442 

This work is licensed under a Creative Commons Attribution 4.0 International License 

IHJPAS. 37 (1) 2024 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

PISSN: 1609-4042, EISSN: 2521-3407 

  

 

 

 

    

1Saed M. Turq*       2Emad A. Kuffi   

1 Syedna Ibrahim Secondary School, Ministry of Education ,Hebron, Palestine.   
2 Department of Mathematics, College of Basic Education, Mustansiriyah University, Baghdad, Iraq. 

 

 

*Corresponding Author. saedturq@gmail.com 

 

 

 

 

Abstract   

In this paper we have presented a comparison between two novel integral transformations that are 

of great importance in the solution of differential equations. These two transformations are the 

complex Sadik transform and the KAJ transform. An uncompressed forced oscillator, which is an 

important application, served as the basis for comparison. The application was solved and exact 

solutions were obtained. Therefore, in this paper, the exact solution was found based on two 

different integral transforms: the first integral transform complex Sadik and the second integral 

transform KAJ. And these exact solutions obtained from these two integral transforms were new 

methods with simple algebraic calculations and applied to different problems. The main purpose 

of this comparison is the exact solutions, and until we show the importance of the diversity and 

difference of the kernel of the integral transform by keeping the period t between 0 and infinity. 

Keywords : Complex Sadik Transform, (KAJ) Kuffi Abbass Jawad Transform, Response of An 

Undamped uncompressed forced oscillator, Ordinary Differential Equations. 

1. Introduction 

First, one should know the importance of integral transforms in solving differential equations of 

all kinds, both ordinary and partial, where these differential equations are transformed into 

algebraic equations that are easier to compute by relying on the integral transform and then taking 

the inverse of the transform [2, 15, 27 and 29]. 

Integral transforms are currently also of great importance in applied mathematics, as they are used 

in the coding of images, among other things. They are of great importance for modern technical 

applications, such as in genetic engineering, and facilitate the solution of complex differential 

systems, such as colon cancer and drug concentration, regardless of whether they are partial or 

ordinary systems [13,14,27,30]. 

Received 11 March 2023, Received 14 May 2023, Accepted 29 May 2023, Published 20 January 2024 

 

 

Comparison of Complex Sadik and KAJ Transforms for Ordinary Differential 

Equations to the Response of an Uncompressed Forced Oscillator 

 

doi.org/10.30526/37.1.3 

https://creativecommons.org/licenses/by/4.0/
https://jih.uobaghdad.edu.iq/index.php/j/index#1609-4042
https://jih.uobaghdad.edu.iq/index.php/j/index#2521-3407
mailto:saedturq@gmail.com
https://orcid.org/0000-0003-4664-1295
mailto:saedturq@gmail.com
https://orcid.org/0009-0004-5254-674X
mailto:bemad.kuffi@uomustansiriyah.edu.iq


IHJPAS. 37 (1) 2024 

443 
 

Due to the importance of integral transformations in many vital applications, many papers have 

appeared on this topic and in various fields [1,3-11,16-26,28] 

Definition 1.1. [29] The Complex Sadik Transform (CST) is denoted by the operator 

𝐒𝑎
𝑐{.}, thetransform form is as follows: 

𝐒𝑎
𝑐 [𝑔(𝑡)] = 𝐅𝑐(𝑠𝛼, 𝛽) =

1

𝑠𝛽
∫  

∞

0

𝑔(𝑡)𝑒−𝑖𝑠
𝛼𝑡𝑑𝑡. 

Where 𝑠 is a complex variable, 𝛼 is  any nonzero real number, and 𝛽 is any real number. 

Definition 1.2. [2] The "Kuffi-Abbas-Jawad"(KAJ) Transform denoted by the operator 

𝐒𝑚{.}, 𝑡ℎ𝑒 transform form is as follows: 

𝐒𝑚[𝑔(𝑡)] = 𝐾(𝑣) =
1

𝑣𝑛
∫  

∞

0

𝑔 (
𝑡

𝑣
) 𝑒−𝑡𝑑𝑡. 

Where 𝑛 is any integer number, and 0 < 𝑙1 ≤ 𝑣 ≤ 𝑙2, where 𝑙1 and 𝑙2 are either finite or infinite. 

