1. Introduction
By [1, 2] we know that some polynomial differential system(1)u˙=Uu,v,v˙=Vu,v,where U(u,v) and V(u,v) are polynomial functions, can be transformed to an equation of the form(2)dxdt=p0t+p1tx+⋯+pntxn=Pt,x,t,x∈R,where pi(t) are polynomial in cost, sint. The fact that systems with a homogeneous nonlinearity can be transformed to (2) with n=3 has been exploited in a number of previous papers [1–4]. The limit cycles of (1) correspond to 2π-periodic solutions of (2). This fact has been used to facilitate certain computations and to provide some information about the global phase portrait of the system. In this paper we will use the method of Mironenko [5] to construct some differential equations which are equivalent to (2) in the sense of having the same reflecting function.

Now, we simply introduce the concept of the reflecting function.

Consider differential system(3)x′=Xt,x,t∈R, x∈D⊂Rn,which has a continuously differentiable right-hand side and general solution φ(t;t0,x0). For each such system, the reflecting function is defined as F(t,x)≔φ(-t,t,x) [5].

If system (3) is 2π-periodic with respect to t, then Tx≔F-π,x is the Poincaré mapping of (3) over the period [-π,π]. Thus, the solution x=φ(t;-π,x0) of (3) defined on [-π,π] is 2π-periodic if and only if x0 is a fixed point of T(x).

A differentiable function F(t,x) is a reflecting function of system (3) if and only if it is a solution of the Cauchy problem(4)Ftt,x+Fxt,xXt,x+X-t,Ft,x=0,F0,x=x.

If the reflecting functions of two differential systems coincide in their common domain, then these systems are said to be equivalent [5].

If F(t,x) is the reflecting function of (3), then it is also the reflecting function of any system(5)x′=Xt,x+Fx-1Rt,x-R-t,Ft,x,where R(t,x) is an arbitrary vector function such that the solutions of every system above are uniquely determined by its initial conditions. The systems (3) and (5) have the same operators of translation along solutions on the symmetric time interval [-ω,ω], and therefore the initial data x(-ω) of solutions of the boundary value problems of the form Φ(x(ω),x(-ω))=0, where Φ is an arbitrary function, coincide for such systems.

In general, it is very difficult to find out the reflecting function of (3), so to writ out system (5) is difficult, too. How to judge if two systems are equivalent when we do not know their reflecting function? This is a very interesting problem! Mironenko in [5–7] has studied it and obtained some good results.

Lemma 1 (see [<xref ref-type="bibr" rid="B6">6</xref>]).
If continuously differentiable vector functions Δi(t,x) (i=1,2,…,m) are solutions of differential system(6)Δtt,x+Δxt,xXt,x=Xxt,xΔt,x,then all perturbed systems of the form(7)x′=Xt,x+∑i=1mαitΔit,x,where αi(t) are arbitrary continuous scalar odd functions, are equivalent to each other and to system (3).

The theory of reflecting function has been used for studying the qualitative behavior of solutions of differential systems by many authors [5–13].

Now, for a given equation (2), we present a method for constructing other first-order differential equations which are equivalent to (2), that is, to find out the solution of (6) with X(t,x)=P(t,x). However, it is impossible to find all the solutions in most cases. Therefore, we take only polynomial solutions of (6), that is, its solutions of the form(8)Δ=a0t+a1tx+⋯+antxn,where the coefficients ai(t) (i=0,1,2,…,n) are assumed to be differentiable functions on R and an(t) is not to be zero identically.

Bel’skii in papers [8–10] has discussed, respectively, when a first-order polynomial differential equation is equivalent to a linear equation (n=1) and Riccati equation (n=2) and Abel equation (n=3). In this paper, we will in detail discuss when is a first-order equation equivalent to (2) with n≥3. We have obtained the sufficient conditions for (6) has solution in form of (8) and we give the explicit expressions of Δ(t,x). At the same time, in addition, we also find out some nonpolynomial solutions of (6). The results of this paper generalize the results of Bel’skii [8–10], and the method of proof in this paper is more simple and efficient than in the previous papers.

2. Main Results
Now, we consider (2) and assume that pn(t) is not to be zero identically in R and n>1. By Bel’skii [8], we know that if pn≠0, then the degree of polynomial Δ(t,x) is equal to n. In the following, we suppose that n≥3 and pn(t)≠0.

