EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 4, 2021, 1184-1199 ISSN 1307-5543 – ejpam.com Published by New York Business Global Analytical Study for Certain Ordinary Differential Equations with Variable Coefficients via Gα-Transform Patarawadee Prasertsang1, Supaknaree Sattaso1,∗, Kamsing Nonlaopon2, Hwajoon Kim3 1 Department of General Science, Kasetsart University, Chalermphrakiat Sakon Nakhon Province Campus, Sakon Nakhon 47000, Thailand 2 Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand 3 Department of IT Engineering, Kyungdong University, Yangju, Gyeonggi, Korea Abstract. Gα-transform, which is a comprehensive and essential form of Laplace-type integral transforms, has both advantages and limitations. The purpose of this study is to consider the applicable range ofGα-transform in finding solutions of ordinary differential equations with variable coefficients. Finally, several examples are given to demonstrate the effectiveness of these results. 2020 Mathematics Subject Classifications: 34A25, 34A26, 44A05 Key Words and Phrases: Laplace transform, Sumudu transform, Elzaki transform, Gα-transform, Ordinary differential equation 1. Introduction The differential equations have played a central role in every aspect of applied math- ematics for a very long time, and their importance has increased further with the advent of computers. Several mathematical methods have been applied by various researchers in various fields of science and engineering to obtain the analytical solutions of differ- ential equations, which appeared in the literature [26, 34, 36]. To solve the differential equations, the integral transforms were extensively used. The Laplace transform is one of many integral transforms in applied mathematics and is often used to solve differential equations. The Laplace transform reduces a linear differential equation to an algebraic equation, which can then be solved using algebra’s formal rules. After that, the differential equation can then be solved by applying the inverse Laplace transform [33]. The Laplace transform is beneficial for finding the solution of the diffusion equation in transient flow [8, 35, 43]. In ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4066 Email addresses: patarawadee.s@ku.th (P. Prasertsang), supaknaree.s@ku.th (S. Sattaso), nkamsi@kku.ac.th (K. Nonlaopon), cellmath@gmail.com (Hj. Kim) http://www.ejpam.com 1184 © 2021 EJPAM All rights reserved. S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1185 addition, many researchers mainly had paid attention to study for theory and applications of Laplace transform, see [9–11, 21, 41] for more details. The Laplace transform is a well-known fact that it converts a function f of a real variable t to a function F of a complex variable s, which is defined by F (s) = L{f(t)} = ∫ ∞ 0 e−stf(t)dt. In addition, if f(t) is a piecewise continuous on [0,∞) and has an