Download ENGR 311 TRANSFORM CALCULUS AND PARTIAL DIFFERENTIAL EQUATIONS and more Exams Advanced Education in PDF only on Docsity!
[ ]
dt
2
dx
+ 2 L { y }= L e
CONCORDIA UNIVERSITY
FACULTY OF ENGINEERING AND COMPUTER SCIENCE
DEPARTMENT OF MECHANICAL AND INDUSTRIAL ENGINEERING
ENGR 311 TRANSFORM CALCULUS AND PARTIAL DIFFERENTIAL EQUATIONS
Mid Term Examination (May 28, 2026 )
Total Marks: 30 Time: 18:30 – 20:
Individual Work – closed book/notes test;
Materials allowed: approved calculator;
You must show all the steps in the calculations(s) to get the full mark assigned to the question.
1 Use the Laplace transform to solve the following differential equations (5)
(a
y
''
− 3 y
'
−
4 t
, y
=5, y
'
Laplace transform of DE yields
{
d
2
} {
dy
}
− 4 t
s
2
Y ( s )− sy ( 0 )− y
'
( 0 )− 3 sY ( s )− y ( 0 ) + 2 sY ( s )=
s + 4
Putting the initial values in the equation and solving for Y ( s )
s
(¿¿ 2 − 3 s + 2 ) Y ( s )= 5 s − 10 +
s + 4
s
(¿¿ 2 − 3 s + 2 )( s + 4 )
Y ( s )=
5 s
s
2
− 3 s + 2 s
2
− 3 s + 2
5 s
2
Y s =
s − 1
s − 2
s + 4
Y ( s )=¿
5 s − 1 6 s − 2 30 s + 4
Taking inverse Laplace transform
y ( t )=
e
t
e
2 t
e
− 4 t
(b
y + 2 y + 2 y =∂( t - u )
Laplace transform of DE yields
[ s
2
Y ( s )− sy ( 0 )− y
'
( 0 )]+ 2 [ sY ( s )− y ( 0 )]+ 2 Y ( s )= e
−
πs
Putting the initial values in the equation and solving for Y ( s )
[ s
2
Y
s
− 0 − 0 ]+ 2 [ sY
s
− 0 ]+ 2 Y ( s )= e
−
πs
[ s
2
+ 2 s + 2 ] Y ( s )= e
−
πs
L − 3 L
{
{
s
2
s
2
s
s ( s + 1 ) s ( s + 1 )
s s + 1 s s + 1
2 (a
e
− πs
Y ( s )=
( s + 1 )
2
Taking inverse Laplace transform
y
t
= e
−(
t
−
π
)
sin( t − π ) u ( t − π )
Express the following function in terms of unit step functions and
evaluate
L { f ( t )}.
f ( t )= 20 t − 20 t u ( t − 5 )
F ( s )=
− 20 e
−
5 s
(
)
t
f ( t )=
20 t , 0 ≤t ∈ 5
0, t ≥ 5
(b Find the inverse Laplace transform of:
F ( s ) =
2 - 3 se
s ( s +1)
F ( s )=
(
3 s − 4
)
e
− s
(
)
e
− s
f
t
= 2 − 2 e
−
t
t − 1
− 7 e
−(
t −
1 )
u ( t − 1 )
3 Find the Laplace transform of the given periodic function.
f (t)
a 2a 3a 4a
t
The function f ( t ) has amplitude 1 and interval 2a. On interval 0 ≤t ∈ 2 a , f ( t ) can be
defined by
f ( t )=
1, 0 ≤ t ∈ a
−1, a≤ t ∈ 2 a
Using the theorem of a periodic function, the Laplace transform of the function can be
written as
f (t)
− 3
− 3
− 3
2 2
− 3
− 3
∫
(− 2 x ) dx + ∫
( 3 ) dx
− 3 0
a
n
∫
(− 2 x ) cos
− 3
x dx + ∫
( 3 )cos
0
x dx
b
n
∫(− 2 x ) sin
− 3
x dx + ∫
( 3 ) sin
0
x dx
3 nπ 3
− 3
nπ nπ 3 nπ 3
f ( x )=
− 2 x , − 3 ∈ x ∈ 0
3, 0 ≤ x ∈ 3
In the given problem, p = 3, we have from the formula of Fourier series
3
a
0
∫
f ( x ) dx
[
0 3
]
[
(− x
2
)∣
0
+( 3 x )
3
]
− 3 0
(
)
3
nπ
a
n
∫
f ( x )cos
[
0
x dx
nπ
3
nπ
1 ]
[
0 0
nπ 3 (− 2 )
nπ
nπ
]
(− 2 x )
nπ
sin
[
x
0
∫
sin
− 3
x dx +
0
nπ
sin
x
]
1 − 6 x
sin
nπ
x
−cos
nπ
x
sin
nπ
x
3 nπ
[
− 3
0
nπ nπ 3 − 3
nπ 3 0
0
]
1 − 6 x
3 nπ
sin
nπ
x
− 3
n π
cos
nπ
x
− 3
nπ
sin
nπ
x
0
Putting the limits in the above equation, we get
n
n
2
π
2
n
n
2
π
2
3
nπ
b
n
∫
f ( x )sin
[
0
x dx
nπ
3
nπ
]
[
0 0
nπ 3 (− 2 )
nπ
nπ
3
]
(− 2 x )
nπ
−cos
[
0
x
∫
(−cos
− 3
0
x ) dx +
nπ
−cos
]
x
1 6 x
cos
nπ
x
sin
nπ
x
cos
nπ
x
3
nπ
3
3
nπ
3
− 3 0
0
0
]
3 nπ 3 n
2
π
2
3 nπ 3
nπ
nπ
f ( x )= 3 + ∑
n = 1
n
2
π
2
cos
x +
nπ
sin
x
[
6 x
cos
nπ
x
sin
nπ
x
3
cos
nπ
x
Putting the limits in the above equation, we get
n
9 n
n
n
nπ
n
nπ
nπ
nπ
Therefore the function can be written as
∞
n
nπ
n
nπ
0
0
− 3 − 3
0
