EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 3, 2023, 1389-1405
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Using Gα-transform to study higher-order differential
equations with polynomial coefficients

Supaknaree Sattaso1, Patarawadee Prasertsang1,∗, Torsak Prasertsang1

1 Faculty of Science and Engineering, Kasetsart University, Chalermphrakiat Sakon Nakhon
Province Campus, Sakon Nakhon, Thailand

Abstract. In this study, solutions of higher-order differential equations with polynomial coeffi-
cients (HODEPCs) were obtained by applying the Gα-transform. Based on some characterizations,
the solutions of HODEPCs were investigated. With the general solution of the HODEPCs, the
curves of the general solution can be shown in several examples.

2020 Mathematics Subject Classifications: 34A25, 34A26, 34C20, 44A10

Key Words and Phrases: Laplace-type integral transforms, Gα-transform, Higher-order differ-
ential equation with polynomial coefficients

1. Introduction

Differential equations can be used to express a wide range of physical laws and rela-
tionships. Consequently, differential equations play a vital role in a variety of complicated
events that occur all over the world. Mathematics can be used to model any physical phe-
nomenon. Modeling is a generic method used in engineering, science, and other professions
to convert physical situations or other data into mathematical models. Subsequently, the
differential equations in the models must be solved.

In recent years, many authors have studied solutions to differential equations using
various methods. In general, it is still very difficult to obtain closed-form solutions for
differential equations for most models of real-life problems, but several techniques have
been developed to make it easier to find these solutions. Integral transforms have been
widely applied to solve several different types of differential equations. There are many
publications in the literature on the theory and application of integral transform for solving
differential equations, including contributions by Laplace [3, 4, 6, 14, 20, 21, 27], Sumudu
[1, 5, 7, 15, 28–30], and Elzaki [8–13, 26].

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i3.4792

Email addresses: supaknaree.s@ku.th (S. Sattaso),
patarawadee.s@ku.th (P. Prasertsang), torsak.p@ku.th (T. Prasertsang)

https://www.ejpam.com 1389 © 2023 EJPAM All rights reserved.



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1390

Recently, an extended Laplace transform, called the Laplace-typed integral transform,
or the Gα-transform, or the generalized Laplace-typed integral transforms, has been in-
troduced in [16] and some of its properties have been investigated. The Gα-transform is
defined by the formula

Gα{f(τ)} = wα

∫ ∞

0
e−τ/wf(τ)dτ,

where α ∈ Z and w is a complex variable. By selecting the appropriate α, theGα-transform
can be applied immediately to any situations. Table 1 lists some of the transforms along
with their definitions, and we use α to convert the Gα-transform as appropriate.

Table 1: Some integral transform definitions

Transform Definition Gα-Transform

Laplace
∫∞
0 e−τ/wf(τ)dτ α = 0 and 1/s = w

Sumudu 1
w

∫∞
0 e−τ/wf(τ)dτ α = −1

Elzaki w
∫∞
0 e−τ/wf(τ)dτ α = 1

Furthermore, the Laplace transform is well-known with a strong application in deriva-
tive transforms. To select a transform that provides a simple tool for integral transforms,
one has the option to choose α = −2 and obtain

G−2{f(τ)} =
1

w2

∫ ∞

0
e−τ/wf(τ)dτ ;

see [17] for more details. Kim Hj. [18] used theG−2-transform to solve Laguerre’s equation.
Recently, Sattaso S. et al. [24] explored the properties of the Gα-transform and offered

various examples to demonstrate its usefulness. Some examples can be easily solved with
the Gα-transform but not with the Sumudu or Elzaki transforms.

Furthermore, the application of the n-th partial derivatives to the Gα-transform in
partial differential equations was presented by Kim Hj. et al. [22]. Kim Hj. [2] inves-
tigated the existence and uniqueness of theorems for a variant of a generalized Laplace
transform represented by a logarithmic function. In addition, Kim Hj. [19] considered
an argument based on the rigor of mathematical induction for the Laplace transform of
the n-th derivative of any order. The results can be extended to the generalized Laplace
transform.

Geum Y. H. et al. [25] showed the matrix representation of convolution and related
the mathematical notion of convolution to the concept of convolution in a convolutional
neural network.

Most recently, the range of Gα-transforms that can be used to solve second-order
and third-order ordinary differential equations with variable coefficients was addressed by
Prasertsang P. et al. [23]. Motivated by this discussion, the current paper will extend the
variable coefficients in general form and identify some characterizations among them.



