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The Numerical Solutions of Partial Differential Equations using Two-

Three-Dimensions Differential Transform Method 

 

Rohit Nagargoje1, Avinash Khambayat2 

1. Research Scholar, Department of Mathematics, School of Science, Sandip University Nasik, Maharashtra 

Email: rnagargoje92@gmail.com 

2.  Professor, Department of Mathematics, School of Science, Sandip University 

Nasik, Maharashtra 

Email: avinash.khambayat@sandipuniversity.edu.in 

  

Article History: 

Received: 12-01-2025 

Revised: 15-02-2025 

Accepted: 01-03-2025 

Abstract: Differential Transform Method is the one of the novel and unique methods which 

provides series solution. The approach mainly rests on the Differential transform method 

(DTM) which is one of the approximate methods to achieved exact solution of one 

dimensional up to multidimensional.  The method can easily be applied on many problems in 

the field of Differential Equations and Partial differential equations (PDE). DTM is used to 

reduces the size of calculus work and gives the solution of higher order two-dimensional, 

three-dimensional, multi-dimensional differential equation. 

Keywords: Differential transformed method, two-dimensional, three-dimensional, multi-

dimensional partial differential equation. 

1. Introduction 

The analytical approaches of the Differential Transform Method have been succeeded to solve 

differential equations and integral equations in many fields. The basic concept of the Differential 

Transform Method is based on the Taylor series. The Differential Transform Method was first 

introduced by J.K. Zhou in a study about electrical circuits. The Differential Transform Method obtains 

an analytical solution in the form of a polynomial equations [1]. Furthermore, A variety 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. Furthermore, the differential transformation technique is one of 

the numerical methods for ordinary differential equations. The concept of differential transformation 

was first proposed by Zhou [2]. It is different from the traditional high order Taylor’s series method, 

which requires symbolic competition of the necessary derivatives of the data functions. The Taylor 

series method has computationally taken a long time for big orders; by using this method, it is possible 

to obtain highly accurate results (up to 3th order) or exact solutions for differential equations [3]. With 

this technique the given partial differential equation and related initial conditions are transformed into 

a recurrence equation that finally leads to the solution of a system of algebraic equations as coefficients 

of a power series solution [4]. This method is useful for obtaining exact and approximate solutions of 

linear and nonlinear ordinary and partial differential equations [5-6]. In this method there is no need 

for linearization or perturbations, large computational work and round-off errors are avoided. It has 

mailto:rnagargoje92@gmail.com
mailto:avinash.khambayat@sandipuniversity.edu.in


Communications on Applied Nonlinear Analysis 

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been used to solve effectively, easily and accurately a large class of linear and nonlinear problems with 

approximations. It is possible to solve system of differential equations, differential algebraic equations, 

difference equations, differential difference equations, partial differential equations [7]. Moreover, the 

applying of DTM is effective for solving two-dimensional and three-dimensional PDEs with initial 

value problems [8-10]. It causes scientific findings in the for mentioned study fields. It is critical to 

investigate different techniques for integrating these PDEs. The DTM is a highly successful and 

efficient instrument for addressing both one-dimensional and multidimensional problems [11]. This 

method uses a sequential method to create analytical solutions in the form of polynomials, which is 

based on the Taylor series expansion [12]. The DTM has been used to solving both linear and nonlinear 

differential equations, including the KdV and MKdV equations given by Angalgil & Ayaz, 2009 in 

[13]. The two-point boundary value problem [14]. (Chenand & Liu, 1998), the linear parabolic-

hyperbolic partial differential equations [15]. (Biazar et al., 2010), the two-dimensional nonlinear Gas 

dynamic, and the Klien-Gordon equations [16]. 

