Steps of Exact and Analytic Solutions of Ordinary Differential
Equations using MAHA Integral Transform and Its Applications
MAHA ALSAOUDI1, AHMAD SHARIEH2, AHLAM GUIATNI3,*, GHARIB GHARIB4
1Applied Science Private University,
JORDAN
2University of Jordan,
JORDAN
3The University of Jijel,
Algeria
4Zarqa University,
JORDAN
*Corresponding Author
Abstract: - Transformation such as integral transform is needed to obtain the exact solutions for linear ordinary
differential equations (ODEs) with constant coefficients of higher orders. MAHA transformation exact solution
of ODEs is simpler and easier than the previous with two parameters. The major steps of this transform are
applying the MAHA transform on the given equation followed by
taking t h e inverse transform. The
general steps in numerical solutions involve defining the ODE as a function, defining initial conditions and the
range of the independent variable, using an appropriate ODE solver function, and calling the solver and plotting
the solution using a programming language. The exact and analytical solutions are validated. Both methods are
easy and simple to be deployed in computing scientific applications such as nuclear physics and medical
applications.
Key-Words: - MAHA transform, Integral transform of two parameters, Inverse of MAHA transform, Medical
Applications, Nuclear Physics, Ordinary differential equations.
Received: July 9, 2024. Revised: November 13, 2024. Accepted: December 8, 2024. Published: December 31, 2024.
1 Introduction
One of the most important aspects of mathematics is
differential and integral equations. These provide
tools to solve problems in our world. Thus, this
paper explains the applicability of the MAHA
transform in solving ordinary differential equations
(ODEs) and validation of the solutions by using a
programming language.
The MAHA transform was introduced to
address common and intermediate differential
equations in the time domain. Transformations such
as [1], [2], [3], [4], [5], serve as valuable numerical
tools for solving differential equations. Similarly,
the MAHA transform and its essential properties are
applied to tackle differential equations. Algorithms
are needed to implements these transforms and their
applications, [6]. Thus, we aim to show how MAHA
transform solves ODEs and validate the solution
using programming language such MATLAB.
The MAHA transform for capacities in the set A
is characterized as in formula (1).
A
=
{
f
(
t
)
:
ther
e
exist
m
,
l
1
,
l
2
>
0
.
|
f
(
t
)
|
<
m
e
l
i
|
t
|
,
if
t
∈
(
−
1)
i
x
[0
,
∞
)
}
(1)
where m is constant limited number, and l1, l2 might
be limited or boundless. MAHA fundamental
change is signified by the administrator M(.), which
is characterized by the vital condition as in formula
(2).
(2)
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DOI: 10.37394/23206.2024.23.100
Maha Alsaoudi, Ahmad Sharieh,
Ahlam Guiatni, Gharib Gharib
E-ISSN: 2224-2880
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Volume 23, 2024
The parameters u and v in this essential
transform are employed to compute the variable t,
the argument of the function f. This indispensable
transform exhibits additional connections with [7],
[8].
The purpose of this study is to demonstrate the
significance of this intriguing novel transform and
its efficiency in solving linear differential equations,
[9], [10], [11].
The MAHA transform is used in this work to
solve ODEs, as well as several practical applications
which are considered basic in mathematical physical
fields. The main contributions of this paper are: a
review of MAHA transformation and its properties
to tackle differential equations, design numerical
algorithm to solve the differential equations, and
validate the exact solution by MAHA
transformation and the numerical analysis solution.
The programming codes are written in MATLAB.
In the methodology section, the steps of MAHA
transformation and algorithms of exact and
analytical solutions with their implementation are
presented. The exact and the numerical analysis
solution are validated on some sample of ordinary
differential equations.
Exact solution of nuclear physic and medical
applications are presented. The exact and numerical
solutions are validated.
In section 2, a related work is presented. In
section 3, the methodology presents the steps of
MAHA transformation for exact solution and
analysis methods for the selected equations. Section
4 presents results of solving sample of equations and
applications by deploying MATLAB. Section 5
concludes the manuscript.
2 Related Works
Through studying and researching integral
transformations and their applications in life, we
found that there are many integral transformations
with two parameters. The most important of these,
according are ZZ transformations, [12].
The value of , in formula (2), belongs to the set
of integers, gives transformation the generality and
the preference by using it in important applications.
