# Solutions of Nonlinear Integro-Partial Differential Equations by the Method ofG′/G,1/G

Authors: Daba Meshesha Gusu; Chala Bulo
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this article, a special expansion method is implemented in solving nonlinear integro-partial differential equations of2+1-dimensional using a special expansion method of G′/G,1/G. We obtained the solutions for 2+1-dimensional nonlinear integro-differential equations in real physical phenomena. The method is applied on 2+1-dimensional space time and solved in three different cases: hyperbolic, trigonometric, and rational functions. The obtained solutions for each result were illustrated by graphical plots using Wolfram Mathematica 9.0 software packages. Furthermore, the obtained results are exactly fit with exact solutions which solves the complicity of finding the solution for nonlinear integro-partial differential equations. Finally, the method is powerful and effective to solve partial differential equations of nonlinear integro form.


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## Body

## 1. Introduction

Nonlinear differential equations (NLDEs) are differential equations (DEs) that are not a linear equation in the unknown functions and their derivatives. Many models in mathematics and physics are described by nonlinear partial differential equations (NLPDEs). And they are widely used to describe complex phenomena in various fields of applied sciences, especially in physics and engineering. The investigation of searching solutions for nonlinear integro differential equations plays an important role in nonlinear physical science, because the solutions can describe various natural phenomena of the problems such as wave traveling, vibrations, solitons, and propagation with a finite speed. Different researchers studied the practical and theoretical investigations of applications of partial differential equations [1–5].Integro-differential equations (IDEs) are differential functional equations involving unknown functionfx together with both differential and integral operations on f. The integro-partial differential equation (IPDE) is an IDE such that the unknown function depends on more than one independent variables. The IPDE is divided into linear and nonlinear. We model real-life problems usually result in mathematical form as follows: functional equations, ordinary differential equations (ODEs) or partial differential equations (PDEs), integral or IDEs, and stochastic equations. Most mathematical modeling of physical phenomena contains IDEs [6–11]. The theory and application of integro-differential equations play an important role in many fields like engineering sciences, fluid dynamics, and nonlinear optics [12–14].Ito equation (IE) is one form of NLPDE which describes many branches of physics such as condensed matter physics, fluid dynamics, and optics. The KdV equation was investigated experimentally and theoretically to find its solution [15]. [16] discovered numerically that its solutions seemed to decompose at large times into a collection of solitons: well-separated solitary waves [17]. Ito [18, 19] obtained the well-known generalized 1+1-dimensional and generalized 2+1-dimensional Ito equations by generalization of the bilinear KdV equation. There are two essential types of solutions for NLPDEs, which are analytical (exact solution) and numerical solutions. The methods of solving exact solution of nonlinear integro-partial differential equations (NLIPDEs) are used in differential transform method [20], reliable method [21], Kudryashov method [22], generalized Kudryashov method [23], auxiliary equation method [24, 25], extended auxiliary equation method [26], G′/G-expansion method [27], inverse scattering method [28], modified simple equation method [29–31], G`/G,1/G-expansion method [32, 33], and extended simple equation method [34]. NLIPDEs can be observed in various scientific fields, such as plasma physics, optical fibers, fluid dynamics, and chemical physics [35]. The last decades have been witnessed for the discovery of a number of new techniques to solve the nonlinear differential system [36, 37]. Exact solutions for the nonlinear integro-partial differential equations have fundamental importance, since most of the complex phenomena are modeled mathematically by nonlinear integro-partial differential equations.The nonlinear evolution equations (NLEEs) play a key role in describing a scientific phenomena. Among these, the Korteweg-de Vries (KdV) equation models the shallow water wave dynamics near ocean shore and beaches, the nonlinear Schrodinger’s equation describes the dynamics of propagation of solitons through optical fibers, and the Schrodinger-Hirota equation describes the dispersive soliton propagation through optical fibers [35, 38–40]. The investigation of finding exact solution of many kinds of Ito equation has been conducted via different peoples and in different times [41–45].In some fields such as nuclear reactor dynamics and thermoelasticity, we need to reflect the effect memory of the systems in model. If such systems are modeled using partial differential equations, which involves functions at a given space and time, the effect of past history is ignored. Therefore, in order to incorporate the memory effect in such systems, an integral term in the basic partial differential equation is introduced and this leads to an integro-partial differential equation.The interaction process of two internal long waves is also described by Ito equation [19] which has the single-soliton solution and periodic solution of Equation (1) by using the Hirota bilinear method. Different scholars have been obtained different solutions. New exact solutions to the 2+1-dimensional Ito equation, extended homoclinic test technique, has been studied [46]. Moreover, [47] constructed the generalized solitary wave solutions by using the Exp-function method. Extended three-wave method for the 1+2-dimensional Ito equation has been studied by [48]. [46] investigated multiperiodic wave solutions of Equation (1) by using Riemann theta function, and [49] discussed the solutions having the nature of breathers, rogue waves, and solitary waves by applying the homoclinic breather limit method. Also, [50] has studied multiple kink solutions and multiple singular kink solutions for 2+1-dimensional nonlinear models generated by the Jaulent-Miodek hierarchy. Very recently, it has also been shown that there are diverse interaction solutions to Equation (1). Finding the exact solution of NLPDEs has advantages better than numerical solutions, since they can reduce the error term.The Ito equation is the Korteweg-de Vries (KdV) equation type equation and its bilinear transformation. The rolling behavior of ships in the regular sea and interaction process of two internal long waves are often predicted by Ito equation and gave the single-soliton solution and periodic solution of Equation (1) [19]. Generally, in this article, we consider NLPDE of 2+1-dimensional of the following form:
(1)Wtt+Wxxxt+32WxWt+WWxt+3Wxx∫∞xWtdx+αWyt+ωWxt=0,where Wx,y,t is an analytic function, ∂x−1=∫dx, and α and ω are two auxiliary constants.Ifα=0 and ω=0, then Equation (1) can be reduced to 1+1-dimensional Ito equation which was first proposed by [36].This article focuses on2+1-dimensional NLIPD equations when α and ω are different from zero.Then, by settingW=ux from Equation (1), we get the following:
(2)uxtt+uxxxxt+32uxxuxt+uxuxxt+3uxxx∫∞xuxtdx+αuxyt+ωuxxt=0.Large varieties of physical, chemical, and biological phenomena are governed by NLPDEs. One of the most exciting advances of nonlinear science and theoretical physics has been the development of methods to look for the solutions of NLPDEs. Finding the solutions to NLPDEs plays an important role in nonlinear science like nonlinear wave phenomena of dispersion, dissipation, diffusion, reaction, convection, and generally in nonlinear physical science, since they provide much physical information and more insight into the physical aspects of the problem and thus lead to further applications. Some NLPDE are Ito equation, Sawada-Kotera equation, Burgers-Fisher equation, KP-hierarchy equation, Jaulent–Miodek equation, etc. In this paper, we aim to solve NLPDE using the special expansion method and to illustrate the nature of the obtained solutions in geometrical models.


## 2. Mathematical Formulation and Results

### 2.1. Special Expansion Method

In this section, we explain the steps to be followed for the methods ofG′/G,1/G which is to determine the solutions of travelling waves of nonlinear physical phenomena. To apply the method of G′/G,1/G, we describe the main steps based on [32] as follows:Remark 1.
Let the second-order linear ordinary differential equations (ODEs) of the form(3)G′′ξ+λGξ=η.
Setϕ=G′/G and β=1/G, and then, we get
(4)ϕ′=−ϕ2+ηβ−λ,β′=−ϕβ.Remark 2.
Ifλ<0, then the general solution of Equation (3) has the following form:
(5)Gξ=A1sinhξ−λ+A2coshξ−λ+ηλ,where A1 and A2 are arbitrary constants.Consequently, we have(6)β2=−ϕ2λ+2ηβλ−λ2λ2σ+η2,where σ=A12−A22.Remark 3.
Ifλ>0, then the general solution of Equation (3) has the following form:
(7)Gξ=A1sinξλ+A2cosξλ+η2.Remark 4.
Ifλ=0, then the general solution of Equation (3) has the following form: G′′+λG=η which becomes G′′=η. Then, integrating both sides with respect to ξ, it becomes the following: G′=ηξ+A1, where A1 is constant of integration.
Again integrating with respect toξ, we get the following:
(8)Gξ=η2ξ2+A1ξ+A2,where A1 and A2 are constants.And subsequently, it establishes the relation betweenϕ and β.


#### 2.1.1. The Main Steps of Special Expansion Method

In this section, we describe the main steps of the special expansion methods by using two variablesG′/G,1/G expansion method.Step 1.
First, we change the given nonlinear integro-partial differential equation (NLIPDE) to nonlinear partial differential equations (NLPDEs).
Suppose we have the following NLPDEs in the form(9)Gu,ut,ux,uy,uxx,uyy,uxt,uxy,⋯=0,where G is a polynomial of unknown function u and its total derivatives with respect to ξ. And its various partial derivatives in which the highest order derivatives and nonlinear terms are involved.Step 2.
We use wave transformation to convert NLPDEs to ODEs:(10)ux,y,t=Uξ,ξ=x+y−kt,where ω is arbitrary constant.
Apply the traveling wave transformation Equation (10) and integrate the resulting equation with respect to ξ as many as possible. And hence, Equation (9) is reduced to the following ODEs:
(11)PU,U′,U′′,U′′′,U′′′′,⋯=0.Step 3.
Assume that the solution of Equation (11) can be expressed by a polynomial in the two variables ϕ and β as follows:
(12)Uξ=∑i=0Nciϕi+∑i=1Nbiϕi−1β,where cii=0,1,2,3,⋯N and bii=1,2,3,⋯,N are arbitrary constants. And ϕ and β satisfy the following condition.
(13)ϕξ=G′ξGξandβξ=1Gξ.Step 4.
Determine the positive integerN in Equation (12) by using the homogeneous balance equation between the highest order derivatives and the nonlinear term appearing in Equation (11). More precisely, if the degree of Uξ is deg Uξ=N, then the degree of other term will be expressed as follows [51]:
(14)degdqUξdξq=N+qanddegUξpdrUξdξrs=Np+sN+r,where N is the degree of the highest order of the function, q is the order of the highest order, p is the degree of coefficient of nonlinear term, s is the degree of nonlinear term, and r is the order of nonlinear term which can be expressed as follows:
(15)N+q=Np+sN+r.Step 5.
From the above steps, we get the value ofN and applying Equation (12), we obtain the form of the solution of Equation (11). Then, substitute Equation (12) into Equation (11) along with Equation (4); the left-hand side Equation (11) can be converted into a polynomial in ϕ and β in which the degree of β is not longer than 1. Equating each coefficients of this polynomial to zero yields a system of algebraic equation which can be solved by using the Wolfram Mathematica 9.0 software package to get the values ci, bi, k, and η. Proceeding to these steps, we will concentrate on the following applications.


