4 The Laplace Transform
EXERCISES 4.1
Definition of the Laplace Transform
1
∞
1 −st 1 −st 1.
{f (t)} =
−e dt +
e dt = e
−se
s
0
1
0
1
2 −s 1
1 −s
1 −s 1
= e − , s>0
= e − − 0− e
s
s
s
s
s
2
2
4
4
2.
{f (t)} =
4e−st dt = − e−st = − (e−2s − 1), s > 0
s
s
0
0
∞
1
1
∞
1 −st
1 −st 1 −st −st
−st
3.
{f (t)} =
te dt +
e dt = − te
− 2e
−se
s
s
0
1
0
1
1
1
1 −s
1 −s
1
−s
−s
− 0 − 2 − (0 − e ) = 2 (1 − e ), s > 0
= − e − 2e
s
s
s
s
s
1
1
2
2
1
4.
{f (t)} =
(2t + 1)e−st dt = − te−st − 2 e−st − e−st s
s
s
0
0
2
1
2
2
1
1
2
= (1 − 3e−s ) + 2 (1 − e−s ),
= − e−s − 2 e−s − e−s − 0 − 2 −
s
s
s
s
s
s
s
π
π
s
1
5.
{f (t)} =
e−st sin t − 2
e−st cos t (sin t)e−st dt = − 2
s +1
s +1
0
0
1
1
1
= 0+ 2
e−πs − 0 − 2
= 2
(e−πs + 1), s > 0
s +1
s +1
s +1
∞
∞
s
1
−st
−st
−st
6.
{f (t)} =
e
e
(cos t)e dt = − 2
cos t + 2
sin t s
+
1
s
+
1
π/2
π/2
1
1
e−πs/2 = − 2
e−πs/2 , s > 0
=0− 0+ 2
s +1
s +1
0, 0 < t < 1
7. f (t) =
t, t > 1
∞
∞
1 −st
1 −st 1 −s
1 −s
−st
{f (t)} =
te
dt = − te
− 2e
= s e + s2 e , s > 0
s
s
1
1
0,
0<t<1
8. f (t) =
2t − 2, t > 1
∞
∞
1
1
2
{f (t)} = 2
(t − 1)e−st dt = 2 − (t − 1)e−st − 2 e−st = 2 e−s , s > 0
s
s
s
1
−st
∞
−st
1
1
198
s>0
4.1
Definition of the Laplace Transform
1 − t, 0 < t < 1
so
0,
t>1
1
1
∞
1
1
1 −st −st
−st
−st
−st
{f (t)} =
(1 − t)e
dt +
0e
dt =
(1 − t)e
dt = − (1 − t)e
+ 2e
s
s
0
1
0
0
1 −s 1
1
= 2e + − 2, s>0
s
s s
9. The function is f (t) =

 0, 0 < t < a
10. f (t) = c, a < t < b ;

0, t > b
11.
{f (t)} =
∞
b
{f (t)} =
e
7
∞
dt = e
{f (t)} =
(1−s)t
e
0
12.
0
∞
−2t−5 −st
e
e
−5
∞
dt = e
{f (t)} =
4t −st
te e
dt =
∞
te
=
15.
{f (t)} =
−
dt =
∞
e−5 −(s+2)t e−5
dt = −
=
e
,
s+2
s+2
0
1
1
e(4−s)t
te(4−s)t −
4−s
(4 − s)2
t2 e−2t e−st dt =
∞
1 2 −(s+2)t
2
2
−
te−(s+2)t −
e−(s+2)t
t e
s+2
(s + 2)2
(s + 2)3
e−t (sin t)e−st dt =
∞
0
−(s + 1) −(s+1)t
1
e
e−(s+1)t cos t
sin t −
(s + 1)2 + 1
(s + 1)2 + 1
1
1
= 2
, s > −1
=
(s + 1)2 + 1
s + 2s + 2
{f (t)} =
∞
et (cos t)e−st dt =
0
∞
1−s
1
e(1−s)t cos t +
e(1−s)t sin t
(1 − s)2 + 1
(1 − s)2 + 1
1−s
s−1
=−
= 2
, s>1
(1 − s)2 + 1
s − 2s + 2
=
17.
{f (t)} =
0
∞
2
=
,
(s + 2)3
0
∞
0
(cos t)e(1−s)t dt
0
∞
∞
(sin t)e−(s+1)t dt
0
=
16.
s > −2
t2 e−(s+2)t dt
0
∞
∞
0
t(cos t)e−st dt
0
=
=
st
s2 − 1
− 2
−
s + 1 (s2 + 1)2
s2 − 1
2 ,
(s2 + 1)
s>1
s>4
0
(4−s)t
0
1
=
,
(4 − s)2
{f (t)} =
∞
0
14.
−(s+2)t
0
∞
b
c −st c −sa
dt = − e
− e−sb ), s > 0
= s (e
s
a
∞
e7
e7 (1−s)t e7
e
=
,
dt =
=
0
−
1−s
1−s
s−1
0
e
0
13.
ce
a
t+7 −st
e
−st
−st
(cos t)e
+
2s
t
+
s2 + 1 (s2 + 1)2
s>0
199
∞
−st
(sin t)e
0
s > −2
4.1
18.
Definition of the Laplace Transform
{f (t)} =
∞
t(sin t)e−st dt
0
t
2s
− 2
−
2
s + 1 (s + 1)2
=
=
2s
2 ,
4!
{2t4 } = 2 5
s
21.
{4t − 10} =
23.
{t2 + 6t − 3} =
25.
{t3 + 3t2 + 3t + 1} =
27.
{1 + e4t } =
29.
{1 + 2e2t + e4t } =
31.
33.
34.
35.
36.
37.
38.
(cos t)e
−
s2 − 1
st
+
2
s + 1 (s2 + 1)2
∞
−st
(sin t)e
0
s>0
(s2 + 1)
19.
−st
4
10
−
s2
s
2
6
3
+ 2−
s3
s
s
3!
2
3
1
+3 3 + 2 +
s4
s
s
s
1
1
+
s s−4
1
2
1
+
+
s s−2 s−4
5!
s6
20.
{t5 } =
22.
{7t + 3} =
24.
2
16 9
{−4t2 + 16t + 9} = −4 3 + 2 +
s
s
s
26.
3!
2
6
1
{8t3 − 12t2 + 6t − 1} = 8 4 − 12 3 + 2 −
s
s
s
s
28.
{t2 − e−9t + 5} =
30.
{e2t − 2 + e−2t } =
3
7
+
s2
s
5
2
1
+
−
s3
s+9 s
1
2
1
− +
s−2 s s+2
2
3
s
2
{4t2 − 5 sin 3t} = 4 3 − 5 2
32.
{cos 5t + sin 2t} = 2
+
s
s +9
s + 25 s2 + 4
1
1
1
1
k
kt
−kt
{sinh kt} =
{e − e } =
−
= 2
2
2 s−k s+k
s − k2
s
s2 − k 2
t
−t
1 2t 1
1
1
t
t e −e
{e sinh t} =
e
=
e −
=
−
2
2
2
2(s − 2) 2s
t
−t
e +e
1
1 1 −2t
1
{e−t cosh t} =
=
e−t
=
+ e
+
2
2 2
2s 2(s + 2)
1
2
{sin 2t cos 2t} =
sin 4t = 2
2
s + 16
1 1
1
1 s
{cos2 t} =
+ cos 2t =
+
2 2
2s 2 s2 + 4
{cosh kt} =
1
2
{ekt + ekt } =
39. From the addition formula for the sine function, sin(4t + 5) = sin 4t cos 5 + cos 4t sin 5 so
{sin(4t + 5)} = (cos 5)
{sin 4t} + (sin 5)
{cos 4t} = (cos 5)
4
s2 + 16
+ (sin 5)
s
s2 + 16
40. From the addition formula for the cosine function,
√
π
3
π
π
1
cos t −
= cos t cos + sin t sin =
cos t + sin t
6
6
6
2
2
so
√
π 3
1
cos t −
=
{cos t} +
{sin t}
6
2
2
√
√
1 1
1 3s + 1
3 s
+
=
.
=
2 s2 + 1 2 s2 + 1
2 s2 + 1
200
=
4 cos 5 + (sin 5)s
.
s2 + 16
4.1
Definition of the Laplace Transform
41. (a) Using integration by parts for α > 0,
∞
∞
∞
α −t
α −t Γ(α + 1) =
t e dt = −t e + α
tα−1 e−t dt = αΓ(α).
0
0
0
(b) Let u = st so that du = s dt. Then
∞
∞
u α 1
1
α
−st α
{t } =
e t dt =
e−u
du = α+1 Γ(α + 1),
s
s
s
0
0
42. (a)
−1/2
{t
Γ(1/2)
} = 1/2 =
s
π
s
{t1/2 } =
(b)
√
π
Γ(3/2)
=
3/2
3/2
s
2s
(c)
α > −1.
{t3/2 } =
√
Γ(5/2)
3 π
=
s5/2
4s5/2
43. Let F (t) = t1/3 . Then F (t) is of exponential order, but f (t) = F (t) = 13 t−2/3 is unbounded near t = 0 and
hence is not of exponential order. Let
2
2
f (t) = 2tet cos et =
2
d
sin et .
dt
This function is not of exponential order, but we can show that its Laplace transform exists. Using integration
by parts we have
∞
a
a
d
t2
t2
−st
t2
−st
t2 −st
t2
{2te cos e } =
dt = lim e
e
sin e + s
e
sin e dt
sin e
a→∞
dt
0
0
0
∞
2
2
= − sin 1 + s
e−st sin et dt = s {sin et } − sin 1.
0
2
Since sin et is continuous and of exponential order,
2
{sin et } exists, and therefore
2
2
{2tet cos et } exists.
44. The relation will be valid when s is greater than the maximum of c1 and c2 .
2
45. Since et is an increasing function and t2 > ln M + ct for M > 0 we have et > eln M +ct = M ect for t sufficiently
2
large and for any c. Thus, et is not of exponential order.
46. Assuming that (c) of Theorem 4.1 is applicable with a complex exponent, we have
{e(a+ib)t } =
1
1
(s − a) + ib
s − a + ib
.
=
=
s − (a + ib)
(s − a) − ib (s − a) + ib
(s − a)2 + b2
By Euler’s formula, eiθ = cos θ + i sin θ, so
{e(a+ib)t } =
=
=
{eat eibt } =
{eat cos bt} + i
{eat (cos bt + i sin bt)}
{eat sin bt}
s−a
b
+i
.
(s − a)2 + b2
(s − a)2 + b2
Equating real and imaginary parts we get
{eat cos bt} =
s−a
(s − a)2 + b2
and
{eat sin bt} =
b
.
(s − a)2 + b2
47. We want f (αx + βy) = αf (x) + βf (y) or
m(αx + βy) + b = α(mx + b) + β(my + b) = m(αx + βy) + (α + β)b
for all real numbers α and β. Taking α = β = 1 we see that b = 2b, so b = 0. Thus, f (x) = mx + b will be a
linear transformation when b = 0.
201
4.1
Definition of the Laplace Transform
{tn−1 } = (n − 1)!/sn . Then, using the definition of the Laplace transform and integration by
48. Assume that
parts, we have
∞
∞
1
n ∞ −st n−1
e−st tn dt = − e−st tn +
e t
dt
s
s 0
0
0
n
n (n − 1)!
n!
=0+
= n+1 .
{tn−1 } =
s
s
sn
s
{tn } =
EXERCISES 4.2
1.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
=
1
s4
1
2
2
s3
=
1 2
t
2
1
1
3!
= t3
6
s4
6
1
1
48
48 4!
=
= t − 2t4
−
−
·
s2
s5
s2
24 s5
2 1
1
5!
2
2
4 3!
1
1 5
=
4· 2 − · 4 +
− 3
· 6 = 4t − t3 +
t
s s
s
6 s
120 s
3
120
3
1
(s + 1)3
3 2
1 3!
1
1
=
= 1 + 3t + t2 + t3
+
+
+
3
·
·
·
4
2
3
4
s
s
s
2 s
6 s
2
6
1
(s + 2)2
2
1
=
+ 4 · 2 + 2 · 3 = 1 + 4t + 2t2
3
s
s
s
s
1
1
1
= t − 1 + e2t
− +
2
s
s s−2
4
1
1
6
1 1 4!
1
+ 5−
=
4· + · 5 −
= 4 + t4 − e−8t
s s
s+8
s 4 s
s+8
4
1
1
1
1
=
= e−t/4
4s + 1
4
s + 1/4
4
1
1
1
1
=
·
= e2t/5
5s − 2
5 s − 2/5
5
5
7
5
5
=
·
= sin 7t
s2 + 49
7 s2 + 49
7
10s
= 10 cos 4t
s2 + 16
4s
s
1
=
= cos t
4s2 + 1
s2 + 1/4
2
1
1
1/2
1
1
=
·
= sin t
4s2 + 1
2 s2 + 1/4
2
2
2.
1
s3
The Inverse Transform and
Transforms of Derivatives
=
202
4.2
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
s
3
−
2
·
= 2 cos 3t − 2 sin 3t
s2 + 9
s2 + 9
√
√ √
√
s+1
2
2
s
1
=
+√ 2
= cos 2t +
sin 2 t
s2 + 2
s2 + 2
s
+
2
2
2
1
1 1 1
1
1 1
=
·
−
·
= − e−3t
s2 + 3s
3 s 3 s+3
3 3
s+1
1 1 5
1
1 5
=
− · + ·
= − + e4t
2
s − 4s
4 s 4 s−4
4 4
s
3
1
1
3
1
1
=
·
+ ·
= et + e−3t
2
s + 2s − 3
4 s−1 4 s+3
4
4
1
1
1
1
1
1
1
=
·
− ·
= e4t − e−5t
s2 + s − 20
9 s−4 9 s+5
9
9
1
1
0.9s
=
(0.3) ·
+ (0.6) ·
= 0.3e0.1t + 0.6e−0.2t
(s − 0.1)(s + 0.2)
s − 0.1
s + 0.2
√ √
√
√
√
3
s−3
s
√
√
− 3· 2
= cosh 3 t − 3 sinh 3 t
=
2
s −3
s −3
(s − 3 )(s + 3 )
s
1
1
1
1
1
1
1
=
·
−
+ ·
= e2t − e3t + e6t
(s − 2)(s − 3)(s − 6)
2 s−2 s−3 2 s−6
2
2
s2 + 1
1 1
1
1
1
5
1
=
· −
− ·
+ ·
s(s − 1)(s + 1)(s − 2)
2 s s−1 3 s+1 6 s−2
1
5
1
= − et − e−t + e2t
2
3
6
√
1
1
1 1 1 s
1 1
5t
=
=
·
−
=
−
cos
s3 + 5s
s(s2 + 5)
5 s 5 s2 + 5
5 5
s
1
s
1
2
1
1
1
1
1
=
·
+
·
−
·
= cos 2t + sin 2t − e−2t
(s2 + 4)(s + 2)
4 s2 + 4 4 s2 + 4 4 s + 2
4
4
4
2s − 4
2s − 4
4
3
s
3
=
=
−
+
+
+
(s2 + s)(s2 + 1)
s(s + 1)(s2 + 1)
s s + 1 s2 + 1 s2 + 1
2s − 6
s2 + 9
The Inverse Transform and Transforms of Derivatives
=
2·
= −4 + 3e−t + cos t + 3 sin t
√
√ √
√
1
3
3
1
1
1
1
√ · 2
28.