Note that when = 0, the MAHA transform
becomes Shehu transform. When = -1, the MAHA
transform becomes KKAT transform and Quideen
transform. The Maha integral transformation is
more general transformation than the previous
transformations with two parameters. Through this
, we can choose the appropriate number for the
application of physics, engineering, or a life
application to convert the ordinary differential
equation into an easy and simple algebraic equation.
Then, by taking the inverse of this transform, we
can obtain the exact solution.
The value of can be referred to as, for
example, the number of samples taken in the
application. There are many mathematical models in
the linear modeling that depend on this , for
example, drug concentration problems, chemical
problems, and others.
Through our study and our knowledge of many
integral transforms with one parameter or two
parameters or more, there is no preference between
one transform and another, but there is an important
matter, which is that each transform has a use for
some applications through after converting
differential equations, integral equation or systems
into algebraic equations or algebraic system.
Conversion is made easier by simplifying that
application, [13], [14], [15], [16]. Also, algorithms
are essential to compute values in such systems.
MATLAB programming language can be
utilized to solve ordinary differential equations
analytically and numerically. It offers various
numerical methods to solve linear ordinary
differential equations (ODEs). The primary function
for this purpose is ode45, but there are other
functions which support in solving different types of
ODEs and specific requirements. The ode45 being
the most commonly used for non-stiff problems.
The ode45 uses the Dormand-Prince method, a
variable-step, variable-order (Runge-Kutta) method.
To numerically solve a 2nd ode, for example, you
first convert it to a system of first-order ODE.
The general steps to solve ODEs numerically in
MATLAB are:
1. Define the ODE as a function.
2. Define initial conditions and the range of the
independent variable.
3. Use an appropriate ODE solver function
(ode45, etc.).
4. Call the solver and plot the solution.
These steps will be followed when the
MATLAB code is deployed to solve odes in the
methodology and applications sections.
3 Methodology
3.1 MAHA Integral Transform of Some
Functions
Assuming the existence of the fundamental data (2)
for any function f(t), the sufficient data for the
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presence of the MAHA integral transform are that,
for t t greater than or equal to 0, f(t) should be
piecewise continuous and of exceptional order.
Otherwise, the MAHA transform may or may not
exist. The MAHA integral transform has basic
capacities:
1) If f(t) = k, k is an arbitrary constant
function, then via the definition we get:
2) If f(t) = t so:
So, by integration by parts,
Also:
(i)
(ii)
(iii) In general, if n positive integer number, then
, and if n > -1, then
, where is
Gamma function.
3) If f(t) = , where a is an arbitrary
constant number, so:
,
also
4) If f(t) = sin(at), where a is an arbitrary
constant number, so:
5)
If f(t)
=
cos(at), a is
an arbitrary
constant
number, so:
After simple computations, we get:
6)
If f(t)
=
sinh(at), a is
an arbitrary
constant
number, so:
After simple computations, we get:
7)
If f(t)
=
cosh(at), a is
an arbitrary
constant
number, so:
8)
Shifting property of MAHA integral
transform
If MAHA integral transform of f(t) is F(u,v),
then MAHA transform of function
is
given by
.
Proof
dt
Theorem (3.1)
(i) +
F(u,v)
(ii)
+
F(u,v)
(iii)
+
F(u,v)
(iv)
+
F(u,v)
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(iiv)
+
F(u,v)
Proof
(i) By the definition, we get:
Integration by parts method, we have:
(ii)
Also, by the definition, we get:
Also, Integration by parts method, we obtain:
The proof of (iii) and (iv) is similar to (ii).
(iiv) We can confirm by Mathematical
Induction.
3.1.1 The Inverse of MAHA Integral
Transform
In this part, we present the inverse of MAHA
transform technique of basic functions:
(1)
, where n > 0 integer
number.
, where a is a constant
number.
(4)
(5)
(6)
(7)
(8)
3.2 Application of MAHA Integral
Transform of Ordinary Differential
Equations (ODEs)
The MAHA necessary transform can be utilized as
a successful device to solve ordinary differential
equations. The steps of an exact solution of
differential equation using MAHA transform are:
1)
Apply MAHA transform on the given
equation utilizing the initial conditions.
2) Take inverse transform, you will get the
exact solution.
3) Write a programming code to find values of
the dependent variables based on the values
of independent variable.
The steps of a numerical analysis solution of a
differential equation are:
1. Define the differential equation as a system
of first-order ODE.
2. Fill in the initial conditions.
3. Set the range values of the independent
variable as needed.
4. Solve the differential equation (you can use
a built in function of a programming
language).