exponential order k, then the Laplace transform F (s) = L{f(t)} exists for s > k. For s = 1/u, the Laplace transform L{f(t)} can be rewritten as L{f(t)} = ∫ ∞ 0 e−t/uf(t)dt. In the last two decades, many integral transforms in the class of Laplace-typed integral transform are introduced, such as Sumudu transform, Elzaki transform, natural transform, Aboodh transform, Mohand transform, Gα-transform, HY-transform, and Kamal trans- form. These transforms have been used for solving different types of integral equations, ordinary differential equations, partial differential equations, and fractional differential equations, see [1, 3, 15, 22, 37, 42, 45] for more details. Since the Laplace transform is not suitable for solving some differential equations, in 1993, G. Watugala [44] introduced a new transform, named Sumudu transform, and shown that Sumudu transform has fascinating properties, making it easy to visualize and apply it for finding the solution of ordinary differential equations in control engineering problems. Thus, the Sumudu transform is an ideal transform for control engineering and applied mathematics. In 2010, H. Eltayeb and A. Kilicman [14] introduced some relationships between Sumudu transform and Laplace transform. They showed that the solution which is given by Laplace transform into a complex domain and given by Sumudu transform into a real domain. Thus, this leads them to consider that if the solution exists by Sumudu trans- form, then the solution also exists by Laplace transform. Moreover, they showed a strong relationship between Sumudu transform and other integral transforms, see A. Kilicman et al.[13]. Many researchers applied Sumudu transform to solve the system of dynamic equations, partial differential equations with variable coefficient, a semi-infinite string, an integro- differential equation, the fractional neutron transport equation, see [2, 4–6, 12, 20, 23– 25, 27, 28] for more details. The Sumudu transform converts a function f of a real variable t to a function of a complex variable u, which is defined by S{f(t)} = 1 u ∫ ∞ 0 e−t/uf(t)dt. In addition, if f(t) is a piecewise continuous on [0,∞) and has an exponential order k, then the Sumudu transform S{f(t)} exists for u < 1/k. S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1186 Elzaki transform is the modified version of Laplace transform and Sumudu transform, which was first introduced by T.M. Elzaki [16] in 2011. Elzaki transform was then pre- sented when Sumudu transform failed to solve some differential equations with variable coefficients [19]. T.M. Elzaki et al. [17, 18] showed that Elzaki transform provides a method for analyzing ordinary differential equations such as linear dynamic systems equa- tion, signals-delay differential equation, and the renewal equation in statistics. The Elzaki transform converts a function f of a real variable t to a function of a complex