Theorem 2.
For (2), suppose that n≥3 and pn(t)≠0. Then the function Δ(t,x) of the form (8) is a solution of (6), if and only if(9)a0′=p1a0-a1p0,(10)ak′=k+1pk+1a0-ak+1p0+k-1pka1-p1ak,k=1,2,…,n-2,(11)ai=pipnan,i=2,3,…,n-1,(12)a1=p1pnan+an′n-1pn,(13)a0=p0pnan+annpnpn-1pn′+an′pn-1nn-1pn2.Therefore, Δ can be represented in the form(14)Δ=anpnPt,x+an′n-1pnx+pn-1npn+annpnpn-1pn′.

Proof.
Let us substitute the function Δ(t,x) of the form (8) into (6); we get(15)a0′+a1′x+⋯+an′xn+a1+2a2x+⋯+nanxn-1·p0+p1x+⋯+pnxn=p1+2p2x+⋯+npnxn-1·a0+a1x+⋯+anxn.Equating the coefficients of the like power of x, we obtain(16)ak′=∑i=1k+1ipiak+1-i-aipk+1-i,k=0,1,2,…,2n-1,where ak=0 and pk=0, when k>n.

From (16) when k=2n-1 we get nanpn=npnan. Taking k=2n-2 in (16), we have an-1=(pn-1/pn)an. Similarly for k=2n-3, 2n-4,…,n+1, we get(17)ai=pipnan,i=2,3,…,n-1;that is, relation (11) is true.

Substituting (11) into (16) and simplifying we get (10).

Using relation (11) and taking k=n in (16), we get(18)an′=n-1a1pn-p1an;it implies relation (12).

Using relation (12) and taking k=n-1 in (16), we have(19)an-1′=npna0-p0an+n-2n-1pn-1pnan′;substituting an-1=(pn-1/pn)an into the above, we get that relation (13) is held.

Substituting (11)–(13) into (8), we get(20)Δt,x=anpnp0+p1x+⋯+pnxn+an′n-1pnx+pn-1npn+annpnpn-1pn′=anpnPt,x+an′n-1pnx+pn-1npn+annpnpn-1pn′,which completes the proof.

We rewrite P(t,x) as follows:(21)Pt,x=pnx+pn-1npnn+δn-2x+pn-1npnn-2+⋯+δ1x+pn-1npn+δ0,in which(22)δk=1k!pn∂kPt,x∂xkx=-pn-1/npn,k=0,1,…,n-2, δn-1≡0.

Lemma 3.
If δi=0, i=k+1,k+2,…,n-2, k≥0, then(23)δk=pkpn-Cnkpn-1npnn-k,(24)pk+1pn=Cnk+1pn-1npnn-k-1.

Proof.
As δi=0, i=k+1,k+2,…,n-1, so (21) goes to(25)Pt,xpn=x+pn-1npnn+δkx+pn-1npnk+δk-1x+pn-1npnk-1+⋯+δ0,equating the coefficients of xk and xk+1; from here we get relations (23) and (24).

Theorem 4.
Suppose that δi=0, i=k+1,k+2,…,n-2, δk≠0(1<k<n-1), and(26)m+1pm+1μk0+m-1pmμk1=pmpn′-n-1n-kpmpnδk′δk,m=2,3,…,k-1,(27)a0′=p1a0-a1p0,(28)a1′=2p2a0-2p0a2,(29)am=pmpnan,m=2,3,…,n-1,where(30)μk0=1npnpn-1pn′-pn-1nn-kpn2δk′δk,μk1=-δk′n-kpnδk,(31)a0=p0pnan+μk0an,(32)a1=p1pnan+μk1an,(33)an=δk-n-1/n-k.Then the function Δ(t,x) in (8) is a solution of (6); that is, (2) is equivalent to equation(34)x′=Pt,x+αtΔt,x,where α(t) is an arbitrary continuous odd function.