5. Extract and plot the solution.
Example 1:
1
st
order linear ODE
Consider the differential equation:
(3)
Exact Solution:
1 ) Take MAHA integral transform to this
condition, you get:
So
2) Take inverse to both sides, you get:
The exact solution:
The MATLAB code Solution:
3) % Define the range of x
x = linspace(0, 5);% as needed
% Compute y(x)
y = exp(-x);
% Plot the function
figure; plot(x, y, 'LineWidth', 2);
xlabel('x');ylabel('y'); title('y(x) = e^{-
x}');grid on;
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Fig. 1: On the interval [0, 5], plot exact solution of
Numerical Analysis Solution using the built-in
function ode45:
% Define the differential equation function: y’ = -
y
ode = @(x, y) -y;
% Define the initial condition
y0 = 1;
% Define the range of x values over which to solve
the equation
xspan = [0 5]; %adjust as needed
% Solve the differential equation
[x, y] = ode45(ode, xspan, y0);
% Plot the solution
plot(x, y, '-o'); xlabel('x');
ylabel('y(x)');
title('Solution of dy/dx + y = 0');grid on;
Fig. 2: Numerical solution of dy/dx +y =0, y(0)
=1 on the interval [0, 5]
Figure 1 and Figure 2 show the plot of the
exact and numerical solutions, respectively.
These visually validate the solutions by the two
methods.
Example 2: 2
nd
DE
Find the solution of the second-order differential
equation y′′+y=0, with initial conditions y(0)=1 and
y′(0)=1
:
′′ ′ (4)
Exact Solution:
1 ) Take MAHA transform to this differential
condition, you will get:
So
+
2) Take the inverse MAHA transform of this
equation to get the exact solution:
The MATLAB code Solution:
3) Use MATLAB code, and compute:
y = sin(x)+ cos(x);
This is validated by the following analytical
algorithm (Figure 3).
Numerical Analysis Solution Using the built-in
function ode45:
In this case, the equation y′′+y=0 is rewritten as a
system of first-order ODEs: y1′=y2, y2′=−y1.Here,
y1=y, and y2=y′. Therefore, ode is defined as a
function that takes x (the independent variable) and
y (a dependent variable as a vector containing y and
y′) and returns the derivatives [y'(x); y''(x)]. The
built-in function ode45 is used.
% Rewrite the differential equation y'' + y = 0 as a
system of first-order ODEs:
% y1 =y and y2 =y' % y1' =y2 and y2' = -y1
% Define the system of ODEs
dydt = @(t, y) [y(2); -y(1)];
y0 = [1; 1]; % y(0) = 1, y'(0) = 1
% Define the time span for the solution
tspan = [0 10]; % adjust as needed
% Solve the system of ODEs using ode45
[t, y] = ode45(dydt, tspan, y0);
% Plot the solution
figure;plot(t, y(:,1), '-o');
xlabel('Time t');ylabel('Solution y');
title('Solution of the differential equation y'''' + y =
0');
grid on;
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Fig. 3: Virtual Validation of exact and numerical
solution of
Example 3:
2
nd
request ODE
Consider
the
2
nd
request
di
ff
eren
tial
equation:
′′ ′′ (5)
Exact Solution:
1) Take MAHA transform to above
differential equation, we get:
So
Now
After simple computations, we get: A=3, B=-2.
2)Apply inverse
to get general exact solution as:
3) Use MATLAB and compute:
y = 3*exp(2*x)- 2*exp(x);
This is validated by the following analytical
algorithm (Figure 4).
Numerical Analysis Solution Using the built-in
function ode45:
To solve y′′−3y′+2y=0, with initial conditions
y(0)=1 and y′(0)=4, for example, in MATLAB, you
can use the ode45 function as follows. Let y1′=y2,
y2′=3y2−2y1; where y1′ =y and y2=y′. Therefore,
ode is defined as a function that takes x and y (a
vector containing y and y′) and returns the
derivatives [y'(x); y''(x)].
