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S. S. Nourazar et al.

We construct a homotopy for Equation (3.19) in the following form:

 ∂υ ∂u 
H (υ , p ) =(1 − p )  − 0  +
∂t 
 ∂t

 ∂υ ∂ 2υ

∂υ
p  − 2 − 2υ
− υ (1 − υ )(υ − 3) 
∂x
 ∂t ∂x


(3.21)

The solution of Equation (3.19) can be written as a power series in p as:

υ =υ0 + pυ1 + p 2υ2 +

(3.22)

Substituting Equation (3.22) and Equation (3.20) into Equation (3.21) and equating the terms with identical
powers of p:
−3

∂υ
p : 0
∂t
0

∂u0
υ 0 ( x, 0 )
=
,
∂t

3

(

e
∂υ
∂υ1 ∂u0 ∂ υ0
+
=
+ 2υ0 0 + υ0 (1 − υ0 )(υ0 − 3) ,
2
∂t
∂t
∂x
∂x
2
∂υ



υ
υ
υ
p 2 : 2 = 21 + 2υ0 1 + 2υ1 0 + 8υ0υ1 − 3υ1 − 3υ1υ02 ,
∂t
∂x
∂x
∂x
2
υ
υ


υ
υ


∂υ
p 3 : 3 = 22 + 2υ0 2 + 2υ2 0 + 2υ1 1 + 8υ0υ2 − 3υ2
∂t
∂x
∂x
∂x
∂x
2
2
2
− 3υ2υ0 + 4υ1 − 3υ0υ1 ,
2

p1 :

3e

)x

(

3 −1
4

)x

3 −1
4

+e



3

(

)x

3 −1
4

υ1 ( x, 0 ) = 0,

,

(3.23)

υ 2 ( x, 0 ) =
0,
υ3 ( x, 0 ) = 0.

Using the Maple package to solve recursive sequences, Equation (3.23), we obtain the followings:

(

−3

υ 0 ( x, t ) =

(

3

e

υ1 ( x, t ) = −

)x

)x

3 −1
4

3e
3 −1
4

+e

9
2  3(
e





(3
)x

(

3

4

3−4

3 −1
4

)x

3 −1

+e



(

3

,

)
)x 

3 −1
4

2

t,





 3( 3 −1) x − 3( 3 −1) x 
e 4
− e 4  43 − 24 3


27

υ 2 ( x, t ) = 
t2,
3
8
 3( 3 −1) x − 3( 3 −1) x 
e 4
+e 4 





(




27 
υ3 ( x, t ) = 
16  3(

 e



)



   3(
1

e
4 
3 −1)
3( 3 −1) 
 

x
x

4
+ e 4  
 
 

4

 − 3(
 − 4 +e





)x 

3 −1

2

2
)x  

3 −1
4





(

)


3
 389 − 225 3 t .



(3.24)

By setting p = 1 in Equation (3.22) the solution of Equation (3.19) can be obtained as υ = υ0 + υ1 + υ2 + υ3 +
Thus the solution of Equation (3.19) can be written as:

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