Numerical Solution of Linear Ordinary
Differential Equations of Higher Order by
Differential Transformation Method
Mukesh Grover Dr. Arun Kumar Tomer
Assistant Professor Professor
Department of Mathematics Department of Mathematics GZSCCET, MRSST University, Bathinda, India S.M.D.R.S.D College Pathankot, India
Abstract
Differential transformation method is used to find the solution of the higher order differential equations. The approximate solution of the problem is calculated in the form of a rapid convergent series. Four numerical examples have been considered to illustrate the efficiency and implementation of the method and the results are compared with the exact solutions. Thereafter, in the form of results are shown graphically. The numerical result obtain by DTM are compared with the solution which are obtain by Method of variation of parameters.
Keywords: Linear Ordinary Differential Equations, Differential Transformation Method, Method of variation of parameters, Taylor Series, Recurrence Relation
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I. INTRODUCTION
The classical Taylor series is one of the earliest analytic ways to solve many problems, like ordinary differential equations, Partial differential equations, Linear Ordinary Differential Equations of Higher Order and integral equations. However, since it requires a lot of calculation for the derivatives of functions. It takes a lot of computational time to higher order derivatives. We have introduce Modification in Taylor series method which is called the differential transform method(DTM).in this method to find the coefficients of the Taylor series of the given function by solving the induced recursive equation from the given differential equation. Since proposed in (Zhou, 1986), there are good interest on the applications of DTM to solve various scientific problems. The DTM is an approximation to exact solution of the functions which are differentiable in the form of polynomials. This method is different with the traditional higher order Taylor series since the series needs more time in computation and requires the computation of the necessary derivatives.
The DTM is an alternative procedure for getting Taylor series solution of the differential equations. This method reduces the size of computational domain and is easily applicable to many problems. Large list of methods, exact, approximate and purely numerical are available for the solution of differential equations. Most of these methods are computationally intensive because they are trial-and error in nature, or need complicated symbolic computations. The differential transformation technique is one of the numerical methods for ordinary differential equations. This method constructs a semi-analytical numerical technique that uses Taylor series for the solution of differential equations in the form of a polynomial. It is different from the high-order Taylor series method which requires symbolic computation of the necessary derivatives of the data functions. The Taylor series method is computationally time-consuming especially for high order equations. The differential transform is an iterative procedure for obtaining analytic Taylor series solutions of differential equations. The main advantage of this method is that it can be applied directly to nonlinear ODEs without requiring linearization, perturbation. This method will not consume too much computer time when applying to nonlinear or parameter varying systems. This method gives an analytical solution in the form of a polynomial. But, it is different from Taylor series method that requires computation of the high order derivatives.
The differential transform method is an iterative procedure that is described by the transformed equations of original functions for solution of differential equations. Chen and Liu have applied this method to solve two-boundary-value problems [6]. Jang, Chen and Liu apply the two-dimensional differential transform method to solve partial differential equations. Yu and Chen apply the differential transformation method to the optimization of the rectangular fins with variable thermal parameters [7]. Unlike the traditional high order Taylor series method which requires a lot of symbolic computations, the differential transform method is an iterative procedure for obtaining Taylor series solutions. The analytical results of the boundary value problems have been obtained in terms of a convergent series with easily computable components.
II. DIFFERENTIAL TRANSFORMATION METHOD
The differential transformation of the kth derivative of a function f(x) is defined by
2.2 ) ( ) ( ) ( 1 . 2 ) ( ! 1 ) ( 0 0 0 k k x x k k x x k F x f dx x f d k k F
and the inverse differential transformation of F(k) is defined by
k k
t
k
F
t
y
0)
(
)
(
(2.3)The combination result of all above equations is called approximate solution of the functions y (t). Serial No’s. The operation Properties of Differential Transform Method (DTM)
1 z(t)u(t)v(t) Z(k)U(k)V(k) 2 z(t) u(t) Z(k)U(k)
3
dt t du t
z( ) ( ) Z(k) (k 1)U(k 1)
4 2 2 ) ( ) ( dt t u d t
z Z(k) (k1)(k 2)U(k 2)
5 m m dt t u d t
z( ) ( ) Z(k) (k 1)(k 2)...(k m)U(k m)
6 z(t) u(t)v(t)
k n n k U n V k Z 0 ) ( ) ( ) ( 7 m t t z( )
m k if m k if m k m k k Z , 0 , 1 ) ( ), ( ) (
8 z(t)exp(t)
! ) ( k k Z k 9 m t t z()(1 )
! ) 1 )....( 1 ( ) ( k k m m m k
Z
10 ) cos( ) ( ) sin( ) ( t t z and t t z and k k k Z k 2 sin ! ) ( 2 cos ! ) ( k k k Z k
III. ILLUSTRATED EXAMPLES
In this section, the DTM has been successfully used to study the higher order linear ordinary differential equations. Choosing examples with known solutions allows for a more complete error analysis as shown in all examples.
Example 4.1: Solve the second order linear ordinary differential equation
) cosh( 2 ) ( 3 4 2 2 t t y dt dy dt y d
(4.1)
with the initial conditions y(0)1,y'(0) 1 (4.2) Now, find the exact solution of equation (4.1) using the method of variation of parameters.
