On the Derivation of a Closed-Form Expression for the Solutions of a Subclass of Generalized Abel Differential Equations

Abstract

We investigate the properties of a general class of differential equations described by 𝑑𝑦(𝑡)/𝑑𝑡=𝑓_𝑘+1(𝑡)𝑦(𝑡)^𝑘+1+𝑓_𝑘(𝑡)𝑦(𝑡)^𝑘+⋯+𝑓₂(𝑡)𝑦(𝑡)²+𝑓₁(𝑡)𝑦(𝑡)+𝑓₀(𝑡), with 𝑘>1 a positive integer and 𝑓_𝑖(𝑡),0≤𝑖≤𝑘+1, with 𝑓_𝑖(𝑡), real functions of 𝑡. For 𝑘=2, these equations reduce to the class of Abel differential equations of the first kind, for which a standard solution procedure is available. However, for 𝑘>2 no general solution methodology exists, to the best of our knowledge, that can lead to their solution. We develop a general solution methodology that for odd values of 𝑘 connects the closed form solution of the differential equations with the existence of closed-form expressions for the roots of the polynomial that appears on the right-hand side of the differential equation. Moreover, the closed-form expression (when it exists) for the polynomial roots enables the expression of the solution of the differential equation in closed form, based on the class of Hyper-Lambert functions. However, for certain even values of 𝑘, we prove that such closed form does not exist in general, and consequently there is no closed-form expression for the solution of the differential equation through this methodology.

1. Introduction

In [1], a differential equation was derived, in the context of the theoretical analysis of a security protocol, that can be written in the following form: ()(𝑡)1−𝑡𝑑𝑦=𝑘𝑑𝑡𝑘−12𝑦(𝑡)−1^{𝑘−⌈𝑘/2⌉+1}1−𝑦(𝑡)^⌈𝑘/2⌉. (1) In [2], the differential equations (1) are written as partial fractions in such a way that the form of their solution matches the recursive pattern of the definition of the Hyper-Lambert functions that were proposed in [3] as a generalization of the well-known Lambert 𝑊 function (see [4] for an introduction to this function and its properties). The differential equations defined by (1) can be seen as belonging to the general differential equation form 𝑑𝑦(𝑡)𝑑𝑡=𝑓_𝑘+1(𝑡)𝑦(𝑡)^𝑘+1+𝑓_𝑘(𝑡)𝑦(𝑡)^𝑘+⋯+𝑓₂(𝑡)𝑦(𝑡)²+𝑓₁(𝑡)𝑦(𝑡)+𝑓₀(𝑡), (2) with 𝑓_𝑖(𝑡), 0≤𝑖≤𝑘+1 real functions. The class defined by (2) appears to generalize, naturally, the Abel class of differential equations (see [5]). Unfortunately, no general solution strategy exists (to the best of our knowledge) for the solution of this general class for 𝑘>2, in analogy with the methodology that exists for the Abel class. Thus, we could not reduce the problem of finding the solution of (1) into the problem of solving (2) for various values of 𝑘.

In this paper, we propose a general methodology which for odd values of 𝑘>2 provides a closed-form expression of the solution of (1) based on closed-form solutions of the roots of the equation 𝑔(𝑦(𝑡))−𝑠 where 𝑔(𝑦(𝑡))=^𝑘−1_{⌈𝑘/2⌉−1}𝑦(𝑡)^{𝑘−⌈𝑘/2⌉+1}(1−𝑦(𝑡))^⌈𝑘/2⌉ (the polynomial on the right-hand side of (1)) and 𝑠≠0. Moreover, this closed-form solutions are linked with the generalized Hyper-Lambert functions.

However, for certain even values of 𝑘>2 (at least for values of 𝑘 where 𝑘+1 is prime), we prove that the roots of 𝑔(𝑦(𝑡))−𝑠 cannot be expressed by radicals, and, thus, no closed-form solution can be obtained using this methodology. This latter result may be of independent interest.

2. Properties of the Target Class of Differential Equations

The target differential equation class is the following (see [1] for its derivation): 𝑑𝑥(𝑡)𝑑𝑡=−^𝑘_{⌈⌉𝑗=𝑘/2}𝑘𝑗𝑥(𝑡)1−(1−𝑡)^𝑗𝑥(𝑡)1−𝑡^𝑘−𝑗×𝑗1−𝑘+−𝑥(𝑡)()×1−𝑡_{⌈𝑘/2⌉−1𝑗=0}𝑘𝑗×𝑥(𝑡)1−(1−𝑡)^𝑗𝑥(𝑡)1−𝑡^𝑘−𝑗. (3) If we set 𝑦(𝑡)=𝑥(𝑡)/(1−𝑡), (3) becomes 𝑑𝑦(𝑡)=𝑦(𝑡)𝑑𝑡×1−𝑡1−^{⌈𝑘/2⌉−1}_𝑗=0𝑘𝑗1−𝑦(𝑡)^𝑗𝑦(𝑡)^𝑘−𝑗−1×1−𝑡_{𝑘⌈⌉𝑗=𝑘/2}𝑘𝑗𝑗1−𝑘1−𝑦(𝑡)^𝑗𝑦(𝑡)^𝑘−𝑗. (4) After some algebraic manipulations, (4) becomes ()(𝑡)1−𝑡𝑑𝑦=𝑑𝑡_{𝑘⌈⌉𝑗=𝑘/2}𝑘𝑗1−𝑦(𝑡)^𝑗𝑦(𝑡)^𝑘−𝑗𝑗1−𝑘−𝑦. (5) Let 𝑔(𝑦(𝑡)) be the polynomial, in 𝑦(𝑡), at the right-hand side of the differential equation (5). Throughout this paper, when we refer to this polynomial, we will always mean that the polynomial is in 𝑦(𝑡), treating 𝑦(𝑡) as an independent variable. Based on the following two lemmas, whose proofs follow by straightforward algebraic manipulations, we will present the properties of the class of ordinary differential equations defined by (5).

Lemma 1. For any 𝑙, 0≤𝑙≤𝑘, and independently of 𝑦, it holds ^𝑘_𝑗=𝑙𝑘𝑗𝑦^{𝑙−1−𝑗}1−𝑦^𝑗−𝑙𝑗1−𝑘−𝑦=−𝑘−1𝑙−1. (6) Based on Lemma 1, the following can be proved.

Lemma 2. The right-hand side of the differential equation (5) can be written as follows: 𝑔𝑦(𝑡)=𝑘𝑘−12𝑦(𝑡)−1^{𝑘−⌈𝑘/2⌉+1}1−𝑦(𝑡)^⌈𝑘/2⌉. (7) From Lemma 2, the following property of the polynomial 𝑔(𝑦(𝑡)) can be readily derived.

