8_xxx_erturk.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 2, No. 3, 2009, (426-447) ISSN 1307-5543 – www.ejpam.com Solutions of Different Types of the linear and Non- linear Higher-Order Boundary Value Problems by Dif- ferential Transformation Method I. H. Abdel-Halim Hassan1, Vedat Suat Ertürk2∗ 1 Department of Mathematics, Faculty of Science, Zagazig University, Zagazig, Egypt 2 Department of Mathematics, Faculty of Arts and Sciences, Ondokuz Mayıs University, 55139, Samsun, Turkey Abstract. In [12], a numerical comparison between the differential transform method and Adomian decomposition method for solving fourth-order boundary value problems was pre- sented. In this article, we use the differential transformation method (DTM) to solve the linear and non-linear higher-order boundary value problems (HOBVPs). The method proved to be very successful and powerful in computing such elements. The specific problems chosen for this purpose is that of the different types of higher order (e.g. fifth, sixth, ninth, tenth and twelfth ) boundary value problems. The differential transformation (DT) solutions are com- pared with the theoretical solution. It is shown that the solutions obtained from the technique have a very high degree of accuracy. 2000 Mathematics Subject Classifications: 35C10, 74S30, 65L10, 34B05, 34B15. ∗Corresponding author. Email addresses: ismhalim�hotmail. om (I. Abdel-Halim), vserturk�omu.edu.tr (V. Ertürk) http://www.ejpam.com 426 c© 2009 EJPAM All rights reserved. I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 427 Key Words and Phrases: Higher-order boundary value problem, Differential transformation, Taylor’s series expansion. 1. Introduction Recently a great deal of interest has been focused on the applications of the (DTM) to solve various scientific models, see Refs.[2],[3],[4],[5-6],[7-8],[9],[13-15], [16- 17], [18] and [22]. The (DTM) has also been applied to solve linear and non-linear higher-order initial value problems, for example [15] and [16]. In this paper, we are interested in the differential transformation method (DTM) to solve linear and non-linear higher-order boundary value problems (HOBVPs). Many numerical techniques, such as finite difference method [10] and decomposition method [19-21] have been implemented to solve (HOBVPs) numerically. The differential transformation method (DTM) is a numerical method for solving boundary value problem [17]. The concept of differential transformation was first proposed by Zhou [23] in 1986, and it was applied to solve linear and non-linear initial value problems in electric circuit analysis. The method can be used to evaluates the approximat- ing solution by the finite Taylor series and by an iteration procedure described by the transformed equations obtained from the original equation using the operations of differential transformation. The basic definitions of the differential transforma- tion are introduced in Section 2. The mathematical background of the higher-order boundary value problems is described in Section 3. Analysis of higher-order boundary value problems is illustrated by differential transformation method (DTM) in Section 4. Numerical examples are used to illustrate the effectiveness of the proposed method in Section 5. I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 428 2. The Differential Transformation Method (DTM) An kth- order differential transformation (DT) of