Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 383 https://internationalpubls.com Mathematical Analysis of Some (3+1)-D Models via Rangaig Transform 1Sandeep Sharma, 2Inderdeep Singh 1,2Sant Baba Bhag Singh University, Jalandhar-144030, Punjab, India Email: 1 sandeepsharma200@gmail.com, 2inderdeeps.ma.12@gmail.com Article History: Received: 05-08-2024 Revised: 24-09-2024 Accepted: 07-10-2024 Abstract: In this research, we suggest a hybrid method to solve (3+1)-D PDEs arising in various applications of sciences and engineering. The plausibility of the proposed method is demonstrated by considering the “Rangaig Transform” and the classical “Homotopy Analysis Technique”. Some experimental work has been performed to demonstrate the accuracy and simplicity of the proposed hybrid scheme. Keywords: The “Rangaig Transform”, HAM, “(3+1)-D Telegraph Equation”, “(3+1)-D Diffusion Equation”, “(3+1)-D Schrodinger Equations”, Test Examples. 1. INTRODUCTION: - The remaining higher-dimensional partial differential equations (PDEs) are in great demand in the areas of mathematical physics, engineering, and many other applied branches. In general, such equations are derived from some problems that include three spatial and one temporal coordinate, such as fluid dynamics, quantum, electromagnetic, or wave theory. In less trivial (significantly non-linear) situations, what these equations look like can be quite complicated indeed, and getting hold of exact or approximate solutions is relatively difficult. This is achieved using advanced mathematics, which writes these equations in a form that is much more accessible for some math to be performed. Both the Rangaig Transform and Homotopy Analysis Method (HAM) are very potent in manipulating the solution of (3+1)-dimensional PDEs. The Rangaig Transform is an integral transform that goes beyond the conventional traditional employs of Fourier and Laplace type transforms and can be used to offer a new perspective at linear, nonlinear PDEs putting them in simpler algebraical forms. On the other hand, HAM [15-17] is an analytic method that converges rapidly to the approximate solutions for the nonlinear PDEs, where it can easily deform continuously from a simple problem to another complex problem and control the convergence of the solution. From the papers cited, as well as from the applications seen in this section, it can be ascertained that both methods are efficacious for nonlinear (3+1)-D PDEs; they could be wave equations, heat equations, fluid dynamic schemes, and others. The use of the concept can be followed from abstract theoretical mathematical physics right through to engineering divisions, including machine learning, as well as industrial divisions, such as financial and computational sciences. Applying the “Rangaig Transform” with the help of HAM can offer exact & approximate sophisticated multi-dimensional global problems. It is possible, and such highly nontrivial, (3+1)-dimensional systems, sample methods can be formulated and employed here, encouraging their possible use in the numerical solution of some actual complex physical partial differential equation. