Iteratioal Joural of Computatioal Sciece ad Mathematics ISSN 0974-3189 Volume 3, Number 1 (2011), pp 121-125 Iteratioal Research Publicatio House http://wwwirphousecom Solutio of Differetial Equatio from the Trasform Techique S Vishwa Prasad Rao 1, PS Rama Chadra Rao 2 ad C Prabhaara Rao 3 1 Departmet of Mathematics, New Sciece Post-Graduate College, Huter Road, Haamoda, Waragal, Adhra Pradesh, Idia E-mail: vishwa72@yahoocom 2 Departmet of Mathematics, Kaatiya Istitute of Techology ad Sciece, Haamoda, Waragal, Adhra Pradesh, Idia Email:patibada20@yahoocoi 3 Departmet of Mathematics, Vasavi Egieerig College, Hyderabad, Adhra Pradesh, Idia Abstract I this article differetial trasform method is cosider to solve secod order differetial equatios The aalytical ad umerical results of the equatios have bee obtaied i terms of coverget series with easily comparable Three examples are give to illustrate the efficiecy of the preset method Differetial trasform techique may be cosidered as alterative ad efficiet for fidig the approximate solutios of the boudary values problems Keywords: Differetial trasform method, Secod order differetial equatio, Laplace trasform Mathematical subject classificatio (2010): 65Mxx, 65M99 Itroductio The Differetial trasform method has bee successfully used by Zhou[6] to solve a liear ad oliear iitial value problems i electric circuit aalysis This method costructs a aalytical solutio i the form of a polyomial The differetial trasform method is a alterative method for fidig the aalytic solutio of the differetial equatios I this paper, we apply the differetial trasform method which is based o Taylor expasio to costruct aalytical approximate solutios of the iitial value problem
122 S Vishwa Prasad Rao et al This paper is orgaized as follows: I Sectio 2, the differetial trasformatio method is described I Sectio 3, the method is implemeted to three examples, ad coclusio is give i Sectio 4 Differetial trasformatio method Differetial trasformatio of fuctio y( x ) is defied as follows 1 d y( x) Y ( ) =! dx x= 0 (1) I (1), y( x ) is the origial fuctio ad Y ( ) is the trasformed fuctio Differetial iverse trasform of Yis ( ) defied as follows y( x) = x Y( ) (2) = 0 I fact, from (1) ad (2), we obtai x d y( x) yx ( ) = = 0! dx (3) Eq (3) implies that the cocept of differetial trasformatio is derived from the Taylor series expasio From the defiitios (1) ad (2), it is easy to obtai the followig mathematical operatios: 1 If f ( x) = g( x) ± h( x) the F( ) = G( ) ± H( ) 2 If f ( x) = cg( x) the F( ) = cg( ), where c is costat x 1 3 If f ( x) = e the F ( ) =! d g( x) ( + )! 4 If f( x) = the F( ) = G( + ) dx! 5 If f ( x) = g( x) h( x) the F( ) = G( ) F( r) 6 If f ( x) = x the F( ) = δ ( ), where δ the Kroecer delta r= 0 Numerical Examples To demostrate the method itroduced i this study, three examples are solved here Example 1 11 1 y + 5y + 6y= 5e t (4)
Solutio of Differetial Equatio 123 y (0) = 2 ad y 1 (0) = 1 (5) Taig the differetial trasform both sides of (4), we obtai ( 2)( 1) ( 2) ( 1) ( 1) 6 ( ) 5 1 + + Y + + + Y + + Y = (6)! where Y ( ) is the differetial trasform From the iitial coditio give by Eq(5) we have Y (0) = 2 ad Y (1) = 1 (7) Taig Eq(7) i Eq(6) ad by recursive method, we have 59 09 247 Y (2) = 6Y (3) = Y (4) = Y (5) = 6 12 40 59 109 247 yx xy x x x x x 2 3 4 5 ( ) = ( ) = 2 + 6 + + + = 0 6 12 40 5 x 16 2x 15 3x ( ) = +, y x e e e 12 3 4 which is exact solutio of the equatio (4) Now applyig laplace trasform to the equatio (4), we get the same solutio Example 2 11 1 y 2y + 2y = 0 (8) y (0) = 1 ad y 1 (0) = 1 (9) Taig the differetial trasform both sides of (8), we obtai ( + 2)( + 1) Y( + 2) 2( + 1) Y( + 1) + 2 Y( ) = 0 (10) where Y ( ) is the differetial trasform From the iitial coditio give by Eq(9) we have Y (0) = 1 ad Y (1) = 1 (11) Taig Eq(11) i Eq(10) ad by recursive method, we have Y (2) = 0 Y (3) = Y (4) = Y (5) = 3 6 30
124 S Vishwa Prasad Rao et al 1 3 1 4 1 5 yx ( ) = xy ( ) = 1 + x x x x+ 2 6 30 = 0 x y( x) = e cosx, which is exact solutio of the equatio (8) Now applyig laplace trasform to the equatio (8) we get the same solutio Example 3 11 1 2 y + y = x + 2x (12) y (0) = 4 ad y 1 (1) = 2 (13) Taig the differetial trasform both sides of (12), we obtai ( + 2)( + 1) Y( + 2) + ( + 1) Y( + 1) = δ ( 2) + 2 δ ( 1) (14) where Y ( ) is the differetial trasform From the iitial coditio give by Eq(13) we have Y (0) = 4 ad Y (1) = 2 (15) Taig Eq(15) i Eq(14) ad by recursive method, we have 1 Y (2) = 1 Y (3) = 0 Y (4) = Y (5) = 12 60 1 1 yx xy x x x x 2 4 5 ( ) = ( ) = 4 2 + + + = 0 12 60 x 1 3 ( ) = 2+ 2 +, which is exact solutio of the equatio (12) y x e x 3 Now applyig laplace trasform to the equatio (12), we get the same solutio Coclusio I this paper, we have show that the differetial trasform method ca be used successfully for fidig the solutio of secod order differetial equatios ad we compare the solutio with other alterate method It may be cocluded that this techique is very powerful ad efficiet i fidig solutios
Solutio of Differetial Equatio 125 Refereces [1] Abdel-Halim Hassa I H, Vedat Suat Ertr, Applyig differetial trasformatio method to the oe-dimesioal plaar bratu problem, It J Cotemp math Scieces, 2(30) (2007) 1493-1504 [2] Fatma Ayaz Applicatios of differetial trasform method to differetialalgebraic equatios Applied Mathematics ad Computatio, 152(2004): 649-657 [3] Jag, MJ, Che, CL, Liu, YC, O solvig the iitial value problem usig the differetial trasformatio method, Appl Math Comput, 115 (2000), 145-160 [4] Rama Chadra Rao, P S Special Multistep Methods Based o Numerical Differetiatio for Solvig the Iitial Value Problem, Applied Mathematics ad Computatio Elsvier, 181, p500-510, 2006 [5] Vishwa Prasad Rao S ad Rama Chadra Rao P S Trasform Techique For The Solutio Of A Class Of Differetial Equatio, Iteratioal Joural Of Applied Mathematical Aalysis ad Applicatios Vol 3,No2,pp 161-166 [6] Zhou J K Differetial trasformatio ad its applicatio for electrical circuits Huarjug Uiversity Press, Wu-uhah, Chia,(i Chiese) 1986