Liouvillian integrability of some quadratic Liénard polynomial differential systems

Abstract

In this paper we characterize all the Liouvillian first integrals of some quadratic Liénard polynomial differential systems. In addition we also find Liénard polynomial differential systems which admit a transcendental first integral.

1 Introduction and statement of the main results

A Liénard polynomial differential system is a system of the form

$$\begin{aligned} x' =y, \quad y' = -\,g_1(x) -f_1(x) y \end{aligned}$$

(1)

where $g_1,f_1 \in \mathbb {C}[x]$. It is named like this after the work of the French physicist Alfred-Marie Liénard which obtained it in [13] as a second order differential equation. This system has been intensively studied since the development of radio and vacuum technology, mainly due to the fact that this system can be used to model many phenomena such as oscillating circuits (where it appears as the Raylegh or van der Pol equation), the electric heart activity in cardiology, the neurons activity in neurology and it also appears in biology, mechanics, seismology, chemistry, physics and cosmology (see for instance, [2, 9,10,11, 15, 16, 18, 19, 22,23,24] and the references therein).

There are many results regarding the dynamics of the Liénard polynomial differential systems such as the existence of limit cycles, or of invariant algebraic curves, or of analytic first integrals, to cite just a few. However, almost nothing is known regarding its Liouvillian integrability. The main issue of this paper is to fill in this gap in the case of quadratic Liénard polynomial differential systems.

More concretely, we consider the quadratic Liénard polynomial differential system of the form

$$\begin{aligned} \dot{x} =y,\quad \dot{y}=a_{00} +a_{10} x +a_{20} x^2 + y(a_{01} +a_{11} x), \end{aligned}$$

(2)

where $a_{ij} \in \mathbb {C}$ for $i=0,1,2$; $j=0,1$ and the dot denotes derivative with respect to the time $t \in \mathbb {R}$. We assume that it is quadratic, i.e., $|a_{20}|^2 + |a_{11}|^2 \ne 0$ and that $|a_{00}|^2 +|a_{10}|^2 +|a_{20}|^2 \ne 0$ (otherwise with a reparameterization of time the system becomes linear).

We also assume that:

• (H1) either $a_{20} a_{11}=0$, or if $a_{20} a_{11} \ne 0$ then $a_{10} a_{11}^2 -2 a_{01} a_{11} a_{20} -4 a_{20}^2 \ne 0$.

The first main result of the paper is the following in which we are able to find a transcendental first integral for system (2) as stated in the following theorem.

Theorem 1

If $a_{20}=a_{10}=0$, system (2) is transcendentally integrable with a first integral given by

$$\begin{aligned} I= T\frac{T x \, _0{\tilde{F}}_1\left( ;\frac{4}{3};-\frac{T^3}{18}\right) +6 \, _0{\tilde{F}}_1\left( ;\frac{1}{3};-\frac{T^3}{18}\right) }{ \, _0{\tilde{F}}_1\left( ;\frac{2}{3};-\frac{T^3}{18}\right) (T x+2)+6 \, _0{\tilde{F}}_1\left( ;-\frac{1}{3};-\frac{T^3}{18}\right) }, \end{aligned}$$

where $_0{\tilde{F}}_1$ is the regularized confluent hypergeometric function which is a transcendental function (see [1]) and

$$\begin{aligned} T= Y- \frac{X^2}{2}, \quad X=(a_{00} a_{11})^{1/3} (a_{01} + a_{11} x ), \quad Y= \frac{1}{a_{00}^{4/3}a_{11}^{1/3}} y. \end{aligned}$$

Theorem 1 is proved in Sect. 3. The second main result of the paper characterizes the Liouvillian first integrals of system (2) (see the definition of Liouvillian integrability in Sect. 2) under the assumptions that it is quadratic and hypothesis (H1) holds. We emphasize that the case in which hypothesis (H1) does not hold the problem of characterizing all the Liouvillian first integrals of system (2) remains open.

Theorem 2

System (2) with $|a_{20}|^2 +|a_{10}|^2 \ne 0$, $|a_{20}|^2 +|a_{11}|^2 \ne 0$ and satisfying hypothesis (H1) is Liouvillian integrable if and only if either $a_{20} \ne 0$ and $a_{11}=a_{01}=0$, or $a_{20}=0$, $a_{11}a_{10} \ne 0$ and $a_{01} a_{10}=a_{00}a_{11}$. In the first case a Liouvillian first integral is

$$\begin{aligned} I=a_{00} x +\frac{a_{10}}{2} x^2 -\frac{1}{2} y^2+ \frac{a_{20}}{3} x^3, \end{aligned}$$

while in the second case, a Liouvillian first integral is

$$\begin{aligned} I=(a_{10}+a_{11}y) \exp \left( \frac{a_{11}^2 \big (\frac{a_{00}}{a_{10}}+x \big )^2 -2 (a_{11}y +a_{10})}{2a_{10}}\right) . \end{aligned}$$

