MATH 215/255
Fall 2015
Assignment 6
Due date: November 06, 2015
1. Let F (s) =
Answer.
1
, find L−1 [F (s)].
(2s − 1)(2s − 3)
Notice that
F (s) =
The partial fraction of
1
1
1
= ·
.
1
(2s − 1)(2s − 3)
4 (s − 2 )(s − 32 )
1
(s −
1
2 )(s
− 32 )
is:
1
(s −
1
2 )(s
−
3
2)
=
1
s−
3
2
−
1
.
s − 12
By the First shifting property, we know that
i
t
1 h 3t
L−1 [F (s)](t) =
e2 − e2 .
4
2. Use the Laplace transform to solve the initial-value problem y 00 + y = t, y(0) = 2 and
y 0 (0) = 2.
Answer.
Let Y (s) = L[y(t)](s), apply the Laplace transform on the both sides of
00
3
y + y = t , then
L[y 00 ] + L[y] = L[t].
Since y(0) = 2 and y 0 (0) = 0, by the transform of derivatives, we have
L[y 00 ] = s2 L[y] − sy(0) − sy 0 (0)
= s2 Y (s) − 2s − 2.
So we get
s2 Y (s) − 2s − 2 + Y (s) =
1
.
s2
So we get
1
1
1
2s
2
Y (s) = 2
+ 2s + 2 = 2 2
+ 2
+ 2
.
2
s +1 s
s (s + 1) s + 1 s + 1
For
1
s2 (s2
+ 1)
, we have
1
1
1
= 2− 2
.
s2 (s2 + 1)
s
s +1
Hence we know that
Y (s) =
1
1
2s
2
1
2s
1
+ 2
+ 2
= 2+ 2
+ 2
.
− 2
2
s
s +1 s +1 s +1
s
s +1 s +1
So we get
y(t) = L−1 [Y (s)](t)
s
1
−1 1
−1
−1
= L
+ 2L
+L
s2
s2 + 1
s2 + 1
= t + 2 cos(t) + sin(t).
3. Let F (s) =
Answer.
1
, find L−1 [F (s)].
s(s + 1)2
By the transform of integral, we know that
Z t
1
−1
−1
(τ ) dτ
L [F (s)](t) =
L
(s + 1)2
0
Z t
e−τ τ By the First shifting property
=
0
= −te−t − e−t + 1.
4. Let F (s) =
Answer.
3s
, find L−1 [F (s)].
(s + 1)4
Notice that
F (s) =
3s
3(s + 1 − 1)
3(s + 1) − 3
3
3
=
=
=
−
.
4
4
4
3
(s + 1)
(s + 1)
(s + 1)
(s + 1)
(s + 1)4
So we get
−1
L
1
1
−1
[F (s)](t) = 3L
− 3L
(s + 1)3
(s + 1)4
t2
t3
= 3e−t · − 3e−t ·
2!2
3!
3
t
t
= 3e−t
.
−
2
6
−1
5. Find the Laplace transform of the solution to the problem:
x00 + 2x0 + x = f (t),
x(0) = x0 (0) = 0,
where
f (t) =
2, if 0 ≤ t ≤ 2,
t, if t > 2.
Answer.
Let X(s) = L[x(t)], apply the Laplace transform on the both sides of
00
0
x + 2x + x = f (t), then
L[x00 ] + 2L[x0 ] + L[x] = L[f (t)].
2
Since x(0) = x0 (0) = 0, by the transform of derivatives, then
L[x00 ] = s2 L[x] − sx(0) − x0 (0) = s2 X(s),
and L[x0 ] = sL[x] − x(0) = sX(s).
So we get
s2 X(s) + 2sX(s) + X(s) = L[f (t)](s).
By the definition of f (t), we know that
f (t) = 2[u(t − 0) − u(t − 2)] + t[u(t − 2) − u(t − ∞)]
= 2[1 − u(t − 2)] + tu(t − 2)
= 2 + (t − 2)u(t − 2).
So we get
L[f (t)](s) = 2L[1] + L[(t − 2)u(t − 2)]
2
=
+ e−2s L[t](s) By the second shifting property
s
2 e−2s
=
+ 2 .
s
s
So we get
(s + 1)2 X(s) =
2 e−2s
+ 2 .
s
s
So we get
L[x(t)](s) = X(s) =
1
2 e−2s
+
.
(s + 1)2 s
s2

 1, if 0 ≤ t < 1,
t2 , if 1 ≤ t < 2,
6. Find L[f (t)], where f (t) =

4, if t ≥ 2.
Answer.
For f (t), we have
f (t) = 1 · [u(t − 0) − u(t − 1)] + t2 [u(t − 1) − u(t − 2)] + 4[u(t − 2) − u(t − ∞)]
= 1 − u(t − 1) + t2 u(t − 1) − t2 u(t − 2) + 4u(t − 2)
= 1 + (t2 − 1)u(t − 1) − (t2 − 4)u(t − 2).
