Abstract

This paper presents a simple and efficient method for determining the rational solution of Riccati differential equation with coefficients rational. In case the differential Galois group of the differential equation (𝐸𝑙)∶𝑦=𝑟𝑦, 𝑟∈ℂ(𝑥) is reducible, we look for the rational solutions of Riccati differential equation 𝜃+𝜃2=𝑟, by reducing the number of checks to be made and by accelerating the search for the partial fraction decomposition of the solution reserved for the poles of 𝜃 which are false poles of 𝑟. This partial fraction decomposition of solution can be used to code 𝑟. The examples demonstrate the effectiveness of the method.

1. Introduction

The quadratic Riccati differential equation:𝐸𝑅∶𝜎=𝑝2𝜎2+𝑝1𝜎+𝑝0,(1.1) where 𝑝0,𝑝1, and 𝑝2 are in a differential field 𝕂,𝑝2≠0. The quadratic Riccati differential equation is first converted to a reduced Riccati differential equation:𝐸𝑟∶𝜃+𝜃2=𝑟,(1.2) where 𝜃=−𝑝2𝜎−(1/2)𝑎, with 𝑎=(𝑝2/𝑝2)+𝑝1 and 𝑟=(1/4)𝑎2−(1/2)𝑎−𝑝2𝑝0.

Furthermore, we put 𝑦/𝑦=𝜃, reduced Riccati differential equation (1.2) is converted to a second-order linear ordinary differential equation𝐸𝑙∶𝑦=𝑟𝑦.(1.3) If we have a particular solution nonzero of (𝐸𝑙) then general solution is 𝑦=𝑐𝑢 where 𝑐=𝜆/𝑢2, 𝜆 constant (see [1–5]).

In the paper, we base ourselves mainly on the work of Kovacic [2] where the differential Galois group of the differential equation (𝐸𝑙) is reducible and we take 𝕂=ℂ(𝑋).

The case where every solution of (𝐸𝑙) is Liouvillian corresponds to the case where reduced Riccati differential equation (𝐸𝑟) has algebraic solution over 𝕂. The case where differential Galois group is reducible corresponds to the case where the Riccati differential equation (𝐸𝑟) has the rational solution (𝑢/𝑢)𝑢 solution of (𝐸𝑙). The solution 𝑢 of (𝐸𝑙) is rational fraction if and only if 𝑢/𝑢 is rational fraction with simples poles, integers residues and negative degree.

The field ℂ(𝑋)[𝑢] is differential extension of ℂ(𝑋) by exponential of an integral and if 𝑣=1/𝑢2 then (𝑢,𝑣) two solutions of (𝐸𝑙) are linearly independent over field of constants ℂ. The ordinary extension ℂ(𝑋)[𝑢,𝑣] is differential extension of ℂ(𝑋)[𝑢], by an integral. ℂ(𝑋)[𝑢,𝑣] is Picard-Vessiot extension of ℂ(𝑋)[𝑢] for the differential equation (𝐸𝑙) (see [2, 3, 6–8]). The existence of rational solution 𝑢/𝑢 of Riccati differential equation (𝐸𝑟) and research primitive of 1/𝑢2give all solutions of (𝐸𝑟).

This paper presents a simple and efficient method for determining the solution of Riccati differential equation with coefficients rational. In case the differential Galois group of the differential equation (𝐸𝑙)∶𝑦=𝑟𝑦, 𝑟∈ℂ(𝑥) is reducible, we look for the rational solutions of Riccati differential equation 𝜃+𝜃2=𝑟, by reducing the number of checks to be made and by accelerating the search for the partial fraction decomposition of the solution reserved for the poles of 𝜃 which are false poles of 𝑟. This partial fraction decomposition of solution can be used to code 𝑟. The examples demonstrate the effectiveness of the method.

2. Form of Rational Solution of Equation: (𝐸𝑟)

Let 𝑟∈ℂ(𝑥),𝑟≠0 be rational fraction and let 𝜃∈ℂ(𝑥) be the rational solution of Riccati differential equation (𝐸𝑟)∶𝜃+𝜃2=𝑟.

2.1. Study in the Pole 𝑐 of Multiplicity 𝜈 of 𝜃

We put 𝜏𝜃=(𝑥−𝑐)𝜈,where𝜏(𝑐)≠0.(2.1) We have𝑟=𝜃+𝜃2=(𝑥−𝑐)−2𝜈𝜈𝜏−2(𝑥−𝑐)𝜈−12−𝜈24(𝑥−𝑐)2𝜈−2+𝜏(𝑥−𝑐)𝜈.(2.2) Thus 𝜈𝜏−2(𝑥−𝑐)𝜈−12=(𝑥−𝑐)2𝜈𝑟+(𝑥−𝑐)𝜈𝜈24(𝑥−𝑐)𝜈−2−𝜏.(2.3)

Case 1 (𝜈≥2). The function (𝑥−𝑐)2𝜈𝑟 defines and is equal to 𝜏(𝑐)2 at 𝑐.
Thus 𝑐 is pole of multiplicity 2𝜈 of 𝑟 where lim𝑥→𝑐(𝑥−𝑐)2𝜈𝑟=lim𝑥→𝑐(𝑥−𝑐)𝜈𝜃2.(2.4)

Case 2 (𝜈=1). We have 1𝜏−22=(𝑥−𝑐)21𝑟+4−(𝑥−𝑐)𝜏.(2.5)
The function (𝑥−𝑐)2𝑟 defines and is equal to 𝜏(𝑐)(𝜏(𝑐)−1) at 𝑐.

Situation 1 (lim𝑥→𝑐(𝑥−𝑐)2𝑟≠0,−1/4)
c is double pole of 𝑟 and the residue 𝜏(𝑐) of 𝜃 at 𝑐 have tow possibility values as follows: 1𝜏(𝑐)−22=lim𝑥→𝑐(𝑥−𝑐)21𝑟+4.(2.6) Thus, 𝑐 is double pole of 𝑟 and the residue of 𝜃 at simple pole 𝑐 equals 𝜏(𝑐)=𝛼𝑐+12,(2.7) where 𝛼2𝑐=lim𝑥→𝑐(𝑥−𝑐)21𝑟+4.(2.8)

Situation 2 (lim𝑥→𝑐(𝑥−𝑐)2𝑟=−1/4)
𝑐 is double pole of 𝑟 and the residue of 𝜃 at 𝑐 is 1/2.

Situation 3 (lim𝑥→𝑐(𝑥−𝑐)2𝑟=0)
𝑐 is a simple pole of 𝑟 or is not pole of 𝑟 and the residue of 𝜃 at simple pole 𝑐 equals 1.

Proposition 2.1. Let 𝜃∈ℂ(𝑥) such as 𝜃′+𝜃2=𝑟.
(1)The fraction: 𝑟=𝑁/𝐷 with 𝑁 and 𝐷 being polynomials relatively prime 𝐷=𝐷1𝐷22𝐷23𝐷24,(2.9) where 𝐷1,𝐷2,𝐷3, and 𝐷4 are polynomials relatively prime pairwise. 𝐷1,𝐷2, and 𝐷3 are with simple roots, and 𝐷4 is without simple root 𝐷∀𝑐∈Root2lim𝑥→𝑐(𝑥−𝑐)21𝑟=−4,𝐷∀𝑐∈Root3lim𝑥→𝑐(𝑥−𝑐)21𝑟≠−4.(2.10)(2)(a)Let 𝜈≥2. 𝑐 pole of multiplicity 𝜈 of 𝜃⇔𝑐∈Root(𝐷4).(b)𝑐 simple pole of 𝜃 with residue ≠1,1/2⇔𝑐∈Root(𝐷3), the residue of 𝜃 where 𝑐 equals 𝛼𝑐+(1/2) where 𝛼2𝑐=lim𝑥→𝑐(𝑥−𝑐)21𝑟+4.(2.11)(c)𝑐 simple pole of 𝜃 with residue =1/2⇔𝑐∈Roots(𝐷2).(d)𝑐 simple pole of 𝜃 with residue =1⇔𝑐∈Roots(𝐷1) or 𝑐 pole of 𝜃 and false pole of 𝑟.

Corollary 2.2. One assumes that 𝑟=𝑁/𝐷 with 𝑁 and 𝐷 polynomials being relatively prime, 𝐷=𝐷1𝐷22𝐷23𝐷24 where 𝐷1,𝐷2,𝐷3, and 𝐷4 are polynomials relatively prime pairwise, 𝐷1,𝐷2, and 𝐷3 are with simple roots, and 𝐷4 is without simple root 𝐷∀𝑐∈Root2lim𝑥→𝑐(𝑥−𝑐)21𝑟=−4,𝐷∀𝑐∈Root3lim𝑥→𝑐(𝑥−𝑐)21𝑟≠−4.(2.12) a rational fraction 𝜃. Verify 𝜃+𝜃2=𝑟 is the shape 𝜃=𝐸(𝜃)+𝑐∈Roots(𝐷4)𝜃𝑐+𝑐∈Roots(𝐷3)𝛼𝑐+1/2+1𝑥−𝑐2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0(2.13) with 𝐷0 being monic polynomial with simple roots and the roots are false poles of 𝑟.

2.2. Study in the Infinity

Let 𝑡∈ℂ(𝑥),𝑡≠0 rational fraction such as (𝑑𝑡/𝑑𝑥)+𝑡2=0.

Case 1. We assume that 𝑑𝑜(𝜃)<0.
We have 𝑑𝑜(𝜃)<0, and 𝑑𝑜(𝜃2)<0 thus 𝑑𝑜(𝑟)<0. We put 𝜃=𝑡𝜎(𝑡)with𝜎beingrationalfractiondefinedat0.(2.14) We have 𝜎(0)=lim𝑥→∞𝑥𝜃=sumtheresiduesof𝜃,𝑟=𝜃+𝜃2=𝜎2𝑡−𝜎2−𝜎𝑡3,lim𝑥→∞𝑥2𝑟=𝜎(0)2−𝜎(0).(2.15) Thus 𝑑𝑜𝑟≤−2 and (𝜎(0)−(1/2))2=lim𝑥→∞𝑥2𝑟+(1/4).
If: lim𝑥→∞𝑥2𝑟=−1/4 then the sum of residues of 𝜃: 𝜎(0)=1/2.
If: lim𝑥→∞𝑥2𝑟≠−1/4 then the sum of residues of 𝜃: 𝜎(0)=𝛼∞+(1/2) where 𝛼2∞=lim𝑥→∞𝑥21𝑟+4.(2.16)

Case 2. We assume that 𝑑𝑜(𝜃)=0.
𝐸(𝜃) constant ≠0. We put 𝜃=𝐸(𝜃)+𝑡𝜎(𝑡),with𝜎beingrationalfractiondefinedat0.(2.17) We have 𝜎(0)=lim𝑥→∞𝑥(𝜃−𝐸(𝜃))=thesumofresiduesof𝜃,𝑟=𝜃+𝜃2=(𝐸(𝜃))2𝜎+2𝐸(𝜃)𝑡𝜎+2𝑡−𝜎2−𝜎𝑡3.(2.18) So 𝐸(𝑟) constant equals 𝐸(𝜃)2: []2𝐸(𝜃)sumtheresiduesof𝜃=sumtheresiduesof𝑟.(2.19)

Case 3. We assume that 𝑑𝑜(𝜃)>0.
We put 𝜈=𝑑𝑜(𝜃)≥1,𝜃=𝑡−𝜈𝜎(𝑡)(2.20) with 𝜎 being rational fraction defined at 0. The scalar 𝜎(0) is the dominant coefficient of 𝐸(𝜃). We have 𝑟=𝜃+𝜃2=𝑡−2𝜈𝜎2+𝜈𝑡𝜈+1𝜎−𝑡𝜈+2𝜎.(2.21) Thus, 𝑡2𝜈𝑟=𝜎(0)2+𝑜(𝑡).(2.22) So 𝑑𝑜𝑟=2𝜈=2𝑑𝑜𝜃 and 𝜎2(𝑜) the dominant coefficient of 𝐸(𝑟).