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 384 https://internationalpubls.com Integral transforms have significantly evolved over the years, providing powerful methods for solving differential equations. The Rangaig Transform, introduced by Rangaig et al. (2017), is one such innovative tool that simplifies the resolution of partial differential equations ([1], [22]). Similarly, Aboodh's work on the Aboodh Transform offers an effective alternative for integral equations, advancing the computational efficiency of these solutions [2]. The Homotopy Analysis Method (HAM) has also been utilized to solve nonlinear Schrödinger equations, with significant contributions from Alomari, Noorani, and Nazar [3]. Eltayeb and Kilicman further explored the utility of the Sumudu Transform in differential equations, expanding the scope of applicable mathematical problems [4]. On a parallel track, Elzaki’s contributions introduced the Elzaki Transform, applied to both linear and nonlinear equations, underscoring the versatility of integral transforms in mathematical modelling ([5], [6], [24]). The Homotopy Analysis Method (HAM) continues to be an effective technique in solving nonlinear problems, as evidenced by the research of Ganjiani [7], Gupta and Kumar [8], and Jafari and Seifi [9]. Furthermore, new developments, such as the Shehu Transform proposed by Maitama and Zhao [18] and generalizations of the Rangaig Transform by Mansour and Kuffi ([19], [20]), have opened new possibilities for addressing complex mathematical challenges. In this study, the potential of the “Rangaig Transform”-based “Homotopy Analysis Method” (RT- HAM) to solve some examples in (3+1) dimensions PDEs is investigated. In the end, we will scrutinize its possibility of use in various physical problems, comparing it with previous methods that solve versatility and convergence control over various linear and nonlinear complex scenario problems. The remainder of this paper is organized as follows: Section 2 gives the basic definitions and concepts utilized in the “Rangaig Transform”. Within Section 3, the rebuilt resilient homotopy examination system has been mentioned. The “Homotopy Analysis” and “Rangaig Transform Method” for solving (3+1)-D PDEs are described in Section 4. We provide Test problems of (3+1)-D “Telegraph Equation”, “Diffusion Equation”, and “Klein Gordan Equation” to be solved in section 5. The discussion of a conclusion is in section 6. 2. BASIC CONCEPT OF “RANGAIG TRANSFORM” [19-22] The following part introduces the fundamental principles and features of another transform, the “Rangaig Transform”, which will be introduced in this work. In this study, we introduce the “Rangaig Transform” as a novel transformation for exponentially ordered functions within the set H. 