We emphasized explicitly on the derivation and implementation of a new numerical algorithm scheme which gave stable results that show the applicability of the method. In this paper, we aimed to solve some second order initial value problems of ordinary differential equations and compare the results with the theoretical solution. Using this method to solve some initial value problems of second order ordinary differential equations, we discovered that the results compared favorably with the theoretical solution which led to the conclusion that the new numerical algorithm scheme derived in the research is approximately correct and can be prescribed for any related ordinary differential equations.
Numerical methods are methods that are constructed through a given interval. The methods start with an initial point and then take a short step forward in time to find the next solution point. The process then continues with subsequent steps to map out the solution. There are two main numerical methods of solving initial value problems of ordinary differential equations. They are single step methods, also known as one step method and multistep methods. The single-step methods are the method that uses information about the solution at one point xn, to advance it to the next point xn+1. The single step methods have certain advantages which include, being self-starting and having the flexibility to change step size from one step to the next.
Various numerical methods have been developed for the solution of some initial value problems of ordinary differential equations. Some of the numerical analysts who have worked extensively on the development on numerical methods are: [
Recently [
In this paper, a new one-step numerical method is developed with the above mentioned characteristics in mind to solve some initial value problems of ordinary differential equations which were based on the local representation of the theoretical solution to initial value problem of the form y ′ = f ( x , y ) ; y ( a ) = η . In the interval ( x n , x n + 1 ) by interpolating function F ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + b x R e ( e k x + μ ) , a 0 , a 1 , a 2 , a 3 and b are real undetermined coefficients and k , μ are complex parameters. But in this paper, we shall be using the same assumptions but different interpolating functions such as:
F ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + b x R e ( e k x + μ ) , where a 0 , a 1 , a 2 , a 3 and b are real undetermined coefficients and k , μ are complex parameters.
Considering an interpolating function:
f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + b x R e ( e k x + μ ) (1)
where a 0 , a 1 , a 2 , a 3 and b are real undetermined coefficients and k , μ are complex parameters.
Since k and μ are complex parameters, then we have:
k = k 1 + i k 2 (2)
Also, μ = i θ , where i 2 = − 1 , therefore putting this together with (2) in (1), we have the Interpolating function to be:
f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + b x e k 1 x cos ( k 2 x + σ ) (3)
Let us define R ( x ) and θ ( x ) as follows:
R ( x ) = x e k 1 x θ ( x ) = k 2 x + σ } (4)
Putting (4) in (3), we have:
f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + b R ( x ) cos θ ( x ) (5)
By assumption, y n is a numerical estimate to the theoretical solution y ( x n ) and also f n = f ( x n , y n ) . Let our mesh points (self length) be define as follows:
x n = a + n h ; n = 0 , 1 , 2 , ⋯ , a = 0 , x n = n h . x n + 1 = ( n + 1 ) h (6)
Imposing the following constraints on the interpolating function (5), we have:
1) The interpolating function must coincide with the theoretical solution at x = x n and x = x n + 1 . This required that:
f ( x n + 1 ) = a 0 + a 1 x n + 1 + a 2 x n + 1 2 + a 3 x n + 1 3 + b R ( x n + 1 ) cos θ ( x n ) . (7)
That is, f ( x n ) = y ( x n ) and
f ( x n + 1 ) = a 0 + a 1 x n + 1 + a 2 x n + 1 2 + a 3 x n + 1 3 + b R ( x n + 1 ) cos θ ( x n + 1 ) (8)
It implies that f ( x n + 1 ) = y ( x n + 1 ) .
2) The first, second, third and fourth derivatives with respect to x of the interpolating function respectively coincide with the differential equation as well as its first, second, third and fourth derivatives with respect to x at xn, i.e.