By the second shifting property, we have
L[f (t)](s) = L[1] + L[(t2 − 1)u(t − 1)] − L[(t2 − 4)u(t − 2)]
1
=
+ e−s L[(t + 1)2 − 1] − e−2s L[(t + 2)2 − 4]
s
1
=
+ e−s L[t2 + 2t] − e−2s L[t2 + 4t]
s
1
2
4
−s 2
−2s 2
=
+e
+
−e
+
.
s
s3 s2
s3 s2
3
7. Use the Laplace transform to solve the initial-value problem: x00 + 4x0 + 5x = 1 and
x(0) = x0 (0) = 0.
Answer.
Let X(s) = L[x](s), apply the Laplace transform on the both sides of
x00 + 4x0 + 5x = t, then
L[x00 ] + 4L[x0 ] + 5L[x] = L[1].
Since x(0) = x0 (0) = 0, by the transform of derivatives, then
L[x00 ] = s2 L[x] − sx(0) − x0 (0) = s2 X(s),
and L[x0 ] = sL[x] − x(0) = sX(s).
So we get
1
s2 X(s) + 4sX(s) + 5X(s) = .
s
That is,
X(s) =
s(s2
1
.
+ 4s + 5)
Since s2 + 4s + 5 = (s + 2)2 + 1, let the partial fraction of
s(s2
s(s2
1
be
+ 4s + 5)
A
Bs + C
1
= + 2
.
+ 4s + 5)
s
s + 4s + 5
So we get
1
A= ,
5
1
B=− ,
5
4
and C = − .
5
So we get
X(s) =
=
=
=
1
s(s2 + 4s + 5)
1 1 1
s+4
· − · 2
5 s 5 s + 4s + 5
1 1 1
s+2+2
· − ·
5 s 5 (s + 2)2 + 1
1 1 1
s+2
2
1
· − ·
− ·
.
5 s 5 (s + 2)2 + 1 5 (s + 2)2 + 1
By the first shifting property, we have
x(t) = L−1 [X(s)](t)
1 −1 1
1 −1
s+2
2 −1
1
=
L
− L
− L
5
s
5
(s + 2)2 + 1
5
(s + 2)2 + 1
−2t
−2t
1 e
s
2e
1
=
−
L−1 2
−
L−1 2
5
5
s +1
5
s +1
1 1 −2t
2
=
− e cos(t) − e−2t sin(t).
5 5
5
4
−1
8. Find L
Answer.
9. Let F (s) =
s
.
(s2 + 1)2
By the transform of convolution, then
1
s
s
−1
−1
−1
= L
∗L
L
(s2 + 1)2
s2 + 1
s2 + 1
Z t
sin(τ ) cos(t − τ ) dτ
=
0
Z τ
1
=
[sin(t − sin(2τ − t)] dτ
2 0
t
1
1
sin(t)τ − cos(2τ − t) =
2
2
0
1
1
1
=
t sin(t) − cos(t) + cos(t)
2
2
2
1
t sin(t).
=
2
s
, find L−1 [F (s)].
(s + 1)(s2 + 1)
Answer.
By the transform of convolution, then
s
L−1 [F (s)](t) = L−1
(s + 1)(s2 + 1)
1
s
−1
−1
= L
∗L
(s + 1)
s2 + 1
Z t
=
e−τ cos(t − τ ) dτ By the First shifting property
0
=
1
[sin(t) + cos(t) − e−t ].
2
1
10. (a) Use the transform of convolution to find L
.
(s2 + 1)2
Answer. By the transform of convolution, then
1
1
1
−1
−1
−1
L
= L
∗L
(s2 + 1)2
s2 + 1
s2 + 1
Z t
=
sin(τ ) sin(t − τ ) dτ
0
Z
1 t
=
[cos(t − 2τ ) − cos(t)] dτ
2 0
Z
1 t
=
[cos(2τ − t) − cos(t)] dτ
2 0
t
1 1
=
sin(2τ − t) − cos(t)τ 2 2
0
1
=
[sin(t) − t cos(t)]
2
−1
5
(b) By differentiating under the integral sign, one can get the differentiation property
of the transform:
d
L[−tf (t)](s) = L[f (t)](s).
ds
1
−1
Use the differentiation property of the Laplace transform to find L
.
(s2 + 1)2
d
s
1 − s2
1
2
Hint:
=
=− 2
.
+
ds s2 + 1
(s2 + 1)2
s + 1 (s2 + 1)2
Answer. Notice that
s
−1
L
(t) = cos(t),
s2 + 1
d
ds
and
s
s2 + 1
=
1 − s2
1
2
=− 2
+
.
(s2 + 1)2
s + 1 (s2 + 1)2
Then
1
1
−1
(t)
+
2L
(t)
s2 + 1
(s2 + 1)2
1
−1
= − sin(t) + 2L
(t).
(s2 + 1)2
−t cos(t) = −L−1
So we know that
L
−1
1
(t) =
(s2 + 1)2
6
1
[−t cos(t) + sin(t)].
2