𝐻 = {𝜂(𝑡) ∃𝑁, 𝐴1, 𝐴2 > 0, |𝜂(𝑡)| > 𝑁𝑒𝐴𝑖|𝑡|, 𝑡 ∈ (−1)𝑖−1 × (−∞, 0)} (1) 𝑁, the arbitrary constant One can have an infinite or endlessly finite value for the arbitrary constant 𝐴1, 𝐴2. We now introduce a new transformation that may be included in (1) as follows: ℛ[𝜂(𝑡)] = 𝑇(𝜛) = 1 𝜛 ∫ 𝑐𝜛𝑡 0 −∞ ℎ(𝑡)𝑑𝑡, 1 𝐴1 ≤ 𝜛 ≤ 1 𝐴2 , (2) There is a transformation process known as the “Rangaig Transform”. The definition is really a statement that factor 𝜛 takes the place of the variable 𝑡 in the function ℎ. On the other hand, it is possible to claim that there is a transition to a description of the function 𝜂(𝑡) in the Pi-variant Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 385 https://internationalpubls.com space 𝑇(𝜛). The following demonstrates the application of the “Rangaig Transform” in obtaining results for a certain type of function [19-22]. The following “Rangaig Transform” is accomplished in some core forms. The general function is: - 𝜂(𝑡) = ℛ{𝜂(𝑡)} i. ℛ{𝜂(𝑡)} = ℛ{1} = 1 𝜛2 ii. ℛ{1} = − 1 𝜛3 iii. ℛ{𝑡} = − 1 𝜛3 iv. ℛ{𝑡𝑛 , 𝑛 ≥ 0} = (−1)2𝑛! 𝜛𝑛+2 v. ℛ{𝑠𝑖𝑛(𝑡)} = − 1 𝜛(𝜛2+1) vi. ℛ{𝑐𝑜𝑠(𝑡)} = 1 (𝜛2+1) vii. ℛ{𝑡𝑛 , 𝑛 ≤ 0} = (−1)𝑛+1Γ(−𝑛) 𝜛𝑛 viii. ℛ{𝑒𝑎𝑡} = 1 𝜛(𝜛+𝑎) ix. ℛ{𝑀(𝑡 − 𝑎)} = 1 𝜛2 𝑒𝑎𝑡 Theorem 1: - [22] (Transformation of Rangaig Derivatives) If 𝜂(𝑡), 𝜂1(𝑡), 𝜂𝑛(𝑡) ∈ 𝐻, then ℛ[𝜂𝑛(𝑡)] = 𝑇(𝜛) = (−1)𝑛𝜛𝑛𝑇(𝜛) + (−1)𝑛+1 ∑( 𝑛−1 𝓀=0 − 1)𝓀𝜛𝑛−2−𝓀𝜂(𝓀)(0) (3) Theorem 2: - [22] (Transformation of Rangaig Integrals), If 𝑚𝑛(𝑡) = ∫ ∫ ∫ ⋯ 𝑡3 0 ∫ 𝜂(𝜏)(𝑑𝜏)𝑛 𝑡𝑛+1 0 𝑡2 0 , 𝑡1 0 so that 𝑚(t) ∈ H. Next, we define the Rangaig transform of 𝑚n(t) as follows: ℛ[𝑚𝑛(𝑡)] = 𝑇𝑛(𝜛) = ( −1 𝜛 ) 𝑛 𝑇(𝜛) Theorem 3: - [22] The convolution identity's Rangaig transform is provided by: ℛ[(𝜂 ∗ 𝜅)(𝑡)] = −𝜛𝑇1(𝜛)𝑇2(𝜛) , Where (𝜂 ∗ 𝜅)(𝑡) = ∫ 𝜂 𝑡 0 (𝑡 − 𝜏)𝜅(𝜏)𝑑𝜏, 𝑇1(𝜛) and 𝑇2(𝜛) is, respectively, the Rangaig transform of 𝜂(𝑡) and 𝜅(𝑡). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 386 https://internationalpubls.com Theorem 4: - [22] (Duality relation of R-Transform and L-Transform) The transformation relation between “Rangaig Transform” 𝑇(𝜛) and “Laplace Transform” 𝐹(𝜛) of 𝜂(𝑡) if 𝜂(𝑡) and 𝜂(−𝑡) exist over 𝐻. 𝑇(𝜛) = 1 𝜛 𝐹(−𝜛) Proposition 1. If 𝜕𝑤(𝑥,𝑡) 𝜕𝑡 exist, and we can apply integration by parts, we get the following: ℛ [ 𝜕𝑤(𝑥, 𝑡) 𝜕𝑡 ] = −𝜛𝑇(𝑥,𝜛) + 1 𝜛 𝑤(𝑥, 0) (4) Proof. To illustrate it, we use the integration by parts and formula (2). Proposition 2. If we assume that 𝑇(𝑥, 𝜛) is the “Rangaig Transform” of 𝜇(𝑥, 𝑡), we obtain: ℛ [ 𝜕𝑤(𝑥, 𝑡) 𝜕𝑡 ] = (−1)𝑛𝜛𝑛𝑇(𝑥,𝜛) + (−1)𝑛+1 ∑( 𝑛−1 𝓀=0 − 1)𝓀𝜛𝑛−2−𝓀 𝜕𝓀𝑤(𝑥, 0) 𝜕𝑡𝓀 (5) Proof: - We demonstrate mathematical induction to show that (5) is valid. Using the formula (5) and assuming 𝑛 = 1, we get: [ 𝜕𝑤(𝑥, 𝑡) 𝜕𝑡 ] = −𝜛𝑇(𝑥,𝜛) + 1 𝜛 𝑤(𝑥, 0) (6) Thus, we observe that the formula holds when 𝑛 = 1 based on (4). Make the inductive assumption that the formula is valid for 𝑛 so that ℛ [ 𝜕𝑤(𝑥, 𝑡) 𝜕𝑡 ] = (−1)𝑛𝜛𝑛𝑇(𝑥,𝜛) + (−1)𝑛+1 ∑( 𝑛−1 𝓀=0 − 1)𝓀𝜛𝑛−2−𝓀 𝜕𝓀𝑤(𝑥, 