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1391

The remaining sections of the paper are organized as follows. Section 2 introduces a
definition and lemmas to prove the theorem. Section 3 derives the solutions of HODEPCs
via the Gα-transform and obtains some theorem and corollary. Some applications and
conclusions are given in sections 4 and 5, respectively.

2. Preliminaries

To analyze the study for HODEPCs via the Gα-transform, a definition and lemmas
are given, as follows

Definition 1. [24] Let f(τ) be a piecewise continuous function on τ ≥ 0 and has an
exponential order k. The Gα-transform of f(τ), briefly Gα{f(τ)}, is characterized with
the formula

Gα{f(τ)} = wα

∫ ∞

0
e−τ/wf(τ)dτ,

where α ∈ Z and w > 0 is a complex variable and Gα{f(τ)} exists for w < 1/k.

Lemma 1. [24] If y(m)(τ) is a piecewise continuous function on [0,∞) for m ∈ N ∪ {0}
and has an exponential order k for w < 1

k , then

Gα{τny(m)(τ)} = w2nd
nGα{y(m)(τ)}

dwn
−
(
n

1

)
[α− (n− 1)]w2n−1d

n−1Gα{y(m)(τ)}
dwn−1

+ · · · −
(

n

n− 1

)
[α− (n− 1)] [α− (n− 2)] · · · (α− 1)wn+1dGα{y(m)(τ)}

dw

+ [α− (n− 1)] [α− (n− 2)] · · ·αwnGα{y(m)(τ)}. (1)

From Lemma 1, Eq. (1) can be rewritten as the following,

Gα{τny(m)(τ)} =

n∑
l=0

(−1)n−l

(
n

l

) n−1∏
s=l

(m+ α− s)
Y (l)(w)

wm−n−l

−
m−1∑
k=0

n∏
L=1

(−m+ k + L)wα−m+k+L+1y(k)(0). (2)

If m = 0 in Eq. (2),

Gα{τny(τ)} =
n∑

l=0

(−1)n−l

(
n

l

) n−1∏
s=l

(α− s)
Y (l)(w)

w−n−l
.

If n = 0 in Eq. (2),

Gα{y(m)(τ)} =
Y (w)

wm
−

m−1∑
k=0

wα−m+k+1y(k)(0),

where Y (w) = Gα{y(τ)}.



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1392

Lemma 2. [24] Assume that y(τ) =
∑∞

n=0 anτ
n is a piecewise continuous function on

[0,∞) and has an exponential order at infinity with the function on |f(τ)| ≤ Mekτ for
τ ≥ C̄ where C̄ is a constant, then

Gα{f(τ)} =
∞∑
n=0

n!anw
α+n+1.

3. Analytical study for HODEPCs via Gα-transform

Denote n,m ∈ N ∪ {0} and m ≥ n,

ϱ = 2, 3, 4, . . . , n, ϱ1 = 0, 1, 2, . . . , n− 1, ϱ2 = 0, 1, 2, . . . , n, ρ = 0, 1, 2, . . . ,m,

ρ1 = m− (n− ϱ1 − 1),m− (n− ϱ1 − 2),m− (n− ϱ1 − 3), . . . ,m− 2,m− 1,m,

ρ2 = ϱ1 + 1, ϱ1 + 2, ϱ1 + 3, . . . , n− 2, n− 1, n,

ρ3 = 0, 1, 2, . . . ,m− (n− ϱ1 + 2),m− (n− ϱ1 + 1),m− (n− ϱ1),

υ(u) = aρ,0

ρ−1∑
k=0

uα−ρ+k+1y(k)(0) + aρ,1

ρ−1∑
k=0

(−ρ+ k + 1)uα−ρ+k+2y(k)(0)

+aρ,ϱ

ρ−1∑
k=0

ϱ∏
L=1

(−ρ+ k + L)uα−ρ+k+L+1y(k)(0),

Θϱ2(u) =

m∑
ρ=0

aρ,ϱ2
uρ−2ϱ2

−
(
ϱ2 + 1

ϱ2

) m∑
ρ=0

(ρ+ α− ϱ2)
aρ,ϱ2+1

uρ−ϱ2−(ϱ2+1)

+

(
ϱ2 + 2

ϱ2

) m∑
ρ=0

ϱ2+1∏
s=ϱ2

(ρ+ α− s)
aρ,ϱ2+2

uρ−ϱ2−(ϱ1+2)

