On The Relationship Between Aboodh Transform and New Integral Transform " ZZ Transform"

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1 On The Relationhip Between Aboodh Tranform and New Integral Tranform " ZZ Tranform" 1,2 Mohand M. Abdelrahim Mahgob and 1,3 Abdelbagy A. Alhikh 1Mathematic Department Faclty of Science and Art-Almikwah -Albaha Unierity- Sadi Arabia 2Mathematic Department Faclty of Science-Omdrman Ilamic Unierity, Khartom -Sdan 3Mathematic Department Faclty of Edcation-Alzaeim Alazhari Unierity, Khartom -Sdan Abtract: In thi paper we dice ome relationhip between Aboodh tranform and the new integral tranform called ZZ tranform, we ole firt and econd order ordinary differential eqation with contant and non-contant coefficient, ing both tranform,and howing ZZ tranform i cloely connected with Aboodh tranform. Keyword: Aboodh Tranform, ZZ Tranform, Differential eqation 1.Introdction In the literatre there are nmero integral tranform [10] and widely ed in phyic, atronomy a well a in engineering. In order to ole the differential eqation, the integral tranform were exteniely ed and th there are eeral work on the theory and application of integral tranform ch a the Laplace, Forier, Mellin, and Hankel, Forier Tranform, Smd Tranform, Elzaki Tranform and Aboodh Tranform. Aboodh Tranform [1,2] wa introdced by Khalid Aboodh in 2013, to facilitate the proce of oling ordinary and partial differential eqation in the time domain. Thi tranformation ha deeper connection with the Laplace and Elzaki Tranform. [3,4, 5]. New integral tranform, named a ZZ Tranformation [6-9] intrpdce by Zain Ul Abadin Zafar [2016 ], ZZ tranform wa cceflly applied to integral eqation, ordinary differential eqation. the main objectie i to introdce a comparatie tdy to ole differential eqation by ing Aboodh tranform and ZZ tranform. The plane of the paper i a follow: In ection 2, we introdce the baic idea of Aboodh tranform, and ZZ Tranform, Application in 3 and conclion in 4, repectiely. 2.Definition and tandard Relt : 2.1. Aboodh tranform : Definition : A new tranform called the Aboodh tranform defined for fnction of exponential order we conider fnction in the et A, defined by: A = {f(t): M, k 1, k 2 > 0, f(t) < Me t For a gien fnction in the et M mt be finite nmber,k 1, k 2 may be finite or infinite. Aboodh tranform which i defined by the integral eqation A[f(t)] = K() = 1 f(t)e t dt t 0, k 1 k 2 (1) 0 71

2 Aboodh tranform of ome fnction : A(1) = 1, 2 A(tn ) = n!, n+2 A(eat ) = 1 2 a a 1 A(in(at)) = ( 2 +a 2, A(co(at)) = ) ( 2 +a 2 ).Aboodh tranform of deriatie : A[f (t)] = K() f(0), A[f (t)] = 2 K() f (0) n 1 f (k) (0) 2 n+k f(0) A[f (n) (t)] = n K() k=0. ` A{tf(t)} = d k() 1 k() d A{tf (t)} = d d f(0) [k() ] 1 f(0) [k() ], A{tf (t)} = d d [2 k() f (0) f(0)] 1 [2 k() f (0) f(0)] 2.2 The ZZ Tranform: Definition: Let f(t) be a fnction defined for all t 0. The ZZ tranform of f(t) i the fnction Z(, ) defined by Z(, ) = H{f(t)} = 0 f(t)e t dt proided the integral on the right ide exit. The niqe fnction f(t) in (2) i called the inere tranform of Z(, ) i indicated by f(t) = H 1 {Z(, )} Eqation (2) can be written a H{f(t)} = 0 f(t)e t dt ZZ tranform of ome fnction : H{1} = 1, H{t n } = n! n n, H{eat } = a a 2 E(in(at)) =, E(co(at)) = 2 +a a 2 2. ZZ tranform of deriatie : 1) let H{f(t)} = Z(, ) then H{f (n) (t)} = n n 1 n Z(, ) k=0 f (k) (0) d 2) (i) H{tf(t)} = 2 (Z(, )) + Z(, ) d (ii)) H{tf (t)} = 2 d d ( Z(, )) + Z(, ) n k n k (iii) H{tf (t)} = d d (Z(, )) Z(, ) + f(0) 3 Application : Example 3.1 conider the firt order differential eqation dy dx + y = 0 (4) (2) (3) 72