( 4 ( 3) 3)8
1 3 2 4
t t t
e t e e t y
Where Y(K) is the differential transformation of y(t) and the transformation of initial conditions are Y(0)=1,Y(1)=-1. Using the both conditions are apply in equation (4.3) and put the values of k = 0,1,2,3,4…… and we get distinct finite values of y(2), y(3), y(4), y(5), y(6), y(7)……so on
91445760 544446 ) 9 ( , 1209690 1437 ) 8 ( , 1008 163 ) 7 ( , 720 269 ) 6 ( , 40 29 ) 5 ( , 8 9 ) 4 ( , 6 7 ) 3 ( , 2 1 ) 2
( y y y y y y y
y
the inverse transformation in equation (2.2), we get
) 4 . 4 ( ... 91445760 544446 1209690 1437 1008 163 720 269 40 29 8 9 6 7 2 1 1 )
(t t t2 t3 t4 t5 t6 t7 t8 t9
y
That is the approximate solution of equation (4.4) and shown in Figure 1. Example 4.2: Solve the Third order linear ordinary differential equation
t e dt dy dt y d dt y d 3 2 2
2 3 3
(4.5)
with the initial conditions y(0) 1,y'(0) 1,y''(0) 1 (4.6) Now, find the exact solution of equation (4.5) using the method of variation of parameters.
) 1 9 3 ( 6 1 )
(t e2t et e2t e3t y
Taking differential transformation of equation (4.5) using properties of DTM and the following recurrence relation is obtained as follows: ) 3 )( 2 )( 1 ( ! 1 ) 1 ( ) 1 ( 2 ) 2 ( ) 2 )( 1 ( 3 ) 3 ( k k k k k Y k k Y k k k
Y (4.7)
Where Y(K) is the differential transformation of y(t) and the transformation of initial conditions are Y(0)=1,Y(1)=1, Y(2)=-1. Using the both conditions are apply in equation (4.3) and put the values of k = 0,1,2,3,4…… and we get distinct finite values of y(3), y(4), y(5), y(6), y(7)……so on
17282160 16017 ) 9 ( , 1024128 4267 ) 8 ( , 1016 17 ) 7 ( , 120 7 ) 6 ( , 120 21 ) 5 ( , 12 5 ) 4 ( , 6 5 ) 3
( y y y y y y
y the inverse transformation
in equation (2.2), we get
) 8 . 4 ( ... 17282160 16017 1024128 4267 1016 17 120 7 120 21 12 5 6 5 1 )
(t tt2 t3 t4 t5 t6 t7 t8 t9
y That is the approximate
solution of equation (4.8) and shown in Figure 2. Example 4.3: Solve the Second order linear ordinary differential equation
) cos( ) ( 2 2 t t y dt y d
(4.9)
with the initial conditions y(0) 1,y'(0)1 (4.10) Now, find the exact solution of equation (4.9) using the method of variation of parameters.
t t t
t
y ( 2)sin cos 2
1 )
(
Taking differential transformation of equation (4.9) using properties of DTM and the following recurrence relation is obtained as follows ) 2 )( 1 ( 2 cos ! 1 ) ( ) 2 ( k k k k k Y k Y k (4.11)
Where Y(K) is the differential transformation of y(t) and the transformation of initial conditions are Y(0)=1,Y’(0)=-1. Using the both conditions are apply in equation (4.11) and put the values of k = 0,1,2,3,4…… and we get distinct finite values of y(2), y(3), y(4), y(5), y(6), y(7)……so on
0 ) 9 ( , 40320 1 ) 8 ( , 5040 1 ) 7 ( , 0 ) 6 ( , 120 1 ) 5 ( , 24 1 ) 4 ( , 6 1 ) 3 ( , 0 ) 2
( y y y y y y y
y
the inverse transformation in equation (2.2), we get
) 12 . 4 ...( 5040 1 5040 1 120 1 24 1 6 1 1 )
(t t t3 t4 t5 t7 t8
y
That is the approximate solution of equation (4.12) and shown in Figure 3.
2 2
2
3 3
4 4
6 5
2 t
dt dy
dt y d
dt y d
dt y d
(4.13)
with the initial conditions ,y(0) 1,y'(0) 2,y''(0) 1,y'''(0) 1 (4.14) Now, find the exact solution of equation (4.13) using the method of variation of parameters.
) 1613 88
1296 243
222 90
36 ( 648
1 )
(t t3 t2 t e2t et e3t
y
Taking differential transformation of equation (4.13) using properties of DTM and the following recurrence relation is obtained as follows:
) 15 . 4 ( )
4 )( 3 )( 2 )( 1 (
) 2 ( ) 1 ( ) 1 ( 6 ) 2 ( ) 2 )( 1 ( 5 ) 3 ( ) 1 )( 2 )( 3 ( 2 ) 4 (
k k k k
k k
Y k k
Y k k k
Y k k k k
Y
Where Y(K) is the differential transformation of y(t) and the transformation of initial conditions are Y(0)=1,Y(1)=-2, Y(2)=1,Y(3)=-1. Using the both conditions are apply in equation (4.11) and put the values of k = 0,1,2,3,4…… and we get distinct finite values of y(4), y(5), y(6), y(7), y(8), .……so on
, 5240320
31 )
8 ( , 2520
41 )
7 ( , 180
11 ) 6 ( , 60 11 ) 5 ( , 12
5 ) 4
( y y y y
y
the inverse transformation in equation (2.2), we get
) 16 . 4 ...( 5240320
31
2520 41
180 11
60 11
12 5 2
1 )
(t tt2t3 t4 t5 t7 t7 t8 y
IV. CONCLUSION
The comparison in between the exact solution and its approximate solution in Examples 4.1,4.2, 4.3, 4.4 obtained with the help of Method of variation of parameters and DTM. From the numerical results, it is clear that the DTM is efficient and accurate. By increasing the order of approximation more accuracy can be obtained. The results are also expressed graphically in Figures. The Blue line represents the curve corresponding to the exact solution whereas the Red line corresponds to the approximate solution.
REFERENCES
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