The differential transform approach is also taken into consideration for solving the three-dimensional 

linear Helmholtz problem in the following form: 

𝑙
𝜕𝑧2

𝜕𝑝2
+𝑚

𝜕𝑧2

𝜕𝑞2
+ 𝑛

𝜕𝑧2

𝜕𝑟2
+ 𝜆𝑧 = 𝐺(𝑝, 𝑞, 𝑟) 

With the initial conditions: 

𝑧(𝑜, 𝑞, 𝑟) = 𝑔1(𝑞, 𝑟) = 𝑧𝑝(𝑜, 𝑞, 𝑟) = 𝑔2(𝑞, 𝑟) 

𝑧(𝑝, 𝑜, 𝑟) = 𝑔3(𝑝, 𝑟) = 𝑧𝑞(𝑝, 𝑜, 𝑟) = 𝑔4(𝑝, 𝑟) 

𝑧(𝑝, 𝑞, 𝑜) = 𝑔5(𝑝, 𝑞) = 𝑧𝑟(𝑝, 𝑞, 𝑜) = 𝑔6(𝑝, 𝑞) 

Where 𝑔1(𝑞, 𝑟), 𝑔2(𝑞, 𝑟), 𝑔3(𝑝, 𝑟), 𝑔4(𝑝, 𝑟), 𝑔5(𝑝, 𝑞), 𝑔6(𝑝, 𝑞) and 𝑙,𝑚, 𝑛, 𝜆 are given function and 

constant respective [17]. 

This equation has wide applications in various filed such as electrical and mechanical engineering and 

physics. These equations the reader is referred to (Zwilinger, 1992 Burdenand and Faires, 1993) [18]. 

Jafari and Zabini solved the above equations by homotope perturbation method and homotop analysis 

method respectively [19]. (Jafari.et.al.2010b and 2010c). In this paper we apply DTM for Helmholtz 

equation and Schrodinger equations, Partial Differential Equation [20-21]. 

2. Methods 

In this research article we proposed, the Differential Transformation Method (DTM) has been 

successfully applied to find exact and approximate solution of the second order differential equations. 

The method was used in a direct way without using linearization, perturbation or restrictive 

assumptions. And solving Partial differential equations in two, three-dimensional linear and nonlinear, 

have been solved effectively using the Differential Transform Method.  Therefore, it is not affected by 

computation round off errors and one is not faced with the necessities of large computer memory and 

time. 

 



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Definitions of Multi-dimensional DTM:  

We define - dimensional differential transform and fundamental operation of the function ),...,( 21 mxxxz  

as, 

)0.....0,0(

2

21

21

...

21

21
,...,

),...,,(

!!...!

1
),...,(

1

21













=

+++

m

m

k

m

kk

m

kkk

m

m
xxx

xxxz

kkk
kkkZ

                                                              (1) 

Where ),...,( 21 mxxxz is original function and ),...,( 21 mkkkZ is transformed function. 

 The differential inverse transform of ),...,( 21 mxxxz  is defined as follows 

m

m

k

m

k

k k k

k

mm xxxkkkZxxxz ...),...,(...),...,( 2

1 2

1

2

0 0 0

12121  


=



=



=

=                                                                           (2) 

and from equation (1) and (2) we can conclude  

m

m

m

m

k

m

k

k k k

k

k

m

kk

m

kkk

m

m xxx
xxx

xxxz

kkk
xxxz ...

,...,

),...,(

!!...!

1
...),...,( 2

1 2

1

1

21

2

0 0 0

1

)0.....0,0(

2

21

21

...

21

21  


=



=



=

+++













=

   

                     (3) 

Theorem 1: If 𝑧(𝑝1, 𝑝2… , 𝑝𝑚) = 𝜆𝑓 (𝑝1, 𝑝2… , 𝑝𝑚) then, 

𝑍(𝑘1, 𝑘2… , 𝑘𝑚) = 𝜆𝑓 (𝑘1, 𝑘2, … , 𝑘𝑚) 

                       Theorem 2:  If 𝑧(𝑝1, 𝑝2… , 𝑝𝑚) =
𝜕𝑓(𝑝1,𝑝2,…,𝑝𝑚) 

