Differential Transform Method For Ordinary Differential Equations
NARHARI PATIL
1and AVINASH KHAMBAYAT
21
Professor & Head Department of Mathematics
Shri Sant Gajanan Maharaj College of Engineering, Shegaon, Maharashtra, INDIA.
2
Department of Mathematics
G. H. Raisoni Institute of Engineering and Management, Jalgaon, Maharashtra, INDIA
(Received on: May 24, 2012) ABSTRACT
In this paper, we study Differential Transform Method for solving Ordinary Differential Equations. The approximate solution of the equation is calculated in the form of series with easily computable componants.This Powerful method catches the exact solution.
Some types of Ordinary Differential Equations are solved as numerical examples.
Keywords: Series solutions, Differential transform, System of Ordinary Differential Equations.
1. INTRODUCTION
Consider the system of Ordinary Differential Equations of the first order with initial conditions,
dx dy
1,
,
, … ,
,
,
, 1, 2, … , , 20
y
2
,
,
, … ,
,
= y
n(1.1) Where each equation represents the first
derivative of unknown function as a mapping depending on the independent variable and n unknown functions
, , . . .2. THE DIFFERENTIAL TRANSFORM METHOD
The transformation of the
derivative of a function with one variable is
follows
!
ೖೖ
,
and the inverse transformation is defined by,
∞
In this section we shall give basic theorem of one dimensional transform method.
Definition1. If u (t) is analytic in domain in the domain T then let,
k k
dt t u k d
t ( )
) ,
( =
φ for all t ∈ T (2.1) for
,
,, where
k belong to the set of non negative integers, denoted as integer K domain. Therefore (2.1) can be written as,
k all for t t at
dt t u d k k
k t k
U
kk i
0
) (
! ) 1 ,
! ( ) 1 (
=
=
= φ
(2.2)
Where U (K) is called spectrum of u(t) at
in the k –domain and the inverse differential transform of U( k ) is defined as follows
∑
∞=
−
=
o k
t
kt k U t
u ( ) ( )(
0) (2.3)
In the real application, the function u(t) is expressed by a finite series and equation (2.3) can be written as,
2.4
The particular case of equation (2.4) when
0 is referred to as the Maclaurins series of u (t), and is express as,
2.5
The following theorems that can be deduced from equations (2.1) and (2.4) are given below
Theorem 1
If !" # $ %
!& # $' Theorem 2
If
% ( ) *
) * +1, - * 0 , - . * / Theorem 3
If
%
!Theorem 4
If u t g t h t then
∑ &5' 5
Theorem 5
If
%
భ
Theorem 6
If
భ,
%
! !6 %
Theorem 7
If y
%
!ೖ, 7 -8 9:%8%
Theorem 8
If y sin 6 ! %
!ೖ
sin
6 ! , Where !, are constant Theorem 9
If cos 6 ! %
!ೖcos
6 ! , Where !, are constant
3. NUMERICAL APPLICATION
Here the Differential Transform Method applied on some numerical examples to obtaining exact solution of ordinary differential equation. We considered linear and non linear system of ordinary differential equation of first order.
′,
0 0
′@
6 2
8-% 6 4 8-%2A ,
0 1
′,
0 0
′@
2
8-% 6 4 8-%2A ,
0 2
By using above theorems and basic properties of the Differential Transform Method,
′6 1 6 1
6 1 1
6 1
Also,
sin 1 6 0, (
!ೖsin 6 0
!sin B
C
sin 2 % 2
! sin E 2 The system of (3.1) is transformed as,
6 1 1
6 1
6 1 1 6 1 F 1
2
6 2
1 2 1
! 8-% E 2
2 2
! sin E 2 G
6 1 1
( 6 1
6 1
@
2
!