Proposition 2.3. Let 𝜃∈ℂ(𝑥) such as 𝜃+𝜃2=𝑟.
(1)𝑑𝑜(𝜃)<0⇔𝑑𝑜(𝑟)<0. Thus 𝑑𝑜(𝑟)≤−2 and ⎧⎪⎨⎪⎩1thesumofresiduesof𝜃=2iflim𝑥→∞𝑥21𝑟=−4,𝛼∞+12iflim𝑥→∞𝑥21𝑟≠−4.(2.23)(2)𝑑𝑜(𝜃)=0⇔𝑑𝑜(𝑟)=0. In the case𝐸(𝜃) is square root of 𝐸(𝑟): 2𝐸(𝜃)(sumofresiduesof𝜃)=sumofresiduesof𝑟=lim𝑥→∞𝑥(𝑟−𝐸(𝑟)).(2.24)(3)We have 𝑑𝑜(𝜃)>0⇔𝑑𝑜(𝑟)>0. In the case(a)𝑑𝑜(𝑟)=2𝑑𝑜(𝜃); (b)the dominant coefficient of 𝐸(𝜃) is square root of 𝐸(𝑟).

2.3. Determination of 𝐸(𝜃); 𝑑𝑜(𝑟)=2𝜈>0

We assume that 𝑟 is a rational fraction of degree 2𝜈>0 and 𝜃∈ℂ(𝑥) such as 𝜃+𝜃2=𝑟.

Let 𝑎 be the dominant coefficient of 𝐸(𝑟). Thus 𝑟∼𝑎𝑥2𝜈 if 𝑥 tends to ∞: 𝑡2𝜈𝐸(𝑟)𝑎=1+𝑎1𝑡+⋯+𝑎2𝜈𝑡2𝜈.(2.25) The Taylor's expansion of order 𝜈+1 at 0 𝑡2𝜈𝐸(𝑟)𝑎1/2=1+𝑠1𝑡+⋯+𝑠𝜈+1𝑡𝜈+1𝑡+𝑜𝜈+1.(2.26) We have 𝑡2𝜈𝑟=𝑡2𝜈𝑡𝐸(𝑟)+𝑜2𝜈=𝑡2𝜈𝐸𝑡(𝑟)+𝑜𝜈+1𝑡=𝑎2𝜈𝐸(𝑟)𝑎1/22𝑡+𝑜𝜈+1.(2.27) We have 𝜃=𝑡−𝜈𝜎(𝑡) with 𝜎 rational fraction defined at 0, 𝜎(0)2=𝑎 and 𝜈𝜎+2𝑡𝜈+12=𝑡2𝜈𝜈𝑟+24𝑡2𝜈+2+𝑡𝜈+2𝜎=𝑎1+𝑠1𝑡+⋯+𝑠𝜈+1𝑡𝜈+12𝑡+𝑜𝜈+1𝜈𝜎+2𝑡𝜈+1=𝜎(0)1+𝑠1𝑡+⋯+𝑠𝜈+1𝑡𝜈+1𝑡+𝑜𝜈+1.(2.28) Thus𝜃=𝑡−𝜈𝑡𝜎=𝜎(0)−𝜈+𝑠1𝑡−(𝜈−1)+⋯+𝑠𝜈+1𝑡−𝜈2𝑡+𝑜(𝑡).(2.29) Imply𝑡𝐸(𝜃)=𝜎(0)−𝜈+𝑠1𝑡−(𝜈−1)+⋯+𝑠𝜈,𝜎(0)𝑠𝜈+1−𝜈2=sumofresiduesof𝜃.(2.30)

Proposition 2.4. Let 𝑟 be a rational fraction of degree 2𝜈>0, 𝜃∈ℂ(𝑥) such as 𝜃+𝜃2=𝑟 and 𝑎 the dominant coefficient of 𝐸(𝑟). If 𝑡2𝜈𝐸(𝑟)𝑎1/2=1+𝑠1𝑡+⋯+𝑠𝜈+1𝑡𝜈+1𝑡+𝑜𝜈+1.(2.31) Then 𝑡𝐸(𝜃)=𝛼−𝜈+𝑠1𝑡−(𝜈−1)+⋯+𝑠𝜈,𝛼𝑠𝜈+1−𝜈2=sumofresiduesof𝜃,(2.32) where 𝛼2=𝑎.(2.33)

3. Determination of Partial Fraction Decomposition

Let 𝑟=𝑁/𝐷 be rational fraction with 𝑁 and 𝐷 polynomials being relatively prime, 𝐷=𝐷1𝐷22𝐷23𝐷24 where 𝐷1,𝐷2,𝐷3, and 𝐷4 polynomials being relatively prime pair-wise, 𝐷1,𝐷2, and 𝐷3 with simple roots and 𝐷4 without simple root𝐷∀𝑐∈Root2lim𝑥→𝑐(𝑥−𝑐)21𝑟=−4,𝐷∀𝑐∈Root3lim𝑥→𝑐(𝑥−𝑐)21𝑟≠−4.(3.1) Let 𝜃∈ℂ(𝑥) be rational fraction. Verify 𝜃+𝜃2=𝑟.

3.1. Case 𝑑𝑜𝐷3=0 and 𝑑𝑜𝐷4=0

We have 𝑟=𝑁/𝐷1𝐷22.

This case corresponds to the fact that one pole 𝑐 of 𝑟 is simple or double withlim𝑥→𝑐(𝑥−𝑐)21𝑟=−4.(3.2)

Proposition 3.1. One assumes 𝑑𝑜𝑟<0 and 𝑑𝑜𝐷3=𝑑𝑜𝐷4=0. One has: lim𝑥→∞𝑥21𝑟+4=𝑞22(3.3) with 𝑞 being positive integer of parity against that of 𝑑𝑜𝐷21𝜃=2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0(3.4) with 𝐷0 polynomial of degree 𝑑𝑜𝐷0=12𝑞+1−𝑑𝑜𝐷2−𝑑𝑜𝐷1.(3.5)

Proof. We have 1𝜃=2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0.(3.6) The sum of residues of 𝜃 equal (1/2)𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷0=𝛼∞+(1/2) with 𝛼2∞=lim𝑥→∞𝑥2𝑟+(1/4). In particular 𝛼∞=𝑞/2 with 𝑞∈ℕ.
Remark 3.2. If lim𝑥→∞𝑥2𝑟=−1/4 then 𝑑𝑜𝐷2=1, 𝑑𝑜𝐷1=𝑑𝑜𝐷0=0, 𝜃=(1/2)(1/(𝑥−𝑐)), and 𝑟=−1/4(𝑥−𝑐)2.

Proposition 3.3. One assumes 𝑑𝑜𝑟=0 and 𝑑𝑜𝐷3=𝑑𝑜𝐷4=0.
(1)𝐸(𝜃) square root of 𝐸(𝑟) such as 𝑝=(1/𝐸(𝜃))lim𝑥→∞𝑥[𝑟−𝐸(𝑟)] is positive integer of same parity as 𝑑𝑜𝐷2.(2)One has 1𝜃=𝐸(𝜃)+2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0(3.7) with 𝐷0 being polynomial of degree 𝑑𝑜𝐷0=12𝑝−𝑑o𝐷1−12𝑑𝑜𝐷2.(3.8)

Proof. We have 1𝜃=𝐸(𝜃)+2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0.(3.9) We have 2𝐸(𝜃)sumofresidues(𝜃)=sumofresidues(𝑟)=lim𝑥→∞𝑥[],1𝑟−𝐸(𝑟)2𝐸(𝜃)2𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷0=lim𝑥→∞𝑥[].𝑟−𝐸(𝑟)(3.10) Thus 𝑑𝑜𝐷2+2𝑑𝑜𝐷1+2𝑑𝑜𝐷0=1𝐸(𝜃)lim𝑥→∞𝑥[].𝑟−𝐸(𝑟)(3.11) If 𝑟 is constant then 𝐷1=𝐷2=𝐷0=1 and 𝜃 are constant.
If 𝑟 is nonconstant then 𝑟 is not polynomial thus 1𝑝=𝐸(𝜃)lim𝑥→∞𝑥[]𝑟−𝐸(𝑟)(3.12) is a positive integer of same parity as 𝑑𝑜𝐷2.

Proposition 3.4. We assume 𝑑𝑜𝑟=2𝜈>0 and 𝑑𝑜𝐷3=𝑑𝑜𝐷4=0.
Let 𝑎 be the dominant coefficient of 𝐸(𝑟) and one considers the Taylor's expansion at infinity: 𝑡2𝜈𝐸(𝑟)𝑎1/2=1+𝑠1𝑡+⋯+𝑠𝜈+1𝑡𝜈+1𝑡+𝑜𝜈+1.(3.13)(1)4𝑎𝑠2𝜈+1=𝑝2,(3.14) where 𝑝 is a positive integer, 𝑝≥𝑑𝑜𝐷2+2𝑑𝑜𝐷1+𝜈, and same parity of 𝑑𝑜𝐷2+2𝑑𝑜𝐷1+𝜈.(2)One has 1𝜃=𝐸(𝜃)+2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0(3.15) with 𝐷0 being polynomial of degree 𝑑𝑜𝐷0=𝑝−𝜈−𝑑𝑜𝐷22−𝑑𝑜𝐷1(3.16) and 𝛼 the dominant coefficient of 𝐸(𝜃) where 𝑝=2𝛼𝑠𝜈+1.(3.17)

Proof. We have 𝑡𝐸(𝜃)=𝛼−𝜈+𝑠1𝑡−(𝜈−1)+⋯+𝑠𝜈,𝛼𝑠𝜈+1−𝜈21=sumofresiduesof𝜃,𝜃=𝐸(𝜃)+2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0.(3.18) Thus 12𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝜈2(3.19)𝛼𝑠𝜈+1=𝑝/2 with 𝑝 positive integer. Thus 𝑝≥𝑑𝑜𝐷2+2𝑑𝑜𝐷1+𝜈, 𝑝 and 𝑑𝑜𝐷2+2𝑑𝑜𝐷1+𝜈 have the same parity.

3.2. Case: 𝐷3=𝑋−𝑐 and 𝑑𝑜𝐷4=0

This case corresponds to the fact that a pole of 𝑟 is simple or double with only double pole 𝑐 such as:lim𝑥→𝑐(𝑥−𝑐)21𝑟≠−4,𝑁(3.20)𝑟=𝐷1𝐷22(𝑥−𝑐)2.(3.21)

Proposition 3.5. Consider (3.21) and let 𝜃∈ℂ(𝑥) be a rational fraction. Verify 𝜃+𝜃2=𝑟. One assumes 𝑑𝑜𝑟<0. Accordingly, in view of (2.11) and (2.16) one has 𝛼𝜃=𝑐+(1/2)+1𝑥−𝑐2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0(3.22) with 𝐷0 being polynomial of degree 𝑑𝑜𝐷01=𝜆−2𝑑𝑜𝐷2−𝑑𝑜𝐷1,(3.23) where 𝜆=𝛼∞−𝛼𝑐(3.24) one half positive integer of same parity as 𝑑𝑜𝐷2.

Proof. We have lim𝑥→∞𝑥𝜃=𝛼𝑐+12+12𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷0,𝛼𝑐+12𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷02=lim𝑥→∞𝑥21𝑟+4=𝛼2∞.(3.25) Thus 12𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷0=𝛼∞−𝛼𝑐.(3.26)

Proposition 3.6. Consider (3.21) and let 𝜃∈ℂ(𝑥) be a rational fraction. Verify: 𝜃′+𝜃2=𝑟. One assumes 𝑑𝑜𝑟=0. One has: 𝛼𝜃=𝐸(𝜃)+𝑐+(1/2)+1𝑥−𝑐2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0(3.27) with 𝐸2(𝜃)=𝐸(𝑟) and 𝐷0 being polynomial of degree 𝑑𝑜𝐷0=121𝜆−2𝑑𝑜𝐷2−𝑑𝑜𝐷1−12,(3.28) where 1𝜆=𝐸(𝜃)lim𝑥→∞𝑥[]𝑟−𝐸(𝑟)−2𝛼𝑐(3.29)𝜆 is positive integer of parity against that of 𝑑𝑜𝐷2.