𝜕𝑝1
 then, 

𝑍(𝑘1, 𝑘2… , 𝑘𝑚) = (𝑘𝑖 + 1)𝐹 (𝑘1, 𝑘2, … , (𝑘𝑖 + 1),… , 𝑘𝑚) 

                       Theorem 3: If 𝑧(𝑝1, 𝑝2… , 𝑝𝑚) = 𝑝1
ℎ1 , 𝑝2

ℎ2…𝑝𝑚
ℎ𝑚 then, 

𝑍(𝑘1, 𝑘2… , 𝑘𝑚) = 𝛿(𝑘1 − ℎ1)𝛿(𝑘2 − ℎ2)…𝛿(𝑘𝑚 − ℎ𝑚)  

                       where, 

𝛿(𝑘𝑖 − ℎ𝑖) = {
1
0
                𝑘𝑖 = ℎ𝑖 

                                                                                                              𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 

Theorem 4: If 𝑧(𝑝1, 𝑝2… , 𝑝𝑚) =  𝑝1
ℎ1 , 𝑝2

ℎ2…sin(𝑎𝑥𝑖 + 𝑏)…𝑝𝑚
ℎ𝑚 𝑡ℎ𝑒𝑛 𝑍(𝑘1, 𝑘2… , 𝑘𝑚) 

       = 𝛿(𝑘1 − 𝑘2)…
𝑎𝑘𝑖

𝑘𝑖!
sin (

𝑘𝑖𝜋

2
+ 𝑏)…𝛿(𝑘𝑚 − ℎ𝑚)  

 

Theorem 5: If 𝑧(𝑝1, 𝑝2… , 𝑝𝑚) =  𝑝1
ℎ1 , 𝑝2

ℎ2…cos(𝑎𝑥𝑖 + 𝑏)…𝑝𝑚
ℎ𝑚 𝑡ℎ𝑒𝑛 𝑍(𝑘1, 𝑘2… , 𝑘𝑚) 

= 𝛿(𝑘1 − 𝑘2)…
𝑎𝑘𝑖

𝑘𝑖!
cos (

𝑘𝑖𝜋

2
+ 𝑏)…𝛿(𝑘𝑚 − ℎ𝑚) 

 

 



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3. Results 

  Solving two numerical using DTM: 

Example [1] Consider the following Two-Dimensional Schrodinger equations: 

 
𝜕2𝑧

𝜕𝑝2
+

𝜕2𝑧

𝜕𝑞2
− 4𝑧 = (24𝑝2 − 5𝑝4) cos(q)                                                                                                        (4) 

With the initial condition 𝑧(0, 𝑞) = 0 , 𝑧𝑝(0, 𝑞) = 0                                                                                     (5) 

The exact solution can be expressed as 𝑧(𝑝, 𝑡) = 𝑝4cos (q) 

Taking the differential transform of (4) 

(𝑘1 + 2) (𝑘1 + 1) 𝑍 (𝑘1 + 2𝑘2) + (𝑘2 + 2) (𝑘2 + 1) 𝑍 (𝑘1, 𝑘2 + 2) − 4𝑍 (𝑘1, 𝑘2)

= 24 𝛿(𝑘1 − 2)
1

𝑘2!
cos (

𝑘2𝜋

2
) − 5𝛿(𝑘1 − 4)

1

𝑘2!
cos (

𝑘2𝜋

2
)                                         (6) 

From the initial condition given by equation (5)  

𝑍(0, 𝑘2,) = 0  

𝑍(1,  𝑘2  ) = 0 

𝑘2 = 0,1,2,…                                                                                                                                                        (7) 

Substituting equation (6) into equation (7) 

By means of recursive method, the results are  

  𝑍(𝑘1,𝑘2,)  =                             {
1

𝑘2!
cos (

𝑘2𝜋

2
)                                                    , 𝑖𝑓 𝑘1 = 4                   

0                                        , 𝑜. 𝑤.
 

