Download Differential Equations - Solved Assignments - Semester Spring 2005 PDF

Page 1

Assignment 1 (Spring 2005)


Maximum Marks 60

Due Date 07, April 2005

Assignment Weight age 2%





Question 1



(a) Define initial value and boundary value problem elaborate with the help of examples (at least one).

05


Solution



INITIAL VALUE PROBLEM



Differential equation of first order or greater in which the dependent variable y or its derivative are

specified at one points such as

 

 0 0

,
dy

f x y
dx

y x y







If equation of the second order

       

   

2

2 1 02

'

0 0 0 1
, '

d y dy
a x a x a x y g x

dx dx

y x y y x y

  

 



Where
0

y and
1
'y are arbitrary constants

Is called the initial value problem and the  0 0y x y  
'

0 1
, 'y x y is called the initial conditions



BOUNDARY VALUE PROBLEM



Differential equation of order two or greater in which the dependent variable y or its derivative are

specified at different points such as



       

   

2

2 1 02

0 0 1 1
,

d y dy
a x a x a x y g x

dx dx

y x y y x y

  

 



Where
0

y and
1
'y are arbitrary constants

Is called the boundary value problem and the    0 0 1 1,y x y y x y  is called the boundary conditions





NOT THE PART OF THE SOLUTION (MOTIVATION OF THE CONDITIONS)

Page 30

Assignment 4 (Spring 2005)


Maximum Marks 40

Due Date 12, May 2005

Assignment Weight age 2%





Question 1



Solve the differential equation
/// //

6 1 cosr r    .find complimentary
c

r and particular solution (
p

r )

by the undetermined coefficient (superposition approach). 10



Solution
/// //

6 1 cosr r   
Complementary function

To find
c

r , we solve the associated homogeneous differential equation


/// //

6 0r r 

Put
n

r e


 ,
2 3

' , r'' , '''
n n n

r ne n e r n e
  

  

Substitute in the given differential equation to obtain the auxiliary equation

 

/// //

3 2 2

6 0

6 0 6 0

0,0,6

r r

n n n n

n

 

    





Hence, the auxiliary equation has complex roots. Hence the complementary function is


1 2 3

6
c

r c c c e


  

Particular Integral
Corresponding to cos( )g   :

1
cos sin

p
r A B  

Corresponding to   1f   :
2p

r C



Therefore, the normal assumption for the particular solution is


1 2p p p

r r r 


1

cos sin
p

r A B   +C

Clearly there is duplication of

(i) The constant function between
c

r and
2p

r .

To remove this duplication, we multiply
2p

r with
2

 . This duplication can’t be removed by multiplying

with . Hence, the correct assumption for the particular solution
2p

r is


2

cos sin
p

r A B C    