F 1 ( x n ) = f n F 2 ( x n ) = f n 1 F 3 ( x n ) = f n 2 F 4 ( x n ) = f n 3 } (9)
From Equation (9) implies:
f ′ ( x n ) = f n + 2 a 2 x n + 3 a 3 x n 2 + [ cos θ ( x n ) d d x ( b R ( x n ) ) + b R ( x n ) d d x ( cos θ ( x ) ) ] (10)
where
d d x ( b R ( x n ) ) = d d x ( b x e k 1 x ) = e k 1 x ⋅ b + b x ⋅ k 1 e k 1 x = b e k 1 x + b k 1 x e k 1 x = b e k 1 x + b k 1 R ( x n ) (11)
d d x cos θ ( x n ) d d x [ cos ( k 2 x + σ ) ] = − k 2 sin θ ( x n ) (12)
Putting (11) & (12) in (10) we have:
f ′ ( x n ) = f n = a 1 + 2 a 2 x n + 3 a 3 x n 2 [ cos θ ( x n ) { b e k 1 x + b k 1 R ( x n ) } + b R ( x n ) ( − k 2 sin θ ( x n ) ) ] = a 1 + 2 a 2 x n + 3 a 3 x n 2 + [ b e k 1 x cos θ ( x n ) + b k 1 R ( x n ) cos θ ( x n ) − b R ( x n ) ( − k 2 sin θ ( x n ) ) ]
f n = a 1 + 2 a 2 x n + 3 a 3 x n 2 + b [ e k 1 x n cos θ ( x n ) + k 1 R ( x n ) cos θ ( x n ) − k 2 R ( x n ) sin θ ( x n ) ] (13)
That is, F ′ ( x n ) = f n
F ″ ( x n ) = f ′ n = 2 a 2 x n + 6 a 3 x n + b [ ( e k 1 x n d d x cos θ ( x n ) + cos θ ( x n ) d d x e k 1 x n ) + ( k 1 R ( x n ) d d x cos θ ( x n ) + cos θ ( x n ) d d x k 1 R ( x n ) − ( R ( x n ) d d x sin θ ( x n ) + sin θ ( x n ) d d x R ( x n ) ) ) ] (14)
where
k 1 R ( x n ) = k 1 [ x e k 1 x n ] = k 1 ( x d d x e k 1 x n + e k 1 x n d d x x ) = k 1 2 x e k 1 x n + k 1 e k 1 x n = k 1 2 R ( x n ) + k 1 e k 1 x n (15)
Since R ( x n ) = x n e k 1 x n = e k 1 x n + k 1 R ( x n ) .
Putting (15) in (14)
F ″ ( x n ) = f ′ n = 2 a 2 x n + 6 a 3 x n + b [ { e k 1 x n ( − sin θ ( x n ) ) + cos θ ( x n ) ( k 1 e k 1 x n ) } + k 1 R ( x n ) ( − sin θ ( x n ) ) + cos θ ( x n ) ( k 1 2 R ( x n ) + k 1 e k 1 x n ) − { R ( x n ) cos θ ( x n ) + sin θ ( x n ) ( e k 1 x n + k 1 R ( x n ) ) } ]
F ″ ( x n ) = 2 a 2 x n + 6 a 3 x n + b [ { − e k 1 x n sin θ ( x n ) + k 1 e k 1 x n cos θ ( x n ) } − k 1 R ( x n ) sin θ ( x n ) + k 1 2 R ( x n ) cos θ ( x n ) + k 1 e k 1 x n cos θ ( x n ) − R ( x n ) cos θ ( x n ) − e k 1 x n sin θ ( x n ) − k 1 R ( x n ) sin θ ( x n ) ]
f ′ n = 2 a 2 x n + 6 a 3 x n + b { − 2 e k 1 x n sin θ ( x n ) + 2 k 1 e k 1 x n cos θ ( x n ) − 2 k 1 R ( x n ) sin θ ( x n ) + k 1 2 R ( x n ) cos θ ( x n ) − R ( x n ) cos θ ( x n ) } (16)
F ‴ ( x n ) = f ″ n = 6 a 3 + b [ − { 2 e k 1 x n d d x sin θ ( x n ) + sin θ ( x n ) d d x 2 e k 1 x n } + { 2 k 1 e k 1 x n d d x cos θ ( x n ) + cos θ ( x n ) d d x 2 k 1 e k 1 x n } − { 2 k 1 R ( x n ) d d x sin θ ( x n ) + sin θ ( x n ) d d x 2 k 1 R ( x n ) } + { k 1 2 R ( x n ) d d x cos θ ( x n ) + cos θ ( x n ) d d x k 1 2 R ( x n ) } − { R ( x n ) d d x cos θ ( x n ) + cos θ ( x n ) d d x R ( x n ) } ]
