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© 2015, IRJET ISO 9001:2008 Certified Journal Page 128

Differential Transform Technique for Second Order Differential Equations: A

Comparative Approach.

Adio, A.K Ph.D

Lecturer, Department of Basic Sciences

Babcock

University, Illishan-Remo, Ogun State, Nigeria.

---***---Abstract: In this paper, one dimensional Differential Transform Method (DTM) is applied on a class of second order differential equations leading to derivation of closed form solutions. Comparison is made between numerical solutions obtained using DTM and exact solutions derivable from application of Laplace transform method. This comparison buttresses the efficiency of the discussed method. Five examples are presented to illustrate the efficiency and simplicity of the method.

Keywords:

Differential Transform Method, Laplace Transform Method, Ordinary differential equations, Taylor Series, Recurrence relation.

1. INTRODUCTION

The Differential transform method (DTM) is a semi analytical numerical method that uses Taylor series for the solution of differential equations.

It is an alternative procedure for obtaining the Taylor series solution of the given differential equation and is promising for various other types of differential equations.

By application of the method, it is possible to obtain highly accurate results or exact solutions for differential equations. The concept of the differential transform method was first proposed by Zhou[8], who solved linear and nonlinear initial value problems in electric circuit analysis. In recent years, Abdel-HalimHassan[1] used differential transform method to solve higher order initial value problems.

Ayaz[2] used DTM to find the series solution of system of differential equations.

Naharil and Avinash[5] applied DTM to system of linear differential equations. Hassan[3] compared series solution obtained by DTM with decomposition method for linear and

nonlinear initial value problems and prove that DTM is a reliable tool to find the numerical solutions.

2. THE DIFFERENTIAL TRANSFORM METHODS

The transformation of the

k

th derivative of a function with one variable is

)

)

(

(

!

1

)

(

k

k

dx

x

u

d

k

k

U

at

x

x

0(1)

Where u(x)is the original function and U(k)is the transformed function and the differential inverse transformation u(k)is defined by

 

0

0)

)( ( )

(

k

k

x

x k u x

u

) 2 (

When

x

0

0

, the function u(x)defined in (2) is expressed as

 0

) ( )

(

k

k

x k u x

u (3)

Equation (3) shows the similarity between one dimensional differential transform and one dimensional Taylor series expansion.

The following fundamental theorems on differential transform method are handy:

Theorem 1:

(2)

International Research Journal of Engineering and Technology (IRJET)

e-ISSN: 2395 -0056 Volume: 02 Issue: 07 | Oct-2015 www.irjet.net p-ISSN: 2395-0072

© 2015, IRJET ISO 9001:2008 Certified Journal Page 129

If

u

(

t

)

t

n, then U(k)

(kn)where

0

1

)

(

k

n

, if

n k n k   Theorem 3:

If u(t) t

e , then

! 1 ) ( k k UTheorem 4:

If u(t)g(t)h(t), then

   k l l k H l G k U 0 ) ( ) ( ) ( Theorem 5:

If

x

(

t

)

x

1

(

t

)

x

2

(

t

)

, then

(

)

(

1

)

2

(

1

)

0 1 1

k

k

X

k

X

k

X

k k

Theorem 6:

If m

m

dt

t

x

d

t

x

(

)

1

(

)

, then ( )

! )! (

)

( X1 k m

k m k k

X   

If x(t)sin(

t

), then

)

2

sin(

!

)

(

k

k

k

X

k

where

and

are constants.

Theorem 9:

If x(t)cos(

t

), then

)

2

cos(

!

)

(

k

k

k

X

k

where

and

are constants.

2.1 Numerical Applications Example 1:

Consider the second order differential equation:

t

e

x

dt

dx

dt

x

d

3 2 2

8

6

(4)

With initial conditions x(0)0;x(0)2.

Applying differential transform method with initial conditions

2 ) 1 ( ; 0 ) 0

(  u

u to equation(4),using the above mentioned

theorem, we have:

)}

(

8

)

1

(

)

1

(

6

!

3

{

)

2

)(

1

(

1

)

2

(

k

u

k

u

k

k

k

k

k

u

k

(5)

Put

k

0

, we have .

2 13 ) 2 (  u , 1  k . 6 65 ) 3 (  u , 2  k . 44 295 ) 4 (  u , 3  k . 120 1277 ) 5 ( 

u and so on .

Thus the closed form of the solution when

n

5

(number of terms), using equation (3) can be written easily as

. 120 1277 24 295 6 65 2 13 2 ) ( )

( 2 3 4 5

3 0 t t t t t t k u t x k

k     

) 6 (

Using the Laplace transform method, the exact solution of example 1 is

. 2 1 2 3 )

(t e4t e3t e2t

x   

Example 2.

Consider the second order differential equation:

t

x

dt

x

d

2

cos

9

2 2

(7)

with initial conditionsx(0)1;x(0)3.

Applying differential transform method with initial conditions

3 ) 1 ( ; 1 ) 0

(  u

u to equation(7)using the above mentioned

theorem, we obtain

)}

(

9

)

2

cos(

!

2

{

)

2

)(

1

(

1

)

2

(

k

u

k

k

k

k

k

u

k

(8)

(3)

© 2015, IRJET ISO 9001:2008 Certified Journal Page 130 , 3  k 40 81 ) 5 ( 

u and so on

Thus the closed form of the solution, when

n

5

(number of terms), using equation (3) can be written easily as

5 4 3 2 3 0 40 81 6 17 2 9 4 3 1 ) ( )

(t u k t t t t t t

x

k

k      

) 9 (

Using the Laplace transform method, the exact solution of example 2is

 ) (t

x cos2.