### 2.2. Application of Special Expansion Method on2+1-Dimensional Ito Equation

From the traveling wave transformation of Equation (10), we have the following:
(16)ux,y,t=Uξ,ξ=x+y−kt.Inserting Equation (10) into Equation (2), we get the following ODE form:
(17)k−α−ωU′−U3−3U′2+d=0,where d is constant of integration.Let settingU′=V; then, Equation (17) becomes
(18)k−α−ωV−V′′−3V2+d=0.Balancing the highest order derivativeV′′ with the nonlinear term V2 of Equation (18) by Equation (14), we get the following:
(19)N+2=N×0+2×N+0,and we get N=2.Consequently, the solution of Equation (18) has the following form:
(20)Vξ=∑i=02ciϕi+∑i=12biϕi−1β=co+c1ϕ+b1β+c2ϕ2+b2ϕβ,wherec0,c1,c2,b1,b2 are constants.To solve for Equation (18), first we have to find V, V′, V′′, and V2 from Equation (20). We have Vξ=co+c1ϕ+b1β+c2ϕ2+b2ϕβ. Now, by using Equation (4), we obtain
(21)V′=−c1ϕ2+c1ηβ−c1λ−b1ϕβ−2c2ϕ3+2c2ϕηβ−2c2ϕλ−b2βϕ2+b2ηβ2−b2λβ+b2ϕ2β.Differentiating Equation (21), we get the following:
(22)V′′=−2c1ϕ−ϕ2+ηβ−λ+c1η−ϕβ−b1β−ϕ2+ηβ−λ−b1ϕ−ϕβ−6c2ϕ2−ϕ2+ηβ−λ+2c2ηβ−ϕ2+ηβ−λ+2c2ηϕ−ϕβ−2c2λ−ϕ2+ηβ−λ−2b2βϕ−ϕ2+ηβ−λ−b2ϕ2−ϕβ+2b2ηβ−ϕβ−b2λ−ϕβ+2b2βϕ−ϕ2+ηβ−λ+b2ϕ2−ϕβ.Hence, we get(23)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1β2η+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2β2η2−4c2ηβλ+2c2λ2−2b2ϕηβ2+b2βϕλ.From Equation (20),
(24)V2=b12β2+2ϕb1b2β2+ϕ2b22β2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.


#### 2.2.1. Case I: Ifλ<0 (Hyperbolic Function Solutions)

Substituting Equation (6) into Equation (23), we get the following:
(25)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1η−ϕ2λ+2ηβλ−λ2λ2σ+η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2−ϕ2λ+2ηβλ−λ2λ2σ+η2−2c2ηβλ+2c2λ2−2b2ϕη−ϕ2λ+2ηβλ−λ2λ2σ+η2+b2βϕλ=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2+b1ηϕ2λλ2σ+η2−2b1η2βλλ2σ+η2+b1ηλ2λ2σ+η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ−2c2η2ϕ2λλ2σ+η2+4c2η3βλλ2σ+η2−2c2η2λ2λ2σ+η2−2c2ηβλ+2c2λ2+2b2ηϕ3λλ2σ+η2−4b2ϕη2βλλ2σ+η2+2b2ϕηλ2λ2σ+η2+b2βϕλ.Again, putting Equation (6) into Equation (24), we obtain the following:
(26)V2=−b12ϕ2λλ2σ+η2+2b12ηβλλ2σ+η2−b12λ2λ2σ+η2−2b1b2ϕ3λλ2σ+η2+4ϕb1b2ηβλλ2σ+η2−2b1b2ϕλ2λ2σ+η2−ϕ4b22λλ2σ+η2+2ϕ2b22ηβλλ2σ+η2−ϕ2b22λ2λ2σ+η2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.By using Equations (20), (25), and (26), we obtain Equation (18) equals to
(27)cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3+3c1ηβϕ−2c1λϕ−2b1βϕ2−b1ηϕ2λλ2σ+η2+2b1η2βλλ2σ+η2−b1ηλ2λ2σ+η2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−8c2ϕ2λ−2c2η2ϕ2λλ2σ+η2−4c2η3βλλ2σ+η2+2c2η2λ2λ2σ+η2+2c2ηβλ−2c2λ2−2b2ηϕ3λλ2σ+η2+4b2ϕη2βλλ2σ+η2−2b2ϕηλ2λ2σ+η2−b2βϕλ+3b12ϕ2λλ2σ+η2−6b12ηβλλ2σ+η2+3b12λ2λ2σ+η2+6b1b2ϕ3λλ2σ+η2−12ϕb1b2ηβλλ2σ+η2+6b1b2ϕλ2λ2σ+η2+3ϕ4b22λλ2σ+η2−6ϕ2b22ηβλλ2σ+η2+3ϕ2b22λ2λ2σ+η2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc0c1−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22+d=0.Setting the coefficients of polynomialϕ and β zero from Equation (27), we obtain the following sets of algebraic equations:
(28)ϕ4:−6c2−3c22+3b22λλ2σ+η2=0,ϕ3β:−6b2c2=0,ϕ3:−2c2−6c2c1−2b2ηλλ2σ+η2+6b1b2λλ2σ+η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1−6b22ηλλ2σ+η2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2λλ2σ+η2+3b12λλ2σ+η2+3b22λ2λ2σ+η2−b1ηλλ2σ+η2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1+4b2η2λλ2σ+η2−12b2b1ηλλ2σ+η2=0,ϕ:c1k−α−ω−2c1λ−6c0c1−2b2ηλ2λ2σ+η2+6b1b2λ2λ2σ+η2,β:b1k−α−ω−b1λ−6b1c0+2c2ηλ+2b1η2λλ2σ+η2−4η3c2λλ2σ+η2−6ηλb12λ2σ+η2,ϕ0:c0k−α−ω+2c2λ2−3c02−b1ηλ2λ2σ+η2+2c2η2λ2λ2σ+η2+3b12λ2λ2σ+η2+d=0.Now, by solving the above algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following sets of results.Result 1(29)c0=16−13λ±97λ2−12d,c1=0,c2=−2,b1=±i10σλ=2η,b2=0,k=α±97λ2−12d+ω,η=±i52σλ.Now by using the above values of variables and putting into Equation (20), we get the following:
(30)V11=16−13λ±97λ2−12d+2ηβ−2ϕ2.Then, by using Equation (13), the solution of Equation (2) is as follows:
(31)V11=16−13λ±97λ2−12d+2η1G−2G′G2.But, when we put the value ofG from Equation (5), we get the following:
(32)V11=16−13λ±97λ2−12d+2ηA1sinhξ−λ+A2coshξ−λ+η/λ−2A1sinhξ−λ+A2coshξ−λ+η/λ′A1sinhξ−λ+A2coshξ−λ+η/λ2.Then, the exact solution of2+1-dimensional Ito equation of the form (2) is as follows:
(33)V11=16−13λ±97λ2−12d+2ηA1sinhξ−λ+A2coshξ−λ+η/λ+2λA1coshξ−λ+A2sinhξ−λ2A1sinhξ−λ+A2coshξ−λ+η/λ2.By lettingA1=0, η=0, and A2>0 in Equation (33), the particular solution of Equation (2) is as follows:
(34)V11a=16−13λ±97λ2−12d+2λtanh2ξ−λ.Then, by lettingλ=−0.3, d=0.4, t=0, and y=3, the graph of Equation (34) has the following forms.From Figure1, we see the solution of 2+1-dimensional Ito equation with respect to the result 1 and λ=−0.3, d=0.4, t=0, y=3, A1=0, η=0, and A2>0, and we obtain y≤0, when −10≤x≤10.Figure 1
Graphical solution of Equation (34) when k=α±97λ2−12d+ω in 3D and 2D, respectively.Again lettingA2=0, η=0, and A1>0 in Equation (34), we get another particular solution of Equation (2) as follows:
(35)V11b=16−13λ±97λ2−12d+2λcoth2ξ−λ.Then, by lettingλ=−0.3, d=0.4, t=0, and y=3, the graph of solution of Equation (35) has the following forms.Figure2 of the solution (35) concludes the solution of 2+1-dimensional Ito equation of result 1 when λ=−0.3, d=0.4, t=0, y=3, A2=0, η=0, A1>0, and for −10≤x≤10, it has no solution in real number on the domain y>3, and the solution is always y<3 in the range of −10≤x≤10.Figure 2
Graphical solution of Equation (35) when k=α±97λ2−12d+ω in 3D and 2D, respectively.Result 2(36)c0=13−4λ±28λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±228λ2−3d,η=0.Now by using the above values of variables and putting into Equation (20), we get the following:
(37)V1,1ξ=13−4λ±28λ2−3d−2ϕ2.Then, by using Equation (13), the solution of Equation (2) has the form 1/3−4λ±28λ2−3d−2G′/G2. Now, when we put the value of G from Equation (5), we get the following:
(38)V12=13−4λ±28λ2−3d−2A1−λcoshξ−λ+A2−λsinhξ−λA1sinhξ−λ+A2coshξ−λ2.Then, the exact solution of2+1-dimensional Ito equation of (2), if λ<0 with respect to result 2 based Equation (36), is as follows:
(39)13−4λ±28λ2−3d−2A1−λcoshξ−λ+A2−λsinhξ−λ2A1sinhξ−λ+A2coshξ−λ2.By lettingA1=0 and A2>0 in Equation (39), the particular solution of Equation (2) is as follows:
(40)V12a=13−4λ±28λ2−3d−2A2−λsinhξ−λ2A2coshξ−λ2.Then, the general solution of2+1-dimensional Ito equation for hyperbolic function with respect to result on Equation (36), when A1=0 and A2>0, has the form of
(41)=13−4λ±28λ2−3d+2λtanh2ξ−λ.The graph of the particular solution whenλ=−0.3, d=0.4, t=0, and y=3 is as follows.From Figure3, the solution of 2+1-dimensional Ito equation with respect to result 2 and λ=−0.3, d=0.4, t=0, y=3, A1=0, η=0, and A2>0 implies that condensed form around x=−3.Figure 3
Graphical solution of Equation (41) when k=α±228λ2−3d in 3D and 2D, respectively.Again lettingA2=0 and A1>0 in Equation (39), we get another particular solution of Equation (2) as follows:
(42)V12b=13−4λ±28λ2−3d−2A1−λcoshξ−λ2A1sinhξ−λ2.Then, the general solution of2+1-dimensional Ito equation for hyperbolic function with respect to result on Equation (36) when A2=0 and A1>0 has the form of
(43)V12b=13−4λ±28λ2−3d+2λcoth2ξ−λ.The graph of the particular solution whenλ=−0.3, d=0.4, t=0, and y=0 as follows.From Figure4 above, we see the solution of Equation (43) when λ=−0.3, d=0.4, t=0, y=3, A2=0, and A1>0, −10≤x≤10 has the solution between −10≤x≤10, and it bends downward.Figure 4
Graphical solution of Equation (43) when k=α±228λ2−3d in 3D and 2D, respectively.


