 Open Access
 Total Downloads : 415
 Authors : Adesanya, A. Olaide, Odekunle, M. Remilekun, Udoh, M. Mfon
 Paper ID : IJERTV1IS5489
 Volume & Issue : Volume 01, Issue 05 (July 2012)
 Published (First Online): 04082012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Starting Two Stepsfour Off Steps Method Accurately for the Solution of Second Order Initial Value Problems
Starting two stepsfour off steps method accurately for the solution of second order initial value problems
lAdesanya, A. Olaide, lOdekunle, M. Remilekun, 2Udoh, M. Mfon
lDepartment of Mathematics, Madibbo Adama University of Technology, Yola, Adamawa State, Nigeria
2Department of Mathematics and Computer Science, Cross River State
University of Technology, Cross River State, Nigeria
Abstract
We derive an order five hybrid method through collocation of the differential system and interpolation of the approximate solution which is implemented in predictor corrector mode. Continuous block method was used to generate the independent solution which served as predictor. The efficiency of our method was tested on some second order initial value problem and was found to give better approximation than the existing methods.
Keywords hybrid method, collocation, differential system, interpolation, approximate solution, continuous block method, independent solution, predictor corrector mode
A.M.S Subject Classification 65L05, 65L06, 65D30

Introduction
This paper considers the approximate solution to second order initial value problem of the form
n
Y11 = f(x, Y, Y1), Yk(xO) = Yk, k = 0, 1 (1)
1
where f is continuous and differentiable within the interval of integration.
Convectionally, higher order ordinary differential equation are solved by method of reduction to syatem of first order ordinary differential equation. This method is extensively discussed by Awoyemi and Kayode[5], Adesanya, Anake and Udoh [4], Kayode and Awoyemi [14], Jator [11], Awoyemi and Idowu [7] to mention few. These authors suggested that the direct method for solving higher order ordinary differential equation are more efficient since the method of reduction increased the dimension of the resulting system fo first order ordinary differential equation; hence it waste alot of computer and human effort.
Many scholars have proposed method implemented in predictor corrector mode for the direct solution of (1), among them are Awoyemi [6], Kayode [13], Olabode [16], Adesanya, Anake and Oghoyon [3], Kayode and Adeyeye [12]. These authors proposed an implicit multistep method in which seperate predictors are needed to implement the corrector. The major setback of this method is that the predictors are reducing order of accuracy, therefore it has an great effect on the accuracy of the method.
Scholars later proposed block method to cater for some setbacks of the predictor corrector method. Block method has the properties for Runge kutta method of being self starting and does not require developing seperate predictors and eval uate fewer function per step. Among scholars that proposed block method are Jator [10], Jator and Li [9], Awoyemi et al. [8], Adesanya et al. [1], Majid,Azmi and Suleiman [15], Adesanya et al. [2], Omar et al. [17],Siamak [19], Omar and Suleiman [18]. It was observwd that in block method, the number of interpolation points must be equal to the order of the differential equation, hence this method does not exhaust all possible interpolation points therefore a method of lower order
2
is developed.
In this paper, we developed a method which is implemented in predictor corrector method but the predictors are constant order predictors hence cater for the setbacks of the convectional predictor corrector method. The predictors are developed adopting block method hence these methods combine the properties of both predictorcorrector method and block method, therefore address some of the setbacks of the method.

