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Echi, 2:1http://dx.doi.org/10.4172/scientificreports.607

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Volume 2 • Issue 1 • 2013

Keywords: Second-order linear differential equation; Volterra integral equation; Taylor’s expansion; Error analysis

Introduction Some second-order differential equations with variable coefficients

can be solved analytically by various methods [1]. For general cases, one must appeal to numerical techniques or approximate approaches for getting its solutions [2,3]. The Adomian decomposition method for solving differential and integral equations, linear or nonlinear, has been subject of extensive analytical and numerical studies. In particular Adomian’s decomposition method has been proposed for solving second-order linear differential equation.

In the present paper, we give an approach for determining the solution of second-order linear differential equation. The second-order linear ordinary differential equation is first converted to a Volterra integral equation. By solving the resulting Volterra equation by means of Taylor’s expansion, different approaches based on differentiation and integration methods are employed to reduce the resulting integral equation to a system of linear equation for the unknown and its derivatives the approximate solution of second-order linear differential equation is obtained. By studying the estimation of the error give the efficiency and high accuracy of the proposed method [4-6].

Volterra Integrals Equations We consider the following second-order differential equation : (E) : y’’(t)+p(t) y’(t) y(t) = g(t) (1)

with p,q and g are infinitely differential functions in open interval I ⊂ R. We fix a point a of the interval I. We have, ∀ x ∈ I,

1 1 1( ) ( ) ( ) ( )( ) = ( ) ( ) ( )( 1)! ! ( 1)! ( 1)!

i i i ix x

a a

t x a x a x t xy t dt y a y a y t dti i i i

+ + −− − − −′′ ′− ++ + −∫ ∫ (2)

and1 1( ) ( )( ) ( ) = ( ) ( )

( 1)! ( 1)!

i ix

a

t x a xp t y t dt p a y ai i

+ +− −′ −+ +∫

1( ) ( )[ ( ) ( )] ( )

! ( 1)!

i ix

a

t x t xp t p t y t dti i

+− − ′− ++∫ (3)

The differential equation (E) equivalent at integral equation :

,( ) : , ( ) ( ) = ( )x

i i x iaE x I h t y t dt f x∀ ∈ ∫ (4)

with 1 1

,( ) ( ) ( )( ) = ( ) ( ( ) ( ))( 1)! ! ( 1)!

i i i

i xt x t x t xh t p t q t p ti i i

− +− − − ′− + −− +

(5)

*Corresponding author: Nadhem Echi, Department of Mathematics, El Manar University, Tunis 1002, Tunisia, E-mail: nadhemechi_fsg@yahoo.fr

Received October 02, 2012; Published February 12, 2013

Citation: Echi N (2013) Approximate Solution of Second-Order Linear Differential Equation. 2: 607 doi:10.4172/scientificreports.607

Copyright: © 2013 Echi N. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

AbstractThis paper presents an efficient approach for determining the solution of second-order linear differential equation.

The second-order linear ordinary differential equation is first converted to a Volterra integral equation. By solving the resulting Volterra equation by means of Taylor’s expansion, different approaches based on differentiation and integration methods are employed to reduce the resulting integral equation to a system of linear equation for the unknown and its derivatives the approximate solution of second-order linear differential equation is obtained. Test example demonstrates the effectiveness of the method and gives the efficiency and high accuracy of the proposed method.

Mathematics subject classification: 34-XX; 41-XX

Approximate Solution of Second-Order Linear Differential EquationNadhem Echi*Department of Mathematics, El Manar University , Tunis 1002, Tunisia

1 1( ) ( ) ( )( ) = ( ) [ ( ) ( ) ( )] ( )! ( 1)! ( 1)!

i i ix

i a

a x a x t xf x y a y a p a y a g t dti i i

+ +− − −′− + + ++ +∫

(6)

Linear System and Approximate Solution Next following the method suggested in [1,6]. We fix one positive

integer n ≥ 1. We have:

1( )

,=0

( )( ) = ( ) ( )!

knk

n xk

t xy t y x R tk

− −+∑ (7)

where Rn,x(t) denotes integral remainder 1

( ),

( )( ) = ( )( 1)!

nt nn x x

t sR t y s dsn

−−−∫ (8)

In particular, if the desired solution y(t) is a polynomial of degree equal to or less than n-1, then Rn,x(t)=0.