#### 2.2.2. Case II: Ifλ>0 (Trigonometric Function Solutions)

Substituting (6) into (23), we get
(44)V′′=2c1ϕ3−2c1ηβϕ+2c1λϕ−c1ηϕβ+2b1βϕ2−b1ηϕ2λλ2σ−η2+2b1η2βλλ2σ−η2−b1ηλ2λ2σ−η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2ϕ2λλ2σ−η2−4c2η3βλλ2σ−η2+2c2η2λ2λ2σ−η2−4c2ηβλ+2c2λ2−2b2ηϕ3λλ2σ−η2+4b2ϕη2βλλ2σ−η2−2b2ϕηλ2λ2σ−η2+b2λϕβ.Using Equation (6), Equation (24) becomes
(45)V2=b12ϕ2λλ2σ−η2−2b12ηβλλ2σ−η2+b12λ2λ2σ−η2+2b1b2λϕ3λ2σ−η2−4b1b2ϕηβλλ2σ−η2+2ϕb1b2λ2λ2σ−η2+ϕ4b22λλ2σ−η2+2b22ηβλϕ2λ2σ−η2+ϕ2b22λ2λ2σ−η2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.Again from Equations (44), (45), and (20), Equation (18) becomes
(46)cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3−2c1λϕ+3c1ηϕβ−2b1βϕ2+b1ηϕ2λλ2σ−η2−2b1η2βλλ2σ−η2+b1ηλ2λ2σ−η2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−6c2ϕ2λ−2c2η2ϕ2λλ2σ−η2+4c2η3βλλ2σ−η2−2c2η2λ2λ2σ−η2+4c2ηβλ−2c2λϕ2−2c2λ2+2b2ηϕ3λλ2σ−η2−4b2ϕβη2λλ2σ−η2+2b2ϕηλ2λ2σ−η2−b2λϕβ−3b12ϕ2λλ2σ−η2+6b12ηβλλ2σ−η2−3b12λ2λ2σ−η2−6b1b2λϕ3λ2σ−η2+12b1b2ϕηβλλ2σ−η2−6ϕb1b2λ2λ2σ−η2−3ϕ4b22λλ2σ−η2−6b22ηβλϕ2λ2σ−η2−3ϕ2b22λ2λ2σ−η2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc0c1−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22+d=0.Equating the coefficients of the powers ofβ and ϕ to be zero, we obtain the following system of algebraic equations:
(47)ϕ4:−6c2−3c22−3b22λλ2σ−η2=0,ϕ3β:−6b2c2=0,ϕ3:−2c1−6c2c1+2b2ηλλ2σ−η2−6b1b2λλ2σ−η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1−6b22ηλλ2σ−η2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2λλ2σ−η2−3b12λλ2σ−η2−3b22λ2λ2σ−η2+b1ηλλ2σ−η2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1+4b2η2λλ2σ−η2+12b2b1ηλλ2σ−η2=0,ϕ:c1k−α−ω−2c1λ−6c0c1+2b2ηλ2λ2σ−η2−6ϕb1b2λ2λ2σ−η2,β:b1k−α−ω−b1λ−6b1c0+4c2ηλ−2b1η2λλ2σ−η2+4η3c2λλ2σ−η2−6ηλb12λ2σ+η2=0,ϕ0:c0k−α−ω−2c2λ2−3c02+b1ηλ2λ2σ−η2−2c2η2λ2λ2σ−η2−3b12λ2λ2σ−η2+d=0.Then, by solving the above algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following results.Result 3(48)c0=1613λ±−12d+47λ2,c1=0,c2=−2,b1=±λ7σ=2η,b2=0,k=±−12d+47λ2+ω,η=±12λ7σ.Setting the above values of variables into Equation (20), we can get the following form of solution:
(49)Vξ=1613λ±−12d+47λ2+2ηβ−2ϕ2.Then, by using Equation (13), the solution of Equation (2) is a follows:Then, the exact solution of2+1-dimensional Ito equation of the form Equation (2) is as follows:
(50)V21=1613λ±−12d+47λ2+2ηA1sinξλ+A2cosξλ+η/2+2A1λcosξλ−A2λsinξλ2A1sinξλ+A2cosξλ+η/22.By lettingA1=0, η=0, and A2>0 in Equation (50), the particular solution of Equation (2) is
(51)V21a=1613λ±−12d+47λ2+2λtan2ξλ.Then, whenλ=0.7, d=0.4, and y=3, the graph of Equation (51) has the following forms:Figure5 shows the exact solution of 2+1-dimensional Ito equation with respect to result 3 when A1=0, η=0, A2>0, λ=0.7, d=0.4, y=3, and −10≤x≤10. This figure implies that the solutions are upward parabolic, and it repeated above x-axis.Figure 5
Graphical solution of Equation (51) when ±−12d+47λ2+ω in 3D and 2D, respectively.Again lettingA2=0, η=0, and A1>0 in Equation (50), we get another particular solution of Equation (6) as follows:
(52)V21b=1613λ±−12d+47λ2+2λcot2ξλ.Then, the graph of Equation (52) has the following form:From the above Figure6, we see that it is the exact solution of 2+1-dimensional Ito equation with respect to result 3 of Equation (52), when A1>0, η=0, A2=0, λ=0.7, d=0.4, y=3, and −10≤x≤10. And the result of the solution shows it has the form a parabolic upward and crosses the x-axis.Figure 6
Graphical solution of Equation (52) when ±−12d+47λ2+ω in 3D and 2D, respectively.Result 4(53)c0=13−4λ±4λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±4λ2−3d+ω,η=0.Then, setting the above values of variables into Equation (20), we can get the following form of solution for Ito equation:
(54)Vξ=13−4λ±4λ2−3d−2ϕ2.Then, by using Equation (13) for ϕ, the solution of Equation (2) is a follows:
(55)Vξ=13−4λ±4λ2−3d−2G′G2.But, when we put the value ofF of the form Equation (7), we get the following:
(56)Vξ=13−4λ±4λ2−3d−2A1sinξλ+A2cosξλ′A1sinξλ+A2cosξλ2.Then, finally we get the exact solution of2+1-dimensional Ito equation with respect to the second case result of the form (2) as follows:
(57)V22=13−4λ±4λ2−3d−2A1λcosξλ−A2λsinξλ2A1sinξλ+A2cosξλ2.By lettingA1=0 and A2>0 in Equation (57), the particular solution of Equation (2) is as follows:
(58)V22a=13−4λ±4λ2−3d−2λtan2ξλ.Figure7 shows the exact solution of 2+1-dimensional Ito equation with respect to result 4 when A1=0, A2>0, η=0, λ=0.7, d=0.4, y=3, and at −10≤x≤10, and it makes parabolic bend downward on y-axis.Figure 7
Graphical solution of Equation (58) when k=α±4λ2−3d+ω in 3D and 2D, respectively.Again lettingA2=0 and A1>0 in Equation (57), we get another particular solution of Equation (2) as follows:
(59)V22b=13−4λ±4λ2−3d−2A1λcosξλ2A1sinξλ2,V22b=13−4λ±4λ2−3d−2λcot2ξλ.Figure8 shows the exact solution of 2+1-dimension Ito equation with respect to result 4 when A2=0, A1>0, η=0, λ=0.7, d=0.4, y=3, and at −10≤x≤10, and it lies always below x-axis in the form of parabolic.Figure 8
Graphical solution of (44) when k= in 3D and 2D, respectively.


#### 2.2.3. Case III: Whenλ=0 (Rational Function Solutions)

Substituting Equation (6) into Equation (23), we get
(60)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1ϕ2A12−2ηA2+2b1ηβA12−2ηA2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2ϕ2A12−2ηA2−4c2η3βA12−2ηA2−4c2ηβλ+2c2λ2−2b2ηϕ3A12−2ηA2+4b2ϕη2βA12−2ηA2+b2λϕβ.Again also substituting Equation (6) into Equation (24), we get
(61)V2=b12ϕ2−2ηβA12−2ηA2+2ϕb1b2ϕ2−2ηβA12−2ηA2+ϕ2b22ϕ2−2ηβA12−2ηA2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc1c0+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22(62)=b12ϕ2A12−2ηA2−2b12ηβA12−2ηA2+2b1b2ϕ3A12−2ηA2−4ϕb1b2ηβA12−2ηA2+ϕ4b22A12−2ηA2−2ϕ2b22ηβA12−2ηA2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc1c0+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.From Equation (20), (60), and (62), Equation (18) becomes
(63)Vξ=cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3+3c1ηβϕ−2c1λϕ−2b1βϕ2+b1ηϕ2A12−2ηA2−2b1ηβA12−2ηA2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−8c2ϕ2λ−2c2η2ϕ2A12−2ηA2+4c2η3βA12−2ηA2+4c2ηβλ−2c2λ2+2b2ηϕ3A12−2ηA2−4b2ϕη2βA12−2ηA2−b2λϕβ−3b12ϕ2A12−2ηA2+6b12ηβA12−2ηA2−6b1b2ϕ3A12−2ηA2+12ϕb1b2ηβA12−2ηA2−3ϕ4b22A12−2ηA2+6ϕ2b22ηβA12−2ηA2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc1c0−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22.Then, by setting the coefficients of the polynomial to be zero, we obtain the following sets of algebraic.(64)ϕ4:−6c2−3c22−3b22A12−2ηA2=0,ϕ3β:−6b2c2=0,ϕ3:−2c1−6c2c1+2b2ηA12−2ηA2−6b1b2λ2σ−η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1+6b22ηA12−2ηA2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2A12−2ηA2−3b12A12−2ηA2+b1ηA12−2ηA2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1−4b2η2A12−2ηA2+12b2b1ηA12−2ηA2=0,ϕ:c1k−α−ω−2c1λ−6c0c1=0,β:b1k−α−ω−b1λ−6b1c0+4c2ηλ−2b1ηA12−2ηA2+4η3c2A12−2ηA2+6ηb12A12−2ηA2,ϕ0:c0k−α−ω−2c2λ2−3c02+d=0.Then, by solving previous algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following results.Result 5(65)c0=13−4λ±4λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±24λ2−3d+ω,η=0.Then, setting Equation (65) values of variables into Equation (20), we can get the following form of solution for Ito equation:
(66)V31=13−4λ±4λ2−3d−2η/2ξ2+A1ξ+A2′η/2ξ2+A1ξ+A22.When the value ofη=0, then the solution of Equation (66) is as follows:
(67)V31=13−4λ±4λ2−3d−2A1A1ξ+A22.Finally, we get the exact solution of2+1-dimensional space time equation with respect to the first result of case 3 in Equation (65) as follows:
(68)V31=13−4λ±4λ2−3d−2A12A1ξ+A22.Whenλ=0, we have
(69)V31=±−3d−2A12A1ξ+A22.Using the parametersd=−0.5, A1=1, A2=2, and t=0, the graph of Equation (69) is as follows:Figure9 represents the exact solution of 2+1-dimensional space time with respect to the parameters d=−0.5, A1=1, A2=2, t=0, and y=3 which has maximum limit on the y-axis.Figure 9
Graphical solution of (69) when k=α±24λ2−3d+ω in 3D and 2D, respectively.Result 6(70)c0=16−11λ±73λ2−12d,c1=0,c2=−2,b1=2λA2±λλA22−A12=2η,(71)b2=0,k=αA12±73λ2−12dA12+ωA12A12,η=λA2±λλA22−A12.Then, setting the Equation (70) values of variables into Equation (20), we can get the following form of solution for Ito equation:
(72)V32=16−11λ±73λ2−12d+2ηβ−2ϕ2.Then by using Equation (13) for ϕ and β, the simplification of Equation (72) leads to
(73)V32=16−11λ±73λ2−12d+2η1G−2G′G2.When we put the value ofG from Equation (8) in Equation (73),
(74)V32=16−11λ±73λ2−12d+2ηη/2ξ2+A1ξ+A2−2ηξ+A1η/2ξ2+A1ξ+A22.Hence, we get the exact solution of2+1-dimensional space time with respect to result of (70) as follows:
(75)V32=16−11λ±73λ2−12d+2ηη/2ξ2+A1ξ+A2−2ηξ+A12η/2ξ2+A1ξ+A22.By lettingη=0, A1>0, and A2>0 in Equation (75), the particular solution of Equation (2) with respect to (70) is as follows:
(76)V32a=16−11λ±73λ2−12d−2A12A1ξ+A22.Whenλ=0, we get
(77)V32a=±−12d−2A12A1ξ+A22.Then, by settingd=−0.5, A1=1, A2=2, and t=0, the graph of (77) is a follows:Figure10 reveals the exact solution of 2+1-dimensional space time with respect to result 6 when the parameters d=−0.5, A1=1, A2=2, and t=0 and in similar way as figure 9 fixes its maximum from below on y-axis.Figure 10
Graphical solution of (77) when k=αA12±73λ2−12dA12+ωA12/A12 in 3D and 2D, respectively.


