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Applied Mathematics, 2016, 7, 2165-2173
http://www.scirp.org/journal/am
ISSN Online: 2152-7393
ISSN Print: 2152-7385
DOI: 10.4236/am.2016.717172 November 17, 2016
The Zhou’s Method for Solving the Euler
Equidimensional Equation
Pedro Pablo Cárdenas Alzate1, Jhon Jairo León Salazar2, Carlos Alberto Rodríguez Varela2
1Department of Mathematics and GEDNOL, Universidad Tecnológica de Pereira, Pereira, Colombia
2Department of Mathematics, Universidad Tecnológica de Pereira, Pereira, Colombia
Abstract
In this work, we apply the Zhou’s method [1] or differential transformation met
hod
(DTM) for solving the Euler equidimensional equation. The Zhou’
s method may be
considered as alternative and efficient for finding the approximate solutions of initial
values problems. We prove superiority of this method by applying them on the some
Euler type equation, in this case of order 2 and 3 [2].
The power series solution of the
reduced equation transforms into an approximate implicit solution of the original
equations. The results agreed with the exact solution obtaine
d via transformation to a
constant coefficient equation.
Keywords
Zhou’s Method, Equidimensional Equation, Euler Equation, DTM
1. Introduction
We know that when the coefficients
( )
px
and
( )
qx
are analytic functions on a
given domain, then the equation
( ) ( )
0y pxy qxy
′′ ′
+ +=
has analytic fundamental
solution. We want to study equations with coefficients
p
and
q
having singularities, for
this reason we study in this paper with one of the simplest cases,
Euler
’
s equidimen-
sional equation.
This is an important problem because many differential equations in
physical sciences have coefficients with singularities [3]. One of the special features of
the equidimensional equation is that order of each derivative is equal to the power of
the independent variable. This means that this type of equations can be reduced to
linear equation with constant coefficient by using a change of the form
e
t
x=
.
Many numerical methods were developed for this type of equations, specifically on
Euler’s equations such that Laplace transform method and Adomian method [4]. The
How to cite this paper:
Cárdenas Alzate
,
P
.P., Salazar, J.J.L. and Varela, C.A.R. (2016
)
The Zhou’s Method for Solving the Euler
Equidimensional Equation
.
Applied M
a-
thematics
,
7
, 2165-2173.
http://dx.doi.org/10.4236/am.2016.71717
2
Received:
September 15, 2016
Accepted:
November 14, 2016
Published:
November 17, 2016
Copyright © 201
6 by authors and
Scientific
Research Publishing Inc.
This work is licensed under the Creative
Commons Attribution International
License
(CC BY 4.0).
http://creativecommons.org/licenses/by/4.0/
Open Access
P. P. Cárdenas Alzate et al.
2166
method proposed in this paper was first established by Zhou to solve problems in
electric circuits analysis. In this work, the differential transformation method is applied
to solver the Euler equidimensional equations and to illustrate this method, several
equations of this type are solved [5] [6].
2. The Euler Equidimensional Equation
A Euler equidimensional equation is a differential equation of the form
( )
12
12
1 2 10
12
d d dd
d
dd d
nn
nn
nn
nn
y y yy
ax a x ax ax ay g x
x
xx x
−
−
−−
+ ++ + + =
(1)
where
0
,,
n
aa
are constants and
d
d
n
n
y
x
is an
n-th
derivative of the function
( )
yx
and
( )
gx
is a continuous function.
Now, we consider a second order differential equation (homogeneous Euler equidi-
mensional) of the form
2
22
dd 0, 0
d
dyy
ax bx cy x
x
x+ += >
(2)
The solution can be obtained by using the change of variables
et
x=
(3)
where
d1
dt
xx
=
. In fact, for
0x>
, we introduce
et
x=
, therefore
( )
lntx=
. Then, the
first and second derivatives of
( )
yx
are related by the chain rule,
22
2 2 22
d d d 1d d d 1d 1 d 1 d
and
d dd d d d d
dd
y yt y y y y y
x tx xt xxt t
x x xt
= = = =−+
(4)
Now, substituting (4) in (2) yields a second order differential equation with constant
coefficients,
i.e
.,
2
222
1 d d 1d 0
dd
dyy y
ax bx cy
t xt
xt
− + +=
2
2
d dd 0
dd
dyyy
a abcy
tt
t− + +=
( )
2
2
dd
0
d
dyy
a b a cy
t
t+− +=
(5)
Equation (5) can be solved using the characteristic polynomial
( )
2
0am b a m c+ − +=
(6)
where roots are
1
m
and
2
m
which give the general solution but depending on the
type of roots it has,
i.e
.,
a) If
12
mm≠
, real or complex, then the general solution of the Equation (2) is given
by
( )
