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 – 20:30
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' +2 y=e−4t , y (0)=5, y' ( 0 )=5
) Laplace transform of DE yields
L { } {}
d2
dt2
−3 L
dy
dx
+2 L { y }=L {e −4 t }
1
s Y ( s)−sy (0)− y (0)−3 [sY (s )− y (0) ] +2 sY ( s )=
2 '
s+ 4
Putting the initial values in the equation and solving for Y(s)
s
1
(¿¿ 2−3 s+2)Y ( s)=5 s−10+
s +4
¿
s
(¿¿ 2−3 s +2)( s+ 4)
Y (s )= 2 5 s − 2 10 +1
s −32s +2 s −3 s+ 2 ¿
5 s +10 s−39
Y (s )=
( s−1 ) ( s−2 ) ( s +4 )
24 1 1 1 1 1
Y (s )=¿ + +
5 s−1 6 s−2 30 s+ 4
Taking inverse Laplace transform
24 1 1
y (t )= et+ e2t+ e−4 t
5 6 30
(b y + 2 y + 2 y =∂(t - u ) (5)
) 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 ]+ 2Y ( s)=e−πs
[ s 2 +2 s+ 2 ] Y ( s)=e−πs
ENGR 311- Page 1 of 5
, e−πs
Y ( s)=
( s+1)2+ 1
Taking inverse Laplace transform
y (t )=e−(t−π) sin(t −π )u (t −π )
2 (a Express the following function in terms of unit step functions and (2.5
. ) evaluate )
L{f (t)}.
f (t)
100
t
5
{
f (t )= 20 t , 0 ≤t ∈5
0, t ≥ 5
f (t )=20 t −20 t u(t −5)
(b
20
( ) 1 5
F ( s)= s2 −20 e − 5s s2 + s
Find the inverse Laplace transform of: (5)
)
2 - 3se- s + 4e- s
F (s) =
s(s +1)
2
(
3 s−4
F ( s)= s( s+1) − s( s+1) e−s )
2 2 4 7
¿ s − s +1+ s − s+1 e−s( )
f ( t )=2−2e−t+ 4 u ( t −1)−7 e−(t−1)u( t−1)
3 Find the Laplace transform of the given periodic function. (2.5
)
f (t)
1
t
a 2a 3a 4a
–1
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
ENGR 311- Page 2 of 5
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 – 20:30
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' +2 y=e−4t , y (0)=5, y' ( 0 )=5
) Laplace transform of DE yields
L { } {}
d2
dt2
−3 L
dy
dx
+2 L { y }=L {e −4 t }
1
s Y ( s)−sy (0)− y (0)−3 [sY (s )− y (0) ] +2 sY ( s )=
2 '
s+ 4
Putting the initial values in the equation and solving for Y(s)
s
1
(¿¿ 2−3 s+2)Y ( s)=5 s−10+
s +4
¿
s
(¿¿ 2−3 s +2)( s+ 4)
Y (s )= 2 5 s − 2 10 +1
s −32s +2 s −3 s+ 2 ¿
5 s +10 s−39
Y (s )=
( s−1 ) ( s−2 ) ( s +4 )
24 1 1 1 1 1
Y (s )=¿ + +
5 s−1 6 s−2 30 s+ 4
Taking inverse Laplace transform
24 1 1
y (t )= et+ e2t+ e−4 t
5 6 30
(b y + 2 y + 2 y =∂(t - u ) (5)
) 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 ]+ 2Y ( s)=e−πs
[ s 2 +2 s+ 2 ] Y ( s)=e−πs
ENGR 311- Page 1 of 5
, e−πs
Y ( s)=
( s+1)2+ 1
Taking inverse Laplace transform
y (t )=e−(t−π) sin(t −π )u (t −π )
2 (a Express the following function in terms of unit step functions and (2.5
. ) evaluate )
L{f (t)}.
f (t)
100
t
5
{
f (t )= 20 t , 0 ≤t ∈5
0, t ≥ 5
f (t )=20 t −20 t u(t −5)
(b
20
( ) 1 5
F ( s)= s2 −20 e − 5s s2 + s
Find the inverse Laplace transform of: (5)
)
2 - 3se- s + 4e- s
F (s) =
s(s +1)
2
(
3 s−4
F ( s)= s( s+1) − s( s+1) e−s )
2 2 4 7
¿ s − s +1+ s − s+1 e−s( )
f ( t )=2−2e−t+ 4 u ( t −1)−7 e−(t−1)u( t−1)
3 Find the Laplace transform of the given periodic function. (2.5
)
f (t)
1
t
a 2a 3a 4a
–1
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
ENGR 311- Page 2 of 5