## 2.1. Special Expansion Method

In this section, we explain the steps to be followed for the methods ofG′/G,1/G which is to determine the solutions of travelling waves of nonlinear physical phenomena. To apply the method of G′/G,1/G, we describe the main steps based on [32] as follows:Remark 1.
Let the second-order linear ordinary differential equations (ODEs) of the form(3)G′′ξ+λGξ=η.
Setϕ=G′/G and β=1/G, and then, we get
(4)ϕ′=−ϕ2+ηβ−λ,β′=−ϕβ.Remark 2.
Ifλ<0, then the general solution of Equation (3) has the following form:
(5)Gξ=A1sinhξ−λ+A2coshξ−λ+ηλ,where A1 and A2 are arbitrary constants.Consequently, we have(6)β2=−ϕ2λ+2ηβλ−λ2λ2σ+η2,where σ=A12−A22.Remark 3.
Ifλ>0, then the general solution of Equation (3) has the following form:
(7)Gξ=A1sinξλ+A2cosξλ+η2.Remark 4.
Ifλ=0, then the general solution of Equation (3) has the following form: G′′+λG=η which becomes G′′=η. Then, integrating both sides with respect to ξ, it becomes the following: G′=ηξ+A1, where A1 is constant of integration.
Again integrating with respect toξ, we get the following:
(8)Gξ=η2ξ2+A1ξ+A2,where A1 and A2 are constants.And subsequently, it establishes the relation betweenϕ and β.


### 2.1.1. The Main Steps of Special Expansion Method

In this section, we describe the main steps of the special expansion methods by using two variablesG′/G,1/G expansion method.Step 1.
First, we change the given nonlinear integro-partial differential equation (NLIPDE) to nonlinear partial differential equations (NLPDEs).
Suppose we have the following NLPDEs in the form(9)Gu,ut,ux,uy,uxx,uyy,uxt,uxy,⋯=0,where G is a polynomial of unknown function u and its total derivatives with respect to ξ. And its various partial derivatives in which the highest order derivatives and nonlinear terms are involved.Step 2.
We use wave transformation to convert NLPDEs to ODEs:(10)ux,y,t=Uξ,ξ=x+y−kt,where ω is arbitrary constant.
Apply the traveling wave transformation Equation (10) and integrate the resulting equation with respect to ξ as many as possible. And hence, Equation (9) is reduced to the following ODEs:
(11)PU,U′,U′′,U′′′,U′′′′,⋯=0.Step 3.
Assume that the solution of Equation (11) can be expressed by a polynomial in the two variables ϕ and β as follows:
(12)Uξ=∑i=0Nciϕi+∑i=1Nbiϕi−1β,where cii=0,1,2,3,⋯N and bii=1,2,3,⋯,N are arbitrary constants. And ϕ and β satisfy the following condition.
(13)ϕξ=G′ξGξandβξ=1Gξ.Step 4.
Determine the positive integerN in Equation (12) by using the homogeneous balance equation between the highest order derivatives and the nonlinear term appearing in Equation (11). More precisely, if the degree of Uξ is deg Uξ=N, then the degree of other term will be expressed as follows [51]:
(14)degdqUξdξq=N+qanddegUξpdrUξdξrs=Np+sN+r,where N is the degree of the highest order of the function, q is the order of the highest order, p is the degree of coefficient of nonlinear term, s is the degree of nonlinear term, and r is the order of nonlinear term which can be expressed as follows:
(15)N+q=Np+sN+r.Step 5.
From the above steps, we get the value ofN and applying Equation (12), we obtain the form of the solution of Equation (11). Then, substitute Equation (12) into Equation (11) along with Equation (4); the left-hand side Equation (11) can be converted into a polynomial in ϕ and β in which the degree of β is not longer than 1. Equating each coefficients of this polynomial to zero yields a system of algebraic equation which can be solved by using the Wolfram Mathematica 9.0 software package to get the values ci, bi, k, and η. Proceeding to these steps, we will concentrate on the following applications.


## 2.1.1. The Main Steps of Special Expansion Method

In this section, we describe the main steps of the special expansion methods by using two variablesG′/G,1/G expansion method.Step 1.
First, we change the given nonlinear integro-partial differential equation (NLIPDE) to nonlinear partial differential equations (NLPDEs).
Suppose we have the following NLPDEs in the form(9)Gu,ut,ux,uy,uxx,uyy,uxt,uxy,⋯=0,where G is a polynomial of unknown function u and its total derivatives with respect to ξ. And its various partial derivatives in which the highest order derivatives and nonlinear terms are involved.Step 2.
We use wave transformation to convert NLPDEs to ODEs:(10)ux,y,t=Uξ,ξ=x+y−kt,where ω is arbitrary constant.
Apply the traveling wave transformation Equation (10) and integrate the resulting equation with respect to ξ as many as possible. And hence, Equation (9) is reduced to the following ODEs:
(11)PU,U′,U′′,U′′′,U′′′′,⋯=0.Step 3.
Assume that the solution of Equation (11) can be expressed by a polynomial in the two variables ϕ and β as follows:
(12)Uξ=∑i=0Nciϕi+∑i=1Nbiϕi−1β,where cii=0,1,2,3,⋯N and bii=1,2,3,⋯,N are arbitrary constants. And ϕ and β satisfy the following condition.
(13)ϕξ=G′ξGξandβξ=1Gξ.Step 4.
Determine the positive integerN in Equation (12) by using the homogeneous balance equation between the highest order derivatives and the nonlinear term appearing in Equation (11). More precisely, if the degree of Uξ is deg Uξ=N, then the degree of other term will be expressed as follows [51]:
(14)degdqUξdξq=N+qanddegUξpdrUξdξrs=Np+sN+r,where N is the degree of the highest order of the function, q is the order of the highest order, p is the degree of coefficient of nonlinear term, s is the degree of nonlinear term, and r is the order of nonlinear term which can be expressed as follows:
(15)N+q=Np+sN+r.Step 5.
From the above steps, we get the value ofN and applying Equation (12), we obtain the form of the solution of Equation (11). Then, substitute Equation (12) into Equation (11) along with Equation (4); the left-hand side Equation (11) can be converted into a polynomial in ϕ and β in which the degree of β is not longer than 1. Equating each coefficients of this polynomial to zero yields a system of algebraic equation which can be solved by using the Wolfram Mathematica 9.0 software package to get the values ci, bi, k, and η. Proceeding to these steps, we will concentrate on the following applications.


## 2.2. Application of Special Expansion Method on2+1-Dimensional Ito Equation

From the traveling wave transformation of Equation (10), we have the following:
(16)ux,y,t=Uξ,ξ=x+y−kt.Inserting Equation (10) into Equation (2), we get the following ODE form:
(17)k−α−ωU′−U3−3U′2+d=0,where d is constant of integration.Let settingU′=V; then, Equation (17) becomes
(18)k−α−ωV−V′′−3V2+d=0.Balancing the highest order derivativeV′′ with the nonlinear term V2 of Equation (18) by Equation (14), we get the following:
(19)N+2=N×0+2×N+0,and we get N=2.Consequently, the solution of Equation (18) has the following form:
(20)Vξ=∑i=02ciϕi+∑i=12biϕi−1β=co+c1ϕ+b1β+c2ϕ2+b2ϕβ,wherec0,c1,c2,b1,b2 are constants.To solve for Equation (18), first we have to find V, V′, V′′, and V2 from Equation (20). We have Vξ=co+c1ϕ+b1β+c2ϕ2+b2ϕβ. Now, by using Equation (4), we obtain
(21)V′=−c1ϕ2+c1ηβ−c1λ−b1ϕβ−2c2ϕ3+2c2ϕηβ−2c2ϕλ−b2βϕ2+b2ηβ2−b2λβ+b2ϕ2β.Differentiating Equation (21), we get the following:
(22)V′′=−2c1ϕ−ϕ2+ηβ−λ+c1η−ϕβ−b1β−ϕ2+ηβ−λ−b1ϕ−ϕβ−6c2ϕ2−ϕ2+ηβ−λ+2c2ηβ−ϕ2+ηβ−λ+2c2ηϕ−ϕβ−2c2λ−ϕ2+ηβ−λ−2b2βϕ−ϕ2+ηβ−λ−b2ϕ2−ϕβ+2b2ηβ−ϕβ−b2λ−ϕβ+2b2βϕ−ϕ2+ηβ−λ+b2ϕ2−ϕβ.Hence, we get(23)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1β2η+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2β2η2−4c2ηβλ+2c2λ2−2b2ϕηβ2+b2βϕλ.From Equation (20),
(24)V2=b12β2+2ϕb1b2β2+ϕ2b22β2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.