Proof. We have 𝛼2𝐸(𝜃)𝑐+12+12𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷0=lim𝑥→∞𝑥[].𝑟−𝐸(𝑟)(3.30) Thus 1+𝑑𝑜𝐷2+2𝑑𝑜𝐷1+2𝑑𝑜𝐷0=1𝐸(𝜃)lim𝑥→∞𝑥[]𝑟−𝐸(𝑟)−2𝛼𝑐.(3.31)

Proposition 3.7. Consider (3.21) and let 𝜃∈ℂ(𝑥) be a rational fraction. Verify 𝜃+𝜃2=𝑟. One assumes 𝑑𝑜𝑟=2𝜈>0. Let 𝑎 be the dominant coefficient of 𝐸(𝑟), let 𝛼 be the dominant coefficient of 𝐸(𝜃), 𝑡=1/𝑥 and Taylors expansion: 𝑡2𝜈𝐸(𝑟)𝑎1/2=1+𝑠1𝑡+⋯+𝑠𝜈+1𝑡𝜈+1𝑡+𝑜𝜈+1.(3.32) One has 𝛼𝜃=𝐸(𝜃)+𝑐+1/2+1𝑥−𝑐2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0,(3.33) with 𝐷0 being polynomial of degree: 𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝛼𝑐−1+𝜈+𝑑𝑜𝐷22−𝑑𝑜𝐷1,𝛼2=𝑎.(3.34)

Proof. We have 𝑡𝐸(𝜃)=𝛼−𝜈+𝑠1𝑡−(𝜈−1)+⋯+𝑠𝜈,𝛼𝑠𝜈+1−𝜈2=sumofresiduesof𝜃.(3.35) Thus 𝛼𝜃=𝐸(𝜃)+𝑐+(1/2)+1𝑥−𝑐2𝐷2𝐷2+𝐷1𝐷1+𝐷0𝐷0𝛼𝑠𝜈+1−𝜈2=𝛼𝑐+12+12𝑑𝑜𝐷2+𝑑𝑜𝐷1+𝑑𝑜𝐷0.(3.36) Thus 𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝛼𝑐−1+𝜈+𝑑𝑜𝐷22−𝑑𝑜𝐷1.(3.37)

3.3. Case: (𝑑𝑜𝐷3≠0 and 𝑑𝑜𝐷4≠0) or (𝑑𝑜𝐷3≥2 and 𝑑𝑜𝐷4=0)

𝐷1𝐷22 and 𝐷3𝐷4 are polynomials being relatively prime. Thus there are only two polynomials 𝑁1 and 𝑁2 such as 𝐷1𝐷22𝐷23𝐷24(𝑟−𝐸(𝑟))=𝑁1𝐷1𝐷22+𝑁2𝐷3𝐷4,𝑑𝑜𝑁1<𝑑𝑜𝐷3𝐷4.(3.38) Thus𝑁𝑟=𝐸(𝑟)+1𝐷23𝐷24+𝑁2𝐷1𝐷22𝐷3𝐷4.(3.39) We have𝑑𝑜𝑁1𝐷23𝐷24=𝑑𝑜𝑁1𝐷3𝐷41𝐷3𝐷4<𝑑𝑜1𝐷3𝐷4≤−2.(3.40) Because 𝐷4 does not have simple roots verify 𝑑𝑜𝐷4=0 or 𝑑𝑜𝐷4≥2. Thus 𝑑𝑜(𝐷3𝐷4)≥2. Thuslim𝑥→∞𝑥2𝑁1𝐷23𝐷24𝑑=0,𝑜𝑑(𝑟−𝐸(𝑟))<0,𝑜𝑁2𝐷1𝐷22𝐷3𝐷4<0.(3.41) We consider the rational fraction𝑁𝐹=𝐸(𝑟)+1𝐷23𝐷24−𝛿,(3.42) where1𝛿=4⎡⎢⎢⎣1𝑘𝐷3𝐷32+1𝑘𝐷+13𝐷3⎤⎥⎥⎦,𝑘=𝑑𝑜𝐷3≠0.(3.43)

Proposition 3.8. One assumes 𝑑𝑜𝐷3𝑑≠0𝑜𝐷4≠0 or 𝑑𝑜𝐷3𝑑≥2𝑜𝐷4=0. (1)If 𝑑𝑜𝑟≥0 then: 𝑑𝑜𝐹=𝑑𝑜𝑟 and 𝐸(𝐹)=𝐸(𝑟).(2)If 𝑑𝑜𝑟<0 then: 𝑑𝑜𝐹=−2 where lim𝑥→∞𝑥2𝐹=1/4.(3)For all 𝑐 root of 𝐷3 one has lim𝑥→𝑐(𝑥−𝑐)2𝐹=lim𝑥→𝑐(𝑥−𝑐)21𝑟+4.(3.44)(4)For all 𝑐 root of 𝐷4 of multiplicity 𝜈 one has (𝑥−𝑐)2𝜈𝑟=(𝑥−𝑐)2𝜈𝐹+𝑜(𝑥−𝑐)𝜈−1.(3.45)

Proof. (1) 𝑟−𝐹=𝑁2/𝐷1𝐷22𝐷3𝐷4+𝛿 is from negative degree.
(2) lim𝑥→∞𝑥2𝐹=lim𝑥→∞(𝑥2𝑁1/𝐷23𝐷24)−lim𝑥→∞𝑥2𝛿=−lim𝑥→∞𝑥2𝛿=1/4.
(3) Let 𝑐 be root of 𝐷3lim𝑥→∞(𝑥−𝑐)2(𝑟−𝐹)=lim𝑥→𝑐𝑁2𝐷1𝐷22𝐷4(𝑥−𝑐)2𝐷3+lim𝑥→𝑐(𝑥−𝑐)2𝛿=lim𝑥→𝑐(𝑥−𝑐)21𝛿=−4.(3.46)
(4) Let 𝑐 be root of 𝐷4(𝑥−𝑐)2𝜈𝑁(𝑟−𝐹)=2𝐷1𝐷22𝐷3(𝑥−𝑐)2𝜈𝐷4+(𝑥−𝑐)2𝜈𝛿=𝑜(𝑥−𝑐)𝜈−1.(3.47)

Lemma 3.9. Let 𝑍 be nonzero rational fraction. Σ is a finished set such as Σ∩Roots(𝑍)∪poles(𝑍)=∅.(3.48)(1)There exists 𝑂open connected set on which one has a square root holomorphic of 𝑍, containing for every 𝑐∈Σ a half-right closed by origin 𝑐. (2)If, besides, 𝑍 is from even degree then one can choose 𝑂 of complementary compact.

Proof. We fix 𝑐0 is not an element of Σ. Let Σ be a finished set containing roots, poles of 𝑍 and 𝑐0. They consider all rights linked to the different pairs of points of Σ∪Σ. They choose a point 𝑤 not being on these rights. For all 𝑐∈Σ∪Σ, straight line which joins right 𝑤 and 𝑐 does not contain any other point of Σ∪Σ.
Case 1 (𝑑𝑜𝑍 even). Replacing 𝑍 by rational fraction 𝑍/(𝑥−𝑐0)𝑑𝑜𝑍.
Assume 𝑑𝑜𝑍=0. We put ⋃𝐾=𝑐∈Σ′[𝑤,𝑐]. 𝐾 is compact-connected.
We put 𝑂=ℂ⧵𝐾. 𝑂 is open at infinity such as for all 𝑐∈Σ reaching right 𝑤 in 𝑐 private of 𝑤 is contained in 𝑂.
If 𝛾 is a shoelace of 𝑂then 𝐾 is included in one connected component of ℂ⧵𝛾. Thus:12𝑖𝜋𝛾𝑍𝑍(𝑥)𝑑𝑥=±𝑐∈Σ′𝑍residue𝑍,𝑐=±𝑑𝑜𝑍=0.(3.49) Thus there exist the primitive of 𝑍/𝑍 and the determination of logarithm of 𝑍 and the square root of 𝑍 in 𝑂.
Case 2 (𝑑𝑜𝑍 odd). They use the previous case in replacing 𝑍 by rational fraction 𝑍/(𝑥−𝑐0).

Notation 3.3. We choose a square root of the polynomial of even degree 𝑁1+𝐷23𝐷24[𝐸(𝑟)−𝛿] on a connected open at infinity, roots of 𝐷3 and root of 𝐷4. We put (𝐹)1/2=1𝐷3𝐷4𝑁1+𝐷23𝐷24[]𝐸(𝑟)−𝛿1/2.(3.50)

Proposition 3.10. One assumes 𝑑𝑜𝐷3𝑑≠0𝑜𝐷4≠0 or 𝑑𝑜𝐷3𝑑≥2𝑜𝐷4=0.
One has 𝜃=𝐸(𝜃)+𝑐∈Root𝐷3∪Root𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷3𝐷4𝐷2𝐷3𝐷4+𝐷1𝐷1+𝐷0𝐷0,(3.51) where 𝜀𝑐=±1.

Proof. For 𝑐 root of multiplicity 𝜈≥2 of 𝐷4(𝑥−𝑐)𝜈𝜈𝜃−2(𝑥−𝑐)𝜈−12=(𝑥−𝑐)𝜈(𝐹)1/22+𝑜(𝑥−𝑐)𝜈−1.(3.52) Thus (𝑥−𝑐)𝜈𝜈𝜃−2(𝑥−𝑐)𝜈−1=𝜀𝑐(𝑥−𝑐)𝜈(𝐹)1/2+𝑜(𝑥−𝑐)𝜈−1,𝜈𝜃−2(𝑥−𝑐)=𝜀𝑐(𝐹)1/21+𝑜.𝑥−𝑐(3.53) Partial fraction of 𝜃, associated in root 𝑐 of (𝐷4), minus 𝜈/2(𝑥−𝑐), is in sign meadows that of (𝐹)1/2.
For 𝑐 root of 𝐷31(residueof𝜃at𝑐)−22=lim𝑥→𝑐(𝑥−𝑐)(𝐹)1/22.(3.54) Thus 1(residueof𝜃at𝑐)−2=𝜀𝑐residueof(𝐹)1/2.at𝑐(3.55) Partial fraction of 𝜃, associated in root 𝑐 of (𝐷3), minus 1/2(𝑥−𝑐), is in sign meadows that of (𝐹)1/2. Thus 𝜃=𝐸(𝜃)+𝑐∈Root𝐷3∪Root𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷3𝐷4𝐷2𝐷3𝐷4+𝐷1𝐷1+𝐷0𝐷0.(3.56)

Proposition 3.11. One assumes 𝑑𝑜𝑟<0 and 𝑑𝑜𝐷3𝑑≠0𝑜𝐷4≠0 or 𝑑𝑜𝐷3𝑑≥2𝑜𝐷4=0.
One has 𝜃=𝑐∈Root𝐷3∪Root𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷3𝐷4𝐷2𝐷3𝐷4+𝐷1𝐷1+𝐷0𝐷0,(3.57) where 𝐷0 being a polynomial of degree ⎡⎢⎢⎣𝑐∈Roots𝐷3∪Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0−12⎤⎥⎥⎦2=lim𝑥→∞𝑥21𝑟+4.(3.58)

Proof. We have lim𝑥→∞𝑥𝜃=𝑐∈Roots𝐷3∪Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0,lim𝑥→∞𝑥21𝑟+4=lim𝑥→∞1𝑥𝜃−22.(3.59) Thus ⎡⎢⎢⎣𝑐∈Roots𝐷3∪Zéros𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0−12⎤⎥⎥⎦2=lim𝑥→∞𝑥21𝑟+4.(3.60)