Theorem 2 is proved in Sect. 5. To prove it we will need to characterize the invariant algebraic curves of system (2) with $|a_{20}|^2 +|a_{10}|^2 \ne 0$ (see the definition of invariant algebraic curve in Sect. 2). The result is important by its own because the interest on the invariant algebraic curves of polynomial Liénard systems has a long history. If we write system (1) in the form

$$\begin{aligned} \dot{x} =y, \quad \dot{y} =-\,g_n(x) -y f_m(x) \end{aligned}$$

(3)

where $f_m$ and $g_n$ are polynomials of degrees m and n, respectively, and $f_m g_n \not \equiv 0$ then Odani in [17], showed that system (3) does not have invariant algebraic curves when $n \le m$ and $g_n/f_m \not \equiv$ constant. In [12], Hayashi stated that if $n=m+1$ then the only possible invariant algebraic curves of system (3) are of the form $F(x,y)=0$ with $F(x,y)=y-P(x)$, where P(x) satisfy certain conditions. However it was proved recently by Demina [8] that the result stated in Hayashi is wrong in general. We will show in Sect. 2 that in the case of quadratic systems the result in [12] is indeed right. Further research on invariant algebraic curves of Liénard polynomial differential systems has been done in [6]. However, all of them are partial results and there are no results that characterize all the invariant algebraic curves for the case in which (3) is quadratic. This is the content of the following result.

Theorem 3

The unique irreducible invariant algebraic curves of system (2) satisfying $|a_{20}|^2 +|a_{11}|^2 \ne 0$, $|a_{10}|^2 +|a_{20}|^2 \ne 0$ and hypothesis (H1) are:

• (a) $a_{10}+a_{11} y=0$ if $a_{20}=0$, $a_{11} \ne 0$ and $a_{01} a_{10}=a_{00}a_{11}$. The cofactor is $a_{01} + a_{11} x$;

• (b) $-\frac{a_{01} a_{11}a_{20} +a_{20}^2-a_{10} a_{11}^2}{a_{11}^3} +\frac{a_{20}}{a_{11}} x+y=0$ if $a_{20} a_{11} \ne 0$ and $a_{00} =\frac{(a_{01}a_{11}+a_{20})(a_{10} a_{11}^2 -a_{01} a_{11}a_{20}-a_{20}^2)}{a_{11}^4}$. The cofactor is $\frac{a_{01} a_{11} + a_{20} + a_{11}^2 x}{a_{11}}$;

• (c) $- \frac{a_{00}a_{11}}{2a_{20}} -\frac{a_{01} a_{11} +2 a_{20}}{a_{11}} x+y-\frac{a_{11}}{2} x^2=0$ if $a_{20}a_{11} \ne 0$ and $a_{10} =\frac{2(a_{01}a_{11}a_{20}+2 a_{20}^2)}{a_{11}^2}$. The cofactor is $-\frac{2 a_{20}}{a_{11}}$;

• (d) $a_{00} x +\frac{a_{10}}{2} x^2 -\frac{1}{2} y^2+ \frac{a_{20}}{3} x^3=k$ for any $k\in \mathbb {C}$ if $a_{20} \ne 0$ and $a_{11}=a_{01}=0$. The cofactor is zero;

• (e)
  $$\begin{aligned} \begin{aligned}&\frac{ (6 a_{01}^2 - 25 a_{10}) (6 a_{01}^2 + 25 a_{10})^2 }{375000 a_{01}a_{20}} + \frac{12 a_{01}^4-100 a_{01}^2 a_{10}-625 a_{10}^2}{2500 a_{01}} x -\frac{6 a_{01}^2 + 25 a_{10}}{125 } y\\&-\,\frac{(2 a_{01}^2 + 25 a_{10}) a_{20} }{50 a_{01}} x^2 -\frac{2 a_{20}}{5 } x y + \frac{a_{20}}{2 a_{01}} y^2 -\frac{a_{20}^2}{3a_{01}} x^3 =0 \end{aligned} \end{aligned}$$

  if $a_{20} a_{01} \ne 0$, $a_{11}=0$ and $a_{00}= -\,\frac{(6 a_{01}^2 - 25 a_{10}) (6 a_{01}^2 + 25 a_{10})}{2500 a_{20}}$. The cofactor is 6 / 5.

Theorem 3 is proved in Sect. 4. Note that the invariant algebraic curves in the case $a_{20} \ne 0$ and $a_{11}=a_{01}=0$ [see statement (d)], or in the case $a_{20}=0$, $a_{11} \ne 0$ and $a_{01} a_{10}=a_{00}a_{11}$ [see statement (a)] follow directly from Theorem 2, taking into account the form of the first integrals. In the paper we also introduce in Sect. 2 some definitions and results that we may use during the proof of the main results.