Corollary 3. The roots of the polynomial 𝑔(𝑦(𝑡)) are 0 with multiplicity ⌊𝑘/2⌋+1 and 1 with multiplicity ⌈𝑘/2⌉.

Lemma 4. The solution 𝑦(𝑡) of the differential equation (5) with 𝑦(0)∈[0,1) is a function that has the following two properties: (i) all its values are real numbers in the interval (0,1), and (ii) it is monotonically increasing.

Proof. Since 𝑦(0)∈[0,1), because 0≤𝑥(0)<1 and 𝑡=0, and the polynomial 𝑔(𝑦(𝑡)) contains the factors 𝑦(𝑡) and (1−𝑦(𝑡)) and its roots are 0 and 1, the solution function 𝑦(𝑡) of the differential equation (5) cannot get values below 0 or above 1. Thus, statement (i) holds.
Since 𝑑𝑦(𝑡)/𝑑𝑡 is always positive, as 𝑔(𝑦(𝑡)) is always positive for 𝑦(𝑡)∈(0,1) and 1−𝑡>0, then 𝑦(𝑡) is a monotonically increasing function, thus proving statement (ii).

Lemma 5. The roots of the first derivative of 𝑔(𝑦(𝑡)) as a function of 𝑦(𝑡) are 𝑤₁=0 with multiplicity ⌊𝑘/2⌋, 𝑤₂=1 with multiplicity ⌈𝑘/2⌉, and 𝑤₃=(𝑘+1−⌈𝑘/2⌉)/(𝑘+1), while 𝑤₃ is a maximum point of 𝑔(𝑦(𝑡)).

Proof. Since 𝑔(𝑦(𝑡)) is continuous on [0,1], 𝑔(0)=𝑔(1)=0, and it is differentiable on (0,1), there is 𝑤∈(0,1) where 𝑑𝑔(𝑦(𝑡))/𝑑𝑦(𝑡)|_{𝑦(𝑡)=𝑤}=0 and 𝑔(𝑦(𝑡)) attains minimum and maximum values on [0,1]. After some algebraic manipulations we obtain a closed formula for the first derivative of 𝑔(𝑦(𝑡)): 𝑑𝑔𝑦(𝑡)(𝑡)=𝑑𝑦𝑘𝑘−12𝑘−1𝑘+1−21(𝑡)(𝑡)1−𝑦×𝑦^{𝑘−⌈𝑘/2⌉}1−𝑦(𝑡)^⌈𝑘/2⌉. (8) From (8), it is readily derived that the roots of the first derivative of 𝑔(𝑦(𝑡)) are 𝑤₁=0 with multiplicity ⌊𝑘/2⌋, 𝑤₂=1 with multiplicity ⌈𝑘/2⌉, and 𝑤₃=(𝑘+1−⌈𝑘/2⌉)/(𝑘+1).
The second derivative of 𝑔(𝑦(𝑡)) is given by the following formula: _{⌈⌉𝑘−}1_1𝑘/2−1𝑑²𝑔𝑦(𝑡)(𝑡)=𝑘𝑑𝑦𝑘−2𝑤(𝑡)𝑦(𝑡)^{𝑘−⌈𝑘/2⌉−1}×1−𝑦(𝑡)^⌈𝑘/2⌉−𝑘2𝑤(𝑡)𝑦(𝑡)^{𝑘−⌈𝑘/2⌉}×1−𝑦(𝑡)^{⌈𝑘/2⌉−1}−𝑘2𝑦(𝑡)^{𝑘−⌈𝑘/2⌉}×1−𝑦(𝑡)^{⌈𝑘/2⌉−2}, (9) where 𝑤(𝑡)=𝑘+1−⌈𝑘/2⌉(1/(1−𝑦(𝑡))). For 𝑦(𝑡)=𝑤₃, the factor 𝑤(𝑡) of (9) becomes 0, and thus a negative term remains. Thus 𝑑²𝑔(𝑦(𝑡))/𝑑𝑦(𝑡)<0 on [0,1] which means that 𝑤₃ is a maximum point of 𝑔(𝑦(𝑡)).

Lemma 6. The function 𝑔(𝑦(𝑡)) as function of 𝑦(𝑡) is not monotonic on [0,1], and it is an increasing function on (0,𝑤₃) and a decreasing function on (𝑤₃,1).

Proof. Since 𝑦(𝑡)∈(0,1), the sign of the first derivative of 𝑔(𝑦(𝑡)) in (8) is determined by the factor (𝑘+1−⌈𝑘/2⌉(1/(1−𝑦(𝑡)))). Specifically, it holds that (𝑑𝑔(𝑦(𝑡))/𝑑𝑦(𝑡))>0 for 𝑦(𝑡)∈(0,𝑤₃) and (𝑑𝑔(𝑦(𝑡))/𝑑𝑦(𝑡))<0 for 𝑦(𝑡)∈(𝑤₃,1). Since the 𝑑𝑔(𝑦(𝑡))/𝑑𝑦(𝑡) changes sign on [0,1], the function 𝑔(𝑦(𝑡)) is not monotonic, but it is an increasing function on (0,𝑤₃) and a decreasing function on (𝑤₃,1).

Lemma 7. For odd 𝑘, 𝑤₃=1/2 and 𝑔(𝑤₃)=^𝑘−1_{⌈𝑘/2⌉−1}(1/2)^{2⌈𝑘/2⌉}, while for even 𝑘,  𝑤₃=1−(𝑘/(2(𝑘+1)))>1/2 and lim_𝑘→∞𝑤₃=1/2.

Proof. For an odd 𝑘, it holds ⌈𝑘/2⌉=(𝑘+1)/2 and 𝑘−⌈𝑘/2⌉+1=⌈𝑘/2⌉. Thus, 𝑤₃=𝑘+1−⌈𝑘/2⌉=1𝑘+12, (10) and by applying 𝑦(𝑡)=𝑤₃ in (7), we obtain the following:𝑔𝑤₃=𝑘𝑘−121−12^{2⌈𝑘/2⌉}. (11) For even 𝑘, it holds ⌈𝑘/2⌉=𝑘/2. Consequently, we obtain 𝑤₃𝑘=1−2(𝑘+1) (12) which means that 𝑤₃>1/2. Applying L’Hospital’s rule, for large even 𝑘, we have that lim_𝑘→∞𝑤₃=1/2.

The following theorem gives the specific form of the polynomial function 𝑔(𝑦(𝑡)) of (5).