=
− √ · 2
= √ sinh 3 t − √ sin 3 t
4
s −9
6 3 s −3 6 3 s +3
6 3
6 3
1
1
1
1
1
1
1
1
2
29.
=
·
−
·
=
·
−
·
(s2 + 1)(s2 + 4)
3 s2 + 1 3 s2 + 4
3 s2 + 1 6 s2 + 4
1
1
= sin t − sin 2t
3
6
6s + 3
s
1
s
1
2
30.
+
−2· 2
− ·
=
2· 2
(s2 + 1)(s2 + 4)
s + 1 s2 + 1
s + 4 2 s2 + 4
1
= 2 cos t + sin t − 2 cos 2t − sin 2t
2
31. The Laplace transform of the initial-value problem is
1
s {y} − y(0) −
{y} = .
s
203
4.2
The Inverse Transform and Transforms of Derivatives
Solving for
{y} we obtain
1
1
{y} = − +
.
s s−1
Thus
y = −1 + et .
32. The Laplace transform of the initial-value problem is
2s
Solving for
{y} − 2y(0) +
{y} = 0.
{y} we obtain
6
3
=
.
2s + 1
s + 1/2
{y} =
Thus
y = 3e−t/2 .
33. The Laplace transform of the initial-value problem is
s
Solving for
{y} − y(0) + 6
{y} =
1
.
s−4
{y} we obtain
{y} =
1
2
1
1
19
1
+
=
·
+
·
.
(s − 4)(s + 6) s + 6
10 s − 4 10 s + 6
Thus
y=
1 4t 19 −6t
e + e .
10
10
34. The Laplace transform of the initial-value problem is
s
Solving for
{y} −
{y} =
2s
s2 + 25
.
{y} we obtain
{y} =
2s
1
1
1
s
5
5
=
·
−
+
· 2
.
2
2
(s − 1)(s + 25)
13 s − 1 13 s + 25 13 s + 25
Thus
y=
1 t
1
5
e −
cos 5t +
sin 5t.
13
13
13
35. The Laplace transform of the initial-value problem is
s2
Solving for
{y} − sy(0) − y (0) + 5 [s
{y} − y(0)] + 4
{y} = 0.
{y} we obtain
{y} =
s+5
4 1
1 1
=
−
.
s2 + 5s + 4
3 s+1 3 s+4
Thus
y=
4 −t 1 −4t
e − e .
3
3
36. The Laplace transform of the initial-value problem is
s2
{y} − sy(0) − y (0) − 4 [s
{y} − y(0)] =
204
3
6
−
.
s−3 s+1
4.2
Solving for
The Inverse Transform and Transforms of Derivatives
{y} we obtain
{y} =
=
6
3
s−5
−
+ 2
2
2
(s − 3)(s − 4s) (s + 1)(s − 4s) s − 4s
5 1
2
3
1
11
1
· −
− ·
+
·
.
2 s s − 3 5 s + 1 10 s − 4
Thus
y=
5
3
11
− 2e3t − e−t + e4t .
2
5
10
37. The Laplace transform of the initial-value problem is
s2
Solving for
{y} − sy(0) +
{y} =
2
.
s2 + 2
{y} we obtain
{y} =
2
10s
10s
2
2
+
= 2
+
−
.
(s2 + 1)(s2 + 2) s2 + 1
s + 1 s2 + 1 s2 + 2
Thus
y = 10 cos t + 2 sin t −
√
√
2 sin 2 t.
38. The Laplace transform of the initial-value problem is
{y} + 9
s2
Solving for
{y} =
1
.
s−1
{y} we obtain
{y} =
1
1
1
1
1
1
s
=
·
−
· 2
−
· 2
.
2
(s − 1)(s + 9)
10 s − 1 10 s + 9 10 s + 9
Thus
y=
1 t
1
1
e −
sin 3t −
cos 3t.
10
30
10
39. The Laplace transform of the initial-value problem is
2 s3
{y} − s2 (0) − sy (0) − y (0) + 3 s2
Solving for
{y} − sy(0) − y (0) − 3[s
{y} − y(0)] − 2
{y} =
1
.
s+1
{y} we obtain
{y} =
2s + 3
1 1
5
1
8
1
1 1
=
+
−
+
.
(s + 1)(s − 1)(2s + 1)(s + 2)
2 s + 1 18 s − 1 9 s + 1/2 9 s + 2
Thus
y=
1 −t
5
8
1
e + et − e−t/2 + e−2t .
2
18
9
9
40. The Laplace transform of the initial-value problem is
s3
{y} − s2 (0) − sy (0) − y (0) + 2 s2
Solving for
{y} − sy(0) − y (0) − [s
{y} − y(0)] − 2
{y} we obtain
{y} =
=
s2 + 12
(s − 1)(s + 1)(s + 2)(s2 + 9)
13 1
13 1
16 1
3
s
1
3
−
+
+
−
.
60 s − 1 20 s + 1 39 s + 2 130 s2 + 9 65 s2 + 9
205
{y} =
3
.
s2 + 9
4.2
The Inverse Transform and Transforms of Derivatives
Thus
y=
13 t 13 −t 16 −2t
1
3
e − e + e
cos 3t −
sin 3t.
+
60
20
39
130
65
41. The Laplace transform of the initial-value problem is
{y} +
s
Solving for
{y} =
s+3
.
s2 + 6s + 13
{y} we obtain
s+3
1
1
1
s+1
= ·
− · 2
2
(s + 1)(s + 6s + 13)
4 s + 1 4 s + 6s + 13
1
1
s+3
2
1
−
−
.
= ·
4 s + 1 4 (s + 3)2 + 4 (s + 3)2 + 4
{y} =
Thus
y=
1 −t 1 −3t
1
e − e
cos 2t + e−3t sin 2t.
4
4
4
42. The Laplace transform of the initial-value problem is
s2
Solving for
{y} − s · 1 − 3 − 2[s
{y} − 1] + 5
{y} = (s2 − 2s + 5)
{y} − s − 1 = 0.
{y} we obtain
{y} =
s+1
s−1+2
s−1
2
=
=
+
.
s2 − 2s + 5
(s − 1)2 + 22
(s − 1)2 + 22
(s − 1)2 + 22
Thus
y = et cos 2t + et sin 2t.
43. (a) Differentiating f (t) = teat we get f (t) = ateat + eat so
{ateat + eat } = s
{teat }, where we have used
f (0) = 0. Writing the equation as
{teat } +
a
and solving for
{eat } = s
{teat }
{teat } we get
{teat } =
1
s−a
{eat } =
1
.
(s − a)2
(b) Starting with f (t) = t sin kt we have
f (t) = kt cos kt + sin kt
f (t) = −k 2 t sin kt + 2k cos kt.
Then
{−k 2 t sin t + 2k cos kt} = s2
{t sin kt}
where we have used f (0) = 0 and f (0) = 0. Writing the above equation as
−k 2
and solving for
{t sin kt} + 2k
{cos kt} = s2
{t sin kt}
{t sin kt} gives
{t sin kt} =
44. Let f1 (t) = 1 and f2 (t) =
2k
s2 + k 2
1,
t ≥ 0, t = 1
0,
t=1
{cos kt} =
. Then
s
2k
2ks
= 2
.
s2 + k 2 s2 + k 2
(s + k 2 )2
{f1 (t)} =
206
{f2 (t)} = 1/s, but f1 (t) = f2 (t).
4.3 Translation Theorems
45. For y − 4y = 6e3t − 3e−t the transfer function is W (s) = 1/(s2 − 4s). The zero-input response is
s−5
5 1 1
1
5 1
y0 (t) =
=
·
−
·
= − e4t ,
s2 − 4s
4 s 4 s−4
4 4
and the zero-state response is
y1 (t) =
=
=
6
3
−
(s − 3)(s2 − 4s) (s + 1)(s2 − 4s)
27
1
2
5 1 3
1
·
−
+ · − ·
20 s − 4 s − 3 4 s 5 s + 1
27 4t
5 3
e − 2e3t + − e−t .
20
4 5
46. From Theorem 4.4, if f and f are continuous and of exponential order,
{f (t)} = sF (s) − f (0). From
{f (t)} = 0 so
Theorem 4.5, lims→∞
lim [sF (s) − f (0)] = 0
and
s→∞
For f (t) = cos kt,
lim sF (s) = lim s
s→∞
s→∞
lim F (s) = f (0).
s→∞
s
s2 + k 2
= 1 = f (0).
EXERCISES 4.3
Translation Theorems
1.
te10t =
1
(s − 10)2
2.
3.
3 −2t =
t e
5.
2 t et + e2t
=
6.
2t
e (t − 1)2 =
7.
t
e sin 3t =
9.
{(1 − et + 3e−4t ) cos 5t} =
10.
11.
3!
(s + 2)4
4.
te2t + 2te3t + te4t =
2 2t
t e − 2te2t + e2t =
3
(s − 1)2 + 9
te−6t =
1
(s + 6)2
10 −7t =
t e
10!
(s + 7)11
2
1
1
+
+
2
2
(s − 2)
(s − 3)
(s − 4)2
2
1
2
−
+
(s − 2)3
(s − 2)2
s−2
8.
−2t
cos 4t =
e
{cos 5t − et cos 5t + 3e−4t cos 5t} =
s+2
(s + 2)2 + 16
s
s−1
3(s + 4)
−
+
s2 + 25 (s − 1)2 + 25 (s + 4)2 + 25
t
t
5
9
4
e3t 9 − 4t + 10 sin
+
=
9e3t − 4te3t + 10e3t sin
=
−
2
2
s − 3 (s − 3)2
(s − 3)2 + 1/4
1
2
1
1
=
= t2 e−2t
(s + 2)3
2 (s + 2)3
2
207
4.3
Translation Theorems
12.
13.
1
(s − 1)4
=
1
s2 − 6s + 10
1
2
s + 2s + 5
1
6
3!
(s − 1)4
=
=
1 3 t
t e
6
1
(s − 3)2 + 12
= e3t sin t
1 −t
e sin 2t
2
s
1
s+2
= e−2t cos t − 2e−2t sin t
−2
=
s2 + 4s + 5
(s + 2)2 + 12
(s + 2)2 + 12
2s + 5
1
1
(s + 3)
5
= 2e−3t cos 5t − e−3t sin 5t
−
=
2
2
2
2
2
2
s + 6s + 34
(s + 3) + 5
5 (s + 3) + 5
5
s
s+1−1
1
1
=
=
= e−t − te−t
−
(s + 1)2
(s + 1)2
s + 1 (s + 1)2
5(s − 2) + 10
5
5s
10
=
=
= 5e2t + 10te2t
+
(s − 2)2
(s − 2)2
s − 2 (s − 2)2
5
2s − 1
5
3
3
1
4
2
=
= 5 − t − 5e−t − 4te−t − t2 e−t
−
−
−
−
s2 (s + 1)3
s s2
s + 1 (s + 1)2
2 (s + 1)3
2
1
(s + 1)2
2
1
1
3!
=
= te−2t − t2 e−2t + t3 e−2t
−
+
(s + 2)4
(s + 2)2
(s + 2)3
6 (s + 2)4
6
14.
15.
16.
17.
18.
19.
20.
=
1
2
2 (s + 1)2 + 22
=
21. The Laplace transform of the differential equation is
s
Solving for
{y} − y(0) + 4
{y} =
1
.
s+4
{y} we obtain
{y} =
1
2
+
.
2
(s + 4)
s+4
Thus
y = te−4t + 2e−4t .
22. The Laplace transform of the differential equation is
s
Solving for
{y} −
{y} =
1
1
+
.
s (s − 1)2
{y} we obtain
{y} =
1
1
1
1
1
=− +
.
+
+
3
s(s − 1) (s − 1)
s s − 1 (s − 1)3
Thus
1
y = −1 + et + t2 et .
2
23. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) + 2 s
{y} − y(0) +
{y} we obtain
{y} =
s+3
2
1
+
=
.
(s + 1)2
s + 1 (s + 1)2
Thus
y = e−t + 2te−t .
208
{y} = 0.
4.3 Translation Theorems
24. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) − 4 [s
s2
{y} − y(0)] + 4
{y} =
6
.
(s − 2)4
1
1 5 2t
5!
. Thus, y =
t e .
20 (s − 2)6
20
25. The Laplace transform of the differential equation is
Solving for
{y} we obtain
{y} − sy(0) − y (0) − 6 [s
s2
Solving for
{y} =
{y} − y(0)] + 9
{y} =
1
.
s2
{y} we obtain
{y} =
1 + s2
1
10
1
2 1 1 1
2
+
+
=
−
.
2
2
2
s (s − 3)
27 s 9 s
27 s − 3
9 (s − 3)2
Thus
y=
2
1
2
10
+ t − e3t + te3t .
27 9
27
9
26. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) − 4 [s
{y} − y(0)] + 4
{y} =
6
.
s4
{y} we obtain
{y} =
s5 − 4s4 + 6
13
1
3 1 9 1
3 2
1 3!
1 1
+
−
=
+
+
+
.
s4 (s − 2)2
4 s 8 s2
4 s3
4 s4
4 s−2
8 (s − 2)2
Thus
y=
3 9
1
1
13
3
+ t + t2 + t3 + e2t − te2t .
4 8
4
4
4
8
27. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) − 6 [s
{y} − y(0)] + 13
{y} = 0.
{y} we obtain
3
3
2
{y} = − 2
=−
.
s − 6s + 13
2 (s − 3)2 + 22
Thus
3
y = − e3t sin 2t.
2
28. The Laplace transform of the differential equation is
2 s2 {y} − sy(0) + 20 s
Solving for
{y} − y(0) + 51
{y} = 0.
{y} we obtain
{y} =
4s + 40
2s2 + 20s + 51
Thus
=
2s + 20
2(s + 5)
10
=
+
.
2
2
(s + 5) + 1/2
(s + 5) + 1/2 (s + 5)2 + 1/2
√
√
√
y = 2e−5t cos(t/ 2 ) + 10 2 e−5t sin(t/ 2 ).
29. The Laplace transform of the differential equation is
s2
{y} − sy(0) − y (0) − [s
{y} − y(0)] =
209
s−1
.
(s − 1)2 + 1
4.3
Translation Theorems
Solving for
{y} we obtain
{y} =
1
s(s2 − 2s + 2)
Thus
=
1 1 1
s−1
1
1
−
+
.
2
2 s 2 (s − 1) + 1 2 (s − 1)2 + 1
1 1 t
1
− e cos t + et sin t.
2 2
2
y=
30. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) − 2 [s
s2
Solving for
{y} − y(0)] + 5
{y} =
1
1
.
+
s s2
{y} we obtain
{y} =
4s2 + s + 1
−7s/25 + 109/25
7 1 1 1
+
=
+
s2 (s2 − 2s + 5)
25 s 5 s2
s2 − 2s + 5
=
7 1 1 1
7
51
s−1
2
−
+
.
+
25 s 5 s2
25 (s − 1)2 + 22
25 (s − 1)2 + 22
Thus
y=
7
51
1
7
+ t − et cos 2t + et sin 2t.