2 Preliminary results

Consider the system

$$\begin{aligned} x'=P(x,y), \quad y'=Q(x,y) \end{aligned}$$

(4)

with $P,Q \in \mathbb {C}[x,y]$ (here $\mathbb {C}[x,y]$ denotes the ring of polynomials in the variables x, y with coefficients in $\mathbb {C}$) with $\text {degree} \, (P,Q) \le m$. We recall that a non-constant function $I :\mathbb {C}^2 \rightarrow \mathbb {C}$ is a first integral of system (4) on an open subset $D \subset \mathbb {C}^2$ if I(x(t), y(t)) is constant for all values of t such that the solution (x(t), y(t)) of (4) is defined on D. We say that Eq. (2) is Liouvillian integrable if there exists a Liouvillian first integral I. Furthermore, we say that Eq. (2) is transcendentally integrable if there exists a transcendental first integral I.

Given an irreducible polynomial, g(x, y), we have that $g(x,y)=0$ is an invariant algebraic curve for system (4) if there exists a polynomial $K=K(x,y)$ of degree at most $m-1$, called the cofactor of g such that

$$\begin{aligned} P(x,y) \frac{\partial g}{\partial x} + Q(x,y) \frac{\partial g}{\partial y} =K g. \end{aligned}$$

(5)

It is well known that to determine all the invariant algebraic curves of a polynomial differential system, it is enough to know the irreducible ones.

The first result based on the previous works of Seidenberg [20] were stated and proved in [7]. In the next theorem we included only the results from [7] that will be used in the paper.

Theorem 4

Let $g(x,y)=0$ be an invariant algebraic curve of a planar system with corresponding cofactor K(x, y). Assume that $p=(x_0,y_0)$ is one of the critical points of the system. If $g(x_0,y_0) \ne 0$, then $K(x_0,y_0)=0$. Moreover, assume that $\lambda$ and $\mu$ are the eigenvalues of such critical point. If either $\mu \ne 0$ and $\lambda$ and $\mu$ are rationally independent or $\lambda \mu < 0$, or $\mu =0$, then $K(x_0,y_0) \in \{\lambda ,\mu , \lambda +\mu \}$.

In [12], Hayashi, the author characterized the irreducible invariant algebraic curves of all systems of the form (1) such that $\text {deg} (g_1)=\text {deg} (f_1) +1$. More precisely he proved the following result

Theorem 5

System (1) with $\text {deg} (g_1)=\text {deg} (f_1) +1$ has an invariant algebraic curve $g(x,y)=0$ if and only if $g(x,y)= y -P(x)$ where P(x) satisfies

$$\begin{aligned} g_1(x)=-\,(f_1(x)+P'(x)) P(x) \end{aligned}$$

(6)

and

• (1) either P(x) has degree one;

• (2) or P(x) is such that $P(x) +\int f_1(x) \, d x$ is a polynomial of degree one.

It was stated by Demina in [8] that the result stated in Proposition 5 is wrong in general and she gives a counterexample with degree three. We will show that for quadratic polynomial Liénard differential systems satisfying hypothesis (H1), Proposition 5 is right (the case in which quadratic polynomial Liénard differential systems do not satisfy hypothesis (H1) is still open). This is the content of the Theorem 7 below. To state it we will use the following result in [8]. Consider system (2) as in (1). Then

$$\begin{aligned} f_1= -\,a_{01} -a_{11}x, \quad g_2=-\,a_{00} -a_{10} x -a_{20} x^2 \end{aligned}$$

with $a_{20}a_{11} \ne 0$. Define

$$\begin{aligned} \rho = -\,\frac{2a_{20}}{a_{11}}, \quad h(x)=\int (f_1(x)+\rho )\, d x =-\,a_{01} x -\frac{2 a_{20}}{a_{11}} x -\frac{a_{11}}{2} x^2, \end{aligned}$$

and consider system the change of variables

$$\begin{aligned} s= x, \quad z=y + h(x). \end{aligned}$$

In the new variables (s, z) system (2) becomes

$$\begin{aligned} \begin{aligned} \dot{s}&= z+\frac{a_{01} a_{11}+2 a_{20}}{a_{11}} s +\frac{a_{11}}{2} s^2, \\ \dot{z}&= a_{00} -\frac{2 a_{20}}{a_{11}} z+ \frac{a_{10} a_{11}^2 - 2 a_{01} a_{11} a_{20} - 4 a_{20}^2}{a_{11}^2}s. \end{aligned} \end{aligned}$$

(7)

The following proposition was proved in [8]

Proposition 6

Any invariant algebraic curve of system (7) has degree either 0 or 2 in the variable s.