Methodology
2.1 Development of the corrector
We consider a power series approximate solution of the form
Y(x) =
r+sl
j=O
ajxj (2)
The second derivatives of (2) gives
Y11 =
r+sl
j=2
j(j – 1)ajxjl (3)
Substituting (3) into (1) gives
f(x, Y, Y1) =
r+sl
j=2
j(j – 1)ajxjl (4)
3
3
3
where r and s are the number of interpolation and collocation points respec tively. Interpolating (2) at xn+r, r = 0 (2 ) 4 and collocating (4) at xn+s, s = 0 (2 ) 2,
3
gives a non linear system of equation
AX = U (5)
where
1 xn x2 3 4 5 6
n
x
x
x
x
I
n
n
n n
1 xn+ 2
3
1 x 4
x2 2
3
n+ 3
x3 2
3
n+ 3
x4 2
n+ 3
x5 2
n+ 3
x6 2 I
3
n+ 3
n+ n+
I
1 x 4
n+ 3
n+ 3
n+ 3
n+ 3
n+ 3
n+ 3
n+
I
I
I
n+ 3
I n+
x2 4
x3 4
x4 4
x5 4
x6 4
X = 2
3
I 0 0 2 6xn 12x
20×3
n+ 3
30×4
n+ 3
n
I
3
n+ 3
3
n+ 3
0 0 2 6xn+2 12×2
I 0 0 2 6xn+ 2 12×2 2
n
20×3 2
n+ 3
n
I
30×4 2
I
n+ 3
0 0 2 6xn+ 4 12×2 4
20×3 4
30×4 4
n+2
20x
3
n+2
30x
4
n+2
A = aO al a2 a3 a4 a5 a6
U =
Yn Yn+ 2
Yn+ 4
fn fn+ 2
fn+ 4
fn+2
3
3
3
3
solving (5) for a1j s and substituting into (2) gives a continuous hybrid linear multistep method in the form
Where
2
Y(x) = o:O + o: 2 Yn+ 2 + o: 4 Yn+ 4 + h
2
/32j fn+2j + /3 2 fn+ 2 + /3 4 fn+ 4 \
(6)
3 3 3 3
j=O
3 3 3 3
o:O
= 1 (243t6 – 1458t5 + 2970t4 – 2160t3 + 432t – 64)
64
4
o: 2
3
o: 4
3
= 1 32
(
(
= 1 64
243t6
243t6
– 1458t
5
5
– 1458t
+ 2970t4
+ 2970t4
– 2160t
– 2160t
+ 384t)
3
3
( ) –
+ 336t)
/3O
= 1 1215t6 7533t5 + 16470t4 14760t3 + 4320t2 + 128t 8640
/3 2
3
/3 4
3
= 1
(
960
( ) –
=
960
1 (
1350t6
135t
– 891t
6
– 8019t
5
+ 2010t
5
+ 16050t4
– 1560t
4
– 11280t
3
+ 1664t)
3
+ 256t
)
/32
= 1 243t5 810t4 + 720t3 128"t 8640
h
t = xxn , Yn+j = Y(xn + jh), fn+j = f (xn, Y (xn + jh) , Y1 (xn + jh))
evaluating (6) at t = 2 gives a discrete scheme
(
3
3
p Yn+2 = 3Yn+ 4 – 3Yn+ 2 + Yn + 27
fn+2 + 9fn+ 2 – 9fn+ 2 – fn
(7)
3
3
equation (7) is our corrector
2.2 Development of the predictor
3
Interpolating (2) at xn+r, r = 0, 1 and collocating (4) at xn+s, s = 0 (l ) 2 gives
equation (5) where
A = aO al a2 a3 a4 a5 a6 a7 a8
3
3
3
3
U =
Yn Yn+l fn fn+ l
fn+ 2
fn+l fn+ 4
fn+ 5
fn+2
U =
Yn Yn+l fn fn+ l
fn+ 2
fn+l fn+ 4
fn+ 5
fn+2
5
1 xn x2 x3 x4
x5 x6 x7 x8
n n n
n+l
n+l
n+l
I
0
2
0
2
0
2
0
2
0
2
0
2
0
2
I 1 xn+l x2 x3 x4
0
n n n n
n+l
n+l
n+l
n+l
x5 x6 x7 x8
n
n
n
n
n
I
I
I
20×3
30×4
42×5
56×6
0
6xn 12×2
20×3
30×4
42×5
56×6
2
I 0 6xn+ l 12x l
3
20×3 l 30×4 l 42×5 l 56×6 l
3
n+ 3
n+ 3
n+ 3
n+ 3
n+ 3
n+ 3
I 0
I
n+ 3
n+ 3
n+ 3
n+ 3
n+ 3
n+ 3
n+ 3
I
I
I
n+ 3
X = 0
6xn+ 2
12×2 2
n++l
20×3 2
20x
3
n++l
30×4 2
30x
4
n++l
42×5 2
42x
5
n+l
56×6 2
56x
6
n++l
6xn+l 12×2
20x
3
n++l
30x
4
n++l
42x
5
n+l
56x
6
n++l
0
I
6xn+ 4 12×2 4 20×3 4
3
30×4 4
n+ 3
42×5 4
n+ 3
56×6 4 I
3
n+ 3
n+ n+
I
0
n+2
3
n+ 3
n+ 3
0
3
3
I
I
n+ 3
n+ 3
n+ 3
n+ 3
n+
I
I
n+ 3
6xn+ 5 12×2 5 20×3 5
30×4 5
42×5 5
56×6 5
6xn+2 12×2
20x
3
n+2
30x
4
n+2
42x
5
n+2
56x
6
n+2
solving for a1j s using Guassian elimination method and substituting into (2) givesa continuous hybrid linear multistep method in the form
l
Y(x) =
o:jYn+j+p
2
/3j fn+j + /3 l fn+ l + /3 2 fn+ 2 + /3 4 fn+ 4 + /3 5 fn+ 5 \
(8)
j=O
j=O
3 3 3 3 3 3 3 3
h
where t = xxn , Yn+j = Y(xn + jh), fn+j = f (xn, Y (xn + jh) , Y1 (xn + jh))
o:O = 1 – t o:l = t
1 /I 243t8 – 2268t7 + 8820t6 – 18522t5 + 22736t4 – 16464t3 \I
/3O =
13440
1 ( 8
+6720t2
7 6
– 1265t
5 4 3 )
/3 l 3
/3 2
3
=
2240