F ‴ ( x n ) = 6 a 3 + b [ − ( 2 e k 1 x n cos θ ( x n ) + 2 k 1 e k 1 x n sin θ ( x n ) ) + 2 k 1 e k 1 x n ( − sin θ ( x n ) ) + 2 k 1 2 e k 1 x n cos θ ( x n ) − 2 k 1 R ( x n ) cos θ ( x n ) + [ 2 k 1 e k 1 x n + 2 k 1 2 R ( x n ) ] sin θ ( x n ) + k 1 2 R ( x n ) ( − sin θ ( x n ) ) + k 1 2 e k 1 x n + k 1 3 R ( x n ) cos θ ( x n ) − R ( x n ) ( − sin θ ( x n ) ) + ( e k 1 x n + k 1 R ( x n ) ) cos θ ( x n ) ]
F ‴ ( x n ) = f ″ n = 6 a 3 + b [ − 3 e k 1 x n cos θ ( x n ) − 6 k 1 e k 1 x n sin θ ( x n ) + 3 k 1 2 R ( x n ) sin θ ( x n ) − 2 k 1 R ( x n ) cos θ ( x n ) + k 1 3 R ( x n ) cos θ ( x n ) + R ( x n ) sin θ ( x n ) − k 1 e k 1 x n cos θ ( x n ) ] (17)
F ′ v ( x n ) = f ‴ ( x n ) = b [ − 3 { e k 1 x n d d x cos θ ( x n ) + cos θ ( x n ) d d x e k 1 x n } − 6 { e k 1 x n d d x sin θ ( x n ) + sin θ ( x n ) d d x e k 1 x n } + 3 k 1 2 { e k 1 x n d d x cos θ ( x n ) + cos θ ( x n ) d d x e k 1 x n } − 3 k 1 2 { R ( x n ) d d x sin θ ( x n ) + sin θ ( x n ) d d x R ( x n ) }
− 2 k 1 { R ( x n ) d d x cos θ ( x n ) + cos θ ( x n ) d d x R ( x n ) } + k 1 3 { R ( x n ) d d x cos θ ( x n ) + cos θ d d x R ( x n ) } + { R ( x n ) d d x sin θ ( x n ) + sin θ ( x n ) d d x R ( x n ) } − k 1 { e k 1 x n d d x cos θ ( x n ) + cos θ ( x n ) d d x e k 1 x n }
F I V ( x n ) = b [ 4 e k 1 x n sin θ ( x n ) − 11 k 1 e k 1 x n cos θ ( x n ) − 12 k 1 2 e k 1 x n sin θ ( x n ) + 4 k 1 3 e k 1 x n cos θ ( x n ) − 5 k 1 2 R ( x n ) cos θ ( x n ) − 4 k 1 3 R ( x n ) sin θ ( x n ) + 3 k 1 R ( x n ) sin θ ( x n ) + R ( x n ) cos θ ( x n ) + k 1 e k 1 x n sin θ ( x n ) − k 1 2 e k 1 x n cos θ ( x n ) + k 1 4 R ( x n ) cos θ ( x n ) ] (18)
b = f ‴ n e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } (19)
From (17), we have:
a 3 = 1 6 [ f n 2 − b { − 3 e k 1 x n cos θ ( x n ) − 6 k 1 e k 1 x n sin θ ( x n ) + 3 k 1 2 e k 1 x n cos θ ( x n ) − 3 k 1 2 R ( x n ) sin θ ( x n ) − 2 k 1 R ( x n ) cos θ ( x n ) + k 1 3 R ( x n ) cos θ ( x n ) + R ( x n ) sin θ ( x n ) − k 1 e k 1 x n cos θ ( x n ) } ] (20)
Putting (19) in (20), we have:
a 3 = 1 6 [ f n 2 − ( f n 3 e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ) × { − 3 e k 1 x n cos θ ( x n ) − 6 k 1 e k 1 x n sin θ ( x n ) + 3 k 1 2 e k 1 x n cos θ ( x n ) − 3 k 1 2 R ( x n ) sin θ ( x n ) − 2 k 1 R ( x n ) cos θ ( x n ) + k 1 3 R ( x n ) cos θ ( x n ) + R ( x n ) sin θ ( x n ) − k 1 e k 1 x n cos θ ( x n ) } ] (21)
Let:
v = − 3 e k 1 x n cos θ ( x n ) − 6 k 1 e k 1 x n sin θ ( x n ) + 3 k 1 2 e k 1 x n cos θ ( x n ) − 3 k 1 2 R ( x n ) sin θ ( x n ) − 2 k 1 R ( x n ) cos θ ( x n ) + k 1 3 R ( x n ) cos θ ( x n ) + R ( x n ) sin θ ( x n ) − k 1 e k 1 x n cos θ ( x n ) (22)