1.1. The KAJ and Complex Sadik Integral Transforms for Some Basic Functions 

In this part, we present the KAJ integral transform and the novel complex integral transform for 

some important basic functions in the following table: 

 

Table 1: KAJ transform and the complex Sadik integral transform for some basic functions 

Functions 

𝑔(𝑡) 

𝐒𝑚{𝑔(𝑡)} = 𝐾(𝑣) 

"Kuffi Abbass Jawad (KAJ) Transform" 

𝐒𝑎
𝑐{𝑔(𝑡)} = 𝐅𝑐(𝑠) 

"Complex Sadik Transform" 

1 
1

𝑣𝑛+1
 

−𝑖

𝑠(𝛼+𝛽)
 

𝑡𝑟 , 𝑟 ∈ ℕ 
𝑟!

𝑣𝑛+𝑟
 (−𝑖)𝑟+1

𝑟!

𝑠𝑟𝛼+(𝛼+𝛽)
 

𝑒𝑎𝑡 , 𝑎constant 
𝑣

𝑣𝑛(𝑣 − 𝑎)
 

−1

𝑠𝛽
[

𝑎

(𝑠2𝛼 + 𝑎2)
+ 𝑖

𝑠𝛼

(𝑠2𝛼 + 𝑎2)
 

sin⁡(𝑎𝑡) 
𝑎

𝑣𝑛−1(𝑣2 + 𝑎2)
 

−𝑎

𝑠𝛽(𝑠2𝛼 − 𝑎2)
 

cos⁡(𝑎𝑡) 
1

𝑣𝑛−2(𝑣2 + 𝑎2)
 

−𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑎2)
 

sinh⁡(𝑎𝑡) 
𝑎

𝑣𝑛−1(𝑣2 − 𝑎2)
 

−𝑎

𝑠𝛽(𝑠2𝛼 + 𝑎2)
 

cosh⁡(𝑎𝑡) 
1

𝑣𝑛−2(𝑣2 − 𝑎2)
 

−𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 + 𝑎2)
 

 



IHJPAS. 37 (1) 2024 

444 
 

2. Complex Sadik Integral and KAJ Transforms of Derivatives: 

Theorem 2.1. [29] Let 𝐅𝑐(𝑠) be the complex Sadik integral transform of 𝑓(𝑡)(𝐅𝑐(𝑠) = 𝐒𝑎
𝑐 [𝑓(𝑡)]), 

then 

𝐒𝑎
𝑐 [𝑓(𝑛)(𝑡)] = (𝑖𝑠𝛼)𝑛𝐅𝑐(𝑠) −

1

𝑠𝛽
[∑  

𝑛

𝑘=1

  (𝑖𝑠𝛼)𝑘−1𝑓(𝑛−𝑘)(0)]. 

In this paper, we want to generalize KAJ transform of 𝑚𝑡ℎ-derivative and prove that by 

mathematical induction.  

Theorem 2.2. Let 𝐾(𝑣) be the Kuffi-Abbas-Jawad (KAJ) transform of 𝑓(𝑡)(𝐾(𝑣) = 𝐒𝑚[𝑓(𝑡)]), 

then 

𝐒𝑚[𝑓
(ℎ)(𝑡)] = 𝑣ℎ𝐾(𝑣) −

1

𝑣𝑛
[∑  

ℎ

𝑘=1

 𝑣𝑘𝑓(ℎ−𝑘)(0)]. 

Proof. by Mathematical Induction 

1 For ℎ = 1, 

𝐒𝑚[𝑓
′(𝑡)] = 𝑣𝐾(𝑣) −

𝑣𝑓(0)

𝑣𝑛
. 

Thus true for ℎ = 1. 

2 Assume that, true for ℎ = 𝑟 that means: 

𝐒𝑚[𝑓
(𝑟)(𝑡)] = 𝑣𝑟𝐾(𝑣) −

1

𝑣𝑛
[∑  

𝑟

𝑘=1

 𝑣𝑘𝑓(𝑟−𝑘)(0)]. 

3 We want to prove for ℎ = 𝑟 + 1 



IHJPAS. 37 (1) 2024 

445 
 

𝐒𝑚[𝑓
(𝑟+1)(𝑡)] = 𝐒𝑚 [(𝑓(𝑟)(𝑡))

′
] ⁡= 𝑣𝐒𝑚[𝑓

(𝑟)(𝑡)] −
𝑣𝑓(𝑟)(0)

𝑣𝑛
,

⁡= 𝑣 [𝑣𝑟𝐾(𝑣) −
1

𝑣𝑛
[∑  

𝑟

𝑘=1

 𝑣𝑘𝑓(𝑟−𝑘)(0)]] −
𝑣𝑓(𝑟)(0)

𝑣𝑛
,

= 𝑣𝑟+1𝐾(𝑣) −
1

𝑣𝑛
[∑  

𝑟

𝑘=1

 𝑣𝑘+1𝑓(𝑟−𝑘)(0)]] −
𝑣𝑓(𝑟)(0)