We put for all i and j positive integer ≥ 1: 1

,( )( ) = ( )( 1)!

jx

ij i xa

t xb x h t dtj

−−−∫ (9)

For an integer i ≥ 1, the function y is solution of (E) if and only if we have:

( 1)

=1( ) ( ) = ( )

nj

ij ij

b x y x f x−∑ (10)

We consider the matrix:

Bn(x)=(bij(x))i,j=1,…,n (11)

and the column

Citation: Echi N (2013) Approximate Solution of Second-Order Linear Differential Equation. 2: 607 doi:10.4172/scientificreports.607

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Volume 2 • Issue 1 • 2013

Thus, for all k = 0,…,+ j-2, we have: ( ) ( ) = 0kijb a (21)

So we have: ( 1) 1

2( ) = ( 1)i j i j jij i jb a C+ − + −

+ −− (22)( ) 1 1

1( ) = ( 1) ( )i j i j jij i jb a C p a+ + − −

+ −− (23)

The explore Taylor’s approximations, in a of ( 1)i jijb + + at order n-2,

give that of bij at order n+i+j-1 following :

bij(x)=b(n)ij(x)+o((x-a)n+i+j-1) (24)with

1

( )( ) ( )( ) = ( )

( 1)( 1)!( 1)! ( ) !( 1)!

i j i j

n ija x a xb x p a

i j i j i j i j

+ − +− −− −

+ − − − + − (25)

For i ≥ 1 and j ≥ 1, we put :1( )( ) ( )

( 1)!( 1)!

i j

ij ija xb x b x

i j

+ −−=

− − (26)

We have:

( )( ) ( ) (( ) )nij n ijb x b x o x a= + − (27)

with:

( )

( 1)

1=2 1

( 2)

2=2 1

1 ( )( ) = ( )1 ( )

1 1 ( )[ ] ( )( 1)( ) ( 1)!

1 1 ( )[ ] ( )( 1)( 1) ( 1)!

n ij

lnl

ll i j l

lnl

ll i j l

p ab x x ai j i j i

j p a x ai i i j lC

q a x ai i i j lC

−

−+ + −

−

−+ + −

− −+ − +

−+ −

+ + −

+ −+ + + −

∑

∑

(28)So the approximation in a, at order n, of ijb use the approximation

at order n-1 of p and the approximation at order n-2 of q.

We put: , =1,...,( ) = ( ( ))n ij i j nB x b x and ( ) ( ) , =1,...,( ) = ( ( ))n n ij i j nB x b x

Let Dn(x) the diagonal matrix having 1( )

( 1)!

ia xi

−−−

as i-th diagonal

coefficient. For all : i=1,…,n.

1

1 0 . . 00 ( ) 0 . .. . . . .

( ) = . . . . 0( )0 . . 0( 1)!

nn

a x

D xa xn

−

− −

−

(29)

We have : ( ) ( ) ( ) ( ) ( )n n n nB x x a D x B x D x= − (30)

( ) ( )( ) ( ) ( ) ( ) ( )n n n nB x x a D x B x D x= − (31)

( )( ) ( ) (( ) )nn nB x B x o x a− = − (32)

Approximation of Fn We have:

1 1( ) ( ) ( )( ) = ( ) [ ( ) ( ) ( )] ( )! ( 1)! ( 1)!

i i ix

i a

a x a x t xf x y a y a p a y a g t dti i i

+ +− − −′− + + ++ +∫

(33)We are going to use the explore Taylor’s approximations, in a of fi

(x) at order i+n. For k ≤ i+1, we have :

1

( 1)

( )( )

( ).

( ) = , ( ) = ..

.( )

( )

n n

n n

y xf x

y xF x Y x

f xy x−

′

(12)

For all positive integer n ≥ 2, the equation (E) equivalent at

∀ x ∈ I, Bn(x)Yn(x)=Fn(x) (13)

Application of Cramer’s rule to the resulting system yields an approximate solution of equation (1). It is also noted that not only y(x) but also y(i) (x) (1 ≤ i ≤ n-1) are determined via solving the resulting system. The solution y(x) can be approximately obtained to be:

1det( ( ))( )

det( ( ))nn

B xy x

B x≡ (14)

where

1 12 1

2 22 2

1

2

( ) ( ) ... ( )( ) ( ) ... ( ). . .