12
1 2 12
,, ,0
mm
yx x x x
α α αα
= + ∈>
b) If
12
mm=
, then the general solution of the Equation (2) is given by
P. P. Cárdenas Alzate et al.
2167
( ) ( )
11
1 2 12
ln , , , 0
mm
yx x x x x
α α αα
= + ∈>
3. The Zhou’s Method or DTM
Differential transformation method (DTM) of the function
( )
yx
is defined as
( )
0
1d
!d
k
kxx
y
Yk kx
=
=
(7)
In (7), we have that
( )
yx
is the original function and
( )
Yk
is the transformed
function. The inverse differential transformation is defined as
( ) ( )
0
,
k
k
yx Ykx
∞
=
=
∑
(8)
but in real applications, function
( )
yx
is expressed by a finite series and Equation (8)
can be written as
( ) ( )
0
,
nk
k
yx Ykx
=
=
∑
(9)
which implies that
( )
1
k
kn
Ykx
∞
= +
∑
is negligibly small where
n
is decided by the convergence of natural frequency in this
study.
The following theorems that can be deduced from Equations (7) and (9) and the
proofs are available in [4] [5] [6].
Theorem 1
If
( ) ( ) ( )
yx f x gx= ±
, then
() ( ) ( )
Yk Fk Gx
= ±
.
Theorem 2
If
( ) ( )
1
yx f x
α
=
, then
( ) ( )
1
Yk Fk
α
=
with
1
α
constant.
Theorem 3
If
( )
d
d
n
n
f
yx x
=
, then
( ) ( ) ( )
!
!
kn
Yk Fk n
k
+
= +
.
Theorem 4
If
( ) ( ) ( )
yx f xgx=
, then
( ) ( ) ( )
1
11
0
k
k
Yk Fk Gk k
=
= −
∑
.
Theorem 5
If
( )
n
yx x=
, then
( ) ( )
Yk k n
δ
= −
, where
( )
1,
0,
kn
kn kn
δ
=
−=
≠
Theorem 6 (
Cárdenas, P
)
. If
( ) ( )
n
yx xf x=
, then
()( )
0,
,
kn
Yk Fk n k n
<
=−≥
with
n∈
.
4. Numerical Results
To illustrate the ability of the Zhou’s method [2] [7] for the Euler equidimensional
equation, the next problem is provided and the results reveal that this method is very
effective.
P. P. Cárdenas Alzate et al.
2168
Example 1 (
Homogeneous case
)
. To begin, we consider the initial value problem
( ) ( )
2
4 40
12,111
x y xy y
yy
′′ ′
− +=
′
=−=−
(10)
Using the substitution (3) and (4), the IVP (10) is transformed to a second order
differential equation with constant coefficients,
i.e
.,
( ) ( )
( )
( ) ( )
22
11
4 40x yt yt x yt yt
x
x
′′ ′ ′
− − +=
( ) ( ) ( ) ( )
440yt yt yt yt
′′ ′ ′
−− + =
( ) ( ) ( )
540y t y t yt
′′ ′
−+=
(11)
Now, of the initial conditions we have that as
1
x=
, then
0t=
and therefore
( )
02y= −
and
( )
0 11y′= −
. So, the new IVP is given by
() ()
540
02,011
yyy
yy
′′ ′
−+=
′
=−=−
(12)
The exact solution of the problem (12) is
( )
4
3
yx x x= −
. Taking the differential
transformation of this problem we obtain
( ) ( ) ( ) ( ) ( )
2! 1!
25 14 0
!!
kk
Yk Yk Yk
kk
++
+ − ++ =
or
()()( )( ) ( ) ( )
1
2 5 1 14
21
Yk k Yk Yk
kk
+ = + +−
++
(13)
where
( )
02
Y= −
and
( )
1 11Y= −
. Therefore, the recurrence Equation (13) gives:
•
0k=
,
( ) ( ) ( )
( )
( )
1 1 47
2 5 1 4 0 55 8
2 22
Y YY= − = −+=−
•
1k=
,
( ) ( ) ( )
( )
( )
1 1 191
3 10 2 4 1 235 44
6 66
Y YY= − =−+ =−
•
2k=
,
( ) ( ) ( )
( )
1 1 955 767
4 15 3 4 2 94
12 12 2 24
Y YY
= − =−+=−
Therefore, using (9), the closed form of the solution can be easily written as
( ) ( ) ( ) ( ) ( ) ( )
23
0
23 4
01 2 3
47 191 767
2 11 2 6 24
nk
k
yt Ykt Y Y t Y t Y t
tt t t
=
= =++ + +
=−− − − − −
∑
(14)
but since
( )
lntx=
, then we obtain (see Figure 1)
( ) ( ) ( )
( )
( )
( )
( )
( )
23 4
47 191 767
2 11ln ln ln ln
2 6 24
yx x x x x≈− − − − − −
P. P. Cárdenas Alzate et al.
2169
Figure 1. The Zhou’s method vs. exact solution.