𝑣𝑛
,

= 𝑣𝑟+1𝐾(𝑣) −
1

𝑣𝑛
[∑  

𝑟

𝑘=1

 𝑣𝑘+1𝑓(𝑟−𝑘)(0)] + 𝑣𝑓(𝑟)(0)] ,

= 𝑣𝑟+1𝐾(𝑣) −
1

𝑣𝑛
[∑  

𝑟

𝑘=0

 𝑣𝑘+1𝑓(𝑟−𝑘)(0)]] ,

= 𝑣𝑟+1𝐾(𝑣) −
1

𝑣𝑛
[∑  

𝑟+1

𝑘=1

 𝑣(𝑘−1)+1𝑓(𝑟−(𝑘−1))(0)]] ,

= 𝑣𝑟+1𝐾(𝑣) −
1

𝑣𝑛
[∑  

𝑟+1

𝑘=1

 𝑣𝑘𝑓(𝑟+1−𝑘)(0)]] ,

⁡= 𝐒𝑚[𝑓
(𝑟+1)(𝑡)].

 

So theorem is true for 𝑛 ∈ ℕ. 

3. Main Results: 

In this part, we present two real life problems, response of an undamped forced mechanical 

oscillator and response of an undamped forced electrical oscillator. 

Example 3.1. (Response of an Undamped Forced Mechanical Oscillator) 

Consider the differential equation of the forced mechanical oscillator: 

�̈�(𝑡) + 𝑤0
2𝑋(𝑡) =

𝐹

𝑚
cos(𝑤𝑡). 

Where 𝑤0 = √
𝑘

𝑚
,represents the natural frequency of the oscillator with initial boundary conditions, 

as follows: 

𝑋(0) = 0,

�̇�(0) = 0.
 

 Complex Sadik Transform  



IHJPAS. 37 (1) 2024 

446 
 

𝐒𝑎
𝑐 {�̈�(𝑡) + 𝑤0

2𝑋(𝑡) =
𝐹

𝑚
cos⁡(𝑤𝑡)} ,

(𝑖𝑠𝛼)2𝐗𝑐(𝑠) −
�̇�(0)

𝑠𝐵𝛽
−
𝑖𝑠𝛼𝑋(0)

𝑠𝛽
+ 𝑤0

2𝐗𝑐(𝑠) =
𝐹

𝑚

−𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

(𝑖𝑠𝛼)2𝐗𝑐(𝑠) + 𝑤0
2𝐗𝑐(𝑠) =

𝐹

𝑚

−𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

𝑖2(𝑠𝛼)2𝐗𝑐(𝑠) + 𝑤0
2𝐗𝑐(𝑠) =

−𝑖𝐹𝑠𝛼

𝑚𝑠𝛽(𝑠2𝛼 −𝑤2)
,

⁡−𝑠2𝛼𝐗𝑐(𝑠) + 𝑤0
2𝐗𝑐(𝑠) =

−𝑖𝐹𝑠𝛼

𝑚𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

(𝑤0
2 − 𝑠2𝛼)𝐗𝑐(𝑠) =

−𝑖𝐹𝑠𝛼

𝑚𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

𝐗𝑐(𝑠) =
−𝑖𝐹𝑠𝛼

𝑚𝑠𝛽(𝑠2𝛼 − 𝑤2)(𝑤0
2 − 𝑠2𝛼)

,

𝐗𝑐(𝑠) =
𝑖𝐹𝑠𝛼

𝑚𝑠𝛽
[

−1

(𝑠2𝛼 − 𝑤2)(𝑤0
2 − 𝑠2𝛼)

] ,

𝐗𝑐(𝑠) =
𝑖𝐹𝑠𝛼

𝑚𝑠𝛽
[

1

(𝑠2𝛼 − 𝑤2)(𝑤2 − 𝑤0
2)

+
1

(𝑤0
2 − 𝑠2𝛼)(𝑤2 −𝑤0

2)
]

𝐗𝑐(𝑠) =
𝑖𝐹𝑠𝛼

𝑚𝑠𝛽(𝑤2 − 𝑤0
2)
[

1

(𝑠2𝛼 − 𝑤2)
+

1

(𝑤0
2 − 𝑠2𝛼)

] ,

𝐗𝑐(𝑠) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
{
𝑖𝑠𝛼

𝑠𝛽
[

1

(𝑠2𝛼 − 𝑤2)
+

1

(𝑤0
2 − 𝑠2𝛼)