=
4480
1215t
1 (
243t
– 10260t
8
– 2160t
7
+ 7872t
+ 14616t
+ 34524t
– 58086t
6
+ 14616t
+ 49140t
5
– 6720t
– 16800t
4
+ 267t
+ 267t
3 )
/3l
= 1 (1215t8 – 9720t7 + 30492t6 – 46872t5 + 35560t4 – 11200t3 + 525t)
3360
6
/3 4
3
= 1
(
7
4480
1215t8 – 9180t
+ 26964t6 – 38682t
+ 27320t4 – 8400t
+ 363t)
/3 5
3
= 1
7
5
3
( ) – –
2240
243t8 – 1728t
– 4788t – 6552t
+ 2536t4 – 1344t
+ 57t)
(
6
5
3
/32
= 1 243t8 1620t7 + 4284t6 5670t5 + 3836t4 1120t3 + 47t 13440
3
Solving for the independent solution yn+s, s = 1 (l ) 2, gives a continuous hybrid
block formula
l
(jh)m
(m)
2 \
2
Y(x) =
Yn +h
\Jjfn+j + \J l fn+ l + \J 2 fn+ 2 + \J 4 fn+ 4 + \J 5 fn+ 5
m!
j=O
j=O
3 3 3 3
3 3 3
3
(9)
where
\JO
= 1 243t8 2268t7 + 8820t6 18522t5 + 22736t4 16464t3 + 6720t2 13440
\J l
3
\J 2
3
( ) – –
= 1
(
2240
=
4480
1215t
1 (
243t8
– 2160t
7
– 10260t
8
+ 7872t6
+ 34524t
7
+ 14616t5
– 58086t
6
+ 14616t4
5
– 6720t3)
+ 49140t
– 16800t
4 3)
(
7
5
\Jl
= 1 (1215t8 – 9720t7 + 30492t6 – 46872t5 + 35560t4 – 11200t3)
\J 4
3
\J 5
3
3360
= 1
4480
( ) – –
=
2240
1 (
1215t8
243t
– 1728t
8
– 9180t
7
+ 26964t6
– 4788t
– 6552t
6
– 38682t
+ 2536t
5
+ 27320t4
4
– 8400t3)
– 1344t
3)
\J2
= 1 243t8 1620t7 + 4284t6 5670t5 + 3836t4 1120t3 13440
7
3
evaluating (9) at t = 0 (l ) 2, gives a discrete block formula in the form
m
A(O)Ym = eYn + pdf (Yn) + pbF(Y ) (10)
where A(O) = 6 x 6 identity matrix
Ym =
Yn+ l Yn+ 2 Yn+l Yn+ 4 Yn+ 5 Yn+2
3
3
3
3
f (Yn) =
Ynl Yn2 Yn3 Yn4 Yn5 Yn
3
3
3
3
F(Ym) =
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
I
e = I
I
fn+ l
fn+ 2
fn+l fn+ 4
I
I
I
fn+ 5
fn+2
d =
28549
lO8864O
lO27
l7OlO
253
2688
lO88
85O5
35225
2l7728
4l
2lO
8
I
b = I
I
275 5l84
–
57l7 l2O96O
lO62l 272l6O
–
77O3
36288O
4O3 6O48O
–
l99 2l7728
l94 945
–
8
8l
788 85O5
–
97 l89O
46 2835
–
l9 85O5
l65 448
–
267
448O
5 32
–
363
448O
57 224O
–
47 l344O
l5O4 2835
–
8
945
2624
85O5
–
8
8l
32 945
–
8
l7Ol
8375 l2O96
3l25 72576
3l25 72576
–
625
24l92
275 5l84
–
l375 2l7728
6
7
I
3 35
68 lO5
I
3 7O
6
35 I
0
3
3
Substituting (10) into the first derivative of (9) at t = l (l ) 2 gives
3
( )
3
( )
3
( )
Y1
= Y1
+ h
l8l44O
756O
3
6O48O
3
2835
I ( l9O87 ) fn + (27l3 ) fn+ l – (l5487 ) fn+ 2 + ( 586 ) fn+l I
n+ l
n
–
6737
6O48O
fn+ 4 +
263
756O
fn+ 5 –
863
l8l44O
fn+2
n+ l
n
–
6737
6O48O
fn+ 4 +
263
756O
fn+ 5 –
863
l8l44O
fn+2
3
( )
3
( )
3
( )
Y1
= Y1
+ h
ll34O
l89
3
378O
3
2835
I ( ll39 ) fn + ( 94 ) fn+ l + ( ll ) fn+ 2 + ( 332 ) fn+l I
n+ 2
n
–
269
378O
fn+ 4 +
22
945
fn+ 5 –
37
ll34O
fn+2
n+ 2
n
–
269
378O
fn+ 4 +
22
945
fn+ 5 –
37
ll34O
fn+2
( )
3
( )
3
( )
Y1
= Y1
+ h
l344
56
3
224O
3
lO5
I ( l37 ) fn + (27 ) fn+ l + ( 387 ) fn+ 2 + ( 34 ) fn+l I
n+l
n
–
243 224O
fn+ 4 +
9 28O
fn+ 5 –
29 672O
fn+2
n+l
n
–
243 224O
fn+ 4 +
9 28O
fn+ 5 –
29 672O
fn+2
3
( )
3
( )
3
( )
Y1
= Y1
+ h
2835
945
3
945
3
2835
I ( 286 ) fn + (464 ) fn+ l + (l28 ) fn+ 2 + (l5O4 ) fn+l I
n+ 4
n
+
58
945
fn+ 4 +
l6
945
fn+ 5 –
8
2835
fn+2
n+ 4
n
+
58
945
fn+ 4 +
l6
945
fn+ 5 –
8
2835
fn+2
( )
( )
( )
Yn1 + 5
= Yn1
+ h
36288
+
l5l2
fn+ 4 +
3
235
l5l2
l2O96
fn+ 5 –
3
275
36285
567
fn+2
I ( 37l5 ) fn + ( 725 ) fn+ l + ( 2l25 ) fn+ 2 + (25O ) fn+l I
3
3875
l2O96
3
3875
l2O96
3
3
( )
( )
( )
Yn1 +2
= Yn1
+ h
42O
35
fn+ 4 +
3
l8 35
l4O
fn+ 5 +
3
4l 42O
lO5
fn+2
I ( 4l ) fn + (l8 ) fn+ l + ( 9 ) fn+ 2 + ( 68 ) fn+l I
+
9 l4O
+
9 l4O
3
3
9