−
(
ϱ2 + 3

ϱ2

) m∑
ρ=0

ϱ2+2∏
s=ϱ2

(ρ+ α− s)
aρ,ϱ2+3

uρ−ϱ2−(ϱ2+3)
+ · · ·

+(−1)n−2−ϱ2

(
n− 2

ϱ2

) m∑
ρ=ϱ2

n−3∏
s=ϱ2

(ρ+ α− s)
aρ,n−2

uρ−ϱ2−n+2

+(−1)n−1−ϱ2

(
n− 1

ϱ2

) m∑
ρ=ϱ2

n−2∏
s=ϱ2

(ρ+ α− s)
aρ,ϱ2+n−1

uρ−ϱ2−n+1

+(−1)n−ϱ2

(
n

ϱ2

) m∑
ρ=ϱ1

n−1∏
s=ϱ2

(ρ+ α− s)
aρ,n

uρ−ϱ2−n
.

Theorem 1. Consider the higher-order differential equation in the form

m∑
i=0

(

n∑
j=0

ai,jτ
j)y(i)(τ) = Φ(τ), (3)



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1393

where
∑n

j=0 ai,jτ
j are polynomial functions with degree n in terms of τ where i =

0, 1, 2, . . . ,m, j = 0, 1, 2, . . . , n, where m ≥ n and ai,j are polynomial coefficients with
am,n ̸= 0 and Φ(τ) is an unknown function. If Eq. (3) satisfies the following conditions

aρ1,ϱ1 = 0, (4)

n∑
j=ρ2

(−1)j−ϱ1

(
j

ϱ1

) j−1∏
s=ϱ1

(−ρ2 + j + α− s)a−ρ2+j,j = 0, (5)

aρ3,ϱ1 +
n∑

j=ϱ1+1

(−1)j−ϱ1

(
j

ϱ1

) j−1∏
s=ϱ1

(ρ3 − ϱ1 + j + α− s)aρ3−ϱ1+j,j = 0, (6)

then, it is appropriately solved using the Gα-transform.

Proof. Taking the Gα-transform along both sides of (3) and applying Eq. (2), it follows
that,

m∑
ρ=0

aρ,ϱ2

ϱ2∑
l=0

(−1)n−l

(
ϱ2
l

) ϱ2−1∏
s=l

(ρ+ α− s)
Y (l)(u)

uρ−ϱ2−l

−
m∑
ρ=0

aρ,ϱ2

ρ−1∑
k=0

ϱ2∏
L=1

(−ρ+ k + L)uα−ρ+k+L+1y(k)(0) = Gα [Φ(τ)] , (7)

for ϱ2 = 0, 1, 2, . . . , n. Therefore, Eq. (7) can be represented in terms of all derivatives of
Y (u) as follows

n∑
ϱ2=0

Y (ϱ2)(u)Θϱ2(u) = Gα [Φ(τ)] + υ(u). (8)

Solving the equation (8) using the Gα-transform method, means that the coefficients
of Y (u), Y ′(u), Y ′′(u), · · · , Y (n−3)(u), Y (n−2)(u), Y (n−1)(u) are equal to zero. That
is, Θϱ2(u) = 0 for all ϱ2 except ϱ2 = n.

Let us consider the coefficient of Y (u) is zero, or Θ0(u) = 0, by setting γ1,0 =
1, 2, 3, . . . , n− 1, γ2,0 = n, n+ 1, n+ 2, . . . ,m, and γ3,0 = m+ 1,m+ 2,m+ 3, . . . ,m+ n,
yields

um → am,0 = 0, (9)

um−γ1,0 → am−γ1,0,0 +

γ1,0∑
j=1

(−1)j−0

(
j

0

) j−1∏
s=0

(m− γ1,0 + j + α− s)am−γ1,0+j,j = 0, (10)

um−γ2,0 → am−γ2,0,0 +

n∑
j=1

(−1)j−0

(
j

0

) j−1∏
s=0

(m− γ2,0 + j + α− s)am−γ2,0+j,j = 0, (11)



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1394

um−γ3,0 →
n∑

j=γ3,0−m

(−1)j−0

(
j

0

) j−1∏
s=0

(m− γ3,0 + j + α− s)am−γ3,0+j,j = 0. (12)

