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1. UtilitasMathematica
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99
Exact Solutions to a Mathematical Model Representing Avascular Tumor
Growth Via Exponential Function Method
Afaf N. Yousif1, Ahmed Farooq Qasim2
1
Department of mathematics, College of Computer Sciences and Mathematics, University of
Mosul, Mosul, Iraq.
2
Department of mathematics, College of Computer Sciences and Mathematics, University of
Mosul, Mosul, Iraq, ahmednumerical@uomosul.edu.iq
Abstract
In this paper, conducted a study on avascular tumor growth of partial differential system. As has
been suggested a modification of this mathematical model was proposed, then we found the exact
solutions for this model based on the new exponential-function method Without the need to
convert the system from partial differential equations to ordinary differential equations. The
solution of the system provides us with the possibility to study the influence of parameters on the
spread of disease and how to control and then eliminate it. The effect of parameters on the spread
and growth of disease was studied, after obtaining a general formula for exact solutions for
nonlinear system, and then studying the effect of increasing or decreasing these parameters and
the effect of their absence on the disease growth.
Keywords: Exact solutions, Avascular tumor growth, Exponential-function method.
1. Introduction
Mathematical modeling is a powerful tool for testing hypotheses, confirming experiments, and
simulating the dynamics of complex systems as well as helping to understand the mechanistic
underpinnings of complex systems in a relatively quick time without the huge costs of laboratory
experiments and biological changes, particularly in oncology. It should also be noted that the
tumor is a mass of tissue formed by the division of cells at an accelerated rate [1,2,3,4,5]. There
are several mathematical models that describe natural tumor growth, scientist stein A. described
in [6] a linear model describing the natural tumor growth of renal cell carcinoma based on certain
measurements. In [7] Claret L. presented a new exponential model that assumed that the tumor
growth rate is proportional to the tumor burden as it was adopted in the process of tumor
suppression. Also Gardner S. N. in [8] relied on partial differential equations to describe the
growth of brain tumors. Among the wide applications in this field is the proliferation-invasion
model, which assumes that net proliferation and invasion contributes to tumor growth. There are
many methods that find us exact traveling wave solutions, including the tanh method, the tan-
expansion method, the exponential-function method and the 𝐺′
/𝐺 -expansion method. Malik A.

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et al found in [9] exact traveling wave solutions Using the (
𝐺′
𝐺
)-expansion method of the
Bogoyavlenskii equation, In [10] Raslan K. R. et al., a traveling wave solution was created for
various types of nonlinear partial differential equations using the modified extended tanh method,
reduced to ordinary differential. dish Ic U. in [11] used tan expansion method On the equation
(2 + 1) -dimensional burgers equation to find traveling wave solutions. In [12] Javeed S. et. al.
Use the exponential function method to find the exact solutions of burger’s equation, zakharov-
kuznetsov and kortewegede vries Equation. It was found that this method is useful and effective
in solving nonlinear conformable (PDEs) and simpler and more efficient than other methods. The
Exp-function method has been successfully applied to many kinds of NPDEs, such as, Kawahara
equation [13], Boussinesq equations [14], Double Sine-Gordon equation [15], Fisher equation
[16], and the other important nonlinear partial differential equations [17]. In this paper we apply
the Exp-function method to obtain exact solutions of nonlinear partial differential equations,
namely, of avascular tumor growth.
2. Mathematical Model
we will express a multicellular model of avascular tumor growth by a nonlinear system of
partial differential equations as follows: [18]
𝜕𝑝
𝜕𝑡
=
𝜕
𝜕𝑥
[
𝑝
𝑝+𝑞+𝑠
𝜕
𝜕𝑥
(𝑝 + 𝑞 + 𝑠)] + 𝑔(𝑐)𝑝(1 − (𝑝 + 𝑞 + 𝑠 + 𝑛)) − 𝑓(𝑐)𝑝 (1)
𝜕𝑞
𝜕𝑡
=
𝜕
𝜕𝑥
[
𝑞
𝑝+𝑞+𝑠
𝜕
𝜕𝑥
(𝑝 + 𝑞 + 𝑠)] + 𝑓(𝑐)𝑝 − ℎ(𝑐)𝑞 (2)
𝜕𝑠
𝜕𝑡
=
𝜕
𝜕𝑥
[
𝑠
𝑝+𝑞+𝑠
𝜕
𝜕𝑥
(𝑝 + 𝑞 + 𝑠)] + 𝑔(𝑐)𝑠(г − (𝑝 + 𝑞 + 𝑠 + 𝑛)) (3)
𝜕𝑛
𝜕𝑡
= ℎ(𝑐)𝑞 (4)
Where
𝑐(𝑥, 𝑡) =
𝑐0𝛾
𝑝+𝛾
(1 − 𝛼(𝑝 + 𝑞 + 𝑠 + 𝑛)) (5)
Where the model consists of four equations with four variables as 𝑝(𝑥, 𝑡) represents proliferating
(living) cells, 𝑞(𝑥, 𝑡) quiescent cells (live but not proliferating), 𝑠(𝑥, 𝑡) surrounding cells, and 𝑛
necrotic cells. According to the model presented by Sherratt & Chaplain [19] , the mitosis rate
𝑔(𝑐) of proliferating cells is directly proportional to the concentration of nutrients 𝑐(𝑥, 𝑡), and is
limited by the total number of cells in the body and cells that do not receive sufficient nutrients
turn in to quiescent cells. Many inactive cells undergo necrosis when kept away from nutrients at
a rate of 𝑓(𝑐) towards the depth of the tumor and tumor growth and development is restricted by
adjacent epithelial cells depend for mitosis on the concentration of nutrients 𝑐(𝑥, 𝑡) as well as
ℎ(𝑐) represents the rate of rotating quiescent cells to necrosis and the proliferating cells become
quiescent at a rate of 𝑔(𝑐), 𝑐0 is a nutrient deliberation in lack of a tumor cell population. Also 𝛼,
𝛾, г are dimensionless parameters the 𝑥 denotes the coordinates of place and 𝑡 of time as well as
the 𝛼 ∈ [0,1]. With the following initial and boundary conditions [18]:
Initial conditions:
𝑝(𝑥, 0) = 𝑝0(𝑥) , 𝑞(𝑥, 0) = 𝑞0(𝑥) , 𝑠(𝑥, 𝑡) = 𝑠0(𝑥) , 𝑛(𝑥, 0) = 𝑛0(𝑥).