variable u, which is defined by E{f(t)} = u ∫ ∞ 0 e−t/uf(t)dt. In particular, if f(t) is a piecewise continuous on t ≥ 0 and has an exponential order k, then the Elzaki transform E{f(t)} exists for u < 1/k. Recently, Hj. Kim [29] introduced the intrinsic structure and some properties of Gα- transform, which is defined by F (u) = Gα{f(t)} = uα ∫ ∞ 0 e−t/uf(t)dt, where α ∈ Z and u is a complex variable. The Gα-transform can be applied directly to any situation by choosing α appropriately. In addition, if f(t) is a piecewise continuous on t ≥ 0 and has an exponential order k, then the Gα-transform Gα{f(t)} exists for u < 1/k. The Gα-transform is a Laplace-type integral transform can be reduced to the Laplace transform, Sumudu transform, and Elzaki transform for α = 0,−1, 1, respectively. Moreover, we know that the Laplace transform has a strong point in the transforms of derivatives. If we set α = −2, then we obtain a simple tool for transforms of integral, which can be rewritten as G−2{f(t)} = 1 u2 ∫ ∞ 0 e−t/uf(t)dt, see [30]. Further, Hj. Kim [31] also solved Laguerre’s equation by the G−2-transform. In 2019, S. Sattaso et al. [39] studied the properties of Gα-transform and presented an example that cannot be solved by the Sumudu and Elzaki transforms, but it can be solved by the Gα-transform. Furthermore, Hj. Kim et al. [38] considered an application of Gα-transform in partial differential equations by using the n-th partial derivatives, and Hj. Kim [7, 32] also considered a proof concerning the Laplace transform of the n-th derivative of any order by mathematical induction and considered a variant of Gα-transform represented by a logarithmic function. The connection of this transform to the convolutional neural network can be found in [40]. In this paper, we give some conditions of certain ordinary differential equations that can be solved by Gα-transform. Furthermore, we include examples to demonstrate the effectiveness of these results. S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1187 2. Preliminaries In this section, we give some basic properties of the Gα-transform, which would appear in this study quite frequently. The proofs of the following properties are given in [29, 39]. Lemma 1. [29] (Gα-transform of derivatives) If f(t), f ′(t), . . . , f (m−1)(t) are continuous and f (m)(t) is a piecewise continuous function on [0,∞) for m ∈ N ∪ {0} and has an exponential order k for u < 1/k, then the following properties hold: (i) Gα{f ′(t)} = F (u) u − uαf(0); (ii) Gα{f ′′(t)} = F (u) u2 − uα−1f(0)− uαf ′(0); (iii) Gα{f (m)(t)} = F (u) um − m−1∑ k=0 uα−m+(k+1)f (k)(0), where F (u) = Gα{f(t)}. Lemma 2. [39] (Gα-transform of multiplication by power of t) If f(t) is a piecewise continuous function on [0,∞) and has an exponential order k for u < 1/k, then the following properties hold: (i) Gα{tf(t)} = u2F ′(u)− αuF (u); (ii) Gα{t2f(t)} = u4F ′′(u)− 2(α− 