% Define the differential equation as a system of
first-order ODEs.
ode = @(x, y)[y(2);3*y(2)- 2*y(1)];
% Initial conditions
y0 = [1; 4]; % [y(0); y'(0)]
xspan = [0 10]; % Adjust as needed
% Solve the differential equation
[x, sol] = ode45(ode, xspan, y0);
y = sol(:, 1); % y(x)
yp = sol(:, 2); % y'(x)
%Plot the solutions
figure; subplot(2, 1, 1);
plot(x, y, '-o'); xlabel('x');
ylabel('y(x)');
title('Solution of y'''' - 3y'' + 2y = 0'); grid on;
subplot(2, 1, 2);plot(x, yp, '-o');
xlabel('x');ylabel('y''(x)');
title('Derivative of y(x)');grid on;
Fig. 4: Virtual validation of exact
and analytical solution of
Example 4:
2nd order linear nonhomogeneous
Consider 2nd order linear nonhomogeneous request
differential equation:
′′
(6)
Exact Solution:
1) Since y’ (0) is unknown, let y’(0)=a.
Take
MAHA transform of this condition and utilizing
beginning conditions, you will have:
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So
After simple computations, you will get:
2)Take inverse of MAHA transform, then the
exact solution is:
To find a, note that
Then, we find a =12/5.
Then, the exact solution is
3) Use MATLAB and compute:
y= 0.8*sin(3*x)+0.2*cos(2*x)+0.8*cos(3*x)
This is validated by the following analytical
algorithm (Figure 5).
Numerical Analysis Solution using the built-in
function ode45:
To solve the differential equation y′′+9y=cos(2x)
with the boundary conditions y(0)=1 and y(π/5)=−1
in MATLAB, you can use the bvp4c function which
is designed for boundary value problems.
syms y(x)
Dy = diff(y);
D2y = diff(y, 2);
% Define the differential equation
ode = D2y + 9*y == cos(2*x);
cond1 = y(0) == 1; cond2 = y(pi/5) == -1;
sol = dsolve(ode, [cond1, cond2]);
x_vals = linspace(0, 2*pi, 1000);
y_vals = double(subs(sol, x, x_vals));
figure; plot(x_vals, y_vals, '-o');
xlabel('x');
ylabel('y');
title('Solution of the differential equation y'''' + 9y =
cos(2x)');
grid on;
Fig. 5: A solution of
′′
Example 5:
2nd ode
Consider the
di
ff
eren
tial
equation
′′ ′′(7)
Exact Solution:
1) Take MAHA technique of this differential
problem and appling the initial data:
So
+
]
We get: A = 2, B = 4, C = −9
2)Take inverse transform, we obtain:
3) Use MATLAB and compute:
y= 2*exp(3*x)+4*exp(2*x)-9*exp(x)
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Plot the function on the interval [0, 1], for example
(Figure 6).
Numerical Analysis Solution using the built-in
function ode45:
To solve the 2nd ode y′′−3y′+2y=4e3x, with initial
conditions y(0)=−3 and y′(0)=5, the non-
homogeneous term 4e3x needs to be handled
separately from the homogeneous equation. In
MATLAB, the analytical solution can be
implemented using ode45 as follows.
% Define the differential equation as a system of 1st
ODEs
ode = @(x, y) [y(2); 3*y(2) - 2*y(1) +
4*exp(3*x)];
% Initial conditions
y0 = [-3; 5]; % [y(0); y'(0)]
% Define the range of x values
xspan = [0 1];
% Solve the differential equation
[x, sol] = ode45(ode, xspan, y0);
% Extract solutions
y = sol(:, 1); % y(x)
yp = sol(:, 2); % y'(x)
% Plot the solutions
figure;
subplot(2, 1, 1);plot(x, y, '-o');
xlabel('x');
ylabel('y(x)');
title(' solution of y′′−3y′+2y=4e3x ');
grid on;
subplot(2, 1, 2);
plot(x, yp, '-o');xlabel('x');
ylabel('y''(x)');
title(' y=2e^(3x)');
grid on;
Fig. 6: Visualize validation of exact and numerical
solutions, on the interval [0, 1], of
′′ ′′
3.3 Applications
3.3.1 Maha Integral Transform on ”Nuclear
Physics”
The following problem is based on nuclear
physics fundamentals. Consider the 1st order linear
ordinary differential equation:
(8)
The essential relationship describing
radioactive decay is given in this equation, where
N(t) during time t denotes the number of not
decayed atoms left in a sample of radioactive
isotope, and α is the decay constant. Apply the
Maha integral transform M, to set:
′ (9)
Therefore
So
(10)
Take the inverse to both sides, we get the exact
solution: (11)
This is the proper type of radioactive decay.
You can solve dN(t)/dt =−αN(t) analytically in
MATLAB, where α (alpha) is a constant, using
ode45 function as follows. The plot in Figure 7
visualizes how the population N(t) evolves over
time t.
