=
,
dt = lim −
b→∞
s − ia 0 s − ia
∞
−st
ate
ct −st
+ be e
dt = a
0
Z
∞
−st
te
0
dt + b
Z
∞
ect e−st dt .
0
We can then use the results from (ii) and (iii) to get
Z ∞
Z ∞
b
a
ct
−st
.
L{at + be } = a
te dt + b
ect e−st dt = 2 +
s
s−c
0
0
(vi) Since
10t
L{e
sin 2t} =
Z
∞
10t
e
−st
sin 10t e
0
dt =
Z
∞
sin 10t e−(s−10)t ,
0
to get the solution it suffices to compute the integral
Z ∞
sin 2t e−pt dt ,
0
and replace p with s − 10.
We integrate by parts with u = sin 2t, v ′ = e−pt , u′ = 2 cos 2t, v = −e−pt /p
∞
Z ∞
Z
e−pt
2 ∞
−pt
sin at e dt = sin 2t
+
cos 2t e−pt dt
−p 0
p 0
0
We integrate the last term in (S1.1) by parts again (u = cos 2t, v ′ = e−pt )
∞
Z ∞
Z
e−pt
2 ∞
−pt
cos 2t e dt = cos 2t
−
sin 2t e−pt dt .
−p
p
0
0
0
1
(S1.1)
We substitute the above in (S1.1) and get
∞
Z ∞
∞
2 e−pt
−pt
− 2 cos 2t e−pt 0
sin 2t e dt = sin 2t
p
0
Z −p 0
4 ∞
− 2
sin 2t e−pt dt .
p 0
We take the last term to the right-hand side and get
∞
Z
2 e−pt
p2 + 4 ∞
−pt ∞
−pt
−
.
cos
2t
e
sin
2t
e
dt
=
sin
2t
0
p2
−p 0
p2
0
By computing the limits we obtain
∞
e−pt
sin 2t
= 0−0 = 0,
−p 0
and
Therefore
and hence
cos 2t e−pt
p2 + 4
p2
Z
∞
0
Z
= 0 − cos 0 e−0 = −1 .
∞
sin 2t e−pt dt =
0
∞
sin 2t e−pt dt =
0
p2
2
,
p2
a
.
+4
Finally, to get the solution we substitute p by s − 10,
Z ∞
2
2
= 2
.
sin 2t e−(s−10)t dt =
2
(s − 10) + 4
s − 20s + 104
0
1.2.
(a)
L[5 − 3t] = L[5] − 3L[t] =
5
3
− 2.
s s
(b)
L[7t3 − 2 sin 3t] = 7L[t3 ] − 2L[sin 3t] = 42/s4 − 6/(s2 + 9) .
(c)
L[4t e−2t ] = 4L[t]s→s+2 = 4/(s + 2)2 .
(d)
L[t2 e−4t ] = L[t2 ]s→s+4 = 2/(s + 4)3 .
(e)
L[(t + 1)2 ] = L[t2 + 2t + 1] = L[t2 ] + L[2t] + L[1] = 2/s3 + 2/s2 + 1/s .
2
(f)
1
1
L[sin2 t] = L[1 − cos 2t] = (L[1] − L[cos 2t]) = 1/2(1/s − s/(s2 + 4)) .
2
2
1.3.
at
L{e f (t)} =
Z
∞
at
−st
e f (t)e
0
dt =
Z
∞
f (t)e−(s−a)t dt = L[f (t)]s→s−a .
0
1.4.
L{f (t)} = F (s) =
Z
∞
f (t)e−st dt .
0
We differentiate the above with respect to s and get (note that we treat t as a constant
when we differentiate with respect to s)
Z ∞
dF (s)
=
−tf (t)e−st dt = −L{tf (t)} .
ds
0
1.5.
(a) From Problem 1.1 (iv)
L[eiat ] =
1
.
s − ia
Also
eiat = cos at + i sin at .