Let us consider the coefficient of Y ′(u) is zero, or Θ1(u) = 0, by setting γ1,1 =
1, 2, 3, . . . , n−2, γ2,1 = n−1, n, n+1, . . . ,m, and γ3,1 = m+1,m+2,m+3, . . . ,m+n−1,
we obtain

um−2 → am,1 = 0, (13)

um−γ1,1−2 → am−γ1,1,1

+

γ1,1+1∑
j=2

(−1)j−1

(
j

1

) j−1∏
s=1

(m− 1− γ1,1 + j + α− s)× (14)

am−1−γ1,1+j,j = 0, (15)

um−γ2,1−2 → am−γ2,1,1

+

n∑
j=2

(−1)j−1

(
j

1

) j−1∏
s=1

(m− 1− γ2,1 + j + α− s)× (16)

am−1−γ2,1+j,j = 0, (17)

um−γ3,1−2 →
n∑

j=γ3,1−m+1

(−1)j−1

(
j

1

) j−1∏
s=1

(m− 1− γ3,1 + j + α− s)× (18)

am−1−γ3,1+j,j = 0. (19)

Let us consider the coefficient of Y ′′(u) is zero, or Θ2(u) = 0, by setting γ1,2 =
1, 2, 3, . . . , n−3, γ2,2 = n−2, n−1, n, . . . ,m, and γ3,2 = m+1,m+2,m+3, . . . ,m+n−2,
we have

um−4 → am,2 = 0, (20)

um−γ1,2−4 → am−γ1,2,2

+

γ1,2+2∑
j=3

(−1)j−2

(
j

2

) j−1∏
s=2

(m− 2− γ1,2 + j + α− s)am−2−γ1,2+j,j = 0, (21)

um−γ2,2−4 → am−γ2,2,2

+
n∑

j=3

(−1)j−2

(
j

2

) j−1∏
s=2

(m− 2− γ2,2 + j + α− s)am−2−γ2,2+j,j = 0, (22)

um−γ3,2−4 →
n∑

j=γ3,2−m+2

(−1)j−2

(
j

2

) j−1∏
s=2

(m− 2− γ3,2 + j + α− s)am−2−γ3,2+j,j = 0. (23)

Similarly, the coefficients of Y ′′′(u), Y (4)(u), Y (5)(u), . . . , Y (n−3)(u), Y (n−2)(u),
Y (n−1)(u) are equal to zero. Next, we will state the following equations from the co-
efficient of Y (n−3)(u) = Y (n−2)(u) = Y (n−1)(u) = 0, by letting γ1,n−3 = 1, 2, γ2,n−3 =



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1395

3, 4, 5, . . . ,m, γ3,n−3 = m+1,m+2,m+3, γ2,n−2 = 2, 3, 4, . . . ,m, γ3,n−2 = m+1,m+2,
and γ2,n−1 = 1, 2, 3, . . . ,m, it follows that

um−2(n−3) → am,n−3 = 0, (24)

um−γ1,n−3−2(n−3) → am−γ1,n−3,n−3 +

γ1,n−3+n−3∑
j=n−2

(−1)j−(n−3)

(
j

n− 3

)
×

j−1∏
s=n−3

(m− (n− 3)− γ1,n−3 + j + α− s)×

am−(n−3)−γ1,n−3+j,j = 0, (25)

um−γ2,n−3−2(n−3) → am−γ2,n−3,n−3 +
n∑

j=n−2

(−1)j−(n−3)

(
j

n− 3

)
×

j−1∏
s=n−3

(m− (n− 3)− γ2,n−3 + j + α− s)×

am−(n−3)−γ2+j,j = 0, (26)

um−γ3,n−3−2(n−3) →
n∑

j=γ3,n−3−m+(n−3)

(−1)j−(n−3)

(
j

n− 3

)
×

j−1∏
s=n−3

(m− (n− 3)− γ3,n−3 + j + α− s)×

am−(n−3)−γ3,n−3+j,j = 0, (27)

um−2(n−2) → am,n−2 = 0, (28)

um−2(n−2)−1 → am−1,n−2 +
n−1∑

j=n−1

(−1)j−(n−2)

(
j

n− 2

)
×

j−1∏
s=n−2

(m− (n− 1) + j + α− s)am−(n−1)+j,j = 0, (29)

um−γ2,n−2−2(n−2) → am−γ2,n−2,n−2 +
n∑

j=n−1

(−1)j−(n−2)