% Define the parameters
alpha = 0.1; % for example
% Define the differential equation function
ode = @(t, N) -alpha * N;
N0 = 100;% Initial population size
tspan = [0 10];% Adjust as needed
% Solve the differential equation
[t, N] = ode45(ode, tspan, N0);
% Plot the solution
plot(t, N, '-o');
xlabel('Time (t)');
ylabel('N(t)');
title(['Solution of dN(t)/dt = -\alpha N(t), \alpha = ',
num2str(alpha)]);
grid on;
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Fig. 7: Plot of the numerical solution of
, with initial value N(0) =100
3.3.2 Blood Glucose Concentration
During continuous intravenous glucose injection,
the concentration of glucose in the blood is G(t)
exceeding the baseline value at the start of the
infusion. The function G(t) satisfies the initial
value problem (I, V, P).
The variables k, α and γ in equation ( 1 2 )
represent the constant
velocity of elimination, the
rate of infusion, and the volume, respectively, in
which glucose is distributed. The Maha integral
transform technique can be utilized to solve the
equation in (12). Then, the
concentration of
glucose presents in the blood stream at time t is
G(t)=α/(γk)(1−e−kt)
.
Upon bilateral application of the Maha
integral transform on (12), the resulting
expression is obtained as in equation (13).
(13)
Let M { G(t)}
= F (u, v). By utilizing
the initial value problem (I, V, P) and the
integral
transform outlined in section 3.1, the
equation (13) can be rearranged with the aid of
equation
(12) as:
So
(14)
After simple computation and using inverse
of Maha transform to this expression, we get the
concentration of glucose in the blood as in (15).
(15)
A MATLAB code can be utilized to produce a
plot showing the exact solution of G(t) as a function
of time t (Figure 8). The exact formula
G(t)=α/(γk)(1−e−kt) is directly used to compute G(t),
ensuring accuracy and validation of the solution.
The parameters alpha, gamma, k, and the range of
t values (tspan) can be adjusted as needed for a
specific scenario. Note that the “%” in the
programming code is for comments and not for
execution statements.
% Define the parameters
alpha = 1.0; gamma = 1.0;k = 0.1;
% Define the function for G(t)
G_exact = @(t)(alpha/(gamma*k)) *(1-exp(-k*t));
tspan = [0 50]; % example time span
G0 = 0; % initial value of G
G = G_exact(t);
% Plot the exact solution
plot(t,G,'-o'); xlabel('Time (t)');
ylabel('G(t)');
title('G(t)=\alpha/(\gamma k)(1-e^{-kt})');
grid on;
This code is self-contained and can be executed
directly in MATLAB. Adjust the parameter values
as needed to fit your specific problem. The
following code solves the given differential equation
analytically and plots the solution (Figure 8).
% Define parameters
alpha = 1;
gamma = 1;
k = 0.1;
tspan = [0 50]; % example time span
G0 = 0; % initial value of G
% Define the differential equation as a function
dGdt = @(t, G) (alpha / gamma) - k * G;
% Solve the differential equation using ode45
[t, G] = ode45(dGdt, tspan, G0);
% Plot the solution
figure;
plot(t, G, 'LineWidth', 2);
title('Solution of dG/dt = (\alpha / \gamma) - k \cdot
G');
xlabel('Time (t)');
ylabel('G(t)');
grid on;
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Fig. 8: Plotted solution of G(t) with alpha = 1.0;
gamma = 1.0;k = 0.1
3.3.3 Aorta Pressure
The heart’s contraction facilitates the
transportation of blood into the aorta. The
initial value problem is concerned with the aortic
pressure function P(t) as in equation (16).
′
(16)
where P(0)
=
p
0
, and
c, k, A, w are constants.
The Maha integral transform technique is
utilized to derive
the pressure in the aorta.
With
the bilateral application of the Maha integral
transform to equation (16). The resulting
expression is as in (17).
′
(17)
By applying the initial value problem and
utilizing the transform outlined in section 3.1, the
rearrangement of equation (17) can be expressed
as:
(18)
After preforming basic calculations and applying
partition fractions along with the inverse of the
Maha integral transform to the given expression,
the resultant value obtained represents the amount
of pressure in the aorta as in function (19).
(19)
Figure 9 shows visualization solution, by the
following MATLAB code to analytically solve and
plot the given function P(t). Figure 10 shows a
numerical solution for the same ODE.
