Higher College of Technology: Engineering Math Quiz 2 Assignment

This document presents the solutions to a Quiz 2 assignment in Engineering Mathematics (MATH3120N) from the Higher College of Technology. The assignment covers two main parts. Part A includes solving a partial differential equation using the D'Alembert method and finding the Laplace transform of a function involving cosine. Part B involves solving an Euler-Cauchy differential equation using a transformation method and method of undetermined coefficients. The solutions are presented step-by-step, detailing the application of relevant formulas and techniques. This assignment is designed to test students' understanding of differential equations, Laplace transforms, and related mathematical concepts commonly used in engineering applications.

Note –
(i) Use equation editor (which is available in Word) to type Mathematics
(ii) Save the file in PDF format and Submit this assignment in e-learning
(iii) Not allowed submission by email
Part A (2x2.5 = 5 Marks)
1. Solve the partial differential equation 2
2
2
2
6 t
z
x
z




 [2.5]
Solution: The characteristic equation is 2 6 6m m i     . Then by using D’Alembert method,
the general solution is
     , 6 6z x t f x i t g x i t   
HIGHER COLLEGE OF TECHNOLOGY
DEPARTMENT OF ENGINEERING
Section Electrical & Electronics Engineering
Student Name
I.D. Number
Quiz II-ASSIGNMENT Semester-II AY 2019 - 20
Max.
Marks:
10 Course Name: Engineering Mathematics
Level: A.DIPLOM
A
Course Code: MATH3120N
Section No. Specialization Common
Date Of
Submission
:
Lecturer’s/Course
Coordinator Name:
Dr.Ganesh Venkataraman

2. Find  ttL 22 cos . [2.5]
Solution: We know that 2 cos 2 1
cos 2
t
t 
 . So
 
 
   
2
2 2
cos 2 1
cos 2
1 cos 2 1
2
1 cos 2 1
2
1 1
2 2
t
L t L
L t
L t L
s
s s
 
  
 
 
   
 
   
We know that         1
n
nn
n
d
L t f t L f t
ds
  , so
      
 
 
2
22 2 2
2
2
6 2
33 2
2 2 2
cos 1 cos
1 1
2 2
2 24 32
4
d
L t t L t
ds
d s
d
s s
s
s s
s
s
 
  
   

 




Hence,    
 
6 2
33
2
2
2 2 2 3
4
c 2
s 4
oL s s
s s
t t   


Part B (1x5 = 5 Marks)
3. Solve the Euler Cauchy differential equation xxy
dx
dy
x
dx
yd
x 22
2
2
2 ln52  . [5]
Solution: Let z
x e , then
 
2
2
2 1 ,
d y dy
x D D y x Dy
dx dx
   , Where d
dx . Then the given differential equation converted into
  
   
2 2
2 2 2
1 2 5
3 5 ...
z
z
D D D y e z
D D y e z i
   
   .
The characteristic equation is 2 3 5 0m m   , the roots of the equation is
3 11
2 2
m i  . Then the complementary equation is
3
2 1 2
11 11
cos sin
2 2
t
c cy z z
e c
    
        


   
. By method of undetermined coefficient, the particular
solution is of the form    2 2 ...z
py Az Bz C e ii   . Differentiating (ii) we get
      2 2 2 2 2
' 2 2 2 2 2 2z z z
py Az B e Az Bz C e Az A B z C e        
          2 2 2 2 2
'' 4 2 2 2 2 2 2 2 4 8 4 2 2 4z z z
py Az A B e Az A B z C e Az A B z A B C e            
We know that particular solution satisfied the differential equation, i.e
2 2
'' 3 ' 5 z
p p py y y e z   . Substitute the values, we get

          
 
   
2 2 2 2 2 2
2 2 2 2
2 2
4 8 4 2 2 4 3 2 2 2 2 5
4 8 4 2 2 4 6 6 6 6 5 5 5
3 2 3 2 2 3
z z
Az A B z A B C Az A B z C Az Bz C e z e
Az Az Bz A B C Az Az Bz C Az Bz C z
Az A B z A B C z
            
            
     
Comparing both sides we get
3 1
2 3 0
2 2 3 0
A
A B
A B C

 
  
Solving above equations we get 1 2 4
, ,
3 9 27
A B C    
Hence, the particular solution is
2 2 2
22 4
3 9 27
x
p
x x
e x e
y x e 
Then the general solution is c py y y 
3 2 2 2
22 1 2
11 11 2 4
cos sin
2 2 3 9 27
z z z
zz z e e
y z
e z
c c e
    
            
    
Putting the value of z, we get
       2 2 23
22 1 2
11 ln 11 ln ln 2 ln 4
cos sin
2 2 3 9 27
x x x x x x
y x c c x
    
        
    
    