We obtained the series solution as 

𝑍(𝑝, 𝑞) = ∑ ∑ 𝑍(𝑘1, 𝑘2)𝑝
𝑘1𝑞𝑘2 = 

∞

𝑘2=0

∞

𝑘1=0

𝑝4 cos(q) 

Which is exact solution. 

Example [2] Consider the following Three-Dimensional Helmholtz equations: 

𝜕2𝑧

𝜕𝑝2
+
𝜕2𝑧

𝜕𝑞2
−
𝜕2𝑧

𝜕𝑟2
− 8𝑧 = (24𝑝2 − 8𝑝4) cos (q)sin (𝑟)                                                                            (8) 

With the initial condition 𝑧(0, 𝑞, 𝑟) = 0 , 𝑧𝑝(0, 𝑞, 𝑟) = 0                                                                             (9) 

The exact solution can be expressed as 𝑧(𝑝, 𝑡) = 𝑝4cos (q)sin (𝑟) 

Taking the differential transform of (8) 

(k1+2) (k1+1) Z (k1+2, k2, k3) + (k2+2) (k2+1) Z (k1, k2+2, k3)- (k3+2) (k3+1) Z (k1, k2, k3+2)-   8Z (k1, 

k2, k3) 



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= 24𝛿(𝑘1 − 2)
1

𝑘2!
cos (

𝑘2𝜋

2
)
1

𝑘3!
sin (

𝑘3𝜋

2
) − 8𝛿(𝑘1 − 4)

1

𝑘2!
cos (

𝑘2𝜋

2
)
1

𝑘3!
sin (

𝑘3𝜋

2
)                               (10) 

From the initial condition given by equation (9)  

𝑍(0, 𝑘2, 𝑘3) = 0  

𝑍(1, 𝑘2, 𝑘3) = 0        

𝑘2,𝑘3 = 0,1,2,…                                                                                                                                                (11) 

Substituting equation (11) into equation (10) 

By means of recursive method, the results are  

Z(𝑘1,𝑘2,𝑘3,) =  0                                             𝑖𝑓 𝑘1 ≠ 8 & 𝑘2,𝑘3, = 0,1,2,…              

𝑍(8 𝑘2,𝑘3 =
1

𝑘2!
cos (

𝑘2𝜋

2
)
1

𝑘3!
sin (

𝑘3𝜋

2
)      𝑖𝑓 𝑘2,𝑘3, = 0,1,2, …                                    

We obtained the series solution as 

𝑍(𝑝, 𝑞, 𝑟) = ∑ ∑ ∑ 𝑍(𝑘1𝑘2, 𝑘3)

∞

𝑘3=0

𝑝𝑘1𝑞𝑘2𝑟𝑘3 = 

∞

𝑘2=0

∞

𝑘1=0

𝑝4 cos(q) sin(𝑟) 

Which is exact solution. 

 

Example [3] Consider PIDE, 

𝑥𝑢𝑥 = 𝑢𝑡𝑡 + 𝑥𝑠𝑖𝑛𝑡 − ∫ 𝑠𝑖𝑛(𝑡 − 𝑧) 𝑢(𝑥, 𝑧) 𝑑𝑧
𝑡

0

 

With initial conditions 𝑢(𝑥, 0) = 0, 𝑢𝑡(𝑥, 0) = 𝑥  and boundary condition 𝑢(1, 𝑡) = 𝑡 

Solution: Given, 

 𝑥𝑢𝑥 = 𝑢𝑡𝑡 + 𝑥𝑠𝑖𝑛𝑡 − ∫ 𝑠𝑖𝑛(𝑡 − 𝑧)𝑢(𝑥, 𝑧)𝑑𝑧
𝑡

0

 