25 5
25
25
31. Taking the Laplace transform of both sides of the differential equation and letting c = y(0) we obtain
{y } +
{y} − sy(0) − y (0) + 2s
2
s
s2
{2y } +
{y} = 0
{y} − 2y(0) +
{y} = 0
{y} − cs − 2 + 2s
{y} − 2c +
2
s + 2s + 1
{y} = 0
{y} = cs + 2c + 2
2c + 2
cs
+
{y} =
2
(s + 1)
(s + 1)2
=c
=
Therefore,
y(t) = c
1
s+1
+ (c + 2)
1
(s + 1)2
s+1−1
2c + 2
+
(s + 1)2
(s + 1)2
c
c+2
.
+
s + 1 (s + 1)2
= ce−t + (c + 2)te−t .
To find c we let y(1) = 2. Then 2 = ce−1 + (c + 2)e−1 = 2(c + 1)e−1 and c = e − 1. Thus
y(t) = (e − 1)e−t + (e + 1)te−t .
32. Taking the Laplace transform of both sides of the differential equation and letting c = y (0) we obtain
{y } +
2
s
{8y } +
{20y} = 0
{y} − y (0) + 8s
{y} + 20
{y} = 0
{y} − c + 8s
{y} + 20
{y} = 0
(s + 8s + 20)
{y} = c
s2
2
{y} =
210
c
c
=
.
s2 + 8s + 20
(s + 4)2 + 4
4.3 Translation Theorems
Therefore,
y(t) =
c
(s + 4)2 + 4
=
c −4t
e
sin 2t = c1 e−4t sin 2t.
2
To find c we let y (π) = 0. Then 0 = y (π) = ce−4π and c = 0. Thus, y(t) = 0. (Since the differential equation
is homogeneous and both boundary conditions are 0, we can see immediately that y(t) = 0 is a solution. We
have shown that it is the only solution.)
33. Recall from Section 3.8 that mx = −kx − βx . Now m = W/g = 4/32 = 18 slug, and 4 = 2k so that k = 2 lb/ft.
Thus, the differential equation is x + 7x + 16x = 0. The initial conditions are x(0) = −3/2 and x (0) = 0.
The Laplace transform of the differential equation is
s2
Solving for
3
{x} + s + 7s
2
{x} +
21
+ 16
2
{x} = 0.
{x} we obtain
√
√
−3s/2 − 21/2
15/2
3
s + 7/2
7 15
√
√
{x} = 2
=−
−
.
s + 7s + 16
2 (s + 7/2)2 + ( 15/2)2
10 (s + 7/2)2 + ( 15/2)2
Thus
√
√
√
3 −7t/2
15
15
7 15 −7t/2
x=− e
cos
sin
t−
e
t.
2
2
10
2
34. The differential equation is
d2 q
dq
+ 20 + 200q = 150,
2
dt
dt
The Laplace transform of this equation is
s2
Solving for
{q} + 20s
q(0) = q (0) = 0.
{q} + 200
150
.
s
{q} =
{q} we obtain
{q} =
150
s(s2 + 20s + 200)
Thus
q(t) =
=
3 1 3
s + 10
10
3
−
−
.
2
2
4 s 4 (s + 10) + 10
4 (s + 10)2 + 102
3 3 −10t
3
cos 10t − e−10t sin 10t
− e
4 4
4
and
i(t) = q (t) = 15e−10t sin 10t.
35. The differential equation is
d2 q
dq
E0
+ 2λ + ω 2 q =
,
2
dt
dt
L
The Laplace transform of this equation is
s2
or
Solving for
{q} + 2λs
{q} + ω 2
2
s + 2λs + ω 2
q(0) = q (0) = 0.
{q} =
{q} =
E0 1
L s
E0 1
.
L s
{q} and using partial fractions we obtain
E0 1/ω 2
(1/ω 2 )s + 2λ/ω 2
s + 2λ
E0 1
{q} =
− 2
−
=
.
L
s
s + 2λs + ω 2
Lω 2 s s2 + 2λs + ω 2
211
4.3
Translation Theorems
For λ > ω we write s2 + 2λs + ω 2 = (s + λ)2 − λ2 − ω 2 , so (recalling that ω 2 = 1/LC)
1
s+λ
λ
{q} = E0 C
−
−
.
s (s + λ)2 − (λ2 − ω 2 ) (s + λ)2 − (λ2 − ω 2 )
Thus for λ > ω,
λ
q(t) = E0 C 1 − e−λt cosh λ2 − ω 2 t − √
sinh λ2 − ω 2 t .
λ2 − ω 2
For λ < ω we write s2 + 2λs + ω 2 = (s + λ)2 + ω 2 − λ2 , so
s+λ
λ
1
{q} = E0 C
−
−
.
s (s + λ)2 + (ω 2 − λ2 ) (s + λ)2 + (ω 2 − λ2 )
Thus for λ < ω,
λ
q(t) = E0 C 1 − e−λt cos ω 2 − λ2 t − √
sin ω 2 − λ2 t .
ω 2 − λ2
For λ = ω, s2 + 2λ + ω 2 = (s + λ)2 and
E0
E0 1
E0 1/λ2
1
1/λ2
1/λ
1
λ
{q} =
=
.
=
−
−
−
−
L s(s + λ)2
L
s
s + λ (s + λ)2
Lλ2 s s + λ (s + λ)2
Thus for λ = ω,
q(t) = E0 C 1 − e−λt − λte−λt .
36. The differential equation is
dq
1
+ q = E0 e−kt , q(0) = 0.
dt
C
The Laplace transform of this equation is
R
Rs
Solving for
{q} +
1
C
{q} = E0
1
.
s+k
{q} we obtain
{q} =
E0 C
E0 /R
=
.
(s + k)(RCs + 1)
(s + k)(s + 1/RC)
When 1/RC = k we have by partial fractions
E0 1/(1/RC − k) 1/(1/RC − k)
E0
1
1
1
{q} =
−
=
−
.
R
s+k
s + 1/RC
R 1/RC − k s + k s + 1/RC
Thus
q(t) =
E0 C −kt
− e−t/RC .
e
1 − kRC
When 1/RC = k we have
{q} =
Thus
q(t) =
37.
38.
(t − 1)
2−t
e
e−s
(t − 1) = 2
s
(t − 2) =
e−(t−2)
E0
1
.
R (s + k)2
E0 −kt
E0 −t/RC
=
.
te
te
R
R
e−2s
(t − 2) =
s+1
212
4.3 Translation Theorems
39.
t
(t − 2) =
{(t − 2)
(t − 2) + 2
(t − 2)} =
e−2s
2e−2s
+
2
s
s
Alternatively, (16) of this section could be used:
{t
40.
(t − 1) = 3
(3t + 1)
−2s
−2s
(t − 2)} = e
(t − 1)
{t + 2} = e
(t − 1) + 4
1
2
+
2
s
s
.
3e−s
4e−s
(t − 1) = 2 +
s
s
Alternatively, (16) of this section could be used:
41.
cos 2t
(t − π) =
{(3t + 1)
(t − 1)} = e−s
{cos 2(t − π)
(t − π)} =
{3t + 4} = e−s
3
4
+
s2
s
.
se−πs
s2 + 4
Alternatively, (16) of this section could be used:
{cos 2t
42.
sin t
t−
π =
2
(t − π)} = e−πs
π
cos t −
2
{cos 2(t + π)} = e−πs
t−
44.
45.
46.
47.
48.
49. (c)
57.
{cos t} = e−πs/2
s
.
s2 + 1
−2s 1 2 −2s
1
e
=
= (t − 2)2 (t − 2)
e
·
s3
2 s3
2
(1 + e−2s )2
1
2e−2s
e−4s
=
+
+
= e−2t + 2e−2(t−2) (t − 2) + e−2(t−4) (t − 4)
s+2
s+2
s+2
s+2
−πs e
= sin(t − π) (t − π) = − sin t (t − π)
s2 + 1
−πs/2 se
π π
π
=
cos
2
t
−
t
−
=
−
cos
2t
t
−
s2 + 4
2
2
2
e−s
e−s
e−s
=
−
= (t − 1) − e−(t−1) (t − 1)
s(s + 1)
s
s+1
−2s
e−2s
e−2s
e
e−2s
=
−
− 2 +
= − (t − 2) − (t − 2) (t − 2) + et−2 (t − 2)
s2 (s − 1)
s
s
s−1
43.
56.
s
.
s2 + 4
π se−πs/2
= 2
2
s +1
Alternatively, (16) of this section could be used:
π π sin t
t−
sin t +
= e−πs/2
= e−πs/2
2
2
55.
{cos 2t} = e−πs
50. (e)
51. (f )
52. (b)
2 4
(t − 3) = − e−3s
s s
1 e−4s
e−5s
1 − (t − 4) + (t − 5) = −
+
s
s
s
2
t
(t − 1) =
(t − 1)2 + 2t − 1
(t − 1) =
2
2
1 −s
=
+
+
e
s3
s2
s
53. (a)
54. (d)
2−4
213
(t − 1)2 + 2(t − 1) − 1
(t − 1)
4.3
Translation Theorems
Alternatively, by (16) of this section,
(t − 1)} = e−s
{t2 + 2t + 1} = e−s
3π
− cos t −
2
{t2
58.
59.
60.
61.
62.
sin t
(t − 2) =
t−t
t−
3π
2
=
f (t) =
f (t) =
3π
2
=−
2
2
1
+ 2+
s3
s
s
.
se−3πs/2
s2 + 1
1
e−2s
2e−2s
(t − 2) = 2 − 2 −
s
s
s
−2πs
1
e
sin t − sin(t − 2π) (t − 2π) = 2
−
s + 1 s2 + 1
t − (t − 2)
(t − 2π) =
sin t − sin t
t−
(t − 2) − 2
(t − a) −
e−as
e−bs
(t − b) =
−
s
s
(t − 1) +
(t − 2) +
e−s
e−2s
e−3s
1 e−s
(t − 3) + · · · =
+
+
+ ··· =
s
s
s
s 1 − e−s
63. The Laplace transform of the differential equation is
{y} − y(0) +
s
Solving for
{y} we obtain
{y} =
5 −s
e .
s
5e−s
1
−s 1
{y} =
= 5e
−
.
s(s + 1)
s s+1
Thus
y=5
(t − 1) − 5e−(t−1)
(t − 1).
64. The Laplace transform of the differential equation is
s
Solving for
{y} − y(0) +
{y} =
1 2 −s
− e .
s s
{y} we obtain
1
2e−s
1
1
1
−s 1
{y} =
−
= −
− 2e
−
.
s(s + 1) s(s + 1)
s s+1
s s+1
Thus
y = 1 − e−t − 2 1 − e−(t−1) (t − 1).
65. The Laplace transform of the differential equation is
s
Solving for
{y} − y(0) + 2
{y} =
1
s+1
− e−s 2 .
2
s
s
{y} we obtain
{y} =
1
1 1
1 1
1 1 1 1
1 1
−s s + 1
−s 1 1
+
−
−
e
=
−
+
−
e
+
.
s2 (s + 2)
s2 (s + 2)
4 s 2 s2
4 s+2
4 s 2 s2
4 s+2
Thus
1 1
1 1
1
1
y = − + t + e−2t −
+ (t − 1) − e−2(t−1)
4 2
4
4 2
4
(t − 1).
66. The Laplace transform of the differential equation is
s2
{y} − sy(0) − y (0) + 4
214
{y} =
1 e−s
−
.
s
s
4.3 Translation Theorems
Solving for
{y} we obtain
{y} =
1−s
1
1 1 1 s
1 2
1 s
−s
−s 1 1
−
e
=
−
−
−
e
−
.
s(s2 + 4)
s(s2 + 4)
4 s 4 s2 + 4 2 s2 + 4
4 s 4 s2 + 4
Thus
y=
1 1
1
1 1
− cos 2t − sin 2t −
− cos 2(t − 1) (t − 1).
4 4
2
4 4
67. The Laplace transform of the differential equation is
s2
Solving for
1
{y} = e−2πs 2
.
s +1
{y} − sy(0) − y (0) + 4
{y} we obtain
{y} =
Thus
y = cos 2t +
s
1
1 2
−2πs 1
+
e
−
.
s2 + 4
3 s2 + 1 6 s2 + 4
1
1
sin(t − 2π) − sin 2(t − 2π) (t − 2π).
3
6
68. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) − 5 [s
s2
Solving for
{y} − y(0)] + 6
{y} =
e−s
.
s
{y} we obtain
1
1
+
s(s − 2)(s − 3) (s − 2)(s − 3)
1 1
1 1
1
1
−s 1 1
=e
−
+
−
+
.
6 s 2 s−2 3 s−3
s−2 s−3
{y} = e−s
Thus
1 1 2(t−1) 1 3(t−1)
y=
− e
+ e
6 2
3
(t − 1) − e2t + e3t .
69. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) +
{y} we obtain
−πs
{y} = e
{y} =
e−2πs
e−πs
−
.
s
s
1
s
s
1
−2πs 1
−
−e
−
+ 2
.
s s2 + 1
s s2 + 1
s +1
Thus
y = [1 − cos(t − π)]
(t − π) − [1 − cos(t − 2π)]
(t − 2π) + sin t.
70. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) + 4 s
{y} − y(0) + 3
{y} we obtain
{y} =
1 e−2s
e−4s
e−6s
−
−
+
.
s
s
s
s
1 1 1 1
1 1
1 1
1 1
−2s 1 1
{y} =
−
+
−e
−
+
3 s 2 s+1 6 s+3
3 s 2 s+1 6 s+3
1 1
1 1
1 1 1 1
1 1 1 1
− e−4s
−
+
+ e−6s
−
+
.
3 s 2 s+1 6 s+3
3 s 2 s+1 6 s+3
215
4.3
Translation Theorems
Thus
y=
1 1 −t 1 −3t
1 1 −(t−2) 1 −3(t−2)
(t − 2)
−
+ e
− e + e
− e
3 2
6
3 2
6
1 1 −(t−4) 1 −3(t−4)
1 1 −(t−6) 1 −3(t−6)
−
− e
− e
(t − 4) +
(t − 6).
+ e
+ e
3 2
6
3 2
6
71. Recall from Section 3.8 that mx = −kx + f (t). Now m = W/g = 32/32 = 1 slug, and 32 = 2k so that
k = 16 lb/ft. Thus, the differential equation is x + 16x = f (t). The initial conditions are x(0) = 0, x (0) = 0.
Also, since
f (t) =
20t, 0 ≤ t < 5
t≥5
0,
and 20t = 20(t − 5) + 100 we can write
f (t) = 20t − 20t
(t − 5) = 20t − 20(t − 5)
(t − 5) − 100
(t − 5).
The Laplace transform of the differential equation is
{x} + 16
s2
Solving for
{x} =
20 20 −5s 100 −5s
− 2e
−
e .
s2
s
s
{x} we obtain
20
100
20
−
e−5s −
e−5s
s2 (s2 + 16) s2 (s2 + 16)
s(s2 + 16)
5
5 1
25 1 25
4
s
−5s
=
−
−
·
·
1−e
· −
·
e−5s .