]} ,

𝐗𝑐(𝑠) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
{
𝑖𝑠𝛼

𝑠𝛽
[

1

(𝑠2𝛼 − 𝑤2)
−

1

(𝑠2𝛼 − 𝑤0
2)
]} , 

𝐗𝑐(𝑠) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[

𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤2)
−

𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤0
2)
] ,

 

inverse , then 

𝑋(𝑡) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[−cos⁡(𝑤𝑡) + cos⁡(𝑤0𝑡)],

𝑋(𝑡) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[cos⁡(𝑤0𝑡) − cos⁡(𝑤𝑡)],

𝑋(𝑡) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[−2sin⁡(

(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)] ,

𝑋(𝑡) =
2𝐹

𝑚(𝑤0
2 − 𝑤2)

sin⁡(
(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
) ,

 

or because sin⁡(−𝑥) = −sin⁡(𝑥) 

𝑋(𝑡) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[−2sin⁡(

(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)] ,

𝑋(𝑡) =
2𝐹

𝑚(𝑤2 − 𝑤0
2)
sin (

(𝑤 + 𝑤0)𝑡

2
) sin (

(𝑤 − 𝑤0)𝑡

2
) .

 



IHJPAS. 37 (1) 2024 

447 
 

Kuffi Abbass Jawad (AKJ) Transform 

𝐒𝑚 {�̈�(𝑡) + 𝑤0
2𝑋(𝑡) =

𝐹

𝑚
cos⁡(𝑤𝑡)} ,

𝑣2𝐾(𝑣) +
1

𝑣2
[−𝑣2𝑋(0) − 𝑣�̇�(0)] + 𝑤0

2𝐾(𝑣) =
𝐹

𝑚
[

1

𝑣𝑛−2(𝑣2 + 𝑤2)
] ,

(𝑣2 + 𝑤0
2)𝐾(𝑣) =

𝐹

𝑚
[

1

𝑣𝑛−2(𝑣2 + 𝑤2)
] ,

𝐾(𝑣) =
𝐹

𝑚
[

1

𝑣𝑛−2(𝑣2 + 𝑤2)(𝑣2 + 𝑤0
2)
] ,

𝐾(𝑣) =
𝐹

𝑚𝑣𝑛−2
[

1

(𝑣2 + 𝑤2)(𝑣2 + 𝑤0
2)
] ,

𝐾(𝑣) =
𝐹

𝑚𝑣𝑛−2
[

1

(𝑣2 + 𝑤0
2)(𝑤2 − 𝑤0

2)
−

1

(𝑣2 + 𝑤2)(𝑤2 − 𝑤0
2)
] ,

𝐾(𝑣) =
𝐹

𝑚𝑣𝑛−2(𝑤2 − 𝑤0
2)
[

1

(𝑣2 + 𝑤0
2)

−
1

(𝑣2 + 𝑤2)
] ,

𝐒𝑚[𝑋(𝑡)] = 𝐾(𝑣) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[

1

𝑣𝑛−2(𝑣2 +𝑤0
2)

−
1

𝑣𝑛−2(𝑣2 +𝑤2)
] ,

 

inverse , then 

𝑋(𝑡) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[cos⁡(𝑤0𝑡) − cos⁡(𝑤𝑡)],

𝑋(𝑡) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[−2sin⁡(

(𝑤0 +𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)]

𝑋(𝑡) =
2𝐹

𝑚(𝑤0
2 − 𝑤2)

sin⁡(
(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
) ,

 

or because sin⁡(−𝑥) = −sin⁡(𝑥) 

𝑋(𝑡) =
𝐹

𝑚(𝑤2 − 𝑤0
2)
[−2sin⁡(

(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)] ,

𝑋(𝑡) =
2𝐹

𝑚(𝑤2 − 𝑤0
2)
sin (

(𝑤 + 𝑤0)𝑡

2
) sin (

(𝑤 − 𝑤0)𝑡

2
) .

 

Example 3.2. (Response of an Undamped Forced Electrical Oscillator) 

Consider the differential equation of the forced electrical oscillator: 

𝐿�̈�(𝑡) + 𝑅�̇�(𝑡) +
1

𝐶
𝑄(𝑡) ⁡= 𝑉 cos(𝑤𝑡) ,

�̈�(𝑡) +
𝑅

𝐿
�̇�(𝑡) + 𝑤0

2𝑄(𝑡) ⁡=
𝑉

𝐿
cos(𝑤𝑡) .

 

For an undamped forced electrical oscillator, resistance 𝑅 = 0 



IHJPAS. 37 (1) 2024 

448 
 

�̈�(𝑡) + 𝑤0
2𝑄(𝑡) =

𝑉

𝐿
cos(𝑤𝑡). 