Theorem 8. The right-hand side 𝑔(𝑦(𝑡)) of differential equation (5) is a polynomial function of one argument of (𝑘+1) degree, and the minimum degree of its terms is 𝑘−⌈𝑘/2⌉+1; that is, it has the form: ()(𝑡)1−𝑡𝑑𝑦𝑦(𝑡=𝑑𝑡=𝑔)^⌈𝑘/2⌉_𝑗=0𝑎_𝑗⋅𝑦(𝑡)^𝑠+𝑗, (13) where 𝑠=𝑘−⌈𝑘/2⌉+1 and 𝑎_𝑗=(−1)^𝑗𝑘2𝑗𝑘𝑘−12−1. (14)

Proof. Based on (7) and the binomial theorem, the results of the theorem can be readily deduced.

Lemma 9. The function 𝑔 is concave on [0,1].

Proof. The interval [0,1] is convex, since every interval in ℝ¹ is convex. Thus, for 0≤𝜆≤1, the convex combination of any 𝑦(𝑡₁) and 𝑦(𝑡₂), 𝑦(𝑡₃)=𝜆𝑦(𝑡₁)+(1−𝜆)𝑦(𝑡₂) is in the interval [0,1]. If 𝜆 is close to 0, the convex combination is close to 𝑦(𝑡₂). Based on Lemma 6, it is obvious that 𝑔(𝑦(𝑡₃))≥𝜆𝑔(𝑦(𝑡₁))+(1−𝜆)𝑔(𝑦(𝑡₂)), which means that the function 𝑔 is concave.

Table 1 presents the polynomial function of the right-hand side of the differential equation using Theorem 8 for various values of 𝑘.

Table 1 

The polynomial of the right-hand side of (13) for various values of 𝑘.

Some plots of the polynomial function 𝑔(𝑦(𝑡)) as a function of 𝑦(𝑡) in the same graph for various odd values of 𝑘 appear in Figure 1. It is easy to verify the result of Lemma 7 that the local maxima of 𝑔 appear at 𝑦(𝑡)=0.5 independently of 𝑘 value. Moreover, the function 𝑔 is symmetric around the axis 𝑦(𝑡)=0.5; that is, for every value 𝑔(𝑦(𝑡))≤𝑔(0.5), there are 𝑦(𝑡₁)<0.5, 𝑦(𝑡₂)>0.5, and |𝑦(𝑡₁)−0.5|=|𝑦(𝑡₂)−0.5| where 𝑔(𝑦(𝑡₁))=𝑔(𝑦(𝑡₂)). By solving the differential equation (5) for a certain odd value of 𝑘 where 𝑔(𝑦(𝑡))∈(0,𝑔(0.5)), we can calculate 𝑡 that 𝑔(𝑦(𝑡)) gets its maximum value; that is, the first derivative of 𝑦(𝑡) becomes maximum. This value of 𝑡 along with the fact that 𝑔(𝑦(𝑡)) is symmetric around the axis 𝑦(𝑡)=0.5 can be used to calculate the time value at which the sought function attains a specific target value, which can be of use in determining a particular stopping time for the protocol given in [1].

Figure 1 

A plot of 𝑔(𝑦(𝑡)) as a function of 𝑦(𝑡) for odd values of 𝑘.

Figure 2 presents some plots of 𝑔 as a function of 𝑦(𝑡) in the same graph for various even values of 𝑘. In these plots, the local maxima of 𝑔 appear at 𝑦(𝑡)=0.66,0.6,0.57, and 0.55 for 𝑘=2,4,6,8 correspondingly. Moreover, as 𝑘 increases, the local maxima of 𝑔 are obtained at a 𝑦(𝑡) closer to 0.5. But in practice, 𝑘 takes small values. Moreover, the range (0,𝑔(𝑤₃)) where the function 𝑔 is an increasing function is longer than the range (𝑔(𝑤₃),1) where the function 𝑔 is a decreasing function; that is, the values of 𝑔 are not equally balanced around axis 𝑦(𝑡)=𝑤₃.

Figure 2 

A plot of 𝑔(𝑦(𝑡)) as a function of 𝑦(𝑡) for even values of 𝑘.

Theorem 10. For 𝑠∈(0,𝑔(𝑤₃)), the equation 𝑔(𝑦(𝑡))=𝑠 has always one real root 𝑟₁(𝑠) within the interval (0,𝑤₃) and one real root 𝑟₂(𝑠) within the interval (𝑤₃,1). In addition, the root 𝑟₁(𝑠) is a monotonically increasing function of 𝑠, while 𝑟₂(𝑠) is a monotonically decreasing function of 𝑠.

Proof. Let 𝑝(𝑦(𝑡))=𝑔(𝑦(𝑡))−𝑠. Since the first derivative of 𝑝(𝑦(𝑡)) is equal to the first derivative of 𝑔(𝑦(𝑡)), from Lemma 5, it follows that 𝑤₃ is also a root of 𝑑𝑝(𝑦(𝑡))/𝑑𝑡. By Lemmas 5 and 6 we conclude that the polynomial 𝑝(𝑦(𝑡)) attains its maximum at 𝑤₃. Moreover, it is monotone increasing for 𝑦(𝑡)<𝑤₃ and monotone decreasing for 𝑦(𝑡)>𝑤₃.
It holds 𝑝(0)=𝑝(1)=−𝑠<0, while 𝑝(𝑤₃)=𝑔(𝑤₃)−𝑠>0 since 𝑠∈(0,𝑔(𝑤₃)). Thus, since 𝑝(𝑦(𝑡)) as a function of 𝑦(𝑡) is continuous on (0,1) and it has opposite signs at the endpoints of intervals (0,𝑤₃) and (𝑤₃,1), it has one real solution 𝑟₁(𝑠) in the first interval and one real solution 𝑟₂(𝑠) in the second one.
Finally, due to the monotonicity of 𝑝(𝑦(𝑡)) in each of these intervals, the root 𝑟₁(𝑠) monotonically increases as 𝑠 increases, while the root 𝑟₂(𝑠) monotonically decreases.

Lemma 11. For odd 𝑘, it holds 𝑟₁(𝑠)+𝑟₂(𝑠)=1.

Proof. From Lemma 7, 𝑤₃=1/2. Thus 𝑟₁(𝑠)∈(0,1/2) and 𝑟₂(𝑠)∈(1/2,1). Since 𝑝(𝑦(𝑡)) as a function of 𝑦(𝑡) is continuous on (0,1) and it has opposite signs at the endpoints of intervals (0,1/2) and (1/2,1), the roots 𝑟₁(𝑠) and 𝑟₂(𝑠) can be approximated by the use of the bisection method.
After 𝑛 iterations of the bisection method on these intervals, the calculated roots 𝑟₁(𝑠) and 𝑟₂(𝑠) satisfy the identity 𝑟₁(𝑠)+𝑟₂(𝑠)=1 since the intervals are balanced, 𝑔(𝑟₁(𝑠))=𝑔(𝑟₂(𝑠))=𝑠, and in each iteration the method operates in one of the two halves of the two intervals.