### 2.2.1. Case I: Ifλ<0 (Hyperbolic Function Solutions)

Substituting Equation (6) into Equation (23), we get the following:
(25)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1η−ϕ2λ+2ηβλ−λ2λ2σ+η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2−ϕ2λ+2ηβλ−λ2λ2σ+η2−2c2ηβλ+2c2λ2−2b2ϕη−ϕ2λ+2ηβλ−λ2λ2σ+η2+b2βϕλ=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2+b1ηϕ2λλ2σ+η2−2b1η2βλλ2σ+η2+b1ηλ2λ2σ+η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ−2c2η2ϕ2λλ2σ+η2+4c2η3βλλ2σ+η2−2c2η2λ2λ2σ+η2−2c2ηβλ+2c2λ2+2b2ηϕ3λλ2σ+η2−4b2ϕη2βλλ2σ+η2+2b2ϕηλ2λ2σ+η2+b2βϕλ.Again, putting Equation (6) into Equation (24), we obtain the following:
(26)V2=−b12ϕ2λλ2σ+η2+2b12ηβλλ2σ+η2−b12λ2λ2σ+η2−2b1b2ϕ3λλ2σ+η2+4ϕb1b2ηβλλ2σ+η2−2b1b2ϕλ2λ2σ+η2−ϕ4b22λλ2σ+η2+2ϕ2b22ηβλλ2σ+η2−ϕ2b22λ2λ2σ+η2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.By using Equations (20), (25), and (26), we obtain Equation (18) equals to
(27)cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3+3c1ηβϕ−2c1λϕ−2b1βϕ2−b1ηϕ2λλ2σ+η2+2b1η2βλλ2σ+η2−b1ηλ2λ2σ+η2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−8c2ϕ2λ−2c2η2ϕ2λλ2σ+η2−4c2η3βλλ2σ+η2+2c2η2λ2λ2σ+η2+2c2ηβλ−2c2λ2−2b2ηϕ3λλ2σ+η2+4b2ϕη2βλλ2σ+η2−2b2ϕηλ2λ2σ+η2−b2βϕλ+3b12ϕ2λλ2σ+η2−6b12ηβλλ2σ+η2+3b12λ2λ2σ+η2+6b1b2ϕ3λλ2σ+η2−12ϕb1b2ηβλλ2σ+η2+6b1b2ϕλ2λ2σ+η2+3ϕ4b22λλ2σ+η2−6ϕ2b22ηβλλ2σ+η2+3ϕ2b22λ2λ2σ+η2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc0c1−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22+d=0.Setting the coefficients of polynomialϕ and β zero from Equation (27), we obtain the following sets of algebraic equations:
(28)ϕ4:−6c2−3c22+3b22λλ2σ+η2=0,ϕ3β:−6b2c2=0,ϕ3:−2c2−6c2c1−2b2ηλλ2σ+η2+6b1b2λλ2σ+η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1−6b22ηλλ2σ+η2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2λλ2σ+η2+3b12λλ2σ+η2+3b22λ2λ2σ+η2−b1ηλλ2σ+η2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1+4b2η2λλ2σ+η2−12b2b1ηλλ2σ+η2=0,ϕ:c1k−α−ω−2c1λ−6c0c1−2b2ηλ2λ2σ+η2+6b1b2λ2λ2σ+η2,β:b1k−α−ω−b1λ−6b1c0+2c2ηλ+2b1η2λλ2σ+η2−4η3c2λλ2σ+η2−6ηλb12λ2σ+η2,ϕ0:c0k−α−ω+2c2λ2−3c02−b1ηλ2λ2σ+η2+2c2η2λ2λ2σ+η2+3b12λ2λ2σ+η2+d=0.Now, by solving the above algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following sets of results.Result 1(29)c0=16−13λ±97λ2−12d,c1=0,c2=−2,b1=±i10σλ=2η,b2=0,k=α±97λ2−12d+ω,η=±i52σλ.Now by using the above values of variables and putting into Equation (20), we get the following:
(30)V11=16−13λ±97λ2−12d+2ηβ−2ϕ2.Then, by using Equation (13), the solution of Equation (2) is as follows:
(31)V11=16−13λ±97λ2−12d+2η1G−2G′G2.But, when we put the value ofG from Equation (5), we get the following:
(32)V11=16−13λ±97λ2−12d+2ηA1sinhξ−λ+A2coshξ−λ+η/λ−2A1sinhξ−λ+A2coshξ−λ+η/λ′A1sinhξ−λ+A2coshξ−λ+η/λ2.Then, the exact solution of2+1-dimensional Ito equation of the form (2) is as follows:
(33)V11=16−13λ±97λ2−12d+2ηA1sinhξ−λ+A2coshξ−λ+η/λ+2λA1coshξ−λ+A2sinhξ−λ2A1sinhξ−λ+A2coshξ−λ+η/λ2.By lettingA1=0, η=0, and A2>0 in Equation (33), the particular solution of Equation (2) is as follows:
(34)V11a=16−13λ±97λ2−12d+2λtanh2ξ−λ.Then, by lettingλ=−0.3, d=0.4, t=0, and y=3, the graph of Equation (34) has the following forms.From Figure1, we see the solution of 2+1-dimensional Ito equation with respect to the result 1 and λ=−0.3, d=0.4, t=0, y=3, A1=0, η=0, and A2>0, and we obtain y≤0, when −10≤x≤10.Figure 1
Graphical solution of Equation (34) when k=α±97λ2−12d+ω in 3D and 2D, respectively.Again lettingA2=0, η=0, and A1>0 in Equation (34), we get another particular solution of Equation (2) as follows:
(35)V11b=16−13λ±97λ2−12d+2λcoth2ξ−λ.Then, by lettingλ=−0.3, d=0.4, t=0, and y=3, the graph of solution of Equation (35) has the following forms.Figure2 of the solution (35) concludes the solution of 2+1-dimensional Ito equation of result 1 when λ=−0.3, d=0.4, t=0, y=3, A2=0, η=0, A1>0, and for −10≤x≤10, it has no solution in real number on the domain y>3, and the solution is always y<3 in the range of −10≤x≤10.Figure 2
Graphical solution of Equation (35) when k=α±97λ2−12d+ω in 3D and 2D, respectively.Result 2(36)c0=13−4λ±28λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±228λ2−3d,η=0.Now by using the above values of variables and putting into Equation (20), we get the following:
(37)V1,1ξ=13−4λ±28λ2−3d−2ϕ2.Then, by using Equation (13), the solution of Equation (2) has the form 1/3−4λ±28λ2−3d−2G′/G2. Now, when we put the value of G from Equation (5), we get the following:
(38)V12=13−4λ±28λ2−3d−2A1−λcoshξ−λ+A2−λsinhξ−λA1sinhξ−λ+A2coshξ−λ2.Then, the exact solution of2+1-dimensional Ito equation of (2), if λ<0 with respect to result 2 based Equation (36), is as follows:
(39)13−4λ±28λ2−3d−2A1−λcoshξ−λ+A2−λsinhξ−λ2A1sinhξ−λ+A2coshξ−λ2.By lettingA1=0 and A2>0 in Equation (39), the particular solution of Equation (2) is as follows:
(40)V12a=13−4λ±28λ2−3d−2A2−λsinhξ−λ2A2coshξ−λ2.Then, the general solution of2+1-dimensional Ito equation for hyperbolic function with respect to result on Equation (36), when A1=0 and A2>0, has the form of
(41)=13−4λ±28λ2−3d+2λtanh2ξ−λ.The graph of the particular solution whenλ=−0.3, d=0.4, t=0, and y=3 is as follows.From Figure3, the solution of 2+1-dimensional Ito equation with respect to result 2 and λ=−0.3, d=0.4, t=0, y=3, A1=0, η=0, and A2>0 implies that condensed form around x=−3.Figure 3
Graphical solution of Equation (41) when k=α±228λ2−3d in 3D and 2D, respectively.Again lettingA2=0 and A1>0 in Equation (39), we get another particular solution of Equation (2) as follows:
(42)V12b=13−4λ±28λ2−3d−2A1−λcoshξ−λ2A1sinhξ−λ2.Then, the general solution of2+1-dimensional Ito equation for hyperbolic function with respect to result on Equation (36) when A2=0 and A1>0 has the form of
(43)V12b=13−4λ±28λ2−3d+2λcoth2ξ−λ.The graph of the particular solution whenλ=−0.3, d=0.4, t=0, and y=0 as follows.From Figure4 above, we see the solution of Equation (43) when λ=−0.3, d=0.4, t=0, y=3, A2=0, and A1>0, −10≤x≤10 has the solution between −10≤x≤10, and it bends downward.Figure 4
Graphical solution of Equation (43) when k=α±228λ2−3d in 3D and 2D, respectively.