We state our main result when $a_{20} a_{11} \ne 0$ showing that the results in [12] are right when we are under the assumptions (H1).

Theorem 7

The unique irreducible invariant algebraic curves of system (2) with the assumptions $a_{20} a_{11} \ne 0$ and $a_{10}a_{11}^2 -2 a_{01} a_{11} a_{20} -4 a_{20}^2 \ne 0$ are:

• (i) $-\frac{a_{01} a_{11}a_{20} +a_{20}^2-a_{10} a_{11}^2}{a_{11}^3} +\frac{a_{20}}{a_{11}} x+y=0$ if $a_{20} a_{11} \ne 0$ and $a_{00} =\frac{(a_{01}a_{11}+a_{20})(a_{10} a_{11}^2 -a_{01} a_{11}a_{20}-a_{20}^2)}{a_{11}^4}$. The cofactor is $\frac{a_{01} a_{11} + a_{20} + a_{11}^2 x}{a_{11}}$;

• (ii) $- \frac{a_{00}a_{11}}{2a_{20}} -\frac{a_{01} a_{11} +2 a_{20}}{a_{11}} x+y-\frac{a_{11}}{2} x^2=0$ if $a_{10} =\frac{2(a_{01}a_{11}a_{20}+2 a_{20}^2)}{a_{11}^2}$. The cofactor is $-\frac{2 a_{20}}{a_{11}}$.

These invariant algebraic curves are precisely the ones provided using Proposition 5 in [12].

Proof of the theorem

Let F(x, y) be an invariant algebraic curve of system (2). It is clear that there is a one-to-one correspondence between invariant algebraic curves F(x, y) of system (2) and invariant algebraic curves G(s, z) of system (7), so we will work with system (7). In view of Proposition 6 the invariant algebraic curve G(s, z) has degree either 0 or 2 in the variable s. Moreover it was proved in [14] that the cofactor of the invariant curve F(x, y) cannot depend on y and so the cofactor of G(s, z) cannot depend on z. So, we can write it as $K(s,z)=k_0 + s k_1$ with $k_i \in \mathbb {C}$. We consider both cases separately.

Case 1: G(s, z) has degree 0 ins. We thus have that $G(s,z)=G(z)$ and G(z) must satisfy

$$\begin{aligned} \left( a_{00} -\frac{2 a_{20}}{a_{11}} z+ \frac{a_{10} a_{11}^2 - 2 a_{01} a_{11} a_{20} - 4 a_{20}^2}{a_{11}^2}s\right) \frac{d G(z)}{d z}= (k_0 + k_1 s) G(z). \end{aligned}$$

Solving this linear differential equation we get

$$\begin{aligned} G(z)=\alpha (a_{00} a_{11}^2 + (a_{10} a_{11}^2 s -2 a_{01} a_{11} a_{20} -4 a_{20}^2)s -2 a_{11} a_{20} z)^{-a_{11} (k_0+k_1 s)/(2 a_{20})}, \quad \alpha \in \mathbb {C}{\setminus } \{0\}. \end{aligned}$$

Since G cannot depend on s we must have

$$\begin{aligned} a_{10} a_{11}^2 s -2 a_{01} a_{11} a_{20} -4 a_{20}^2=0 \quad \text {and} \quad k_1=0. \end{aligned}$$

Moreover since G must be irreducible then $k_0=-\,2 a_{20}/a_{11}$. Now passing to the variables (x, y), this provides the invariant algebraic curve in statement (ii).

Case 2: G(s, z) has degree 2 ins. We can write $G(s,z)=P_2(z) s^2 +P_1(z) s +P_0(z)$ where $P_2(z) \ne 0$. Clearly G(s, z) satisfies

$$\begin{aligned} \begin{aligned}&\left( z+\frac{a_{01} a_{11}+2 a_{20}}{a_{11}} s +\frac{a_{11}}{2} s^2\right) (2 s P_2(z) +P_1(z))\\&\quad + \left( a_{00} -\frac{2 a_{20}}{a_{11}} z+ \frac{a_{10} a_{11}^2 - 2 a_{01} a_{11} a_{20} - 4 a_{20}^2}{a_{11}^2}s\right) (P_2'(z) s^2 + P_1'(z) s +P_0'(z))\\&\quad \quad =(k_0 +k_1 s) (P_2(z) s^2 +P_1(z) s +P_0(z)). \end{aligned} \end{aligned}$$

(8)

Computing the coefficient of $s^{3}$ in (8) we get

$$\begin{aligned} a_{11}^2 (a_{11} -k_1) P_2(z) + (a_{10} a_{11}^2 -2 a_{01} a_{11} a_{20} -4 a_{20}^2) P_2'(z)=0. \end{aligned}$$