4 s2
16 s2 + 16
4 s
4 s2 + 16
{x} =
Thus
5
5
5
5
25 25
x(t) = t −
sin 4t − (t − 5) −
sin 4(t − 5)
(t − 5) −
−
cos 4(t − 5)
(t − 5)
4
16
4
16
4
4
=
5
5
5
t−
sin 4t − t
4
16
4
(t − 5) +
5
sin 4(t − 5)
16
(t − 5) +
25
cos 4(t − 5)
4
(t − 5).
72. Recall from Section 3.8 that mx = −kx + f (t). Now m = W/g = 32/32 = 1 slug, and 32 = 2k so that
k = 16 lb/ft. Thus, the differential equation is x + 16x = f (t). The initial conditions are x(0) = 0, x (0) = 0.
Also, since
f (t) =
sin t, 0 ≤ t < 2π
0,
t ≥ 2π
and sin t = sin(t − 2π) we can write
f (t) = sin t − sin(t − 2π) (t − 2π).
The Laplace transform of the differential equation is
s2
Solving for
{x} + 16
{x} =
1
1
−
e−2πs .
s2 + 1 s2 + 1
{x} we obtain
1
1
−
e−2πs
(s2 + 16) (s2 + 1) (s2 + 16) (s2 + 1)
−1/15
1/15
−1/15
1/15
= 2
+ 2
− 2
+ 2
e−2πs .
s + 16 s + 1
s + 16 s + 1
{x} =
216
4.3 Translation Theorems
Thus
x(t) = −
=
1
1
1
sin 4t +
sin t +
sin 4(t − 2π)
60
15
60
(t − 2π) −
1
sin(t − 2π)
15
(t − 2π)
1
1
− 60
sin 4t + 15
sin t, 0 ≤ t < 2π
t ≥ 2π.
0,
73. The differential equation is
2.5
dq
+ 12.5q = 5
dt
(t − 3).
{q} + 5
2 −3s
e .
s
The Laplace transform of this equation is
s
Solving for
{q} =
{q} we obtain
2
e−3s =
s(s + 5)
{q} =
Thus
q(t) =
2
5
2 1 2
1
· − ·
5 s 5 s+5
2
(t − 3) − e−5(t−3)
5
e−3s .
(t − 3).
74. The differential equation is
dq
+ 10q = 30et − 30et (t − 1.5).
dt
The Laplace transform of this equation is
10
Solving for
{q} we obtain
{q} =
Thus
s
{q} − q0 +
q0 −
3
2
q(t) =
q0 −
3
2
·
{q} =
3
3e1.5 −1.5s
.
−
e
s − 1 s − 1.5
1
3
1
+ ·
− 3e1.5
s+1 2 s−1
−2/5
2/5
+
s+1
s − 1.5
3
6
e−t + et + e1.5 e−(t−1.5) − e1.5(t−1.5)
2
5
e−1.5s .
(t − 1.5).
75. (a) The differential equation is
di
3π
3π
+ 10i = sin t + cos t −
t−
,
dt
2
2
i(0) = 0.
The Laplace transform of this equation is
s
Solving for
{i} + 10
{i} =
1
se−3πs/2
+
.
s2 + 1
s2 + 1
{i} we obtain
1
s
+
e−3πs/2
(s2 + 1)(s + 10) (s2 + 1)(s + 10)
1
s
10
1
−10
10s
1
1
−
+
+
+
+
e−3πs/2 .
=
101 s + 10 s2 + 1 s2 + 1
101 s + 10 s2 + 1 s2 + 1
{i} =
Thus
i(t) =
1 −10t
− cos t + 10 sin t
e
101
1
3π
3π
3π
+
−10e−10(t−3π/2) + 10 cos t −
+ sin t −
t−
.
101
2
2
2
217
4.3
Translation Theorems
(b)
i
0.2
1
2
5
4
3
t
6
-0.2
The maximum value of i(t) is approximately 0.1 at t = 1.7, the minimum is approximately −0.1 at 4.7.
76. (a) The differential equation is
dq
1
+
q = E0 [
dt
0.01
(t − 1) −
(t − 3)],
q(0) = 0
dq
+ 100q = E0 [
dt
The Laplace transform of this equation is
(t − 1) −
(t − 3)],
q(0) = 0.
{q} + 100
{q} = E0
50
or
50
50s
Solving for
1 −s 1 −3s
.
e − e
s
s
{q} we obtain
E0
1 1
e−s
e−3s
E0 1 1
1
1
−s
−3s
{q} =
.
−
=
−
e −
−
e
50 s(s + 2) s(s + 2)
50 2 s s + 2
2 s s+2
Thus
q(t) =
(b)
E0 (t − 1) − 1 − e−2(t−3)
(t − 3) .
1 − e−2(t−1)
100
q
1
1
2
4
3
5
6
t
The maximum value of q(t) is approximately 1 at t = 3.
77. The differential equation is
d4 y
= w0 [1 − (x − L/2)].
dx4
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
w0 1 s4 {y} − sy (0) − y (0) =
1 − e−Ls/2 .
EI s
Letting y (0) = c1 and y (0) = c2 we have
c1
c2
w0 1 −Ls/2
{y} = 3 + 4 +
1
−
e
s
s
EI s5
so that
4 1
L
1
1 w0
L
2
3
4
y(x) = c1 x + c2 x +
x − x−
.
x−
2
6
24 EI
2
2
EI
To find c1 and c2 we compute
y (x) = c1 + c2 x +
2
1 w0
L
x2 − x −
2 EI
2
and
218
x−
L
2
4.3 Translation Theorems
y (x) = c2 +
w0
L
L
x− x−
x−
.
EI
2
2
Then y (L) = y (L) = 0 yields the system
1 w0
c1 + c2 L +
L2 −
2 EI
2
L
3 w0 L2
= c1 + c2 L +
=0
2
8 EI
w0
c2 +
EI
1 w0 L
L
= c2 +
= 0.
2
2 EI
Solving for c1 and c2 we obtain c1 = 18 w0 L2 /EI and c2 = − 12 w0 L/EI. Thus
y(x) =
w0
EI
1
1
1
1 2 2
L x − Lx3 + x4 −
16
12
24
24
x−
L
2
4
x−
L
2
.
78. The differential equation is
d4 y
= w0 [ (x − L/3) − (x − 2L/3)].
dx4
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
w0 1 −Ls/3
s4 {y} − sy (0) − y (0) =
− e−2Ls/3 .
e
EI s
EI
Letting y (0) = c1 and y (0) = c2 we have
{y} =
c1
c2
w0 1 −Ls/3
−2Ls/3
e
+
+
−
e
s3
s4
EI s5
so that
y(x) =
1
1
1 w0
c1 x2 + c2 x3 +
2
6
24 EI
x−
L
3
4
x−
L
3
4
2L
− x−
3
x−
2L
3
.
To find c1 and c2 we compute
y (x) = c1 + c2 x +
and
w0
y (x) = c2 +
EI
1 w0
2 EI
x−
L
3
2
x−
L
3
2
2L
− x−
3
x−
2L
3
L
L
2L
2L
x−
x−
− x−
x−
.
3
3
3
3
Then y (L) = y (L) = 0 yields the system
2 2
2L
1 w0
L
1 w0 L2
c1 + c2 L +
= c1 + c2 L +
=0
−
2 EI
3
3
6 EI
c2 +
w0
EI
1 w0 L
2L L
−
= 0.
= c2 +
3
3
3 EI
Solving for c1 and c2 we obtain c1 = 16 w0 L2 /EI and c2 = − 13 w0 L/EI. Thus
y(x) =
w0
EI
1
1
1 2 2
L x − Lx3 +
12
18
24
79. The differential equation is
EI
x−
L
3
4
x−
d4 y
2w0 L
L
=
−
x
+
x
−
dx4
L
2
2
219
L
3
4
2L
− x−
3
x−
L
2
.
x−
2L
3
.
4.3
Translation Theorems
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
{y} − sy (0) − y (0) =
s4
2w0 L
1
1
− 2 + 2 e−Ls/2 .
EIL 2s s
s
Letting y (0) = c1 and y (0) = c2 we have
{y} =
so that
c1
c2
2w0
1
1
L
+ 4+
− 6 + 6 e−Ls/2
3
5
s
s
EIL 2s
s
s
5 1
L
1
2w0 L 4
1 5
1
L
x−
c1 x2 + c2 x3 +
x −
x +
x−
2
6
EIL 48
120
120
2
2
5 1
5L 4
L
1
w0
L
= c1 x2 + c2 x3 +
x − x5 + x −
.
x−
2
6
60EIL 2
2
2
y(x) =
To find c1 and c2 we compute
y (x) = c1 + c2 x +
and
y (x) = c2 +
3
w0
L
30Lx2 − 20x3 + 20 x −
60EIL
2
2
w0
L
60Lx − 60x2 + 60 x −
60EIL
2
x−
x−
L
2
L
2
.
Then y (L) = y (L) = 0 yields the system
c1 + c2 L +
w0
5
5w0 L2
30L3 − 20L3 + L3 = c1 + c2 L +
=0
60EIL
2
24EI
w0
w0 L
c2 +
[60L2 − 60L2 + 15L2 ] = c2 +
= 0.
60EIL
4EI
Solving for c1 and c2 we obtain c1 = w0 L2 /24EI and c2 = −w0 L/4EI. Thus
5
w0 L2 2
5L 4
w0 L 3
w0
L
y(x) =
x −
x +
x − x5 + x −
48EI
24EI
60EIL 2
2
x−
80. The differential equation is
d4 y
= w0 [1 − (x − L/2)].
dx4
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
w0 1 s4 {y} − sy (0) − y (0) =
1 − e−Ls/2 .
EI s
EI
Letting y (0) = c1 and y (0) = c2 we have
{y} =
so that
y(x) =
c1
c2
w0 1 −Ls/2
1
−
e
+
+
s3
s4
EI s5
4
1
1
1 w0
L
c1 x2 + c2 x3 +
x4 − x −
2
6
24 EI
2
x−
L
2
.
To find c1 and c2 we compute
2
1 w0
L
2
x − x−
y (x) = c1 + c2 x +
2 EI
2
220
L
x−
2
.
L
2
.
4.3 Translation Theorems
Then y(L) = y (L) = 0 yields the system
1
1
1 w0
c1 L2 + c2 L3 +
L4 −
2
6
24 EI
4
1
L
1
5w0
= c1 L2 + c2 L3 +
L4 = 0
2
2
6
128EI
1 w0
c1 + c2 L +
L2 −
2 EI
2
L
3w0 2
= c1 + c2 L +
L = 0.
2
8EI
9
57
Solving for c1 and c2 we obtain c1 = 128
w0 L2 /EI and c2 = − 128
w0 L/EI. Thus
4 w0
L
19
1
1
9 2 2
L
y(x) =
x−
L x −
Lx3 + x4 −
x−
.
EI 256
256
24
24
2
2
81. (a) The temperature T of the cake inside the oven is modeled by
dT
= k(T − Tm )
dt
where Tm is the ambient temperature of the oven. For 0 ≤ t ≤ 4, we have
Tm = 70 +
Hence for t ≥ 0,
Tm =
300 − 70
t = 70 + 57.5t.
4−0
70 + 57.5t, 0 ≤ t < 4
t ≥ 4.
300,
In terms of the unit step function,
Tm = (70 + 57.5t)[1 −
(t − 4)] + 300
(t − 4) = 70 + 57.5t + (230 − 57.5t)
(t − 4).
The initial-value problem is then
dT
= k[T − 70 − 57.5t − (230 − 57.5t)
dt
(b) Let t(s) =
or
(t − 4)],
T (0) = 70.
{T (t)}. Transforming the equation, using 230 − 57.5t = −57.5(t − 4) and Theorem 4.7, gives
70 57.5 57.5 −4s
st(s) − 70 = k t(s) −
− 2 + 2 e
s
s
s
t(s) =
70
70k
57.5k
57.5k
−
− 2
+ 2
e−4s .
s − k s(s − k) s (s − k) s (s − k)
After using partial functions, the inverse transform is then
1
1
1 kt
1 k(t−4)
T (t) = 70 + 57.5
− 57.5
+t− e
+t−4− e
k
k
k
k
(t − 4).
Of course, the obvious question is: What is k? If the cake is supposed to bake for, say, 20 minutes, then
T (20) = 300. That is,
1
1
1 20k
1 16k
300 = 70 + 57.5
− 57.5
.
+ 20 − e
+ 16 − e
k
k
k
k
But this equation has no physically meaningful solution. This should be no surprise since the model predicts
the asymptotic behavior T (t) → 300 as t increases. Using T (20) = 299 instead, we find, with the help of a
CAS, that k ≈ −0.3.
82. In order to apply Theorem 4.7 we need the function to have the form f (t − a)
rewrite the functions given in the forms shown below.
221
(t − a). To accomplish this
4.3
Translation Theorems
(a) 2t + 1 = 2(t − 1 + 1) + 1 = 2(t − 1) + 3
(b) et = et−5+5 = e5 et−5
(c) cos t = − cos(t − π)
(d) t2 − 3t = (t − 2)2 + (t − 2) − 2
{tekti } = 1/(s − ki)2 . Then, using Euler’s formula,
83. (a) From Theorem 4.6 we have
{tekti } =
=
{t cos kt + it sin kt} =
{t cos kt} + i
{t sin kt}
1
(s + ki)2
s2 − k 2
2ks
= 2
= 2
+i 2
.
2
2
2
(s − ki)
(s + k )
(s + k 2 )2
(s + k 2 )2
Equating real and imaginary parts we have
{t cos kt} =
s2 − k 2
(s2 + k 2 )2
{t sin kt} =
and
2ks
(s2 + k 2 )2
.
(b) The Laplace transform of the differential equation is
s2
{x} we obtain
Solving for
{x} + ω 2
{x} =
s
s2 + ω 2
.
{x} = s/(s2 + ω 2 )2 . Thus x = (1/2ω)t sin ωt.
EXERCISES 4.4
Additional Operational Properties
−10t
d
}=−
ds
1.
{te
3.
{t cos 2t} = −
5.
{t2 sinh t} =
6.
7.
8.
d
ds
1
s + 10
s
s2 + 4
1
=
(s + 10)2
=
s2 − 4
2
(s2 + 4)
3 d
2.
{t e } = (−1)
4.
{t sinh 3t} = −
3 t
3
ds3
d
ds
1
s−1
3
s2 − 9
=
=
6
(s − 1)4
6s
1
d2
6s2 + 2
=
3
2
2
ds
s −1
(s2 − 1)
2s s2 − 3
s
d2
d
1 − s2
2
=
{t cos t} = 2
=
3
ds
s2 + 1
ds (s2 + 1)2
(s2 + 1)
2t
6
12(s − 2)
d
=
te sin 6t = −
2
ds (s − 2)2 + 36
[(s − 2)2 + 36]
−3t
d
s+3
(s + 3)2 − 9
te
cos 3t = −
=
2
2
ds (s + 3) + 9
[(s + 3)2 + 9]
9. The Laplace transform of the differential equation is
s
Solving for
{y} +
{y} =
2s
.
(s2 + 1)2
{y} we obtain
{y} =
2s
1 1
1 s
1
1 1
s
−
+
+
=−
+ 2
.
(s + 1)(s2 + 1)2
2 s + 1 2 s2 + 1 2 s2 + 1 (s2 + 1)2
(s + 1)2
222
2
(s2 − 9)
4.4
Thus
Additional Operational Properties
1
1
1
1
1
y(t) = − e−t − sin t + cos t + (sin t − t cos t) + t sin t
2
2
2
2
2
1 −t 1
1
1
= − e + cos t − t cos t + t sin t.