Analysis of the basic properties of the block

Order of the method
We defined a linear operator on (6) to give
2 \l
2
Â£yY(x) h = Y(x) o:O + o: 2 Yn+ 2 + o: 4 Yn+ 4 + h
/32j fn+2j + /3 2 fn+ 2 + /3 4 fn+ 4
3 3 3 3
j=O
3 3 3 3
(11)
Expanding Yn+j and fn+j in Taylor series and comparing the coefficient of h
gives
Â£yY(x) h = COY(x) + ClhY1(x) + … + CphpYp(x) + Cp+lhp+lYp+l(x)
+Cp+2hp+2Yp+2(x) + … (12)
Definition 1 Order
The difference operator Â£ and the associated continuous linear multistep method
(6) are said to be of order p if CO = Cl = … = Cp = Cp+l = 0 and Cp+2 is called the error constant and implies that the local truncation error is given by tn+k = Cp+2h(p+2)Y(p+2)(x) + 0 (hp+3)
328O5
The order of our discrete scheme is 5, with error constant Cp+2 = 8

Consistency
A linear multistep method (6) is said to be consistent if it has order p 1 and if p(1) = p1(1) = 0 and p11(1) = 2!o(1) where p(r) is the first characteristic polyno mial and o(r) is the second characteristic polynomial.
10
For our method,
3
p(r) = r2 + 3r 4 – 3r 2 + 1
3
27
3
3
and o(r) = l (r2 + 9r 4 – 9r 2 – 1 .
Clearly p(1) = p1(1) = 0 and p11(1) = 2!o(1).
Hence our method is consistent

