Dennis G. Zill - Differential Equations with
Boundary-Value Problems, Exercise 7.1. From 11 to 18
Leonardo J. Perez Gomez
14 de agosto de 2022
1. Let the figure 1
2.00
1.75
1.50
1.25
f(t)
1.00
0.75
0.50
0.25
0.00
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
t
Figura 1: Problem 7
0 0≤t<1
∴f (t) =
f1 (t) t≥1
If f1 (t) = t
0 0≤t<1
∴f (t) =
t t≥1
Z ∞ Z 1 Z ∞
∴ L {f (t)} = −st
e f (t)dt = −st
e (0)dt + e−st (t)dt
Z0 ∞ 0 1
= e−st (t)dt
1
1
, e−st (t)dt
R
1.
dv = e−st dt e−st t e−st
Z Z
u=t −st
−st ∴ e dt = − − − dt
du = dt v = − e s s s
e−st t 1 e−st t 1
Z −st
−st e
=− + e dt = − + −
s s s s s
−st −st
e t e t 1
=− − 2 = e−st − − 2
s s s s
b
t 1
∴ L {f (t)} = lı́m e − − 2 −st
b→∞ s s 1
−s 1 1
=e − − 2
s s
2. Let the figure 2
4.0
3.5
3.0
2.5
f(t)
2.0
1.5
1.0
0.5
0.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
t
Figura 2: Problem 8
0 0≤t<1
∴ f (t) =
f1 (t) t≥1
If f1 (t) = 2t − 2
0 0≤t<1
∴ f (t) =
2t − 2 t≥1
Z 1 Z ∞ Z ∞
∴ L {f (t)} = e −st
(0)dt + e−st
(2(t − 1))dt = e−st (2(t − 1))dt
0 1 1
2
Boundary-Value Problems, Exercise 7.1. From 11 to 18
Leonardo J. Perez Gomez
14 de agosto de 2022
1. Let the figure 1
2.00
1.75
1.50
1.25
f(t)
1.00
0.75
0.50
0.25
0.00
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
t
Figura 1: Problem 7
0 0≤t<1
∴f (t) =
f1 (t) t≥1
If f1 (t) = t
0 0≤t<1
∴f (t) =
t t≥1
Z ∞ Z 1 Z ∞
∴ L {f (t)} = −st
e f (t)dt = −st
e (0)dt + e−st (t)dt
Z0 ∞ 0 1
= e−st (t)dt
1
1
, e−st (t)dt
R
1.
dv = e−st dt e−st t e−st
Z Z
u=t −st
−st ∴ e dt = − − − dt
du = dt v = − e s s s
e−st t 1 e−st t 1
Z −st
−st e
=− + e dt = − + −
s s s s s
−st −st
e t e t 1
=− − 2 = e−st − − 2
s s s s
b
t 1
∴ L {f (t)} = lı́m e − − 2 −st
b→∞ s s 1
−s 1 1
=e − − 2
s s
2. Let the figure 2
4.0
3.5
3.0
2.5
f(t)
2.0
1.5
1.0
0.5
0.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
t
Figura 2: Problem 8
0 0≤t<1
∴ f (t) =
f1 (t) t≥1
If f1 (t) = 2t − 2
0 0≤t<1
∴ f (t) =
2t − 2 t≥1
Z 1 Z ∞ Z ∞
∴ L {f (t)} = e −st
(0)dt + e−st
(2(t − 1))dt = e−st (2(t − 1))dt
0 1 1
2