1)u3F ′(u) + (α− 1)αu2F (u); (iii) Gα{tnf(t)} = u2nF (n)(u)− ( n 1 ) (α− (n− 1))u2n−1F (n−1) + · · · − ( n n− 1 ) (α− (n− 1)) (α− (n− 2)) · · · (α− 1)un+1F ′(u) + (α− (n− 1)) (α− (n− 2)) · · ·αunF (u), where F (u) = Gα{f(t)}. Lemma 3. [39] If f (m)(t) is a piecewise continuous function on [0,∞) for m ∈ N ∪ {0} and has an exponential order k for u < 1/k, then Gα{tnf (m)(t)} = u2n dnGα{f (m)(t)} dun − ( n 1 ) [α− (n− 1)]u2n−1d n−1Gα{f (m)(t)} dun−1 + · · · − ( n n− 1 ) [α− (n− 1)] [α− (n− 2)] · · · (α− 1)un+1dGα{f (m)(t)} du + [α− (n− 1)] [α− (n− 2)] · · ·αunGα{f (m)(t)}. (1) Lemma 4. [29] If f(t) = tn for n ∈ N ∪ {0}, then Gα{tn} = n!un+α+1. S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1188 Remark 1. By using Lemma 3, substituting n = 1, 2, and 3 in (1) and derivatives, after some simplification, we obtain (i) Gα{tf (m)(t)} = F ′(u) um−2 − (m+ α) F (u) um−1 − m−1∑ k=0 (1 + k −m)u2+k+α−mf (k)(0); (ii) Gα{t2f (m)(t)} = F ′′(u) um−4 − 2(m+α− 1) F ′(u) um−3 + [m(m+ 1) + 2(α− 1)m+ (α− 1)α] ×F (u) um−2 − m−1∑ k=0 [(α−m+ k + 1)(α−m+ k)− 2(α− 1)(α−m+ k + 1) +(α− 1)α]u3+k+α−mf (k)(0); (iii) Gα{t3f (m)(t)} = F ′′′(u) um−6 − 3(m+ α− 2) F ′′(u) um−5 + [3m(m+ 1) + 6(α− 2)m +3(α− 2)(α− 1)] F ′(u) um−4 − [m(m+ 1)(m+ 2) + 3(α− 2)m(m+ 1) +3(α− 2)(α− 1)m+ (α− 2)(α− 1)α] F (u) um−3 − m−1∑ k=0 [(α−m+ k + 1)(α−m+ k)(α−m+ k − 1) −3(α−2)(α−m+k+1)(α−m+k)+3(α−2)(α−1)(α−m+k+1) −(α− 2)(α− 1)α]u4+k+α−mf (k)(0), where F (u) = Gα{f(t)}. 3. Main Results In this section, we show some conditions of certain ordinary differential equations to ensure that those ordinary differential equations can be solved by Gα-transform. Theorem 1. Consider the m-th order ordinary differential equation of the form( amt2 + bmt+ cm ) y(m)(t) + ( am−1t 2 + bm−1t+ cm−1 ) y(m−1)(t) + · · ·+ ( a0t 2 + b0t+ c0 ) y(t) = g(t), (2) where aj , bj , cj are constants, j = 0, 1, 2, . . . ,m and g(t) is an unknown function. The Gα-transform is a suitable method for solving (2), if the following conditions are satisfies cm = bm = cm−1 = (α− 1)αa0 = 2(α− 1)a0 = 0, [2 + 2(α− 1) + (α− 1)α]a1 − αb0 = bi−1 − 2(α+ i− 1)ai = 0 for i = 1, 2, 3, . . . ,m, and [i(i+ 1) + 2(α− 1)i+ (α− 1)α]ai − (i+ α− 1)bi−1 + ci−2 = 0 for i = 2, 3, 4, . . . ,m. S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1189 Proof. By using Remark 1(1-2) and taking Gα-transform of both sides to (2), we obtain[ am um−4 + am−1 um−5 + · · ·+ a1 u−3 + a0 u−4 ] F ′′(u) + [ −2(α+m− 1) am um−3 − 2(α+m− 2) am−1 um−4 − · · · − 2α a1 u−2 − 2(α− 1) a0 u−3 + bm um−2 + bm−1 um−3 + · · ·+ b1 u−1 + b0 u−2 ] F ′(u) + [( m(m+ 1) + 2(α− 1)m+ (α− 1)α ) am um−2 + ( (m− 1)m+ 2(α− 1)(m− 1) + (α− 1)α )am−1 um−3 + · · ·+ ( 2 + 2(α− 1) + (α− 1)α ) a1 u−1 + (α− 1)α a0 u−2 − (α+m) bm um−1 −(α+m− 1) bm−1 um−2 − · · · − (α+ 1)b1 − α b0 u−1 + cm um + cm−1 um−1 + · · ·+ c1 u + c0 ] F (u) = Gα{g(t)} − q(u), (3) where q(u) be contained in some expressions that are started by summation and