Example 2 (
Non-homogeneous case
)
. We consider the following IVP
( )
( ) ( )
2
4 2 ln
1 2, 1 0
x y xy y x
yy
′′ ′
+ +=
′
= =
(15)
Then, problem (15) is transformed to a second order differential equation with con-
stant coefficient by using (3) and (4),
i.e
.,
( ) ( )
( )
( ) ( )
22
11
42x yt yt x yt yt t
x
x
′′ ′ ′
− + +=
( ) ( ) ( )
4 () 2y t y t yt yt t
′′ ′ ′
−+ + =
( ) ( ) ( )
32y t y t yt t
′′ ′
++=
(16)
We know that of the initial conditions
1x=
and therefore
0t=
, so we obtain
( )
02y=
and
( )
00y′=
. Then, the IVP is given by
( ) ( )
32
0 2, 0 0
y y yt
yy
′′ ′
++=
′
= =
(17)
The exact solution of the problem (15) is
( ) ( )
12
91 3
5 ln
42 4
yx x x x
−−
=−+ −
. Now,
the DTM of (17) is
( ) ( ) ( ) () ( ) ( )
2! 1!
23 12 1
!!
kk
Yk Yk Yk k
kk
δ
++
+ + ++ = −
or
( ) ( )( ) ( ) ( ) ( ) ( )
1
2 3 1 12 1
21
Yk k Yk Yk k
kk
δ
+ = − + +− + −
++
(18)
with
( )
02Y=
and
( )
10Y=
. So, the recurrence Equation (18) gives:
•
0k=
,
P. P. Cárdenas Alzate et al.
2170
( ) ( ) ( ) ( )
( )
( )
11
2 31 2 0 1 4 2
22
Y YY
δ
= − − + − = −=−
•
1k=
,
( ) ( ) ( ) ( ) ( )
( )
( )
1 1 13
3 3 2 2 2 1 0 12 1
6 66
Y YY
δ
= − − + = +=
•
2k=
,
( ) ( ) ( ) ( ) ( )
( )
1 1 39 31
4 33 3 2 2 1 4
12 12 2 24
Y YY
δ
= − − + = −+=−
•
3k=
,
()()( ) ( )
( )
1 1 31 13 67
5 12 4 2 3 2
20 20 2 3 120
Y YY
δ
=− − + = −=
Therefore, using (9), the closed form of the solution can be easily written as
( ) ( ) ( ) ( ) ( ) ( )
23
0
23 4 5
01 2 3
13 31 67
20 2 6 24 120
nk
k
yt Ykt Y Y t Y t Y t
tt t t t
=
= =++ + +
=+− + − + +
∑
(19)
But since
( )
lntx=
, then we obtain (see Figure 2)
( ) ( )
( )
()
( )
( )
()
()
()
23 4 5
13 35 67
2 2 ln ln ln ln
6 24 120
yx x x x x
≈− + − + +
Example 3 (
Third order Euler
’
s equation
)
. Consider the following IVP
()( ) ( )
32
10 20 20 0
1 0, 1 1, 1 1
x y x y xy y
yy y
′′′ ′′ ′
+ − +=
′ ′′
= =−=
(20)
Now, to find
( )
yx
′′′
we use the chain rule. In fact we obtain
Figure 2. The Zhou’s method vs. exact solution.
P. P. Cárdenas Alzate et al.
2171
3 2 2 32
3 22 32 2 3 2
32
33 32 3
d d1dd 2dd 11dd
dd d
d d d dd
1d 3d 2d
d
dd
y yy yy y y
x t tx
x xt xt x t t
y yy
t
xt xt x
= − =− −+ −
=−+
(21)
Therefore, using (3), (4) and (21) we have
( ) ( )
32
32
1 11
3 2 10 20 20 0x y y y x yy xy y
x
xx
′′′ ′′ ′ ′′ ′ ′
−+ + − − + =
3 2 10 10 20 20 0yyyyyyy
′′′ ′′ ′ ′′ ′ ′
−++ − − + =
7 28 20 0yy y y
′′′ ′′ ′
+− + =
(22)
Now, as in the previous example
1x=
and then
0t=
. So, the new initial con-
ditions are given by
()( )
0 0, 0 1yy
′
= = −
and
()
01y′′ =
. Using (7) we find that
( ) ( )
0 0, 1 1YY= = −
and
( )
1
22!