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Boundary conditions:
𝜕𝑝
𝜕𝑥
= 0,
𝜕𝑞
𝜕𝑥
= 0, and
𝜕𝑠
𝜕𝑥
= 0 at 𝑥 = 0.
Where 𝑓(𝑐) , ℎ(𝑐) and 𝑔(𝑐) are given functions and г, 𝛾 and are parameter values also given
[18].
Fig.1: the interaction between the proliferating, quiescent, surrounding and necrotic cell
densities.
3. Exponential-function method
Suppose the general form of a nonlinear partial differential equation with two independent
variables 𝑥 and 𝑡 has the following form [20]:
𝑝(𝑢, 𝑢𝑡, 𝑢𝑥, 𝑢𝑡𝑡, 𝑢𝑥𝑡, 𝑢𝑥𝑥, … ) = 0. (6)
Where 𝑢 = 𝑢(𝑥, 𝑡), unknown function and 𝑝 is a polynomial in 𝑢(𝑥, 𝑡). and consider the
traveling wave transformation:
𝑢(𝑥, 𝑡) = 𝑢(𝜉) , 𝜉 = 𝑥 + 𝑠𝑡. (7)
Where 𝜉 is the combination of 𝑥 and 𝑡, and 𝑠 is the wave speed by using equation (7), the
equation (6) becomes an ordinary differential equation:
𝑄(𝑢, 𝑠𝑢′
, 𝑢′
, 𝑠2
𝑢″
, 𝑠𝑢″
, 𝑢″
, … ) = 0 (8)
the wave solution to equation (8) is as follows:
𝑢(𝜉) =
∑ 𝑎𝑛exp (𝑛𝜉)
𝑑
𝑛=−𝑐
∑ 𝑏𝑚exp (𝑚𝜉)
𝑞
𝑚=−𝑝
=
𝑎−𝑐 exp(−𝑐𝜉)+⋯+𝑎𝑑exp (𝑑𝜉)
𝑏−𝑝 exp(−𝑝𝜉)+⋯+𝑏𝑞exp (𝑞𝜉)
. (9)
Where 𝑎𝑛 and 𝑏𝑛 are unknown constants and 𝑐, 𝑑, 𝑝 and q are positive integers. Equation (9) can
be reformulated as:

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𝑢(𝜉) =
𝑎𝑐 exp(𝑐𝜉)+⋯+𝑎−𝑑exp (−𝑑𝜉)
𝑏𝑝 exp(𝑝𝜉)+⋯+𝑏−𝑞exp (−𝑞𝜉)
, (10)
by balance the linear and nonlinear higher order terms in equation (8) to get the values of 𝑐 and
𝑝, and balance the linear and nonlinear lowest order terms in equation (8) in order to get the
values of 𝑑 and 𝑞 , we replace equation (10) with equation (8) after substituting the values of 𝑐,
𝑑, 𝑝 and 𝑞, and simplifying to get:
∑ 𝑐𝑘 exp(±𝑘𝜉) = 0
𝑘 , 𝑘 = 0,1,2, … (11)
solving the algebraic equations (11) after setting each coefficient of the 𝑐𝑘 = 0, the known 𝑎𝑐
and 𝑏𝑝 substituted in the equation(10) to obtain on solution for the equation (6).
4. Modification method
The system (1-4) is difficult to solve analytically because it contains nonlinear fractions, so
we will suggest a simpler system based on some biological constraints, the aim of which is to
treat and simplify the nonlinear fractions [21].
We will replace
𝑝
𝑝+𝑞+𝑠
,
𝑞
𝑝+𝑞+𝑠
and
𝑠
𝑝+𝑞+𝑠
with 𝑘1, 𝑘2 and 𝑘3 respectively, so that we get:
𝜕𝑝
𝜕𝑡
=
𝜕
𝜕𝑥
[𝑘1
𝜕
𝜕𝑥
(𝑝 + 𝑞 + 𝑠)] + 𝑔(𝑐)𝑝(1 − (𝑝 + 𝑞 + 𝑠 + 𝑛)) − 𝑓(𝑐)𝑝 (12)
𝜕𝑞
𝜕𝑡
=
𝜕
𝜕𝑥
[𝑘2
𝜕
𝜕𝑥
(𝑝 + 𝑞 + 𝑠)] + 𝑓(𝑐)𝑝 − ℎ(𝑐)𝑞 (13)
𝜕𝑠
𝜕𝑡
=
𝜕
𝜕𝑥
[𝑘3
𝜕
𝜕𝑥
(𝑝 + 𝑞 + 𝑠)] + 𝑔(𝑐)𝑠(г − (𝑝 + 𝑞 + 𝑠 + 𝑛)) (14)
𝜕𝑛
𝜕𝑡
= ℎ(𝑐)𝑞 (15)
𝑐(𝑥, 𝑡) =
𝑐0𝛾
𝑝+𝛾
(1 − 𝛼(𝑝 + 𝑞 + 𝑠 + 𝑛)) (16)
We will find a specific exact solution that fulfills the following hypotheses:
𝑝 = 𝑘1(𝑝 + 𝑞 + 𝑠), 𝑞 = 𝑘2(𝑝 + 𝑞 + 𝑠), 𝑠 = 𝑘3(𝑝 + 𝑞 + 𝑠), 𝑛 = 𝑘4𝑛. (17)
We replace the previous (17) assumptions with the system (12-16):
𝑘1
𝜕
𝜕𝑡
(𝑝 + 𝑞 + 𝑠) =
𝜕
𝜕𝑥
[𝑘1
𝜕
𝜕𝑥
(
𝑝
𝑘1
)] + 𝑔(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠)(1 − ((𝑘1 + 𝑘2 + 𝑘3)(𝑝 + 𝑞 + 𝑠) +