do not influence the proof steps. If the Gα-transform is suitable method for solving (2), then the coefficient of F (u) and F ′(u) in (3) should be equal to zero. Thus, if the coefficient of F (u) = 0, then um → cm = 0; um−1 → cm−1 − (m+ α)bm = 0; um−2 → cm−2 − (m+ α− 1)bm−1 + (m(m+ 1) + 2(α− 1)m+ (α− 1)α) am = 0; ... u0 → c0 − (α+ 1)b1 + (6 + 4(α− 1) + (α− 1)α) a2 = 0; u−1 → −αb0 + ( 2 + 2(α− 1) + (α− 1)α ) a1 = 0; u−2 → (α− 1)αa0 = 0. And if the coefficient of F ′(u) = 0, then um−2 → bm = 0 um−3 → bm−1 − 2(m+ α− 1)am = 0 um−4 → bm−2 − 2(m+ α− 2)am−1 = 0 ... u−1 → b1 − 2(α+ 1)a2 = 0 u−2 → b0 − 2αa1 = 0 u−3 → 2(α− 1)a0 = 0. In general, we can show that cm = bm = cm−1 = (α− 1)αa0 = 2(α− 1)a0 = 0, S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1190 [2 + 2(α− 1) + (α− 1)α]a1 − αb0 = bi−1 − 2(α+ i− 1)ai = 0 for i = 1, 2, 3, . . . ,m, and [i(i+ 1) + 2(α− 1)i+ (α− 1)α]ai − (i+ α− 1)bi−1 + ci−2 = 0 for i = 2, 3, 4, . . . ,m. This completes the proof. Remark 2. From Theorem 1, if g(t) = 0, we can just set the coefficient of F (u) equal to zero to reduce conditions. Therefore, the Gα-transform is a suitable method for solving equation (2), if cm = cm−1 − (m+ α)bm = [2 + 2(α− 1) + (α− 1)α]a1 − αb0 = (α− 1)αa0 = 0, and [i(i+ 1) + 2(α− 1)i+ (α− 1)α]ai − (i+ α− 1)bi−1 + ci−2 = 0 for i = 2, 3, 4, . . . ,m. Theorem 2. Consider the m-th order ordinary differential equation of the form( amt3 + bmt2 + cmt+ dm ) y(m)(t) + ( am−1t 3 + bm−1t 2 + cm−1t+ dm−1 ) y(m−1)(t) + · · ·+ ( a0t 3 + b0t 2 + c0t+ d0 ) y(t) = g(t), (4) where aj , bj , cj , dj are constants, j = 0, 1, 2, . . . ,m and g(t) is an unknown function. The Gα-transform is a suitable method for solving (4), if the following conditions are satisfies dm = cm = bm = dm−1 = cm−1 = dm−2 = 0, (α− 2)(α− 1)αa0 = 3(α− 2)(α− 1)a0 = 3(α− 2)a0 = 0, αc0 − [2 + 2(α− 1) + (α− 1)α]b1 +[24 + 18(α− 2) + 6(α− 2)(α− 1) + (α− 2)(α− 1)α]a2 = 0, (α− 1)αb0 − [6 + 6(α− 2) + 3(α− 2)(α− 1) + (α− 2)(α− 1)α]a1 = 0, 2(α− 1)b0 − [6 + 6(α− 2) + 3(α− 2)(α− 1)]a1 = 0, di−3 − (α+ i− 2)ci−2 + [(i− 1)i+ 2(α− 1)(i− 1) + (α− 1)α]bi−1 −[i(i+ 1)(i+ 2) + 3(α− 2)i(i+ 1) + 3(α− 2)(α− 1)i+ (α− 2)(α− 1)α]ai = 0 for i = 3, 4, 5, . . . ,m, ci−2 − 2(α+ i− 2)bi−1 + [3i(i+ 1) + 6(α− 2)i+ 3(α− 2)(α− 1)]ai = 0 for i = 2, 3, 4, . . . ,m, and bi−1 − 3(α+ i− 2)ai = 0 for i = 1, 2, 3, . . . ,m. Proof. By using Remark 1 and taking Gα-transform of both sides to (4), we obtain[ am um−6 + am−1 um−7 + · · ·+ a1 u−5 + a0 u−6 ] F ′′′(u) + [ −3(m+ α− 2) am um−5 − 3(m+ α− 3) am−1 um−6 S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1191 − · · · − 3(α− 1) a1 u−4 − 3(α− 2) a0 u−5 + bm um−4 + bm−1 um−5 + · · ·+ b1 u−3 + b0 u−4 ] F ′′(u) + [( 3m(m+ 1) + 6(α− 2)m+ 3(α− 2)(α− 1) ) am um−4 + ( 3(m− 1)m+ 6(α− 2)(m− 1) + 3(α− 2)(α− 1) )am−1 um−5 + · · ·+ ( 6 + 6(α− 2) + 3(α− 2)(α− 1) ) a1 u−3 + 3(α− 2)(α− 1) × a0 u−4 − 2(α+m− 1) bm um−3 − 2(α+m− 2) bm−1 um−4 − · · · − 2α b1 u−2 − 2(α− 