Since by assumptions, $a_{10} a_{11}^2 -2 a_{01} a_{11} a_{20} -4 a_{20}^2 \ne 0$, solving the linear differential equation in $P_2(z)$ we obtain

$$\begin{aligned} P_2(z)=\beta \exp \left( \frac{a_{11}^2 (a_{11}-k_1)z}{a_{10} a_{11}^2 -2 a_{01} a_{11} a_{20} -4 a_{20}^2}\right) , \quad \beta \in \mathbb {C}\setminus \{0\}. \end{aligned}$$

Taking into account that $P_2$ must be a polynomial we get $k_1 =a_{11}$ and $P_2$ must be a constant that we can set equal to 1. Now, computing the coefficient of $s^2$ in (8) we get the linear differential equation

$$\begin{aligned} 2 a_{11} (2 a_{01} a_{11} +4 a_{20} -a_{11}k_0) -a_{11}^3 P_1(z) + 2 (a_{10} a_{11}^2 -2 a_{01} a_{11} a_{20} -4 a_{20}^2) P_1'(z)=0. \end{aligned}$$

Solving it, and using that $P_1(z)$ must be a polynomial we obtain

$$\begin{aligned} P_1(z)=\frac{4 a_{01} a_{11} +8 a_{20} -2 a_{11}k_0}{a_{11}^2}. \end{aligned}$$

Next computing the coefficient of s in (8) we get the linear differential equation

$$\begin{aligned} \begin{aligned}&4 (a_{01} a_{11} + 2 a_{20})^2 - 2 a_{11} k_0 (3 a_{01} a_{11} + 6 a_{20} - a_{11} k_0) + 2 a_{11}^3 z -a_{11}^4 P_0(z) \\&\quad +a_{11} (a_{10} a_{11}^2 -2 a_{01} a_{11} a_{20} -4 a_{20}^2) P_0'(z)=0. \end{aligned} \end{aligned}$$

Solving it, and using that $P_0(z)$ must be a polynomial we obtain

$$\begin{aligned} P_0(z)=\frac{4 a_{01}^2 a_{11}^2 + 8 a_{20}^2 - 6 a_{01} a_{11} (-2 a_{20} + a_{11} k_0) + 2 a_{11} (-6 a_{20} k_0 + a_{11} (a_{10} + k_0^2 + a_{11} z))}{a_{11}^4}. \end{aligned}$$

Finally, the independent term in the variable s in Eq. (8) gives

$$\begin{aligned} \begin{aligned}&\frac{4 (a_{01} a_{11}+a_{20}-a_{11}k_0)}{a_{11}^2} z + \frac{2}{a_{11}^4} (a_{00} a_{11}^3 - (2 a_{01}^2 a_{11}^2 - a_{10} a_{11}^2 - 6 a_{01} a_{11} a_{20} + 4a_{20}^2) k_0 \\&\quad + 3 a_{11} (a_{01} a_{11} +2 a_{20}) k_0^2 -a_{11}^2 k_0^3 =0. \end{aligned} \end{aligned}$$

Solving it in $a_{00}$ and $k_0$ we get

$$\begin{aligned} k_0=\frac{a_{00}a_{11}+a_{20}}{a_{11}}, \quad a_{00} =\frac{(a_{01}a_{11}+a_{20})(a_{10} a_{11}^2 -a_{01} a_{11}a_{20}-a_{20}^2)}{a_{11}^4}. \end{aligned}$$

Now passing to the variables (x, y), this provides the invariant algebraic curve in statement (i). This concludes the proof. $\square$

Now we continue stating auxiliary notions, results and tools that will be used in the proof of Theorem 3.

A polynomial g(x, y) is said to be a weight homogeneous polynomial if there exist $s=(s_1,s_2) \in \mathbb {N}^2$ and $m \in \mathbb {N}$ such that for all $\mu \in \mathbb {R}\setminus \{0\}$,

$$\begin{aligned} g(\mu ^{s_1} x, \mu ^{s_2} y)=\mu ^m g(x,y), \end{aligned}$$

where $\mathbb {R}$ denotes the set of real numbers, and $\mathbb {N}$ the set of positive integers. We shall refer to $s=(s_1,s_2)$ as the weight of g, m as the weight degree and $x=(x_1,x_2) \mapsto (\mu ^{s_1} x, \mu ^{s_2} y)$ as the weight change of variables.

Now we recall the method of characteristic curves for solving linear partial differential equations (see, for instance, Bleecker and Csordas [3]).

Consider the following fist-order linear partial differential equation

$$\begin{aligned} \alpha (x,y) A_x + \beta (x,y)A_y =f(x,y), \end{aligned}$$

(9)

where $A=A(x,y)$, and $\alpha ,\beta ,f$ are $C^1$ maps.