Proof.
For k≤m≤n-2, by Lemma 3, substituting (12) and (13) into (10), we obtain(35)am′=m+1pm+1a0-am+1p0+m-1pma1-amp1=m+1pm+1npnpn-1pn′an+pm+1nn-1pn-1pn2an′+m-1pman′n-1pn.Substituting (24) into this relation, we obtain(36)am′=m+1n-mCnm+1pn-1npnn-m′an+m+1n-1Cnm+1pn-1npnn-man′+m-1n-1pmpnan′.By (11) and (23), we get(37)anδm′+n-mn-1δman′=0.As δm=0 (m=k+1,k+2,…,n-1), (37) is an identity when m=k+1,k+2,…,n-1. For m=k, solving (37), we have(38)an=δk-n-1/n-k.Substituting (38) into (11)–(13), we obtain (29)–(32).

On the other hand, by relation (10), for 1<m<k, we have(39)am′=m+1pm+1a0-am+1p0+m-1pma1-amp1.Substituting (29)–(32) and (37) into the above, we obtain(40)pmpn′an-n-1n-kpmanδk′pnδk=m+1pm+1μk0an+m-1pmμk1an,which deduces that relation (26) is true. By Theorem 2 and in (10) taking k=1 we get that relations (27) and (28) are held.

By Lemma 1 and Theorem 2, the proof is completed.

Theorem 5.
Supposing that n>3,(41)δi=0,i=2,3,…,n-2,δ1=p1pn-Cn1pn-1npnn-1≠0,η′an+nn-1ηan′=0,ak=pkpnan,k=2,3,…,n-1(42)a0=p0pn-pn-1δ1npn+1npnpn-1pn′an+λ0pn-1npn,(43)a1=npn-1npnn-1an+λ0,where(44)η≔p0pn-pn-1npnn+1pnpn-1npn′-pn-1npnδ1,(45)an=e-n-1∫δ1pndtλ1+λ0n-1∫pnen-1∫δ1pndtdt,where λ0 and λ1 are arbitrary constants. Then the function Δ(t,x) of the form (8) is a solution of (6); that is, (2) is equivalent to (34).

Proof.
As δm=0, m=2,3,…,n-1 and δ1≠0, by Lemma 3, we have(46)δ1=p1pn-Cn1pn-1npnn-1,p2pn=Cn2pn-1npnn-2.Using δm=0 (m=2,3,…,n-1), it is not difficult to check that (37) is an identity when m=2,3,…,n-2. For m=1, in relation (10) taking k=1 and computing, it follows that(47)a1′=nanpn-1npnn-1′,which implies that(48)a1=npn-1npnn-1an+λ0.Combining relation (12), it yields(49)an′=-n-1δ1pnan+n-1λ0pn;solving this equation we get the expression (45).

Substituting (49) into (13) implies (42). Substituting (42)–(45) into equation a0′=p1a0-p0a1 and simplifying we get η′an+(n/(n-1))ηan′=0. By Theorem 2, the proof is finished.

Similarly, we get the following.

Theorem 6.
Supposing that δm=0, m=1,2,…,n-2 and δ0=p0/pn-(pn-1/npn)n≠0,(50)δ0+1npnpn-1pn′′an+nn-1an′δ0+1npnpn-1pn′=0,ak=pkpnan,k=2,3,…,n-1,a0=p0pnan+annpnpn-1pn′+pn-1npnλ0,a1=p1pnan+λ0,where(51)an=n-1λ0∫pndt+λ1;λ0 and λ1 are arbitrary constants. Then (2) is equivalent to (34).

Theorem 7.
Supposing that δm=0, m=0,1,2,…,n-2,(52)anpn-1pn′+λ0pn-1=λ2npn,a0=p0pnan+λ2,a1=p1pnan+λ0,ak=pkpnan,k=2,3,…,n-1,where (53)an=n-1λ0∫pndt+λ1;λi (i=0,1,2) are arbitrary constants. Then (2) is equivalent to (34).

By the literature [5], we know the following.

Corollary 8.
If all the conditions of one of the above four theorems (Theorems 4–7) are satisfied and (2) is 2π-periodic in t, then the initial data x(-π) of solutions of the boundary value problems of the form Φ(x(π),x(-π))=0, where Φ is an arbitrary function, coincide for systems (2) and (34).