1) b0 u−3 + cm um−2 + cm−1 um−3 + · · ·+ c1 u−1 + c0 u−2 ] F ′(u) + [ − ( m(m+ 1)(m+ 2) + 3(α− 2)m(m+ 1) + 3(α− 2)(α− 1)m+ (α− 2)(α− 1)α ) am um−3 − ( (m− 1)m(m+ 1) + 3(α− 2)(m− 1)m + 3(α− 2)(α− 1)(m− 1) + (α− 2)(α− 1)α )am−1 um−4 − · · · − ( 6 + 6(α− 2) + 3(α− 2)(α− 1) + (α− 2)(α− 1)α ) a1 u−2 − (α− 2)(α− 1)α a0 u−3 − (m(m+ 1) +2(α− 1)m+ (α− 1)α) bm um−2 + ( (m− 1)m+ 2(α− 1)(m− 1) + (α− 1)α ) bm−1 um−3 + · · ·+ ( 2 + 2(α− 1) + (α− 1)α ) b1 u−1 + (α− 1)α b0 u−2 − (m+ α) cm um−1 − (m+ α− 1) cm−1 um−2 − · · · − (α+ 1) c1 u0 − α c0 u−1 + dm um + dm−1 um−1 + · · ·+ d1 u1 + d0 u0 ] F (u) = Gα{g(t)} − r(u), where r(u) be contained in some expressions that are started by summation and do not influence the proof steps. By using the previous results, which similar to the Theorem 1, we know that the coefficients of F (u), F ′(u) and F ′′(u) should be equal to zero, by the same process as Theorem 1, we can show that dm = cm = bm = dm−1 = cm−1 = dm−2 = 0, (α− 2)(α− 1)αa0 = 3(α− 2)(α− 1)a0 = 3(α− 2)a0 = 0, αc0 − [2 + 2(α− 1) + (α− 1)α]b1 +[24 + 18(α− 2) + 6(α− 2)(α− 1) + (α− 2)(α− 1)α]a2 = 0, (α− 1)αb0 − [6 + 6(α− 2) + 3(α− 2)(α− 1) + (α− 2)(α− 1)α]a1 = 0, 2(α− 1)b0 − [6 + 6(α− 2) + 3(α− 2)(α− 1)]a1 = 0, di−3 − (α+ i− 2)ci−2 + [(i− 1)i+ 2(α− 1)(i− 1) + (α− 1)α]bi−1 −[i(i+ 1)(i+ 2) + 3(α− 2)i(i+ 1) + 3(α− 2)(α− 1)i+ (α− 2)(α− 1)α]ai = 0 for i = 3, 4, 5, . . . ,m, ci−2 − 2(α+ i− 2)bi−1 + [3i(i+ 1) + 6(α− 2)i+ 3(α− 2)(α− 1)]ai = 0 for i = 2, 3, 4, . . . ,m, and bi−1 − 3(α + i − 2)ai = 0 for i = 1, 2, 3, . . . ,m. The proof is completed. S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1192 Remark 3. From Theorem 2, if g(t) = 0, we can just set the coefficient of F (u) equal to zero and F ′(u) equal to zero to reduce conditions. Therefore, the Gα-transform is a suitable method for solving equation (4), if dm = cm = dm−1 = cm−1 − 2(α+m− 1)bm = 0, dm−2 − (α+m− 1)cm−1 + [m(m+ 1) + 2(α− 1)m+ (α− 1)α]bm = 0, (α− 2)(α− 1)αa0 = 3(α− 2)(α− 1)a0 = 0, αc0 − [2 + 2(α− 1) + (α− 1)α]b1+ [24 + 18(α− 2) + 6(α− 2)(α− 1) + (α− 2)(α− 1)α]a2 = 0, (α− 1)αb0 − [6 + 6(α− 2) + 3(α− 2)(α− 1) + (α− 2)(α− 1)α]a1 = 0, 2(α− 1)b0 − [6 + 6(α− 2) + 3(α− 2)(α− 1)]a1 = 0, di−3 − (α+ i− 2)ci−2 + [(i− 1)i+ 2(α− 1)(i− 1) + (α− 1)α]bi−1 −[i(i+ 1)(i+ 2) + 3(α− 2)i(i+ 1) + 3(α− 2)(α− 1)i+ (α− 2)(α− 1)α]ai = 0 for i = 3, 4, 5, . . . ,m, and ci−2 − 2(α+ i− 2)bi−1 + [3i(i+ 1) + 6(α− 2)i+ 3(α− 2)(α− 1)]ai = 0 for i = 2, 3, 4, . . . ,m. 4. Examples In this section, we show the usage of Gα-transform for solving the ordinary differential equations with variable coefficients that according to Theorem 1 and Theorem 2 via some examples. Example 1. Consider the ordinary differential equation with variable coefficients of the form t2y′′(t) + 4ty′(t) + 2y(t) = t3. (5) From (2) and (5), we have a2 = 1, b1 = 4, c0 = 2, a0 = a1 = 0, b0 = b2 = 0, c1 = c2 = 0, and we define α = 1 to satisfy with the conditions of Theorem 1, so