Then, (21) becomes:
a 3 = 1 6 [ f n 2 − ( f n 3 e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ) v ] (23)
From (16), we have:
a 2 = 1 2 [ f n 1 − ( f n 2 − ( f n 3 e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ) v ) x n − b { − 2 e k 1 x n sin θ ( x n ) − 2 k 1 e k 1 x n cos θ ( x n ) − 2 k 1 R ( x n ) sin θ ( x n ) + k 1 2 R ( x n ) cos θ ( x n ) − R ( x n ) cos θ ( x n ) } ] (24)
Putting (19) and (23) in (24) we have:
a 2 = 1 2 [ f n 1 − ( f n 2 − ( f n 3 e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ) v ) x n − ( f n 3 e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ) × { − 2 e k 1 x n sin θ ( x n ) − 2 k 1 e k 1 x n cos θ ( x n ) − 2 k 1 R ( x n ) sin θ ( x n ) + k 1 2 R ( x n ) cos θ ( x n ) − R ( x n ) cos θ ( x n ) } ] (25)
From (13) we have:
a 1 = f n − 2 a 2 x n − 3 a 3 x n 2 − b [ e k 1 x n cos θ ( x n ) + k 1 R ( x n ) cos θ ( x n ) − R ( x n ) sin θ ( x n ) ] (26)
Putting (19), (23) and (25) in (26), we have:
a 1 = f n − ( [ − [ f n 2 − { − 3 e k 1 x n cos θ ( x n ) − 6 k 1 e k 1 x n sin θ ( x n ) + 3 k 1 2 e k 1 x n cos θ ( x n ) − 3 k 1 2 R ( x n ) sin θ ( x n ) − 2 k 1 R ( x n ) cos θ ( x n ) } f n 3 ] e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ] x n − { − 2 e k 1 x n sin θ ( x n ) − 2 k 1 e k 1 x n cos θ ( x n ) − 2 k 1 R ( x n ) sin θ ( x n ) + k 1 2 R ( x n ) cos θ ( x n ) − R ( x n ) cos θ ( x n ) } e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ) x n
− 1 2 [ f n 2 − { − 3 e k 1 x n cos θ ( x n ) − 6 k 1 e k 1 x n sin θ ( x n ) + 3 k 1 2 e k 1 x n cos θ ( x n ) − 3 k 1 2 R ( x n ) sin θ ( x n ) − 2 k 1 R ( x n ) cos θ ( x n ) } e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ] x n 2 − ( f n 3 e k 1 x n { 4 sin θ ( x n ) − 11 k 1 cos θ ( x n ) − 12 k 1 2 sin θ ( x n ) + 4 k 1 3 cos θ ( x n ) + k 1 sin θ ( x n ) − k 1 2 cos θ ( x n ) } + R ( x n ) { − 5 k 1 2 cos θ ( x n ) − 4 k 1 3 sin θ ( x n ) + 3 k 1 sin θ ( x n ) + cos θ ( x n ) + k 1 4 cos θ ( x n ) } ) × [ e k 1 x n cos θ ( x n ) + k 1 R ( x n ) cos θ ( x n ) − R ( x n ) sin θ ( x n ) ] (27)
For preservative of the scheme, then we can write the new scheme in a more compact form as:
y n + 1 = y n + a 1 h + a 2 a 2 ( 2 n + 1 ) + h 3 a 3 ( 3 n 2 + 3 n + 1 ) + b R n [ h e k 1 h ( cos θ n cos k 2 h − sin θ n sin k 2 h ) − cos θ n ]
Putting a 1 , a 2 and a 3 as derived above, we arrived at a new scheme. But to test the scheme, we shall proceed to write a programme which will command the scheme to solve some firs order differential equations.