3. The Solution of the Class of the Differential Equations for Odd 𝑘

By the class of differential equations defined by (13), it is easy to see that it has the form 𝑑𝑦(𝑡)/𝑑𝑡=∑^𝑘+1_𝑖=0𝑓_𝑖(𝑡)𝑦(𝑡)^𝑖, with 𝑘>1 and 𝑓_𝑖(𝑡), 0≤𝑖≤𝑘+1, specific real functions of 𝑡. This form appears to generalize the class of Abel differential equations of the first kind in that it involves the sought function to powers greater than 3, for 𝑘≥3. To the best of our knowledge, no general solution strategy exists for generalized differential equations of the Abel type that can be exploited to solve (13). In this section we will describe such a solution strategy for odd 𝑘 that makes use of the result of Lemma 4 and the properties of the polynomial function 𝑔(𝑦(𝑡)) that have been proved in Section 2.

Definition 12. Let 𝐼 be an index set, and let 𝑓_𝑖∶ℂ→ℂ be arbitrary complex functions which do not vanish identically. Assuming 𝑚,𝑛∈ℕ such that 𝑚≥𝑛, we define 𝐹_𝑛,𝑚(𝑧)∶ℕ²×ℂ→ℂ as follows: 𝐹_𝑛,𝑚(𝑧)=𝑒^𝑧𝑒if𝑛=1,_{𝑓𝑚−(𝑛−1)(𝑧)𝐹𝑛−1,𝑚(𝑧)}if𝑛>1. (15)

Definition 13. Let 𝑓_𝑖 be as in Definition 12 and 𝑧∈ℂ. Then we define 𝐺𝑓₁(𝑧),𝑓₂(𝑧),…,𝑓_𝑗(𝑧);𝑧=𝑧𝐹_{𝑗+1,𝑗+1}(𝑧). (16)

Definition 14. Let 𝑓_𝑖 be as in Definition 12, 𝐺 as in Definition 13, and 𝑦∈ℂ. Then the function HW({𝑓_𝑖}_𝑖∈𝐼;𝑦) is the function which satisfies the following equation: 𝐺𝑓_{𝑖𝑖∈𝐼};HW𝑓_{𝑖𝑖∈𝐼};𝑦=𝑦. (17) If we set 𝑠∈ℝ₊ at the right side of (1), we obtain the following functional equation: ()(𝑡)1−𝑡𝑑𝑦𝑑𝑡=𝑠. (18) Our approach relies on the following fact for the solution 𝑡(𝑠) of the functional equation (18).

Claim 1. Let 𝑟(𝑠) be a real function in ℝ, such that each value of 𝑡⁻¹ belongs to the domain of 𝑟(𝑠) for every value 𝑠∈ℝ₊. Assuming that 𝑏=𝑡(𝑠), then 𝑦(𝑏)=𝑦(𝑡(𝑠))=𝑟(𝑡⁻¹(𝑏)).

Lemma 15. The solution 𝑡(𝑠), 𝑠∈ℝ₊, of the functional equation ()(𝑡)1−𝑡𝑑𝑦𝑑𝑡=𝑠 (19) forms a solution of the equation 𝑔(𝑦(𝑡))=𝑠.

Proof. Given 𝑠∈ℝ₊ and the solution 𝑡(𝑠) to (18), it follows from (13) that 𝑦(𝑡(𝑠)) is a root of the polynomial equation 𝑔(𝑦(𝑡))=𝑠.

Theorem 16. For odd 𝑘 and 𝑠∈ℝ₊, the solution 𝑡(𝑠) of the functional equation (18) is equal to 𝑡(𝑠)=1+𝑐𝑒^{−∫(1/𝑠)(𝑑𝑟(𝑠)/𝑑𝑠)𝑑𝑠} (20) with 𝑟(𝑠)=(1±1−4ℎ(𝑠))/2, where ℎ(𝑠)=(𝑠/^𝑘−1_{⌈𝑘/2⌉−1})^{1/⌈𝑘/2⌉} and the real positive solution of the equation 𝑔(𝑦(𝑡))=𝑠.

Proof. For odd 𝑘 the exponents of 𝑦(𝑡) and (1−𝑦(𝑡)) of the polynomial 𝑔(𝑦(𝑡)) in (7) become equal, that is, 𝑘−⌈𝑘/2⌉+1=⌈𝑘/2⌉=𝑙, and the polynomial 𝑔(𝑦(𝑡)) could be written as follows: 𝑔𝑦(𝑡)=𝑘𝑘−12𝑦(𝑡)−1^𝑙1−𝑦(𝑡)^𝑙. (21) Based on (21), the polynomial equation 𝑔(𝑦(𝑡))=𝑠 becomes 𝑦(𝑡)1−𝑦(𝑡)=^𝑙_{⌈⌉𝑘−1𝑘}𝑠_/2−1. (22) Let ℎ(𝑠)=^𝑙𝑠/^𝑘−1_{⌈𝑘/2⌉−1}, the right-hand side of (22). By Lemmas 5 and 7, we have 𝑠≤^𝑘−1_{⌈𝑘/2⌉−1}(1/(2^{2⌈𝑘/2⌉})) which leads after some algebraic manipulations to 4ℎ(𝑠)≤1. This means that the discriminant of the quadratic equation (22) Δ=1−4ℎ(𝑠) is positive, and consequently by applying the closed-form formula for the solutions of the quadratic polynomial equation (22), we find that the solutions are the following: 𝑟₁(𝑠)=1+1−4ℎ(𝑠)2,𝑟₂(𝑠)=1−1−4ℎ(𝑠)2. (23) It is obvious that 𝑟₁(𝑠)+𝑟₂(𝑠)=1. Thus, 𝑟₁(𝑠) and 𝑟₂(𝑠) are real valued functions, where 𝑟₁(𝑠) is a monotonically decreasing function with 𝑤₃≤𝑟₁(𝑠)≤1 and 𝑟₂(𝑠) is a monotonically increasing function with 0≤𝑟₂(𝑠)≤𝑤₃. By Theorem 10 and Lemma 11, the roots 𝑟₁(𝑠) and 𝑟₂(𝑠) are the appropriate roots, and consequently we set 𝑟(𝑠)=𝑟₂(𝑠) for 𝑦(𝑡)∈[0,0.5] and 𝑟(𝑠)=𝑟₁(𝑠) for 𝑦(𝑡)∈(0.5,1].
Using the chain differentiation rule, we obtain the following: 𝑦(𝑡(𝑠(𝑠)⟹(𝑡)))=𝑟𝑑𝑦𝑑𝑡_{𝑡=𝑡(𝑠)}𝑑𝑡(𝑠)=(𝑠)𝑑𝑠𝑑𝑟.𝑑𝑠 (24) However, since (1−𝑡)(𝑑𝑦(𝑡)/𝑑𝑡)=𝑠 for 𝑡=𝑡(𝑠), it follows that 𝑑𝑦(𝑡)𝑑𝑡_{𝑡=𝑡(𝑠)}=𝑠(𝑠).1−𝑡 (25) Thus, (24) becomes 𝑑𝑦(𝑡)𝑑𝑡_{𝑡=𝑡(𝑠)}𝑑𝑡(𝑠)=(𝑠)𝑑𝑠𝑑𝑟⟹𝑠𝑑𝑠(𝑠)(𝑠)1−𝑡𝑑𝑡=(𝑠)𝑑𝑠𝑑𝑟.𝑑𝑠 (26) Solving the differential equation in (26), we obtain the solution shown in (20) as required.