Proposition 3.12. One assumes 𝑑𝑜𝑟=0. One has 𝜃=𝐸(𝜃)+𝑐∈Root𝐷3∪Root𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷3𝐷4𝐷2𝐷3𝐷4+𝐷1𝐷1+𝐷0𝐷0,(3.61) where 𝐸2(𝜃)=𝐸(𝑟) and 𝐷0 being a polynomial of degree: 𝑑𝑜𝐷0=12𝐸(𝜃)lim𝑥→∞𝑥[]−⎡⎢⎢⎣𝑟−𝐸(𝑟)𝐷𝑐∈𝑅3𝐷∪𝑅4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1⎤⎥⎥⎦.(3.62)

Proof. We have 𝐸2(𝜃)=𝐸(𝑟),2𝐸(𝜃)𝑐residueof𝜃at𝑐=𝑐residueof𝑟at𝑐.(3.63) Thus ⎡⎢⎢⎣2𝐸(𝜃)𝐷𝑐∈𝑅3𝐷∪𝑅4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0⎤⎥⎥⎦=lim𝑥→∞𝑥(𝑟−𝐸(𝑟)).(3.64)

Proposition 3.13. One assumes 𝑑𝑜𝑟=2𝜈>0. One has 𝜃=𝐸(𝜃)+𝑐∈Root𝐷3∪Root𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷3𝐷4𝐷2𝐷3𝐷4+𝐷1𝐷1+𝐷0𝐷0,(3.65) where 𝐷0 being a polynomial of degree: 𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝜈2−⎡⎢⎢⎣𝐷𝑐∈𝑅3𝐷∪𝑅4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1⎤⎥⎥⎦.(3.66)

Proof. Let 𝑎 be dominant coefficient of 𝐸(𝑟), and we have 𝛼𝑠𝜈+1−(𝜈/2)= sum of residues of 𝜃, 𝜃=𝐸(𝜃)+𝑐∈Zéros𝐷3∪Zéros𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷3𝐷4𝐷2𝐷3𝐷4+𝐷1𝐷1+𝐷0𝐷0.(3.67) Thus 𝛼𝑠𝜈+1−𝜈2=⎡⎢⎢⎣𝐷𝑐∈𝑅3𝐷∪𝑅4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0⎤⎥⎥⎦.(3.68) Thus 𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝜈2−⎡⎢⎢⎣𝐷𝑐∈𝑅3𝐷∪𝑅4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷3𝐷4+𝑑𝑜𝐷1⎤⎥⎥⎦.(3.69)

3.4. Case 𝑑𝑜𝐷3=0 and 𝑑𝑜𝐷4≠0

𝐷1𝐷22 and 𝐷4 are polynomials being relatively prime. Thus there are only two polynomials 𝑁1 and 𝑁2 such as 𝐷1𝐷22𝐷23𝐷24(𝑟−𝐸(𝑟))=𝑁1𝐷1𝐷22+𝑁2𝐷4,𝑑𝑜𝑁1<𝑑𝑜𝐷4.(3.70) We have𝑁𝑟=𝐸(𝑟)+1𝐷24+𝑁2𝐷1𝐷22𝐷4,lim𝑥→∞𝑥2𝑁1𝐷24=0.(3.71) We consider the rational fraction:𝑁𝐹=𝐸(𝑟)+1𝐷24−14𝑑𝑜𝐷4𝐷4𝐷4.(3.72)

Proposition 3.14. One assumes 𝑑𝑜𝐷3𝑑=0𝑜𝐷4≠0. (1)If 𝑑𝑜𝑟≥0 then: 𝑑𝑜𝐹=𝑑𝑜𝑟 and 𝐸(𝐹)=𝐸(𝑟).(2)If 𝑑𝑜𝑟<0 then: 𝑑𝑜𝐹=−2 where lim𝑥→∞𝑥2𝐹=1/4. (3)For all 𝑐 root of 𝐷4 of multiplicity 𝜈 one has (𝑥−𝑐)2𝜈𝑟=(𝑥−𝑐)2𝜈𝐹+𝑜(𝑥−𝑐)𝜈−1.(3.73)

Proof. (1) 𝑟−𝐹=(𝑁2/𝐷1𝐷22𝐷4)+(1/4𝑑𝑜𝐷4)(𝐷4/𝐷4) is from negative degree.
(2) lim𝑥→∞𝑥2𝐹=lim𝑥→∞𝑁1𝐷24−lim𝑥→∞𝑥214𝑑𝑜𝐷4𝐷4𝐷4=−lim𝑥→∞𝑥214𝑑𝑜𝐷4𝐷4𝐷4=14.(3.74)
(3) Let 𝑐 be root of 𝐷4: (𝑥−𝑐)2𝜈𝑁(𝑟−𝐹)=2𝐷1𝐷22(𝑥−𝑐)2𝜈𝐷4+(𝑥−𝑐)2𝜈14𝑑𝑜𝐷4𝐷4𝐷4=𝑜(𝑥−𝑐)𝜈−1.(3.75)

Notation 3.4. We choose a square root of the polynomial of even degree: 𝑁1+𝐷24[𝐸(𝑟)−(1/4𝑑𝑜𝐷4)(𝐷4/𝐷4)] on a connected open at infinity, root of 𝐷4. We put (𝐹)1/2=1𝐷4𝑁1+𝐷241𝐸(𝑟)−4𝑑𝑜𝐷4𝐷4𝐷41/2.(3.76)

Proposition 3.15. One assumes 𝑑𝑜𝑟<0. One has 𝜃=𝑐∈Roots𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷4𝐷2𝐷4+𝐷1𝐷1+𝐷0𝐷0,(3.77) where 𝜀𝑐=±1 and 𝐷0 being a polynomial of degree: ⎡⎢⎢⎣𝑐∈Root𝐷4𝜀𝑐residue𝑜𝑓(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0−12⎤⎥⎥⎦2=lim𝑥→∞𝑥21𝑟+4.(3.78)

Proof. lim𝑥→∞𝑥𝜃=𝑐∈Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0,lim𝑥→∞𝑥21𝑟+4=lim𝑥→∞1𝑥𝜃−22.(3.79) Thus ⎡⎢⎢⎣𝑐∈Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0−12⎤⎥⎥⎦2=lim𝑥→∞𝑥21𝑟+4.(3.80)

Proposition 3.16. One assumes 𝑑𝑜𝑟=0. One has 𝜃=𝐸(𝜃)+𝑐∈Roots𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷4𝐷2𝐷4+𝐷1𝐷1+𝐷0𝐷0,(3.81) where 𝜀𝑐=±1 and 𝐷0 being a polynomial of degree 𝑑𝑜𝐷0=12𝐸(𝜃)lim𝑥→∞𝑥[]−⎡⎢⎢⎣𝑟−𝐸(𝑟)𝑐∈Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1⎤⎥⎥⎦.(3.82)

Proof. We have 𝐸2(𝜃)=𝐸(𝑟),2𝐸(𝜃)𝑐residueof𝜃at𝑐=𝑐residueof𝑟at𝑐.(3.83) Thus ⎡⎢⎢⎣2𝐸(𝜃)𝑐∈Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0⎤⎥⎥⎦=lim𝑥→∞𝑥(𝑟−𝐸(𝑟)).(3.84)

Proposition 3.17. One assumes 𝑑𝑜𝑟=2𝜈>0. One has 𝜃=𝐸(𝜃)+𝐷𝑐∈𝑅𝑜𝑜𝑡𝑠4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷4𝐷2𝐷4+𝐷1𝐷1+𝐷0𝐷0,(3.85) where 𝜀𝑐=±1 and 𝐷0 being a polynomial of degree 𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝜈2−⎡⎢⎢⎣𝐷𝑐∈𝑅𝑜𝑜𝑡𝑠4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1⎤⎥⎥⎦.(3.86)

Proof. Let 𝑎 dominant coefficient of 𝐸(𝑟). 𝐸(𝜃)=𝛼[𝑡−𝜈+𝑠1𝑡−(𝜈−1)+⋯+𝑠𝜈]. 𝛼𝑠𝜈+1−(𝜈/2)=sumresidueof𝜃, where 𝛼2=𝑎. Thus 𝜃=𝐸(𝜃)+𝑐∈Roots𝐷4𝜀𝑐partialfractionof(𝐹)1/2+1at𝑐2𝐷2𝐷4𝐷2𝐷4+𝐷1𝐷1+𝐷0𝐷0.(3.87) Thus 𝛼𝑠𝜈+1−𝜈2=⎡⎢⎢⎣𝑐∈Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1+𝑑𝑜𝐷0⎤⎥⎥⎦.(3.88) Thus 𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝜈2−⎡⎢⎢⎣𝑐∈Roots𝐷4𝜀𝑐residueof(𝐹)1/2+1at𝑐2𝑑𝑜𝐷2𝐷4+𝑑𝑜𝐷1⎤⎥⎥⎦.(3.89)

4. Recurrent Method at Infinity

4.1. Presentation of 𝐷0 as Determinant

Proposition 4.1. Let 𝑐1,…,𝑐𝑚 be complex constants. We put 𝑃(𝑥)=𝑥−𝑐1⋯𝑥−𝑐𝑚=𝑥𝑚−𝑝1𝑥𝑚−1+𝑝2𝑥𝑚−2−⋯+(−1)𝑚𝑝𝑚.(4.1) For 𝑗∈ℕ, we put 𝜎𝑗=𝑐𝑗1+⋯+𝑐𝑗𝑚 and |||||||||||||Δ=1𝑐1⋅⋅𝑐1𝑚−11𝑐2⋅⋅𝑐2𝑚−1⋅⋅⋅⋅1𝑐𝑚⋅⋅𝑐𝑚𝑚−1|||||||||||||.(4.2)(1)One has Δ2=||||||||||||𝜎0𝜎1⋅⋅𝜎𝑚−1𝜎1𝜎2⋅⋅𝜎𝑚𝜎⋅⋅⋅⋅𝑚−1𝜎𝑚⋅⋅𝜎2𝑚−2||||||||||||.(4.3)(2)One has Δ2𝑃||||||||||||𝜎(𝑥)=0⋅⋅𝜎𝑚−11𝜎⋅⋅𝑥⋅⋅⋅⋅⋅⋅𝑚⋅⋅𝜎2𝑚−1𝑥𝑚||||||||||||.(4.4)