Where 𝑤0 = √
1

𝐶𝐿
, and 𝑄(𝑡) is the instantaneous charge. 

With the initial boundary conditions are: 

𝑄(0) = 0,

�̇�(0) = 0.
 

𝐒𝑎
𝑐 {�̈�(𝑡) + 𝑤0

2𝑄(𝑡) =
𝑉

𝐿
cos⁡(𝑤𝑡)} ,

(𝑖𝑠𝛼)2𝐐𝑐(𝑠) −
�̇�(0)

𝑠𝐵𝛽
−
𝑖𝑠𝛼𝑄(0)

𝑠𝛽
+ 𝑤0

2𝐐𝑐(𝑠) =
𝑉

𝐿

−𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

(𝑖𝑠𝛼)2𝐐𝑐(𝑠) + 𝑤0
2𝐐𝑐(𝑠) =

𝑉

𝐿

−𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

𝑖2(𝑠𝛼)2𝐐𝑐(𝑠) + 𝑤0
2𝐐𝑐(𝑠) =

−𝑖𝑉𝑠𝛼

𝐿𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

⁡−𝑠2𝛼𝐐𝑐(𝑠) + 𝑤0
2𝐐𝑐(𝑠) =

−𝑖𝑉𝑠𝛼

𝐿𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

(𝑤0
2 − 𝑠2𝛼)𝐐𝑐(𝑠) =

−𝑖𝑉𝑠𝛼

𝐿𝑠𝛽(𝑠2𝛼 − 𝑤2)
,

𝐐𝑐(𝑠) =
−𝑖𝑉𝑠𝛼

𝐿𝑠𝛽(𝑠2𝛼 −𝑤2)(𝑤0
2 − 𝑠2𝛼)

,

𝐐𝑐(𝑠) =
𝑖𝑉𝑠𝛼

𝐿𝑠𝛽
[

−1

(𝑠2𝛼 − 𝑤2)(𝑤0
2 − 𝑠2𝛼)

] ,

𝐐𝑐(𝑠) =
𝑖𝑉𝑠𝛼

𝐿𝑠𝛽
[

1

(𝑠2𝛼 − 𝑤2)(𝑤2 − 𝑤0
2)

+
1

(𝑤0
2 − 𝑠2𝛼)(𝑤2 −𝑤0

2)
]

𝐐𝑐(𝑠) =
𝑖𝑉𝑠𝛼

𝐿𝑠𝛽(𝑤2 − 𝑤0
2)
[

1

(𝑠2𝛼 − 𝑤2)
+

1

(𝑤0
2 − 𝑠2𝛼)

] ,

𝐐𝑐(𝑠) =
𝑉

𝐿(𝑤2 − 𝑤0
2)
{
𝑖𝑠𝛼

𝑠𝛽
[

1

(𝑠2𝛼 −𝑤2)
+

1

(𝑤0
2 − 𝑠2𝛼)

]} ,

𝐐𝑐(𝑠) =
𝑉

𝐿(𝑤2 − 𝑤0
2)
{
𝑖𝑠𝛼

𝑠𝛽
[

1

(𝑠2𝛼 −𝑤2)
−

1

(𝑠2𝛼 − 𝑤0
2)
]} ,

𝐐𝑐(𝑠) =
𝑉

𝐿(𝑤2 − 𝑤0
2)
[

𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤2)
−

𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 𝑤0
2)
] , 

 

inverse , then 



IHJPAS. 37 (1) 2024 

449 
 

𝑄(𝑡) =
𝑉

𝐿(𝑤2 − 𝑤0
2)
[−cos⁡(𝑤𝑡) + cos⁡(𝑤0𝑡)]

𝑄(𝑡) =
𝑄

𝐿(𝑤2 − 𝑤0
2)
[cos⁡(𝑤0𝑡) − cos⁡(𝑤𝑡)]

𝑄(𝑡) =
𝑉

𝐿(𝑤2 −𝑤0
2)
[−2sin⁡(

(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)]

𝑄(𝑡) =
2𝑉

𝐿(𝑤0
2 − 𝑤2)

sin⁡(
(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
) ,

 

or because sin⁡(−𝑥) = −sin⁡(𝑥) 

𝑄(𝑡) =
𝑉

𝐿(𝑤2 −𝑤0
2)
[−2sin⁡(

(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)]

𝑄(𝑡) =
2𝑉

𝐿(𝑤2 − 𝑤0
2)
sin⁡(

(𝑤 + 𝑤0)𝑡

2
) sin⁡(

(𝑤 − 𝑤0)𝑡

2
)