% Define the parameters
P0 = 0; % Initial value
c = 1; k = 1; A = 1; w = 1;
t = linspace(0, 100, 1000);
% Define the function P(t)
P = @(t) P0*exp(-c/k * t)+(c*A*w*k^2)/
(w^2*k^2+c^2)*((c/(w*k))*sin(w*t)-
cos(w*t)+exp(-c/k*t));
% Evaluate the function
P_vals =P(t);
plot(t, P_vals, 'LineWidth', 2);
title('Solution to P(t)');
xlabel('Time t');ylabel('P(t)');
grid on;
Fig. 9: Plot of exact solution of P(t)’
% Define the parameters
c = 1;k = 1; A = 1;w = 1;
% Define the differential equation
ode = diff(P, t) + (c/k) * P == c * A * sin(w * t);
% Define the initial condition
cond = P(0) == 0;
% Solve the differential equation
sol = dsolve(ode, cond);
% Convert the symbolic solution to a MATLAB
function
P_sol = matlabFunction(sol);
t_vals = linspace(0, 100, 1000);
P_vals = P_sol(t_vals);
% Plot the solution
figure; plot(t_vals, P_vals, '-o');
xlabel('Time t'); ylabel('P(t)');
title('Solution of P''(t) + c/k P(t) = cA sin(wt)');grid
on;
WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2024.23.100
Maha Alsaoudi, Ahmad Sharieh,
Ahlam Guiatni, Gharib Gharib
E-ISSN: 2224-2880
979
Volume 23, 2024
Fig. 10: Plot of numerical solution of P(t) with P(0)
=0, and c=k =A=w =1
t = linspace(0, 100, 1000);
The results show that the MATLAB codes for
the solutions are matched and agreed with the
theoretical approach. The programming codes
present examples and will help users in writing
computing solutions to problems similar to the
presented ones.
4 Conclusions
MAHA integral transform with two parameters
conversion of linear ordinary differential equations
with constant coefficients and higher orders are
extended. The correctness of the transform is
proved in the methodology section. Steps to solve
ODEs using MAHA integral transform are
presented. The steps are applied to find the exact
solutions of five different examples of ODEs and
three different examples of applications. The steps
to solve DOEs numerically and validate the exact
solution are presented. MATLAB codes are
deployed to show exact (direct) solutions and
analytical solutions for the selected eight ODEs.
The exact solutions and the numerical ones for
given functions are validated and plotted. The two
methods are applied to find exact solutions and
numerical ones of nuclear physics and two medical
applications. It is found that the exact solution is
simpler and easier than the previous two
parameters, and it can be numerically validated.
The presented programming code will be helpful
for users interested in computing scientific
numerical applications.
As a future work, we intend to investigate the
performance of Maha transform in enhancing the
security of image encryption/ decryption. The
image encryption process starts with representing
the image as numerical data. Then, applying Maha
transformation on these values through ODE
results in an encrypted version of the image. For
decryption, if the correct initial conditions and
ODE system are given, then the original image can
be retrieved by reversing the transformation.
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Belafhal, R. EL Aitouni, T. Usman,
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WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2024.23.100
Maha Alsaoudi, Ahmad Sharieh,
Ahlam Guiatni, Gharib Gharib
E-ISSN: 2224-2880
980
Volume 23, 2024
models
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Contribution of Individual Authors to the
Creation of a Scientific Article (Ghostwriting
Policy)
Conceptualization, Alsaoudi and Sharieh;
methodology, Sharieh; validation Alsaoudi and
Guiatni; formal analysis, Alsaoudi; investigation,
Alsaoudi; resources, Gharib; data curation, Sharieh;
writing/original draft preparation, Guiatni;
writing/review and editing, Guiatni; supervision,
Gharib; project administration, Gharib; steps of the
solutions and programming language code, Sharieh.
Sources of Funding for Research Presented in a
Scientific Article or Scientific Article Itself
No funding was received for conducting this study.
Conflict of Interest
The authors have no conflicts of interest to declare.
Creative Commons Attribution License 4.0
(Attribution 4.0 International, CC BY 4.0)
This article is published under the terms of the
Creative Commons Attribution License 4.0
https://creativecommons.org/licenses/by/4.0/deed.en
_US
WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2024.23.100
Maha Alsaoudi, Ahmad Sharieh,
Ahlam Guiatni, Gharib Gharib
E-ISSN: 2224-2880
981
Volume 23, 2024
