Variational Iteration and Homotopy Perturbation Methods for…
y ( x) = y1 ( x) = ai = a a1 a2
(22)
i =0
using the initial/boundary conditions, the unknown constants can be evaluated, the values of which are then substituted back into the approximate solutions earlier obtained. 5. Applications We illustrate the ability and reliability of the methods with the following examples. Example 1: Consider the Volterra-Fredholm integro differential equations (Biazer, J and Eslami, M. (2010)). x
1
1
1
y( x) xy( x) xy ( x) = f ( x) ( x 2t )( y (t )) 2 dt xty(t )dt where
f ( x) =
2 6 1 4 23 5 x x x3 2 x 2 x 25 3 15 3
with condition y(0) = 1, y(0) = 0 and exact solution y ( x) = x 1 . Now using the transformation in equation (4.4), we have 2
0 y1 0 1 y1 x 1 = 2 f ( x ) ( x 2 t ) y ( t ) dt xty ( t ) dt y y x x 1 1 2 2 and
5 2 7 1 5 1 4 2 3 1 23 y2,n 1 ( x) = y2,n x x x x x y1,n y2,n x 2 3 175 15 4 3 2 15
y1,2 n ( x 2 1) x
1 2 y1,n ( x 1) x 2 1 (2x 1) 2
Example 2: 1
y( x) xy( x) sin xy ( x) = e x (1 x sin x) 2 e 2t y 2 (t )dt 1
with conditions
y(0) = y(0) = y(0) = 1
Exact solution for the problem is y ( x) = e
x
The Newton’s linearization scheme of order three is given as
G
G G G G y k yk yk yk= 0 yk yk yk yk
where y k = y k 1 y k i
i
i
and 1
G = y( x) xy( x) sin xy ( x) e x (1 x sin x) 2 e 2t y 2 (t )dt 1
Thus, 1
1
1
1
y( x) xy( x) sin xy ( x) 2 yy e 2t y 2 (t )dt = e x (1 x sin x) 2 e 2t y 2 (1 2 y)dt www.ijmsi.org
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