a function y(x) = f (x) is de- fined about a point x = x0 as: Y (k) = 1 k! h dk y(x) d xk i x=x0 , (2.1) where k belongs to the set of non-negative integers, denoted as the K-domain. The function y(x) may be expressed in terms of the differential transforms (DT), Y (k) as: y(x) = ∑∞ k=0 h (x−x0) k k! i Y (k). (2.2) Upon combining (2.1) and (2.2), we obtain y(x) = ∑∞ k=0(x − x0) k 1 k! h dk y(x) d xk i x=x0 , (2.3) which is actually the Taylor’s series for y(x) about x = x0. From the basic definition of the differential transforms (DT), one can obtain cer- tain laws of transformational operations, some of these, are listed in the following. (1): If z(x) = u(x)± v(x) then, Z(k) = U(k)± V (k). (2): If z(x) = αu(x) then, Z(k) = αU(k). Here α is a constant. (3): If z(x) = du(x) d x then, Z(k) = (k+ 1)U(k+ 1). (4): If z(x) = d2u(x) d x2 then, Z(k) = (k+ 1)(k+ 2)U(k+ 2). (5): If z(x) = dmu(x) d xm then, Z(k) = (k+ 1)(k+ 2) · · · (k+m)U(k+m). (6): If z(x) = u(x)v(x) then, Z(k) = ∑k l=0 V (l)U(k− l). (7): If z(x) = x m then, Z(k) = δ (k− n)where,δ (k− n) =    1 k = n 0 k 6= n. (8): If z(x) = exp(λx) then, Z(k) = λk k! . (9): If z(x) = (1+ x)m then, Z(k) = m(m−1)···(m−k+1) k! . (10): If z(x) =sin(ωx +α) then, Z(k) = ωk k! sin(πk 2 +α). (11): If z(x) = cos(ωx +α) then, Z(k) = ωk k! cos(πk 2 +α). I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 429 3. The Higher-order Boundary Value Problems (i) even-order boundary value problems Consider the special (2m)-order BVP of the form y(2m)(x) = f (x , y), 0< x < b, (3.1) with boundary conditions y(2 j)(0) = α2 j; j = 0, 1, 2, . . . , (m− 1), (3.2) y(2 j)(b) = β2 j; j = 0, 1, 2, . . . , (m− 1), (3.3) (ii) odd-order boundary value problems Consider the special (2m+ 1)-order BVP of the form y(2m+1)(x) = f (x , y), 0< x < b, (3.4) with boundary conditions y(2 j+1)(0) = γ2 j+1; j = 0, 1, 2, . . . , m, (3.5) y(2 j+1)(b) = γ2 j+1; j = 0, 1, 2, . . . , m, (3.6) It is interesting to point out that y(x) and f (x , y) are assumed real and as many times differentiable as required for x ∈ [0, b] and α2 j and β2 j , j = 0, 1, 2, . . . (m− 1) describe the even-order are real finite constants [11], moreover, the conditions α2 j , j = 0, 1, 2, . . . , (m− 1) describe the even-order derivatives at the boundary x = 0, while γ2 j+1 and γ2 j+1, j = 0, 1, 2, . . . , m describe the odd-order are real finite constants and the conditions γ2 j+1, j = 0, 1, 2, . . . m describe the odd-order derivatives at the boundary x = b. Theorems which list the conditions for existence and uniqueness solution of such problems are contained in a comprehensive survey in a book by Agarwal [1], though no numerical methods are contained therein for solving HOBVP’s of higher order. I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 430 4. Analysis of Higher-order Boundary Value Problems by Differential Transformation Let the differential transform of the deflection function y(x) be defined from Eq. (2.1) as: Y (k) = 1 k! h dk y(x) d xk i x=x0 , (4.1) where x0 = 0. Also the deflection function may be expressed in terms of Y (k) from Eq. (2.2) as: y(x) = ∑∞ k=0 h xk k! i Y (k). (4.2) Now, using the transformational operations which has been formed in Sec.2, one can obtain by taking the differential transform of Eq. (3.1) and (3.4) respectively and some simplification, the following recurrence equations as m= 0, 1, 2, . . .. Y (2m+ k) = ∑∞ k=0 h (2m)! (2m+k)! i Y (., .), (4.3) Y (2m+ k+ 