Corollary 17. Let 𝑎(𝑠)=1−4ℎ(𝑠). Then the solution to (18) is equal to 𝑡(𝑠)=𝑎(𝑠)1+𝑐+1𝑎(𝑠)𝑒−1^𝑝1(𝑠),𝑖𝑓𝑟(𝑠)=𝑟₁(𝑠),𝑎(𝑠)1+𝑐−1𝑎(𝑠)𝑒+1^𝑝2(𝑠),𝑖𝑓𝑟(𝑠)=𝑟₂(𝑠), (27) where 𝑝₁(𝑠) and 𝑝₂(𝑠) are given by 𝑝₁(𝑠)=^⌈𝑘/2⌉_𝑖=22^{𝑖−12⌈⌉}_⌈⌉^{𝑘/2−𝑖−1}_{𝑘/2−𝑖}(𝑖−1)_⌈⌉^𝑘−1_𝑘/2−1×(−1)^𝑖−1(𝑎(𝑠)−1)^𝑖−1−1(𝑎(𝑠)+1)^𝑖−1,𝑝₂(𝑠)=^⌈𝑘/2⌉_𝑖=22^{𝑖−12⌈⌉}_⌈⌉^{𝑘/2−𝑖−1}_{𝑘/2−𝑖}(𝑖−1)_⌈⌉^𝑘−1_𝑘/2−1×(−1)^𝑖(𝑎(𝑠)−1)^𝑖−1+1(𝑎(𝑠)+1)^𝑖−1. (28)

Proof. We compute the integral in (20) for 𝑟(𝑠)=𝑟₁(𝑠). Let 𝑙=⌈𝑘/2⌉, 𝑚=^𝑘−1_{⌈𝑘/2⌉−1}, and 𝑎(𝑠)=𝑠1−4𝑚^1/𝑙1/2,𝑟(𝑠)=12+12𝑎(𝑠). (29)(30) After some algebraic manipulations, we obtain the derivative 𝑟^′(𝑠) of 𝑟(𝑠) as follows: 𝑟^′(𝑠)=12𝑎^′(𝑠). (31) By (29), we obtain the following expression for 𝑠: (𝑠)𝑠=1−𝑎^2𝑙𝑚4^𝑙. (32) Multiplying (31) by 1/𝑠 and after some algebraic manipulations, we obtain 1𝑠𝑟^′(𝑠)=4^𝑙𝑎2𝑚^′(𝑠)1−𝑎(𝑠)^2𝑙. (33) Based on (32) and (33), the integral in (20) becomes 1𝑠𝑟^′(𝑠)=(−1)^𝑙4^𝑙𝑎2𝑚^′(𝑠)𝑎(𝑠)²−1^𝑙𝑑𝑠. (34) Setting 𝑎(𝑠)=𝑤 for convenience, we can form the rational parts expansion of the rational part of the integral in (34) as follows: 1𝑤²−1^𝑙=1(𝑤−1)^𝑙(𝑤+1)^𝑙=^𝑙_𝑖=1𝐴_𝑖(𝑤−1)^𝑖+^𝑙_𝑖=1𝐵_𝑖(𝑤+1)^𝑖 (35) with the coefficients given, by standard calculus techniques, by the following: 𝐴_𝑖=1()!𝑑𝑙−𝑖𝑑𝑤^{(𝑙−𝑖)(𝑙−𝑖)}(𝑤−1)^𝑙1(𝑤−1)^𝑙(𝑤+1)^𝑙_𝑤=1=1()!𝑑𝑙−𝑖𝑑𝑤^{(𝑙−𝑖)(𝑙−𝑖)}1(𝑤+1)^𝑙_𝑤=1,𝐵_𝑖=1()!𝑑𝑙−𝑖𝑑𝑤^{(𝑙−𝑖)(𝑙−𝑖)}(𝑤+1)^𝑙1(𝑤−1)^𝑙(𝑤+1)^𝑙_𝑤=−1=1()!𝑑𝑙−𝑖𝑑𝑤^{(𝑙−𝑖)(𝑙−𝑖)}1(𝑤−1)^𝑙_𝑤=−1 (36) for 𝑖=1,…,𝑙.
Since 𝑑^{(𝑙−𝑖)}𝑑𝑤^{(𝑙−𝑖)}1(𝑤±1)^𝑙=(−1)^𝑙−𝑖1(𝑤±1)^2𝑙−𝑖(2𝑙−𝑖−1)!()!,𝑙−1 (37) the coefficients 𝐴_𝑖 and 𝐵_𝑖 become 𝐴_𝑖=1()!()𝑙−𝑖−1^𝑙−𝑖12^2𝑙−𝑖(2𝑙−𝑖−1)!()!=()𝑙−12𝑙−𝑖−1𝑙−𝑖−1^𝑙−𝑖12^2𝑙−𝑖,𝐵_𝑖=1()!()𝑙−𝑖−1^𝑙−𝑖1(−2)^2𝑙−𝑖(2𝑙−𝑖−1)!()!=1𝑙−12𝑙−𝑖−1𝑙−𝑖(−1)^𝑙2^2𝑙−𝑖. (38) Based on (34), (35), and (38), the integral in (20) becomes 1𝑠𝑟^′(𝑠)=(−1)^𝑙4^𝑙2𝑚^𝑙_𝑖=1𝑎^′(𝑠)𝐴_𝑖(𝑎(𝑠)−1)^𝑖+^𝑙_𝑖=1𝑎^′(𝑠)𝐵_𝑖(𝑎(𝑠)+1)^𝑖=^{2𝑙−2𝑙−1}𝑚𝑎(𝑠)ln+1𝑎(𝑠)+()−1−1^𝑙4^𝑙×2𝑚_𝑙𝑖=2𝑎^′(𝑠)𝐴_𝑖(𝑎(𝑠)−1)^𝑖+^𝑙_𝑖=2𝑎^′(𝑠)𝐵_𝑖(𝑎(𝑠)+1)^𝑖=^{2𝑙−2𝑙−1}𝑚𝑎(𝑠)ln+1𝑎(𝑠)+−1_𝑙𝑖=22^{𝑖−12𝑙−𝑖−1}_𝑙−𝑖(−1)^𝑖𝑚𝑎^′(𝑠)(𝑎(𝑠)−1)^𝑖+2^{𝑖−12𝑙−𝑖−1}_𝑙−𝑖𝑚𝑎^′(𝑠)(𝑎(𝑠)+1)^𝑖. (39) Since ^2𝑙−2_𝑙−1=𝑚, after some algebraic manipulations (39) is rewritten as follows: 1𝑠𝑟^′(𝑠)𝑎(𝑠)=ln+1𝑎(𝑠)−1+𝑝₁(𝑠), (40) where 𝑝₁(𝑠)=^𝑙_𝑖=22^{𝑖−12𝑙−𝑖−1}_𝑙−𝑖()𝑚×()𝑖−1−1^𝑖−1(𝑎(𝑠)−1)^𝑖−1−1(𝑎(𝑠)+1)^𝑖−1. (41) For 𝑟(𝑠)=𝑟₂(𝑠), the derivative 𝑟^′(𝑠) of 𝑟(𝑠) becomes 𝑟^′(𝑠)1=−2𝑎^′(𝑠). (42) Applying the previous proof method for 𝑟^′(𝑠) given by (42), the integral in (20) becomes 1𝑠𝑟^′(𝑠)𝑎(𝑠)𝑑𝑠=ln−1𝑎(𝑠)+1+𝑝₄(𝑠), (43) where 𝑝₂(𝑠)=^𝑙_𝑖=22^{𝑖−12𝑙−𝑖−1}_𝑙−𝑖()𝑚×()𝑖−1−1^𝑖(𝑎(𝑠)−1)^𝑖−1+1(𝑎(𝑠)+1)^𝑖−1. (44) Thus, from (20), we obtain the following: 𝑡(𝑠)=𝑎(𝑠)1+𝑐+1𝑎(𝑠)𝑒−1^𝑝1(𝑠),if𝑟(𝑠)=𝑟₁(𝑠),𝑎(𝑠)1+𝑐−1𝑎(𝑠)𝑒+1^𝑝2(𝑠),if𝑟(𝑠)=𝑟₂(𝑠), (45) completing the proof of the theorem.