### 2.2.2. Case II: Ifλ>0 (Trigonometric Function Solutions)

Substituting (6) into (23), we get
(44)V′′=2c1ϕ3−2c1ηβϕ+2c1λϕ−c1ηϕβ+2b1βϕ2−b1ηϕ2λλ2σ−η2+2b1η2βλλ2σ−η2−b1ηλ2λ2σ−η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2ϕ2λλ2σ−η2−4c2η3βλλ2σ−η2+2c2η2λ2λ2σ−η2−4c2ηβλ+2c2λ2−2b2ηϕ3λλ2σ−η2+4b2ϕη2βλλ2σ−η2−2b2ϕηλ2λ2σ−η2+b2λϕβ.Using Equation (6), Equation (24) becomes
(45)V2=b12ϕ2λλ2σ−η2−2b12ηβλλ2σ−η2+b12λ2λ2σ−η2+2b1b2λϕ3λ2σ−η2−4b1b2ϕηβλλ2σ−η2+2ϕb1b2λ2λ2σ−η2+ϕ4b22λλ2σ−η2+2b22ηβλϕ2λ2σ−η2+ϕ2b22λ2λ2σ−η2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.Again from Equations (44), (45), and (20), Equation (18) becomes
(46)cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3−2c1λϕ+3c1ηϕβ−2b1βϕ2+b1ηϕ2λλ2σ−η2−2b1η2βλλ2σ−η2+b1ηλ2λ2σ−η2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−6c2ϕ2λ−2c2η2ϕ2λλ2σ−η2+4c2η3βλλ2σ−η2−2c2η2λ2λ2σ−η2+4c2ηβλ−2c2λϕ2−2c2λ2+2b2ηϕ3λλ2σ−η2−4b2ϕβη2λλ2σ−η2+2b2ϕηλ2λ2σ−η2−b2λϕβ−3b12ϕ2λλ2σ−η2+6b12ηβλλ2σ−η2−3b12λ2λ2σ−η2−6b1b2λϕ3λ2σ−η2+12b1b2ϕηβλλ2σ−η2−6ϕb1b2λ2λ2σ−η2−3ϕ4b22λλ2σ−η2−6b22ηβλϕ2λ2σ−η2−3ϕ2b22λ2λ2σ−η2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc0c1−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22+d=0.Equating the coefficients of the powers ofβ and ϕ to be zero, we obtain the following system of algebraic equations:
(47)ϕ4:−6c2−3c22−3b22λλ2σ−η2=0,ϕ3β:−6b2c2=0,ϕ3:−2c1−6c2c1+2b2ηλλ2σ−η2−6b1b2λλ2σ−η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1−6b22ηλλ2σ−η2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2λλ2σ−η2−3b12λλ2σ−η2−3b22λ2λ2σ−η2+b1ηλλ2σ−η2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1+4b2η2λλ2σ−η2+12b2b1ηλλ2σ−η2=0,ϕ:c1k−α−ω−2c1λ−6c0c1+2b2ηλ2λ2σ−η2−6ϕb1b2λ2λ2σ−η2,β:b1k−α−ω−b1λ−6b1c0+4c2ηλ−2b1η2λλ2σ−η2+4η3c2λλ2σ−η2−6ηλb12λ2σ+η2=0,ϕ0:c0k−α−ω−2c2λ2−3c02+b1ηλ2λ2σ−η2−2c2η2λ2λ2σ−η2−3b12λ2λ2σ−η2+d=0.Then, by solving the above algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following results.Result 3(48)c0=1613λ±−12d+47λ2,c1=0,c2=−2,b1=±λ7σ=2η,b2=0,k=±−12d+47λ2+ω,η=±12λ7σ.Setting the above values of variables into Equation (20), we can get the following form of solution:
(49)Vξ=1613λ±−12d+47λ2+2ηβ−2ϕ2.Then, by using Equation (13), the solution of Equation (2) is a follows:Then, the exact solution of2+1-dimensional Ito equation of the form Equation (2) is as follows:
(50)V21=1613λ±−12d+47λ2+2ηA1sinξλ+A2cosξλ+η/2+2A1λcosξλ−A2λsinξλ2A1sinξλ+A2cosξλ+η/22.By lettingA1=0, η=0, and A2>0 in Equation (50), the particular solution of Equation (2) is
(51)V21a=1613λ±−12d+47λ2+2λtan2ξλ.Then, whenλ=0.7, d=0.4, and y=3, the graph of Equation (51) has the following forms:Figure5 shows the exact solution of 2+1-dimensional Ito equation with respect to result 3 when A1=0, η=0, A2>0, λ=0.7, d=0.4, y=3, and −10≤x≤10. This figure implies that the solutions are upward parabolic, and it repeated above x-axis.Figure 5
Graphical solution of Equation (51) when ±−12d+47λ2+ω in 3D and 2D, respectively.Again lettingA2=0, η=0, and A1>0 in Equation (50), we get another particular solution of Equation (6) as follows:
(52)V21b=1613λ±−12d+47λ2+2λcot2ξλ.Then, the graph of Equation (52) has the following form:From the above Figure6, we see that it is the exact solution of 2+1-dimensional Ito equation with respect to result 3 of Equation (52), when A1>0, η=0, A2=0, λ=0.7, d=0.4, y=3, and −10≤x≤10. And the result of the solution shows it has the form a parabolic upward and crosses the x-axis.Figure 6
Graphical solution of Equation (52) when ±−12d+47λ2+ω in 3D and 2D, respectively.Result 4(53)c0=13−4λ±4λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±4λ2−3d+ω,η=0.Then, setting the above values of variables into Equation (20), we can get the following form of solution for Ito equation:
(54)Vξ=13−4λ±4λ2−3d−2ϕ2.Then, by using Equation (13) for ϕ, the solution of Equation (2) is a follows:
(55)Vξ=13−4λ±4λ2−3d−2G′G2.But, when we put the value ofF of the form Equation (7), we get the following:
(56)Vξ=13−4λ±4λ2−3d−2A1sinξλ+A2cosξλ′A1sinξλ+A2cosξλ2.Then, finally we get the exact solution of2+1-dimensional Ito equation with respect to the second case result of the form (2) as follows:
(57)V22=13−4λ±4λ2−3d−2A1λcosξλ−A2λsinξλ2A1sinξλ+A2cosξλ2.By lettingA1=0 and A2>0 in Equation (57), the particular solution of Equation (2) is as follows:
(58)V22a=13−4λ±4λ2−3d−2λtan2ξλ.Figure7 shows the exact solution of 2+1-dimensional Ito equation with respect to result 4 when A1=0, A2>0, η=0, λ=0.7, d=0.4, y=3, and at −10≤x≤10, and it makes parabolic bend downward on y-axis.Figure 7
Graphical solution of Equation (58) when k=α±4λ2−3d+ω in 3D and 2D, respectively.Again lettingA2=0 and A1>0 in Equation (57), we get another particular solution of Equation (2) as follows:
(59)V22b=13−4λ±4λ2−3d−2A1λcosξλ2A1sinξλ2,V22b=13−4λ±4λ2−3d−2λcot2ξλ.Figure8 shows the exact solution of 2+1-dimension Ito equation with respect to result 4 when A2=0, A1>0, η=0, λ=0.7, d=0.4, y=3, and at −10≤x≤10, and it lies always below x-axis in the form of parabolic.Figure 8
Graphical solution of (44) when k= in 3D and 2D, respectively.


### 2.2.3. Case III: Whenλ=0 (Rational Function Solutions)

Substituting Equation (6) into Equation (23), we get
(60)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1ϕ2A12−2ηA2+2b1ηβA12−2ηA2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2ϕ2A12−2ηA2−4c2η3βA12−2ηA2−4c2ηβλ+2c2λ2−2b2ηϕ3A12−2ηA2+4b2ϕη2βA12−2ηA2+b2λϕβ.Again also substituting Equation (6) into Equation (24), we get
(61)V2=b12ϕ2−2ηβA12−2ηA2+2ϕb1b2ϕ2−2ηβA12−2ηA2+ϕ2b22ϕ2−2ηβA12−2ηA2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc1c0+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22(62)=b12ϕ2A12−2ηA2−2b12ηβA12−2ηA2+2b1b2ϕ3A12−2ηA2−4ϕb1b2ηβA12−2ηA2+ϕ4b22A12−2ηA2−2ϕ2b22ηβA12−2ηA2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc1c0+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.From Equation (20), (60), and (62), Equation (18) becomes
(63)Vξ=cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3+3c1ηβϕ−2c1λϕ−2b1βϕ2+b1ηϕ2A12−2ηA2−2b1ηβA12−2ηA2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−8c2ϕ2λ−2c2η2ϕ2A12−2ηA2+4c2η3βA12−2ηA2+4c2ηβλ−2c2λ2+2b2ηϕ3A12−2ηA2−4b2ϕη2βA12−2ηA2−b2λϕβ−3b12ϕ2A12−2ηA2+6b12ηβA12−2ηA2−6b1b2ϕ3A12−2ηA2+12ϕb1b2ηβA12−2ηA2−3ϕ4b22A12−2ηA2+6ϕ2b22ηβA12−2ηA2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc1c0−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22.Then, by setting the coefficients of the polynomial to be zero, we obtain the following sets of algebraic.(64)ϕ4:−6c2−3c22−3b22A12−2ηA2=0,ϕ3β:−6b2c2=0,ϕ3:−2c1−6c2c1+2b2ηA12−2ηA2−6b1b2λ2σ−η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1+6b22ηA12−2ηA2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2A12−2ηA2−3b12A12−2ηA2+b1ηA12−2ηA2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1−4b2η2A12−2ηA2+12b2b1ηA12−2ηA2=0,ϕ:c1k−α−ω−2c1λ−6c0c1=0,β:b1k−α−ω−b1λ−6b1c0+4c2ηλ−2b1ηA12−2ηA2+4η3c2A12−2ηA2+6ηb12A12−2ηA2,ϕ0:c0k−α−ω−2c2λ2−3c02+d=0.Then, by solving previous algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following results.Result 5(65)c0=13−4λ±4λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±24λ2−3d+ω,η=0.Then, setting Equation (65) values of variables into Equation (20), we can get the following form of solution for Ito equation:
(66)V31=13−4λ±4λ2−3d−2η/2ξ2+A1ξ+A2′η/2ξ2+A1ξ+A22.When the value ofη=0, then the solution of Equation (66) is as follows:
(67)V31=13−4λ±4λ2−3d−2A1A1ξ+A22.Finally, we get the exact solution of2+1-dimensional space time equation with respect to the first result of case 3 in Equation (65) as follows:
(68)V31=13−4λ±4λ2−3d−2A12A1ξ+A22.Whenλ=0, we have
(69)V31=±−3d−2A12A1ξ+A22.Using the parametersd=−0.5, A1=1, A2=2, and t=0, the graph of Equation (69) is as follows:Figure9 represents the exact solution of 2+1-dimensional space time with respect to the parameters d=−0.5, A1=1, A2=2, t=0, and y=3 which has maximum limit on the y-axis.Figure 9
Graphical solution of (69) when k=α±24λ2−3d+ω in 3D and 2D, respectively.Result 6(70)c0=16−11λ±73λ2−12d,c1=0,c2=−2,b1=2λA2±λλA22−A12=2η,(71)b2=0,k=αA12±73λ2−12dA12+ωA12A12,η=λA2±λλA22−A12.Then, setting the Equation (70) values of variables into Equation (20), we can get the following form of solution for Ito equation:
(72)V32=16−11λ±73λ2−12d+2ηβ−2ϕ2.Then by using Equation (13) for ϕ and β, the simplification of Equation (72) leads to
(73)V32=16−11λ±73λ2−12d+2η1G−2G′G2.When we put the value ofG from Equation (8) in Equation (73),
(74)V32=16−11λ±73λ2−12d+2ηη/2ξ2+A1ξ+A2−2ηξ+A1η/2ξ2+A1ξ+A22.Hence, we get the exact solution of2+1-dimensional space time with respect to result of (70) as follows:
(75)V32=16−11λ±73λ2−12d+2ηη/2ξ2+A1ξ+A2−2ηξ+A12η/2ξ2+A1ξ+A22.By lettingη=0, A1>0, and A2>0 in Equation (75), the particular solution of Equation (2) with respect to (70) is as follows:
(76)V32a=16−11λ±73λ2−12d−2A12A1ξ+A22.Whenλ=0, we get
(77)V32a=±−12d−2A12A1ξ+A22.Then, by settingd=−0.5, A1=1, A2=2, and t=0, the graph of (77) is a follows:Figure10 reveals the exact solution of 2+1-dimensional space time with respect to result 6 when the parameters d=−0.5, A1=1, A2=2, and t=0 and in similar way as figure 9 fixes its maximum from below on y-axis.Figure 10
Graphical solution of (77) when k=αA12±73λ2−12dA12+ωA12/A12 in 3D and 2D, respectively.