A curve x(t), y(t) in the xy plane is a characteristic curve of (9), if at each point $(x_0,y_0)$ on the curve, the vector $(\alpha (x_0,y_0), \beta (x_0,y_0))$ is tangent to the curve. So, a characteristic curve is a solution of the system

$$\begin{aligned} \frac{dx}{dy}= \frac{\alpha (x,y)}{\beta (x,y)}. \end{aligned}$$

(10)

This ordinary differential equation is known as the characteristic equation of system (9).

Assume that system (10) has a solution in the implicit form $h(x,y)=c_1$ where $c_1$ is an arbitrary constant. We consider the change of variables

$$\begin{aligned} u=h(x,y), \quad v=y \end{aligned}$$

(11)

and write its inverse transformation as $x=p(u,v)$ and $y=q(u,v)$ (of course sometimes the explicit inverse transformation cannot be obtained, or it is not well defined). Then the linear partial differential equation (9) becomes the following ordinary differential equation in v (for fixed u) of the form

$$\begin{aligned} {\overline{\beta }} (u,v) {\overline{A}}_v ={\overline{f}}(u,v) \end{aligned}$$

(12)

where ${\overline{\beta }}, {\overline{A}}$ are $\beta$ and A written in terms of u and v.

If ${\overline{A}}={\overline{A}}(u,v)$ is a solution of (12), then by the transformation (11) we get that

$$\begin{aligned} A (x,y)={\overline{A}} (h(x,y),y) \end{aligned}$$

is a solution of the linear partial differential equation (9). Moreover, the general solution of (12) is also the general solution of (9) when, using (11), we write u and v as a function of x and y.

A function $E=\exp (f/g) \not \in \mathbb {C}$ with coprime polynomials $g,f \in \mathbb {C}[x,y]$ is an exponential factor of (4) if there exists a polynomial $L=L(x,y)$ of degree at most $m-1$, called the cofactor of E such that

$$\begin{aligned} P(x,y) \frac{\partial E}{\partial x} + Q(x,y) \frac{\partial E}{\partial y} =L E. \end{aligned}$$

(13)

It is well-known that if $E=\exp (f/g)$ contains a non-constant polynomial g, then $g=0$ must be an invariant algebraic curve.

A non-constant function $R :\mathbb {C}^2 \rightarrow \mathbb {C}$ is an integrating factor of system (4) on an open subset $D \subset \mathbb {C}^2$ if

$$\begin{aligned} P(x,y) \frac{\partial R}{\partial x} + Q(x,y) \frac{\partial R}{\partial y} =R \Biggl (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \Biggr ). \end{aligned}$$

The problem of finding Liouvillian first integrals of a planar polynomial differential system (4) is reduced to the problem of constructing invariant algebraic curves and exponential factors. We have the following powerful theorems.

Theorem 8

Let $F_j=0$ for $j=1,\ldots ,r$ with $r \in \mathbb {N}_0$ be irreducible invariant algebraic curves of system (4) and $E_k=\exp (g_k/f_k)$ for $k=1,\ldots ,s \in \mathbb {N}_0$ be exponential factors of system (4). The polynomial differential system (4) has a Liouvillian first integral if and only if it has an integrating factor of the form in

$$\begin{aligned} F_1^{d_1}\cdots F_r^{d_r} E_1^{e_1} \cdots E_r^{e_r}, \quad d_1,\ldots ,d_r, e_1 ,\ldots , e_s \in \mathbb {C}. \end{aligned}$$

(14)

Theorem 8 was proved by Singer [21] (see also [4]).

Theorem 9

Under the assumptions of Theorem 8, the function given in (14) is an integrating factor of the polynomial differential system (4) if and only if the following condition holds:

$$\begin{aligned} \sum _{j=1}^r d_j K_j + \sum _{k=1}^s e_k L_k =-\, \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\right) . \end{aligned}$$

Theorem 9 follows from classical Darboux theory of integrability (see [5, 21, 25]).

3 Proof of Theorem 1

Under the assumptions of Theorem 1 system (2) becomes

$$\begin{aligned} \dot{x} =y,\quad \dot{y}=a_{00} + y(a_{01} +a_{11} x), \end{aligned}$$

(15)

with $a_{00} a_{11} \ne 0$. We introduce the change of variables

$$\begin{aligned} x_1 =a_{01} +a_{11} x, \quad y_1=y \end{aligned}$$

and system (15) in these new variables is

$$\begin{aligned} \dot{x}_1 =a_{11} y_1,\quad \dot{y}_1=a_{00} + x_1 y_1. \end{aligned}$$

(16)

Now with the change of variables and parameterization of time in the form

$$\begin{aligned} X= (a_{00}a_{11})^{1/3} x_1, \quad Y=\frac{1}{a_{00}^{4/3}a_{11}^{1/3}} y_1 \end{aligned}$$

system (16) can be written as

$$\begin{aligned} \dot{X}= Y, \quad \dot{Y}= 1 + XY. \end{aligned}$$

We introduce the new variable

$$\begin{aligned} T=Y-\frac{X^2}{2} \end{aligned}$$

and system (16) becomes

$$\begin{aligned} \dot{T}= 1, \quad \dot{X}= T+\frac{X^2}{2}. \end{aligned}$$

A first integral of the above system can be obtained as the constant of motion of the linear equation

$$\begin{aligned} \frac{d X}{d T} = T+\frac{X^2}{2}. \end{aligned}$$

Doing so we get that in fact a first integral is the one given in the statement of Theorem 1. Note that it is clearly transcendental. This concludes the proof of the theorem.