Therefore, using the linearity of the Laplace transform,
Also
L{eiat } = L{cos at + i sin at} = L{cos at} + iL{sin at} .
(S1.2)
1
s + ia
s + ia
s
a
=
= 2
= 2
+i 2
2
2
s − ia
(s − ia)(s + ia)
s +a
s +a
s + a2
(S1.3)
Therefore by comparing the real and imaginary parts of (S1.2) and (S1.3) we find
the desired Laplace transforms.
(b) From Problem 1.1 (ii) and the first shift theorem
L[teiat ] =
1
(s − ia)2
Also
teiat = t cos at + it sin at
3
Therefore, using the linearity of the Laplace transform
L{teiat } = L{t cos at + it sin at} = L{t cos at} + iL{t sin at} .
(S1.4)
Similarly to part (a)
(s + ia)2
(s + ia)2
1
=
=
(s − ia)2
(s − ia)2 (s + ia)2
(s2 + a2 )2
.
(S1.5)
(s2 + 2ias − a2 )2
s2 − a2
2as
=
+
i
=
(s2 + a2 )2
(s2 + a2 )2
s2 + a2
Therefore by comparing the real and imaginary parts of (S1.4) and (S1.5) we find
the desired Laplace transforms.
1.6.
1!
. We now assume that the statement is
For n = 1, i.e., f (t) = t we have F (s) = s12 = s1+1
true for n − 1, i.e. that
(n − 1)!
(n − 1)!
L[tn−1 ] = (n−1)+1 =
.
(S1.6)
s
sn
Next, we can compute the Laplace transform by direct integration as
Z ∞
n
L[t ] =
tn e−st dt .
0
After integration by parts with u = tn , u′ = ntn−1 and v ′ = e−st , v = − e s we get
Z ∞
Z ∞
−st ∞
e−st
n −st
ne
t e dt = −t
+
ntn−1
dt.
s 0
s
0
0
i∞
h
n e−st
The first term in the RHS above is −t s
= 0. For the second term we can use
−st
(S1.6) to show that
Z
n
L[t ] =
0
∞
nt
0
n
dt =
s
s
−st
n−1 e
Z
∞
tn−1 e−st dt =
0
n (n − 1)!
n!
=
.
s sn
sn+1
1.7.
(a)
1
1
1
=−
+
.
(s + 3)(s + 7)
4(s + 7) 4(s + 3)
From the table of Laplace transforms
1
f (t) = (e−3t − e−7t ) .
4
4
(b)
s
2
2s + 6
=
2
+
3
s2 + 4
s2 + 4
s2 + 4
From the table of Laplace transforms we get
f (t) = 2 cos 2t + 3 sin 2t .
(c)
4s
1
1
2
=
−
+
2
(s − 1)(s + 1)
s − 1 s + 1 (s + 1)2
so
(d)
f (t) = et − e−t + 2te−t .
1
3
1
3s
=−
+
−
2
(s − 1)(s − 4)
(s − 1) 2 (s − 2) 2 (s + 2)
We then deduce from the table of Laplace transforms that
3
1
f (t) = −et + e2t − e−2t .
2
2
(e) Use partial fractions:
s
(s −
1)2 (s2
+ 4)
=
3
1
1 (−3s − 8) 1
1
+
+
.
2
25 (s − 1) 25 s + 4
5 (s − 1)2
Then, from the table of Laplace transforms and the first shift theorem, we get
f (t) =
(f)
3 t 1 t
4
3
e + te −
cos 2t −
sin 2t .
25
5
25
25
s
s
.
=
s2 + 4 s + 8
(s + 2)2 + 4
Next
L
−1
s
(s + 2)2 + 4
=L
−1
s−2
s2 + 4
s→s+2
From the table of Laplace transforms,
s−2
−1
L
= cos 2t − sin 2t .
s2 + 4
Then, from the first shift theorem, we get
f (t) = e−2t cos 2t − e−2t sin 2t .
5
.