𝑘4𝑛)) − 𝑓(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) (18)
𝑘2
𝜕
𝜕𝑡
(𝑝 + 𝑞 + 𝑠) =
𝜕
𝜕𝑥
[𝑘2
𝜕
𝜕𝑥
(
𝑞
𝑘2
)] + 𝑓(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) − ℎ(𝑐)𝑘2(𝑝 + 𝑞 + 𝑠) (19)

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𝑘3
𝜕
𝜕𝑡
(𝑝 + 𝑞 + 𝑠) =
𝜕
𝜕𝑥
[𝑘3
𝜕
𝜕𝑥
(
𝑠
𝑘3
)] + 𝑔(𝑐)𝑘3(𝑝 + 𝑞 + 𝑠)(г − ((𝑘1 + 𝑘2 + 𝑘3)(𝑝 + 𝑞 + 𝑠) +
𝑘4𝑛)) (20)
𝑘4
𝜕𝑛
𝜕𝑡
= ℎ(𝑐)𝑘2(𝑝 + 𝑞 + 𝑠) (21)
𝑐(𝑥, 𝑡) =
𝑐0𝛾
𝑘1(𝑝+𝑞+𝑠)+𝛾
(1 − 𝛼((𝑘1 + 𝑘2 + 𝑘3)(𝑝 + 𝑞 + 𝑠) + 𝑘4𝑛)) (22)
Simplify the system (18-21):
𝑘1
𝜕
𝜕𝑡
(𝑝 + 𝑞 + 𝑠) =
𝜕
𝜕𝑥
[
𝜕
𝜕𝑥
(𝑝)] + 𝑔(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) (1 − ((𝑘1 + 𝑘2 + 𝑘3)(𝑝 + 𝑞 + 𝑠) +
𝑘4𝑛)) − 𝑓(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) (23)
𝑘2
𝜕
𝜕𝑡
(𝑝 + 𝑞 + 𝑠) =
𝜕
𝜕𝑥
[
𝜕
𝜕𝑥
(𝑞)] + 𝑓(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) − ℎ(𝑐)𝑘2(𝑝 + 𝑞 + 𝑠) (24)
𝑘3
𝜕
𝜕𝑡
(𝑝 + 𝑞 + 𝑠) =
𝜕
𝜕𝑥
[
𝜕
𝜕𝑥
(𝑠)] + 𝑔(𝑐)𝑘3(𝑝 + 𝑞 + 𝑠)(г − ((𝑘1 + 𝑘2 + 𝑘3)(𝑝 + 𝑞 + 𝑠) + 𝑘𝑛𝑛))
(25)
𝑘4
𝜕𝑛
𝜕𝑡
= ℎ(𝑐)𝑘2(𝑝 + 𝑞 + 𝑠) (26)
For solving the system (23-26) using Exp-Function method, above equations are transformation
to nonlinear ordinary differential equation using the hypothesis (7) to become the system in the
form:
𝑘1𝑠 ( 𝑝́ + 𝑞́ + 𝑆
́) = 𝑝́
́ + 𝑔(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) (1 − ((𝑘1 + 𝑘2 + 𝑘3)(𝑝 + 𝑞 + 𝑠) + 𝑘4𝑛)) −
𝑓(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) (27)
𝑘2 𝑠 ( 𝑝́ + 𝑞́ + 𝑆
́) = 𝑞́
́ + 𝑓(𝑐)𝑘1(𝑝 + 𝑞 + 𝑠) − ℎ(𝑐)𝑘2(𝑝 + 𝑞 + 𝑠) (28)
𝑘3𝑠 ( 𝑝́ + 𝑞́ + 𝑆
́) = 𝑠́
́ + 𝑔(𝑐)𝑘3(𝑝 + 𝑞 + 𝑠)(г − ((𝑘1 + 𝑘2 + 𝑘3)(𝑝 + 𝑞 + 𝑠) + 𝑘𝑛𝑛)) (29)
𝑘4 𝑠 ( 𝑛́) = ℎ(𝑐)𝑘2(𝑝 + 𝑞 + 𝑠) (30)
Note from the system (27-30), The highest derived in the nonlinear system is 𝑝́
́ and the highest
order of nonlinear terms are 𝑝2
and it is the same in other equations of the system.

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To determine values 𝑐, 𝑑 𝑝, and 𝑞, we derivative the equation (4) twice for 𝜂:
𝑝́
́(𝜉) =
𝑐1 𝑒𝑥𝑝((3𝑝+𝑐)𝜉)+⋯
𝑐2 𝑒𝑥𝑝(4 𝑝 𝜉)+⋯
(31)
and
𝑝2(𝜉) =
𝑐3 𝑒𝑥𝑝((4 𝑐) 𝜉)+⋯
𝑐4 𝑒𝑥𝑝(4 𝑝 𝜉)+⋯
(32)
Where 𝑐𝑖 are constants. The exp function method is assumes the equality of the denominator in
the two equations. In addition, the method depends on equal to the least term in the numerator for
the equation (31) With the least term of the numerator in the equation (32) and the largest term of
the numerator in the equation (31) With the largest term to numerator in the equation (32), then
3𝑝 + 𝑐 = 4 𝑐 and 3𝑞 + 𝑑 = 4 𝑑 that mean 𝑝 = 𝑐 and 𝑞 = 𝑑. As a special case, supposed for
equation (4):
𝑏𝑚 = 0 Ɐm = −p … … q/{0}
that mean 𝑏0 = 1, and 𝑎𝑛 = 0 Ɐn = −c … … d/{0} , that mean 𝑎0 = 𝑏 𝑎𝑛𝑑 𝑎1 = 𝑎 , then
the general solution as follows:
𝑝 = 𝛼𝑝𝑒𝜉
+ 𝑐𝑝, 𝑞 = 𝛼𝑞𝑒𝜉
+ 𝑐𝑞, 𝑠 = 𝛼𝑠 𝑒𝜉
+ 𝑐𝑠, 𝑛 = 𝛼𝑛𝑒𝜉
+ 𝑐𝑛. (33)
Where 𝛼𝑝 , 𝛼𝑞 , 𝛼𝑠 , 𝛼𝑛 , 𝑐𝑝 , 𝑐𝑞 , 𝑐𝑠 and 𝑐𝑛 are a constants or functions in terms of 𝑥 and 𝑡.