(
j

n− 2

)
×

j−1∏
s=n−2

(m− (n− 2)− γ2,n−2 + j + α− s)×

am−(n−2)−γ2,n−2+j,j = 0, (30)

um−γ3,n−2−2(n−2) →
n∑

j=γ3,n−2−m+(n−2)

(−1)j−(n−2)

(
j

n− 2

)
×



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1396

j−1∏
s=n−2

(m− (n− 2)− γ3,n−2 + j + α− s)×

am−(n−2)−γ3,n−2+j,j = 0, (31)

um−2(n−1) → am,n−1 = 0, (32)

um−γ2,n−1−2(n−1) → am−γ2,n−1,n−1 − n

j−1∏
s=n−1

(m− (n− 1)− γ2,n−1 + j + α− s)×

am−(n−1)−γ2,n−1+j,j = 0, (33)

um−(m+1)−2(n−1) → −n

j−1∏
s=n−1

(−n+ j + α− s)a−n+j,j = 0. (34)

Hence, according to Eqs. (9), (10), (13), (14), (17), (18), (21), (22), (25), (26), and (29),
it can be reduced to the following forms,

am,ϱ = 0; ϱ = 0, 1, 2, . . . , n− 1,

am−1,ϱ = 0; ϱ = 0, 1, 2, . . . , n− 2,

am−2,ϱ = 0; ϱ = 0, 1, 2, . . . , n− 3,

...

am−(n−3),ϱ = 0; ϱ = 0, 1, 2,

am−(n−2),ϱ = 0; ϱ = 0, 1

am−(n−1),0 = 0,

then, condition (4) becomes true. From Eqs. (12), (16), (20), (24), (28) and (31) can be
rewritten as

n∑
j=γ3−m+ϱ1

(−1)j−ϱ1

(
j

ϱ1

) j−1∏
s=ϱ1

(m− ϱ1 − γ3 + j + α− s)am−ϱ1−γ3+j,j = 0,

for ϱ1 = 0, 1, 2, . . . , n−1 and γ3 = 0, 1, 2, . . . ,m−(n−ϱ1+2),m−(n−ϱ1+1),m−(n−ϱ1).
Thus, condition (5) holds by replacing γ3 −m+ ϱ1 = ρ2. Finally, the conditions as stated
in Eqs. (11), (15), (19), (23), (27), and (30) are properly equated to condition (6). The
proof is completed.

Note that (i) no. COEs is the number of polynomial coefficients in Eq. (3) and
(ii) no. CONs is the number of conditions according to conditions (4)-(6) for solving Eq.
(3) using the Gα-transform.

Corollary 1. Given i = 0, 1, 2, . . . ,m, j = 0, 1, 2, . . . , n, where m ≥ n, ai,j are the
polynomial coefficients of

∑n
j=0 ai,jτ

j with am,n ̸= 0 in Eq. (3), the following statements
hold:

(I) no. COEs is (m+ 1)(n+ 1),



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1397

(II) no. CONs is mn+ n(n+3)
2 ,

(III) no. COEs = no. CONs iff m = (n−1)(n+2)
2 .

Proof. Assume that i = 0, 1, 2, . . . ,m, j = 0, 1, 2, . . . , n, where m ≥ n, ai,j are the
polynomial coefficients of

∑n
j=0 ai,jτ

j with am,n ̸= 0 in Eq. (3) and by letting ϱ1 =
0, 1, 2, . . . , n− 1, and ρ1 = m− (n− ϱ1 − 1),m− (n− ϱ1 − 2),m− (n− ϱ1 − 3), . . . ,m−
2,m − 1,m, ρ2 = ϱ1 + 1, ϱ1 + 2, ϱ1 + 3, . . . , n − 2, n − 1, n, ρ3 = 0, 1, 2, . . . ,m − (n − ϱ1 +
2),m− (n− ϱ1 + 1),m− (n− ϱ1).
(I) For each i = 0, 1, 2, . . . ,m, the numbers of polynomial coefficients of all orders of
differential equations are equal to n+ 1, then no. COEs is (m+ 1)(n+ 1).
(II) From condition (4), we obtain
if ϱ1 = 0, then ρ1 = m− (n− 1),m− (n− 2),m− (n− 3), . . . ,m− 2,m− 1,m, no. CONs
is n,
if ϱ1 = 1, then ρ1 = m− (n− 2),m− (n− 3),m− (n− 4), . . . ,m− 2,m− 1,m, no. CONs
is n− 1,
if ϱ1 = 2, then ρ1 = m− (n− 3),m− (n− 4),m− (n− 5), . . . ,m− 2,m− 1,m, no. CONs
is n− 2,