Y=
. Therefore, we obtain the IVP
() () ( )
7 28 20 0
0 0, 0 1, 0 1
yy y y
yy y
′′′ ′′ ′
+− + =
′ ′′
= =−=
(23)
Applying DTM to (23) we obtain
( ) ( ) ( ) ( ) ( ) ( ) ( )
3! 2! 1!
3 7 2 28 1 20 0
!! !
kk k
Yk Yk Yk Yk
kk k
++ +
+ + + − ++ =
or
( )
( )( )( ) ( )( ) ( ) ( ) ( ) ( )
3
17 2 1 2 28 1 1 20
321
Yk
k k Yk k Yk Yk
kkk
+
= −++++++−
+++
(24)
So, the recurrence equation (24) gives:
•
0
k=
,
( ) ( ) ( ) ( ) ( )
( )
( )
1 1 35
3 7 2 2 28 1 20 0 7 28
6 66
Y Y YY= − + − = −− =−
•
1k=
,
( ) ( )( ) ( ) ( ) ( ) ( )
( )
( )
1
4 7 3 2 3 28 2 2 20 1
24
1 35 1 293
42 56 20 1
24 6 2 24
Y Y YY=− +−
−
= − + − −=
•
2k=
,
( ) ( )( ) ( ) ( ) ( ) ( )
( )
1
5 7 4 3 4 28 3 3 20 2
60
1 293 35 1 1017
84 84 20
60 24 6 2 40
Y Y YY=− +−
− −
=− + −=
Therefore, using (9), the closed form of the solution can be easily written as
P. P. Cárdenas Alzate et al.
2172
Figure 3. The Zhou’s method vs. exact solution.
( ) ( ) ( ) ( ) ( ) ( )
()
23
0
23 4 5
23 4 5
01 2 3
1 35 293 1017
012 6 24 40
1 35 293 1017
2 6 24 40
nk
k
yt Ykt Y Y t Y t Y t
tt t t t
tt t t t
=
= =++ + +
= +− + − + − +
=−+ − + − +
∑
(25)
But since
( )
lntx=
, then we obtain (see Figure 3)
( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
23 4 5
1 35 293 1017
ln ln ln ln ln
2 6 24 40
yx x x x x x≈− + − + − +
5. Conclusion
In this paper, we presented the definition and handling of one-dimensional differential
transformation method or Zhou’s method. Using the substitutions (3) and (4), Euler’s
equidimensional equations were transformed to a second and third order differential
equations with constant coefficients, next using DTM these equations were transformed
into algebraic equations (iterative equations). The new scheme obtained by using the
Zhou’s method yields an analytical solution in the form of a rapidly convergent series.
This method makes the solution procedure much more attractive. The figures [4] [5]
and [6] clearly show the high efficiency of DTM with the three examples proposed.
Acknowledgements
Foremost, we would like to express my sincere gratitude to the Department of Mathe-
matics of the Universidad Tecnológica de Pereira and group GEDNOL for the support
in this work. In the same way, we would like to express sincere thanks to the anonym-
ous reviewers for their positive and constructive comments towards the improvement
of the article.
References
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P. P. Cárdenas Alzate et al.
2173
Huazhong University Press, Wuhan.
[2] Odibat, Z. (2008) Differential Transform Method for Solving Volterra Integral Equations
with Separable Kernels.
Mathematical and Computer Modelling
, 48, 1144-1146.
http://dx.doi.org/10.1016/j.mcm.2007.12.022
[3] Shawagfeh, N. and Kaya, D. (2004) Comparing Numerical Methods for Solutions of Ordi-
nary Differential Equations.
Applied Mathematics Letters
, 17, 323-328.
http://dx.doi.org/10.1016/S0893-9659(04)90070-5
[4] Ardila, W. and Cárdenas, P. (2013) The Zhou’s Method for Solving White-Dwarfs Equa-
tion.
Applied Mathematics
, 10C, 28-32.
[5] Arikoglu, O. (2006) Solution of Difference Equations by Using Differential Transform Me-
thod.
Applied Mathematics and Computational
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[6] Cárdenas, P. and Arboleda, A. (2012) Resolución de ecuaciones diferenciales no lineales por
el método de transformación diferencial. Universidad Tecnológica de Pereira, Colombia.
Tesis de Maestría en Matemáticas
.
[7] Cárdenas, P. (2012) An Iterative Method for Solving Two Special Cases of Lane-Emden
Type Equations.
American Journal of Computational Mathematics
, 4, 242-253.
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