∫ On the ELzaki Transform and Higher Order Ordinary Differential Equations

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ISSN 0973-4554 Volume 6, Number 1 (2011), pp. 107–113

http://www.ripublication.com/atam.htm

On the ELzaki Transform and Higher Order Ordinary Differential Equations

Tarig M. Elzaki¹ and Salih M. Elzaki²

1,2Math. Dept., Sudan University of Science and Technology, (www.sustech.edu) E-mail: ¹[email protected], ²[email protected]

Abstract

In this work a new integral transform namely ELzaki transform was applied to solve higher order linear ordinary differential equations with constants coefficients.

Keywords: ELzaki transform-Differential equation.

Introduction

The differential equations have played a central role in every aspects of applied mathematics for very long time and with the advent of the computer their importance has increased further.

ELzaki transform method is particularly useful for finding solutions of higher order linear ordinary differential equations.

The method is very effective for the solution of the response of linear differential equations.

Tarig ELzaki introduced a new integral transform and named is as ELzaki transform that is defined by the formula.

( ) ( ) ( )

¹ ²

0

, ^t^v , ( , )

T v f t v v f t e dt v k k

∞ −

⎡ ⎤

= Ε⎣ ⎦=

∫

∈ −

Now let A be the set of ELzaki transformable functions that is

( )

^: ^, ^{1, 2} ^0,

( )

^t ^k^j^,

( )

¹ ^j ^{[0 , ) .}

A=⎧⎪⎨f t ∃M k k > f t <Me if t∈ − × ∞ ⎫⎪⎬

⎪ ⎪

⎩ ⎭

Recall the following theorem that was given by Tarig ELzaki, where they discussed Elzaki transform of derivatives:

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Theorem I

Let T v is the ELzaki transform of

( )

^{f t , or}

( )

^⎡_⎣^{E f t}

( ( ) )

⁼^{T v}

( )

^⎤_⎦^.^Then:

( )

^{I E f}^⎡_⎣ ^′

( )

^t ^⎤_⎦⁼^{T v}_v

( )

⁻^vf

( )

⁰

( )

^{II E f}^⎡_⎣ ^′′

( )

^t ^⎤_⎦⁼^{T v}_v

( )

² ⁻^f

( )

⁰ ⁻^vf ^′

( )

⁰

( )

^{( )}

( ) ( )

¹ ( )

( )

0

2 0

n n

k

n k k

III E f t T v v f v

−

=

⎡ ⎤= − − +

⎣ ⎦

∑

Proof

( ) ( ) ( )

0

tv

I E f t v f t e dt

∞ −

′ ′

⎡ ⎤=

⎣ ⎦

∫

. Integrating by parts to find that:

( )

^{T v}

( ) ( )

⁰

E f t vf

′ v

⎡ ⎤= −

⎣ ⎦

( )

^II ^Let^{g t}

( )

⁼^f ^′

( )

^t ^,^then^{E g t}^⎡_⎣ ^′

( )

^⎤_⎦^{= Ε}_v¹ ^⎡_⎣^{g t}

( )

^⎤_⎦⁻^vg

( )

⁰

By using (I) we find that ^{E f}^⎡_⎣ ^′′

( )

^t ^⎤_⎦⁼^{T v}_v

( )

² ⁻^f

( )

⁰ ⁻^vf ^′

( )

⁰

( )

^III Can be proving by mathematical induction.

Application of ELzaki Transform of Higher order Differential Equations

We solve the linear equation of order n with constant Coefficients as

( ) [ ]

^{( )} ⁿ ¹ ⁿ ¹ ² ⁿ ² ^... ⁿ ^{( ) ,} ⁰

f D x t =D x+a D ⁻ x+a D ⁻ x+ +a x=φ t t>

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With the initial Conditions

( )

⁰^,

( )

¹^, ²

( )

²^...., ⁽ⁿ ¹⁾ ^{( )} ⁿ ¹ ⁰

x t =x Dx t =x D x t =x D ⁻ x t =x₋ at t= (2)

Where D ^d

=dt is the differential operator and x₀,x₁...x_n₋₁ are constants.

We take ELzaki transform of equation (1), we have

( )

² ³ ⁴

( )

³ ⁴

0 1 2... 1 1 1 0 1... 2

n n n n n

n n

T v T v

v x v x v x vx a v x v x vx

v v

− − − − −

− − −

⎛ ⎞ ⎛ ⎞

− − − + − −

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

( ) ( ) ( )

1 0

... _n T v _n

a vx a T v v

− v

⎛ ⎞

+ + ⎜ − ⎟+ = Φ

⎝ ⎠ (3)

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Where ^Φ

( )

^v is ELzaki transform of ^φ

( )

^t ^.