2
2
2
2
10. The Laplace transform of the differential equation is
{y} −
s
Solving for
{y} =
2(s − 1)
.
((s − 1)2 + 1)2
{y} we obtain
{y} =
2
.
((s − 1)2 + 1)2
Thus
y = et sin t − tet cos t.
11. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) + 9
s2
Letting y(0) = 2 and y (0) = 5 and solving for
{y} =
s
.
s2 + 9
{y} =
{y} we obtain
2s + 5s + 19s − 45
5
s
2s
+ 2
+ 2
= 2
.
2
2
(s + 9)
s + 9 s + 9 (s + 9)2
3
2
Thus
5
1
sin 3t + t sin 3t.
3
6
y = 2 cos 3t +
12. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) +
{y} =
1
.
s2 + 1
{y} we obtain
{y} =
s3 − s2 + s
s
1
1
= 2
.
− 2
+ 2
2
2
(s + 1)
s + 1 s + 1 (s + 1)2
Thus
y = cos t − sin t +
1
1
sin t − t cos t
2
2
= cos t −
1
1
sin t − t cos t.
2
2
13. The Laplace transform of the differential equation is
s2
{y} − sy(0) − y (0) + 16
{y} =
{cos 4t − cos 4t
(t − π)}
or by (16) of Section 4.3 in the text,
(s2 + 16)
{y} = 1 +
=1+
s
s2 + 16
s
s2 + 16
Thus
{y} =
and
y=
− e−πs
{cos 4(t + π)}
− e−πs
{cos 4t} = 1 +
s
s2 + 16
−
s
s2 + 16
1
s
s
− 2
e−πs
+
s2 + 16 (s2 + 16)2
(s + 16)2
1
1
1
sin 4t + t sin 4t − (t − π) sin 4(t − π) (t − π).
4
8
8
223
e−πs .
4.4
Additional Operational Properties
14. The Laplace transform of the differential equation is
s2
{y} − sy(0) − y (0) +
(s2 + 1)
or
Thus
{y} =
1−
t−
π
π + sin t
t−
2
2
π 1 1 −πs/2
sin t +
+ e−πs/2
− e
s s
2
1 1 −πs/2
=s+ − e
+ e−πs/2 {cos t}
s s
1 1
s
= s + − e−πs/2 + 2
e−πs/2 .
s s
s +1
{y} = s +
s
1
1
s
+
−
e−πs/2 + 2
e−πs/2
s2 + 1 s(s2 + 1) s(s2 + 1)
(s + 1)2
1
s
1
s
s
s
+ −
−
−
e−πs/2 + 2
= 2
e−πs/2
s + 1 s s2 + 1
s s2 + 1
(s + 1)2
1
s
1
s
= −
e−πs/2
− 2
e−πs/2 + 2
s
s s +1
(s + 1)2
{y} =
and
15.
π π 1 π
π
y = 1 − 1 − cos t −
t−
+
t−
sin t −
2
2
2
2
2
π 1 π
π
= 1 − (1 − sin t)
t−
−
t−
cos t
t−
.
2
2
2
2
16.
y
1
t−
π
2
y
4
0.5
2
1
2
3
4
5
6
t
1
2
3
4
5
6
t
-2
-0.5
-4
-1
17. From (7) of Section 4.2 in the text along with Theorem 4.8,
dY
d
d
{y } = − [s2 Y (s) − sy(0) − y (0)] = −s2
− 2sY + y(0),
ds
ds
ds
so that the transform of the given second-order differential equation is the linear first-order differential equation
{ty } = −
in Y (s):
4
3
4
s2 Y + 3sY = − 3
or
Y+ Y =− 5 .
s
s
s
The solution of the latter equation is Y (s) = 4/s4 + c/s3 , so
y(t) =
{Y (s)} =
2 3 c 2
t + t .
3
2
18. From Theorem 4.8 in the text
dY
d
d
{y } = − [sY (s) − y(0)] = −s
−Y
ds
ds
ds
so that the transform of the given second-order differential equation is the linear first-order differential equation
in Y (s):
3
10
Y+
− 2s Y = − .
s
s
{ty } = −
224
4.4
Additional Operational Properties
Using the integrating factor s3 e−s , the last equation yields
2
c 2
5
+ 3 es .
s3
s
But if Y (s) is the Laplace transform of a piecewise-continuous function of exponential order, we must have, in
view of Theorem 4.5, lims→∞ Y (s) = 0. In order to obtain this condition we require c = 0. Hence
5
5
y(t) =
= t2 .
s3
2
Y (s) =
19.
21.
1 3!
6
1 ∗ t3 =
= 5
4
s s
s
20.
2
t ∗ tet =
−t t
e ∗ e cos t =
22.
2t
e ∗ sin t =
t
τ
23.
e dτ
=
0
t
24.
cos τ dτ
1
s
{et } =
25.
e−τ cos τ dτ
0
t
26.
τ sin τ dτ
0
t
=
=
1
s
1
s
τe
dτ
=
0
{et } =
sin τ cos(t − τ ) dτ
28.
s
s(s2 + 1)
=
1
s2 + 1
1
−t
s+1
s+1
e cos t =
=
s (s + 1)2 + 1
s (s2 + 2s + 2)
1
d
1
1 −2s
2
{t sin t} =
−
=−
=
2
s
ds s2 + 1
s (s2 + 1)2
(s2 + 1)
{t}
t
1
(s − 2)(s2 + 1)
1
s(s − 1)
{cos t} =
t−τ
27.
1
s
=
0
t
s−1
(s + 1) [(s − 1)2 + 1]
2
s3 (s − 1)2
=
1
s2 (s − 1)
{sin t}
{cos t} =
0
s
2
(s2 + 1)
t
t
d
d 1 1
3s2 + 1
29.
t
sin τ dτ = −
sin τ dτ = −
=
2
2
ds
ds s s + 1
s2 (s2 + 1)
0
0
t
t
1
3s + 1
d
d 1
−τ
−τ
= 2
30.
t
τ e dτ = −
τ e dτ = −
2
ds
ds s (s + 1)
s (s + 1)3
0
0
t
1
1/(s − 1)
31.
eτ dτ = et − 1
=
=
s(s − 1)
s
0
t
1
1/s(s − 1)
32.
(eτ − 1)dτ = et − t − 1
=
=
s2 (s − 1)
s
0
t
1
1
1/s2 (s − 1)
33.
(eτ − τ − 1)dτ = et − t2 − t − 1
=
=
3
s (s − 1)
s
2
0
1
34. Using
= teat , (8) in the text gives
(s − a)2
t
1
1
=
τ eaτ dτ = 2 (ateat − eat + 1).
2
s(s − a)
a
0
35. (a) The result in (4) in the text is
F (s) =
{F (s)G(s)} = f ∗ g, so identify
2k 3
(s2 + k 2 )2
and
225
G(s) =
4s
.
s2 + k 2
4.4
Additional Operational Properties
Then
f (t) = sin kt − kt cos kt
so
8k 3 s
(s2 + k 2 )3
and
g(t) = 4 cos kt
t
{F (s)G(s)} = f ∗ g = 4
=
f (τ )g(t − τ )dt
0
t
(sin kτ − kτ cos kτ ) cos k(t − τ )dτ.
=4
0
Using a CAS to evaluate the integral we get
8k 3 s
= t sin kt − kt2 cos kt.
(s2 + k 2 )3
(b) Observe from part (a) that
and from Theorem 4.8 that
t(sin kt − kt cos kt) =
8k 3 s
(s2 + k 2 )3
,
tf (t) = −F (s). We saw in (5) in the text that
{sin kt − kt cos kt} = 2k 3 /(s2 + k 2 )2 ,
so
d
2k 3
8k 3 s
t(sin kt − kt cos kt) = −
= 2
.
2
2
2
ds (s + k )
(s + k 2 )3
36. The Laplace transform of the differential equation is
s2
{y} +
y
1
2s
{y} = 2
.
+
(s + 1) (s2 + 1)2
50
Thus
1
2s
+ 2
(s2 + 1)2
(s + 1)3
and, using Problem 35 with k = 1,
{y} =
y=
5
-50
1
1
(sin t − t cos t) + (t sin t − t2 cos t).
2
4
37. The Laplace transform of the given equation is
{f } +
Solving for
{f } we obtain
{f } =
1
s2 + 1
{t}
{f } =
{t}.
. Thus, f (t) = sin t.
38. The Laplace transform of the given equation is
{f } =
Solving for
{f } we obtain
{f } =
{2t} − 4
{sin t}
{f }.
√
2s2 + 2
5
8
2 1
√
+
=
.
s2 (s2 + 5)
5 s2
5 5 s2 + 5
Thus
f (t) =
39. The Laplace transform of the given equation is
{f } =
√
2
8
t + √ sin 5 t.
5
5 5
tet +
226
{t}
{f }.
10
15
t
4.4
Solving for
Additional Operational Properties
{f } we obtain
{f } =
s2
1 1
3
1
2
1
1 1
=
+
.
+
−
(s − 1)3 (s + 1)
8 s − 1 4 (s − 1)2
4 (s − 1)3
8 s+1
Thus
1 t 3 t 1 2 t 1 −t
e + te + t e − e
8
4
4
8
f (t) =
40. The Laplace transform of the given equation is
{f } + 2
Solving for
{cos t}
{f } = 4
−t e
+
{sin t}.
{f } we obtain
{f } =
4s2 + s + 5
4
2
7
=
+4
.
−
(s + 1)3
s + 1 (s + 1)2
(s + 1)3
Thus
f (t) = 4e−t − 7te−t + 4t2 e−t .
41. The Laplace transform of the given equation is
{f } +
Solving for
{f } we obtain
{f } =
{1}
{f } =
{1}.
1
. Thus, f (t) = e−t .
s+1
42. The Laplace transform of the given equation is
{f } =
Solving for
{cos t} +
−t e
{f }.
{f } we obtain
{f } =
s
1
+
.
s2 + 1 s2 + 1
Thus
f (t) = cos t + sin t.
43. The Laplace transform of the given equation is
{f } =
=
Solving for
{1} +
1
8
1
+ 2+
s s
3
{t} −
{t3 }
t
8
(t − τ )3 f (τ ) dτ
3 0
1
16
1
{f } = + 2 + 4
{f }.
s s
s
{f } we obtain
{f } =
s2 (s + 1)
1 1
3 1
1 2
1 s
=
+
+
+
.
4
2
s − 16
8 s + 2 8 s − 2 4 s + 4 2 s2 + 4
Thus
f (t) =
1 −2t 3 2t 1
1
+ e + sin 2t + cos 2t.
e
8
8
4
2
44. The Laplace transform of the given equation is
{t} − 2
Solving for
{f } we obtain
{f } =
{f } =
t
e − e−t
{f }.
s2 − 1
1 1
1 3!
=
−
.
4
2
2s
2 s
12 s4
227
4.4
Additional Operational Properties
Thus
f (t) =
1
1
t − t3 .
2
12
45. The Laplace transform of the given equation is
s
Solving for
{y} − y(0) =
{1} −
{sin t} −
{1}
{y}.
{f } we obtain
{y} =
s2 − s + 1
1
1
2s
= 2
.
−
(s2 + 1)2
s + 1 2 (s2 + 1)2
Thus
1
t sin t.
2
y = sin t −
46. The Laplace transform of the given equation is
s
Solving for
{f } we obtain
47. The differential equation is
0.1
di
1
+ 3i +
dt
0.05
or
di
+ 30i + 200
dt
{y} − y(0) + 6
{y} =
{y} + 9
{1}
{y} =
1
. Thus, y = te−3t .
(s + 3)2
i
30
t
i(τ )dτ = 100
(t − 2)
(t − 1) −
0
t
i(τ )dτ = 1000
(t − 2) ,
(t − 1) −
0
{i} − y(0) + 30
200
{i} +
s
{i} we obtain
{i} =
1000e−s − 1000e−2s
=
s2 + 30s + 200
i(t) = 100 e−10(t−1) − e−20(t−1)
48. The differential equation is
0.005
di
1
+i+
dt
0.02
or
di
+ 200i + 10,000
dt
0.5 1 1.5 2 2.5 3 t
-10
s
Thus
20
10
where i(0) = 0. The Laplace transform of the differential equation is
Solving for
{1}.
-20
-30
1000 −s
{i} =
(e − e−2s ).
s
100
100
−
(e−s − e−2s ).
s + 10 s + 20
(t − 1) − 100 e−10(t−2) − e−20(t−2)
(t − 2).
i
t
i(τ )dτ = 100 t − (t − 1)
(t − 1)
0
t
2
1.5
i(τ )dτ = 20,000 t − (t − 1)
(t − 1) ,
1
0
where i(0) = 0. The Laplace transform of the differential equation is
1 −s
10,000
1
s {i} + 200 {i} +
{i} = 20,000 2 − 2 e
.
s
s
s
228
0.5
0.5
1
1.5
2 t
4.4
Solving for
Additional Operational Properties
{i} we obtain
20,000
2
2
200
−s
{i} =
(1 − e−s ).
(1 − e ) =
−
−
s(s + 100)2
s s + 100 (s + 100)2
Thus
i(t) = 2 − 2e−100t − 200te−100t − 2
49.
50.
{f (t)} =
{f (t)} =
1
1 − e−2as
1
1 − e−2as
a
e−st dt −
0
a
2a
a
e−st dt =
0
(t − 1) + 2e−100(t−1)
(t − 1) + 200(t − 1)e−100(t−1)
(t − 1).
(1 − e−as )2
1 − e−as
e−st dt =
=
−2as
s(1 − e
)
s(1 + e−as )
1
s(1 + e−as )
51. Using integration by parts,
1
1 − e−bs
{f (t)} =
52.
53.
54.
1
1
{f (t)} =
1 − e−2s
{f (t)} =
{f (t)} =
1
1 − e−πs
−st
te
0
π
−st
(2 − t)e
dt +
0
e−st sin t dt =
π
dt =
1
1
− bs
bs e − 1
.
1 − e−s
s2 (1 − e−2s )
eπs/2 + e−πs/2
πs
1
1
·
coth
= 2
s2 + 1 eπs/2 − e−πs/2
s +1
2
e−st sin t dt =
0
a −st
a
te dt =
b
s
1
0
1
1 − e−2πs
2
b
1
1
·
s2 + 1 1 − e−πs
55. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of the equation is
Ls
From Problem 49 we have
{i} + R
{E(t)}.
{E(t)} = (1 − e−s )/s(1 + e−s ). Thus
(Ls + R)
and
{i} =
{i} =
1 − e−s
s(1 + e−s )
1
1 − e−s
1 1 − e−s
1
=
−s
L s(s + R/L)(1 + e )
L s(s + R/L) 1 + e−s
1
1 1
−
(1 − e−s )(1 − e−s + e−2s − e−3s + e−4s − · · · )
=
R s s + R/L
1 1
1
=
−
(1 − 2e−s + 2e−2s − 2e−3s + 2e−4s − · · · ).
R s s + R/L
{i} =
Therefore,
2 1 (t − 1)
1 − e−Rt/L −
1 − e−R(t−1)/L
R
R
2
2
+
(t − 2) −
1 − e−R(t−2)/L
1 − e−R(t−3)/L
R
R
∞ 2 !