Example 9.
The equation(54)x′=x-sint4+costhas the only one 2π-periodic solution x=sint. For this equation, we have(55)p0=cost+sin4t,p1=-4 sin3t,p2=6 sin2t,p3=-4sint,p4=1,δ3=δ2=δ1=0,δ0=cost.By Theorem 6, we get(56)a4=3λ0t+λ1,where λ0 and λ1 are constants:(57)a3=-4a4sint,a2=6a4 sin2t,a1=-4a4 sin3t+λ0,a0=a4 sin4t-λ0sint.As(58)δ0+14p4p3p4′=cost-cost=0,then(59)δ0+14p4p3p4′′a4+43a4′δ0+14p4p3p4′=0;that is, the first condition of Theorem 6 is satisfied. Thus(60)Δ0=3λ0t+λ1x-sint4+λ0x-sint.Taking λ0=0 and λ1=1, we get(61)Δ1=x-sint4.Taking λ0=1 and λ1=0, we have(62)Δ2=3tx-sint4+x-sint.Therefore, (54) is equivalent to equation(63)x′=x-sint4+cost+α0tΔ0+α1tΔ1+α2tΔ2,where αi(t) (i=0,1,2) are arbitrary continuously odd functions. By Corollary 8, (63) has only one solution such that x(-π)=x(π).

Now, taking n=3 in the above theorems, we get the following corollaries.

Corollary 10.
Suppose that p3≠0; then(64)ζ′a3+32ζa3′=0,ζ≔p0p3-p23p33+1p3p23p3′-p23p3σ1,a0=p0p3-p23p3σ1a3+a33p3p2p3′+p23p3λ0,a1=p223p32a3+λ0,a2=p2p3a3,where(65)a3=e-2∫σ1p3dtλ1+2λ0∫p3e2∫a3σ1dtdt;λ0 and λ1 are arbitrary constants; σ1=p1/p3-p22/3p32. Then the Abel equation(66)x′=p0+p1x+p2x2+p3x3is equivalent to equation(67)x′=p0+p1x+p2x2+p3x3+αta0+a1x+a2x2+a3x3,where α(t) is an arbitrary continuously odd function.

Corollary 11.
Suppose that p3≠0, σ1=p1/p3-p22/3p32=0, and σ0=p0/p3-p23/27p33≠0;(68)σ0+13p3p2p3′′a3+32a3′σ0+13p3p2p3′=0,a0=p0p3a3+a33p3p2p3′+p23p3λ0,a1=p1p3a3+λ0,a2=p2p3a3,where(69)a3=2λ0∫p3dt+λ1;λ0 and λ1 are arbitrary constants. Then Abel equation (66) is equivalent to (67).

Corollary 12.
Suppose that p3≠0, σ1=p1/p3-p22/3p32=0, and σ0=p0/p3-p23/27p33=0;(70)a3p2p3′+λ0p2=3λ2p3,a0=p0p3a3+λ2,a1=p1p3a3+λ0,a2=p2p3a3,where(71)a3=2λ0∫p3dt+λ1;λi (i=0,1,2) are arbitrary constants. Then Abel equation (66) is equivalent to (67).

Example 13.
Abel equation(72)x′=x3+6x2+3x-10=x+23-9x+2has three constant solutions x1(t)=-2, x2(t)=-5, and x3(t)=1. For this equation, we have p0=-10, p1=3, p2=6, p3=1, and σ1=-9.

Thus(73)ζ=p0p3-p23p33+1p3p23p3′-p23p3σ1=-18+18=0,and ζ′a3+(3/2)ζa3′=0.

Solving a3′=18a3+2λ0, we get(74)a3=λ1e18t-19λ0,where λ0, λ1 are arbitrary constants. Therefore(75)a1=3λ1e18t-19λ0+9λ1e18t,a0=-10λ1e18t-19λ0+18λ1e18t,Δ0=λ1e18t-19λ0x3+6x2+3x-10+9λ1e18tx+2.Taking λ1=0 and λ1=-9, we obtain(76)Δ1=x3+6x2+3x-10.Taking λ0=0 and λ1=1, we get(77)Δ2=e18tx3+6x2+12x+8.Consequently, by Lemma 1 and Corollary 10, Abel equation (72) is equivalent to equation(78)x′=x3+6x2+3x-10+α0tΔ0+α1tΔ1+α2tΔ2,where αi(t) (i=0,1,2) are arbitrary continuously odd functions. By Corollary 8, (78) has only three solutions such that x(-a)=x(a) (a is a nonzero constant).

