Download Differential Equations

Assignment 5
Maximum Marks
30
Due Date
30th July, 2004
Assignment Weight age
2%
Question 1
Show that the indicial roots differ by an integer. Use the method of Frobenius to obtain two linearly
independent series solutions about the regular singular point x0  0 . Form the general solution
on  0,   .
xy  xy  y  0
solution
put

y   Cn x n  r
n 0

y '   Cn (n  r ) x n  r 1
n0

y ''   Cn (n  r )(n  r  1) x n  r  2
n 0
then the equation becomes
xy   xy   y  x

 C (n  r )(n  r  1) x
n 0

n

xr [
 C (n  r )(n  r  1) x
n 0
n
n2

nr 2

-x
 C (n  r ) x
n 0
-  Cn ( n  r ) x n +
n 0
n  r 1
n

+
C x
n 0
nr
n

C x
n 0
n
n
]=0
 C0 r (r  1) x 1  0 and
(k  r  1)(k  r  1)Ck 1 - (k  r  1)Ck =0
Ck
 Ck 1 =
(k  r  1)
 r1  0 r2  1
and
but r2  r1  1 is an integer, then there exist two linearly independent solutions of the form
y1 

 cn x
n  r1
n 0
, c0  0
y2  Cy1( x) ln x 

 bn x
n 0
Where C is a constant that could be zero.
then

y1   cn x n , c0  0
n 0
and

y2  Cy1 ( x)ln x   bn x n1 , b0  0
n 0
when r1  0
Ck
Ck 1 =
(k  1)
for k  0,1, 2,3, 4,...
C1  C0
C1 C0

2
2
C
C
C3  2  0
3 3.2
C
C
C4  3  0
4
4.3.2
C0
C
C5  4 
5
5.4.3.2
C
C0
C6  5 
6
6.5.4.3.2
C6
C0
C7 

7
7.6.5.4.3.2
---------------------------------------------------------------------------------------------------C
Cn  0
n!
C2 
So,

C0 n
x , c0  0
n 0 n !
y1  
(3a )
n  r2
, b0  0
(3b )
when r2  1 then
Ck 1 =
Ck
k 1
for k  0,1, 2,3, 4,...
C0
2
C
C
C2  1  0
3 3.2
C
C2
C3 
 0
4
4.3.2
C
C0
C4  3 
5
5.4.3.2
C0
C
C5  4 
6
6.5.4.3.2
C
C0
C6  5  
7
7.6.5.4.3.2
---------------------------------------------------------------------------------------C0
Cn 
(n  1)!
So,
C1 

C1 n1
x , C0  0
n0 (n  1)!
y2  Cy1 ( x)ln x  
Question 2
Use the change of variable y  x1/ 2v  x  to find the general solution of the equation
.
Solution
y  x1/ 2v  x 
1
y '   x 1/ 2 v  x   x 1/ 2 v '  x 
2
1
3
y ''  x 5 / 2 v  x   x 3 / 2 v '  x   x 2 v '' ( x )
4
by substituting values in the given differential equation, we get
1
3
1
0= x 2 [ x 5 / 2 v  x   x 3 / 2 v '  x   x 2 v '' ( x ) ]+2x[  x 1/ 2 v  x   x 1/ 2 v '  x  ]+  2 x 2 [ x1/ 2 v  x  ]
4
2

3
3
1
1
3 12
x v( x)  x 2 v ' ( x)  x 2 v '' ( x)  x 2 v( x)   2 x 2 v( x)  0
4
1
By multiplying the above equation by x 2 , we get
x 2 v'' ( x)  xv' ( x)  ( 2 x 2  1 )v( x)  0
4
By comparing the equation with the general Bessel’s equation , which is
x 2 y '' ( x)  xy ' ( x)  ( x 2 v 2 ) y( x)  0
1
1
 2 x2 
 x  
we get
4
2
So, the solution of our equation is
v  C1 J 1 ( x)  C2 J  1 ( x)
2
2
By putting in y  x1/ 2v  x  , we get
y  C1 x
1
2
J 1 (  x )  C2 x
2
1
2
J  1 ( x )
2
which is the required solution
Question 3
Solve the given differential equations subject to the indicated initial conditions.
d 2 x dx dy


 0        (1)
dt 2 dt dt
d 2 y dy
dx

 4  0         2
2
dt
dt
dt
x  0   1, x   0   0,
y  0   1, y   0   5
Solution:
First we write the differential equations of the system in the differential operator form:
( D 2  D) x  Dy  0
( D 2  D) y  4 Dx  0
Then we eliminate one of the dependent variables, say x . Multiplying first equation with 4 and the
second equation with the operator D+1 and then adding, we obtain
[ D  D  1 +4D] y  0
2
D[( D  1)2  4 D]x  0
or
The auxiliary equation of the differential equation found in the previous step is
m[(m  1)2  4]  0
Therefore, roots of the auxiliary equation are
m1  0,
m2  1  2 i,
m3  1  2 i
So that the complementary function for the retained variable y is
y  c4  et (c5 cos 2t  c6 sin 2t.) ---------(3)
Next we eliminate the variable y from the given system. For this purpose we multiply second equation
with 1 while operate on the first equation with the operator D +1 and then subtracting, we obtain
D[( D  1)2  4 D]x  0
So, c1  0,
y  e  t ( cos 2t  2sin 2t.)
c2  1,
c3  2,
c4  5
since we have given only four initial conditions but there are six constants to be determined, so some
of them must multiple of others, to find out we put x and y in (1) and after simplifying, we get
Coefficients of et cos 2t are
c5  2c6  2c3  4c2  0 ---------(5)
Coefficients of e  t sin 2t are
c6  2c5  2c2  4c3  0 ---------(6)
multiplying (5) by 2 and then subtracting (6)
we get
c6  2c2
multiplying (5) by 2 and then adding (6)
we get c5  2c3
by putting these values in (4) we get
x  c4  e t (2c3 cos 2t  2c2 sin 2t.) -------(7)
after substituting the given initial conditions in (3) and (7) we get the values of constants
c1  0,
c2  1,
c3  2,
then solution becomes
x  5  et (4cos 2t  2sin 2t.)
y  et ( cos 2t  2sin 2t.)
c4  5