Zero Stability
A linear multistep method is said to be zero stable, if the zeros of the first char acteristic polynomial p(r) satisfies I r I:s 1 and for I r I= 1 is simple
Hence our method is not zero stable

Stability region
8y.
The method (13) is said to be absolute stable if for a given h, all roots zs of the characteristic polynomial 1 (z, h) = p (z) + po (z) = 0, satisfies I zs I< 1, s = 1, 2, …, n. where h = ,\2p and ,\ = 8f .
The boundary locus method is adopted to determine the region of absolute stability. Substituting the test equation Y11 = – ,\2p into (6) and writing r = cos 0 + i sin 0 gives the stability region as shown in fig (1)
11
y
3
2
1
9
8
7
6
5
4
3
2
1
x
1
2
3
fig(1)


Numerical Experiments

Test Problems
2
We test our method with second order initial value problems Problem 1: Consider the nonlinear initial value problem (I.V.P) Y11 – x(Y1)2 = 0, Y(0) = 1, Y1(0) = l, h = 0.05
2
2x
Exact solution Y(x) = 1 + l ln (2+x )
This problem was solved by Awoyemi [5] where a method of order 8 is proposed and it is implemented in predictorcorrector mode with h = 1/320. Jator [10] also solved this problem in block method where a block of order 6 and steplength of 5 is proposed with h = 0.05.We compared our result with these two results as shown in table 1. Though the result of Awoyemi [5] was not shown in this paper but
12
Jator [10] has shown that there method is better.
Problem 2: We consider the nonlinear initial value problem (I.V.P)
2y
6
4
6
2
Y11 = (yt)2 – 2Y, Y( 1T ) = l, Y1( 1T ) = 3 .
Exact solution (sin x)2
This problem was solved by Awoyemi [5] where a method of order 8 is proposed and it is implemented in predictorcorrector mode with h = 1/320. Jator [10] also solved this problem in block method where a block of order 6 and steplength of 5 is proposed with h = 0.049213. We compared our result with these two results as shown in table 2.Though the result of Awoyemi [5] was not shown in this paper
but Jator [10] has shown that there method is better.
I I
Error= Exact resultComputed result
Table 1 for problem 1
x Exact result Computed result Error Error in [10] 0.1 1.050041729278914 1.0500417292784907 6.6613(16) 7.1629(12)
0.2 1.1003353477310756 1.1003353477310676 7.9936(15) 1.5091(11)
0.3 1.1511404359364668 1.1511404359364292 3.7525(14) 4.5286(11)
0.4 1.2027325540540821 1.2027325540539671 1.1501(13) 1.0808(10)
0.5 1.2554128118829952 1.2554128118826986 2.9665(13) 1.7818(10)
0.6 1.3095196042031119 1.3095196042024151 6.9677(13) 4.4434(10)
0.7 1.3654437542713964 1.3654437542698326 1.5638(12) 7.4446(10)
0.8 1.4236489301936022 1.4236489301901405 3.4616(12) 1.5009(09)
0.9 1.4847002785940522 1.4847002785863199 7.7322(12) 3.7579(09)
1.0 1.549306144340554 1.5493061443162750 1.7780(11) 4.7410(09)
13
Table 2 for problem 2
x
Exact result
Computed result
Error
Error in [10]
1.1069
0.7998266847638
0.799826684754
9.7840(12)
2.8047(10)
1.2069
0.8733436578646
0.873343657851
1.3090(11)
2.7950(10)
1.3069
0.9319765974804
0.931976597464
1.5780(11)
2.1490(10)
1.4069
0.9733879933357
0.972338799331
1.7581(11)
5.4975(11)
1.5069
0.9959269037589
0.995926903740
1.8360(11)
1.1545(10)
1.6069
0.9986947735170
0.998694773498
1.8103(11)
4.4825(10)
1.7069
0.9815812563774
0.981581256055
1.6880(11)
7.7969(10)
1.8069
0.9452681426358
0.945268614248
1.4862(11)
1.1840(09)
1.9069
0.8912045176254
0.891204517613
1.2232(11)
1.6318(09)
2.0069
0.8215443313867
0.821544331377
9.2473(12)
2.0567(09)


Conclusion
We have proposed a twosteps four off steps method of order five in this paper. Continuous block method which has the properties of evaluation at all points within the interval of integration is adopted to give the independent solution at non overlapping intervals as the predictor. This new method forms a bridge between the predictorcorrector method and block method. Hence it shares the properties of both method. The new method performed better than the existing method

block method and the predictor corrector method as shown in the numerical examples.
References

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