( ) =. . .( ) ( ) ... ( )

n

n

n n nn

f x b x b xf x b x b x

B x

f x b x b x

(15)

Error Analysis Approximation of Bij

We are going to use the explore Taylor’s approximations, in a of bij at order i+ j+ n-1. Let: i ≥ 1 and j ≥ 1 For all k = 0,….,i + j-2 , we have:

2 1( ) 1 1

2 1

1

( ) ( )( ) = ( 1) [ ( )( 2)! ( 1)!

( ) ( ( ) ( ))]( )!

i j k i j kxk i j j jij i j i ja

i j kj

i j

x t x tb x C C p ti j k i j k

x tC p t q t dti j k

+ − − + − −+ − −

+ − + −

+ −−

+

− −− −

+ − − + − −

− ′+ ++ −

∫

(16)In particular, we have :

( 2) 1 12 1

21

( ) = ( 1) [ ( ) ( ) ( )

( ) ( ( ) ( )) ]2

xi j i j j jij i j i j a

xji j a

b x C x a C x t p t dt

x tC p t q t dt

+ − + − −+ − + −

−+

− − − −

− ′+ +

∫

∫ (17)Furthermore, we have:

( 1) 1 1 12 1( ) = ( 1) ( 1) ( )

xi j i j j i j jij i j i j a

b x C C p t dt+ − + − + − −+ − + −− + − ∫

1( 1) ( )( ( ) ( ))xi j j

i j aC x t p t q t dt+ −

+ ′+ − − +∫ (18)( ) 1 1 1

1( ) = ( 1) ( ) ( 1) ( ( ) ( ))xi j i j j i j j

ij i j i j ab x C p x C p t q t dt+ + − − + −

+ − + ′− + − +∫ 1 1 1 1

1= ( 1) [ ] ( ) ( 1) ( ) ( 1) ( )xi j j j i j j i j j

i j i j i j i j aC C p x C p a C q t dt+ − − + − + −

+ + − + +− − − − + − ∫ (19)

( 1) 1 1 11( ) = ( 1) [ ] ( ) ( )]i j i j j j j

ij i j i j i jb x C C p x C q x+ + + − − −+ + − +′− − + (20)

Citation: Echi N (2013) Approximate Solution of Second-Order Linear Differential Equation. 2: 607 doi:10.4172/scientificreports.607

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1 1( ) ( )( ) = ( 1) ( )( 1)! ( 1 )!

k i i kx xkk a a

d t x t xg t dt g t dti i kdx

+ + −− −−

+ + −∫ ∫ (34)

and

fi(i+2) = (-1)i+1 g(x) (35)

Thus we have:

fi(x) = f(n)i (x)+o((x-a)i+n) (36)with

( )( )( ) = ( ) [ ( ) ( ) ( )]

!

i

n ia xf x y a y a p a y a

i− ′− + +

1 1 ( )2

2

=0 2

( ) ( 1) 1 ( ) ( )( 1)! ( 2)! !

i i knk i

kk k i

a x g a x ai i kC

+ + −+ +

+ +

− −+ −

+ + ∑ (37)

We put:

( )1

( )2

( )

( )

( )( )

.( ) =

.( )

n

n

n

n n

f xf x

F x

f x

(38)

We have:

( )( ) ( ) ( ) ( )n n nF x x a D x F x= − (39)with:

( )1

( )1

( )2( )

( )

( )

( )

( )( ) =

.

.

( )

n

n

nn

n n

f x

f x

f xF x

f x

(40)

For i=1…,n, we have: ( 2)

( ) 22 1

1 1 1 1 ( )( ) ( ) [ '( ) ( ) ( )] ( )( 1) ( 1)( 2) ( 2)!

knk

n i kk k

g af x y a y a p a y a x ai i i i i i kC

−

−= +

= + + + −+ + + −∑

(41)The construction of F(n) uses p(a), the approximate of g in a at

order n-2. as well as the initial condition of y(a) and y’(a). We have :

( )( ) ( ) ( )[ ( ) (( ) )]nn n nF x x a D x F x o x a= − + − (42)

Lemma:1. The matrix

, =1, ,

11 i j ni j

+ −

has positive determinant.2. We have :

, =1, ,2

, =1, ,

1det1

=1det

i j n nn

i j n

i jC

i j

+ − +

(43)

Proof: We provide C[X] by scalar product: 1

1, = ( ) ( )v w v x w x dx

−⟨ ⟩ ∫

1. Let ν0, ν1,… the orthogonal basis given by:

2 1=2k k

kv L+

with 21= ( 1)2 !

kk

k k k

dL xk dx

− Legendre polynomial of degree k.