Proof. ||||Δ,𝜎0𝜎1⋅⋅𝜎𝑚−1𝜎1𝜎2⋅⋅𝜎𝑚𝜎⋅⋅⋅⋅𝑚−1𝜎𝑚⋅⋅𝜎2𝑚−2|||| are polynomials at 𝑐1,…,𝑐𝑚 with real coefficients. Thus, Δ2𝑃(𝑥) and |||||𝜎0⋅⋅𝜎𝑚−11𝜎⋅⋅𝑥⋅⋅⋅⋅⋅⋅𝑚⋅⋅𝜎2𝑚−1𝑥𝑚||||| are polynomials at 𝑐1,…,𝑐𝑚,𝑥 with real coefficients. Thus, to have both identities we can assume 𝑐1,…,𝑐𝑚,𝑥 are reals. Furthermore, we can restrict to the open of Zariski 𝑐1,…,𝑐𝑚,𝑥 distinct real nonzero.
We have for all 𝑗=1,…,𝑚,𝑐𝑚𝑗=𝑝1𝑐𝑗𝑚−1−𝑝2𝑐𝑗𝑚−2−⋯−(−1)𝑚𝑝𝑚.
We put |||||||||||||||Δ(𝑥)=1𝑐1⋅⋅𝑐1𝑚−1𝑐𝑚11𝑐2⋅⋅𝑐2𝑚−1𝑐𝑚2⋅⋅⋅⋅⋅1𝑐𝑚⋅⋅𝑐𝑚𝑚−1𝑐𝑚𝑚1𝑥⋅⋅𝑥𝑚−1𝑥𝑚|||||||||||||||,Δ|||||||||||||||(𝑥)=1𝑐1⋅⋅𝑐1𝑚−101𝑐2⋅⋅𝑐2𝑚−10⋅⋅⋅⋅⋅⋅1𝑐𝑚⋅⋅𝑐𝑚𝑚−101𝑥⋅⋅𝑥𝑚−1𝑃|||||||||||||||(𝑥)=Δ𝑃(𝑥).(4.5)
We put 𝑣𝑗=⎛⎜⎜⎜⎜⎜⎝𝑐1𝑗−1..𝑐𝑚𝑗−1⎞⎟⎟⎟⎟⎟⎠,for𝑗=1,…,𝑚.(4.6) The scalar product 𝑣𝑗,𝑣𝑘=𝑐1𝑗+𝑘−2+⋯+𝑐𝑚𝑗+𝑘−2=𝜎𝑗+𝑘−2.(4.7) The Gram matrix of 𝑣1,…,𝑣𝑚 is ⎛⎜⎜⎜⎜⎜⎜⎜⎝𝜎𝐺=0𝜎1⋅⋅𝜎𝑚−1𝜎1𝜎2⋅⋅𝜎𝑚𝜎⋅⋅⋅⋅𝑚−1𝜎𝑚⋅⋅𝜎2𝑚−2⎞⎟⎟⎟⎟⎟⎟⎟⎠⋅(4.8) Thus Δ2=||||||||||||𝜎0𝜎1..𝜎𝑚−1𝜎1𝜎2..𝜎𝑚𝜎....𝑚−1𝜎𝑚..𝜎2𝑚−2||||||||||||.(4.9) We put 𝑣𝑗⎛⎜⎜⎜⎜⎜⎜⎜⎝𝑐(𝑥)=1𝑗−1⋅⋅𝑐𝑚𝑗−1𝑥𝑗−1⎞⎟⎟⎟⎟⎟⎟⎟⎠,𝑗=1,…,𝑚+1.(4.10) We notice 𝑣𝑗(𝑥),𝑣𝑘=𝑣(𝑥)𝑗,𝑣𝑘+𝑥𝑗+𝑘−2=𝜎𝑗+𝑘−2+𝑥𝑗+𝑘−2.(4.11) The 𝑗th column of Gram of matrix which defines Δ(𝑥) is ⎛⎜⎜⎜⎜⎜⎜⎜⎝𝜎𝑗−1𝜎𝑗⋅⋅𝜎𝑗+𝑚−1⎞⎟⎟⎟⎟⎟⎟⎟⎠+𝑥𝑗−1⎛⎜⎜⎜⎜⎜⎜⎜⎝1𝑥⋅⋅𝑥𝑚⎞⎟⎟⎟⎟⎟⎟⎟⎠.(4.12) Thus Δ2||||||||||||𝜎(𝑥)=0𝜎1⋅⋅𝜎𝑚𝜎1𝜎2⋅⋅𝜎𝑚+1𝜎⋅⋅⋅⋅𝑚𝜎𝑚+1⋅⋅𝜎2𝑚||||||||||||+||||||||||||1𝜎1⋅⋅𝜎𝑚𝑥𝜎2⋅⋅𝜎𝑚+1𝑥⋅⋅⋅⋅𝑚𝜎𝑚+1⋅⋅𝜎2𝑚||||||||||||||||||||||||𝜎+⋯+0⋅𝜎𝑚−1𝑥𝑚𝜎1⋅𝜎𝑚𝑥𝑚+1𝜎⋅⋅⋅⋅𝑚⋅𝜎2𝑚−1𝑥2𝑚||||||||||||=𝑖,𝑗𝑥𝑖+𝑗−2cofactor𝑖,𝑗(𝑀),(4.13) where ⎛⎜⎜⎜⎜⎜⎜⎜⎝𝜎𝑀=0𝜎1⋅⋅𝜎𝑚𝜎1𝜎2⋅⋅𝜎𝑚+1𝜎⋅⋅⋅⋅𝑚𝜎𝑚+1⋅⋅𝜎2𝑚⎞⎟⎟⎟⎟⎟⎟⎟⎠.(4.14) We obtain Δ2(𝑥)=Δ2𝑃2(𝑥)=(1𝑥⋯𝑥𝑚⎛⎜⎜⎜⎜⎜⎜⎜⎝1𝑥⋅⋅𝑥)Com(𝑀)𝑚⎞⎟⎟⎟⎟⎟⎟⎟⎠(4.15)
The cofactor (𝑚+1,𝑚+1) of 𝑀 is Δ2. The adjoint of 𝑀 is nonzero. We prove that adjoint of 𝑀 of rank 1. det(𝑀)=0. Because: 𝜎𝑚=𝑝1𝜎𝑚−1−𝑝2𝜎𝑚−2−⋯−(−1)𝑚𝑝𝑚𝜎0 and for all 𝑘≥𝑚;𝜎𝑘=𝑝1𝜎𝑘−1−𝑝2𝜎𝑘−2−⋯−(−1)𝑚𝑝𝑚𝜎𝑘−𝑚. Thus, the (𝑚+1)th column of 𝑀 is is a linear combination of other column.
We consider the matrix 𝐸1,1=⎛⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎠10⋅00⋅⋅⋅0⋅⋅0,(4.16)𝑀+𝐸1,1=⎛⎜⎜⎜⎜⎜⎜⎜⎝𝜎0+1𝜎1⋅⋅𝜎𝑚𝜎1𝜎2⋅⋅𝜎𝑚+1𝜎⋅⋅⋅⋅𝑚𝜎𝑚+1⋅⋅𝜎2𝑚⎞⎟⎟⎟⎟⎟⎟⎟⎠,det𝑀+𝐸1,1||||||||||||=det(𝑀)+1𝜎1⋅⋅𝜎𝑚0𝜎2⋅⋅𝜎𝑚+1⋅⋅⋅⋅0𝜎𝑚+1⋅⋅𝜎2𝑚||||||||||||=||||||||||||𝜎2𝜎3⋅⋅𝜎𝑚+1𝜎3𝜎4⋅⋅𝜎𝑚+2𝜎⋅⋅⋅⋅𝑚+1𝜎𝑚+2⋅⋅𝜎2𝑚||||||||||||⋅(4.17) It is the determinant of Gram of ⎛⎜⎜⎜⎜⎜⎝𝑐1⋅⋅𝑐𝑚⎞⎟⎟⎟⎟⎟⎠⎛⎜⎜⎜⎜⎜⎝𝑐,…,𝑚1⋅⋅𝑐𝑚𝑚⎞⎟⎟⎟⎟⎟⎠.(4.18) Thus det𝑀+𝐸1,1=||||||||||||𝑐1⋅⋅⋅𝑐𝑚1𝑐⋅⋅⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝑐𝑚𝑚||||||||||||2⋅(4.19) But ||||||||||||𝑐1⋅⋅⋅𝑐𝑚1𝑐⋅⋅⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝑐𝑚𝑚||||||||||||=(−1)𝑚−1𝑝𝑚||||||||||||𝑐1⋅⋅⋅𝑐1𝑚−11𝑐⋅⋅⋅⋅1⋅⋅⋅⋅𝑚⋅⋅⋅𝑐𝑚𝑚−11||||||||||||=𝑝𝑚Δ.(4.20) Thus det(𝑀+𝐸1,1)=𝑝2𝑚Δ2. Or 𝑝𝑚=𝑐1⋯𝑐𝑚 is scalar nonzero, 𝑀+𝐸1,1 is invertible matrix.
With 𝑀 symmetric matrix, Com(𝑀) symmetric matrix, we have 𝑀Com(𝑀)=Com(𝑀)𝑀=det(𝑀)𝐼=0,𝐼beingmatriceidentity.(4.21) Thus 𝑀+𝐸1,1Com(𝑀)=𝐸1,1Com(𝑀).(4.22) Or 𝑀+𝐸1,1 is invertible and Com(𝑀)≠0, Com(𝑀) of rank 1.
Thus cofactor𝑖,𝑗(𝑀)=cofactor𝑖,𝑚+1(𝑀)cofactor𝑚+1,𝑚+1(𝑀)cofactor𝑚+1,𝑗(𝑀).(4.23) For 𝑖=1,…,𝑚+1,𝑗=1,…,𝑚+1, we put: 𝐶𝑖,𝑗=cofactor𝑖,𝑗(𝑀), 𝐶𝑖,𝑗Δ2=𝐶𝑖,𝑚+1Δ2𝐶𝑚+1,𝑗Δ2.(4.24) Thus: 𝑃(𝑥)2=1≤𝑖,𝑗≤𝑚+1𝐶𝑖,𝑗Δ2𝑥𝑖+𝑗−2=1≤𝑖,𝑗≤𝑚+1𝐶𝑖,𝑚+1Δ2𝐶𝑗,𝑚+1Δ2𝑥𝑖+𝑗−2=1≤𝑖≤𝑚+1𝐶𝑖,𝑚+1Δ2𝑥𝑖−12.(4.25) Or 𝑃 is monic polynomial and 𝐶𝑚+1,𝑚+1/Δ2=1𝑃(𝑥)=𝑖𝐶𝑖,𝑚+1Δ2𝑥𝑖−1=1Δ2||||||||||||𝜎0⋅⋅⋅𝜎𝑚−11𝜎1⋅⋅⋅𝜎𝑚𝑥𝜎⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝜎2𝑚−1𝑥𝑚||||||||||||.(4.26) We obtain Δ2𝑃||||||||||||𝜎(𝑥)=0⋅⋅⋅𝜎𝑚−11𝜎1⋅⋅⋅𝜎𝑚𝑥𝜎⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝜎2𝑚−1𝑥𝑚||||||||||||⋅(4.27)

4.2. Taylor Expansion of 𝑃/𝑃

Proposition 4.2. Let 𝑐1,…,𝑐𝑚 be pairwise distinct constants.
(1)We put, for 𝑗=0,…,𝑚: 𝜎𝑗=𝑐𝑗1+⋯+𝑐𝑗𝑚 and 𝑃(𝑥)=(𝑥−𝑐1)⋯(𝑥−𝑐𝑚)𝑃𝑃+𝑃𝑃2=2𝑖≠𝑗1𝑐𝑖−𝑐𝑗1𝑥−𝑐𝑖.(4.28)(2)We put 𝑡=1/𝑥. Taylor expansion on the neighborhood of infinity: (a)𝑃𝑃+𝑃𝑃2=∞𝑙=2𝑙−2𝜇=0𝜎𝜇𝜎𝑙−2−𝜇−(𝑙−1)𝜎𝑙−2𝑡𝑙,(4.29)(b)𝑥𝑃′𝑃=∞𝑙=0𝜎𝑙𝑡𝑙.(4.30)

Proof. We have 𝑃∑/𝑃=𝑚𝑖=11/(𝑥−𝑐𝑖).
(1)𝑃𝑃+𝑃𝑃2=𝑖≠𝑗1𝑥−𝑐𝑖1𝑥−𝑐𝑗=𝑖≠𝑗𝑐1/𝑖−𝑐𝑗𝑥−𝑐𝑖+𝑐1/𝑗−𝑐𝑖𝑥−𝑐𝑗=2𝑖≠𝑗1𝑐𝑖−𝑐𝑗1𝑥−𝑐𝑖.(4.31)(2)We put 𝑡=1/𝑥. We have 𝑃𝑃+𝑃𝑃2=2𝑡𝑖≠𝑗1𝑐𝑖−𝑐𝑗11−𝑐𝑖𝑡=2𝑡𝑖≠𝑗1𝑐𝑖−𝑐𝑗∞𝑘=0𝑐𝑖𝑡𝑘=∞𝑘=02𝑖≠𝑗𝑐𝑘𝑖𝑐𝑖−𝑐𝑗𝑡𝑘+1=∞𝑘=1𝑖≠𝑗𝑐𝑘𝑖−𝑐𝑘𝑗𝑐𝑖−𝑐𝑗𝑡𝑘+1.(4.32) as 𝑖≠𝑗𝑐𝑘𝑖−𝑐𝑘𝑗𝑐𝑖−𝑐𝑗=𝑖≠𝑗𝜇+𝜈=𝑘−1,0≤𝜇≤𝑘−1𝑐𝜇𝑖𝑐𝜈𝑗=𝜇+𝜈=𝑘−1,0≤𝜇≤𝑘−1𝜎𝜇𝜎𝜈−𝑚𝑖=1𝑐𝑖𝜇+𝜈=𝜇+𝜈=𝑘−1,0≤𝜇≤𝑘−1𝜎𝜇𝜎𝜈−𝑘𝜎𝑘−1.(4.33) Thus 𝑃𝑃+𝑃𝑃2=∞𝑘=1𝑘−1𝜇=0𝜎𝜇𝜎𝑘−1−𝜇−𝑘𝜎𝑘−1𝑡𝑙=∞𝑙=2𝑙−2𝜇=0𝜎𝜇𝜎𝑙−2−𝜇−(𝑙−1)𝜎𝑙−2𝑡𝑙.(4.34)

4.3. Research of 𝐷0

Let 𝑟=𝑁/𝐷1𝐷22𝐷23𝐷24 be a rational fraction and 𝜃∈ℂ(𝑥) where 𝜃+𝜃2=𝑟.