...
if ϱ1 = n− 3, then ρ1 = m− 2,m− 1,m, no. CONs is 3,
if ϱ1 = n− 2, then ρ1 = m− 1,m, no. CONs is 2,
if ϱ1 = n− 1, then ρ1 = m, no. CONs is 1.
Consequently, the number of conditions in conditon (4) is n(n+1)

2 .
From condition (5), we have
if ϱ1 = 0, then ρ2 = 1, 2, 3, . . . , n− 2, n− 1, n, no. CONs is n,
if ϱ1 = 1, then ρ2 = 2, 3, . . . , n− 2, n− 1, n, no. CONs is n− 1,
if ϱ1 = 2, then ρ2 = 3, . . . , n− 2, n− 1, n, no. CONs is n− 2,

...
if ϱ1 = n− 3, then ρ2 = n− 2, n− 1, n, no. CONs is 3,
if ϱ1 = n− 2, then ρ2 = n− 1, n, no. CONs is 2,
if ϱ1 = n− 1, then ρ2 = n, no. CONs is 1,
Consequently, no. CONs in condition (5) is n(n+1)

2 ,
From Eq. (6), we get
if ϱ1 = 0, then ρ3 = 0, 1, 2, . . . ,m− n− 2,m− n− 1,m− n, no. CONs is m− n+ 1,
if ϱ1 = 1, then ρ3 = 0, 1, 2, . . . ,m− n− 3,m− n− 2,m− n− 1, no. CONs is m− n+ 2,
if ϱ1 = 2, then ρ3 = 0, 1, 2, . . . ,m− n− 4,m− n− 3,m− n− 2, no. CONs is m− n+ 3,

...
if ϱ1 = n− 3, then ρ3 = 0, 1, 2, . . . ,m− 5,m− 4,m− 3, no. CONs is m− 2,
if ϱ1 = n− 2, then ρ3 = 0, 1, 2, . . . ,m− 4,m− 3,m− 2, no. CONs is m− 1,
if ϱ1 = n− 1, then ρ3 = 0, 1, 2, . . . ,m− 3,m− 2,m− 1, no. CONs is m.
Consequently, the number of conditions in Eq. (6) is mn − (n−1)n

2 , it follows that no.

CONs is mn+ n(n+3)
2 .

(III)(⇒) If no. COEs = no. CONs, then form (I) and (II), we have (m + 1)(n + 1) =



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1398

mn+ n(n+3)
2 , it can be rewritten as m = (n−1)(n+2)

2 .

(⇐) Let m = (n−1)(n+2)
2 , suppose that no. COEs is not equal to no. CONs, that is

(m + 1)(n + 1) ̸= mn + n(n+3)
2 , implying that m ̸= (n−1)(n+2)

2 , which is a contradiction.
Therefore, no. COEs = no. CONs.

4. Applications

According to Theorem 1, the solutions of the higher-order differential equations with
polynomial coefficients through the Gα-transform can be solved, as follows:

4.1. The application of fifth-order differential equation with polynomial
coefficients where α = 3

4.1.1. Process of general solution

Example 1. Let us consider the fifth-order differential equation with polynomial coeffi-
cients in the form of

t5y(5)(t) + 20t4y(4)(t) + 120t3y′′′(t) + 240t2y′′(t) + 120ty′(t) = t, t ≥ 0. (35)

From Eqs. (3) and Eq. (35), we have

a5,5 = 1, a4,4 = 20, a3,3 = 120, a2,2 = 240, a1,1 = 120,

and we determine α = 3 according to the conditions of Theorem 1, then applying the G3-
transform leads to finding the solution of (35). By using the G3-transform to (35), we
have

G3{t5y(5)(t)}+G3{20t4y(4)(t)}+G3{120t3y′′′(t)}+G3{240t2y′′(t)}+G3{120ty′(t)}
= G3{t}.

Using Lemma 1 and a little rewriting yields

u5F (5)(u) = u5,

F (5)(u) = 1.

It follows that

F (u) =
u5

120
+ c1

u4

24
+ c2

u3

6
+ c3

u2

2
+ c4u+ c5, (36)

where c1, c2, c3, c4, and c5 are constants.