We put: ( 1)=2 2

k

k k

xe + . The Gram Schmidt matrix of linearly

independent system (e0…,en-1) is , 1,...,

1( )1n

i j n

B ai j =

= + −

.

In particular we have : det( ( )) 0.nB a >

2. We put: 1 , 1,... .nV i j ni j

= = +

For: j=1,…,n-1

0 0 1

1 1 1

< , > < , >

= , ( ) = ,

< , > < , >

j j

n j n j

n j n j

e e e e

V e B a e

e e e e

−

− − −

where ej is the j-th vector of the canonical basis.

Let 1

=0= , = n

n n n n j jjP e e v v x e−

− ⟨ ⟩ ∑ orthogonal projection of en on Cn-1[X].

We have : 1

=0= 0, , 1, , = , = ,n

k n k j k jjk n e e e P x e e−

∀ − ⟨ ⟩ ⟨ ⟩ ⟨ ⟩∑ .

Thus, n-th column of (Vn) equal:

0 0 0

1

0=1

1 0 1

, ,. .. ., ,

j

n

jj

n n j

e e e e

x xe e e e

−

− −

⟨ ⟩ ⟨ ⟩ + ⟨ ⟩ ⟨ ⟩

∑

Thus: 10det( ) ( 1) det( ( )).n

n nV x B a−= − We have :

01( 1) = , ( 1) =2n n nP e v v x− −⟨ ⟩ −

n-th order vector entry of en equal n-th order vector entry of ,n n ne v v⟨ ⟩ . One arrives at :

01 = ( 1)2

= ( 1)

nn

n

n

n th order vector entry of ex v

n th order vector entry of v

n th order vector entry of eQ

n th order vector entry of Q

−− −

−

−− −

−

such as 21= ( 1)!

nn

n

dQ xn dx

− . Taylor’s expansion : Q(-1)=(-2)n. Thus :

10

2

11 1 12 2= ( 2) = ( 1)

(2 )!2 2! !

nn n

nn

xn C

n n

−− − −

Thus:

, =1, ,2

, =1, ,

1det1

=1det

i j n nn

i j n

i jC

i j

+ − +

We have : ( ), 1...,

1( ) ( )1n n

i j n

B a B ai j =

= = + −

.

As, ( )det( ( )) 0nB a ≠ , thus taking x close of a, ( )det( ( )) 0nB x ≠ .Hence, taking x close of a; x ≠ a B(n) (x) ≠ 0. Similarly, defines to be : Y(n)(x)=B(n) (x)-1 F(n) (x) (44) We put :

Citation: Echi N (2013) Approximate Solution of Second-Order Linear Differential Equation. 2: 607 doi:10.4172/scientificreports.607

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Volume 2 • Issue 1 • 2013

( )0

( )1

( )

( ) 1

( )( )

.( ) =

.( )

n

n

n

n n

y xy x

Y x

y x−

(45)

From (38) and (41) we have:

( ) ( ) ( )( ) ( ) ( ) ( )n n n nB x D x Y x F x= (46)

The first element of Dn(x)Y(n)(x) is y(n)0 just what we look for. Theorem:1. Taking k=0,…,n-1, we have : y(n)k (x)-y(k) (x) = 0((x-a)n-k-1) (47)2. We have :

( )

( )02

1 ( )( ) = ( ) ( ) (( ) )!

nn n

n nn

y ay x y x x a o x anC

+ − + − (48)We have: Fn(x) = F(n) (x) + (x-a)Dn(x) o((x-a)n) Taking x close of a, x≠a, we have : Bn(x)Yn(x) = B(n)(x)Y(n)(x)Y(n)(x) + (x-a)Dn(x)o((x-a)n)

one arrives at,

( ) ( )( ) ( ) ( ) ( ) ( ) ( )n n n n n nB x D x Y x B x D x Y x=

( )1

11

12( ) ( ) ( 1) (( ) ).!