𝑢𝑡𝑡 = 𝑥𝑢𝑥 − 𝑥𝑠𝑖𝑛𝑡 − 𝑠𝑖𝑛𝑡 ∫ 𝑐𝑜𝑠𝑧𝑢(𝑥, 𝑧)𝑑𝑧 + 𝑐𝑜𝑠𝑡∫ 𝑠𝑖𝑛𝑧𝑢(𝑥, 𝑧)𝑑𝑧
𝑡

0

𝑡

0

                                               (12) 

With  𝑢(𝑥, 0) = 0,   𝑢𝑡(𝑥, 0) = 𝑥                                         (13) 

And  𝑢(1, 𝑡) = 𝑡                                           (14) 

Appling two dimensional DTM on both sides of equations (12),  



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∴ 𝑈(𝑘, ℎ + 2) =
1

(ℎ + 1)(ℎ + 2)

{
 
 
 
 
 

 
 
 
 
 
∑ 𝛿(𝛼 − 1)(𝑘 − 𝛼 + 1)𝑈(𝑘 − 𝛼 + 1, ℎ)

𝑘

𝛼=0

−
𝛿(𝑘 − 1) 𝑠𝑖𝑛 (

ℎ𝜋
2 )

ℎ!

−∑∑(
1

𝛽
)(

𝑠𝑖𝑛 (
(ℎ − 𝛽)𝜋

2
)

(ℎ − 𝛽)! 
)(

𝑐𝑜𝑠 (
𝑠𝜋
2
)

𝑠!
)𝑈(𝑘, 𝛽 − 𝑠 − 1)

𝛽−1

𝑠=0

ℎ

𝛽=1

+∑∑(
1

𝛽
)(

𝑐𝑜𝑠 (
(ℎ − 𝛽)𝜋

2
)

(ℎ − 𝛽)! 
)(

𝑠𝑖𝑛 (
𝑠𝜋
2 )

𝑠! 
)𝑈(𝑘, 𝛽 − 𝑠 − 1)

𝛽−1

𝑠=0

ℎ

𝛽=0 }
 
 
 
 
 

 
 
 
 
 

 

∵   Using trigonometric identity, 𝑠𝑖𝑛𝐴. 𝑐𝑜𝑠𝐵 − 𝑐𝑜𝑠𝐴. 𝑠𝑖𝑛𝐵 = sin (𝐴 − 𝐵) 

We get,  

𝑈(𝑘, ℎ + 2) = 

1

(ℎ + 1)(ℎ + 2)

{
  
 

  
 
∑𝛿(𝛼 − 1)(𝑘 − 𝛼 + 1)𝑈(𝑘 − 𝛼 + 1, ℎ) −

𝑘

𝛼=0

𝛿(𝑘 − 1) 𝑠𝑖𝑛 (
ℎ𝜋
2 )

ℎ!

 −∑∑(
1

𝛽
)(

𝑠𝑖𝑛 (
(ℎ − 𝛽 − 𝑠)𝜋

2
)

(ℎ − 𝛽)!   𝑠!
)𝑈(𝑘, 𝛽 − 𝑠 − 1)

𝛽−1

𝑠=0

ℎ

𝛽=1 }
  
 

  
 

 

                                             (15) 

Now applying two-dimensional DTM on initial conditions (13) and boundary conditions (14) we get,  

𝑈(𝑘, 0) = 0  ;  ∀𝑘,                                                                                                                                            (16) 

∴ 𝑈(0,0) = 𝑈(1,0) = 𝑈(2,0) = 𝑈(3,0) = ⋯ . .= 0  

𝑈(𝑘, 1) = 𝛿(𝑘 − 1) = {
1;   𝑘 = 1
0;   𝑘 ≠ 1

                                                                                                                (17) 

∴ 𝑈(0,1) = 0,   𝑈(1,1) = 1,   𝑈(2,1) = 𝑈(3,1) = 𝑈(4,1) = ⋯ . .0   

𝑈(1, ℎ) = 𝛿(ℎ − 1) = {
1;   ℎ = 1
0;   ℎ ≠ 1

                                                                                                                (18) 