1 =
1 − e−R(t−n)/L
(t − n).
1 − e−Rt/L +
R
R n=1
i(t) =
229
(t − 3) + · · ·
4.4
Additional Operational Properties
The graph of i(t) with L = 1 and R = 1 is shown below.
i
1
0.5
1
2
3
4
t
-0.5
-1
56. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of the equation is
Ls
From Problem 51 we have
{E(t)} =
{i} + R
{i} =
{E(t)}.
1 1
1 1
1
1
− s
= 2−
.
s s e −1
s
s es − 1
Thus
(Ls + R)
{i} =
1
1 1
−
2
s
s es − 1
and
1
1
1
1
1
−
L s2 (s + R/L) L s(s + R/L) es − 1
−s
1 1
L
1
1 1
1
L 1
=
+
−
−
e + e−2s + e−3s + · · · .
−
2
R s
R s R s + R/L
R s s + R/L
{i} =
Therefore
1 L
L −Rt/L
−
(t − 1)
t− + e
1 − e−R(t−1)/L
R R
R
1 1 −
(t − 2) −
(t − 3) − · · ·
1 − e−R(t−2)/L
1 − e−R(t−3)/L
R
R
∞
1
1 !
L
L
=
1 − e−R(t−n)/L
(t − n).
t − + e−Rt/L −
R
R R
R n=1
1
i(t) =
R
The graph of i(t) with L = 1 and R = 1 is shown below.
i
1
0.5
1
2
3
4
t
-0.5
-1
57. The differential equation is x + 2x + 10x = 20f (t), where f (t) is the meander function in Problem 49 with
230
4.4
Additional Operational Properties
a = π. Using the initial conditions x(0) = x (0) = 0 and taking the Laplace transform we obtain
(s2 + 2s + 10)
Then
{x(t)} =
20
1
(1 − e−πs )
s
1 + e−πs
20
(1 − e−πs )(1 − e−πs + e−2πs − e−3πs + · · ·)
=
s
20
=
(1 − 2e−πs + 2e−2πs − 2e−3πs + · · ·)
s
∞
20 40 !
=
(−1)n e−nπs .
+
s
s n=1
{x(t)} =
∞
!
20
40
(−1)n e−nπs
+
s(s2 + 2s + 10) s(s2 + 2s + 10) n=1
∞
!
2
2s + 4
4s + 8
n 4
= − 2
(−1)
+
−
e−nπs
s s + 2s + 10 n=1
s s2 + 2s + 10
∞
!
(s + 1) + 1
2 2(s + 1) + 2
n 1
+4
−
e−nπs
= −
(−1)
2+9
s
(s + 1)2 + 9
s
(s
+
1)
n=1
and
∞
!
1
(−1)n 1 − e−(t−nπ) cos 3(t − nπ)
x(t) = 2 1 − e−t cos 3t − e−t sin 3t + 4
3
n=1
1
− e−(t−nπ) sin 3(t − nπ)
(t − nπ).
3
The graph of x(t) on the interval [0, 2π) is shown below.
x
3
π
2π
t
−3
58. The differential equation is x + 2x + x = 5f (t), where f (t) is the square wave function with a = π. Using the
initial conditions x(0) = x (0) = 0 and taking the Laplace transform, we obtain
(s2 + 2s + 1)
{x(t)} =
5
5
1
= (1 − e−πs + e−2πs − e−3πs + e−4πs − · · ·)
−πs
s 1+e
s
∞
=
Then
{x(t)} =
5!
(−1)n e−nπs .
s n=0
∞
∞
!
!
5
1
1
1
n −nπs
n
e−nπs
(−1)
e
=
5
(−1)
−
−
2
s(s + 1)2 n=0
s
s
+
1
(s
+
1)
n=0
231
4.4
Additional Operational Properties
and
x(t) = 5
∞
!
(−1)n (1 − e−(t−nπ) − (t − nπ)e−(t−nπ) )
(t − nπ).
n=0
The graph of x(t) on the interval [0, 4π) is shown below.
x
5
2π
4π
t
−5
59. f (t) = −
1
t
d
1
[ln(s − 3) − ln(s + 1)] = −
ds
t
1
1
−
s−3 s+1
=−
1 3t
e − e−t
t
60. The transform of Bessel’s equation is
−
d 2
d
[s Y (s) − sy(0) − y (0)] + sY (s) − y(0) −
Y (s) = 0
ds
ds
or, after simplifying and using the initial condition, (s2 + 1)Y + sY = 0. This equation is both separable and
√
linear. Solving gives Y (s) = c/ s2 + 1 . Now Y (s) =
{J0 (t)}, where J0 has a derivative that is continuous
and of exponential order, implies by Problem 46 of Exercises 4.2 that
1 = J0 (0) = lim sY (s) = c lim √
s→∞
s→∞
s
s2 + k 2
=c
so c = 1 and
Y (s) = √
1
s2 + 1
{J0 (t)} = √
or
1
.
s2 + 1
61. (a) Using Theorem 4.8, the Laplace transform of the differential equation is
−
d 2
d
[s Y − sy(0) − y (0)] + sY − y(0) +
[sY − y(0)] + nY
ds
ds
d
d
= − [s2 Y ] + sY +
[sY ] + nY
ds
ds
dY
dY
= −s2
− 2sY + sY + s
+ Y + nY
ds
ds
dY
2
= (s − s )
+ (1 + n − s)Y = 0.
ds
Separating variables, we find
dY
1+n−s
=
ds =
Y
s2 − s
n
1+n
−
s−1
s
ln Y = n ln(s − 1) − (1 + n) ln s + c
Y = c1
(s − 1)n
.
s1+n
232
ds
4.4
Additional Operational Properties
Since the differential equation is homogeneous, any constant multiple of a solution will still be a solution, so
for convenience we take c1 = 1. The following polynomials are solutions of Laguerre’s differential equation:
1
n = 0 : L0 (t) =
=1
s
1
s−1
1
=
=1−t
n = 1 : L1 (t) =
−
s2
s s2
1
1
2
(s − 1)2
1
n = 2 : L2 (t) =
=
− 2 + 3 = 1 − 2t + t2
3
s
s s
s
2
3
1
3
(s − 1)
3
1
1
3
=
n = 3 : L3 (t) =
+ 3 − 4 = 1 − 3t + t2 − t3
−
s4
s s2
s
s
2
6
4
1
4
(s − 1)
6
4
1
=
−
n = 4 : L4 (t) =
+ 3− 4+ 5
s5
s s2
s
s
s
2
1
= 1 − 4t + 3t2 − t3 + t4 .
3
24
(b) Letting f (t) = tn e−t we note that f (k) (0) = 0 for k = 0, 1, 2, . . . , n − 1 and f (n) (0) = n!. Now, by the first
translation theorem,
t n
e d n −t
1
1
{et f (n) (t)} =
{f (n) (t)} s→s−1
=
t
e
n
n! dt
n!
n!
1 n
s
=
{tn e−t } − sn−1 f (0) − sn−2 f (0) − · · · − f (n−1) (0)
n!
s→s−1
1 n
n −t
=
{t e }
s
n!
s→s−1
n!
1 n
(s − 1)n
=
=
= Y,
s
n+1
n!
(s + 1)
sn+1
s→s−1
where Y =
{Ln (t)}. Thus
Ln (t) =
et dn n −t
(t e ),
n! dtn
n = 0, 1, 2, . . . .
62. The output for the first three lines of the program are
9y[t] + 6y [t] + y [t] == t sin[t]
2s
(1 + s2 )2
−11 − 4s − 22s2 − 4s3 − 11s4 − 2s5
Y →−
(1 + s2 )2 (9 + 6s + s2 )
1 − 2s + 9Y + s2 Y + 6(−2 + sY ) ==
The fourth line is the same as the third line with Y → removed. The final line of output shows a solution
involving complex coefficients of eit and e−it . To get the solution in more standard form write the last line as
two lines:
euler={Eˆ(It)−>Cos[t] + I Sin[t], Eˆ(-It)−>Cos[t] - I Sin[t]}
InverseLaplaceTransform[Y, s, t]/.euler//Expand
We see that the solution is
487 247
1
y(t) =
+
t e−3t +
(13 cos t − 15t cos t − 9 sin t + 20t sin t) .
250
50
250
63. The solution is
√
1
1
y(t) = et − e−t/2 cos 15 t −
6
6
233
√
3/5 −t/2
sin 15 t.
e
6
4.4
Additional Operational Properties
64. The solution is
q(t) = 1 − cos t + (6 − 6 cos t)
(t − 3π) − (4 + 4 cos t)
(t − π).
q
5
-5
Π
3Π
t
5Π
EXERCISES 4.5
The Dirac Delta Function
1. The Laplace transform of the differential equation yields
{y} =
1 −2s
e
s−3
so that
y = e3(t−2)
(t − 2).
2. The Laplace transform of the differential equation yields
{y} =
2
e−s
+
s+1 s+1
so that
y = 2e−t + e−(t−1)
(t − 1).
3. The Laplace transform of the differential equation yields
1 {y} = 2
1 + e−2πs
s +1
so that
y = sin t + sin t
(t − 2π).
4. The Laplace transform of the differential equation yields
1
4
{y} =
e−2πs
2
4 s + 16
so that
1
1
y = sin 4(t − 2π) (t − 2π) = sin 4t
4
4
(t − 2π).
5. The Laplace transform of the differential equation yields
1 −πs/2
{y} = 2
e
+ e−3πs/2
s +1
so that
π π
3π
3π
y = sin t −
t−
+ sin t −
t−
2
2
2
2
π
π
= − cos t
t−
+ cos t
t−
.
2
2
234
4.5
The Dirac Delta Function
6. The Laplace transform of the differential equation yields
{y} =
s
1
+
(e−2πs + e−4πs )
s2 + 1 s2 + 1
so that
(t − 2π) +
y = cos t + sin t[
(t − 4π)].
7. The Laplace transform of the differential equation yields
1
1 1 1 1
−s
{y} = 2
(1 + e ) =
−
(1 + e−s )
s + 2s
2 s 2 s+2
so that
1 1
1 1 −2(t−1)
y = − e−2t +
− e
2 2
2 2
(t − 1).
8. The Laplace transform of the differential equation yields
{y} =
s+1
1
3 1 1 1
1 1 −2s
3 1
1 1
−2s
+
e
−
−
−
e
=
+
s2 (s − 2) s(s − 2)
4 s − 2 4 s 2 s2
2 s−2 2 s
so that
y=
3 2t 3 1
1
1
e − − t + e2(t−2) −
4
4 2
2
2
(t − 2).
9. The Laplace transform of the differential equation yields
{y} =
1
e−2πs
(s + 2)2 + 1
so that
y = e−2(t−2π) sin t
(t − 2π).
10. The Laplace transform of the differential equation yields
{y} =
1
e−s
(s + 1)2
so that
y = (t − 1)e−(t−1)
(t − 1).
11. The Laplace transform of the differential equation yields
{y} =
=
4+s
s2 + 4s + 13
+
e−πs + e−3πs
s2 + 4s + 13
−πs
2
3
3
s+2
1
e
+
+
+ e−3πs
2
2
2
2
2
2
3 (s + 2) + 3
(s + 2) + 3
3 (s + 2) + 3
so that
y=
2 −2t
1
sin 3t + e−2t cos 3t + e−2(t−π) sin 3(t − π)
e
3
3
1
+ e−2(t−3π) sin 3(t − 3π) (t − 3π).
3
(t − π)
12. The Laplace transform of the differential equation yields
{y} =
1
e−2s + e−4s
+
2
(s − 1) (s − 6) (s − 1)(s − 6)
=−
−2s
1
1
1
1
1
1
1 1
1 1
+
+ e−4s
−
+
−
+
e
2
25 s − 1 5 (s − 1)
25 s − 6
5 s−1 5 s−6
235
4.5
The Dirac Delta Function
so that
1
1
1
1
1
y = − et − tet + e6t + − et−2 + e6(t−2)
25
5
25
5
5
1
1
(t − 2) + − et−4 + e6(t−4)
5
5
(t − 4).
13. The Laplace transform of the differential equation yields
{y} =
1 2 1 3! 1 P0 3! −Ls/2
y (0) +
y (0) +
e
3
4
2 s
6 s
6 EI s4
so that
y=
1 1
1 P0
y (0)x2 + y (0)x3 +
2
6
6 EI
x−
L
2
3
x−
L
2
.
Using y (L) = 0 and y (L) = 0 we obtain
3
1 P0 L 2 1 P 0 3 1 P 0
L
y=
x −
x +
x−
4 EI
6 EI
6 EI
2

P0 L 2 1 3
L



 EI 4 x − 6 x , 0 ≤ x < 2
=

P0 L2 1
L
L



x−
,
≤ x ≤ L.
4EI 2
12
2
14. From Problem 13 we know that
1
1
1 P0
y = y (0)x2 + y (0)x3 +
2
6
6 EI
L
x−
2
3
L
x−
2
L
x−
2
.
Using y(L) = 0 and y (L) = 0 we obtain
3 1 P0 L 2
L
L
1 P0 3 1 P0
y=
x−
x −
x +
x−
16 EI
12 EI
6 EI
2
2

P0 L 2
L
1 3


0≤x<

 EI 16 x − 12 x ,
2
=
3

P
1 P0
1
L
L 2
L


 0
,
x − x3 +
x−
≤ x ≤ L.
EI 16
12
6 EI
2
2
15. You should disagree. Although formal manipulations of the Laplace transform lead to y(t) = 13 e−t sin 3t in both
cases, this function does not satisfy the initial condition y (0) = 0 of the second initial-value problem.
236
4.6
Systems of Linear Differential Equations
EXERCISES 4.6
Systems of Linear Differential Equations
1. Taking the Laplace transform of the system gives
s
s
{x} = −
{x} +
{y} − 1 = 2
{y}
{x}
so that
{x} =
1
1 1
1 1
=
−
(s − 1)(s + 2)
3 s−1 3 s+2
{y} =
1
2
2 1
1 1
+
=
+
.
s s(s − 1)(s + 2)
3 s−1 3 s+2
and
Then
x=
1 t 1 −2t
e − e
3
3
and
y=
s
{x} − 1 = 2
{y} +
s
{y} − 1 = 8
2 t 1 −2t
e + e .
3
3
2. Taking the Laplace transform of the system gives
1
s−1
1
{x} − 2
s
so that
{y} =
and
y=
s3 + 7s2 − s + 1
1 1
8
1
173 1
53
1
=
−
+
−
2
s(s − 1)(s − 16)
16 s 15 s − 1
96 s − 4 160 s + 4
1
8
173 4t
53 −4t
− et +
e −
e .
16 15
96
160
Then
x=
1 1
173 4t
53 −4t
1
1
y + t = t − et +
e +
e .
8
8
8
15
192
320
3. Taking the Laplace transform of the system gives
s
{x} + 1 =
{x} − 2
{y}
s
{y} − 2 = 5
{x} −
{y}
so that
{x} =
−s − 5
s
5 3
=− 2
−
2
s +9
s + 9 3 s2 + 9
and
x = − cos 3t −
5
sin 3t.