We put𝐷𝜃=𝑆+0𝐷0,(4.35) where ∑𝑆=𝐸(𝜃)+𝑐polesof𝑟𝜃𝑐 and 𝐷0 monic polynomial of degree 𝑚. We have𝜃+𝜃2=𝑆+𝑆2+𝐷0𝐷0+𝐷0𝐷02𝐷+2𝑆0𝐷0,𝜃+𝜃2𝐷=𝑟⟺0𝐷0+𝐷0𝐷02𝐷+2𝑆0𝐷0=𝑅𝐵(4.36) with 𝐵=𝐷1𝐷2𝐷3𝐷4 and 𝑅 polynomials as 𝑟−𝑆−𝑆2=𝑅/𝐵.

Case 1 (𝑑𝑜𝑟=2𝜈≥0). We have 𝑑𝑜𝑆=𝜈 and 𝑑𝑜𝑅/𝐵=𝜈−1. Taylors expansion of (1/𝑥𝜈)𝑆, (1/𝑥𝜈−1)(𝑅/𝐵) and 𝑥(𝐷0/𝐷0) at order 𝜇.
We obtain the Taylors expansion of (1/𝑥𝜈−1)[(𝑅/𝐵)−2𝑆(𝐷0/𝐷0)] at order 𝜇.
The Taylors expansion equals the Taylors expansion of (1/𝑥𝜈−1)[(𝐷0/𝐷0)+(𝐷0/𝐷0)2]. Let 𝑡=1/𝑥. In Taylors expansion of (1/𝑥𝜈−1)[(𝑃/𝑃)+(𝑃/𝑃)2] (see Proposition 4.2) the coefficient of 𝑡𝜈+1 is: 𝜎20−𝜎0=𝑚2−𝑚. For 𝑙≥2, coefficient of 𝑡𝑙+𝜈−1 use: 𝜎0,…,𝜎𝑙−2.
Besides, we have 𝑥𝐷0𝐷0=𝑚𝑖=111−𝑐𝑖𝑡=𝑚∞𝑖=1𝑘=0𝑐𝑘𝑖𝑡𝑘=∞𝑘=0𝜎𝑘𝑡𝑘.(4.37) Thus, Taylors expansion of 1Φ=𝑥𝜈−1⎡⎢⎢⎣𝐷0𝐷0+𝐷0𝐷02⎤⎥⎥⎦−1𝑥𝜈−1𝑅𝐵𝐷−2𝑆0𝐷0(4.38) and the coefficient of 𝑡𝑘 is −2𝛼𝜎𝑘+polynomialat𝜎0,…,𝜎𝑘−1,(4.39)Φ=0 determine 𝜎𝑘 by recurrence at 𝑘=𝜈.

Case 2 (𝑑𝑜𝑟<0). The coefficient of 𝑡𝑘 in Taylors expansion of Ψ=𝑥2⎡⎢⎢⎣𝐷0𝐷0+𝐷0𝐷02⎤⎥⎥⎦−𝑥2𝑅𝐵𝑥𝐷+2𝑥𝑆0𝐷0(4.40) is 2𝑚𝜎𝑘−(𝑘+1)𝜎𝑘+2lim𝑥→∞𝜎𝑥𝑆𝑘+polynomialat𝜎0,…,𝜎𝑘−1.(4.41) By recurrence 𝜎𝑘 at condition 𝑘≠2lim𝑥→∞𝑥𝑆+2𝑚−1.(4.42) We have lim𝑥→∞𝑥𝑆+𝑚=𝛼2∞+(1/2) with 𝛼2∞=lim𝑥→∞𝑥2𝑟+(1/4).
For 𝑘≠2𝛼∞, to be 𝜎𝑘 at function of 𝜎0,…,𝜎𝑘−1.
If 2𝛼∞∈ℕ then, the coefficient of 𝑡2𝛼∞ is nonzero implied that there is no solution, or the coeffcient of 𝑡2𝛼∞ is zero implied that 𝜎2𝛼∞ is arbitrary and 𝜎𝑘; 𝑘>2𝛼∞ depend in a unique way.
𝐷0 is polynomial determined by 𝜎0,…,𝜎2𝑚−1 by determinant formulae; the problem of nonunique suite (𝜎𝑘)𝑘∈ℕ is put if 2𝛼∞ is positive integer equal to or less than 2𝑚−1, then 4lim𝑥→∞𝑥2𝑟+1 is square of integer and lim𝑥→∞𝑥2𝑟≤𝑚2−𝑚.(4.43)

Example 4.3. In this example we consider the Riccati differential equation (1.2) where 𝑧𝑟=−1+2−(1/4)𝑥2;𝑧∈ℝ∗.(4.44) We have 𝐷1=1, 𝐷2=1, 𝐷3=𝑥 and 𝐸2(𝜃)=−1.
We can assume 𝐸(𝜃)=𝑖, lim𝑥→0𝑥21𝑟+4=𝑧2=𝛼20,𝛼𝜃=𝑖+0+(1/2)𝑥+𝐷0𝐷0,(4.45)2𝑖(𝛼0+(1/2)+𝑑𝑜𝐷0)=0. Thus 𝑚=𝑑𝑜𝐷0=−𝛼0−(1/2)∈ℕ.
Thus 𝛼0=−𝑚−(1/2), 𝑟=−1+((𝑚+(1/2))2−(1/4))/𝑥2=−1+(𝑚2+𝑚)/𝑥2 and 𝜃=𝑖−(𝑚/𝑥)+(𝐷0/𝐷0). 𝜃 is a solution of Riccati equation if and only if 𝐷0 verifies 𝐷0𝐷0+𝐷0𝐷02𝑚+2𝑖−𝑥𝐷0𝐷0𝑚=2𝑖𝑥.(4.46) Thus Φ=0 where ⎡⎢⎢⎣𝐷Φ=𝑥0𝐷0+𝐷0𝐷02⎤⎥⎥⎦𝑚+2𝑖−𝑥𝑥𝐷0𝐷0−2𝑖𝑚.(4.47) Taylors expansion of Φ=0 at 𝑡=1/𝑥Φ=∞𝑘=1𝑘−1𝜇=1𝜎𝜇𝜎𝑘−1−𝜇−𝑘𝜎𝑘−1−2𝑚𝜎𝑘−1+2𝑖𝜎𝑘𝑡𝑘.(4.48) As 𝜎0=𝑚. Thus Φ=0⟺∀𝑘≥1,𝑘−1𝜇=0𝜎𝜇𝜎𝑘−1−𝜇−𝑘𝜎𝑘−1−2𝑚𝜎𝑘−1+2𝑖𝜎𝑘=0.(4.49) For 𝑘=1, 𝜎20−𝜎0−2𝑚𝜎0+2𝑖𝜎1=0 equivalent to 2𝑖𝜎1=𝑚2+𝑚.
For all, 𝑘≥2, 2𝑖𝜎𝑘=𝑘𝜎𝑘−1−∑𝑘−2𝜇=1𝜎𝜇𝜎𝑘−1−𝜇. For example ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩𝑚=3∶2𝑖𝜎1=12,2𝑖𝜎2=2𝜎1,2𝑖𝜎3=3𝜎2−𝜎21,2𝑖𝜎4=4𝜎3−2𝜎1𝜎2,2𝑖𝜎5=5𝜎4−2𝜎1𝜎3−𝜎22.(4.50) Thus, the polynomial 𝐷0 is partner to ||||||||||𝜎0𝜎1𝜎21𝜎1𝜎2𝜎3𝑥𝜎2𝜎3𝜎4𝑥2𝜎3𝜎4𝜎5𝑥3||||||||||=||||||||||3−6𝑖−61−6𝑖−6−9𝑖𝑥−6−9𝑖−54𝑥2−9𝑖−5499𝑖𝑥3||||||||||=135𝑥3+810𝑖𝑥2−2025𝑥−2025𝑖.(4.51) Thus 𝐷0=𝑥3+6𝑖𝑥2−15𝑥−15𝑖.(4.52)

5. Method of Last Minor

Let 𝑟=𝑁/𝐷1𝐷22𝐷23𝐷24 be a rational fraction and 𝜃∈ℂ(𝑥) as: 𝜃+𝜃2=𝑟.

𝜃 solution of (1.2) if and only if 𝐷0 verifies𝐵𝐷0+2𝑆𝐵𝐷0=𝑅𝐷0.(5.1) We choose a complex number 𝑐 not pole of 𝑟 and we use the expression of polynomials following the powers of 𝑥−𝑐.

For the sake of simplicity, in the following we assume that 𝑐=0. Constant coefficient constant of 𝐵 is nonzero.

Denote by 𝑎𝑘,𝑏𝑘, and 𝑟𝑘 coefficients of 𝑥𝑘, in 𝐴,𝐵, and 𝑅, respectively, 𝑘∈ℕ. If 𝑘∈ℤ then 𝑎𝑘=𝑏𝑘=𝑟𝑘=0.

5.1. Case 𝑑𝑜𝑟<0

We put 𝑆=𝐴/𝐵.

If 𝑑𝑜𝐵=1 then 𝑟 are pole unique, it is simple or double. If 𝑐 simple pole of 𝑟 then 𝑟 of degree −1 and (1.2) has no rational solution.

If 𝑐 double pole of 𝑟 then 𝐴=𝛼𝑐+(1/2), 𝑟=(𝛼2𝑐−(1/4))/(𝑥−𝑐)2 and 𝑅=0.

Thus,(5.1)𝐷0𝐷0=𝛼−2𝑐+(1/2).𝑥−𝑐(5.2) The solutions𝐷0=−2𝛼𝑐(𝑥−𝑐)−2𝛼𝑐−1,𝐷0=(𝑥−𝑐)−2𝛼𝑐+𝛽,(5.3) where 𝛽 nonzero constant.

We have, an infinity of the other rational solution when −2𝛼𝑐∈ℕ∗, with the same parity of 𝑆.

We assume 𝑑𝑜𝐵≥2and we consider 𝐷0 a polynomial of degree 𝑚≥1 that verifies (5.1).

Proposition 5.1. One assumes 𝑑𝑜𝑟<0,𝑑𝑜𝑑𝐵≥2𝑜=𝑚≥1 and 𝐷0 verifies (5.1). (1)One has 𝑑𝑜𝐵𝐷0+2𝐴𝐷0≤𝑑𝑜𝑑𝐵+𝑚−2,𝑜𝑅≤𝑑𝑜𝐵−2.(5.4)(2)One has 𝑟𝑑𝑜𝐵−2=2𝑎𝑑𝑜𝐵−1𝑚+𝑚(𝑚−1).(3)One has 𝑎𝑑𝑜𝐵−1=12−𝑚+𝛼∞,𝑟𝑑𝑜𝐵−2=𝑚2𝛼∞.−𝑚(5.5)

Proof. (1) 𝑑𝑜(𝐵𝐷0+2𝐴𝐷0)≤sup(𝑑𝑜(𝐵𝐷0),𝑑𝑜(2𝐴𝐷0)). If 𝑑𝑜𝐷0=1 then 𝑑sup𝑜𝐵𝐷0,𝑑𝑜2𝐴𝐷0=𝑑𝑜2𝐴𝐷0=𝑑𝑜𝐴≤𝑑𝑜𝐵−1=𝑑𝑜𝐵+𝑑𝑜𝐷0−2.(5.6) If 𝑑𝑜𝐷0≥2 then 𝑑𝑜𝐵𝐷0=𝑑𝑜𝑑𝐵+𝑚−2,𝑜2𝐴𝐷0=𝑑𝑜𝐴+𝑚−1≤𝑑𝑜𝐵+𝑚−2.(5.7)
(2) We have 𝑑𝑜(𝑅𝐷0)≤𝑑𝑜𝐵+𝑚−2 and coefficients of 𝑥𝑑𝑜𝐵+𝑚−2 in (5.1).
(3) Accordingly, in view of lim𝑥→∞𝑥𝜃=𝛼∞+(1/2).