∴ 𝑈(1,0) = 0,   𝑈(1,1) = 1,   𝑈(1,2) = 𝑈(1,3) = 𝑈(1,4) = ⋯ . .0   

Put ℎ = 0,1,2,3,4, ….  in equation (15) and using equations (16) to (18), 

If ℎ = 0, 

𝑈(𝑘, 2) =
1

2

{
 
 

 
 ∑𝛿(𝛼 − 1)(𝑘 − 𝛼 + 1) 𝑈(𝑘 − 𝛼 + 1, 0)

𝑘

𝛼=0

−

𝛿(𝑘 − 1) 𝑠𝑖𝑛 0

ℎ!
− 0 }

 
 

 
 

 



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∴ 𝑈(𝑘, 2) =
1

2
{∑ 𝛿(𝛼 − 1)(𝑘 − 𝛼 + 1) 𝑈(𝑘 − 𝛼 + 1,0) − 0 − 0

𝑘

𝛼=0

} 

As 𝑈(𝑘, 0) = 0;  ∀𝑘,        ∴ 𝑈(𝑘 − 𝛼 + 1, 0) = 0   

∴ 𝑈(𝑘, 2) = 0, ∀𝑘 

⟹𝑈(0,2) = 0 , 𝑈(1,2) = 0 , 𝑈(2,2) = 0 , 𝑈(3,2) = 0 , 𝑈(4,2) = 0 , … ..                                         (19) 

If ℎ = 1, 

𝑈(𝑘, 3) =
1

6

{
  
 

  
 ∑𝛿(𝛼 − 1)(𝑘 − 𝛼 + 1) 𝑈(𝑘 − 𝛼 + 1, 1)

𝑘

𝛼=0

−

𝛿(𝑘 − 1) 𝑠𝑖𝑛 (
𝜋
2
)

1!
−∑∑(

1

𝛽
)(

𝑠𝑖𝑛 (
(1 − 𝛽 − 𝑠) 𝜋

2
)

(1 − 𝛽)!   𝑠!
)  𝑈(𝑘, 𝛽 − 𝑠 − 1)

𝛽−1

𝑠=0

1

𝛽=1 }
  
 

  
 

 

                If 𝑘 = 0  ∴ 𝑈(0,3) =
1

6
{0 − 0 − 0} =  0 

                If 𝑘 = 1  ∴ 𝑈(1,3) =
1

6
{∑ 𝛿(𝛼 − 1)(2 − 𝛼) 𝑈(2 − 𝛼, 1) − 𝛿(1 − 1)

1

𝛼=0

} 

                                                    =
1

6
{0 + (1)(1) 𝑈(1,1) − 1} = 0            ∵ 𝑈(1,1) = 1  

              If 𝑘 = 2  ∴ 𝑈(2,3) =
1

6
{∑ 𝛿(𝛼 − 1)(3 − 𝛼) 𝑈(3 − 𝛼, 1) − 𝛿(2 − 1)

2

𝛼=0

} 

                              =
1

6
{0 + (1)(2) 𝑈(2,1) + 0 − 0} = 0           ∵ 𝑈(2,1) = 0 

∴ 𝑈(𝑘, 3) = 0, ∀𝑘 

⟹𝑈(0,3) = 0 , 𝑈(1,3) = 0 , 𝑈(2,3) = 0 , 𝑈(3,3) = 0 , 𝑈(4,3) = 0 , … . .                                         (20) 

If ℎ = 2, 

𝑈(𝑘, 4) = 

1

12

{
  
 

  
 ∑ 𝛿(𝛼 − 1)(𝑘 − 𝛼 + 1)𝑈(𝑘 − 𝛼 + 1,2)

𝑘

𝛼=0

−

𝛿(𝑘 − 1) 𝑠𝑖𝑛(𝜋)

