Differential Transform Method for Solving. Linear and Nonlinear Systems of. Ordinary Differential Equations

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1 Applied Mathematical Sciences, Vol 5, 2011, no 70, Differential Transform Method for Solving Linear and Nonlinear Systems of Ordinary Differential Equations Farshid Mirzaee Department of Mathematics Faculty of Science, Malayer University Malayer, , Iran malayeruacir mirzaee@mailiustacir Abstract In this study,differential transform method (DTM) is applied to linear and nonlinear system of ordinary differential equations If the system considered has a solution in terms of the series expansion of known functions,this powerful method catches the exact solutionso as to show this capability and robustness, some systems of ordinary differential equations are solved as numerical examples Keywords: Differential transform method; System of ordinary differential equations; Series solution

2 3466 F Mirzaee 1 Introductin The standard form of system of ordinary differential equations of the n order with conditions is considered [4] as φ 1 (x, y 1 (x),y 1 (x),, y(n) 1 (x),y 2(x),y 2 (x),, y(n) 2 (x),, y N(x),y N (x),, y(n) N (x)) = 0 φ 2 (x, y 1 (x),y 1 (x),, y(n) 1 (x),y 2 (x),y 2 (x),, y(n) 2 (x),, y N (x),y N (x),, y(n) N (x)) = 0, (1) φ N (x, y 1 (x),y 1 (x),, y(n) 1 (x),y 2 (x),y 2 (x),, y(n) 2 (x),, y N (x),y N (x),, y(n) N (x)) = 0 with initial values given for y 1 (x 0 )=y 10,y 1 (x 0)=y 1 0,, y (n 1) 1 (x 0 )=y (n 1) 1 0 y 2 (x 0 )=y 20,y 2(x 0 )=y 2 0,, y (n 1) 2 (x 0 )=y (n 1) 2 0, (2) y N (x 0 )=y N0,y N (x 0)=y N 0,, y (n 1) N (x 0 )=y (n 1) N 0 where φ i for i = 1(1)N are nonlinear continuous functions of its argumentin this paper by using (DTM) we find an approximate solution for y(x) = (y 1 (x),y 2 (x),, y N (x)) 2 Differential transform method (DTM) The transformation of the kth derivative of a function in on variable is a follows: Y (k) = 1 y k! [dk dx (x)] k x=x 0, (3) and the inverse transformation is defind by, y(x) = Y (k)(x x 0 ) k, (4) k=0 see [8] The following theorems that can be deduced from equations (3) and (4) are given below: Theorem 1If y(x) =y 1 (x) ± y 2 (x),then Y (k) =Y 1 (k) ± Y 2 (k) Theorem 2If y(x) =cy 1 (x), then Y (k) =cy 1 (k),where c is a constant

3 Differential transform method 3467 Theorem 3If y(x) = dn y 1 (x), then Y (k) = (k+n)! Y dx n k! 1 (k + n) k Theorem 4If y(x) =y 1 (x)y 2 (x), then Y (k) = Y 1 (k 1 )Y 2 (k k 1 ) k 1 =0 Theorem 5If y(x) =x n, then Y (k) =δ(k n) where δ(k n) = { 1 k = n 0 k n Theorem 6If y(x) =y 1 (x)y 2 (x)y n (x), then k k n 1 k 3 k 2 Y (k) = Y 1 (k 1 )Y 2 (k 2 k 1 )Y n 1 (k n 1 k n 2 )Y n (k k n 1 ) k n 1 =0 k n 2 =0 k 2 =0 k 1 =0 Theorem 7If y(x) =e λx, then Y (k) = λk,where λ is a constant k! Theorem 8If y(x) = sin(ωx + α), then Y (k) = ωk k! sin( kπ 2 + α), where ω and α constants Theorem 9If y(x) = cos(ωx + α), then Y (k) = ωk k! cos( kπ + α), where ω and 2 α constants The proofs of Theorems 1 6 are available in [1] and the proofs of Theorems 7 9 are available in [6] 3 Numerical results In this section, we use the method discussed of the previous sections for solving some examples Example 1 Consider the following system of non-homogeneous differential equations[2,3,5]: y 1 (x) =y 3(x) cos(x) y 2(x) =y 3 (x) e x, (5) y 3 (x) =y 1(x) y 2 (x) with the conditions y 1 (0) = 0 y 2 (0) = 0 (6) y 3 (0) = 2