 

Kuffi Abbass Jawad (KAJ) Transform 

𝐒𝑚 {�̈�(𝑡) + 𝑤0
2𝑄(𝑡) =

𝑉

𝐿
cos⁡(𝑤𝑡)}

𝑣2𝐾(𝑣) +
1

𝑣2
[−𝑣2𝑄(0) − 𝑣�̇�(0)] + 𝑤0

2𝐾(𝑣) =
𝑉

𝐿
[

1

𝑣𝑛−2(𝑣2 + 𝑤2)
]
 

(𝑣2 + 𝑤0
2)𝐾(𝑣) =

𝑉

𝐿
[

1

𝑣𝑛−2(𝑣2 + 𝑤2)
]

𝐾(𝑣) =
𝑉

𝐿
[

1

𝑣𝑛−2(𝑣2 + 𝑤2)(𝑣2 + 𝑤0
2)
]

𝐾(𝑣) =
𝑉

𝐿𝑣𝑛−2
[

1

(𝑣2 + 𝑤2)(𝑣2 + 𝑤0
2)
]

𝐾(𝑣) =
𝑉

𝐿𝑣𝑛−2
[

1

(𝑣2 + 𝑤0
2)(𝑤2 − 𝑤0

2)
−

1

(𝑣2 + 𝑤2)(𝑤2 − 𝑤0
2)
]

𝐾(𝑣) =
𝑉

𝑣𝑛−2(𝑤2 − 𝑤0
2)
[

1

(𝑣2 + 𝑤0
2)

−
1

(𝑣2 + 𝑤2)
]

𝐒𝑚[𝑄(𝑡)] = 𝐾(𝑣) =
𝑉

𝐿(𝑤2 − 𝑤0
2)
[

1

𝑣𝑛−2(𝑣2 + 𝑤0
2)

−
1

𝑣𝑛−2(𝑣2 + 𝑤2)
]

 

inverse , then 

𝑄(𝑡) =
𝑉

𝐿(𝑤2 − 𝑤0
2)
[cos⁡(𝑤0𝑡) − cos⁡(𝑤𝑡)]

𝑄(𝑡) =
𝑉

𝐿(𝑤2 −𝑤0
2)
[−2sin⁡(

(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)]

𝑄(𝑡) =
2𝑉

𝐿(𝑤0
2 − 𝑤2)

sin⁡(
(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)

 

or because sin⁡(−𝑥) = −sin⁡(𝑥) 



IHJPAS. 37 (1) 2024 

450 
 

𝑄(𝑡) =
𝑉

𝐿(𝑤2 −𝑤0
2)
[−2sin⁡(

(𝑤0 + 𝑤)𝑡

2
) sin⁡(

(𝑤0 − 𝑤)𝑡

2
)]

𝑄(𝑡) =
2𝑉

𝐿(𝑤2 − 𝑤0
2)
sin⁡(

(𝑤 + 𝑤0)𝑡

2
) sin⁡(

(𝑤 − 𝑤0)𝑡

2
)

 

Example 3.3. [12] 

Let a body 𝐴 of mass 1 gram move on the  𝑥-axis. It is attracted towards the origin 𝑂 with a force 

equals to 4𝑥. Also, assume that initially it is at rest when 𝑥 = 5; then, determine its position by 

considering: 

1 No other forces acting on it. 

2 damping: force, or, in other words, resistance to the particle, is equal to 8 times the velocity 

at any instant. 

Solution: From Figure 1, for 𝑥 > 0, the net force towards the left is given by 4𝑥, while for 

 
Figure 1.Body Aof mass 1 

 

𝑥 < 0, the net force towards the right is given by 4𝑥. Thus, for both cases, the net force equals 

4𝑥. 

By Newton's second law of motion, mass × acceleration = net force, 

�̈�(𝑡) = −4𝑋(𝑡)

�̈�(𝑡) + 4𝑋(𝑡) = 0
 

The initial conditions are 

𝑋(0) = 5,

�̇�(0) = 0.
 