2!
− ∑∑(

1

𝛽
)(

𝑠𝑖𝑛 (
(2 − 𝛽 − 𝑠)𝜋

2
)

(2 − 𝛽)!   𝑠!
)𝑈(𝑘, 𝛽 − 𝑠 − 1)

𝛽−1

𝑠=0

2

𝛽=1 }
  
 

  
 

 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 10s (2025) 

  

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∴ 𝑈(𝑘, 4) =
1

12

{
  
 

  
 ∑𝛿(𝛼 − 1)(𝑘 − 𝛼 + 1)𝑈(𝑘 − 𝛼 + 1,2) +

𝛿(𝑘 − 1)𝑠𝑖𝑛𝜋

2

𝑘

𝛼=0

−∑
𝑠𝑖𝑛 (

(1 − 𝑠)𝜋
2

)

1! 𝑠!
𝑈(𝑘,−𝑠)

0

𝑠=0

−∑
𝑠𝑖𝑛 (

−𝑠𝜋
2 )

0! 𝑠!

1

𝑠=0

𝑈(𝑘, 1 − 𝑠)
}
  
 

  
 

 

∴ 𝑈(𝑘, 4) =
1

12
{0 + 0 − 𝑈(𝑘, 0) − 0 + 𝑈(𝑘, 0)} 

∴ 𝑈(𝑘, 4) = 0,   ∀𝑘 ∵ 𝑈(𝑘, 0) = 0 

⟹𝑈(0,4) = 0,𝑈(1,4) = 0,𝑈(2,4) = 0, 𝑈(3,4) = 0,𝑈(4,4) = 0                                                       (21) 

Similarly, we get, 

𝑈(𝑘, 4) = 𝑈(𝑘, 5) = 0 𝑈(𝑘, 6) = 𝑈(𝑘, 7) = ⋯ . .= 0,     ∀𝑘 

We know that, 

𝑢(𝑥, 𝑡)  = ∑∑𝑈(𝑘, ℎ)𝑥𝑘𝑡ℎ
∞

ℎ=0

∞

𝑘=0

 

                 = {𝑈(0,0) + 𝑈(1,0)𝑥 + 𝑈(2,0)𝑥2 + 𝑈(3,0)𝑥3 +⋯ . }

+ {𝑈(0,1)𝑡 + 𝑈(1,1)𝑥𝑡 + 𝑈(2,1)𝑥2𝑡 + ⋯ . }

+ {𝑈(0,2)𝑡2 + 𝑈(1,2)𝑥𝑡2 + 𝑈(2,2)𝑥2𝑡2 +⋯} + ⋯. 

 

∴ 𝒖(𝒙, 𝒕) =  𝒙𝒕 

 

 

 

(Fig. 1.3: Graph of  𝒖(𝒙, 𝒕) = 𝒙𝒕  for  −𝟐𝟎 ≤ 𝒖 ≤ 𝟐𝟎; −𝟓 ≤ 𝒙 ≤ 𝟓 ;  𝟎 ≤ 𝒕 ≤ 𝟒) 

-5

0

5

0
1

2
3

4
5

-20

-10

0

10

20

 

x

t
 

u
(x

,t
)

-20

-10

0

10

20



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4. Conclusion 

In this paper, we proposed the Differential Transformation Method (DTM) has been successfully 

applied to find exact and approximate solution of the second order differential equations. The method 

was used in a direct way without using linearization, perturbation or restrictive assumptions. While, 

this equation solving Partial differential equations in two, three-dimensional linear and nonlinear, have 

been solved effectively using the Differential Transform Method.  Therefore, it is not affected by 

computation round off errors and one is not faced with the necessities of large computer memory and 

time. Moreover, this method provides a closed-series form solutions with the convergence region A 

specific advantage of this method is provided to exact and accurate solutions up to the 3rd dimensional 

DTM over any purely numerical method. It may be concluded that DTM is very powerful and efficient 

in finding analytical as well as numerical solutions for wide classes of differential equations. 

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