We put⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝1𝑥𝑥𝑣(𝑥)=−𝑅2⋅⋅𝑥𝑚⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎛⎜⎜⎜⎜⎜⎜⎜⎝⋅⋅+2𝐴012𝑥𝑚𝑥𝑚−1⎞⎟⎟⎟⎟⎟⎟⎟⎠⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝002⋅⋅+𝐵𝑚(𝑚−1)𝑥𝑚−2⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝𝑣1𝑣(𝑥)2𝑣(𝑥)3(⋅⋅𝑣𝑥)𝑚+1⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠(𝑥).(5.8) The 𝑘th element of 𝑣(𝑥) is𝑣𝑘(𝑥)=𝐵(𝑘−1)(𝑘−2)𝑥𝑘−3+2𝐴(𝑘−1)𝑥𝑘−2−𝑅𝑥𝑘−1,(5.9)𝑣𝑘(𝑥) of degree equal to or less than 𝑑𝑜𝐵+𝑘−3 and 𝑥𝑑𝑜𝐵+𝑘−3 of coefficient(𝑘−1)(𝑘−2)+2𝑎𝑑𝑜𝐵−1(𝑘−1)−𝑟𝑑𝑜𝐵−2𝑟=(𝑘−𝑚−1)𝑑𝑜𝐵−2𝑚+𝑘−1=(𝑘−𝑚−1)2𝑎𝑑𝑜𝐵−1.+𝑚+𝑘−2(5.10) For 𝑘≤𝑚, the coefficient of 𝑥𝑑𝑜𝐵+𝑘−3 is zero if and only if 2𝑎𝑑𝑜𝐵−1=−(𝑚+𝑘−2)∈{−(𝑚−1),−𝑚,…,−2(𝑚−1)},(5.11)𝑣𝑚+1 of degree equal to or less than 𝑑𝑜𝐵+𝑚−3.

For 𝑘≥3 the coefficient of more low degree of 𝑣𝑘 is coefficient of 𝑥𝑘−3 equal 𝑏0(𝑘−1)(𝑘−2).

Let 𝑉 matrix of 𝑚+1 row and 𝑘th row is row 𝑙𝑘 of coefficients of 𝑣𝑘(𝑥) in basis of ℂ𝑑𝑜𝐵+𝑚−3[𝑋].

𝑙3,…,𝑙𝑚+1 is linearly independent system.

Equation (5.1) gives𝐷0=𝑑0+𝑑1𝑥+⋯+𝑥𝑚,(5.12) obtained𝑑0𝑣1(𝑥)+⋯+𝑑𝑚−1𝑣𝑚(𝑥)+𝑣𝑚+1𝑑(𝑥)=0,0𝑙1+⋯+𝑑𝑚−1𝑙𝑚+𝑙𝑚+1=0.(5.13) Thus, the matrix 𝑉 is rank 𝑚−1,𝑚 or 𝑚+1. Accordingly, existence of 𝐷0 correspondent is linearly dependent in row 𝑙𝑚+1 of 𝑙1,…,𝑙𝑚.

Let 𝑐1,…,𝑐𝑚 roots of 𝐷0 distinct.

We put, for 𝑗∈ℕ: 𝜎𝑗=𝑐𝑗1+⋯+𝑐𝑗𝑚 and Δ2=||||||||||||𝜎0𝜎1⋅⋅𝜎𝑚−1𝜎1𝜎2⋅⋅𝜎𝑚𝜎⋅⋅⋅⋅𝑚−1𝜎𝑚⋅⋅𝜎2𝑚−2||||||||||||.(5.14)The 𝑚 column vector 𝜎0⋅⋅𝜎𝑚,…,𝜎𝑚−1⋅⋅𝜎2𝑚−1 of 𝑚+1 column is linearly independent.

The equation of hyperplane 𝐻 engendered by vectors 𝜎0⋅⋅𝜎𝑚,…,𝜎𝑚−1⋅⋅𝜎2𝑚−1 is||||||||||||𝜎0⋅⋅⋅𝜎𝑚−1𝑥1𝜎1⋅⋅⋅𝜎𝑚𝑥2𝜎⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝜎2𝑚−1𝑥𝑚+1||||||||||||=0.(5.15) The coefficient of 𝑥𝑚+1 is Δ2≠0 and 𝐻 as equation𝑥𝑚+1=𝜆1𝑥1+⋯+𝜆𝑚𝑥𝑚,(5.16) where 𝜆1⋯𝜆𝑚 scalar satisfying𝜎𝑚,…,𝜎2𝑚−1=𝜆1𝜎0,…,𝜎𝑚−1+⋯+𝜆𝑚𝜎𝑚−1,…,𝜎2𝑚−2.(5.17) The derivative of||||||||||||𝜎0⋅⋅⋅𝜎𝑚−11𝜎1⋅⋅⋅𝜎𝑚𝑥𝜎⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝜎2𝑚−1𝑥𝑚||||||||||||(5.18) is derived from last column. Equation (5.1) is equivalent to:||||||||||||𝜎0⋅⋅⋅𝜎𝑚−11𝜎1⋅⋅⋅𝜎𝑚𝑣1𝜎(𝑥)⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝜎2𝑚−1𝑣𝑚+1||||||||||||(𝑥)=0.(5.19) Thus, for all 𝑥, 𝑣(𝑥)∈𝐻. With application of Taylor, this is equivalent to columns of matrix 𝑉 in hyperplane 𝐻,𝑙𝑚+1=𝜆1𝑙1+⋯+𝜆𝑚𝑙𝑚,||||||||||||𝜎0⋅⋅⋅𝜎𝑚−11𝜎1⋅⋅⋅𝜎𝑚𝑥𝜎⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝜎2𝑚−1𝑥𝑚||||||||||||=||||||||||||𝜎0⋅⋅⋅𝜎𝑚−11𝜎1⋅⋅⋅𝜎𝑚𝑥𝜎⋅⋅⋅𝑚−1⋅⋅⋅𝜎2𝑚−2𝑥𝑚−10⋅⋅⋅0𝑥𝑚−𝜆1−𝜆2𝑥−⋯−𝜆𝑚𝑥𝑚−1||||||||||||=Δ2𝑥𝑚−𝜆1−𝜆2𝑥−⋯−𝜆𝑚𝑥𝑚−1.(5.20) The polynomial 𝐷0 is partner to ||||||||||||𝜎0⋅⋅⋅𝜎𝑚−11𝜎1⋅⋅⋅𝜎𝑚𝑥𝜎⋅⋅⋅⋅⋅⋅𝑚⋅⋅⋅𝜎2𝑚−1𝑥𝑚||||||||||||.(5.21) Thus𝐷0(𝑥)=𝑥𝑚−𝜆1−𝜆2𝑥−⋯−𝜆𝑚𝑥𝑚−1.(5.22) For 𝑘=1,…,𝑚+1, we put𝑣𝑘(𝑥)=𝜌𝑘(𝑥)+𝑥𝑑𝑜𝐵−2𝑤𝑘,(5.23) where 𝑑𝑜𝜌𝑘(𝑥)≤𝑑𝑜𝐵−3.

The matrix 𝑊 partner to 𝑤1⋅𝑤𝑚 is triangular where 𝑘th entry diagonal equal (𝑘−1−𝑚)(2𝑎𝑑𝑜𝐵−1+𝑚+𝑘−2). The coefficient is nonzero except possibly for a single value of 𝑘. Thus, rank of 𝑤1⋅𝑤𝑚 is 𝑚 or 𝑚−1.

For rank(𝑉)=𝑚, (5.1) as solution is equivalent to 𝑙1,…,𝑙𝑚 which is linearly independent.

Hence, we assume rank(𝑉)=𝑚−1. In that case 𝑙1 and 𝑙2 are linear combinations of 𝑙3,…,𝑙𝑚+1. One of both combinations has to express 𝑙𝑚+1. Thus we have two situations.

Situation 1
𝑙2 linear combination of 𝑙3,…,𝑙𝑚+1 where coefficient of 𝑙𝑚+1 is nonzero. By replacement of 𝑙𝑚+1, we obtain 𝑙1 a linear combination of 𝑙2,…,𝑙𝑚.
Thus, 𝑤1 is linear combination of 𝑤2,…,𝑤𝑚. (𝑤2,…,𝑤𝑚) libre system where 𝑑𝑜𝑤𝑘=𝑘−1 for 0𝑘=2,…,𝑚. Thus: 𝑤1=0.
If 𝑙1=𝑚𝑗=2𝛼𝑗𝑙𝑗,(5.24) where 𝛼𝑗 is constant, then 𝑤1=𝑚𝑗=2𝛼𝑗𝑤𝑗.(5.25) As 𝑤1=0, thus for all 𝑗=2,…,𝑚, 𝛼𝑗=0 equivalent to 𝑙1=0. Thus 𝑅=0.

First Way of Having 𝐷0:
𝐷0𝐷0=−2𝐴𝐵(5.26)−2𝐴/𝐵 has to be the sum of simple elements of the shape 𝜇𝑐/(𝑥−𝑐) where 𝜇𝑐 is greater or equal to 1.
Thus 𝐷4=𝐷2=𝐷1=1 and 𝐷3=𝐵.
For all 𝑐 root of 𝐷3,−2𝛼𝑐=𝜇𝑐+1 where 𝛼2𝑐=lim𝑥→𝑐(𝑥−𝑐)2𝑟+(1/4).Thus 𝛼𝑐<0,4lim𝑥→𝑐(𝑥−𝑐)21𝑟+4=𝜇𝑐+12.(5.27) Thus 𝐷0=𝑚𝑐∈Root(𝐷3)(𝑥−𝑐)𝜇𝑐.(5.28)𝐷0 is primitive nonzero in roots of 𝐷3.

Second Way of Having 𝐷0:
𝑙𝑚+1=𝑚𝑗=2𝜆𝑗𝑙𝑗,(5.29) where 𝜆𝑗 is constant (𝑗=2,…,𝑚). The relations which are linearly dependent between 𝑙𝑚+1 and 𝑙1,…,𝑙𝑚 are 𝑙𝑚+1=𝜆𝑙1+𝑚𝑗=2𝜆𝑗𝑙𝑗.(5.30) Thus we have an infinity of solutions: 𝐷0=𝑥𝑚−𝜆−𝑚𝑗=2𝜆𝑗𝑥𝑗−1,(5.31) where 𝜆 is arbitrarily constant.

Situation 2
𝑙1 is linear combination of 𝑙3,…,𝑙𝑚+1 where coefficient of 𝑙𝑚+1 is nonzero and 𝑙2 is linear combination of 𝑙3,…,𝑙𝑚. Thus, 𝑤2 is linear combination of 𝑤3,…,𝑤𝑚. Thus, 𝑤2=0. 𝑤1,𝑤3,…,𝑤𝑚 is libre system.
If 𝑙2=𝑚𝑗=3𝛼𝑗𝑙𝑗,(5.32) then 𝑤2=𝑚𝑗=3𝛼𝑗𝑤𝑗.(5.33) As 𝑤2=0. Thus, for all 𝑗=3,…,𝑚, 𝛼𝑗=0 equivalent to 𝑙2=0 and 2𝐴−𝑥𝑅=0.