3 With the initial condition;, y(0) = 1 (5) Soltion: 1: Applying the Aboodh tranform of both ide of Eq. (4), A { dy } + A{y} = A{0 } (6) dx Uing the differential property of Aboodh tranform Eq.(6) can be written a: K() y(0) Uing initial condition (5), Eq. (7) can be written a: k(y) = 1 (1+) + K() = 0 (7) The inere Aboodh tranform of thi eqation i imply obtained a y(x) = e x (9) Where K()I the Aboodh tranform of the fnction y (x ) 2: Applying the ZZ tranform of both ide of Eq. (4), H { dy } + H{y} = H{0 }, o (10) dx Uing the differential property of ZZ tranform Eq.(6) can be written a: z(, ) y(0) + z(, ) = 0 (11) Uing initial condition (5), Eq. (11) can be written a z(, ) [ + 1] =, z(, ) = (12) + The inere ZZ tranform of thi eqation i imply obtained a y(x) = e x (13) Where H i the ZZ tranform of the fnction y (x ) Example 3.2 Let conider the econd-order differential eqation y + y = 0, (14) With the initial condition;, y(0) = y (0) = 1 (15) Soltion: 1: Applying the Aboodh tranform of both ide of Eq. (14), A{ y } + A{y} = A{0} (16) Uing the differential property of Aboodh tranform Eq.(16) can be written a 2 K() 1 + K() 1 = 0 (17) Uing initial condition (15), Eq. (17) can be written a k() = (18) 2 +1 ( 2 +1) The inere Aboodh tranform of thi eqation i imply obtained a y(x) = co x + in x (19) 2: Applying the ZZ tranform of both ide of Eq. (14), H{ y } + H{y} = H{0}, o (20) Uing the differential property of ZZ tranform Eq.(20) can be written a: (8) 73

4 2 2 Z(, ) y(0) 2 2 Uing initial condition (15), Eq. (21) can be written a 2 2 Z(, ) Z(, ) = 0, therefore Z(, ) = y`(0) + Z(, ) = 0 (21) (22) The inere ZZ tranform of thi eqation i imply obtained a y(x) = co x + in x (23) Example 3.3 Conider the econd-order differential eqation y 3y + 2y = 0 (24) With the initial condition, y(0) = 1, y (0) = 4 (25) Soltion: 1: Applying the Aboodh tranform of both ide of Eq. (24), A{ y } A{3y } + A{2y} = A{0} (26) Uing the differential property of Aboodh tranform Eq.(26) can be written a 2 K() 4 1 3k() k() = 0 k() = ( 1) ( 2) Then take the inere of Aboodh tranform we get y(t) = -2e t + 3e 2t. (28) 3: Applying the ZZ tranform of both ide of Eq. (24),: H{ y } H{3y } + H{2y} = H{0}, o (29) Uing the differential property of ZZ tranform Eq.(29) can be written a: 2 2 Z(, ) f(0) 2 2 f (0) 3 ( z(, ) Uing initial condition (25), Eq. (30) can be written a 2 2 Z(, ) Z(, ) 3 Z(, ) = +2 ( )( 2) + 2Z(, ) = 0 f(0)) + 2Z(, ) = 0 (30) Sole eqation (31) Then take the inere of ZZ tranform we get y(t) = -2e t + 3e 2t (32) Example 3.4 Conider the initial ale problem (31) y (t) + 2y (t) + 5y(t) = e t in t (33) With the initial condition y(0) = 0, y (0) = 0 (34) Soltion: 1: Applying the Aboodh tranform of both ide of Eq. (33), A{y (t)} + A{2y (t)} + A{5y(t)} = A{e t in t} (35) (27) 74