8-%
2
!ೖsin
A (2.6)
Converting it in to a system of differential equations of the first order differential equations
Example 3.1 consider the following system of order two with initial condition
′′6 H
′′4H
0 , 0 0 ,
′0
1
′6 H
′9:8 6 29:82 , H0
0 , H
′0 2 3.1
Considering the four functions,
,
′
,
H ,
H
′As 0 0 ,
′0 1 -*J5-8
0 0 ,
0 1
and
H 0,
0 2 From initial conditions,
0 0 ,
0 1 ,
0 0 ,
0 2 (3.2) The values of
, - 1,2,3,4 ,
( 0,1,2, … and put (3.2) in (2.6) we get Put k = o,
6 1 1
6 1
"-H8
1
1
6 1 1 6 1 F 1
2
6 2
1 2 1
! 8-% E 2
2 2
! sin E 2 G
"-H8
1 0
6 1 1
6 1
"-H8
1 2
6 1 1
( 6 1 F 1
2
2
1 2 1
! 8-% (E 2
2 2
(! sin (E 2 G
gives
1 0 (3.3) Thus for k = 0,
1 1,
1 0,
1 2,
1 0 (3.4) Put k = 1,
2 0 ,
2
,
2 0 ,
2 4 (3.5) Put k = 2,
3
,
3 0,
3
,
3 0 (3.6) By substituting the values of
%
-% 2.4
0
∞
3.7
Take i = 1,
∞
!
+. . .
8-% (3.8) Take i = 3,
∞
0 6 2
sin2x (3.9) Thus
sin ,
8-%2,
′9:8 , H
′29:82 (3.10) Example 3.2 Consider the following system of linear differential equations,
′
6
ith initial conditions
0 0,
0 1 (3.11) Using above mentioned theorems of differential transform method we transform as 6 1
6 1
6
6 1
M
6
N ,
6 1
M
6
N (3.12) with initial condition
0 0,
0 1 (3.13) Put k = 0,
1 M
0 6
0N = 0 + 1 = 1
1 M
0 6
0N 0 6 1 1 Put k = 1,
2 1 ,
2 0 Put k = 2,
3
,
3
Put k = 3,
4 0,
4
Put k = 4,
5
,
5
We have
∞
take i = 1,
∞
6
6
య ఱ6 . . . take i = 2
∞
1 6
య. . . Example 3.3 consider the following system of non homogenous differential equations
9:8 ,
0 1
,
0 0
,
0 0 3.14
using basic theorem of differential transform method,
6 1 1
6 1 O
1
! cos O E
2 P P
0 1
6 1 1
6 1 O
1
!P ,
0 1
ଷ 1 ଵ
ାଵଵ ଶ, ଷ0 2
(3.15)
Put k = 0 in above equations, we get
1 1 ,
1 1 ,
1 1 Put k = 1,
2
,
2 0 ,
2 0
Put k = 2 ,
3
,
3
,
3 1 6
Put k = 3 ,
4
,
4 0 ,
4 1 12
Put k = 4,
5
!,
5
!,
5 1 5!
Put k = 5 ,
6
!,
6 0 ,
6 0 (3.16) By substituting these value of
,
,
in to (3.6) we obtain
,
,
∞
take i = 1,
1 6 6
6 . . . =
take i = 2 ,
!య
6 . . . 8-%
take i = 1,
2 6 1 6
0
6
య6
ర6 . . . =
6 9:8 3.17
Example 3.4 Consider the following differential equation of first order,
′
2
, ′
, ′
6
With initial conditions
ଵ0 1 , ଶ0 1 , ଷ0 0 3.18
By applying basic theorems of the differential transformation on the system (3.15), (3.16) respectively, we get
6 1 2
6 1
5
5
6 1 1
6 1 1
5
ଷ 1 1
1 ଶ 1 1
!
ୀ
(3.19)
0 1 ,
0 1,
0 0 Gives
0 1 ,
0 1,
0 0 (3.20) take k = 0 in (3.17) we get
1 2 ,
1 1
also 5 0 , k = 0 gives
1 1
Put k = 1, 5 0
2 1 ,
2
,
2
Put k = 2, 5 0
3
,
3
,
3
(3.21) Now for
, - 1 ,2, 3 . . . % 0 , 1 , 2
∞