108-3: Engineering Mathematics (I)
Exam-3
Time: 2020-07-25 Sat. 8:10AM-10:40AM (No extension)
Contact
Name
E-mail
Location
Extension
Lecturer
Wenson Chang
wenson@ee.ncku.edu.tw
92A31
62392
TA1
TA2
程浩偉
king295705@gmail.com
92912
62400-1612
李庚道
philiplovemaster@gmail.com
92912
62400-1612
Problem 1(15%):
Solve the following differential equation using Lapalace transform.
y 00 + 4y 0 + 3y = et ; with y(0) = 0, y 0 (0) = 2 .
Sol:
See Example 3.4 on page 12.
Problem 2(20%):
Find f (t) using Lapalce transform.
Z
t
f (τ ) cos(2(t − τ ))dτ
f (t) = 3 +
0
Sol:
See Problem 19 of Ch3.4.
Problem 3(15%):
Compute L[g], where g(t) = 0 for 0 ≤ t < 2 and g(t) = t2 + 1 for t ≥ 3.
Sol:
L[g]
= L H(t − 3)(t2 + 1)
= L (t − 3)2 H(t − 2) + 6(t − 3)H(t − 3) + 10H(t − 3)
= L (t − 3)2 H(t − 2)] + 6L[(t − 3)H(t − 3)] + 10L[H(t − 3)
n!
−3s
2
−3s
−3s
n
= e L[t ] + 6e L[t] + 10e L[1] t
sn+1
2
6
10
= e−3s 3 + 2 +
.
s
s
s
Problem 4(15%):
Determine f using Laplace transform such that
2
Z
t
f (t − τ )e−2τ dτ .
f (t) = 3t + 2
0
1
(1)
Sol:
Z
2
t
dτ
n!
f (t − τ )e
= L 3t + 2
0
2
F (s) = L 3t + 2(f ∗ e−2t )(t)
6
2
F (s) =
+ F (s)
s3
s+2
12
6
+ 4 .
F (s) =
s3
s
L[f (t)]
f (t)
= L−1
=
6
12
+ 4
3
s
s
tn
−2τ
sn+1
3t2 + 2t3 . Problem 5(20%):
Solve the following differential equation using Lapalace transform.
(1 − t)y 00 + ty 0 − y = 0; y(0) = 3; y 0 (0) = −1.
Sol:
See Problem 2 of Ch3.7.
Problem 6(15%):
Using Lapalace transform to solve
y 00 + 2y 0 + 2y = δ(t − 3); y(0) = y 0 (0) = 0 .
Sol:
See Example 3.18 on page 28.
2