We can write (3) in the form,

( ) ( ) ( )

1 2

1_n a_n a_n .... _n

a T v v v

v v ⁻ v ⁻

⎡ + + + ⎤ = Φ + Ψ

⎢ ⎥

⎣ ⎦

Where ^Ψ

( )

^v is made up of all terms on the right hand side of (3) except ^Φ

( )

^v ^,

and is a polynomial in v of degree

(

ⁿ⁻¹

)

^{, Hence:}

( ) ( ) ( ) ( )

f v T v = Φ v + Ψ v Where

( )

¹1 ²2

1_n a_n a_n .... _n

f v a

v v ⁻ v ⁻

= + + +

Thus ELzaki transform solution ^{T v is}

( )

( ) ( ) ( )

( )

v v

T v f v

Φ + Ψ

= (4)

Inversion of (4) yields

( ) ( )

1 v 1

( )

v

x t E E

f v f v

− ⎡Φ ⎤ − ⎡Ψ ⎤

= ⎢ ⎥+ ⎢ ⎥

⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

The inverse operation on the right can be carried out by partial fraction or any method.

Example I (Third Order Ordinary Differential Equation) Solve the equation

(

^D³⁺^D²⁻⁶^{D y t}

) ( )=0 , D =dtd , t >0 (5)

With the initial data

( )

⁰ ^{1 ,}

( )

⁰ ⁰

y = y′ = And ^y^′′

( )

⁰ ⁼⁵⁽⁶⁾

ELzaki transform of equation (5) gives,

( )

³ ¹

( )

⁰

( )

⁰

( ) ( ) ( )

⁰ ² ⁰

( )

⁰ ⁶

( )

⁶

( )

⁰ ⁰

T v T v T v

y y vy y vy vy

v −v − ′ − ′′ + v − − ′ − v + =

In view of the initial conditions, we find:

( )

^v₁² ^v⁴₆^v²³

T v v v

= − +

+ − Or ^{T v}

( )

⁼ ^v₆² ⁺¹_{3 (1 3 )}₊^v²_v ⁺¹_{2 (1 2 )}₋^v²_v

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Inverting gives the solution in the form:

( )

¹₆ ¹₃ ³^t ¹₂ ²^t

y t = + e⁻ + e

Examples II (Four Order Ordinary Differential Equation) Consider the equations:

( )^iv 2 sin , 0

y +y′′′= t t> (7)

With the initial conditions

( )

⁰

( )

⁰ ⁰ ^,

( )

⁰ ^{1 ,}

( )

⁰ ²

y =y′ = y′′ = y′′′ =− (8)

By using the similar technique, we obtain the solution of equation (7) with initial conditions (8) as

7 4 5

2

( ) 2

( 1)(1 ) (1 ) (1 )

v v v

T v = v v + v − v

+ + + +

And then:

( )

⁴ ² ² ₁^v² ₁^v³ ² ₁^v²²

T v v v

v v v

= − + + +

+ + +

Inverting find the solution

( )

¹₂ ² ² ^t ^sin ^cos

y t = t − +e⁻ + t + t

Examples III

Consider the following equation,

2 2 3 sin cos

y′′′+ y′′+ y′+ y = t + t (9)

( )

⁰ ⁰ ^,

( )

⁰ ^{0 ,}

( )

⁰ ⁰

y = y′ = y′′ = (10)

Applied ELzaki transform to equation (9) we obtain:

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( ) ( ) ( ) ( ) ( ) ( )

³ ²

3 2 2 2

1 1 2 2

0 0 0 2 0 2 0 2 0 3

1 1

y y v v

y y y vy y vy v y y

v −v − ′ − ′′ +v − − ′ + v − + = v + v

+ +

Where y is ELzaki transform of y.

Now apply the initial conditions (10) we have:

3

1 2

y v

= v

+ Inverting gives the solution in the form: ^{y t}

( )

⁼^sin^t

Example V

Consider the following equation, 1

y′′′−y′′+y′− = −y t (11)

( )

⁰ ^{1 ,}

( )

⁰ ² ^,

( )

⁰ ¹

y = y′ = y′′ = (12)

By applying ELzaki transform to equation (11) and using the conditions (12).