by replace the previous assumptions (33) with equations (27-30) with some simplifications to
equations:
𝑓(𝑐)𝑘1𝑐𝑝 + 𝑓(𝑐)𝑘1𝑐𝑞 + 𝑔(𝑐)𝑘1
2
𝑐𝑝
2
+ 𝑔(𝑐)𝑘1
2
𝑐𝑠
2
− 𝑔(𝑐)𝑘1𝑐𝑞 + 𝑓(𝑐)𝑘1𝑐𝑠 − 𝑔(𝑐)𝑘1𝑐𝑠 −
𝑔(𝑐)𝑘1𝑐𝑝 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2𝛼𝑠 +
2𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑠 +
𝑔(𝑐)𝑘1𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘4𝑐𝑛 + 2𝑔(𝑐)𝑘1𝑐𝑞𝑘2𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+
2𝑔(𝑐)𝑘1𝑐𝑞𝑘3𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑔(𝑐)𝑘1𝑐𝑞𝑘4𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑠 +
2𝑔(𝑐)𝑘1𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑠 + 𝑔(𝑐)𝑘1𝛼𝑠𝑒2(𝑎𝑥+𝑏𝑡)
𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘1𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
𝑘4𝑐𝑛 +
𝑔(𝑐)𝑘1𝑐𝑠𝑘4𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑝 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2𝛼𝑞 +
2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑠 +
2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑝 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3𝛼𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑞 +
2𝑔(𝑐)𝑘1𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑠 + 𝑔(𝑐)𝑘1𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘4𝛼𝑛 +
𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘4𝑐𝑛 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+
2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑔(𝑐)𝑘1𝑐𝑝𝑘4𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝑐𝑞𝑘3𝑐𝑠 +

7. UtilitasMathematica
ISSN 0315-3681 Volume 120, 2023
105
2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑞 + 𝑔(𝑐)𝑘1
2
𝑐𝑞
2
+ 𝑔(𝑐 )𝑘1𝑐𝑞𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘1𝛼𝑝
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3 + 𝑔(𝑐)𝑘1𝑐𝑝𝑘4𝑐𝑛 +
𝑔(𝑐)𝑘1𝛼𝑞
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑝
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2 + 𝑔(𝑐)𝑘1𝛼𝑠
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2 +
𝑔(𝑐)𝑘1𝛼𝑠
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑞
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2 + 𝑔(𝑐)𝑘1𝑐𝑠𝑘4𝑐𝑛 + 2𝑔(𝑐)𝑘1
2
𝑐𝑞𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
−
𝛼𝑝𝑎2
𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1
2
𝑐𝑞𝑐𝑠 + 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝑐𝑠 + 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝑐𝑞 + 𝑔(𝑐)𝑘1𝑘2𝑐𝑝
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑝
2
+
𝑔(𝑐)𝑘1𝑘2𝑐𝑞
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑞
2
+ 𝑔(𝑐)𝑘1𝑘2𝑐𝑠
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑠
2
− 𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
−
𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
− 𝑔(𝑐)𝑘1𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑔(𝑐)𝑘1
2
𝛼𝑝
2
𝑒2(𝑎𝑥+𝑏𝑡)
+ 𝑔(𝑐)𝑘1
2
𝛼𝑠
2
𝑒2(𝑎𝑥+𝑏𝑡)
+
𝑔(𝑐)𝑘1
2
𝛼𝑞
2
𝑒2(𝑎𝑥+𝑏𝑡)
+ 𝑓(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
+ 𝑓(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 𝑓(𝑐)𝑘1𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+
2𝑔(𝑐)𝑘1
2
𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
𝑐𝑠 + 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑐𝑝 + 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝛼𝑞 + 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑐𝑞 +
2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝛼𝑠 + 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑐𝑠 + 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+
2𝑔(𝑐)𝑘1
2
𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑐𝑞 + 2𝑔(𝑐)𝑘1
2
𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝛼𝑠 + 2𝑔(𝑐)𝑘1
2
𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑠 +
2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑞𝑘2𝑐𝑠 + 𝑘1𝛼𝑝𝑏𝑒𝑎𝑥+𝑏𝑡
+ 𝑘1𝛼𝑞𝑏𝑒𝑎𝑥+𝑏𝑡
+
𝑘1𝛼𝑠𝑏𝑒𝑎𝑥+𝑏𝑡
= 0
𝑘2𝛼𝑝𝑏𝑒𝑎𝑥+𝑏𝑡
+ 𝑘2𝛼𝑞𝑏𝑒𝑎𝑥+𝑏𝑡
+ 𝑘2𝛼𝑠𝑏𝑒𝑎𝑥+𝑏𝑡
− 𝛼𝑞𝑎2
𝑒𝑎𝑥+𝑏𝑡
− 𝑓(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
−