PROBLEM 1. A spring with a mass of 2 kg has natural length m. A force of 5N is required to maintain it stretched to a length of m. If the spring is stretched to a length of m and then released with initial velocity 0, find the position of the mass at any time.
k ( 0.2 ) = 25.6 (1)
So, k = 25.6 0.2 = 128 . (2)
Using this value of the spring constant k, together with m = 2 then, we have
2 d 2 x d t 2 + 128 x = 0 (3)
x ( t ) = c 1 cos 8 t + c 2 sin 8 t (4)
x ′ ( t ) = − 8 c 1 sin 8 t + 8 c 2 cos 8 t (5)
Since the initial velocity is given as x ′ ( 0 ) = 0 , we have c 2 = 0 and so the solution is
x ( t ) = 1 5 cos 8 t (6)
The result compared favourably with the theoretical solutions.
PROBLEM 2:
Suppose that the spring of Problem 1 is immersed in a fluid with damping constant. Find the position of the mass at any time if it starts from the equilibrium Position and is given a push to start it with an initial velocity of m/s.
Mathematical Interpretation of the Problem
From Problem 1, the mass is m = 2 and the spring constant is k = 128 , so the differential Equation (3) becomes
2 d 2 x d t 2 + 40 d x d t + 128 x = 0 (7)
or
d 2 x d t 2 + 20 d x d t + 64 x = 0 (8)
x ( t ) = c 1 e − 4 t + c 2 e − 16 t (9)
We are given that x ( 0 ) = 0 , c 1 + c 2 = 0 . Differentiating, we get
x ′ ( t ) = − 4 c 1 e − 4 t + 16 c 2 e − 16 t (10)
So
x ′ ( 0 ) = − 4 c 1 + 16 c 2 = 0.6 (11)
Since c 2 = − c 1 , this gives 12 c 1 = 0.6 or c 1 = 0.05 . Therefore
x = 0.05 ( e − 4 t − e − 16 t ) (12)
The result compares favourably with the theoretical solutions.
The result compares favourably with the theoretical solutions.
The procedure for development and implementation of a numerical algorithm has been examined in this work. The method employed a basis function for approximation. The scheme that formed the derived method was implemented to test the accuracy. The method gave a numerical algorithm capable of solving second order problems of ordinary differential equations. The new numerical algorithm scheme was employed to solve second order initial value problems of ordinary differential equations problems. The computations were carried out in computer codes. Tables 1-4 represent the numerical results of the examples at various values of step size (H). From the tables, it could be concluded that a smaller value of H, gives a better result. Figures 1-4 readily show the comparison of the newly derived method and the theoretical solution. Since the results obtained compare favorably with the theoretical, this shows the accuracy of the new method and so can be said to be approximately correct.