We now need to invert the function 𝑡(𝑠) or, equivalently, to solve (27), with unknown 𝑠. By setting 𝑧=1+𝑎(𝑠), (27) for 𝑟(𝑠)=𝑟₂(𝑠) after some algebraic manipulations becomes 𝑐()𝑧=𝑧−2𝑡(𝑠)𝑒−1^𝑄(𝑧), (46) where 𝑄(𝑧) is the rational function of 𝑧 given by 𝑄(𝑧)=^𝑙_𝑖=22^{𝑖−12𝑙−𝑖−1}_𝑙−𝑖()𝑚×()𝑖−1−1^𝑖𝑧^𝑖−1+(𝑧−2)^𝑖−1𝑧^𝑖−1(𝑧−2)^𝑖−1. (47) With regard to the constant 𝑐, its value may be obtained from 𝑥(0) as follows. Since 𝑦(𝑡)=𝑥(𝑡)/(1−𝑡), then 𝑦(0)=𝑥(0). Setting 𝑡=0 in 𝑔(𝑦(𝑡))=𝑠, we obtain the value for 𝑠. Thus, using this value of 𝑠, we have established the relation 𝑡(𝑠)=0, and we can compute 𝑎(𝑠) and 𝑧. Now we can solve (46) for 𝑐.

Setting 𝑓(𝑧)=ln(𝑐(2−𝑧)/(1−𝑡(𝑠)))𝑒^𝑧+𝑄(𝑧)𝑒^𝑧, (48) we rewrite (46) as follows: 𝑧=𝑒^{𝑓(𝑧)𝑒𝑧}. (49) Based on Definitions 12, 13, and 14 for 𝑗=1, we can rewrite (49) as follows: 𝑧=𝑒^{𝑓(𝑧)𝑒𝑧}⟹𝑧𝑒^{−𝑓(𝑧)𝑒𝑧}(𝑧)(𝑧).=1⟹𝐺−𝑓;𝑧=1⟹𝑧=HW−𝑓;1 (50) Let 𝑙=⌈𝑘/2⌉ and 𝑚=^𝑘−1_{⌈𝑘/2⌉−1}. Since 𝑧=1+𝑎(𝑠) by substituting 𝑎(𝑠) and after some algebraic manipulations, we obtain the following formula for 𝑠: 𝑠=𝑚𝑧^𝑙(𝑧−2)^𝑙4^𝑙=𝑚HW−𝑓(𝑧);1^𝑙HW−𝑓(𝑧);1−2^𝑙4^𝑙. (51) Equation (51) gives values of 𝑠 for specific values of 𝑡 through Hyper-Lambert functions. Specifically, for a certain value of 𝑡, the function 𝑓(𝑧) takes a particular form, and through (51), we obtain a value for 𝑠. Thus, we have found 𝑡⁻¹(𝑠). In [6], (51) has been derived for 𝑘=3.

Based on Theorem 16, Lemma 15, (50), and that 𝑟(𝑠)=𝑧/2, a solution of the differential equation (13) can be obtained by 𝑦(𝑡(𝑠(𝑠)=(𝑧)))=𝑟HW−𝑓;12. (52) Equivalently, we can obtain the solution by 𝑦(𝑡(𝑠(𝑧)))=𝑟𝑚HW−𝑓;1^𝑙HW−𝑓(𝑧);1−2^𝑙4^𝑙 (53) with 𝑟(𝑠)=𝑟₂(𝑠).