using the G1-transform leads to find the solution of (5). By applying the G1-transform to (5) and using Lemma 3, we obtain G1{t2y′′(t)}+G1{4ty′(t)}+G1{2y(t)} = G1{t3} u2F ′′(u)− 4uF ′(u) + 6F (u) + 4uF ′(u)− 8F (u) + 2F (u) = 6u5 F ′′(u) = 6u3. Then, we have F (u) = 3 10 u5 + c1u+ c2, S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1193 where c1 and c2 are constants. Letting c1 = c2 = 0, we get F (u) = 3 10 u5. By using Lemma 4 and the inverse G1-transform, thus the inverse of u5 is t3 6 , we obtain y(t) = 1 20 t3 as a solution of (5). It is not difficult to show that y(t) = 1 20 t3 satisfies (5). The next example will show that if the conditions do not satisfy Theorem 1, then it is not suitable to solve by this method as the following. Example 2. Consider the Legendre differential equation of the form (1− t2)y′′(t)− 2ty′(t) = t. (6) From (2) and (6), we have a2 = −1, c2 = 1, b1 = −2, a0 = a1 = 0, b0 = b2 = 0, c0 = c1 = 0, and with respect to the conditions in Theorem 1, c2 should be equal to 0, while c2 is equal to 1. Therefore, the conditions of Theorem 1 are not satisfied. If we take Gα-transform both sides of (6), we obtain Gα{(1− t2)y′′(t)} −Gα{2ty′(t)} = Gα{t} −u2F ′′(u) + 2αuF ′(u) + ( (α− 3)α+ 1 u2 ) F (u) = uα+2. Observe that (6) changed into a second-order ordinary differential equation with variable coefficients. Thus, using Gα-transform did not lead to finding the solution of (6). Example 3. Consider the ordinary differential equation with variable coefficients of the form t2y′′(t) + 2ty′(t)− 2y(t) = 0. (7) From (2) and (7), we have a2 = 1, b1 = 2, c0 = −2, a0 = a1 = 0, b0 = b2 = 0, c1 = c2 = 0, and we define α = 1 to satisfy with the conditions of Remark 2, so using the G1-transform leads to find the solution of (7). By applying the G1-transform to (7) and using Lemma 3, we obtain G1{t2y′′(t)}+G1{2ty′(t)} −G1{2y(t)} = 0 u2F ′′(u)− 4uF ′(u) + 6F (u) + 2uF ′(u)− 2F (u)− 2F (u)− 2F (u) = 0 u2F ′′(u)− 2uF ′(u) = 0. Then, we have F ′′(u) F ′(u) = 2 u . By integration both sides, we obtain lnF ′(u) = ln c1u 2 or F ′(u) = c1u 2, S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1194 and hence F (u) = c1 3 u3 + c2, where c1 and c2 are constants. Letting c2 = 0, we get F (u) = c1 3 u3. By using Lemma 4 and the inverse G1-transform, thus the inverse of u3 is t, we obtain y(t) = c1 3 t as a solution of (7). It is not difficult to show that y(t) = c1 3 t satisfies (7). Example 4. Consider the ordinary differential equation with variable coefficients of the form t3y′′′(t) + 9t2y′′(t) + 18ty′(t) + 6y(t) = t. (8) From (4) and (8), we have a3 = 1, b2 = 9, c1 = 8, d0 = 6, a0 = a1 = a2 = 0, b0 = b1 = b3 = 0, c0 = c2 = c3 = 0, d1 = d2 = d3 = 0, and we define α = 2 to satisfy with the conditions of Theorem 2, so using the G2-transform leads to find the solution of (8). By applying the G2-transform to (8) and using Lemma 3, we obtain