0) 𝜕𝑡𝓀 (7) demonstrate that it remains valid at rank 𝑛 + 1. Assume 𝜕𝑛𝑤(𝑥,𝑡) 𝜕𝑡𝑛 = 𝑣(𝑥, 𝑡) and according to (4) and (7), we have: = ℛ [ 𝜕𝑛+1𝑤(𝑥, 𝑡) 𝜕𝑡𝑛+1 ] = ℛ [ 𝜕𝑣(𝑥, 𝑡) 𝜕𝑡 ] = −𝜛ℛ[𝑣(𝑥, 𝑡)] + 1 𝜛 𝑣(𝑥, 0) = −𝜛 [(−1)𝑛𝜛𝑛𝑇(𝑥, 𝜛) + (−1)𝑛+1 ∑( 𝑛−1 𝓀=0 − 1)𝓀𝜛𝑛−2−𝓀 𝜕𝓀𝑤(𝑥, 0) 𝜕𝑡𝓀 ] + 1 𝜛 𝜕𝑛𝑤(𝑥, 𝑡) 𝜕𝑡𝑛 = (−1)𝑛+1𝜛𝑛+1𝑇(𝑥,𝜛) + (−1)𝑛+2 ∑( 𝑛−1 𝓀=0 − 1)𝓀𝜛𝑛−1−𝓀 𝜕𝓀𝑤(𝑥, 0) 𝜕𝑡𝓀 + 1 𝜛 𝜕𝑛𝑤(𝑥, 𝑡) 𝜕𝑡𝑛 = (−1)𝑛+1𝜛𝑛+1𝑇(𝑥, 𝜛) + (−1)𝑛+2 ∑( 𝑛 𝓀=0 − 1)𝓀𝜛𝑛−1−𝓀 𝜕𝓀𝑤(𝑥, 0) 𝜕𝑡𝓀 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 387 https://internationalpubls.com Therefore, the formula (5) holds for every 𝑛 ≥ 1 according to the principle of mathematical induction. 3. HOMOTOPY ANALYSIS METHOD [13-17] As a subsequent step, the next nonlinear differential equation was considered. When 𝒩 is a nonlinear operator, 𝑤(Υ, 𝑡) is an unknown function, and Υ might be {𝑥, 𝑦, 𝑧}. The observed independent variables for time and space are, in that order, the variables 𝑥, 𝑦, 𝑧 and 𝑡 with Liao's invention, the conventional homotopy technique. First, we use 𝑤0(Υ, 𝑡) as the initial estimate for the equilibrium concentrations of 𝑤(Υ, 𝑡). Lastly, ℎ is an added parameter which is not the zero while 𝓅 is the embedding parameter ranging from 0 𝑡𝑜 1; ℒ is an added linear operator; 𝐻(Υ, 𝑡) = 1 is an auxiliary function; ℎ ≠ 0 is an auxiliary parameter; while the 𝛿( Υ, 𝑡; 𝓅) is an unknown function. If 𝓅 = 0 & 𝓅 = 1, we obtain 𝛿( Υ, 𝑡 ; 0) = 𝑤0(Υ, 𝑡) , and 𝛿( Υ, 𝑡 ; 1) = 𝑤(Υ, 𝑡) , As a consequence, solution 𝛿( Υ, 𝑡; 𝓅) shifts from starting guess 𝑤0(Υ, 𝑡) to exact result 𝑤(Υ, 𝑡) as 𝓅 traverses range [0, 1]. When we expand 𝛿( Υ, 𝑡; 𝓅) in the Taylor series with respect to 𝓅, we obtain Where, 𝑤𝑚(Υ, 𝑡) = 1 𝑚! 𝜕𝑚𝛿(Υ, 𝑡; 𝓅) 𝜕𝓅𝑚 | 𝓅=0 If the auxiliary function, auxiliary linear operator, auxiliary parameter ℎ, and initial estimate are all correctly chosen, then the series (10) converges at 𝓅 = 1, and we obtain This ought to be an acceptable solution for the initial nonlinear equation. By definition (11) the governing equation might be obtained from the 0-order deformation equation (9). Explain the vector. 𝑤𝑛⃗⃗⃗⃗ ⃗ = 𝑤0(Υ, 𝑡), 𝑤1(Υ, 𝑡), 𝑤2(Υ, 𝑡) ………… .𝑤𝑛(Υ, 𝑡) 𝒩[𝑤(Υ, 𝑡)] = 0 (8) (1 − 𝓅)ℒ[𝛿(Υ, 𝑡 ; 𝓅) − 𝑤0(Υ, 𝑡)] = 𝓅ℎ𝐻(Υ, 𝑡)𝒩[𝛿(Υ, 𝑡 ; 𝓅)] (9) 𝛿( Υ, 𝑡; 𝓅) = 𝑤0(Υ, 𝑡) + ∑ 𝑤𝑚 (Υ, 𝑡)𝓅𝑚 ∞ 𝑚=1 (10) 𝑤(Υ, 𝑡) = 𝑤0(Υ, 𝑡) + ∑ 𝑤𝑚 ∞ 𝑚=1 (Υ, 𝑡), (11) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 388 https://internationalpubls.com Considering the embedding parameter 𝓅, differentiating the 0-order deformation equation (9) 𝑚- times. After splitting 𝑚! by 𝓅 = 0, the equation for the 𝑚th-order deformation looks like this: ℒ[𝑤𝑚(Υ, 𝑡) − 𝜒𝑚 𝑤𝑚−1 (Υ, 𝑡)] = ℎℛ𝑚[𝑤𝑚−1 (Υ, 𝑡)] Whereas ℛ𝑚 (𝑤𝑚−1)⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗ = 1 𝑚 − 1 ! 𝜕𝑚−1 𝒩 [𝛿(Υ, 𝑡 ; 𝓅) 𝜕𝓅𝑚−1 | 𝓅 = 0 and 𝜒𝑚 = { 0, 𝑚 ≤ 1 1, 𝑚 > 1 . 4. RANGAIG TRANSFORM BASED HOMOTOPY ANALYSIS METHOD (RT-HAM) Take the following partial differential equation of 3-D. 