J. BIAZAR ET AL.

991

Table 3. Comparisons of of 10th-order HAM solu-

tions for different values of in Example 3.

t

0.2x 0.6x

1.0x 1.4x1.8x

1 [12] 1.5e–2 8.4e–2 6. 4e–3 7.49e–36.8e–3

1c 0.924 7.1e–7 4.3e–7 3.7e–7 2.0e–69.6e–6

=0.57 [12] 1.1e–2 1.2e–2 1.2e–2 1.2e–21.1e–2

5c =0.954 1.3e–4 1.3e–4 1.2e–4 1.2e–41.4e–4

0.634 [12] 1.3e–1 1.5e–1 1.5e–1 1.5e–11.3e–1

12c 0.655 1.0e–2 9.9e–3 9.4e–3 9.9e–31.1e–2

Here we will discuss following three cases of .

c

1) In the first case, we take as an example. For

and in (16), we get into a nonlinear

equation, which root is .

1c

0.924

1c10m

2) In second case, we take as an example.

Equation (16) with and has the real solu-

tion .

5c

105cm

=0.954

3) Finally, we take as an example. Equation

(16) with and , introduces

as a proper value for .

12c 10m

12c 0.655

Since the closed-form solution to the problem (22),

(23) is not available, the numerical solution numer is

calculated via the Runge-Kutta-Fehlberg 4-5 technique,

then, we compare the relative errors of the 10th order

u

HAM approximations at different points in the

appr

u

interval , using the formula

(0,1)

appr numer

numer

uu

tu

for different cases of , are reported in Table 3.

c

5. Conclusions

In this paper the solutions of a new way of finding the

control parameter in the homotopy analysis method is

proposed. It is shown for obtained values of such pa-

rameter, HAM approximation series leads to exact solu-

tion of problems or produces an approximate results

which are in a highly agreement with exact solution of

problems. All computations were done using Maple 13

with 15 digit floating point arithmetics (Digits: = 15).

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