G2{t3y′′′(t)}+G2{9t2y′′(t)}+G2{18ty′(t)}+G2{6y(t)} = G2{t} u3F ′′′(u)− 9u2F ′′(u) + 36uF ′(u)− 60F (u) + 9u2F ′′(u)− 54uF ′(u) + 108F (u) +18uF ′(u)− 54F (u) + 6F (u) = u4. Then, we have F ′′′(u) = u. By integration both sides, we obtain F (u) = 1 24 u4 + c1 2 u2 + c2u+ c3, where c1, c2, and c3 are constants. Letting c1 = c2 = c3 = 0, we get F (u) = 1 24 u4. By using Lemma 4 and the inverse G2-transform, thus the inverse of u4 is t, we obtain y(t) = 1 24 t as a solution of (8). The next example will show that if the conditions do not satisfy Theorem 2, then it is not suitable to solve by this method as the following. Example 5. Consider the ordinary differential equation with variable coefficients of the form (t3 + t)y′′′(t) + 6t2y′′(t) + 6ty′(t) = t2. (9) From (4) and (9), we have a3 = 1, c3 = 1, b2 = 6, c1 = 6, a0 = a1 = a2 = 0, b0 = b1 = b3 = 0, c0 = c2 = 0, d0 = d1 = d2 = d3 = 0, S. Sattaso et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1184-1199 1195 and with respect to the conditions in Theorem 2, c3 should be equal to 0, while c3 is equal to 1. Therefore, the conditions of Theorem 2 are not satisfied. If we take Gα-transform both sides of (9), we obtain Gα{(t3 + t)y′′′(t)}+Gα{6t2y′′(t)}+Gα{6ty′(t)} = Gα{t2} u3F ′′′(u)− [3 + 3(α− 2)]u2F ′′(u) + [ 18− 18(α− 2) + 3(α− 2)(α− 1) −12(α− 1) + 1 u2 ] uF ′(u)− [ 24 + 36(α− 2) + 9(α− 2)(α− 1) +(α− 2)(α− 1)α+ (α+ 3) 1 u2 − 24(α− 1) + 6(α+ 1) ] F (u) = 2uα+3. Observe that (9) changed into a third order ordinary differential equation with variable coefficients. Thus, by using Gα-transform did not lead to find the solution of (9). Example 6. Consider the ordinary differential equation with variable coefficients of the form t3y′′′(t) + 4t2y′′(t)− 2ty′(t)− 4y(t) = 0. (10) From (4) and (10), we have a3 = 1, b2 = 4, c1 = −2, d0 = −4, a0 = a1 = a2 = 0, b0 = b1 = b3 = 0, c0 = c2 = c3 = 0, d1 = d2 = d3 = 0, and we define α = 1 to satisfy with the conditions of Remark 3, so using the G1-transform leads to find the solution of (10). By applying the G1-transform to (10) and using Lemma 3, we obtain G1{t3y′′′(t)}+G1{4t2y′′(t)} −G1{2ty′(t)} −G1{4y(t)} = 0 u3F ′′′(u)− 9u2F ′′(u) + 36uF ′(u)− 60F (u) + 3u2F ′′(u)− 18uF ′(u) + 36F (u) +4u2F ′′(u)− 16uF ′(u) + 24F (u)− 2uF ′(u) + 2F (u) + 2F (u)− 4F (u) = 0. Then, we have F ′′′(u) F ′′(u) = 2 u . By integration both sides, we obtain lnF ′′(u) = ln c1u 2 or F ′′(u) = c1u 2, and hence F (u) = c1 12 u4 + c2u+ c3, where c1, c2, and c3 are constants. Letting c2 = c3 = 0, we get F (u) = c1 12 u4. By using Lemma 4 and the inverse G1-transform, thus the inverse of u4 is t2 2 , we obtain y(t) = c1 24 t2 as a solution of (10). REFERENCES 1196 Remark 4. We can see that Example 1, 3, and 6 can be solved by G1-transform, and Example 4 can be solved by G2-transform, it is clear that Sumudu transform cannot be solved for these ordinary differential equations. Remark 5. 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