𝜇𝑡𝑡(Υ) = {ℒ(𝜇) + 𝒩(𝜇) + 𝑓(Υ)}. (12) Where 𝑓(Υ) refers to the functions of 𝑥, 𝑦, 𝑧, while ℒ indicates the linear and 𝒩 nonlinear parts. respectively. Considering both sides of the “Rangaig Transform”, we obtain ℛ{𝜇𝑡𝑡(Υ)} = ℛ{ℒ(𝜇) + 𝒩(𝜇) + 𝑓(Υ)}. (13) With the help of “Rangaig Transform” and an initial condition, we get ℛ{𝜇(Υ, 𝑡)} = 1 𝑤2 𝜇0(Υ) − 1 𝑤3 𝜕𝜇0 𝜕𝑡 + 1 𝑤2 ℛ{ℒ(𝜇) + 𝒩(𝜇) + 𝑓(Υ)}. (14) Taking the nonlinear part as: 𝒩[𝛿(Υ, 𝑡; 𝓅)] = ℛ{𝜇(Υ, 𝑡)} − [ 1 𝑤2 𝜇0(Υ) − 1 𝑤3 𝜕𝜇0 𝜕𝑡 + 1 𝑤2 ℛ{ℒ(𝜇) + 𝒩(𝜇) + 𝑓(Υ)}]. We begin by setting up the zero-order deformation equation; we have (1 − 𝓅)ℛ{𝛿(Υ, 𝑡) − 𝜇0(Υ, 𝑡)} = 𝓅ℎ𝐻(Υ, 𝑡)𝒩[𝛿(Υ, 𝑡; 𝓅)]. (15) 𝐻(Υ, 𝑡) = 1 is an auxiliary function, ℎ ≠ 0 is an auxiliary parameter, ℛ is an auxiliary linear Rangaig operator. When 𝓅 = 0 & 𝓅 = 1, we get, { 𝛿(Υ, 𝑡; 0) = 𝜇0(Υ, 0) 𝛿(Υ, 𝑡; 1) = 𝜇(Υ, 𝑡). As a result, we have an equation of deformation of order 𝑚. ℛ{𝜇𝑚(Υ, 𝑡) − 𝜒𝑚𝜇𝑚−1 (Υ, 𝑡)} = ℎ ℛ𝑚(𝜇𝑚−1 ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗(Υ, 𝑡)). (16) Applying the “Inverse Rangaig Transform” to both sides of Equation (16), we obtain 𝜇𝑚(Υ, 𝑡) − 𝜒𝑚𝜇𝑚−1 (Υ, 𝑡) = ℛ−1{ℎ ℛ𝑚 (𝜇𝑚−1⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ (Υ, 𝑡))}. (17) From the above Equation we get 𝜇1(Υ, 𝑡) = −ℛ−1{ℛ1(𝜇0⃗⃗⃗⃗ (Υ, 𝑡))}, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 389 https://internationalpubls.com 𝜇2(Υ, 𝑡) = 𝜇1(Υ, 𝑡) − ℛ−1{ℛ2(𝜇1⃗⃗⃗⃗ (Υ, 𝑡))}, 𝜇3(Υ, 𝑡) = 𝜇2(Υ, 𝑡) − ℛ−1{ℛ3(𝜇2⃗⃗⃗⃗ (Υ, 𝑡))}, ⋮ Therefore, the solution is: 𝜇(Υ, 𝑡) = 𝜇0 + 𝜇1 + 𝜇2 + ⋯ (18) 3. TEST EXAMPLES: Within this segment, we will adopt both the “Rangaig Transform” and the traditional “Homotopy Analysis Technique (HAM)” to solve a couple of examples of semi-analytical solutions for (3+1)-D PDEs including the “(3+1)-D Telegraph Equation”, “(3+1)-D Diffusion Equation”, and “(3+1)-D Klein Gordan Equation”. Example 1: Consider the (3+1)-D Telegraph Equation of the form: 𝜇𝑡𝑡 − 2𝜋2𝜇 = 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) (19) With initial condition 𝜇(Υ, 0) = 𝑠𝑖𝑛 𝜋𝑥 𝑠𝑖𝑛 𝜋𝑦 𝑠𝑖𝑛 𝜋𝑧, we use Υ = (𝑥, 𝑦, 𝑧), The exact solution is: 𝜇(Υ, 𝑡) = 𝑒𝜋𝑡 sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) Rephrase the stated issue as follows: 𝜇𝑡𝑡 = 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇 (20) When the “Rangaig Transform” is carried out to both sides of Equation (20), we get, ℛ [𝜇𝑡𝑡] = ℛ [ 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇] This implies (−1)2𝜛2ℛ[𝜇] + (−1)3 ∑ (−1)𝓀 𝜛𝓀 1 𝓀=0 𝜕𝓀𝜇 𝜕𝑡𝓀 = ℛ [ 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇] This expression can be expressed as: 𝜛2ℛ[𝜇] − [ 1 𝜛0 𝜇(Υ, 0) + (−1)1 𝜛1 𝜕 𝜕𝑡 𝜇(Υ, 0)] = ℛ [ 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇] When starting conditions are applied, we get 𝜛2ℛ[𝜇] − (1 − 𝜋 𝜛1 ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) = ℛ [ 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇] That suggests ℛ[𝜇] − ( 1 𝜛2 − 𝜋 𝜛3 ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) = 1 𝜛2 ℛ [ 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇] Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 