We get

( )( )

( ) ( )

2 3

3 2

2 2

5 6 2 3

2

3 2 2

1 1 1 1

1 1

1 1 1

y v v

v v v v

v v v

v v v v

y v

v v v v v

⎡ − + − ⎤ = − + +

⎢ ⎥

⎣ ⎦

⎡ − + + ⎤

− + + ⎢ ⎥

= =

− + − − ⎢⎣ − + ⎥⎦

2

2 1 3

1 1

y v v v v

v v

⎡ ⎤

= ⎢⎣ + − ⎥⎦= + −

We take the inverse of ELzaki transform we obtain the solution in the form:

( )

^t

y t = +t e

Examples IV

Let us consider the higher order differential equation:

( ) ( )¹

... ⁿ ⁿ .... ⁿ , 0

y+ + +y′ y′′ y′′′ +y +y ⁺ + =x n> (13)

( )ⁱ

( )

⁰ ⁰ ^, 0,1, 2,....

y = i = (14)

We can write equation (13) in the form:

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( )

0

i n

i

y x

∞

=

∑

= ⁽¹⁵⁾

Now Applying ELzaki transform to equation (15) we have:

( )¹

( )

²

1

0 1

0 !

i i

n

i i j

i j

y y

v v n v

∞ −

+

= = − −

⎡ ⎤

− =

⎢ ⎥

⎣ ⎦

∑ ∑

^Or ^{( )}¹

( )

¹ ²

0 0 1

1 0

i !

i i

n i j

i i j

y y n v

v v

∞ ∞ −

− − +

= − = = =

∑ ∑ ∑

Now applied the initial conditions (14) in the last equation, we have.

! 2

1 v n

y n v

v

⎡ ⎤= +

⎢ − ⎥

⎣ ⎦ And ^y⁼

( )

^{n v}^! ⁿ⁺²⁻

( )

^{n v}^! ⁿ⁺¹

Take the inverses of ELzaki transform we find the solution in the form:

( )

ⁿ ¹

[ ]

y x =x ⁻ x n−

Conclusion

In this paper, ELzaki transform method for the solution of Higher order Differential Equations is successfully expanded. In observed that the ELzaki transform method is robust and is applicable to various types of differential equations.

References

[1] Tarig M. Elzaki,The New Integral Transform “Elzaki Transform” Global Journal of Pure and Applied Mathematics, ISSN 0973-1768,Number 1(2011), pp. 57-64.

[2] Tarig M. Elzaki & Salih M. Elzaki, Application of New Transform “Elzaki Transform” to Partial Differential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768,Number 1(2011), pp. 65-70.

[3] Tarig M. Elzaki & Salih M. Elzaki,On the Connections Between Laplace and Elzaki transforms, Advances in Theoretical and Applied Mathematics, ISSN 0973-4554 Volume 6, Number 1(2011),pp. 1-11.

[4] Tarig M. Elzaki & Salih M. Elzaki,On the Elzaki Transform and Ordinary Differential Equation With Variable Coefficients, Advances in Theoretical and Applied Mathematics. ISSN 0973-4554 Volume 6, Number 1(2011), pp. 13- 18.

[5] Tarig M. Elzaki, Adem Kilicman, Hassan Eltayeb. On Existence and Uniqueness of Generalized Solutions for a Mixed-Type Differential Equation, Journal of Mathematics Research, Vol. 2, No. 4 (2010) pp. 88-92.

[6] Tarig M. Elzaki, Existence and Uniqueness of Solutions for Composite Type Equation, Journal of Science and Technology, (2009). pp. 214-219.

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[7] Lokenath Debnath and D. Bhatta. Integral Transform and Their Application Second Edition, Chapman & Hall /CRC (2006).

[8] A.Kilicman and H.E.Gadain. An application of double Laplace transform and Sumudu transform, Lobachevskii J. Math.30 (3) (2009), pp.214-223.

[9] J. Zhang, A Sumudu based algorithm for solving differential equations, Comp.

Sci. J. Moldova 15(3) (2007), pp – 303-313.

[10] Hassan Eltayeb and Adem kilicman, A Note on the Sumudu Transforms and Differential Equations, Applied Mathematical Sciences, VOL, 4,2010, no.22,1089-1098

[11] Kilicman A. & H. ELtayeb. A Note on Integral Transform and Partial Differential Equation, Applied Mathematical Sciences, 4(3) (2010), PP.109- 118.

[12] Hassan ELtayeh and Adem kilicman, on Some Applications of a New Integral Transform, Int. Journal of Math. Analysis, Vol, 4, 2010, no.3, 123-132.

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