5 1 3 sin 3 cos 5 4 t t

t 

Example 3:

Consider the second order differential equation

t

x

dt

dx

dt

x

d

4

sin

32

32

8

2 2

(10)

With initial conditions x(0)0;x(0)0.

Applying differential transform method with initial conditions

0 ) 1 ( ) 0

( u

u to equation(10), using the above mentioned

theorem, we obtain

)] ( 32 ) 1 ( ) 1 ( 8 )} 2 sin( ! 4 { 32 [ ) 2 )( 1 ( 1 ) 2

( k k u k u k

k k k k u k          ) 11 (

Put k 0, we have u(2)0. , 1  k . 3 64 ) 3 (  u , 2  k . 3 128 ) 4 (  u , 3  k . 15 206 ) 5 ( 

u and so on

Thus the closed form of the solution when

n

5

(number of terms), using equation (3)can be written easily as

5 4 3 3 0 15 206 3 128 3 64 ) ( )

(t u k t t t t

x

k

k   

) 12 (

Using the Laplace transform method, the exact solutions of example (3)is

}. 4 sin ) 1 ( 4 cos ) 1 ( 2 { 5 2 )

(t e 4 t e 4 t

x   t    t

Example 4:

Consider the second order differential equation

)

sin

(

2

2

2 2

t

t

x

dt

dx

dt

x

d

(13)

With initial conditions x(0)6;x(0)5.

Applying differential transform method with initial conditions

5 ) 1 ( , 6 ) 0

(  u

u to equation(13), using the above mentioned

theorem, we obtain

)}

(

)

1

(

)

1

(

2

)]

2

sin(

!

1

)

1

(

[

2

{

)

2

)(

1

(

1

)

2

(

k

k

u

k

u

k

k

k

k

k

k

u

) 14 (

Put k 0, we have u(2)2. , 1  k . 6 7 ) 3 (  u , 2  k . 12 5 ) 4 (  u , 3  k . 120 11 ) 5 ( 

u and so on

Thus the closed form of the solution when

n

5

(number of terms), using equation (3)can be written easily as

. 120 11 12 5 6 7 2 5 6 ) ( )

( 2 3 4 5

3 0 t t t t t t k u t x k

k      

) 15 (

Using the Laplace transform method, the exact solutions of example (4)is

. cos 4 2 ) 1 2 ( )

(t e t t t

xt    

Example 5:

Consider the second order differential equation

)

5

sin

2

5

(cos

10

25

2 2

t

t

x

dt

x

d

(16)

(4)

International Research Journal of Engineering and Technology (IRJET)

e-ISSN: 2395 -0056 Volume: 02 Issue: 07 | Oct-2015 www.irjet.net p-ISSN: 2395-0072

© 2015, IRJET ISO 9001:2008 Certified Journal Page 131

Applying differential transform method with initial conditions

. 2 ) 1 ( , 1 ) 0

(  u

u to equation(10), using the above mentioned

theorem, we obtain

)}

(

25

)])

2

sin(

!

5

[

2

)

2

cos(

!

5

(

10

{

)

2

)(

1

(

1

)

2

(

k

u

k

k

k

k

k

k

k

u

k k

) 17 (

Put k 0, we have . 2 15 ) 2

( 

u

, 1 

k .

2 50 )

3

( 

u

, 2 

k .

24 125 ) 4

( 

u

, 3 

k .

12 625 ) 5

( 

u and so on

Thus the closed form of the solution when

n

5

(number of terms), using equation (3)can be written easily as

. 12 625 24

125 2

50 2

15 2 1 ) ( )

( 2 3 4 5

3

0

t t

t t

t t

k u t

x

k

k      

 ) 18 (

Using the Laplace transform method, the exact solutions of example (5)is

. 5 sin 5

cos ) 1 2 ( )

(t t t t t

x   

3. CONCLUSIONS

In this paper, approximate solution of second order differential equation was obtained making use of differential transform method. The method was used in a direct way without linearization, perturbation or restrictive assumptions.

A comparison was made between the method and the Laplace transform method which produced the exact solution. The differential transform method unlike most numerical techniques provides a closed form solution which confirms the efficiency and numerical suitability of the method for wide classes of linear differential equations.

REFERENCES

[1] Abdel-Halim, H., Differential transform technique for solving higher order initial value problems. Applied Mathematics and Computation, 154,299-311(2004).[2] Ayaz, F., Solutions of the system of differential equations by differential transform method. AppliedMathematics and Computation, 147, 547-567(2004).

[3] Hassan H.A., Comparison of differential transformation technique with Adomian decomposition method for linear and nonlinear initial value problems, Chaos Solutions Fractals, 36(1),

53-65 (2008)

[4] Kuo,B.L., Applications of the differential transform method to the solutions of the free problem. Applied Mathematics and Computation 165, 63-79(2005).

[5] Narhari,P., Avinashi,K., The Numerical solution of Differential Transform Method and the Laplace Transform Method for second order differential equations. International Journal of Mathematics and Computer Research, vol3(2)871-875

(2015).

[6] Narhari, P., Avinashi,K., Differential transform method for system of linear differential equations. Research journal of Mathematical and Statistical Sciences, vol2(3)4-6 (2014).

[7] Stroud, K.A., Further Engineering Mathematics. Programmes and Problems. Palgrave, New York 1996.

[8] Zhou X., Differential Transformation and its applications for Electrical Circuits, Huazhong University Press, Wuhan, China,

(5)

2395-0072

References

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