## 2.2.1. Case I: Ifλ<0 (Hyperbolic Function Solutions)

Substituting Equation (6) into Equation (23), we get the following:
(25)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1η−ϕ2λ+2ηβλ−λ2λ2σ+η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2−ϕ2λ+2ηβλ−λ2λ2σ+η2−2c2ηβλ+2c2λ2−2b2ϕη−ϕ2λ+2ηβλ−λ2λ2σ+η2+b2βϕλ=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2+b1ηϕ2λλ2σ+η2−2b1η2βλλ2σ+η2+b1ηλ2λ2σ+η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ−2c2η2ϕ2λλ2σ+η2+4c2η3βλλ2σ+η2−2c2η2λ2λ2σ+η2−2c2ηβλ+2c2λ2+2b2ηϕ3λλ2σ+η2−4b2ϕη2βλλ2σ+η2+2b2ϕηλ2λ2σ+η2+b2βϕλ.Again, putting Equation (6) into Equation (24), we obtain the following:
(26)V2=−b12ϕ2λλ2σ+η2+2b12ηβλλ2σ+η2−b12λ2λ2σ+η2−2b1b2ϕ3λλ2σ+η2+4ϕb1b2ηβλλ2σ+η2−2b1b2ϕλ2λ2σ+η2−ϕ4b22λλ2σ+η2+2ϕ2b22ηβλλ2σ+η2−ϕ2b22λ2λ2σ+η2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.By using Equations (20), (25), and (26), we obtain Equation (18) equals to
(27)cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3+3c1ηβϕ−2c1λϕ−2b1βϕ2−b1ηϕ2λλ2σ+η2+2b1η2βλλ2σ+η2−b1ηλ2λ2σ+η2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−8c2ϕ2λ−2c2η2ϕ2λλ2σ+η2−4c2η3βλλ2σ+η2+2c2η2λ2λ2σ+η2+2c2ηβλ−2c2λ2−2b2ηϕ3λλ2σ+η2+4b2ϕη2βλλ2σ+η2−2b2ϕηλ2λ2σ+η2−b2βϕλ+3b12ϕ2λλ2σ+η2−6b12ηβλλ2σ+η2+3b12λ2λ2σ+η2+6b1b2ϕ3λλ2σ+η2−12ϕb1b2ηβλλ2σ+η2+6b1b2ϕλ2λ2σ+η2+3ϕ4b22λλ2σ+η2−6ϕ2b22ηβλλ2σ+η2+3ϕ2b22λ2λ2σ+η2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc0c1−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22+d=0.Setting the coefficients of polynomialϕ and β zero from Equation (27), we obtain the following sets of algebraic equations:
(28)ϕ4:−6c2−3c22+3b22λλ2σ+η2=0,ϕ3β:−6b2c2=0,ϕ3:−2c2−6c2c1−2b2ηλλ2σ+η2+6b1b2λλ2σ+η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1−6b22ηλλ2σ+η2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2λλ2σ+η2+3b12λλ2σ+η2+3b22λ2λ2σ+η2−b1ηλλ2σ+η2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1+4b2η2λλ2σ+η2−12b2b1ηλλ2σ+η2=0,ϕ:c1k−α−ω−2c1λ−6c0c1−2b2ηλ2λ2σ+η2+6b1b2λ2λ2σ+η2,β:b1k−α−ω−b1λ−6b1c0+2c2ηλ+2b1η2λλ2σ+η2−4η3c2λλ2σ+η2−6ηλb12λ2σ+η2,ϕ0:c0k−α−ω+2c2λ2−3c02−b1ηλ2λ2σ+η2+2c2η2λ2λ2σ+η2+3b12λ2λ2σ+η2+d=0.Now, by solving the above algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following sets of results.Result 1(29)c0=16−13λ±97λ2−12d,c1=0,c2=−2,b1=±i10σλ=2η,b2=0,k=α±97λ2−12d+ω,η=±i52σλ.Now by using the above values of variables and putting into Equation (20), we get the following:
(30)V11=16−13λ±97λ2−12d+2ηβ−2ϕ2.Then, by using Equation (13), the solution of Equation (2) is as follows:
(31)V11=16−13λ±97λ2−12d+2η1G−2G′G2.But, when we put the value ofG from Equation (5), we get the following:
(32)V11=16−13λ±97λ2−12d+2ηA1sinhξ−λ+A2coshξ−λ+η/λ−2A1sinhξ−λ+A2coshξ−λ+η/λ′A1sinhξ−λ+A2coshξ−λ+η/λ2.Then, the exact solution of2+1-dimensional Ito equation of the form (2) is as follows:
(33)V11=16−13λ±97λ2−12d+2ηA1sinhξ−λ+A2coshξ−λ+η/λ+2λA1coshξ−λ+A2sinhξ−λ2A1sinhξ−λ+A2coshξ−λ+η/λ2.By lettingA1=0, η=0, and A2>0 in Equation (33), the particular solution of Equation (2) is as follows:
(34)V11a=16−13λ±97λ2−12d+2λtanh2ξ−λ.Then, by lettingλ=−0.3, d=0.4, t=0, and y=3, the graph of Equation (34) has the following forms.From Figure1, we see the solution of 2+1-dimensional Ito equation with respect to the result 1 and λ=−0.3, d=0.4, t=0, y=3, A1=0, η=0, and A2>0, and we obtain y≤0, when −10≤x≤10.Figure 1
Graphical solution of Equation (34) when k=α±97λ2−12d+ω in 3D and 2D, respectively.Again lettingA2=0, η=0, and A1>0 in Equation (34), we get another particular solution of Equation (2) as follows:
(35)V11b=16−13λ±97λ2−12d+2λcoth2ξ−λ.Then, by lettingλ=−0.3, d=0.4, t=0, and y=3, the graph of solution of Equation (35) has the following forms.Figure2 of the solution (35) concludes the solution of 2+1-dimensional Ito equation of result 1 when λ=−0.3, d=0.4, t=0, y=3, A2=0, η=0, A1>0, and for −10≤x≤10, it has no solution in real number on the domain y>3, and the solution is always y<3 in the range of −10≤x≤10.Figure 2
Graphical solution of Equation (35) when k=α±97λ2−12d+ω in 3D and 2D, respectively.Result 2(36)c0=13−4λ±28λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±228λ2−3d,η=0.Now by using the above values of variables and putting into Equation (20), we get the following:
(37)V1,1ξ=13−4λ±28λ2−3d−2ϕ2.Then, by using Equation (13), the solution of Equation (2) has the form 1/3−4λ±28λ2−3d−2G′/G2. Now, when we put the value of G from Equation (5), we get the following:
(38)V12=13−4λ±28λ2−3d−2A1−λcoshξ−λ+A2−λsinhξ−λA1sinhξ−λ+A2coshξ−λ2.Then, the exact solution of2+1-dimensional Ito equation of (2), if λ<0 with respect to result 2 based Equation (36), is as follows:
(39)13−4λ±28λ2−3d−2A1−λcoshξ−λ+A2−λsinhξ−λ2A1sinhξ−λ+A2coshξ−λ2.By lettingA1=0 and A2>0 in Equation (39), the particular solution of Equation (2) is as follows:
(40)V12a=13−4λ±28λ2−3d−2A2−λsinhξ−λ2A2coshξ−λ2.Then, the general solution of2+1-dimensional Ito equation for hyperbolic function with respect to result on Equation (36), when A1=0 and A2>0, has the form of
(41)=13−4λ±28λ2−3d+2λtanh2ξ−λ.The graph of the particular solution whenλ=−0.3, d=0.4, t=0, and y=3 is as follows.From Figure3, the solution of 2+1-dimensional Ito equation with respect to result 2 and λ=−0.3, d=0.4, t=0, y=3, A1=0, η=0, and A2>0 implies that condensed form around x=−3.Figure 3
Graphical solution of Equation (41) when k=α±228λ2−3d in 3D and 2D, respectively.Again lettingA2=0 and A1>0 in Equation (39), we get another particular solution of Equation (2) as follows:
(42)V12b=13−4λ±28λ2−3d−2A1−λcoshξ−λ2A1sinhξ−λ2.Then, the general solution of2+1-dimensional Ito equation for hyperbolic function with respect to result on Equation (36) when A2=0 and A1>0 has the form of
(43)V12b=13−4λ±28λ2−3d+2λcoth2ξ−λ.The graph of the particular solution whenλ=−0.3, d=0.4, t=0, and y=0 as follows.From Figure4 above, we see the solution of Equation (43) when λ=−0.3, d=0.4, t=0, y=3, A2=0, and A1>0, −10≤x≤10 has the solution between −10≤x≤10, and it bends downward.Figure 4
Graphical solution of Equation (43) when k=α±228λ2−3d in 3D and 2D, respectively.


## 2.2.2. Case II: Ifλ>0 (Trigonometric Function Solutions)

Substituting (6) into (23), we get
(44)V′′=2c1ϕ3−2c1ηβϕ+2c1λϕ−c1ηϕβ+2b1βϕ2−b1ηϕ2λλ2σ−η2+2b1η2βλλ2σ−η2−b1ηλ2λ2σ−η2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2ϕ2λλ2σ−η2−4c2η3βλλ2σ−η2+2c2η2λ2λ2σ−η2−4c2ηβλ+2c2λ2−2b2ηϕ3λλ2σ−η2+4b2ϕη2βλλ2σ−η2−2b2ϕηλ2λ2σ−η2+b2λϕβ.Using Equation (6), Equation (24) becomes
(45)V2=b12ϕ2λλ2σ−η2−2b12ηβλλ2σ−η2+b12λ2λ2σ−η2+2b1b2λϕ3λ2σ−η2−4b1b2ϕηβλλ2σ−η2+2ϕb1b2λ2λ2σ−η2+ϕ4b22λλ2σ−η2+2b22ηβλϕ2λ2σ−η2+ϕ2b22λ2λ2σ−η2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc0c1+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.Again from Equations (44), (45), and (20), Equation (18) becomes
(46)cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3−2c1λϕ+3c1ηϕβ−2b1βϕ2+b1ηϕ2λλ2σ−η2−2b1η2βλλ2σ−η2+b1ηλ2λ2σ−η2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−6c2ϕ2λ−2c2η2ϕ2λλ2σ−η2+4c2η3βλλ2σ−η2−2c2η2λ2λ2σ−η2+4c2ηβλ−2c2λϕ2−2c2λ2+2b2ηϕ3λλ2σ−η2−4b2ϕβη2λλ2σ−η2+2b2ϕηλ2λ2σ−η2−b2λϕβ−3b12ϕ2λλ2σ−η2+6b12ηβλλ2σ−η2−3b12λ2λ2σ−η2−6b1b2λϕ3λ2σ−η2+12b1b2ϕηβλλ2σ−η2−6ϕb1b2λ2λ2σ−η2−3ϕ4b22λλ2σ−η2−6b22ηβλϕ2λ2σ−η2−3ϕ2b22λ2λ2σ−η2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc0c1−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22+d=0.Equating the coefficients of the powers ofβ and ϕ to be zero, we obtain the following system of algebraic equations:
(47)ϕ4:−6c2−3c22−3b22λλ2σ−η2=0,ϕ3β:−6b2c2=0,ϕ3:−2c1−6c2c1+2b2ηλλ2σ−η2−6b1b2λλ2σ−η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1−6b22ηλλ2σ−η2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2λλ2σ−η2−3b12λλ2σ−η2−3b22λ2λ2σ−η2+b1ηλλ2σ−η2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1+4b2η2λλ2σ−η2+12b2b1ηλλ2σ−η2=0,ϕ:c1k−α−ω−2c1λ−6c0c1+2b2ηλ2λ2σ−η2−6ϕb1b2λ2λ2σ−η2,β:b1k−α−ω−b1λ−6b1c0+4c2ηλ−2b1η2λλ2σ−η2+4η3c2λλ2σ−η2−6ηλb12λ2σ+η2=0,ϕ0:c0k−α−ω−2c2λ2−3c02+b1ηλ2λ2σ−η2−2c2η2λ2λ2σ−η2−3b12λ2λ2σ−η2+d=0.Then, by solving the above algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following results.Result 3(48)c0=1613λ±−12d+47λ2,c1=0,c2=−2,b1=±λ7σ=2η,b2=0,k=±−12d+47λ2+ω,η=±12λ7σ.Setting the above values of variables into Equation (20), we can get the following form of solution:
(49)Vξ=1613λ±−12d+47λ2+2ηβ−2ϕ2.Then, by using Equation (13), the solution of Equation (2) is a follows:Then, the exact solution of2+1-dimensional Ito equation of the form Equation (2) is as follows:
(50)V21=1613λ±−12d+47λ2+2ηA1sinξλ+A2cosξλ+η/2+2A1λcosξλ−A2λsinξλ2A1sinξλ+A2cosξλ+η/22.By lettingA1=0, η=0, and A2>0 in Equation (50), the particular solution of Equation (2) is
(51)V21a=1613λ±−12d+47λ2+2λtan2ξλ.Then, whenλ=0.7, d=0.4, and y=3, the graph of Equation (51) has the following forms:Figure5 shows the exact solution of 2+1-dimensional Ito equation with respect to result 3 when A1=0, η=0, A2>0, λ=0.7, d=0.4, y=3, and −10≤x≤10. This figure implies that the solutions are upward parabolic, and it repeated above x-axis.Figure 5
Graphical solution of Equation (51) when ±−12d+47λ2+ω in 3D and 2D, respectively.Again lettingA2=0, η=0, and A1>0 in Equation (50), we get another particular solution of Equation (6) as follows:
(52)V21b=1613λ±−12d+47λ2+2λcot2ξλ.Then, the graph of Equation (52) has the following form:From the above Figure6, we see that it is the exact solution of 2+1-dimensional Ito equation with respect to result 3 of Equation (52), when A1>0, η=0, A2=0, λ=0.7, d=0.4, y=3, and −10≤x≤10. And the result of the solution shows it has the form a parabolic upward and crosses the x-axis.Figure 6
Graphical solution of Equation (52) when ±−12d+47λ2+ω in 3D and 2D, respectively.Result 4(53)c0=13−4λ±4λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±4λ2−3d+ω,η=0.Then, setting the above values of variables into Equation (20), we can get the following form of solution for Ito equation:
(54)Vξ=13−4λ±4λ2−3d−2ϕ2.Then, by using Equation (13) for ϕ, the solution of Equation (2) is a follows:
(55)Vξ=13−4λ±4λ2−3d−2G′G2.But, when we put the value ofF of the form Equation (7), we get the following:
(56)Vξ=13−4λ±4λ2−3d−2A1sinξλ+A2cosξλ′A1sinξλ+A2cosξλ2.Then, finally we get the exact solution of2+1-dimensional Ito equation with respect to the second case result of the form (2) as follows:
(57)V22=13−4λ±4λ2−3d−2A1λcosξλ−A2λsinξλ2A1sinξλ+A2cosξλ2.By lettingA1=0 and A2>0 in Equation (57), the particular solution of Equation (2) is as follows:
(58)V22a=13−4λ±4λ2−3d−2λtan2ξλ.Figure7 shows the exact solution of 2+1-dimensional Ito equation with respect to result 4 when A1=0, A2>0, η=0, λ=0.7, d=0.4, y=3, and at −10≤x≤10, and it makes parabolic bend downward on y-axis.Figure 7
Graphical solution of Equation (58) when k=α±4λ2−3d+ω in 3D and 2D, respectively.Again lettingA2=0 and A1>0 in Equation (57), we get another particular solution of Equation (2) as follows:
(59)V22b=13−4λ±4λ2−3d−2A1λcosξλ2A1sinξλ2,V22b=13−4λ±4λ2−3d−2λcot2ξλ.Figure8 shows the exact solution of 2+1-dimension Ito equation with respect to result 4 when A2=0, A1>0, η=0, λ=0.7, d=0.4, y=3, and at −10≤x≤10, and it lies always below x-axis in the form of parabolic.Figure 8
Graphical solution of (44) when k= in 3D and 2D, respectively.