We will now demonstrate the previous methodology for 𝑘=3. Since 𝑙=⌈𝑘/2⌉=2 and 𝑚=^𝑘−1_{⌈𝑘/2⌉−1}=2, (21) becomes 𝑔𝑦(𝑡)=2𝑦(𝑡)²1−𝑦(𝑡)²=2𝑦(𝑡)⁴−4𝑦(𝑡)³+2𝑦(𝑡)². (54) From Theorem 16, the real positive solutions 𝑟₁(𝑠) and 𝑟₂(𝑠) of 𝑔(𝑦(𝑡))=𝑠 for 𝑠≤(1/8) can be written as follows: 𝑟₁(𝑠)=1+1−2√2𝑠2,𝑟₂(𝑠)=1−1−2√2𝑠2. (55) Moreover, the rational function 𝑄(𝑧) of 𝑧 involved in (46) becomes, after some algebraic manipulations, as follows: 𝑄(𝑧)=2(𝑧−1)𝑧().𝑧−2 (56) Thus, the function 𝑓(𝑧) defined by (48) becomes 𝑓(𝑧)=ln(𝑐(2−𝑧)/(1−𝑡(𝑠)))𝑒^𝑧+2(𝑧−1)𝑧(𝑧−2)𝑒^𝑧. (57) Finally, the formula for 𝑠 defined by (51) and the solution of the differential equation (54) defined by (53) can be written as follows: 𝑧𝑠=²(𝑧−2)²8=HW−𝑓(𝑧);1²HW−𝑓(𝑧);1−2²8.𝑦(𝑡(𝑠(𝑧)))=𝑟HW−𝑓;1²HW−𝑓(𝑧);1−2²8, (58) where 𝑟(⋅)=𝑟₂(𝑠).

4. Nonclosed-Form Solvability Results for Even Values of 𝑘

In this section, we consider the case where the parameter 𝑘 is an even integer greater than 2, that is, 𝑘=2𝑚, with 𝑚≥2. This case is of distinct interest because it does not seem to be amenable to the methodology we have followed for 𝑘 odd. In Section 3 where we considered odd values of 𝑘, a key step in linking the solution of the differential equation with the class of Hyper-Lambert functions was to find the closed-form solution 𝑦(𝑡), given 𝑠, of the polynomial equation 𝑔(𝑦(𝑡))=𝑠, with 𝑔(𝑦(𝑡)) given by (21). This closed-form solution was needed in Lemma 15 and Theorem 16. When we attempted to apply the same methodology for even values of 𝑘≥4, we were not able to find a closed-form solution of the equation 𝑔(𝑦(𝑡))=𝑠, even for fixed value of 𝑘 (e.g., 𝑘=4). Thus, we focused our attention on investigating the possibility that the equation does not have a closed-form solution. As it turned out, this was indeed the case, at least for values of 𝑘 such that 𝑘+1 is prime and certain ranges of 𝑠. We believe that this result may be of independent interest.

Before proving it, we will state three fundamental theorems of Algebra.

Theorem 18 (Eisenstein’s irreducibility criterion). Let 𝑞 be a prime number, and let 𝑓(𝑥)=𝑎_𝑛𝑥^𝑛+⋯+𝑎₁𝑥¹+𝑎₀ be a polynomial of degree 𝑛 with integer coefficients satisfying the following properties: (i)𝑞does not divide 𝑎_𝑛; (ii)𝑞 divides 𝑎₀,𝑎₁,…,𝑎_𝑛−1; (iii)𝑞² does not divide 𝑎₀. Then the polynomial 𝑓 is irreducible over the field of ℚ of rational numbers.

Theorem 19. Let 𝐾 be a subfield of ℝ, and let 𝑓∈𝐾[𝑋] be an irreducible polynomial of prime degree 𝑛>2. If the Galois group of 𝑓 over 𝐾 is solvable, 𝑓 has either exactly one root in ℝ or all its roots in ℝ.

Theorem 20. Let 𝑓 be a polynomial with coefficients in a field 𝐾 of characteristic zero. Suppose that 𝑓 is solvable by radicals. Then the Galois group of Γ_𝐾(𝑓) of 𝑓 is a solvable group.

We now state and prove the main result, concerning the unsolvability, in closed form, of the polynomial equation 𝑔(𝑦(𝑡))=𝑠.

Theorem 21. The equation 𝑔(𝑦(𝑡))=𝑠 is not solvable by radicals over ℚ for 𝑘=2𝑚 with 𝑚≥2, 𝑛=𝑘+1 a prime, and 𝑠=𝑙/𝑏 a rational number, given in its lower terms (i.e., gcd(𝑙,𝑏)=1), such that 𝑠∈(0,𝑔(𝑤₃)), 𝑙𝑏≠0, and 𝑏 is a square free integer.