Complex Sadik Transform 

 

 

 



IHJPAS. 37 (1) 2024 

451 
 

 

𝐒𝑎
𝑐{�̈�(𝑡) + 4𝑋(𝑡) = 0}

(𝑖𝑠𝛼)2𝐗𝑐(𝑠) −
�̇�(0)

𝑠𝐵𝛽
−
𝑖𝑠𝛼𝑋(0)

𝑠𝛽
+ 4𝐗𝑐(𝑠) = 0

−𝑠2𝛼𝐗𝑐(𝑠) −
5𝑖𝑠𝛼

𝑠𝛽
+ 4𝐗𝑐(𝑠) = 0

(4 − 𝑠2𝛼)𝐗𝑐(𝑠) =
5𝑖𝑠𝛼

𝑠𝛽

𝐗𝑐(𝑠) =
5𝑖𝑠𝛼

𝑠𝛽(4 − 𝑠2𝛼)

𝐗𝑐(𝑠) =
−5𝑖𝑠𝛼

𝑠𝛽(𝑠2𝛼 − 4)

 

inverse 

𝑋(𝑡) = 5 cos(2𝑡). 

 Kuffi Abbass Jawad (KAJ) Transform  

𝐒𝑚{�̈�(𝑡) + 4𝑋(𝑡) = 0},

𝑣2𝐾(𝑣) +
1

𝑣2
[−𝑣2𝑋(0) − 𝑣�̇�(0)] + 4𝐾(𝑣) = 0,

𝑣2𝐾(𝑣) −
5𝑣2

𝑣𝑛
+ 4𝐾(𝑣) = 0,

𝐾(𝑣) =
5

𝑣𝑛−2
,

𝐒𝑚[𝑋(𝑡)] = 𝐾(𝑣) =
5

𝑣𝑛−2(𝑣2 + 4)

 

inverse 

𝑋(𝑡) = 5 cos(2𝑡). 

4. Conclusions 

We conclude that the complex Sadik transformation and the KAJ transform each other to an 

exact solution, and both are effective solutions. The two new integral transformations, the 

complex Sadik transformation and the Kuffi-Abbas-Jawad transformation, provide an exact 

solution for some mechanical and electrical theorems. 

Acknowledgment 

 Our researcher extends his Sincere thanks to the editor and members of the preparatory committee 

of the Ibn AL-Haitham Journal of Pure and Applied Sciences . 

Conflict of Interest  

   There are no conflicts of interest. 

Funding  

There is no funding for the article. 

 



IHJPAS. 37 (1) 2024 

452 
 

References 

1. Amjed,U.;  Moazzam,A.;  Kashif ,M.; Khawar, M.I.  Development of  a New Transformation  

to Solve a New Type of  Ordinary Linear Differential Equation. Bulletin Of Mathematics And 

Statistics Research, 2021, 9(3),56–60. 

2. Abbas,E.S.; Kuffi,E.A.; Jawad,A.A. New Integral kuffi-abbas-jawad kaj Transform and its 

Application on Ordinary Differential Equations. Journal of Interdisciplinary Mathematics, 

2022, 25(5),1427-1433. 

3. Burqan,A.; Qazza,A.; Saadeh,R. A New Attractive Method in Solving Families of Fractional 

Differential Equations by a New Transform. Mathematics 2021, 9(23),1–14. 

4. Saadeh,R.; Qazza,A.; Amawi,K. A New Approach Using Integral Transform to Solve Cancer 

Models. Fractal and Fractional ,2022, 6(9),1–12.  

5. Saadeh,R.; Qazza,A.; Burqan,A.; Khalil,R. Applications on Double ARA–Sumudu Transform 

in Solving Fractional Partial Differential Equations. Symmetry, 2022,14(9),1–17.  

6. Abbas,M.; Akgul,A.; Gokkaya,Z.; Abdullah,F.A. New Applications of  Fractional Differential 

Equations by General Integral Transforms. Available at: https://www.researchsquare.com/article/rs-

2657260/v1.  

7. Akgül ,A.; Ülgül ,E.; Sakar,N.; Bilgi,B.; Eker,A . New Applications of the New General 

Integral Transform Method with Different Fractional Derivatives. Alexandria Engineering 

Journal, 2023, 80(1),498–505. 

8. Saleem,N.; Moazzam,A.; Anjum,A.; Kuffi,E.A. Study of Telegraph Equation via Hefractional 

Laplace Homotopy Perturbation Technique. Ibn Al-Haitham Journal for Pure and Applied 

Sciences, 2023, 36(3),349–364. 

9. Saadeh,R.; Burqan,A,; Qazza,A. A Novel Numerical Approach in Solving Fractional Neutral 

Pantograph Equations via The ARA Integral Transform. Symmetry ,2021, 14(50),1–20.  

10. Saadeh,R.; Burqan,A.; El-Ajou,A.; Al-Smadi,M. A New Efficient Technique Using Laplace 

Transforms and Smooth Expansions to Construct  a Series Solution to The Time-Fractional 

Navier-Stokes Equations. Alexandria Engineering Journal 2022, 61(2),1069–1077. 