| X(n) | THEORITICAL SOLUTION | OLUBUNMI | TRUN |
|---|---|---|---|
| 0.00000000 | 0.20000000 | 0.20000000 | 0.00000000 |
| 0.01000000 | 0.19936034 | 0.20000000 | 0.00063966 |
| 0.02000000 | 0.19744547 | 0.19999953 | 0.00255406 |
| 0.03000000 | 0.19426760 | 0.19626706 | 0.00199946 |
| 0.04000000 | 0.18984708 | 0.19087657 | 0.00102949 |
| 0.05000000 | 0.18421221 | 0.18522314 | 0.00101093 |
| 0.06000000 | 0.17739899 | 0,17885647 | 0.00145748 |
| 0.07000000 | 0.16945103 | 0.17057349 | 0.00112246 |
| 0.08000000 | 0.16041915 | 0.16179043 | 0.00137128 |
| 0.09000000 | 0.15036115 | 0.15114648 | 0.00078533 |
| 0.09999999 | 0.13934135 | 0.14011432 | 0.00077297 |
| 0.11000000 | 0.12743023 | 0.12878653 | 0.0013563 |
| 0.12000000 | 0.11470401 | 0.11534238 | 0.00063837 |
| 0.13000000 | 0.10124406 | 0.10815221 | 0.00690815 |
| 0.14000000 | 0.08713649 | 0.08937685 | 0.00224036 |
| 0.14999999 | 0.07247157 | 0.07387969 | 0.00140812 |
| 0.16000000 | 0.05734305 | 0.05876521 | 0.00142216 |
| 0.17000000 | 0.04184773 | 0.04497863 | 0.0031309 |
| 0.17999999 | 0.02608475 | 0.02765649 | 0.00157174 |
| 0.19000000 | 0.01015490 | 0.01217869 | 0.00202379 |
| 0.19999999 | 0.00583989 | 0.00592321 | 8.332E−05 |
| X(n) | THEORETICAL SOLUTION | OLUBUNMI | TRUN |
|---|---|---|---|
| 0.00000000 | 0.20000000 | 0.20000000 | 0.00000000 |
| 0.00100000 | 0.19999361 | 0.20000000 | 0.00000639 |
| 0.00200000 | 0.19997440 | 0.19998160 | 0.00000720 |
| 0.00300000 | 0.19994241 | 0.19995000 | 0.00000759 |
| 0.00400000 | 0.19989762 | 0.19990601 | 0.00000839 |
| 0.00500000 | 0.19984002 | 0.19984885 | 0.00000883 |
| 0.00600000 | 0.19976965 | 0.19977901 | 0.00000936 |
| 0.00700000 | 0.19968648 | 0.19969999 | 0.00001351 |
| 0.00800000 | 0.19959055 | 0.19960510 | 0.00001455 |
| 0.00900000 | 0.19948183 | 0.19949996 | 0.00001813 |
| 0.01000000 | 0.19936034 | 0.19937993 | 0.00001959 |
| 0.01100000 | 0.19922610 | 0.19924990 | 0.00002380 |
| 0.01200000 | 0.19907911 | 0.19910485 | 0.00002574 |
| 0.01300000 | 0.19891937 | 0.19894679 | 0.00002742 |
| 0.01400000 | 0.19874692 | 0.19877972 | 0.00003280 |
| 0.01500000 | 0.19856173 | 0.19859963 | 0.00003790 |
| 0.01600000 | 0.19836384 | 0.19840453 | 0.00004069 |
| 0.01700000 | 0.19815326 | 0.19819939 | 0.00004613 |
| 0.01800000 | 0.19792998 | 0.19797923 | 0.00004925 |
| 0.01900000 | 0.19769405 | 0.19774904 | 0.00005499 |
| 0.02000000 | 0.19744545 | 0.19750381 | 0.00005836 |
| X(n) | THEORETICAL SOLUTION | OLUBUNMI | TRUN |
|---|---|---|---|
| 0.00000000 | 0.10000000 | 0.10000000 | 0.0000000 |
| 0.00100000 | 0.09900676 | 0.09900348 | 0.00000328 |
| 0.00200000 | 0.09802692 | 0.09807573 | 0.00004891 |
| 0.00300000 | 0.09706028 | 0.09713825 | 0.00007797 |
| 0.00400000 | 0.09610662 | 0.09618537 | 0.00007875 |