3
Then
y=
1
7
1
x − x = 2 cos 3t − sin 3t.
2
2
3
237
4.6
Systems of Linear Differential Equations
4. Taking the Laplace transform of the system gives
{x} + s
{y} =
1
s
{x} + (s − 1)
{y} =
1
s−1
(s + 3)
(s − 1)
so that
{y} =
5s − 1
4
1
1 1 1 1
+
+
=−
3s(s − 1)2
3 s 3 s − 1 3 (s − 1)2
{x} =
1 − 2s
1
1
1 1 1 1
−
−
=
.
3s(s − 1)2
3 s 3 s − 1 3 (s − 1)2
and
Then
x=
1 1 t 1 t
− e − te
3 3
3
1 1
4
y = − + et + tet .
3 3
3
and
5. Taking the Laplace transform of the system gives
(2s − 2)
(s − 3)
{x} + s
{x} + (s − 3)
1
s
2
{y} =
s
{y} =
so that
{x} =
−s − 3
1 1 5 1
2
=−
+
−
s(s − 2)(s − 3)
2 s 2 s−2 s−3
{y} =
3s − 1
1 1 5 1
8 1
=−
−
+
.
s(s − 2)(s − 3)
6 s 2 s−2 3 s−3
and
Then
1 5
x = − + e2t − 2e3t
2 2
1 5
8
y = − − e2t + e3t .
6 2
3
and
6. Taking the Laplace transform of the system gives
(s + 1)
{x} − (s − 1)
{y} = −1
s
{x} + (s + 2)
{y} = 1
so that
{y} =
and
s + 1/2
s + 1/2
√
=
s2 + s + 1
(s + 1/2)2 + ( 3/2)2
√
√
−3/2
3/2
√
{x} = 2
.
=− 3
2
s +s+1
(s + 1/2) + ( 3/2)2
Then
√
−t/2
y=e
3
cos
t
2
and
√
x = − 3 e−t/2 sin
7. Taking the Laplace transform of the system gives
(s2 + 1)
−
{x} −
{x} + (s2 + 1)
{y} = −2
{y} = 1
so that
{x} =
−2s2 − 1
1 1
3 1
=− 2 −
s4 + 2s2
2 s
2 s2 + 2
and
238
√
3
t.
2
4.6
Systems of Linear Differential Equations
√
1
3
x = − t − √ sin 2 t.
2
2 2
Then
√
1
3
y = x + x = − t + √ sin 2 t.
2
2 2
8. Taking the Laplace transform of the system gives
{x} +
{y} = 1
{x} − (s + 1)
{y} = 1
(s + 1)
4
so that
{x} =
s+2
s+1
2
1
=
+
s2 + 2s + 5
(s + 1)2 + 22
2 (s + 1)2 + 22
and
{y} =
Then
−s + 3
=−
s2 + 2s + 5
1
x = e−t cos 2t + e−t sin 2t
2
s+1
2
+2
.
2
2
(s + 1) + 2
(s + 1)2 + 22
and
y = −e−t cos 2t + 2e−t sin 2t.
9. Adding the equations and then subtracting them gives
d2 x
1
= t2 + 2t
2
dt
2
d2 y
1
= t2 − 2t.
dt2
2
Taking the Laplace transform of the system gives
1
1 3!
1 4!
{x} = 8 +
+
5
s 24 s
3 s4
and
{y} =
so that
x=8+
1 4 1 3
t + t
24
3
1 4!
1 3!
−
5
24 s
3 s4
and
y=
1 4 1 3
t − t .
24
3
10. Taking the Laplace transform of the system gives
6
(s − 4)
{x} + s3
{y} =
(s + 2)
{x} − 2s3
{y} = 0
s2 + 1
so that
{x} =
4
4 1
4 s
8 1
=
−
−
(s − 2)(s2 + 1)
5 s − 2 5 s2 + 1 5 s2 + 1
{y} =
2s + 4
2
1 1
1
2
6 s
8 1
= − 2 −2 3 +
−
+
.
s3 (s − 2)(s2 + 1)
s s
s
5 s − 2 5 s2 + 1 5 s2 + 1
and
Then
x=
and
4 2t 4
8
e − cos t − sin t
5
5
5
1
6
8
y = 1 − 2t − 2t2 + e2t − cos t + sin t.
5
5
5
239
4.6
Systems of Linear Differential Equations
11. Taking the Laplace transform of the system gives
{x} + 3(s + 1)
s2
s2
so that
{x} + 3
{y} = 2
{y} =
1
(s + 1)2
2s + 1
1
1
1 2
1
{x} = − 3
= + 2+
.
−
3
s (s + 1)
s s
2 s
s+1
Then
1
x = 1 + t + t2 − e−t
2
and
y=
1
1
1
1 −t 1 te − x = te−t + e−t − .
3
3
3
3
3
12. Taking the Laplace transform of the system gives
{x} + 2
{y} =
2e−s
s
{x} + (s + 1)
{y} =
1 e−s
+
2
s
(s − 4)
−3
so that
{x} =
−1/2
1
+ e−s
(s − 1)(s − 2)
(s − 1)(s − 2)
1
1 1
1 1
1
=
−
+ e−s −
+
2 s−1 2 s−2
s−1 s−2
and
{y} =
e−s
−s/2 + 2
s/4 − 1
+
+ e−s
s
(s − 1)(s − 2)
(s − 1)(s − 2)
1 1
3 1
1
3 1
−s 1
−
+e
−
+
.
=
4 s−1 2 s−2
s 2 s−1 s−2
Then
x=
1 t 1 2t t−1
+ e2(t−1)
e − e + −e
2
2
y=
3 t 1 2t
3
e − e + 1 − et−1 + e2(t−1)
4
2
2
and
13. The system is
(t − 1)
x1 = −3x1 + 2(x2 − x1 )
x2 = −2(x2 − x1 )
x1 (0) = 0
x1 (0) = 1
x2 (0) = 1
x2 (0) = 0.
Taking the Laplace transform of the system gives
{x1 } − 2
{x2 } = 1
{x1 } + (s2 + 2)
{x2 } = s
(s2 + 5)
−2
240
(t − 1).
4.6
so that
and
Systems of Linear Differential Equations
√
2 s
1 1
2 s
4
6
s2 + 2s + 2
{x1 } = 4
=
+
−
+ √ 2
2
2
2
2
s + 7s + 6
5 s +1 5 s +1 5 s +6 5 6 s +6
√
6
s3 + 5s + 2
4 s
2 1
1 s
2
{x2 } = 2
=
+
+
− √ 2
.
2
2
2
2
(s + 1)(s + 6)
5 s +1 5 s +1 5 s +6 5 6 s +6
Then
x1 =
√
√
2
4
1
2
cos t + sin t − cos 6 t + √ sin 6 t
5
5
5
5 6
x2 =
√
√
4
2
2
1
cos t + sin t + cos 6 t − √ sin 6 t.
5
5
5
5 6
and
14. In this system x1 and x2 represent displacements of masses m1 and m2 from their equilibrium positions. Since
the net forces acting on m1 and m2 are
−k1 x1 + k2 (x2 − x1 )
− k2 (x2 − x1 ) − k3 x2 ,
and
respectively, Newton’s second law of motion gives
m1 x1 = −k1 x1 + k2 (x2 − x1 )
m2 x2 = −k2 (x2 − x1 ) − k3 x2 .
Using k1 = k2 = k3 = 1, m1 = m2 = 1, x1 (0) = 0, x1 (0) = −1, x2 (0) = 0, and x2 (0) = 1, and taking the
Laplace transform of the system, we obtain
{x1 } −
{x2 } = −1
{x1 } − (2 + s2 )
{x2 } = −1
(2 + s2 )
so that
Then
1
{x1 } = − 2
s +3
and
√
1
x1 = − √ sin 3 t
3
and
15. (a) By Kirchhoff’s first law we have i1 = i2 + i3 .
{x2 } =
1
s2 + 3
.
√
1
x2 = √ sin 3 t.
3
By Kirchhoff’s second law, on each loop we have
E(t) = Ri1 + L1 i2 and E(t) = Ri1 + L2 i3 or L1 i2 + Ri2 + Ri3 = E(t) and L2 i3 + Ri2 + Ri3 = E(t).
(b) Taking the Laplace transform of the system
0.01i2 + 5i2 + 5i3 = 100
0.0125i3 + 5i2 + 5i3 = 100
gives
{i2 } + 500
{i3 } =
10,000
s
{i2 } + (s + 400)
{i3 } =
8,000
s
(s + 500)
400
so that
{i3 } =
Then
i3 =
80 80 −900t
− e
9
9
8,000
s2 + 900s
and
=
80 1 80
1
−
.
9 s
9 s + 900
i2 = 20 − 0.0025i3 − i3 =
241
100 100 −900t
.
−
e
9
9
4.6
Systems of Linear Differential Equations
(c) i1 = i2 + i3 = 20 − 20e−900t
16. (a) Taking the Laplace transform of the system
i2 + i3 + 10i2 = 120 − 120
(t − 2)
−10i2 + 5i3 + 5i3 = 0
gives
{i2 } + s
(s + 10)
−10s
so that
and
120 1 − e−2s
s
{i3 } = 0
{i2 } =
48
120(s + 1)
60
12 −2s
=
1
−
e
−
+
1 − e−2s
2
(3s + 11s + 10)s
s + 5/3 s + 2
s
{i3 } =
240
240
240 −2s
=
1
−
e
−
1 − e−2s .
2
3s + 11s + 10
s + 5/3 s + 2
and
Then
{i2 } + 5(s + 1)
{i3 } =
i2 = 12 + 48e−5t/3 − 60e−2t − 12 + 48e−5(t−2)/3 − 60e−2(t−2) (t − 2)
i3 = 240e−5t/3 − 240e−2t − 240e−5(t−2)/3 − 240e−2(t−2) (t − 2).
(b) i1 = i2 + i3 = 12 + 288e−5t/3 − 300e−2t − 12 + 288e−5(t−2)/3 − 300e−2(t−2)
(t − 2)
17. Taking the Laplace transform of the system
i2 + 11i2 + 6i3 = 50 sin t
i3 + 6i2 + 6i3 = 50 sin t
gives
{i2 } + 6
{i3 } =
{i2 } + (s + 6)
{i3 } =
(s + 11)
6
50
s2 + 1
50
s2 + 1
so that
{i2 } =
20 1
375
1
145 s
85
1
50s
=−
+
+
+
.
2
2
2
(s + 2)(s + 15)(s + 1)
13 s + 2 1469 s + 15 113 s + 1 113 s + 1
Then
i2 = −
and
i3 =
85
20 −2t
375 −15t 145
e
e
cos t +
sin t
+
+
13
1469
113
113
25
11
30 −2t
250 −15t 280
1
810
+
−
sin t − i2 − i2 =
e
e
cos t +
sin t.
3
6
6
13
1469
113
113
18. Taking the Laplace transform of the system
0.5i1 + 50i2 = 60
0.005i2 + i2 − i1 = 0
242
4.6
Systems of Linear Differential Equations
gives
s
−200
120
s
{i2 } = 0
{i1 } + 100
{i2 } =
{i1 } + (s + 200)
so that
{i2 } =
24,000
6
6 1 6
s + 100
100
−
.
=
−
s(s2 + 200s + 20,000)
5 s 5 (s + 100)2 + 1002
5 (s + 100)2 + 1002
Then
i2 =
and
6 6 −100t
6
cos 100t − e−100t sin 100t
− e
5 5
5
i1 = 0.005i2 + i2 =
6 6 −100t
cos 100t.
− e
5 5
19. Taking the Laplace transform of the system
2i1 + 50i2 = 60
0.005i2 + i2 − i1 = 0
gives
{i1 } + 50
2s
−200
{i1 } + (s + 200)
60
s
{i2 } = 0
{i2 } =
so that
{i2 } =
6,000
s(s2 + 200s + 5,000)
√
√
6 1 6
6 2
s + 100
50 2
√
√
=
−
.
−
5 s 5 (s + 100)2 − (50 2 )2
5 (s + 100)2 − (50 2 )2
Then
and
√
√
√
6 6 −100t
6 2 −100t
i2 = − e
e
cosh 50 2 t −
sinh 50 2 t
5 5
5
√
√
√
6 6 −100t
9 2 −100t
i1 = 0.005i2 + i2 = − e
e
cosh 50 2 t −
sinh 50 2 t.
5 5
10
20. (a) Using Kirchhoff’s first law we write i1 = i2 + i3 . Since i2 = dq/dt we have i1 − i3 = dq/dt. Using Kirchhoff’s
second law and summing the voltage drops across the shorter loop gives
E(t) = iR1 +
1
q,
C
so that
i1 =
1
1
q.
E(t) −
R1
R1 C
Then
dq
1
1
= i1 − i3 =
q − i3
E(t) −
dt
R1
R1 C
and
R1
dq
1
+ q + R1 i3 = E(t).
dt
C
243
(1)
4.6
Systems of Linear Differential Equations
Summing the voltage drops across the longer loop gives
E(t) = i1 R1 + L
di3
+ R2 i3 .
dt
Combining this with (1) we obtain
i1 R1 + L
or
L
di3
1
+ R2 i3 = i1 R1 + q
dt
C
di3
1
+ R2 i3 − q = 0.
dt
C
(b) Using L = R1 = R2 = C = 1, E(t) = 50e−t
(t − 1) = 50e−1 e−(t−1)
the Laplace transform of the system we obtain
50e−1 −s
e
s+1
(s + 1)
{q} +
{i3 } =
(s + 1)
{i3 } −
{q} = 0,
{q} =
50e−1 e−s
(s + 1)2 + 1
so that
(t − 1), q(0) = i3 (0) = 0, and taking
and
q(t) = 50e−1 e−(t−1) sin(t − 1)
(t − 1) = 50e−t sin(t − 1)
(t − 1).
21. (a) Taking the Laplace transform of the system
4θ1 + θ2 + 8θ1 = 0
θ1 + θ2 + 2θ2 = 0
gives
so that
4 s2 + 2
{θ2 } = 3s
s2
{θ2 } = 0
{θ1 } + s2
{θ1 } + s2 + 2
3s2 + 4 s2 + 4
or
{θ2 } =
Then
θ2 =
θ1 =
(b)
θ1
s
3 s
1
−
.
2
2 s + 4/3 2 s2 + 4
2
1
3
cos √ t − cos 2t
2
2
3
so that
{θ2 } = −3s3
θ1 = −θ2 − 2θ2
and
3
2
1
cos √ t + cos 2t.
4
4
3
θ2
2
2
1
1
−1
3π
6π t
−1
−2
−2
244
3π
6π t
4.6
Systems of Linear Differential Equations
Mass m2 has extreme displacements of greater magnitude. Mass m1 first passes through its equilibrium
position at about t = 0.87, and mass m2 first passes through its equilibrium position at about t = 0.66.
√
√
The motion of the pendulums is not periodic since cos(2t/ 3 ) has period 3 π, cos 2t has period π, and
√
the ratio of these periods is 3 , which is not a rational number.
(c) The Lissajous curve is plotted for 0 ≤ t ≤ 30.