4 Proof of Theorem 3

Note that the case $a_{20} \ne 0$ and $a_{11} \ne 0$ (since we are under the assumptions (H1)) follows directly from Theorem 7. So, we can assume that either $a_{20}=0$ and then $a_{11} \ne 0$, or $a_{11} =0$ and $a_{20} \ne 0$ (otherwise the system is not quadratic). In the first case, if $a_{01} a_{10} = a_{00}a_{11}$ it follows directly from Theorem 2 and in the second case, if $a_{01}=0$ it follows directly again from Theorem 2, taking into account the form of the first integral. Hence, in order to prove Theorem 3 we need to consider the cases: $a_{20} = 0$, $a_{11} \ne 0$, $a_{01} a_{10} \ne a_{00}a_{11}$ and the case $a_{20} \ne 0$, $a_{11}=0$, $a_{01} \ne 0$. We consider both cases in two different cases.

Case 1: $a_{20} = 0$, $a_{11} \ne 0$and$a_{01} a_{10} \ne a_{00}a_{11}$. Since $a_{20}=0$ we have that $a_{10} \ne 0$. We first introduce the new variable

$$\begin{aligned} x_1= x +\frac{a_{00}}{a_{10}}, \quad y_1=y. \end{aligned}$$

In these new variables system (2) becomes

$$\begin{aligned} \dot{x}_1=y_1, \quad \dot{y}_1= a_{10} x_1 +{{\tilde{a}}}_{01} y_1 +a_{11} x_1 y_1, \end{aligned}$$

(17)

where ${{\tilde{a}}}_{01}=(a_{01} a_{10}-a_{00}a_{11})/a_{10}$. In view of the assumptions we have that ${{\tilde{a}}}_{01} \ne 0$. We introduce the change of variables and parametrization of time

$$\begin{aligned} x_2= \frac{a_{11}}{{{\tilde{a}}}_{01}} x_1, \quad y_2= \frac{a_{11}}{{{\tilde{a}}}_{01}^2}y_2, \quad t =\frac{1}{{{\tilde{a}}}_{01}} s \end{aligned}$$

in such a way that system (17) becomes

$$\begin{aligned} x_2 ' = y_2, \quad y_2' = {{\tilde{a}}}_{10} x_2 + y_2 + x_2 y_2, \end{aligned}$$

(18)

where ${{\tilde{a}}}_{10}=a_{10}/{{\tilde{a}}}_{01}$ and the prime means derivative with respect to the new time s. Let $g=g(x_2,y_2)=f(x,y)=0$ be an invariant algebraic curve of system (18) with cofactor ${{\tilde{k}}}_0+{{\tilde{k}}}_1 x_2 +{{\tilde{k}}}_2 y_2$. We introduce the weight-change of variables of the form

$$\begin{aligned} x_2 =\mu ^{-1}X, \quad y_2 = \mu ^{-2} Y, \quad s=\mu \tau . \end{aligned}$$

In this form, system (18) becomes

$$\begin{aligned} \begin{aligned} X'&= Y, \\ Y'&= X Y +\mu Y +\mu ^2 {{\tilde{a}}}_{10} X \end{aligned} \end{aligned}$$

(19)

where the prime denotes derivative in the new time $\tau$. Now we set

$$\begin{aligned} G(X,Y)=\mu ^\ell g(\mu ^{-1} X, \mu ^{-2} Y) \end{aligned}$$

and

$$\begin{aligned} K=\mu ^{-1} {{\tilde{k}}}_2 Y + {{\tilde{k}}}_1 X +\mu {{\tilde{k}}}_0 \end{aligned}$$

where $\ell$ is the highest weight degree in the weight homogeneous components of g in X, Y with weight (1, 2).