First Way of Having 𝐷0:
2𝐴𝑥𝐷0=2𝐴𝐷0+𝐵𝐷0,𝐷2𝐴0−𝑥𝐷0=𝑥𝐵𝐷0.(5.34) We put 𝑄=𝐷0−𝑥𝐷0. 𝑄/𝑄=−2𝐴/𝐵, 𝐷4=𝐷2=𝐷1=1,𝑄=(1−𝑚)𝑐∈Root𝐷3(𝑥−𝑐)𝜇𝑐.(5.35) Thus 𝐷0=𝐶𝑥 with 𝐶=−𝑄/𝑥2; 𝐶 rational function.
The coefficient of 𝑥 in 𝑄 is zero. 𝑄(0)=0 (𝐴(0)=0; 2𝐴=𝑥𝑅).

Second Way of Having 𝐷0
𝑙𝑚+1=𝜆1𝑙1+𝑚𝑗=3𝜆𝑗𝑙𝑗,(5.36) thus 𝐷0=𝑥𝑚−𝑚𝑗=1𝜆𝑗𝑥𝑗−1.(5.37)𝜆2 is arbitrarily constant.

5.2. Case 𝑑𝑜𝑟=2𝜈≥0

We put 𝐸=𝐸(𝜃) and 𝑆=𝐸+(𝐴/𝐵). We have𝑑𝑜𝐵𝐷0≤𝑑𝑜𝑑𝐵+𝑚−2,𝑜(𝐸𝐵+𝐴)𝐷0=𝑑𝑜𝐵+𝑚−1+𝜈.(5.38) Thus𝑑𝑜𝑅𝐷0=𝑑𝑜𝐵+𝑚−1+𝜈.(5.39) Thus, 𝑑𝑜𝑅=𝑑𝑜𝐵+𝜈−1 and 𝑟𝑑𝑜𝐵+𝜈−1=2𝛼𝑚 where 𝛼 is dominant coefficient of 𝐸.

We put⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝1𝑥𝑥𝑣(𝑥)=−𝑅2⋅⋅𝑥𝑚⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝01⋅⋅+2(𝐸𝐵+𝐴)2𝑥𝑚𝑥𝑚−1⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝002⋅⋅+𝐵𝑚(𝑚−1)𝑥𝑚−2⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝𝑣1𝑣(𝑥)2𝑣(𝑥)3(⋅⋅𝑣𝑥)𝑚+1⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠(𝑥)⋅(5.40) The 𝑘th element of 𝑣(𝑥) is𝑣𝑘(𝑥)=𝐵(𝑘−1)(𝑘−2)𝑥𝑘−3+2(𝐸𝐵+𝐴)(𝑘−1)𝑥𝑘−2−𝑅𝑥𝑘−1.(5.41)𝑣𝑘(𝑥) is of degree equal to or less than 𝑑𝑜𝐵+𝜈+𝑘−2 where coefficient of 𝑥𝑑𝑜𝐵+𝜈+𝑘−2 equal 2𝛼(𝑘−1−𝑚). For 𝑘=1,…,𝑚, the 𝑘th element of 𝑣(𝑥) is of degree: 𝑑𝑜𝑅+𝑘−1. The (𝑚+1)th element of 𝑣(𝑥) is of degree equal to or less than 𝑑𝑜𝑅+𝑚−1.

Proposition 5.2. One puts 𝑣𝑘(𝑥)=𝜌𝑘(𝑥)+𝑥𝑑𝑜𝑅𝑤𝑘(𝑥) where 𝑑𝑜𝜌𝑘(𝑥)<𝑑𝑜𝑅. (1)𝑑𝑜𝑤𝑘𝑑(𝑥)=𝑘−1for𝑘=1,…,𝑚,𝑜𝑤𝑚+1(𝑥)≤𝑚−1,(5.42)(2)(𝑤1,…,𝑤𝑚) libre system and 𝑤𝑚+1(𝑥)=𝜆1𝑤1(𝑥)+⋯+𝜆𝑚𝑤𝑚(𝑥),(5.43) where 𝜆1,…,𝜆𝑚 constants. (3)𝐷0=𝑥𝑚−𝜆1−𝜆2𝑥−⋯−𝜆𝑚−1𝑥𝑚−1(5.44) is solution if and only if 𝑙𝑚+1=𝜆1𝑙1+⋯+𝜆𝑚𝑙𝑚.(5.45)

Example 5.3. In this example we consider the Riccati differential equation (1.2) where 1𝑟=(𝑥+1)4−5(𝑥+1)3+74(𝑥+1)2+1𝑥+1+𝑥2+2,(5.46)𝐷1=𝐷2=𝐷3=1,𝐷4=(𝑥+1)2,𝑑𝑜(𝑟)=2,𝜈=1. We have 𝑁1=1−5(𝑥+1),𝐹=𝑥2+2+1−5(𝑥+1)(𝑥+1)4+14(𝑥+1)2.(5.47)

Study at (−1)
Laurent series development at −1: (𝐹)1/2=𝜀−11(𝑥+1)2−52(𝑥+1)+⋯;𝜀−1=±1.(5.48)

Study at Infinity
We have 𝐸(𝑟)/𝑥2=1+𝑡2 where 𝑡=1/𝑥, 1+2𝑡21/2=1+𝑡2𝑡+𝑜2.(5.49) Thus 𝑠𝜈+1=1 and 𝐸(𝜃)=𝛼𝑥, where 𝛼2=1, 𝜃=𝛼𝑥+𝜀−11(𝑥+1)2−5+12(𝑥+1)(+𝐷𝑥+1)0𝐷0,𝑑𝑜𝐷0=𝛼𝑠𝜈+1−𝜈2+5𝜀−12−1=𝛼𝑠𝜈+1+5𝜀−12−32.(5.50) Case 𝜀𝛼=−1−1=−1 and 𝜀𝛼=1−1=−1 are to be rejected because we obtain negative values of 𝐷0. (i)If 𝛼=1 and 𝜀−1=1 then: 𝑑𝑜𝐷0=2.(ii)If 𝛼=−1 and 𝜀−1=1 then: 𝑑𝑜𝐷0=0.

Case 1 (𝛼=1 and 𝜀−1=1). We have the following 1𝜃=𝑥+(𝑥+1)2−3+𝐷2(𝑥+1)0𝐷0,(5.51) where 𝑑𝑜𝐷0=2. Research of coefficients of 𝐷0, 1𝑆=𝑥+(𝑥+1)2−3=𝑥2(𝑥+1)3+2𝑥2−(1/2)𝑥−(1/2)(𝑥+1)2(5.52)𝑟−𝑆′−𝑆2=𝑅/𝐵, where 𝑅=4𝑥2+4𝑥 and 𝐵=(1+𝑥)2⎛⎜⎜⎜⎝1𝑥𝑥𝑣(𝑥)=−𝑅2⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝01⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝002⎞⎟⎟⎟⎠=⎛⎜⎜⎜⎝+2𝑆𝐵2𝑥+𝐵−4𝑥−4𝑥2−1−𝑥−2𝑥32+2𝑥+4𝑥3⎞⎟⎟⎟⎠,⎛⎜⎜⎜⎝⎞⎟⎟⎟⎠,𝑉=0−4−40−1−10−22204(5.53)𝑙3=−2𝑙2. Thus 𝐷0=𝑥2+2𝑥, 1𝜃=𝑥+(𝑥+1)2−3+2(𝑥+1)2(𝑥+1)𝑥21+2𝑥=𝑥+(𝑥+1)2−3+12(𝑥+1)𝑥+1𝑥+2.(5.54)

Case 2 (𝛼=−1 and 𝜀−1=1). We obtain the rational fraction 1𝜃=−𝑥+(𝑥+1)2−32(𝑥+1).(5.55) It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.

Example 5.4. In this example we consider the Riccati differential equation (1.2) where 1𝑟=+116(𝑥−1)8−4(𝑥−1)5−296(𝑥−1)4−89(𝑥−1)3−6427(𝑥−1)2−152+18(𝑥−1)30(𝑥+2)2−10,81(𝑥+2)(5.56)𝐷1=𝐷2=1,𝐷3=𝑥+2 and 𝐷4=(𝑥−1)4. We have 𝑁1=(2418𝑥−2454)(𝑥−1)3+(𝑥+2)21,𝛿=−4(𝑥+2)2,1𝐹=+16(2418𝑥−2454)(𝑥−1)3+(𝑥+2)2(𝑥−1)8(𝑥+2)2+14(𝑥+2)2.(5.57) Laurent series development at 1: (𝐹)1/2=𝜀11(𝑥−1)4−2𝑥−1+⋯;𝜀1=±1.(5.58) Laurent series development at −2: (𝐹)1/2=𝜀−2112(𝑥+2)+⋯;𝜀−2=±1.(5.59) We have 12𝐷2𝐷3𝐷4𝐷2𝐷3𝐷4=1+22(𝑥+2).𝑥−1(5.60) Thus 𝜃=𝐸(𝜃)+𝜀11(𝑥−1)4−2𝑥−1+𝜀−211+12(𝑥+2)+22(𝑥+2)+𝐷𝑥−10𝐷0,(5.61) where 𝐸21(𝜃)=𝐸(𝑟)=,𝑑16𝑜𝐷01=−𝐸(𝜃)+2𝜀1−11𝜀−22−52.(5.62) Cases 𝜀1𝜀=1−2=1 and 𝜀1𝜀=−1−2=1 are to be rejected because we obtain negative values of 𝐷0.

Case 1. If 𝜀1=1,𝜀−2=−1,𝐸(𝜃)=1/4 then 𝑑𝑜𝐷0=1, 1𝜃=4+1(𝑥−1)4−5+𝐷𝑥+20𝐷0.(5.63) Research of coefficients of 𝐷0, 1𝑆=4+1(𝑥−1)4−5𝑥+2,(5.64)𝑟−(𝑆+𝑆2)=𝑅/𝐵, where 𝑅=(1/2)𝑥4−10𝑥3+19𝑥2−18𝑥+(5/2), 𝐵=(𝑥−1)4(𝑥+2), 𝑣1𝑥01=⎛⎜⎜⎝−1(𝑥)=−𝑅+2𝑆𝐵2𝑥4+10𝑥3−19𝑥25+18𝑥−2−𝑥4+20𝑥3−38𝑥2⎞⎟⎟⎠,−5+36𝑥−1𝑉=2118−1910−2,−536−3820−5(5.65)𝑙2=2𝑙1. Thus, 𝐷0=𝑥−2, 𝜃1=14+1(𝑥−1)4−5+1𝑥+2.𝑥−2(5.66)

Case 2. If 𝜀1=1, 𝜀−2=−1, 𝐸(𝜃)=−1/4 then: 𝑑𝑜𝐷0=9, 𝜃=−14+1(𝑥−1)4−5+𝐷𝑥+20𝐷0.(5.67) It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.

Case 3. If 𝜀1=−1, 𝜀−2=−1, 𝐸(𝜃)=−1/4 then 𝑑𝑜𝐷0=5, 𝜃3=−14−1(𝑥−1)4−5+4𝑥+2+𝐷𝑥−10𝐷0.(5.68) Research of coefficients of 𝐷0, 𝑆=−14−1(𝑥−1)4−5+4𝑥+2,𝑥−1𝑅=−𝑥4+36𝑥3−137𝑥2+168𝑥−88,𝐵=(𝑥−1)4⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝5(𝑥+2),𝑉=−37−168137−3620000−32−15123−266180−3720004−78−162−368219−36200012−138−20921−474254−33100024−212−95−584285−28200040−300−1392⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠.−698312−21(5.69) We consider 𝑉𝑝 the minor 6×6 obtained by column vectors 1,2,6,7,8, and 9. In ℤ/5ℤ𝑉𝑝=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠−220000−222000−12−1−10002−121000102−200−222−1(5.70)det𝑉𝑝=1. Accordingly, row 6 is not linear combination of other row and So no solution 𝐷0.