## 2.2.3. Case III: Whenλ=0 (Rational Function Solutions)

Substituting Equation (6) into Equation (23), we get
(60)V′′=2c1ϕ3−3c1ηβϕ+2c1λϕ+2b1βϕ2−b1ϕ2A12−2ηA2+2b1ηβA12−2ηA2+b1βλ+6c2ϕ4−10c2ϕ2ηβ+8c2ϕ2λ+2c2η2ϕ2A12−2ηA2−4c2η3βA12−2ηA2−4c2ηβλ+2c2λ2−2b2ηϕ3A12−2ηA2+4b2ϕη2βA12−2ηA2+b2λϕβ.Again also substituting Equation (6) into Equation (24), we get
(61)V2=b12ϕ2−2ηβA12−2ηA2+2ϕb1b2ϕ2−2ηβA12−2ηA2+ϕ2b22ϕ2−2ηβA12−2ηA2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc1c0+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22(62)=b12ϕ2A12−2ηA2−2b12ηβA12−2ηA2+2b1b2ϕ3A12−2ηA2−4ϕb1b2ηβA12−2ηA2+ϕ4b22A12−2ηA2−2ϕ2b22ηβA12−2ηA2+2βb1c0+2βϕb2c0+c02+2βϕb1c1+2βϕ2b2c1+2ϕc1c0+ϕ2c12+2βϕ2b1c2+2βϕ3b2c2+2ϕ2c0c2+2ϕ3c1c2+ϕ4c22.From Equation (20), (60), and (62), Equation (18) becomes
(63)Vξ=cok−α−ω+c1ϕk−α−ω+b1βk−α−ω+c2ϕ2k−α−ω+b2ϕβk−α−ω−2c1ϕ3+3c1ηβϕ−2c1λϕ−2b1βϕ2+b1ηϕ2A12−2ηA2−2b1ηβA12−2ηA2−b1βλ−6c2ϕ4+10c2ϕ2ηβ−8c2ϕ2λ−2c2η2ϕ2A12−2ηA2+4c2η3βA12−2ηA2+4c2ηβλ−2c2λ2+2b2ηϕ3A12−2ηA2−4b2ϕη2βA12−2ηA2−b2λϕβ−3b12ϕ2A12−2ηA2+6b12ηβA12−2ηA2−6b1b2ϕ3A12−2ηA2+12ϕb1b2ηβA12−2ηA2−3ϕ4b22A12−2ηA2+6ϕ2b22ηβA12−2ηA2−6βb1c0−6βϕb2c0−3c02−6βϕb1c1−6βϕ2b2c1−6ϕc1c0−3ϕ2c12−6βϕ2b1c2−6βϕ3b2c2−6ϕ2c0c2−6ϕ3c1c2−3ϕ4c22.Then, by setting the coefficients of the polynomial to be zero, we obtain the following sets of algebraic.(64)ϕ4:−6c2−3c22−3b22A12−2ηA2=0,ϕ3β:−6b2c2=0,ϕ3:−2c1−6c2c1+2b2ηA12−2ηA2−6b1b2λ2σ−η2=0,ϕ2β:−2b1+10c2η−6b2c1−6c2b1+6b22ηA12−2ηA2=0,ϕ2:c2k−α−ω−8c2λ−3c12−6c0c2−2c2η2A12−2ηA2−3b12A12−2ηA2+b1ηA12−2ηA2=0,ϕβ:b2k−α−ω+3c1η−b2λ−6b2c0−6c1b1−4b2η2A12−2ηA2+12b2b1ηA12−2ηA2=0,ϕ:c1k−α−ω−2c1λ−6c0c1=0,β:b1k−α−ω−b1λ−6b1c0+4c2ηλ−2b1ηA12−2ηA2+4η3c2A12−2ηA2+6ηb12A12−2ηA2,ϕ0:c0k−α−ω−2c2λ2−3c02+d=0.Then, by solving previous algebraic equation using Wolfram Mathematica 9.0 software packages, we get the following results.Result 5(65)c0=13−4λ±4λ2−3d,c1=0,c2=−2,b1=0,b2=0,k=α±24λ2−3d+ω,η=0.Then, setting Equation (65) values of variables into Equation (20), we can get the following form of solution for Ito equation:
(66)V31=13−4λ±4λ2−3d−2η/2ξ2+A1ξ+A2′η/2ξ2+A1ξ+A22.When the value ofη=0, then the solution of Equation (66) is as follows:
(67)V31=13−4λ±4λ2−3d−2A1A1ξ+A22.Finally, we get the exact solution of2+1-dimensional space time equation with respect to the first result of case 3 in Equation (65) as follows:
(68)V31=13−4λ±4λ2−3d−2A12A1ξ+A22.Whenλ=0, we have
(69)V31=±−3d−2A12A1ξ+A22.Using the parametersd=−0.5, A1=1, A2=2, and t=0, the graph of Equation (69) is as follows:Figure9 represents the exact solution of 2+1-dimensional space time with respect to the parameters d=−0.5, A1=1, A2=2, t=0, and y=3 which has maximum limit on the y-axis.Figure 9
Graphical solution of (69) when k=α±24λ2−3d+ω in 3D and 2D, respectively.Result 6(70)c0=16−11λ±73λ2−12d,c1=0,c2=−2,b1=2λA2±λλA22−A12=2η,(71)b2=0,k=αA12±73λ2−12dA12+ωA12A12,η=λA2±λλA22−A12.Then, setting the Equation (70) values of variables into Equation (20), we can get the following form of solution for Ito equation:
(72)V32=16−11λ±73λ2−12d+2ηβ−2ϕ2.Then by using Equation (13) for ϕ and β, the simplification of Equation (72) leads to
(73)V32=16−11λ±73λ2−12d+2η1G−2G′G2.When we put the value ofG from Equation (8) in Equation (73),
(74)V32=16−11λ±73λ2−12d+2ηη/2ξ2+A1ξ+A2−2ηξ+A1η/2ξ2+A1ξ+A22.Hence, we get the exact solution of2+1-dimensional space time with respect to result of (70) as follows:
(75)V32=16−11λ±73λ2−12d+2ηη/2ξ2+A1ξ+A2−2ηξ+A12η/2ξ2+A1ξ+A22.By lettingη=0, A1>0, and A2>0 in Equation (75), the particular solution of Equation (2) with respect to (70) is as follows:
(76)V32a=16−11λ±73λ2−12d−2A12A1ξ+A22.Whenλ=0, we get
(77)V32a=±−12d−2A12A1ξ+A22.Then, by settingd=−0.5, A1=1, A2=2, and t=0, the graph of (77) is a follows:Figure10 reveals the exact solution of 2+1-dimensional space time with respect to result 6 when the parameters d=−0.5, A1=1, A2=2, and t=0 and in similar way as figure 9 fixes its maximum from below on y-axis.Figure 10
Graphical solution of (77) when k=αA12±73λ2−12dA12+ωA12/A12 in 3D and 2D, respectively.


## 3. Conclusion

In this article, the solution of NLPDE of2+1-dimensional space time is analytically solved by special expansion method. The special expansion method is one of the direct method to solve the nonlinear partial differential equations. To apply the method, we used the combined form G′/G,1/G expansion method. The method has a great advantage to solve the NLPDE which leads us to get exact solutions for three different cases of λ. The special expansion method was applied on the proposed equations and leads to find for hyperbolic functions, trigonometric functions, and rational functions. Furthermore, the obtained results are exactly fit exact solutions which solve the complicity of finding the solutions for NLPDE. The obtained solution for each results was illustrated by graphical plots using the Wolfram Mathematica 9.0 software, and also, the surface plane models are constructed side by side to show the physical and geometrical interpretations on the ground. Finally, the method is powerful and effective to solve NLPDE.



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*Source: 1283138-2022-08-31.xml*