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1399

4.1.2. Graphical analysis

From Matlab, Figure 1 shows the curves of general solutions y(t) by applying Lemma 2 and
the inverse G3-transform. For the classical solutions, we have to set c2 = c3 = c4 = c5 = 0
and various cases of c1 in Eq. (36),

(i) when c1 = 1, we get y(t) =
t

120
+

1

24
as a solution to Eq. (35),

(ii) when c1 = 3, we get y(t) =
t

120
+

1

8
as a solution to Eq. (35),

(iii) when c1 = 24, we get y(t) =
t

120
+ 1 as a solution to Eq. (35).

We can summarize that the general solutions are line graphs with intercept y-axis at several
points depending on c1.

0 1 2 3 4 5 6 7 8 9 10

t

0

0.2

0.4

0.6

0.8

1

1.2

y
 
(
t
)

c
1
 = 1,  c

2
 = c

3
 = c

4
 =  c

5
 = 0

c
1
 = 3, c

2
 = c

3
 = c

4
 =  c

5
 = 0

c
1
 = 24, c

2
 = c

3
 = c

4
 =  c

5
 = 0

Figure 1: General solutions of Example 1.

4.2. The application of fifth-order differential equation with polynomial
coefficients where α = −1

4.2.1. Process of general solution

Example 2. Let us consider the fifth-order differential equation with polynomial coeffi-
cients in the form of

t3y(5)(t) + (t3 + 6t2)y(4)(t) + (3t2 + 6t)y′′′(t) =
t2

2
+ t, t ≥ 0. (37)

From Eqs.(3) and (37), we have

a5,3 = 1, a4,3 = 1, a4,2 = 6, a3,2 = 3, a3,1 = 6,

and we determine α = −1 according to the conditions of Theorem 1, so applying the G−1-
transform leads to finding the solution of Eq. (37). By using the G−1-transform to (37),
we have

G−1{t3y(5)(t)}+G−1{t3y(4)(t)}+G−1{6t2y(4)(t)}+G−1{3t2y′′′(t)}+G−1{6ty′′′(t)}

= G−1{
t2

2
}+G−1{t}.



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1400

Using Lemma 1 and simplifying the above equation, we have the following

(u2 + u)F ′′′(u) = u2 + u,

F ′′′(u) = 1.

Then, we have

F (u) =
u3

6
+ c1

u2

2
+ c2u+ c3, (38)

where c1, c2, and c3 are constants.

4.2.2. Graphical analysis

From Matlab, Figure 2 draws the curve of general solution y(t) by applying Lemma 2 and
the inverse G−1-transform. For the classical solutions, we have to set c1 = c2 = c3 = 0 in

Eq. (38). It is straightforward to illustrate that y(t) =
t3

36
satisfies Eq. (37).

0 1 2 3 4 5 6 7 8 9 10

t

0

5

10

15

20

25

30

y
 
(
t
)

c
1
 = c

2
 =  c

3
 = 0

Figure 2: General solutions of Example 2.

4.3. The application of seventh-order differential equation with polyno-
mial coefficients where α = −5

4.3.1. Process of general solution

Example 3. Let us consider the seventh-order differential equation with polynomial coef-
ficients in the form of

t4y(7)(t)− 4t3y(6)(t) + 12t2y(5)(t) + (t4 − 24t)y(4)(t) + (−16t3 + 24)y′′′(t) + 120t2y′′(t)

− 480ty′(t) + 840y(t) = t8 + 336t5, t ≥ 0. (39)

From Eqs. (3) and (39), we have

a7,4 = 1, a6,3 = −4, a5,2 = 12, a4,4 = 1, a4,1 = −24, a3,3 = −16, a3,0 = 24, a2,2 = 120,

a1,1 = −480, a0,0 = 840,



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1401

and we determine α = −5 according to the conditions of Theorem 1, so applying the G−5-
transform leads to finding the solution of Eq. (39). By using the G−5-transform to (39),
we have

G−5{t4y(7)(t)} −G−5{4t3y(6)(t)}+G−5{12t2y(5)(t)}+G−5{t4y(4)(t)} −G−5{24ty(4)(t)}
−G−5{16t3y′′′(t)}+G−5{24y′′′(t)}+G−5{120t2y′′(t)} −G−5{480ty′(t)}+G−5{840y(t)}
= G−5{t8}+G−5{336t5}.