4 3468 F Mirzaee The exact solution of this problem is: y(x) =(y 1 (x),y 2 (x),y 3 (x)) = (e x, sin(x),e x + cos(x)) By using Theorems 1,3,7 and 9 choosing x 0 = 0, equations (5) and (6) are transformed as follows: (k +1)Y 1 (k +1) Y 3 (k) = 1 cos( kπ) k! 2 (k +1)Y 2 (k +1) Y 3 (k) = 1 k!, (k +1)Y 3 (k +1) Y 1 (k)+y 2 (k) =0 Y 1 (0) = 1,Y 2 (0) = 0,Y 3 (0) = 2 consequently, we find Y 1 (1) = 1, Y 2 (1) = 1, Y 3 (1) = 1 Y 1 (2) = 1, 2! Y 2(2) = 0, Y 3 (2) = 0 Y 1 (3) = 1, 3! Y 2(3) = 1, 3! Y 3(3) = 1 3! Y 1 (4) = 1, 4! Y 2(4) = 0, Y 3 (4) = 2 4! Y 1 (5) = 1, 5! Y 2(5) = 1, 5! Y 3(5) = 1 5! Therefore, from(4),the solution of the equation(5) is given by y 1 (x) =1+x + 1 2! x ! x ! x ! x5 + = e x, y 2 (x) =x 1 3! x ! x5 = sin(x), y 3 (x) = (1 + x + 1 2! x ! x3 + )+(1 1 2! x ! x4 + )=e x + cos(x) Example 2 Consider the following system of diferential equations of order two[3]: { y 1(x)+y 1 (x) y 2(x) 4y 2 (x) =0, (7) y 1 (x)+y 2 (x) = cos(x) + 2 cos(2x) with the conditions { y1 (0) = 0,y 1 (0) = 1 y 2 (0) = 0,y 2(0) = 2, (8)

5 Differential transform method 3469 the exact solution of this problem is y(x) =(y 1 (x),y 2 (x)) = (sin(x), sin(2x)) By using Theorems 1,2,3, and 9 choosing x 0 = 0, equations (7) and (8)are transformed as follows: (k + 2)(k +1)Y 1 (k +2)+Y 1 (k) (k + 2)(k +1)Y 2 (k +2) 4Y 2 (k) =0 (k +1)Y 1 (k +1)+(k +1)Y 2 (k +1)= 1 cos( kπ 2k+1 )+ cos( kπ ) k! 2 k! 2 Y 1 (0) = 0,Y 1 (1) = 1 Y 2 (0) = 0,Y 2 (1) = 2, consequently, we find Y 1 (2) = 0, Y 2 (2) = 0 Y 1 (3) = 1, Y 3! 2(3) = 8 3! Y 1 (4) = 0, Y 2 (4) = 0 Y 1 (5) = 1, Y 5! 2(5) = 32 5! Terefore, from(4), the solution of equation(7)is given by y 1 (x) =x 1 3! x ! x5 + = sin(x), y 2 (x) =2x 8 3! x ! x5 = sin(2x) Example 3 Consider the following nonlinear system of differential equations[3]: y 1 (x) =2e4x y4 2(x) y 2(x) =y 1 (x) y 3 (x) + cos(x) e 2x with the conditions y 3 (x) =y 2(x) y 4 (x)+e x sin(x) y 4(x) = e 5x y1(x) 2, (9) { y1 (0) = 1,y 2 (0) = 1 y 3 (0) = 0,y 4 (0) = 1, (10)