θ2
2
1
-1-0.5
0.5 1
θ1
-1
-2
(d)
t=0
t=1
t=2
t=3
t=4
t=5
t=6
t=7
t=8
t
θ1
θ2
1
2
3
4
5
6
7
8
9
10
-0.2111
-0.6585
0.4830
-0.1325
-0.4111
0.8327
0.0458
-0.9639
0.3534
0.4370
0.8263
0.6438
-1.9145
0.1715
1.6951
-0.8662
-0.3186
0.9452
-1.2741
-0.3502
t=9
(e) Using a CAS to solve θ1 (t) = θ2 (t) we see that θ1 = θ2 (so that the double pendulum
t=10
t=0.75
is straight out) when t is about 0.75 seconds.
(f ) To make a movie of the pendulum it is necessary to locate the mass in the plane as a function of time.
Suppose that the upper arm is attached to the origin and that the equilibrium position lies along the
245
4.6
Systems of Linear Differential Equations
negative y-axis. Then mass m1 is at (x, (t), y1 (t)) and mass m2 is at (x2 (t), y2 (t)), where
x1 (t) = 16 sin θ1 (t)
and
y1 (t) = −16 cos θ1 (t)
x2 (t) = x1 (t) + 16 sin θ2 (t)
and
y2 (t) = y1 (t) − 16 cos θ2 (t).
and
A reasonable movie can be constructed by letting t range from 0 to 10 in increments of 0.1 seconds.
CHAPTER 4 REVIEW EXERCISES
1
1.
{f (t)} =
−st
te
∞
dt +
0
4
2.
{f (t)} =
(2 − t)e−st dt =
1
e−st dt =
2
1
2
− 2 e−s
s2
s
1 −2s
− e−4s
e
s
3. False; consider f (t) = t−1/2 .
4. False, since f (t) = (et )10 = e10t .
5. True, since lims→∞ F (s) = 1 = 0. (See Theorem 4.5 in the text.)
6. False; consider f (t) = 1 and g(t) = 1.
7.
8.
9.
10.
11.
12.
−7t e
=
1
s+7
1
(s + 7)2
te−7t =
2
s2 + 4
−3t
2
e
sin 2t =
(s + 3)2 + 4
d
2
4s
{t sin 2t} = −
= 2
ds s2 + 4
(s + 4)2
{sin 2t} =
{sin 2t
13.
14.
15.
16.
(t − π)} =
(t − π)} =
2
s2 + 4
e−πs
1 5!
1
= t5
6 s6
6
1
1
1
1
=
= et/3
3s − 1
3 s − 1/3
3
1
1
2
1
=
= t2 e5t
(s − 5)3
2
(s − 5)3
2
√
1
1 √
1
1
1
1
1
√
√
√
√
=
−
+
= − √ e− 5 t + √ e 5 t
2
s −5
2 5 s+ 5 2 5 s− 5
2 5
2 5
20
s6
{sin 2(t − π)
=
246
CHAPTER 4 REVIEW EXERCISES
s
=
s2 − 10s + 29
1 −5s
= (t − 5)
e
s2
s + π −s
=
e
s2 + π 2
17.
18.
19.
5
s−5
2
+
2
2
(s − 5) + 2
2 (s − 5)2 + 22
21.
22.
s
π
e−s + 2
e−s
s2 + π 2
s + π2
= cos π(t − 1)
1
1 L
= 2
L2 s2 + n2 π 2
L nπ
−5t exists for s > −5.
e
23.
24.
5
= e5t cos 2t + e5t sin 2t
2
(t − 5)
20.
(t − 1) + sin π(t − 1) (t − 1)
1
nπ/L
nπ
=
sin
t
s2 + (n2 π 2 )/L2
Lnπ
L
d
te8t f (t) = − F (s − 8).
ds
{eat f (t − k) (t − k)} = e−ks {ea(t+k) f (t)} = e−ks eak {eat f (t)} = e−k(s−a) F (s − a)
t
1
F (s − a)
eaτ f (τ ) dτ =
{eat f (t)} =
, whereas
s
s
0
t
t
F (s) F (s − a)
at
e
.
f (τ ) dτ =
f (τ ) dτ =
=
s
s−a
0
0
s→s−a
s→s−a
25. f (t)
(t − t0 )
26. f (t) − f (t)
27. f (t − t0 )
(t − t0 )
(t − t0 )
28. f (t) − f (t)
(t − t0 ) + f (t)
(t − t1 )
29. f (t) = t − [(t − 1) + 1] (t − 1) +
(t − 1) −
(t − 4) = t − (t − 1)
(t − 1) −
(t − 4)
1
1
1
− 2 e−s − e−4s
2
s
s
s
t
1
1
1 −4(s−1)
e f (t) =
−
e−(s−1) −
e
2
2
(s − 1)
(s − 1)
s−1
{f (t)} =
(t − π) − sin t (t − 3π) = − sin(t − π) (t − π) + sin(t − 3π)
1
1
{f (t)} = − 2
e−πs + 2
e−3πs
s +1
s +1
t
1
1
e f (t) = −
e−π(s−1) +
e−3π(s−1)
(s − 1)2 + 1
(s − 1)2 + 1
30. f (t) = sin t
31. f (t) = 2 − 2
(t − 2) + [(t − 2) + 2] (t − 2) = 2 + (t − 2)
2
1
{f (t)} = + 2 e−2s
s s
t
2
1
e−2(s−1)
e f (t) =
+
s − 1 (s − 1)2
32. f (t) = t − t
(t − 2)
(t − 1) + (2 − t) (t − 1) − (2 − t) (t − 2) = t − 2(t − 1)
1
2
1
{f (t)} = 2 − 2 e−s + 2 e−2s
s
s
s
t
1
2
1
e f (t) =
−
e−(s−1) +
e−2(s−1)
(s − 1)2
(s − 1)2
(s − 1)2
247
(t − 3π)
(t − 1) + (t − 2)
(t − 2)
CHAPTER 4 REVIEW EXERCISES
33. Taking the Laplace transform of the differential equation we obtain
{y} =
so that
5
2
1
+
(s − 1)2
2 (s − 1)3
1
y = 5tet + t2 et .
2
34. Taking the Laplace transform of the differential equation we obtain
{y} =
=
1
(s − 1)2 (s2 − 8s + 20)
6
6
5
1
1
1
s−4
2
−
+
+
2
2
2
169 s − 1 13 (s − 1)
169 (s − 4) + 2
338 (s − 4)2 + 22
so that
y=
6 t
1
6 4t
5 4t
e + tet −
e cos 2t +
e sin 2t.
169
13
169
338
35. Taking the Laplace transform of the given differential equation we obtain
s3 + 6s2 + 1
1
2
e−2s −
e−2s
s2 (s + 1)(s + 5)
s(s + 1)(s + 5)
6 1 1 1
3
1
13
1
=− · + · 2 + ·
−
·
25 s 5 s
2 s + 1 50 s + 5
6 1 1 1
1
1
1
1
− − · + · 2+ ·
−
·
e−2s
25 s 5 s
4 s + 1 100 s + 5
1
1
1
2 1 1
· − ·
+
·
e−2s
−
5 s 2 s + 1 10 s + 5
{y} =
so that
y=−
s2 (s + 1)(s + 5)
−
6
13
4
1
3
1
+ t + e−t − e−5t −
(t − 2) − (t − 2) (t − 2)
25 5
2
50
25
5
1 −(t−2)
9 −5(t−2)
+ e
(t − 2) −
(t − 2).
e
4
100
36. Taking the Laplace transform of the differential equation we obtain
{y} =
s3 + 2
s3 (s − 5)
=−
−
2 + 2s + s2 −s
e
s3 (s − 5)
2 1
1 2
127 1
1 2
37
2 1
37 1 12 1
1
−
+
−
+
−
−
−
−
e−s
125 s 25 s2
5 s3
125 s − 5
125 s 25 s2
5 s3
125 s − 5
so that
2
127 5t
37
37 5(t−1)
2
1
12
1
y=−
− t − t2 +
e − −
− (t − 1) − (t − 1)2 +
e
125 25
5
125
125 25
5
125
37. Taking the Laplace transform of the integral equation we obtain
{y} =
so that
1
1 2
1
+
+
s s2
2 s3
1
y(t) = 1 + t + t2 .
2
38. Taking the Laplace transform of the integral equation we obtain
(
{f })2 = 6 ·
6
s4
or
248
{f } = ±6 ·
1
s2
(t − 1).
CHAPTER 4 REVIEW EXERCISES
so that f (t) = ±6t.
39. Taking the Laplace transform of the system gives
s
4
1
+1
s2
{x} +
{y} =
{x} + s
{y} = 2
so that
{x} =
s2 − 2s + 1
1 1 1 1
9 1
=−
+
+
.
s(s − 2)(s + 2)
4 s 8 s−2 8 s+2
Then
1 1
9
x = − + e2t + e−2t
4 8
8
and y = −x + t =
9 −2t 1 2t
− e + t.
e
4
4
40. Taking the Laplace transform of the system gives
s2
{x} + s2
2s
{x} + s2
1
s−2
1
{y} = −
s−2
{y} =
so that
{x} =
2
1
1 1 1 1
−
+
=
s(s − 2)2
2 s 2 s − 2 (s − 2)2
{y} =
−s − 2
1
3 1 1 1
3 1
−
−
=−
+
.
s2 (s − 2)2
4 s 2 s2
4 s − 2 (s − 2)2
and
Then
x=
1 1 2t
− e + te2t
2 2
3
3 1
and y = − − t + e2t − te2t .
4 2
4
41. The integral equation is
t
i(τ ) dτ = 2t2 + 2t.
10i + 2
0
Taking the Laplace transform we obtain
4
s
s+2
9
2
9
2
2
45
9
{i} =
= 2
=− + 2 +
=− + 2 +
.
+
s3
s2 10s + 2
s (5s + 2)
s s
5s + 1
s s
s + 1/5
Thus
i(t) = −9 + 2t + 9e−t/5 .
42. The differential equation is
1 d2 q
dq
+ 10 + 100q = 10 − 10
2 dt2
dt
Taking the Laplace transform we obtain
{q} =
=
20
s(s2 + 20s + 200)
1 − e−5s
(t − 5).
1 1
1
1
s + 10
10
−
−
1 − e−5s
2
2
2
2
10 s 10 (s + 10) + 10
10 (s + 10) + 10
249
CHAPTER 4 REVIEW EXERCISES
so that
q(t) =
1
1
1
− e−10t cos 10t − e−10t sin 10t
10 10
10
1
1 −10(t−5)
1 −10(t−5)
−
cos 10(t − 5) − e
sin 10(t − 5) (t − 5).
− e
10 10
10
43. Taking the Laplace transform of the given differential equation we obtain
2w0
c1 2!
1
1
c2 3!
L 4!
5!
5!
{y} =
+
· 5−
· 6+
· 6 e−sL/2 +
·
·
EIL 48 s
120 s
120 s
2 s3
6 s4
so that
y=
2w0 L 4
1 5
1
x −
x +
EIL 48
120
120
x−
L
2
5
x−
L
2
+
c1 2 c2 3
x + x
2
6
where y (0) = c1 and y (0) = c2 . Using y (L) = 0 and y (L) = 0 we find
c2 = −w0 L/4EI.
c1 = w0 L2 /24EI,
Hence
w0
L
L2 3 L3 2 1
1
y=
− x5 + x4 −
x +
x +
12EIL
5
2
2
4
5
L
x−
2
5
L
x−
2
.
44. In this case the boundary conditions are y(0) = y (0) = 0 and y(π) = y (π) = 0. If we let c1 = y (0) and
c2 = y (0) then
s4
{y} − s3 y(0) − s2 y (0) − sy(0) − y (0) + 4
and
{y} =
{y} =
{δ(t − π/2)}
c1
2s
c2
4
w0
4
·
+
·
+
·
e−sπ/2 .
2 s4 + 4
4 s4 + 4 4EI s4 + 4
From the table of transforms we get
c1
c2
y=
sin x sinh x + (sin x cosh x − cos x sinh x)
2
4
w0 π
π
π
π +
sin x −
cosh x −
− cos x −
sinh x −
4EI
2
2
2
2
x−
π
2
x−
π
.
2
Using y(π) = 0 and y (π) = 0 we find
c1 =
Hence
y=
w0 sinh π2
,
EI sinh π
c2 = −
w0 cosh π2
.
EI sinh π
w0 sinh π2
w0 cosh π2
sin x sinh x −
(sin x cosh x − cos x sinh x)
2EI sinh π
4EI sinh π
w0 π
π
π
π +
sin x −
cosh x −
− cos x −
sinh x −
4EI
2
2
2
2
45. (a) With ω 2 = g/l and K = k/m the system of differential equations is
θ1 + ω 2 θ1 = −K(θ1 − θ2 )
θ2 + ω 2 θ2 = K(θ1 − θ2 ).
Denoting the Laplace transform of θ(t) by Θ(s) we have that the Laplace transform of the system is
(s2 + ω 2 )Θ1 (s) = −KΘ1 (s) + KΘ2 (s) + sθ0
(s2 + ω 2 )Θ2 (s) = KΘ1 (s) − KΘ2 (s) + sψ0 .
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CHAPTER 4 REVIEW EXERCISES
If we add the two equations, we get
s
Θ1 (s) + Θ2 (s) = (θ0 + ψ0 ) 2
s + ω2
which implies
θ1 (t) + θ2 (t) = (θ0 + ψ0 ) cos ωt.
This enables us to solve for first, say, θ1 (t) and then find θ2 (t) from
θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt.
Now solving
(s2 + ω 2 + K)Θ1 (s) − KΘ2 (s) = sθ0
−kΘ1 (s) + (s2 + ω 2 + K)Θ2 (s) = sψ0
gives
[(s2 + ω 2 + K)2 − K 2 ]Θ1 (s) = s(s2 + ω 2 + K)θ0 + Ksψ0 .
Factoring the difference of two squares and using partial fractions we get
Θ1 (s) =
θ0 + ψ0
s
s
s(s2 + ω 2 + K)θ0 + Ksψ0
θ0 − ψ0
=
,
+
2
2
2
2
2
2
2
(s + ω )(s + ω + 2K)
2
s +ω
2
s + ω 2 + 2K
so
θ0 + ψ0
θ0 − ψ0
cos ωt +
cos ω 2 + 2K t.
2
2
Then from θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt we get
θ0 + ψ0
θ0 − ψ0
θ2 (t) =
cos ωt −
cos ω 2 + 2K t.
2
2
θ1 (t) =
(b) With the initial conditions θ1 (0) = θ0 , θ1 (0) = 0, θ2 (0) = θ0 , θ2 (0) = 0 we have
θ1 (t) = θ0 cos ωt,
θ2 (t) = θ0 cos ωt.
Physically this means that both pendulums swing in the same direction as if they were free since the spring
exerts no influence on the motion (θ1 (t) and θ2 (t) are free of K).
With the initial conditions θ1 (0) = θ0 , θ1 (0) = 0, θ2 (0) = −θ0 , θ2 (0) = 0 we have
θ1 (t) = θ0 cos ω 2 + 2K t,
θ2 (t) = −θ0 cos ω 2 + 2K t.
Physically this means that both pendulums swing in the opposite directions, stretching and compressing
the spring. The amplitude of both displacements is |θ0 |. Moreover, θ1 (t) = θ0 and θ2 (t) = −θ0 at precisely
the same times. At these times the spring is stretched to its maximum.
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