.1

2

nn n n

n

ny a x a o x an

n

−

+ +

+ − − + −

Thus : ( )[ ( ) ( ) ( ) ( )]n n n n nB x D x Y x D x Y x−

( )

1

11

12( ) ( ) ( 1) (( ) ).!

.12

nn n n

n

ny a x a o x an

n

−

+ +

= − − + −

Furthermore: ( )

( )( )( ) ( ) ( ) ( )!

n

n n n ny aD x Y x D x Y x

n− =

1 1

( )

11

12

( ) ( 1) ( ) (( ) )..1

2

n n nn

n

nx a B a o x a

n

− −

+ +

− − + −

By looking at the first constituent and allowing that

( ), 1,...,

1( )1n

i j n

B ai j =

= + −

, we obtain:

( )1

( )0

1 1 1. .1 2

1 1 12 3 1

. .

. .1 1 1. .2 1 2

( )( ) ( ) = ( ) ( 1) (( ) )1 1! 1 . .2

1 1 12 3 1. .. .1 1 1. .

1 2

nn n n

n

n n

n n

n n ny ay x y x x a o x a

nn

n

n n n

−

+

+ +

+

− − − + −

+

+

Thus: ( )

( )02

1 ( )( ) ( ) = ( ) (( ) )!

nn n

n nn

y ay x y x x a o x anC

− − + −

Example: We consider the following second-order differential equation with variable coefficients

y”(x) - 2y’(x)y(x)=0 (49)under the initial condition

y(0) = 0,y’(0) = 1 (50)This equation has an exact solution y(x) = tan(x) (51)Using the present method, we can evaluate its approximate

solution. For example, when n=3 one can find the corresponding third-order approximations as:

8 6 4

3 9 7 5 3

162 8904 70560( ) =19 1734 32760 70560

x x xy xx x x x

− +− + − +

(52)

Of course, when taking a lager n, it is expected that more accurate approximate results can be given. In order to show the variation of the accuracy of approximations (Table 1).

It indicates that the present method gives a very fast convergence, even for x near some unbounded points.

ConclusionsThis paper proposes a effective method for determining the

approximate solution of second-order differential equation with constant and varying coefficients. The second-order differential equations to transformed a Volterra integral equation. By using

x Exact solution y(x) Approximate solution y3(x)0.1 0.1003347 0.10033940.5 0.5463025 0.54694481 1.5574077 1.5644186

1.57 1:2557656×103 1:2536283 ×103

1.57078 6:1249009 ×104 6:0755230 ×104

Table 1: Indicates that the present method gives a very fast convergence, even for x near some unbounded points.

Citation: Echi N (2013) Approximate Solution of Second-Order Linear Differential Equation. 2: 607 doi:10.4172/scientificreports.607

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Volume 2 • Issue 1 • 2013

Taylor’s expansion of the unknown function, the resulting Volterra equation can be approximately solved [7]. Based on these results, an approximation of second-order differential equations can be directly constructed. Test example is given, and obtained results turn out that the present method is efficient, and simple. Moreover, it can be implemented by symbolic computation at any personal computer.References1. Xie F, Gao X (2004) Exact traveling wave solutions for a class of nonlinear

partial differential equations. Chaos, Solutions & Fractals 19: 1113-1117.2. Abbasbandy S (2006) Homotopy perturbation method for quadratic Riccati

differential equation and comparison with Adomian’s decomposition method. Appl Math Comput 172: 485-490.

3. Li XF (2006) Approximate solution of linear ordinary differential equations with variable coefficients. Math Comput Simul 75: 113-125.

4. Tang BQ, Li XF (2007) A new method for determining the solution of Riccati differential equations. Appl Math Comput 194: 431-440.

5. Adomian G, Rach R (1992) Noise terms in decomposition series solution. Comput Math Appl 11: 61-64.

6. El-Tawil MA, Bahnasawi AA, Abdel-Naby A (2004) Solving Riccati differential equation using Adomian’s decomposition method. Appl Math Comput 157: 503-514.

7. Chen Y, Li Biao (2004) General projective Riccati equation method and exact solutions for generalized KdV-type and KdV-Burgers-type equation with nonlinear terms of any order. Chaos, Solitons Fractals 19: 977-984.