390 https://internationalpubls.com Or ℛ[𝜇] = ( 1 𝜛2 − 𝜋 𝜛3 ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) + 1 𝜛2 ℛ [ 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇] Here is how we define the nonlinear component: 𝒩[𝛿(Υ, 𝑡; 𝓅)] = ℛ[𝜇] − ( 1 𝜛2 − 𝜋 𝜛3 ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) − 1 𝜛2 ℛ [ 1 3 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) + 2𝜋2𝜇] (21) We begin by setting up the 0-order deformation according to the assumption adopted herein, namely, 𝐻(Υ, 𝑡) = 1; we have (1 − 𝓅)ℛ{𝛿(Υ, 𝑡) − 𝜇0(Υ, 𝑡)} = 𝓅ℎ𝒩[𝛿(Υ, 𝑡; 𝓅)] When 𝓅 = 0 & 𝓅 = 1, we have { 𝛿(Υ, 𝑡; 0) = 𝜇0(Υ, 0) 𝛿(Υ, 𝑡; 1) = 𝜇(Υ, 𝑡) So, the mth-order deformation eqn. ℛ{𝜇𝑚(Υ, 𝑡) − 𝜒𝑚𝜇𝑚−1(Υ, 𝑡)} = ℎℛ𝑚(𝜇𝑚−1⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ (Υ, 𝑡)) (22) When the “Inverse Rangaig Transform” is carried out to both sides of Equation (22), we get, 𝜇𝑚(Υ, 𝑡) − 𝜒𝑚𝜇𝑚−1( Υ, 𝑡) = ℛ−1 {ℎℛ𝑚(𝜇𝑚−1⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ (Υ, 𝑡))} (23) With ℎ = −1, we can get from Equation (23) 𝜇1(Υ, 𝑡) = −ℛ−1{ℛ1 (𝜇0⃗⃗⃗⃗ (Υ, 𝑡))}, 𝜇2(Υ, 𝑡) = 𝜇1(Υ, 𝑡) − ℛ−1 {ℛ2(𝜇1⃗⃗⃗⃗ (Υ, 𝑡))}, 𝜇3(Υ, 𝑡) = 𝜇2(Υ, 𝑡) − ℛ−1 {ℛ3(𝜇2⃗⃗⃗⃗ (Υ, 𝑡))}, ⋮ Where” ℛ1(𝜇0⃗⃗⃗⃗ (Υ, 𝑡)) = ( 𝜋 𝜛3 − 𝜋2 𝜛4 ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧), ℛ2(𝜇1⃗⃗⃗⃗ (Υ, 𝑡)) = (− 𝜋 𝜛3 + 𝜋2 𝜛4 + 𝜋3 𝜛5 − 𝜋4 𝜛6 ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧), ℛ3(𝜇2⃗⃗⃗⃗ (Υ, 𝑡)) = (− 𝜋3 𝜛5 + 𝜋4 𝜛6 + 𝜋5 𝜛7 − 𝜋6 𝜛8 ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) , ⋮ Consequently, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 391 https://internationalpubls.com 𝜇1(Υ, 𝑡) = (𝜋𝑡 + (𝜋𝑡)2 2! ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧), 𝜇2(Υ, 𝑡) = ( (𝜋𝑡)3 3! + (𝜋𝑡)4 4! ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) 𝜇3(Υ, 𝑡) = ( (𝜋𝑡)5 5! + (𝜋𝑡)6 6! ) sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) ⋮ Consequently, the solution is: 𝜇(Υ, 𝑡) = 𝜇0 + 𝜇1 + 𝜇2 + ⋯ Or 𝜇(Υ, 𝑡) = {1 + 𝜋𝑡 + (𝜋𝑡)2 2! + (𝜋𝑡)3 3! + (𝜋𝑡)4 4! } sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) Or 𝜇(Υ, 𝑡) = 𝑒𝜋𝑡 sin( 𝜋𝑥) sin( 𝜋𝑦) sin( 𝜋𝑧) Figure 1: Example 1's solutions' physical behavior at 𝒕 = 𝟏, 𝒛 = 𝟏 𝟐 -1 -0.5 0 0.5 1 -1 0 1 -40 -20 0 20 40 x Example 1: For t = 1 y S o lu ti o n s Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 392 https://internationalpubls.com Figure 2: The contour diagram obtained from solving Example 1 at 𝒕 = 𝟏, 𝒛 = 𝟏 𝟐 Figures 1 and 2 depict the physical and dynamic behaviour of the solutions found at 𝑧 = 1 2 and 𝑡 = 1 using the "Homotopy Analysis Method" based on the "Rangaig Transform." Example 2: Consider the (3+1)-D Diffusion Equation of the form: 𝜇𝑡 = 1 3𝜋2 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧) (24) With initial condition 𝜇(Υ, 0) = 𝑒−𝑡 sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧, we use Υ = (𝑥, 𝑦, 𝑧), The exact solution is: 𝜇(Υ, 𝑡) = 𝑒−𝑡 sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧 Applying