11. Moazzam,A.; Amjed,U.; Mubarak,F.; Iqbal, M.Z.; Naeem, M.U.. Substitution Method Using 

The Laplace Transformation for Solving Partial Differential Equations Involving More Than 

Two Independent Variables. Bulletin Of Mathematics And Statistics Research, 2021, 9(3),104–

116. 

12. Iqbal,K.; Kalim,M.; Khan,A. Applications of karry-kalim-adnan transformation (kkat) to 

Mechanics and Electrical Circuits. Journal of Function Spaces 2022, 2022(1),1-11. 

13. Jaabar,S.M.; Hussain,A.H. Solving Volterra Integral Equation by Using a New 

Transformation. Journal of Interdisciplinary Mathematics 2021, 24(3),735–741. 

14. Jabber,A.K; Hanna,E. New  transform Fundamental Properties and its Applications. Ibn AL- 

Haitham Journal For Pure and Applied Science, 2018, 31(2),151–163. 

15. Mehdi,S.A; Kuffi,E.A.; Jasim,J.A. Solving Ordinary Differential Equations Using a New 

General Complex Integral Transform. Journal of Interdisciplinary Mathematics, 2022, 

25(6),1919-1932. 

16. Waseem,M; Ali ,L.; Talib,T.; Ali,M.; Iqba,R.; Khawar,M. Fficacy of Two Approaches of 

Transforaminal Epidural Injections in Patients of Lumbar Radiculopathy. Pakistan Armed 

Forces Medical Journal ,2021, 71(4),1368–1371. 

17. Moazzam,A.;  Ayoub,A.; Kashif,M.; Shehzad,K.; Bibi,M. New Transformation ’AMK-

transformation’ to Solve Ordinary Linear Differential Equation of Moment Pareto 

https://www.researchsquare.com/article/rs-2657260/v1
https://www.researchsquare.com/article/rs-2657260/v1


IHJPAS. 37 (1) 2024 

453 
 

Distribution. International Journal of Multidisciplinary Research and Growth Evaluation, 

2021, 2(5),125– 134. 

18. Moazzam,A.; Waqas,M.; Shehzad,K.; Batool,A. Applications of Kamal Transformation in 

Temperature Problems. Scholars Journal of Engineering and Technology 2022, 10(2),5– 8. 

19. Qazza,A. A New Computation Approach: ARA Decomposition Method. WSEAS Transactions 

on Mathematics 2023, 22(29),245–252. 

20. Qazza,A. Solution of Integral Equations via Laplace ARA Transform. European Journal of 

Pure And Applied Mathematics, 2023, 16(2):,919–933. 

21. Qazza,A.; Saadeh,R.; Burqan,A. A New Integral Transform: ARA Transform and its Properties 

and Applications. Symmetry ,2020, 12(6),:1–15. 

22. Qazza,A.; Saadeh,R.; Burqan,A. On the Double ARA-Sumudu Transform and its 

Applications. Mathematics, 2022, 10(15),1–14. 

23. Sedee,A.K.H. Some Properties and Applications of a New General Triple Integral Transform 

Gamar Transform.Complexity ,2023, 2023(1),1–21, 2023.   

24. Burqan,A.; Saadeh,R.; El-Ajou,A. Reliable Solutions to Fractional Lane-Emden Equations via 

Laplace Transform and Residual Error Function. Alexandria Engineering Journal 2022, 

61(12),10551–10562. 

25. 25.Zenteno,M.Sc; Roberto,J.J. Applications of the ZJ Transform for Differential Equations. 

International Journal of Latest Research in Engineering and Technology, 2023, 9(2),7–12,. 

26. Saadeh,R.Z.; Ghazal,B.F. A New Approach on Transforms: Formable Integral Transform and 

its Applications. Axioms ,2021, 10(4),1–21. 

27. Saxena,H.; Gupta,S. A new integral transform called saxena and gupta transform and relation 

between new transform and other integral transforms. Global Journals, 2023, 23(4),9–23. 

28. Qazza,A.; Ahmed,S.A; Saadeh,R. Exact Solutions of Nonlinear Partial Differential Equations 

via the New Double Integral Transform Combined with Iterative Method. Axioms 2022, 

11(6),1–16. 

29. Turq,S.M.; Kuffi,E.A. The New Complex Integral Transform Complex Sadik Transform and 

it's Applications. Ibn AL- Haitham Journal For Pure and Applied Science ,2022, 35(3),120-

127. 

30. Xhaferraj,E.M. The new integral transform: NE transform and its applications. European 

Journal of Formal Sciences and Engineering 2023, 6(1),22–34.