1) = ∑∞ k=0 h (2m+1)! (2m+k+1)! i Y (., .), (4.4) where Y (., .) denotes the transformed function of linear or nonlinear function f (x , y). It may be noted that Eq. (4.2) is independent of the boundary conditions. The differential transforms of the boundary conditions at x = 0 are obtained from Eqs. (3.2) and (3.5) in the cases even-order (odd-order) boundary value problems respectively, with the definition (4.1) as: Y (2 j) = 1 (2 j)! α2 j, j = 0, 1, 2, . . . , (2m− 1), (4.5) Y (2 j+ 1) = 1 (2 j+1)! γ2 j+1, j = 0, 1, 2, . . . , 2m, (4.6) Substituting from (4.5) and (4.6) into (4.3) or (4.4) and using (4.2), yields for j = 0, 1, 2, . . . , (m− 1), y(x) = ∑∞ k=0 h 1 (2 j)! α2 j i Y (k)x k, (4.7) I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 431 and for j = 0, 1, 2, . . . , m, y(x) = ∑∞ k=0 h 1 (2 j+1)! γ2 j+1 i Y (k)x k. (4.8) Noting that y(2r+1)(0) = Ar , r = 0, 1, 2, . . . , (m−1) , and y(2r)(0) = Br , r = 0, 1, 2, . . . , m, are constants that will be approximated at the end point x = b. 5. Numerical Examples In this section, linear and nonlinear HOBVPs will be tested by using the differential transformation method, (see [12]). Example 5.1. We first consider the following linear fifth-order BVP , which is also solved by Adomian decomposition method (ADM) in the study of [21] y(υ)(x) = y(x)− 15ex − 10xex , 0< x < 1, (5.1) subject to the boundary conditions y(0) = 0, y ′(0) = 1, y ′′(0) = 0, y(1) = 0, y ′(1) =−e. (5.2) Applying the operations of (DT) to Eq. (5.1), the following recurrence relation is ob- tained: Y (k+ 5) = k! h Y (k)− 15 k! −10 � ∑k l=0 h δ(k−l−1) l! i�i (km+5)! (5.3) By using Eqs. (2.1) and (5.2) the following transformed B.C.’s at x = 0 can be obtained: Y (0) = 0, Y (1) = 1, Y (2) = 0, (5.4) where, according to Eq. (2.1), a = y′′′(0) 3! = Y (3) and b = y(iv)(0) 4! = Y (4). Utilizing the recurrence relation in Eq. (5.3) and the transformed B.C.’s in Eq. (5.4), Y (k) for k≥ 5 are easily obtained. I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 432 The constants a and b are evaluated from the B.C.’s given in Eq. (5.2) for x =1, by taking N = 13, to obtain the system: 148284463 148262400 a+ 3632669041 3632428800 b = − 4541061529 5448643200 , 239595841 79833600 a+ 5148284463 37065600 b =−e+ 15028547 129729600 . This in turn gives a = −0.3333315065 and b =−0.5000018268. For N = 24, these values are a = −0.333315040 and b = −0.5000018292. Then, by using the inverse transformation rule in Eq. (4.2), we get the following series solution is evaluated up to N = 24: Example 5.2. We next consider the following non-linear fifth-order BVP y(υ)(x) = e−x y2(x), 0 < x < 1, (5.5) subject to the boundary conditions y(0) = 0= y ′(0) = y ′′(0) = 1, y(1) = y ′(1) = e. (5.6) Applying the operations of (DT) to Eq. (5.6), the following recurrence relation is ob- tained: Y (k+ 5) = k! (k+5)! ∑k l=0 ∑l s=0 (−1)s s! Y (l − s)Y (k− l). (5.7) By using Eqs. (2.1) and (5.7) the following transformed B.C.’s at x = 0 can be obtained: Y (0) = 1, Y (1) = 1, Y (2) = 1 2 , (5.8) I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 433 where, according to Eq. (2.1), a1 = y′′′(0) 3! = Y (3) and a2 = y(iv)(0) 4! = Y (4). Utilizing the recurrence relation in Eq. (5.3) and the transformed B.C.’s in Eq. (5.4), Y (k) for k≥ 5 are easily obtained. The constants a1 and a2 are evaluated from the B.C.’s given in Eq. (5.7) for x = 1, by taking N = 12, to obtain the system: e− 80149541 31933440 = 98589 98560 a1 + 1996097 1996097 a2 + 1 133056 a2 1 + 1 47520 a1a2, e− 5848303 2851200 = 285343 95040 a1 + 665471 166320 a2 + 1 13860 a2 1 + 1 3960 a1a2. This in turn gives a1 = 0.666611767 and a2 = 0.0416703271. For N = 20, these values are a1 = 0.1666611892 and a2 = 0.0416703549. Then, by using the inverse transformation rule in Eq. (4.2), we get the following series solution is evaluated up to N = 20: Numerical results for linear and non-linear of fifth- order BVP’s, the differential trans- formation method (DTM) with comparison to the exact solution are given in Table 1. Example 5.3. Again, following the study of [19], we consider the following linear sixth- I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 434 order BVP , which is also solved by Adomian decomposition method (ADM) y(vi)(x) = y(x)− 6ex , 0< x < 1, (5.9) subject to the boundary conditions y(0) = 1, y ′′(0) = −1, y(iv)(0) =−3, y(1) = 0, y ′′ (1) =−2e, y (iv) (1) = −4e. (5.10) Applying the operations of DT to Eq. (5.11), the following recurrence relation is ob- tained: Y (k+ 6) = k![Y (k)− 6 k!] (k+6)! . (5.11) By using Eqs. (2.1) and (5.12) the following transformed B.C.’s at x = 0 can be ob- tained: Y (0) = 1, Y (2) =−1 2 , Y (4) = −1 8 , (5.12) where, according to Eq. (2.1), a = Y (1), b = y′′′(0) 3! = Y (3) and c = y(v)(0) 5! = Y (5). Utilizing the recurrence relation in Eq. (5.13) and the transformed B.C.’s in Eq. (5.14), Y (k) for k ≥ 6 are easily obtained. The constants a, b and c are evaluated from the B.C.’s given in Eq. (5.12) for x = 1, by taking N = 13, to obtain the system: 889750903 889574400 a+ 60481 60480 b+ 332641 332640 c =− 456655181 1245404160 , 332641 39916800 a+ 5041 840 b+ 60481 3024 c =−2e+ 110549143 39916800 , 60481 362880 a+ 1 20 b+ 5041 42 c =−4e+ 829261 120960 . This in turn gives a = −0.5388992288E−6 , b =−0.3333328229 and c = −0.03333330492. For N = 23, these values are a = −0.4667205875E−6 ,b = −0.3333329279 and c = −0.03333327191. Then, by using the inverse transformation rule in Eq. (4.2), we get the following series solution is evaluated up to N = 23: y(x) = 1− 0.4667205875E−6 − 6x − 0.5x2− 0.3333329279x3 − 0.125x4 I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 435 − 0.03333327191x5 − .6944444444E−2 − 2x6− .1190476283E−2 − 2x7 − .1736111111E−3 − 3x8− .2204584867E−4 − 4x9− .2480158730E−5 − x10 − .2505208992E−6 x11− .2296443269E−7 x12 − .1927085335E−8 x13 − .1491196928E−9 x14− .1070602736E−10 x15− .7169215999E−12 x16 − .4498329534E−13 x17− .2655265185E−14 x18− .1479714382E−15 x19 − .7809603484E−17 x20− .3914587736E−18 x21− .1868326192E−19 x22 − .8509971524E−21 x23. (5.13) Example 5.4. We next consider the following non-linear sixth-order BVP y(υi)(x) = ex y2(x), 0< x < 1, (5.14) subject to the boundary conditions y(0) = 1, y ′(0) = −1, y ′′(0) = 1, y(1) = e−1, y ′(1) = −e−1, y ′′(1) = e−1. (5.15) Applying the operations of DT to Eq. (5.16), the following recurrence relation is ob- tained: Y (k+ 6) = k! (k+6)! ∑k l=0 ∑l s=0 1 s! Y (l − s)Y (k− l). (5.16) By using Eqs. (2.1) and (5.17) the following transformed B.C.’s at x = 0 can be ob- tained: Y (0) = 1, Y (1) =−1, Y (2) = 1 2 , (5.17) where, according to Eq. (2.1), a1 = y′′′(0) 3! = Y (3), a2 = y(iv)(0) 4! = Y (4) and a3 = y(v)(0) 5! = Y (5). Utilizing the recurrence relation in Eq. (5.18) and the transformed B.C.’s in Eq. (5.19), Y (k) for k ≥ 6 are given. Y (6) = 1 720 , Y (7) = − 1 5040 , Y (8) = 1 40320 , Y (9) = 1 30240 a1 + 1 362880 , Y (10) = 1 75600 a2 − 1 3628800 , Y (11) = 1 166320 a3 + 1 39916800 , I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 436 ... and so on. The constants a1, a2 and a3 are evaluated from the B.C.’s given in Eq. (5.17) for x = 1, by taking N = 11, to obtain the system: e−1 − 2000701 3991680 = 30241 30240 a1 + 75601 75600 a2 + 166321 166320 a3, −e−1 − 321 44800 = 10081 3360 a1 + 30241 7560 a2 + 75601 15120 a3, e−1 − 46943 45630 = 2521 420 a1 + 10081 840 a2 + 30241 1512 a3. This in turn gives a1 = −.1666630261, a2 = .04166085689 and a3 =.008330914797. For N = 17, these values are a1 = .1666633333, a2 = .04166152737 and a3 = −.008331283835. Then, by using the inverse transformation rule in Eq. (4.2), we get the following series solution is evaluated up to N = 17: Numerical results for linear and non-linear of sixth- order BVP, the differential trans- formation method DTM with comparison to the exact solution are given in Table 2. I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 437 Example 5.5. We consider the following linear ninth-order BVP [20], y(i x)(x) = y(x)− 9ex , 0< x < 1, (5.18) subject to the boundary conditions y( j)(0) = (1− j), j = 0, 1, 2, 3, 4, y( j)(1) = − je, j = 0, 1, 2, 3. (5.19) Applying the operations of DT to Eq. (5.21), the following recurrence relation is ob- tained: Y (k+ 9) = k![Y (k)− 9 k!] (k+9)! . (5.20) By using Eqs. (2.1) and (5.22) the following transformed B.C.’s at x = 0 can be ob- tained: Y (0) = 1, Y (1) = 0, Y (2) = −1 2 , Y (3) = −1 3 , Y (4) = −1 8 , (5.21) where, according to Eq. (2.1), we have a = y(v)(0) 5! = Y (5), b = y(vi)(0) 6! = Y (6), c = y(vii)(0) 7! = Y (7), d = y(viii)(0) 8! = Y (8). Utilizing the recurrence relation in Eq. (5.23) and the transformed B.C.’s in Eq. (5.24), Y (k) for k ≥ 9 are easily obtained: Y (9) = − 1 45360 , Y (10) = − 1 403200 , Y (11) =− 1 3991680 , Y (12) = − 1 43545600 , Y (13) =− 1 518918400 , Y (14) = 1 726485760 a− 1 726485760 , I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 438 Y (15) = 1 726485760 b− 1 145297152000 , Y (16) = 1 4151347200 c − 1 2324754432000 , Y (17) = 1 8821612800 d − 1 39520825344000 , Y (18) = − 1 376610217984000 , Y (19) = − 1 6758061133824000 , Y (20) = − 1 128047474114560000 , and so on. The constants a, b, c and d are evaluated from the B.C.’s given in Eq. (5.22) for x = 1, by taking N = 17, to obtain the system: This in turn gives a = −.3336167167E−1 , b = −.6870399019E−2 and c = −.1255380280E−2 and d = −.1544141065E−3. For N = 26, these values are a = −.3336167167E−1 ,b = −.6870399019E−2 and c = −.1255380280E−2 and d = −.1544141066E−3. Then, by using the inverse transformation rule in Eq. (4.2), we get the following series solution is evaluated up to N = 26: I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 439 Numerical results for linear ninth- order BVP, the differential transformation method (DTM) with comparison to the exact solution are given in Table 3. Table 3: Comparison of numerical result of BVP’s (5.21)-(5.22) see Fig.3. Example 5.6. We consider the following non-linear tenth-order BVP y(x)(x) = e−x y2(x), 0 < x < 1, (5.22) subject to the boundary conditions y(2 j)(0) = 1, j = 0, 1, 2, 3, 4, . y(2 j)(1) = e, j = 0, 1, 2, 3, 4. (5.23) Applying the operations of DT to Eq. (5.26), the following recurrence relation is obtained Y (k+ 10) = k! (k+10)! ∑k l=0 ∑l s=0 (−1)s s! Y (l − s)Y (k− l). (5.24) By using Eqs. (2.1) and (5.27) the following transformed B.C.’s at x = 0 can be ob- tained: Y (0) = 1, Y (2) = 1 2! , Y (4) = 1 4! , Y (6) = 1 6! , Y (8) = 1 8! , (5.25) where, according to Eq. (2.1), we have a1 = y ′(0) = Y (1), Y (5), a2 = y(iii)(0) 3! = Y (3), a3 = y(v)(0) 5! = Y (5), I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 440 a4 = y(vii)(0) 7! = Y (7), a5 = y(i x)(0) 9! = Y (9), Utilizing the recurrence relation in Eq. (5.28) and the transformed B.C.’s in Eq. (5.29), Y (k) for k ≥ 10 are easily obtained: 1; Y (10) = 1 3628800 , Y (11) = 1 19958400 a1 − 1 39916800 , Y (12) = 1 159667200 + 1 239500800 a2 1 − 1 119750400 a1, Y (13) = 1 518918400 a2 + 1 518918400 a1 − 1 889574400 − 1 1037836800 a2 1, Y (14) = 1 1037836800 + 1 1816214400 a1a2 − 1 1816214400 a2 − 1 2724321600 a1+ 1 7264857600 a2 1 ... and so on. The constants a1,a2 , a3, a4 and a5 are evaluated from the B.C.’s given in Eq. (5.27) for x = 1, by taking N = 12, to obtain : a1 = 1.000000124, a2 = .9999819650, a3 = 1.000157229, a4 = .9985666714, a5 = 1.009946626. For N = 17, these values are a1 = .9999698990, a2 = 1.000278188, a3 = .9973109664, a4 = 1.023991491 and a5 = .8383579606. I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 441 Then, by using the inverse transformation rule in Eq. (4.2), we get the following series solution is evaluated up to N = 17: Numerical results for non-linear tenth- order BVP, the differential transformation method (DTM) with comparison to the exact solution are given in Table 4, and which have been indicated in Fig. 4. Table 4: Comparison of numerical result of BVP’s (5.26)-(5.27), see Fig.4. Example 5.7. Finally, we consider the non-linear twelfth-order BVP [20] y(x ii)(x) = 2ex y2(x) + y(iii)(x), 0< x < 1, (5.26) subject to the boundary conditions y(2 j)(0) = 1, j = 0, 1, 2, 3, 4, 5, y(2 j)(1) = e−1, j = 0, 1, 2, 3, 4, 5. (5.27) Applying the operations of (DT) to Eq. (5.31), the following recurrence relation is I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 442 obtained: Y (k+ 12) = k! (k+12)! h 2 ∑k l=0 ∑l s=0 1 s! Y (l − s)Y (k− l) + (k+ 1)(k+ 2)(k+ 3)Y (k+ 3) i . (5.28) By using Eqs. (2.1) and (5.32) the following transformed B.C.’s at x = 0 can be ob- tained: Y (0) = 1, Y (2) = 1 2! , Y (4) = 1 4! , Y (6) = 1 6! , Y (8) = 1 8! , Y (10) = 1 10! , (5.29) where, according to Eq. (2.1), we have Y (1) = a1 = y ′(0), Y (3) = a2 = y(iii)(0) 3! , Y (5) = a3 = y(v)(0) 5! , Y (7) = a4 = y(vii)(0) 7! , Y (9) = a5 = y(i x)(0) 9! , Y (11) = a6 = y(x i)(0) 11! . Utilizing the recurrence relation in Eq. (5.33) and the transformed B.C.’s in Eq. (5.34), Y (k) for k ≥ 12 are easily obtained: Y (12) = 1 239500800 + 1 79833600 a2, Y (13) = 1 2075673600 + 1 1556755200 a1, Y (14) = 1 14529715200 + 1 10897286400 a1 + 1 21794572800 a2 1 + 1 726485760 a3, Y (15) = 1 87178291200 + 1 54486432000 a1 + 1 108972864000 a2 1 + 1 54486432000 a2, Y (16) = 1 498161664000 + 1 326918592000 a1 + 1 871782912000 a2 1 + 1 217945728000 a2+ 1 217945728000 a1a2 + 1 4151347200 a4, ... I. Abdel-Halim and V. Ertürk / Eur. J. Pure Appl. Math, 2 (2009), (426-447) 443 and so on. The constants a1,a2 , a3, a4 , a5 and a6 are evaluated from the B.C.’s given in Eq. (5.32) for x =1, by taking N = 16, to obtain: a1 =−0.9999999967, a2 = −0.1666666720, a3 = −0.0083333307, a4 =−0.0001984133, a5 = −0.0000027557, a5 = −0.0000000251 Then, by using the inverse transformation rule in Eq. 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