the “Rangaig Transform” to both sides of Equation (24), we obtain, ℛ[𝜇𝑡] = ℛ [ 1 3𝜋2 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧)] This implies (−1)𝜛 ℛ[𝜇] + 1 𝜛 𝜇 (Υ, 0) = ℛ [ 1 3𝜋2 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧)]. This implies ℛ[𝜇] − 1 𝜛2 𝜇 (Υ, 0) = − 1 𝜛 ℛ [ 1 3𝜋2 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧)] When starting conditions are applied, we get ℛ[𝜇] = 1 𝜛2 sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧 − 1 𝜛 ℛ [ 1 3𝜋2 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧)] x y Example 1: For t = 1 -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 393 https://internationalpubls.com Here is how we define the nonlinear component: 𝒩[𝛿(Υ, 𝑡 ; 𝓅)] = ℛ[𝜇] − 1 𝜛2 sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧 + 1 𝜛 ℛ [ 1 3𝜋2 (𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝑧𝑧)] (25) We begin by setting up the 0-order deformation according to the assumption adopted herein, namely, 𝐻(Υ, 𝑡) = 1; we have (1 − 𝓅)ℛ{𝛿(Υ, 𝑡) − 𝜇0(Υ, 𝑡)} = 𝓅ℎ𝒩[𝛿(Υ, 𝑡; 𝓅)] When 𝓅 = 0 & 𝓅 = 1, we have { 𝛿(Υ, 𝑡; 0) = 𝜇0(Υ, 0) 𝛿(Υ, 𝑡; 1) = 𝜇(Υ, 𝑡) So, the mth-order deformation eqn. ℛ{𝜇𝑚(Υ, 𝑡) − 𝜒𝑚𝜇𝑚−1(Υ, 𝑡)} = ℎℛ𝑚(𝜇𝑚−1⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ (Υ, 𝑡)) (26) When the “Inverse Rangaig Transform” is carried out to both sides of Equation (26), we get, 𝜇𝑚(Υ, 𝑡) − 𝜒𝑚𝜇𝑚−1(Υ, 𝑡) = ℛ−1{ℎℛ𝑚(𝜇𝑚−1⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ (Υ, 𝑡))} (27) With ℎ = −1, we can get from Equation (27) 𝜇1(Υ, 𝑡) = −ℛ−1 {ℛ1(𝜇0⃗⃗⃗⃗ (Υ, 𝑡))}, 𝜇2(Υ, 𝑡) = 𝜇1(Υ, 𝑡) − ℛ−1{ℛ2(𝜇1⃗⃗⃗⃗ (Υ, 𝑡))}, 𝜇3(Υ, 𝑡) = 𝜇2(Υ, 𝑡) − ℛ−1 {ℛ3 (𝜇2⃗⃗⃗⃗ (Υ, 𝑡))}, ⋮ Where” ℛ1(𝜇0⃗⃗⃗⃗ (Υ, 𝑡)) = − 1 𝑤3 sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧, ℛ2(𝜇1⃗⃗⃗⃗ (Υ, 𝑡)) = ( 1 𝜛3 − 1 𝜛4 ) sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧, ℛ3(𝜇2⃗⃗⃗⃗ (Υ, 𝑡)) = ( 1 𝜛4 − 1 𝜛5 ) sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧, ⋮ Therefore, 𝜇1(Υ, 𝑡) = − sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧, 𝜇2(Υ, 𝑡) = 𝑡2 2! sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧, 𝜇3(Υ, 𝑡) = − 𝑡3 3! sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧, ⋮ Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 394 https://internationalpubls.com The solution is: 𝜇(Υ, 𝑡) = 𝜇0 + 𝜇1 + 𝜇2 + ⋯ Or 𝜇(Υ, 𝑡) = {1 − 𝑡 + 𝑡2 2! − 𝑡3 3! + ⋯} sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧 = 𝑒−𝑡 sin 𝜋𝑥 sin𝜋𝑦 sin 𝜋𝑧 Figure 3: Example 2's solutions' physical behavior at 𝒕 = 𝟐, 𝒛 = 𝝅 𝟐 Figure 4: The contour diagram obtained from solving Example 2 at 𝒕 = 𝟐, 𝒛 = 𝝅 𝟐 -2 -1 0 1 2 -2 0 2 -0.2 -0.1 0 0.1 0.2 x Example 2: For t = 2, z = pi/2 y S o lu ti o n s x y Example 2: For t = 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 395 https://internationalpubls.com Figures 3 and 4 depict the physical and dynamic behaviour of the solutions found at 𝑧 = 1 and 𝑡 = 𝜋 2 using the "Homotopy Analysis Method" based on the "Rangaig transform." Example 3: Consider the (3+1)-D “Klein Gordan Equation” of the form: 𝜇𝓉𝓉 − (𝜇𝓍𝓍 + 𝜇𝓎𝓎 + 𝜇𝓏𝓏) + 𝜇 = 2(sin 𝓍 + sin𝓎 + sin 𝓏) (28) With initial condition (𝜇(Υ, 0) = sin 𝓍 + sin𝓎 + sin 𝓏, we use Υ = (𝓍,𝓎, 𝓏), The exact solution is: 𝜇(Υ, 𝓉) = sin 𝓍 + sin𝓎 + sin 𝓏 + sin 𝓉 Rewrite the given problem as: 𝜇𝓉𝓉 = (𝜇𝓍𝓍 + 𝜇𝓎𝓎 + 𝜇𝓏𝓏) − 𝜇 + 2(sin𝓍 + sin𝓎 + sin 𝓏) (29) Applying the “Rangaig Transform” to both sides of Equation (29), we obtain, ℛ[𝜇𝓉𝓉] = ℛ[(𝜇𝓍𝓍 + 𝜇𝓎𝓎 + 𝜇𝓏𝓏) − 𝜇 + 2(sin 𝓍 + sin𝓎 + sin 𝓏)] This implies (−1)2𝜛2ℛ[𝜇] + (−1)3 ∑ (−1)𝓀 𝜛𝓀 1 𝓀=0 𝜕𝓀𝜇 𝜕𝓉𝓀 = ℛ[(𝜇𝓍𝓍 + 𝜇𝓎𝓎 + 𝜇𝓏𝓏) − 𝜇 + 2 (sin 𝓍 + sin𝓎 + sin 𝓏)]. This implies 𝜛2ℛ[𝜇] − [ 1 𝜛0 𝜇(Υ, 0) + (−1)1 𝜛1 𝜕 𝜕𝓉 𝜇(Υ, 0)] = 1 𝜛2 ℛ[(𝜇𝓍𝓍 + 𝜇𝓎𝓎 + 𝜇𝓏𝓏) − 𝜇 + 2 (sin 𝓍 + sin𝓎 + sin 𝓏)] Or ℛ[𝜇] − [ 1 𝜛2 𝜇(Υ, 0) + (−1)1 𝜛3 𝜕 𝜕𝓉 𝜇(Υ, 0)] = 1 𝜛2 ℛ[(𝜇𝓍𝓍 + 𝜇𝓎𝓎 + 𝜇𝓏𝓏) − 𝜇 + 2 (sin 𝓍 + sin𝓎 + sin 𝓏)] When the initial conditions are applied, we get ℛ[𝜇] = ( 1 𝜛2 ) (sin 𝓍 + sin𝓎 + sin 𝓏) − 1 𝜛3 + 1 𝜛2 ℛ[(𝜇𝑥𝑥 + 𝜇𝑦𝑦 + 𝜇𝓏𝓏) − 𝜇 + 2(sin 𝓍 + sin𝓎 + sin 𝓏)] We define the nonlinear component as follows: 𝒩[𝛿(Υ, 𝓉; 𝓅)] = ℛ[𝜇] − ( 1 𝜛2 ) (sin 𝓍 + sin𝓎 + sin 𝓏) + 1 𝜛3 (30) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 396 https://internationalpubls.com − 1 𝜛2 ℛ[(𝜇𝓍𝓍 + 𝜇𝓎𝓎 + 𝜇𝓏𝓏) − 𝜇 + 2(sin 𝓍 + sin𝓎 + sin 𝓏)] We begin by setting up the 0-order deformation according to the assumption adopted herein, namely, 𝐻(Υ, 𝓉) = 1; we have (1 − 𝓅)ℛ{𝛿(Υ, 𝓉) − 𝜇0(Υ, 𝓉)} = 𝓅ℎ𝒩[𝛿(Υ, 𝓉; 𝓅)] When 𝓅 = 0 & 𝓅 = 1, we have { 𝛿(Υ, 𝓉; 0) = 𝜇0(Υ, 0) 𝛿(Υ, 𝓉; 1) = 𝜇(Υ, 𝓉) Thus, the equation for mth-order deformation. ℛ{𝜇𝑚(Υ, 𝓉) − 𝜒𝑚𝜇𝑚−1(Υ, 𝓉)} = ℎℛ𝑚(𝜇𝑚−1⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ (Υ, 𝓉)) (31) When the “Inverse Rangaig Transform” is carried out to both sides of Equation (31), we get, 𝜇𝑚(Υ, 𝓉) − 𝜒𝑚𝜇𝑚−1(Υ, 𝓉) = ℛ−1 {ℎℛ𝑚(𝜇𝑚−1⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ (Υ, 𝓉))} (32) With ℎ = −1, we can get from Equation (32) 𝜇1(Υ, 𝓉) = −ℛ−1 {ℛ1(𝜇0⃗⃗⃗⃗ (Υ, 𝓉))}, 𝜇2(Υ, 𝓉) = 𝜇1(Υ, 𝓉) − ℛ−1 {ℛ2(𝜇1⃗⃗⃗⃗ (Υ, 𝓉))}, 𝜇3(Υ, 𝓉) = 𝜇2(Υ, 𝓉) − ℛ−1 {ℛ3(𝜇2⃗⃗⃗⃗ (Υ, 𝓉))}, ⋮ Where” ℛ1(𝜇0⃗⃗⃗⃗ (Υ, 𝓉)) = 1 𝑤3 , ℛ2(𝜇1⃗⃗⃗⃗ (Υ, 𝓉)) = (− 1 𝜛3 − 1 𝜛5 ), ℛ3(𝜇2⃗⃗⃗⃗ (Υ, 𝓉)) = ( 1 𝜛5 + 1 𝜛7 ), ⋮ Therefore, 𝜇1(Υ, 𝓉) = 𝓉, 𝜇2(Υ, 𝓉) = − 𝓉3 3! , 𝜇3(Υ, 𝓉) = 𝓉5 5! , ⋮ Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 397 https://internationalpubls.com The solution is: 𝜇(Υ, 𝓉) = 𝜇0 + 𝜇1 + 𝜇2 + ⋯ Or 𝜇(Υ, 𝓉) = sin 𝓍 + sin𝓎 + sin 𝓏 + (𝓉 − 𝓉3 3! + 𝓉5 5! − ⋯) Or 𝜇(Υ, 𝓉) = sin 𝓍 + sin𝓎 + sin 𝓏 + sin 𝓉 Figure 5: Example 3's solutions' physical behavior at 𝒛 = 𝟏, 𝓽 = 𝝅 𝟐 Figure 6: The contour diagram obtained from solving Example 3 at 𝒛 = 𝟏, 𝓉 = 𝝅 𝟐 -2 -1 0 1 2 -2 0 2 -2 0 2 4 x Example 3: for t = pi/2 y S o lu ti o n s x y -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3 (2025) 398 https://internationalpubls.com Figures 5 and 6 depict the physical and dynamic behaviour of the solutions found at 𝑧 = 1 and 𝓉 = 𝜋 2 using the "Homotopy Analysis Method" based on the "Rangaig Transform." 6. 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