Using Lemma 1 and the above equation, this can be rewritten as

(u4 + u)F (4)(u) = 8!(u4 + u),

F (4)(u) = 8!.

Then, we have

F (u) = 8!
u4

24
+ c1

u3

6
+ c2

u2

2
+ c3u+ c4, (40)

where c1, c2, c3, and c4 are constants.

4.3.2. Graphical analysis

From Matlab, Figure 3 shows the curves of general solutions y(t) by applying Lemma 2
and the inverse G−5-transform. We let c1, c2, c3, and c4 in Eq. (40) in various ways as
follows:

(i) when c1 = c2 = c3 = c4 = 0, we get y(t) =
t8

24
as a solution of Eq. (39),

(ii) when c1 = 7!, c2 = 6!, c3 = 5!, c4 = 4!, we get y(t) =
t8

24
+

t7

6
+

t6

2
+ t5+ t4 as a solution

of Eq. (39),

(iii) when c1 = 6× 7!, c2 = 2× 6!, c3 = 5!, c4 = 4!, we get y(t) =
t8

24
+ t7 + t6 + t5 + t4 as a

solution of Eq. (39).

0 1 2 3 4 5 6 7 8 9 10

t

0

0.5

1

1.5

2

y
 
(
t
)

×10
7

c
1
 = c

2
 =  c

3
 = c

4
 = 0

c
1
 = 7!,  c

2
 = 6!, c

3
 = 5!, c

4
 = 4!

c
1
 = 6*7!,  c

2
 = 2*6!, c

3
 = 5!, c

4
 = 4!

Figure 3: General solutions of Example 3.



S. Sattaso, P. Prasertsang, T. Prasertsang / Eur. J. Pure Appl. Math, 16 (3) (2023), 1389-1405 1402

Remark 1. (I) For m = 5 and n = 5, by Corollary 1, no. COEs = 36 and no. CONs =
45 that is no. COEs ̸= no. CONs. The solutions of HODEPCs under conditions (4)-(5)
by the G3-transform are infinite solutions. In particular, in Example 1., the polynomial
coefficients a5,5 = 1, a4,4 = 20, a3,3 = 120, a2,2 = 240, a1,1 = 120, otherwise, 0 can be
solved.
(II) For m = 5 and n = 3, by Corollary 1, no. COEs = no. CONs = 24. Example 2. is
one of the solutions under conditions (4)-(5) which can be solved by the G−1-transform.
(III) For m = 7 and n = 4, by Corollary 1, no. COEs = 40 and no. CONs = 42 that
is no. COEs ̸= no. CONs. In Example 3., the polynomial coefficients a7,4 = 1, a6,3 =
−4, a5,2 = 12, a4,4 = 1, a4,1 = −24, a3,3 = −16, a3,0 = 24, a2,2 = 120, a1,1 = −480, a0,0 =
840, otherwise, 0 which can be solved using the solutions of HODEPCs under conditions
(4)-(5) by the G−5-transform.

In fact, the solutions of HODEPCs under conditions (4)-(5) by the Gα-transform can
be solved using in various examples. If not, the HODEPCs can not find the solutions.

Remark 2. From Example 2. if a3,1 is equal to 1, then we have

t3y(5)(t) + (t3 + 6t2)y(4)(t) + (3t2 + t)y′′′(t) =
t2

2
+ t, t ≥ 0. (41)

The conditions do not satisfy Theorem 1. If we take G−1-transform both sides of Eq. (41),
we obtain

(u2 + u)Y ′′′(u)− 5
Y ′(u)

u
+ 10

Y (u)

u2
= u2 + u.

Observe that Eq. (41) transformed into a third-order differential equation with variable
coefficients. As a result, setting α = −1 did not lead to the solution of (41), including for
α equaling all other values.

5. Conclusions

Solutions to higher-order differential equations with polynomial coefficients (HODE-
PCs) were introduced using the Gα-transform. The theorem and corollary of the HODE-
PCs were obtained to guarantee that they can be corrected by the Gα-transform. Next, we
showed some examples according to the theorem which is a strength of the Gα-transform
in solving the HODEPCs by selecting an appropriate value for α and proper coefficients
of the polynomial.

Acknowledgements

This research was funded by the following individuals: Research and Academic Service
Division and Faculty of Science and Engineering, Kasetsart University, Chalermprakiat
Sakon Nakhon Province Campus, Sakon Nakhon, Thailand.



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