Page 113 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes x x 2 x 4
y (x) = 0.1 0.1 x 1 dx
3
0 2 2.4
x 2 x 4 x 6
= 0.1 0.1
2 2.4 2.4.6

x 2 x 4 x 6
= 0.1 1
2 2.4 2.4.6
...............................................................
...............................................................
...............................................................

)
x 2 1 (x 2 k
)
y (x) = 0.1 1 (x 2 2 ...... ...(2)
k
k 2 2 .1.2 2 k !
2
So the solution of equation (1) is y(x)
x 2 1 2 2 1 x 2 3
y(x) = lim y k ( ) 0.1 1 (x ) ...... ...(2)
x
2
3
k 2 2 2! 2 3! 2
The above series is a convergent series
Example 2: Solve the following by Picard’s method of integrating by successive
approximation
dy
= z,
dx
dy
3
= x (y + z)
dx
1
where y = 1 and z = when x = 0
2
x 1 x
Here y = 1 z dx and z x 3 (y ) z dx
0 2 0
The first approximation gives us
x 1 x
y = 1 dx 1 ,
0 2 2
1 x 1 1 3 x 4
z = x 3 1 dx .
2 0 2 2 2 4
Second approximation
x 1 3 x 3
y = 1 x 4 dx 1 x 5
0 2 8 2 40
1 x 3 3 x 3 1 3 1 3
z = x x 4 dx x 4 x 5 x 8
2 0 2 2 8 2 8 10 64
Third approximation
x 1 3 1 3
y = 1 x 4 x 5 x 8 dx
0 2 8 10 64

106 LOVELY PROFESSIONAL UNIVERSITY