𝑓(𝑐)𝑘1𝑐𝑝 − 𝑓(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
− 𝑓(𝑐)𝑘1𝑐𝑞 − 𝑓(𝑐)𝑘1𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
− 𝑓(𝑐)𝑘1𝑐𝑠 + ℎ(𝑐)𝑘2𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
+
ℎ(𝑐)𝑘2𝑐𝑝 + ℎ(𝑐)𝑘2𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ ℎ(𝑐)𝑘2𝑐𝑞 + ℎ(𝑐)𝑘2𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ ℎ(𝑐)𝑘2𝑐𝑠 = 0
𝑔(𝑐)𝑘3
2
𝛼𝑠
2
𝑒2(𝑎𝑥+𝑏𝑡)
+ 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑐𝑠 + 2𝑔(𝑐)𝑘3
2
𝑐𝑝𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘3
2
𝑐𝑝𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
−
𝑔(𝑐)𝑘3𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
г − 𝑔(𝑐)𝑘3𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
г − 𝑔(𝑐)𝑘3𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
г + 2𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑞 +
2𝑔(𝑐)𝑘1𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘32𝑔(𝑐)𝑘1𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3𝛼𝑞 +
2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘3𝑐𝑠 +
2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝑐𝑞𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑞 +
𝑔(𝑐)𝑘3𝑐𝑞𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘3𝑐𝑝𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘3𝛼𝑝
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2 + 𝑔(𝑐)𝑘3𝛼𝑠
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2 +
𝑔(𝑐)𝑘3𝛼𝑞
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2 + 𝑔(𝑐)𝑘3𝑐𝑠𝑘4𝑐𝑛 + 2𝑔(𝑐)𝑘3
2
𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑐𝑞 + 2𝑔(𝑐)𝑘3
2
𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝛼𝑠 +
2𝑔(𝑐)𝑘3
2
𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑐𝑠 + 2𝑔(𝑐)𝑘3
2
𝑐𝑞𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘3
2
𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
𝑐𝑠 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑐𝑝 +
2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝛼𝑞 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑐𝑞 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝛼𝑠 +
𝑔(𝑐)𝑘1𝛼𝑝
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑞
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑠
2
𝑒2(𝑎𝑥+𝑏𝑡)
𝑘3 +
2𝑔(𝑐)𝑘3𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘3𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘3𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2𝛼𝑠 +
𝑔(𝑐)𝑘3𝛼𝑞𝑒2(𝑎𝑥+𝑏𝑡)
𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘3𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
𝑘4𝑐𝑛 + 2𝑔(𝑐)𝑘3𝑐𝑞𝑘2𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+
𝑔(𝑐)𝑘3𝑐𝑞𝑘4𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘3𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑠 + 𝑔(𝑐)𝑘3𝛼𝑠𝑒2(𝑎𝑥+𝑏𝑡)
𝑘4𝛼𝑛 +
𝑔(𝑐)𝑘3𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘3𝑐𝑠𝑘4𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘3𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑝 +
2𝑔(𝑐)𝑘3𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2𝛼𝑞 + 2𝑔(𝑐)𝑘3𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘3𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘2𝛼𝑠 +
2𝑔(𝑐)𝑘3𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘2𝑐𝑠 + 𝑔(𝑐)𝑘3𝛼𝑝𝑒2(𝑎𝑥+𝑏𝑡)
𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘3𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
𝑘4𝑐𝑛 +
2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑔(𝑐)𝑘3𝑐𝑝𝑘4𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘3
2
𝑐𝑞𝑐𝑠 +

8. UtilitasMathematica
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106
2𝑔(𝑐)𝑘3
2
𝑐𝑝𝑐𝑞 + 2𝑔(𝑐)𝑘3
2
𝑐𝑝𝑐𝑠 + 𝑔(𝑐)𝑘1𝑘3𝑐𝑝
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑞
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑠
2
+ 𝑔(𝑐)𝑘3𝑘2𝑐𝑝
2
+
𝑔(𝑐)𝑘3𝑘2𝑐𝑞
2
+ 𝑔(𝑐)𝑘3𝑘2𝑐𝑠
2
− 𝑔(𝑐)𝑘3𝑐𝑝г − 𝑔(𝑐)𝑘3𝑐𝑞г − 𝑔(𝑐)𝑘3𝑐𝑠г + 𝑔(𝑐)𝑘3
2
𝛼𝑝
2
𝑒2(𝑎𝑥+𝑏𝑡)
+
𝑔(𝑐)𝑘3
2
𝛼𝑞
2
𝑒2(𝑎𝑥+𝑏𝑡)
+ 𝑘3𝛼𝑝𝑏𝑒𝑎𝑥+𝑏𝑡
+ 𝑘3𝛼𝑞𝑏𝑒𝑎𝑥+𝑏𝑡
+ 𝑘3𝛼𝑠𝑏𝑒𝑎𝑥+𝑏𝑡
+ 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑠 +
2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘3𝑐𝑞𝑘2𝑐𝑠 − 𝛼𝑠𝑎2
𝑒𝑎𝑥+𝑏𝑡
+ 𝑔(𝑐)𝑘3
2
𝑐𝑝
2
+ 𝑔(𝑐)𝑘3
2
𝑐𝑞
2
+
𝑔(𝑐)𝑘3
2
𝑐𝑠
2
= 0
𝑘4𝛼𝑛𝑏𝑒𝑎𝑥+𝑏𝑡
− ℎ(𝑐)𝑘2𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
− ℎ(𝑐)𝑘2𝑐𝑝 − ℎ(𝑐)𝑘2𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
− ℎ(𝑐)𝑘2𝑐𝑞 −
ℎ(𝑐)𝑘2𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
− ℎ(𝑐)𝑘2𝑐𝑠 = 0
𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑝 − 𝑘1(𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑝 + 𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑞 + 𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑠) = 0
𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑞 − 𝑘2(𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑝 + 𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑞 + 𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑠) = 0
𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑠 − 𝑘3(𝛼𝑝𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑝 + 𝛼𝑞𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑞 + 𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑠) = 0
𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑛 − 𝑘4(𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑛) = 0
Setting each coefficient of exp(± 𝑛𝜉) , 𝑛 = 0,1,2,3, …
𝑓(𝑐)𝑘1𝑐𝑝 + 𝑓(𝑐)𝑘1𝑐𝑞 + 𝑔(𝑐)𝑘1
2
𝑐𝑝
2
+ 𝑔(𝑐)𝑘1
2
𝑐𝑠
2
− 𝑔(𝑐)𝑘1𝑐𝑞 + 𝑓(𝑐)𝑘1𝑐𝑠 − 𝑔(𝑐)𝑘1𝑐𝑠 −
𝑔(𝑐)𝑘1𝑐𝑝 + 2𝑔(𝑐)𝑘1𝑐𝑞𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑞 + 𝑔(𝑐)𝑘1
2
𝑐𝑞
2
+ 𝑔(𝑐)𝑘1𝑐𝑞𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘1𝑐𝑝𝑘4𝑐𝑛 +
𝑔(𝑐)𝑘1𝑐𝑠𝑘4𝑐𝑛 + 2𝑔(𝑐)𝑘1
2
𝑐𝑞𝑐𝑠 + 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝑐𝑠 + 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝑐𝑞 + 𝑔(𝑐)𝑘1𝑘2𝑐𝑝
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑝
2
+
𝑔(𝑐)𝑘1𝑘2𝑐𝑞
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑞
2
+ 𝑔(𝑐)𝑘1𝑘2𝑐𝑠
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑠
2
+ 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑠 +
2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑞𝑘2𝑐𝑠 = 0 (34)
−𝑔(𝑐)𝑘1𝛼𝑠 + 𝑔(𝑐)𝑘1𝛼𝑞𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘1𝑐𝑞𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘1𝛼𝑠𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘1𝑐𝑠𝑘4𝛼𝑛 +
𝑔(𝑐)𝑘1𝛼𝑝𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘1𝑐𝑝𝑘4𝛼𝑛 − 𝑔(𝑐)𝑘1𝛼𝑝 + 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑐𝑝 + 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑐𝑞 +
2𝑔(𝑐)𝑘1
2
𝛼𝑝𝑐𝑠 + 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝛼𝑞 + 2𝑔(𝑐)𝑘1
2
𝑐𝑝𝛼𝑠 + 2𝑔(𝑐)𝑘1
2
𝛼𝑞𝑐𝑞 + 2𝑔(𝑐)𝑘1
2
𝛼𝑞𝑐𝑠 +
2𝑔(𝑐)𝑘1
2
𝑐𝑞𝛼𝑠 + 2𝑔(𝑐)𝑘1
2
𝛼𝑠𝑐𝑠 − 𝛼𝑝𝑎2
− 2𝑔(𝑐)𝑘1𝑐𝑞𝑘3𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑠𝑘3𝑐𝑠 + 𝑘1𝛼𝑝𝑏 +
𝑘1𝛼𝑞𝑏 + 𝑘1𝛼𝑠𝑏 + 𝑓(𝑐)𝑘1𝛼𝑝 + 𝑓(𝑐)𝑘1𝛼𝑞 + 𝑓(𝑐)𝑘1𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑘3𝑐𝑞 +
2𝑔(𝑐)𝑘1𝛼𝑝𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝛼𝑞 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘2𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑘2𝑐𝑞 +
2𝑔(𝑐)𝑘1𝛼𝑞𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑞𝑘2𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑠𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘2𝑐𝑝 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑠 +
2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝑐𝑝 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝑐𝑞 − 𝑔(𝑐)𝑘1𝛼𝑞 = 0
(35)
2𝑔(𝑐)𝑘1𝛼𝑞𝑘2𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑘3𝛼𝑠 + 𝑔(𝑐)𝑘1𝛼𝑞𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘1𝛼𝑠𝑘4𝛼𝑛 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘2𝛼𝑞 +
2𝑔(𝑐)𝑘1𝛼𝑝𝑘2𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝛼𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝛼𝑠 + 𝑔(𝑐)𝑘1𝛼𝑝𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘1𝛼𝑝
2
𝑘3 +
𝑔(𝑐)𝑘1𝛼𝑞
2
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑝
2
𝑘2 + 𝑔(𝑐)𝑘1𝛼𝑠
2
𝑘2 + 𝑔(𝑐)𝑘1𝛼𝑠
2