| 0.00500000 | 0.09516575 | 0.09527836 | 0.00011261 |
| 0.00600000 | 0.09423748 | 0.09434861 | 0.00011113 |
| 0.00700000 | 0.09332163 | 0.09343216 | 0.00011053 |
| 0.00800000 | 0.09241800 | 0.09252773 | 0.00010973 |
| 0.00900000 | 0.09152640 | 0.09173281 | 0.00020641 |
| 0.01000000 | 0.09064666 | 0.09081132 | 0.00016472 |
| 0.01100000 | 0.08977859 | 0.08999590 | 0.00021731 |
| 0.01200000 | 0.08892203 | 0.08899999 | 0.00022040 |
| 0.01300000 | 0.08807679 | 0.08835681 | 0.00028002 |
| 0.01400000 | 0.08724272 | 0.08752332 | 0.00028079 |
| 0.01500000 | 0.08641962 | 0.08673274 | 0.00031312 |
| 0.01600000 | 0.08560735 | 0.08570748 | 0.00030213 |
| 0.01700000 | 0.08480573 | 0.08413814 | 0.00043241 |
| 0.01800000 | 0.08401462 | 0.08427880 | 0.00051327 |
| 0.01900000 | 0.08323386 | 0.08382617 | 0.00059231 |
| 0.02000000 | 0.08246327 | 0.08209541 | 0.00063214 |
| X(n) | THEORETICAL SOLUTION | OLUBUNMI | TRUN |
|---|---|---|---|
| 0.00000000 | 0.10000000 | 0.10000000 | 0.00000000 |
| 0.00010000 | 0.09990007 | 0.09992720 | 0.00002713 |
| 0.00020000 | 0.09980027 | 0.09983454 | 0.00003427 |
| 0.00030000 | 0.09970061 | 0.09974292 | 0.00004231 |
| 0.00040000 | 0.09960109 | 0.09965531 | 0.00005422 |
| 0.00050000 | 0.09950170 | 0.09956093 | 0.00005923 |
| 0.00060000 | 0.09940244 | 0.09946347 | 0.00006123 |
| 0.00070000 | 0.09930332 | 0.09937088 | 0.00006676 |
| 0.00080000 | 0.09920434 | 0.09927559 | 0.00007125 |
| 0.00090000 | 0.09910548 | 0.09918759 | 0.00008211 |
| 0.00100000 | 0.09900676 | 0.09909777 | 0.00009101 |
| 0.00110000 | 0.09890819 | 0.09894247 | 0.00010113 |
| 0.00120000 | 0.09880973 | 0.09886919 | 0.00013274 |
| 0.00130000 | 0.09871142 | 0.09877561 | 0.00015777 |
| 0.00140000 | 0.09861323 | 0.09861727 | 0.00016238 |
| 0.00150000 | 0.09851518 | 0.09854087 | 0.00018621 |
| 0.00160000 | 0.09841727 | 0.09845110 | 0.00020000 |
| 0.00170000 | 0.09831949 | 0.09836652 | 0.00022135 |
| 0.00180000 | 0.09822183 | 0.09827469 | 0.00022927 |
| 0.00190000 | 0.09812431 | 0.09830139 | 0.00024221 |
| 0.00200000 | 0.09802692 | 0.09837561 | 0.00027725 |
In our subsequent research, we shall pay more attention on the implementation of this new scheme to solve some higher order initial value problems of ordinary differential equation and also compare the results with the existing methods and thereafter we examine the characteristics properties such as the stability, convergence, accuracy and consistency of the scheme.
The authors declare no conflicts of interest regarding the publication of this paper.
Ogunrinde, R.B. and Olubunmi, J.O. (2019) Solution of Some Second Order Ordinary Differential Equations Using a Derived Algorithm. Applied Mathematics, 10, 312-325. https://doi.org/10.4236/am.2019.105022