Proof. Let 𝑠=𝑙/𝑏∈ℚ, with 𝑙𝑏≠0. Then, our target polynomial is 𝑝(𝑦(𝑡))=𝑔(𝑦(𝑡))−(𝑙/𝑏). We will, first, show that 𝑝(𝑦(𝑡)) is irreducible over the set of rational numbers ℚ using Theorem 18. We will work, in the following, with the polynomial 𝑏𝑝(𝑦(𝑡)) in order to have a polynomial with integer coefficients, as required by the statement of Theorem 18. The degree of this polynomial according to Theorem 8 is 𝑛=𝑘+1, which has been assumed to be a prime.
We will show, however, that Theorem 18 is not, directly, applicable to 𝑏𝑝(𝑦(𝑡)) since the first two conditions of the theorem are not met. The coefficients 𝑎_𝑖, 𝑖=0,…,𝑛 in Theorem 18 are as follows: 𝑎_𝑖𝑚()=𝑏()2𝑚−1𝑚−1𝑖−𝑚+1−1^{𝑖−(𝑚+1)},𝑎𝑖=𝑚+1,…,𝑛=2𝑚+1,_𝑖𝑎=0,𝑖=1,…,𝑚,₀=−𝑙. (59) Since |𝑎_𝑛|=𝑎_𝑚+1=𝑏^2𝑚−1_𝑚−1, the first two conditions of Theorem 18 do not hold. Since 0 is not a root of 𝑝(𝑦(𝑡)), we can set 𝑦(𝑡)→1/𝑥, in order to consider the polynomial 𝑓(𝑥)=𝑏𝑝(1/𝑥), instead of 𝑏𝑝(𝑦(𝑡)). Then each nonzero root 𝑥₀ of the polynomial 𝑓(𝑥) corresponds to a nonzero root 𝑦₀ of 𝑝(𝑦(𝑡)) and vice versa by the correspondence 𝑦₀↔1/𝑥₀. Consequently, 𝑝(𝑦(𝑡)) is solved by radicals if and only if 𝑓(𝑥) is solved by radicals.
Thus, from now on we will consider the polynomial 𝑓(𝑥)=𝑏𝑝(1/𝑥). In order to have the inverse of 1/𝑥, we will use the polynomial ℎ(𝑥)=𝑥^2𝑚+1𝑓(𝑥). We will apply Theorem 18 to this polynomial because if it applies to it, then 𝑓(𝑥) will be irreducible, as well as, in turn, 𝑝(𝑦(𝑡)).
After some algebraic manipulations, the polynomial ℎ(𝑥)=𝑥^2𝑚+1𝑓(𝑥) can be written as follows: ℎ(𝑥)=𝑏2𝑚−1𝑚−1^𝑚𝑖=0𝑚𝑖𝑥^𝑚−𝑖(−1)^𝑖−𝑙𝑥^2𝑚+1. (60) Now the coefficients 𝑎_𝑖, 𝑖=0,…,𝑛=2𝑚+1 of the polynomial ℎ(𝑥) in (60) are as follows: 𝑎_2𝑚+1𝑎=−𝑙,_𝑗𝑎=0,𝑗=𝑚+1,…,2𝑚,_𝑗𝑚=𝑏()2𝑚−1𝑚−1𝑚−𝑗−1^𝑚−𝑗,𝑗=0,…,𝑚. (61) In this way, we have decoupled the coefficient 𝑎_𝑛=𝑎_2𝑚+1 from all the rest so as to be feasible to satisfy the first two conditions of Theorem 18. Since, by assumption, gcd(𝑙,𝑏)=1, in order to satisfy the first two conditions, it suffices to choose as 𝑞 any prime divisor of 𝑏. Since 𝑏 is, by assumption, a square free integer, then the third condition is satisfied too. Thus, Theorem 18 applies to ℎ(𝑥) in (60) and, thus, to 𝑝(𝑦(𝑡)). Therefore, 𝑝(𝑦(𝑡)) is irreducible over ℚ.
We will, now, prove that the number of real roots of ℎ(𝑥) is neither equal to 1 nor its degree, which is 𝑛=𝑘+1. From Theorem 10, since we have assumed that 𝑠∈(0,𝑔(𝑤₃)), it follows that 𝑔(𝑦(𝑡)) has at least two roots and, thus, ℎ(𝑥) has also at least two roots because of the relationship between their roots. We will prove that not all of its roots are real. Let us assume, towards a contradiction, that ℎ(𝑥) has 2𝑚+1 real roots. We can differentiate ℎ(𝑥)𝑚 times to obtain its 𝑚th order derivative, which is equal to 𝑙((2𝑚+1)!/𝑚!)𝑥^𝑚+1+𝑚!𝑏^2𝑚−1_𝑚−1. If ℎ(𝑥) had 2𝑚+1 real roots, then its 𝑚th order derivative would have, at least, 2𝑚+1−𝑚=𝑚+1≥3 (since 𝑚≥2, by assumption) real roots.
The roots of 𝑙((2𝑚+1)!/𝑚!)𝑥^𝑚+1+𝑚!𝑏^2𝑚−1_𝑚−1 are the (𝑚+1)th roots of −𝐵/𝐴, with 𝐴=𝑙((2𝑚+1)!/𝑚!) and 𝐵=𝑚!𝑏^2𝑚−1_𝑚−1. These roots have the form |−𝐵/𝐴|^1/(𝑚+1)(cos(±𝜋+2𝑑𝜋)/(𝑚+1)+𝑖sin(±𝜋+2𝑑𝜋)/(𝑚+1)) with 𝑑=0,…,𝑚 (±𝜋 is 𝜋 if −𝐵/𝐴 is positive and −𝜋 if −𝐵/𝐴 is negative).
A root is real if and only if the imaginary part is 0; that is, (±𝜋+2𝑑𝜋)/(𝑚+1)=𝜋((2𝑑±1)/(𝑚+1)) is an integer multiple 𝜆 of 𝜋. Since 𝑑≤𝑚, we only need to consider 𝜆=1. Solving (2𝑑±1)/(𝑚+1)=1, we obtain 𝑑=(𝑚+1∓1)/2. It is easy to see that this gives us at most two values for 𝑑: (i) two values for even 𝑚: 𝑚/2 (for the “−” sign in (2𝑑±1)/(𝑚+1)) and (𝑚+2)/2 (for the “+” sign in (2𝑑±1)/(𝑚+1)) and (ii) no such value for odd 𝑚. Thus, the 𝑚th order derivative of ℎ(𝑥) has at most 2 real roots for even 𝑚 or no real roots for odd 𝑚, while according to the assumption that ℎ(𝑥) has 2𝑚+1 real roots, it should have at least 𝑚+1≥3 real roots, which is a contradiction.
Therefore, either there are no real roots of ℎ(𝑥), or the number of real roots of ℎ(𝑥) is more than 1 but less than its degree 𝑛=2𝑚+1. Thus, from Theorems 19 and 20, it follows that ℎ(𝑥) is not solvable by radicals over ℚ. Therefore, 𝑝(𝑦(𝑡)) is not solvable by radicals over ℚ too.

5. Conclusions and Future Work

In this paper we studied the properties of the solution of the differential equation ()(𝑡)1−𝑡𝑑𝑦=𝑑𝑡𝑘𝑘−12𝑦(𝑡)−1^{𝑘−⌈𝑘/2⌉+1}1−𝑦(𝑡)^⌈𝑘/2⌉. (62) This differential equation arose out of the theoretical analysis of a security-related bit-agreement protocol (see [1]) and does not appear to have been documented elsewhere. It can be seen as a subclass of a natural generalization of the well-known class of Abel differential equations of the first kind.

In this paper we showed that, for odd 𝑘, the solution of this equation can be expressed in closed form through the roots of the polynomial on its right-hand side. Moreover, the resulting expression can be transformed into a closed-form expression involving the generalized Hyper-Lambert functions that were proposed in [3] as a generalization of the well-known Lambert 𝑊 function.

For even values of 𝑘 it was not possible to follow the same approach since, as we showed, no closed-form expression exists for the roots of the polynomial, at least when 𝑘+1 is prime. One possible direction for future research is to extend the proof to cover all value of 𝑘 or show that closed-form solutions exist for nonprime 𝑘.

Another, more interesting direction would be to see whether the more general class given in (2) can be attacked using the strategy proposed in this paper, that is, based on deriving closed-form solutions of the polynomial on the right-hand side and linking them with the solution of the differential equation, at least for certain polynomials 𝑓_𝑖. There seems to be very scarce research work on the properties of (2), and any result towards the determination of its solution or, at least, the properties of its properties would be of high importance.