HIGHER COLLEGE OF TECHNOLOGY DEPARTMENT OF ENGINEERING Section.

partial derivatives, laplace transforms, euler Cauchy differential equation

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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
    
        
    
    