5 Uing the differential property of Aboodh tranform Eq.(35) can be written a 2 k() f (0) f (0) + 2k() 2 f(0) + 5k() = 1 [(1+) 2 +1] 2 1 k() + 2k() + 5k() = [(1+) 2 +1] k() = 1 3 ( 1 ((1+) 2 +1) 1 ((1+) 2 +4) ) (37) Now applying the inere Aboodh tranform, we get y(t) = 1 3 e t in t e t in 2t y(t) = 1 3 e t (in t + in 2t). (38) 2: Applying the ZZ tranform to both ide of (33) we hae H{ y (t)} + 2H{y (t)} + 5H{y(t)} = H{e t in t} (39) Uing the differential property of ZZ tranform Eq.(29) can be written a: Z(, ) y(0) 2 y (0) + 2 (Z(, ) ( ) Z(, ) = + Z(, ) = 1 3 ( (( +1)2 +1) ( +1)2 +1 (( +1)2 +4) Now applying the inere ZZ tranform, we get y(t) = 1 3 e t in t e t in 2t Example 3.5 Conider the initial ale problem ) + y(0)) + 5Z(, ) = (( +1)2 +4) ( +1)2 +1 (40) (36) (41) y(t) = 1 3 e t (in t + in 2t) (42) ty (t) ty (t) + y(t) = 2 (43) With the initial condition Solton: y(0) = 2, y (0) = 1 (44) 1. Applying the Aboodh tranform to both ide of (43) we hae A{ty (t)} A{ty (t)} + A{y(t)} = A{2} (45) Uing the differential property of Aboodh tranform Eq.(45) can be written a d d [2 k() f (0) f(0)] 1 [2 k() f (0) f(0)] + d d [k() f (0) ] + 1 [k() f (0) ] + K() = 2 A[y(t)] = K() ( 2 )k () + 3(1 )k() = 2 Or (46) K () + 3 K() = 2 3 (1 ) 2 2 (1 ). (47) 75

6 Eqation (47) i a linear differential eqation, which ha oltion in the form k() = c + 2 [ c = contant] 3 2 Now applying the inere Aboodh i tranform, we get y(t) = ct + 2 (48) 2: Applying the ZZ tranform to both ide of (43) we hae H{t y (t)} H{ty (t)} + H{y(t)} = H{2} (49) Uing the differential property of ZZ tranform Eq.(49) can be written a d Z(, ) Z(, ) + f(0) + d d d Z(, ) + Z(, ) + Z(, ) = 2 Z (, ) 2 Z(, ) + 2 ( Z (, )) 2 Z(, ) = 2 Z (, ) 1 Z(, ) = 2 (50) Eqation (50) i a linear differential eqation, which ha oltion in the form Z(, ) = 2 + c Now applying the inere ZZ tranform, we get 4 Conclion y(t) = 2 + ct (51) The main goal of thi paper i to condct a comparatie tdy between Aboodh Tranform and new integral ZZ Tranform. The Tow method are powerfl and efficient. Aboodh tranform and ZZ tranform i a conenient tool for oling differential eqation in the time domain. Reference [1] K. S. Aboodh, The New Integral Tranform Aboodh Tranform Global Jornal of pre and Applied Mathematic, 9(1), 35-43(2013). [2] K. S. Aboodh, Application of New Tranform Aboodh tranform to Partial Differential Eqation, Global Jornal of pre and Applied Math, 10(2), (2014). [3]Tarig M. Elzaki (2011), The New Integral Tranform ELzaki Tranform, Global Jornal of Pre and Applied Mathematic, Vol.7, No.1, pp [4] Tarig M. Elzaki and Salih M. Elzaki (2011), Application of New Tranform ELzaki Tranform to Partial Differential Eqation, Global Jornal of Pre and Applied Mathematic, Vol.7, No.1, pp [5] Abdelbagy A. Alhikh and Mohand M. Abdelrahim Mahgob," On The Relationhip Between Elzaki Tranform And New Integral Tranform "ZZ Tranform", International Jornal of Deelopment Reearch Vol. 06, Ie, 08, pp , Agt, [6] Zain Ul Abadin Zafar, ZZ Tranform method, IJAEGT, 4(1), , Jan (2016).. [7] Zain Ul Abadin Zafar, M.O. Ahmad, A. Peraiz, Nazir Ahmad: ZZ Forth Order Compact BVM for the Eqation of Lateral Heat Lo, Pak. J. Engg. & Appl. Sci. Vol. 11, Jly., 2012 (p ) [8] Zafar, H. Khan, Waqar. A. Khan, N-Tranform- Propertie and Application, NUST Jor. Of Engg. Science, 1(1): , [9] Zain Ul Abadin Zafar, et al., Soltion of Brger Eqation with the help of Laplace Decompoition method. Pak. J. Engg. & Appl. Sci. 12: 39-42, [10] J.W. Mile, Integral Tranform in Applied Mathematic, Cambridge: Cambridge Unierity Pre,