Remark 14.
The conclusions of the above three corollaries have been proven by Bel’skii and Mironenko in [9].

Because (6) is a linear equation, if Δi (i=1,2,…,m) are solutions of (6), then Δ=∑i=1mCiΔi also is solution of (6), where Ci (i=1,2,…,m) are arbitrary constants. In the following, we will find out some solutions Δi of (6) in the other form.

Theorem 15.
For the first-order differential equation x′=X(t,x)(t,x∈R), if Δi (i=1,2,…,m) are solutions of (6), then Δ=Δk1Δk2⋯Δkm, ∑i=1mki=1, is a solution of (6), too.

Proof.
As(79)∂Δi∂t+∂Δi∂xXt,x-∂X∂xΔi=0,i=1,2,…,mand k1+k2+⋯+km=1,(80)∂Δ∂t+∂Δ∂xXt,x-∂X∂xΔ=k1Δ1k1-1Δ2k2⋯Δmkm∂Δ1∂t+∂Δ1∂xXt,x-∂X∂xΔ1+k2Δ1k1Δ2k2-1⋯Δmkm∂Δ2∂t+∂Δ2∂xXt,x-∂X∂xΔ2+⋯+kmΔ1k1Δ2k2⋯Δmkm-1∂Δm∂t+∂Δm∂xXt,x-∂X∂xΔm=0.

Thus, the proof is completed.

By Example 9 and Theorem 15 we get that equation(81)x′=x-sint4+costis also equivalent to equation(82)x′=x-sint4+cost+∑i=1mαitΔ0ki0Δ1ki1Δ21-ki0-ki1,where αi(t) (i=1,2,…,m) are arbitrary continuously odd functions, ki0, ki1 are arbitrary constants, and Δi (i=0,1,2) are the same as in Example 9. This equation has only one solution such that x(-π)=x(π).

Using Theorem 15 and Corollary 10 we get the following.

Corollary 16.
Supposing that p3≠0,(83)ζ≔p0p3-p23p33+1p3p23p3′-p23p3σ1=0.Then Abel equation(84)x′=p0+p1x+p2x2+p3x3is equivalent to equation(85)x′=p0+p1x+p2x2+p3x3+α1Δ+α2Δ1+α3Δ2+α4Δ1kΔ21-k,where αi(t) (i=1,2,3,4) are arbitrary continuously odd functions and k is a constant:(86)Δ=a0+a1x+a2x2+a3x3,in which(87)a0=p0p3-p23p3σ1a3+a33p3p2p3′+p23p3λ0,a1=p223p32a3+λ0,a2=p2p3a3,a3=e-2∫σ1p3dtλ1+2λ0∫p3e2∫a3σ1dtdt;λ0, λ1 are arbitrary constants, and σ1=p1/p3-p22/3p32. Consider(88)Δi=a0i+a1ix+a2ix2+a3ix3,i=1,2aj1=ajλ0=1,λ1=0,aj2=ajλ0=0,λ1=1,j=0,1,2,3.

Theorem 17.
For equation(89)x′=βtfx,t,x∈R, fx≠0,(90)Δ=fxϕ∫1fxdx-∫βtdtis a solution of (6). Thus, (89) is equivalent to equation(91)x′=βtfx+αtΔt,x,where α(t) is an arbitrary continuously odd function, ϕ(u) is an arbitrary differentiable function, and β(t) is a continuous function.

This result is easy to be proven.

Obviously, from Theorem 17, we see that if f(x) is a polynomial of x, then the corresponding equation (6) has at least one polynomial solution Δ=f(x). This is implied by taking ϕ(u)=1 in (90).

By Theorem 17, (72) is also equivalent to(92)x′=x3+6x2+3x-10·1+αtϕx+22-9x+22e-18t.If in (92) we put ϕ(u)=1/(1+u2), then we obtain(93)x′=x3+6x2+3x-10·1+αtx+24e36tx+24e36t+x+22-92,where α(t) is an arbitrary continuously odd function. This equation has only three solutions such that x(-a)=x(a) (a is a nonzero constant).