6 3470 F Mirzaee the exact solution of this problem is y(x) =(y 1 (x),y 2 (x),y 3 (x),y 4 (x)) = (e 2x, sin(x) + cos(x), sin(x),e x ) By using Theorems 1,2,3,4,6,7,8 and 9 choosing x 0 = 0,equations (9) and (10) are transformed as follows: k k 2 4 k 1 (k +1)Y 1 (k +1)=2 k 1! Y 4(K 2 k 1 )Y 4 (k k 2 ) k 2 =0 k 1 =0 (k +1)Y 2 (k +1)=Y 1 (k) Y 3 (k)+ 1 cos kπ 2k k! 2 k! (k +1)Y 3 (k +1)=Y 2 (k) Y 4 (k)+ ( 1)k 1 sin kπ k! k! 2 k k 2 ( 5) k 1 (k +1)Y 4 (k +1)= Y 1 (k 2 k 1 )Y 1 (k k 2 ) k 1! k 2 =0 k 1 =0 Y 1 (0) = 1,Y 2 (0) = 1 Y 3 (0) = 0,Y 4 (0) = 1, consequently, we find Y 1 (1) = 2, Y 2 (1) = 1, Y 3 (1) = 1, Y 4 (1) = 1 Y 1 (2) = 2, Y 2 (2) = 1 2! Y 3(2) = 0, Y 4 (2) = 1 2! Y 1 (3) = 8, 3! Y 2(3) = 1, 3! Y 3(3) = 1, 3! Y 4(3) = 1 3! Y 1 (4) = 16 4! Y 2(4) = 1, 4! Y 3(4) = 0, Y 4 (4) = 1 4! Y 1 (5) = 32, 5! 2(5) = 1, 5! 3(5) = 1, Y 5! 4(5) = 1 5! Therefore, from(4),the solution of equation(9) is given by y 1 (x) =1+2x +2x ! x3 + = e 2x, y 2 (x) =[x 1 3! x ! x5 + ]+[1 1 2! x ! x4 + ] = sin(x) + cos(x), y 3 (x) =x 1 3! x ! x5 + = sin(x), y 4 (x) =1 x + 1 2! x2 1 3! x3 + = e x Example 4 Consider the following system of liner differential equations[5,7]: { y 1 (x) =y 1 (x)+y 2 (x) y 2(x) = y 1 (x)+y 2 (x), (11)

7 Differential transform method 3471 with the conditions { y1 (0) = 0 y 2 (0) = 1 (12) By using Theorems 1,2 and 3 choosing x 0 = 0,equations (11) and (12) are transformed as follows: (k +1)Y 1 (k +1) Y 1 (k) Y 2 (k) =0 (k +1)Y 2 (k +1)+Y 1 (k) Y 2 (k) =0, Y 1 (0) = 0,Y 2 (0) = 1 consequently, we find Y 1 (1) = 1, Y 2 (1) = 1 Y 1 (2) = 1, Y 2 (2) = 0 Y 1 (3) = 1, Y 3 2(3) = 1 3 Y 1 (4) = 0, Y 2 (3) = 1 6 Y 1 (5) = 1, Y 30 2(5) = 1 30 Therefore, from(4),the solution of equation(11) is given by y 1 (x) =x + x x x5 ± O(x 6 ), y 2 (x) =1+x 1 3 x3 1 6 x4 ± O(x 5 ) 4 Conclusion In this work,we successfully apply the DTM to find numerical solutions for linear and nonlinear system of ordinary differential equations It is observed that DTM is an effective and reliable tool for the solution of system of ordinary differential equations The method gives rapidly converging series solutionsthe

8 3472 F Mirzaee accuracy of the obtained solution can be improved by taking more terms in the solution In many cases,the series solutions obtained with DTM can be written in exact closed form The present method reduces the computational difficulties of the other traditional methods and all the calculations can be made simple manipulations Several examples were tested by applying the DTM and the results have shown remarkable performance References [1] AArikoglu,IOzkol, Solution of boundary value problems for integro-differential equations by using transform method, Appl Math Comput, 168(2005) [2] JBiazar,E Babolian,RIslam, Solution of the system of ordinary differential equations by Adomian decomposition method, Appl Math Comput, 147(3)(2004) [3] JBiazar,H Ghazvini Hes variational iteration method for solving linear and nonlinear and system of ordinary differential equations, Appl Math Comput, 191(2007) [4] JCButcher,Numerical methods for ordinary differential equations,john Wiley and Sons, (2003) [5] H Jafari,V Daftardar - Gejji, Revised Adomian decomposition method for solving system of ordinary and fractional differential equations, Appl Math Comput, 181(2006) [6] Z Odibat, Differential transform method for solving Volterra integral equation with separable kernels, Math Comput Model, 48(2008) [7] NT Shawagfeh,D Kaya, Comparing numerical methods for solutions of ordinary differential equations, Appl Math Lett, 17(2004) [8] JK Zhou, Differential transformation and Application for electrical circuits, Huazhong University Press, Wuhan, China, (1986) Received: April, 2011

(Received 13 December 2011, accepted 27 December 2012) y(x) Y (k) = 1 [ d k ] dx k. x=0. y(x) = x k Y (k), (2) k=0. [ d k ] y(x) x k k!

(Received 13 December 2011, accepted 27 December 2012) y(x) Y (k) = 1 [ d k ] dx k. x=0. y(x) = x k Y (k), (2) k=0. [ d k ] y(x) x k k! ISSN 749-3889 (print), 749-3897 (online) International Journal of Nonlinear Science Vol.6(23) No.,pp.87-9 Solving a Class of Volterra Integral Equation Systems by the Differential Transform Method Ercan