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑞
2
𝑘2 + 𝑔(𝑐)𝑘1
2
𝛼𝑝
2
+

9. UtilitasMathematica
ISSN 0315-3681 Volume 120, 2023
107
𝑔(𝑐)𝑘1
2
𝛼𝑠
2
+ 𝑔(𝑐)𝑘1
2
𝛼𝑞
2
+ 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝛼𝑞 + 2𝑔(𝑐)𝑘1
2
𝛼𝑝𝛼𝑠 + 2𝑔(𝑐)𝑘1
2
𝛼𝑞𝛼𝑠 = 0
(36)
−𝑓(𝑐)𝑘1𝑐𝑝 − 𝑓(𝑐)𝑘1𝑐𝑞 − 𝑓(𝑐)𝑘1𝑐𝑠 + ℎ(𝑐)𝑘2𝑐𝑝 + ℎ(𝑐)𝑘2𝑐𝑞 + ℎ(𝑐)𝑘2𝑐𝑠 = 0 (35)
𝑘2𝛼𝑝𝑏 + 𝑘2𝛼𝑞𝑏 + 𝑘2𝛼𝑠𝑏 − 𝛼𝑞𝑎2
− 𝑓(𝑐)𝑘1𝛼𝑝 − 𝑓(𝑐)𝑘1𝛼𝑞 − 𝑓(𝑐)𝑘1𝛼𝑠 + ℎ(𝑐)𝑘2𝛼𝑝 +
ℎ(𝑐)𝑘2𝛼𝑞 + ℎ(𝑐)𝑘2𝛼𝑠 = 0 (37)
2𝑔(𝑐)𝑘1𝑐𝑞𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑞 + 𝑔(𝑐)𝑘3𝑐𝑞𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘3𝑐𝑝𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘3𝑐𝑠𝑘4𝑐𝑛 +
2𝑔(𝑐)𝑘3
2
𝑐𝑞𝑐𝑠 + 2𝑔(𝑐)𝑘3
2
𝑐𝑝𝑐𝑞 + 2𝑔(𝑐)𝑘3
2
𝑐𝑝𝑐𝑠 + 𝑔(𝑐)𝑘1𝑘3𝑐𝑝
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑞
2
+ 𝑔(𝑐)𝑘1𝑘3𝑐𝑠
2
+
𝑔(𝑐)𝑘3𝑘2𝑐𝑝
2
+ 𝑔(𝑐)𝑘3𝑘2𝑐𝑞
2
+ 𝑔(𝑐)𝑘3𝑘2𝑐𝑠
2
− 𝑔(𝑐)𝑘3𝑐𝑝г − 𝑔(𝑐)𝑘3𝑐𝑞г − 𝑔(𝑐)𝑘3𝑐𝑠г +
2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘3𝑐𝑞𝑘2𝑐𝑠 + 𝑔(𝑐)𝑘3
2
𝑐𝑝
2
+
𝑔(𝑐)𝑘3
2
𝑐𝑞
2
+ 𝑔(𝑐)𝑘3
2
𝑐𝑠
2
= 0 (38)
−𝑔(𝑐)𝑘3𝛼𝑞г + 2𝑔(𝑐)𝑘3𝛼𝑞𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘3𝑐𝑞𝑘2𝛼𝑠 − 𝛼𝑠𝑎2
− 𝑔(𝑐)𝑘3𝛼𝑠г + 𝑔(𝑐)𝑘3𝛼𝑞𝑘4𝑐𝑛 +
𝑔(𝑐)𝑘3𝑐𝑞𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘3𝛼𝑠𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘3𝑐𝑠𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘3𝛼𝑝𝑘4𝑐𝑛 + 𝑔(𝑐)𝑘3𝑐𝑝𝑘4𝛼𝑛 −
𝑔(𝑐)𝑘3𝛼𝑝г + 2𝑔(𝑐)𝑘3
2
𝑐𝑞𝛼𝑠 + 2𝑔(𝑐)𝑘3
2
𝛼𝑠𝑐𝑠 + 2𝑔(𝑐)𝑘3𝛼𝑞𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘3
2
𝛼𝑞𝑐𝑠 +
2𝑔(𝑐)𝑘3
2
𝛼𝑞𝑐𝑞 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑐𝑝 + 2𝑔(𝑐)𝑘3𝛼𝑝𝑘2𝑐𝑞 + 2𝑔(𝑐)𝑘3𝛼𝑝𝑘2𝑐𝑠 + 2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝛼𝑞 +
2𝑔(𝑐)𝑘3𝑐𝑝𝑘2𝛼𝑠 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑐𝑞 + 2𝑔(𝑐)𝑘3𝛼𝑝𝑘2𝑐𝑝 + 2𝑔(𝑐)𝑘3
2
𝑐𝑝𝛼𝑞 + 2𝑔(𝑐)𝑘1𝑐𝑞𝑘3𝛼𝑠 +
2𝑔(𝑐)𝑘1𝛼𝑠𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝑐𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑞𝑘3𝑐𝑞 + 2𝑔(𝑐)𝑘3
2
𝑐𝑝𝛼𝑠 +
2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑠 + 2𝑔(𝑐)𝑘1𝑐𝑝𝑘3𝛼𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝑐𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝑐𝑝 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝑐𝑞 +
𝑘3𝛼𝑝𝑏 + 2𝑔(𝑐)𝑘3𝛼𝑠𝑘2𝑐𝑠 + 𝑘3𝛼𝑠𝑏 + 𝑘3𝛼𝑞𝑏 = 0 (39)
𝑔(𝑐)𝑘3
2
𝛼𝑠
2
+ 2𝑔(𝑐)𝑘1𝛼𝑞𝑘3𝛼𝑠 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝛼𝑞 + 2𝑔(𝑐)𝑘1𝛼𝑝𝑘3𝛼𝑠 + 𝑔(𝑐)𝑘3𝛼𝑝
2
𝑘2 +
𝑔(𝑐)𝑘3𝛼𝑠
2
𝑘2 + 𝑔(𝑐)𝑘3𝛼𝑞
2
𝑘2 + 2𝑔(𝑐)𝑘3
2
𝛼𝑞𝛼𝑠 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝛼𝑞 + 2𝑔(𝑐)𝑘3
2
𝛼𝑝𝛼𝑠 +
𝑔(𝑐)𝑘1𝛼𝑝
2
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑞
2
𝑘3 + 𝑔(𝑐)𝑘1𝛼𝑠
2
𝑘3 + 2𝑔(𝑐)𝑘3𝛼𝑞𝑘2𝛼𝑠 + 𝑔(𝑐)𝑘3𝛼𝑞𝑘4𝛼𝑛 +
𝑔(𝑐)𝑘3𝛼𝑠𝑘4𝛼𝑛 + 2𝑔(𝑐)𝑘3𝛼𝑝𝑘2𝛼𝑞 + 2𝑔(𝑐)𝑘3𝛼𝑝𝑘2𝛼𝑠 + 𝑔(𝑐)𝑘3𝛼𝑝𝑘4𝛼𝑛 + 𝑔(𝑐)𝑘3
2
𝛼𝑝
2
+
𝑔(𝑐)𝑘3
2
𝛼𝑞
2
= 0 (40)
𝑔(𝑐)𝑐𝑠г − 𝑔(𝑐)𝑘𝑐𝑝𝑐𝑠 − 𝑔(𝑐)𝑘𝑐𝑞𝑐𝑠 − 𝑔(𝑐)𝑘𝑐𝑠
2
− 𝑔(𝑐)𝑘𝑛𝑐𝑠𝑐𝑛 = 0 (41)
−ℎ(𝑐)𝑘2𝑐𝑝 − ℎ(𝑐)𝑘2𝑐𝑞 − ℎ(𝑐)𝑘2𝑐𝑠 = 0 (42)
𝑘4𝛼𝑛𝑏 − ℎ(𝑐)𝑘2𝛼𝑝 − ℎ(𝑐)𝑘2𝛼𝑞 − ℎ(𝑐)𝑘2𝛼𝑠 = 0 (43)
𝑐𝑝 − 𝑘1(𝑐𝑝 + 𝑐𝑞 + 𝑐𝑠) = 0 (44)
𝛼𝑝 − 𝑘1(𝛼𝑝 + 𝛼𝑞 + 𝛼𝑠) = 0 (45)

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𝑐𝑞 − 𝑘2(𝑐𝑝 + 𝑐𝑞 + 𝑐𝑠) = 0 (46)
𝛼𝑞 − 𝑘2(𝛼𝑝 + 𝛼𝑞 + 𝛼𝑠) = 0 (47)
𝑐𝑠 − 𝑘3(𝑐𝑝 + 𝑐𝑞 + 𝑐𝑠) = 0 (48)
𝛼𝑠 − 𝑘3(𝛼𝑝 + 𝛼𝑞 + 𝛼𝑠) = 0 (49)
𝑐𝑛 − 𝑘4𝑐𝑛 = 0 (50)
𝛼𝑛 − 𝑘4𝛼𝑛 = 0 (51)
Solving nonlinear system of algebraic equations gives the following sets of non-trivial solutions.
The first set:
𝑐𝑝 = 0, 𝑐𝑞 = −
𝑐𝑠(𝑘3−1)
𝑘3
, 𝑐𝑠 = 𝑐𝑠, 𝑐𝑛 = 0, г = г, 𝑘1 = 0, 𝑘2 = −𝑘3 + 1, 𝑘3 = 𝑘3, 𝑘4 = 𝑘4, 𝛼𝑝 =
0, 𝛼𝑞 = −
𝛼𝑠(𝑘3−1)
𝑘3
, 𝛼𝑠 = 𝛼𝑠, 𝛼𝑛 = 0, ℎ = 0, 𝑓 = 𝑓, 𝑔 = 0, 𝑎 = 𝑎, 𝑏 = 𝑎2
.
Substituting the above quantities into equation (33) we obtain:
𝑝(𝑥, 𝑡) = 0 , 𝑞(𝑥, 𝑡) = −
𝛼𝑠(𝑘3−1)
𝑘3
𝑒𝑎𝑥+𝑎2𝑡
−
𝑐𝑠(𝑘3−1)
𝑘3
,
𝑠(𝑥, 𝑡) = 𝛼𝑠𝑒𝑎𝑥+𝑎2𝑡
+ 𝑐𝑠 , 𝑛(𝑥, 𝑡) = 0.
The second set:
𝑐𝑝 = 0, 𝑐𝑞 = −
𝑐𝑠(𝑘3−1)
𝑘3
, 𝑐𝑠 = 𝑐𝑠, 𝑐𝑛 = 𝑐𝑛, г = г, 𝑘1 = 0, 𝑘2 = −𝑘3 + 1, 𝑘3 = 𝑘3, 𝑘4 = 1, 𝛼𝑝 =
0, 𝛼𝑞 = −
𝛼𝑠(𝑘3−1)
𝑘3
, 𝛼𝑠 = 𝛼𝑠, 𝛼𝑛 = 0, ℎ = 0, 𝑓 = 𝑓, 𝑔 = 0, 𝑎 = 𝑎, 𝑏 = 𝑎2
.
Substituting the above quantities into equation (33) we obtain:
𝑝(𝑥, 𝑡) = 0 , 𝑞(𝑥, 𝑡) = −
𝛼𝑠(𝑘3−1)
𝑘3
𝑒𝑎𝑥+𝑎2𝑡
−
𝑐𝑠(𝑘3−1)
𝑘3
,
𝑠(𝑥, 𝑡) = 𝛼𝑠𝑒𝑎𝑥+𝑎2𝑡
+ 𝑐𝑠 , 𝑛(𝑥, 𝑡) = 𝑐𝑛 .
The third set:
𝑐𝑝 = 0, 𝑐𝑞 = 0, 𝑐𝑠 = 0, 𝑐𝑛 = 𝑐𝑛, г = г, 𝑘1 = 0, 𝑘2 = 1, 𝑘3 = 0, 𝑘4 = 1, 𝛼𝑝 = 0, 𝛼𝑞 =
𝑏𝛼𝑛
𝑎2−𝑏
,
𝛼𝑠 = 0, 𝛼𝑛 = 𝛼𝑛, ℎ = 𝑎2
− 𝑏, 𝑓 = 𝑓, 𝑔 = 0, 𝑎 = 𝑎, 𝑏 = 𝑏.
Substituting the above quantities into equation (33) we obtain:
𝑝(𝑥, 𝑡) = 0 , 𝑞(𝑥, 𝑡) =
𝑏𝑎𝑛
𝑎2−𝑏
𝑒𝑎𝑥+𝑏𝑡
,

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𝑠(𝑥, 𝑡) = 0 , 𝑛(𝑥, 𝑡) = 𝛼𝑛𝑒𝑎𝑥+𝑏𝑡
+ 𝑐𝑛.
The fourth set:
assuming that 𝑎, 𝑏 > 0, and 𝛼𝑝, 𝛼𝑞, 𝛼𝑠, 𝛼𝑛 ≠ 0, the solution of the previous system (34)-(51), we
get:
𝑐𝑝 = 0, 𝑐𝑞 = 0, 𝑐𝑠 = 0, 𝑐𝑛 =
1+𝑔(𝑐)𝑘1
𝑔(𝑐)𝑘1
, 𝑘1 = 𝑘1, 𝑘2 =
−1
ℎ(𝑐)
,
𝑘3 = −
−ℎ(𝑐)−1+𝑘1ℎ(𝑐)
ℎ(𝑐)
, 𝑘4 = 𝑘4, 𝑎 = 𝑎, 𝑏 = 𝑏, г = −
1+𝑔(𝑐)𝑘1
𝑔(𝑐)𝑘1
,
𝛼𝑝 = −
𝛼𝑠𝑘1ℎ(𝑐)
−ℎ(𝑐)−1+𝑘1ℎ(𝑐)
, 𝛼𝑞 =
𝛼𝑠
−ℎ(𝑐)−1+𝑘1ℎ(𝑐)
, 𝛼𝑠 = 𝛼𝑠,
𝛼𝑛 =
𝛼𝑠 ℎ(𝑐)
−ℎ(𝑐)−1+𝑘1ℎ(𝑐)
, 𝑔(𝑐) = 𝑔(𝑐) , ℎ(𝑐) = ℎ(𝑐), 𝑓(𝑐) = −
1
𝑘1
.
Substituting the above quantities into equation (33) we obtain:
𝑝(𝑥, 𝑡) = −
𝛼𝑠𝑘1ℎ(𝑐)
−ℎ(𝑐)−1+𝑘1ℎ(𝑐)
𝑒𝑎𝑥+𝑏𝑡
, 𝑞(𝑥, 𝑡) =
𝛼𝑠
−ℎ(𝑐)−1+𝑘1ℎ(𝑐)
𝑒𝑎𝑥+𝑏𝑡
,
𝑠(𝑥, 𝑡) = 𝛼𝑠𝑒𝑎𝑥+𝑏𝑡
, 𝑛(𝑥, 𝑡) =
𝛼𝑠ℎ(𝑐)
−ℎ(𝑐)−1+𝑘1ℎ(𝑐)
𝑒𝑎𝑥+𝑏𝑡
+
1+𝑔(𝑐)𝑘1
𝑔(𝑐)𝑘1
. (52)
Fig (2): exact solution for (a) proliferating and surrounding cells (b) quiescent and necrotic cells
with 𝑎 = 𝑏 = 𝑘1 = 𝑘4 = 1, , 𝛼 = 0.4, г = 2, 𝛾 = 10

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Fig (3): Shows the 𝑓(𝑐) with the proliferating (P), and quiescent (q) cells When 𝛼 = 0.4 , г = 2,
𝛾 = 10 ,𝑐0 = 1 and 𝑡 = 1.
Fig (4): Shows the 𝑔(𝑐) with the proliferating (p) , and surrounding (s) cells When 𝛼 = 0.4 ,
г = 2, 𝛾 = 10 , 𝑐0 = 1 and 𝑡 = 1.

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Fig (5): Shows the ℎ(𝑐) with the quiescent (q) and necrotic (n) cells When 𝛼 = 0.4 , г = 2, 𝛾 = 10 ,
𝑐0 = 1 and 𝑡 = 1.
Consider from figures (3-5), the effect of 𝑓(𝑐),ℎ(𝑐) and 𝑔(𝑐), on the solutions for the system
(1)-(5), where ℎ(𝑐) represent the rate of rotating quiescent cells to necrosis, 𝑔(𝑐) is the mitosis
amount function of the flourishing cells and 𝑓(𝑐) is the rate at which proliferating cells become
quiescent with the following parameter values, г = 0 , 𝛾 = 10 , 𝑐0 = 1 and 𝛼 = 0.4, We notice
from Figure (3) that 𝑝 (proliferating cells) increases steadily when 𝑓(𝑐) (the rate at which
proliferating cells become quiescent) approaches zero (decay) and 𝑞 (quiescent cells) decays for
all values of 𝑓(𝑐) in the period 𝑥. from Figure (4), show that 𝑝 (proliferating cells) and 𝑠
(surrounding cells) increases when the values of 𝑥 and 𝑔(𝑐) (is the mitosis amount function of
the flourishing cells) increase. Figure (5), show that 𝑛 (quiescent cells) increases as ℎ(𝑐)
(represent the rate of rotating quiescent cells to necrosis) approaches zero (decay) 𝑞 (necrotic
cells) approaches zero (decay) in all values of 𝑥 and ℎ(𝑐) (represent the rate of rotating quiescent
cells to necrosis).
5. Conclusion
In this paper, a modification of the avascular tumor growth model represented by the system of
nonlinear PDEs was proposed by simplifying the nonlinear fractions, and the exact solution of
the modified system was found based on the exponential-function method. The exact solution
enables us to study the effect of parameters, (𝑐) , ℎ(𝑐) and 𝑔(𝑐) on the spread of the disease and
how to control it. Every one of the calculations was done with the guide of Maple 18
programming.
Acknowledgments
The research is supported by College of Computer Sciences and Mathematics, University of
Mosul, Republic of Iraq.

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