We note that $G=0$ is an invariant algebraic curve of system (19) with cofactor K. Indeed

$$\begin{aligned} \frac{d G}{d \tau } =\mu ^{\ell +1} \frac{d g}{d \tau } =\mu ^{\ell +1} ({{\tilde{k}}}_2 y_1 + {{\tilde{k}}}_1 x_1 +{{\tilde{k}}}_0) g = (\mu ^{-1} {{\tilde{k}}}_2 Y +{{\tilde{k}}}_1 X +\mu {{\tilde{k}}}_0) G=KG. \end{aligned}$$

Assume that $G=\sum _{i=0}^j G_i$ where $G_i$ is a weight homogeneous polynomial in X, Y with weight degree $\ell -i$ for $i=0,\ldots ,j$ and $\ell \ge j$. Obviously

$$\begin{aligned} g=G|_{\mu =1}. \end{aligned}$$

From the definition of invariant algebraic curve we have

$$\begin{aligned} \begin{aligned}&Y \sum _{i=0}^j \mu ^i \frac{\partial G_i}{\partial X} +\big ( X Y +\mu Y +\mu ^2 {{\tilde{a}}}_{10} X \big ) \sum _{i=0}^j \frac{\partial G_i}{\partial Y} = (\mu ^{-1} {{\tilde{k}}}_2 Y +{{\tilde{k}}}_1 X +\mu {{\tilde{k}}}_0)\sum _{i=0}^j \mu ^i G_i. \end{aligned}\nonumber \\ \end{aligned}$$

(20)

Equating the terms with $\mu ^{-1}$ in (20) we get

$$\begin{aligned} {{\tilde{k}}}_2 Y G_0=0. \end{aligned}$$

Since $G_0 \ne 0$ (otherwise g would be constant) we must have ${{\tilde{k}}}_2=0$.

Equating now the terms in (20) with $\mu ^0$ we get

$$\begin{aligned} L[G_0]=k_1 X G_0, \end{aligned}$$

(21)

where

$$\begin{aligned} L=Y \frac{\partial }{\partial X} - X Y \frac{\partial }{\partial Y}. \end{aligned}$$

The characteristic equation associated with the first linear partial differential equation of system (21) is

$$\begin{aligned} \frac{d X}{d Y} = \frac{1}{X} \end{aligned}$$

whose general solution is $u=X^2/2- Y$.

According with the method of characteristics we make the change of variables

$$\begin{aligned} u=X^2/2- Y, \quad v= X. \end{aligned}$$

(22)

Its inverse transformation is

$$\begin{aligned} Y= v^2/2-u, \quad X=v. \end{aligned}$$

(23)

Under changes (22) and (23), the first equation in (27) becomes the following ordinary differential equation (for fixed u)

$$\begin{aligned} (v^2/2-u) \frac{d {\overline{G}}_0}{d v} = {{\tilde{k}}}_1 v {\overline{G}}_0 \end{aligned}$$

where ${\overline{G}}_0$ is $G_0$ written in the variables u, v. The above equation has the general solution

$$\begin{aligned} {\overline{G}}_0={\overline{F}}_0 (u) v^{ {{\tilde{k}}}_1}, \end{aligned}$$

where ${\overline{F}}_0$ is an arbitrary smooth function in the variable u. In order that $G_0$ be a weight homogenous polynomial with weight degree m, we must have ${{\tilde{k}}}_1 =j$ for some integer j. Since X and Y have weight degrees 1 and 2 respectively we get that $G_0$ should be of weight degree even, i.e., $\ell =2m$ for some convenient $m \in \mathbb {N}$. So, $G_0$ has the form

$$\begin{aligned} G_0= a_\ell Y^j \big (X^2/2-Y\big )^{m-j}, \quad a_\ell \in \mathbb {C}\setminus \{0\}. \end{aligned}$$

The terms in (20) with $\mu ^1$ satisfy

$$\begin{aligned} L[G_1] + Y \frac{\partial G_0}{Y}={{\tilde{k}}}_1 X G_1 +{{\tilde{k}}}_0 G_0. \end{aligned}$$

Substituting $G_0$ in the above equation and doing some computations we obtain

$$\begin{aligned} \begin{aligned} L[G_1]&=j X G_1 + {{\tilde{k}}}_0 a_\ell Y^j (X^2/2- Y)^{m-j} - a_\ell j Y^j (X^2/2- Y)^{m-j}\\&\quad + a_\ell (m-j) Y^{j+1} (X^2/2- Y)^{m-j-1}. \end{aligned} \end{aligned}$$

Using the transformations in (22) and (23) and working in a similar way to solve ${\overline{G}}_0$ we get the following ordinary differential equation

$$\begin{aligned} (v^2/2-u)\frac{d {\overline{G}}_1}{d v} =j v {\overline{G}}_1 - a_\ell ({{\tilde{k}}}_0-j) (v^2/2-u)^j u^{m-j} + a_\ell (m-j) (v^2/2-u)^{j+1} u^{m-j-1}. \end{aligned}$$

This is a linear differential equation whose solution is

$$\begin{aligned} \begin{aligned} {\overline{G}}_1&= {{\tilde{G}}}_1(u) (v^2/2-u)^{j} - a_\ell (v^2/2-u)^{j} \left( ({{\tilde{k}}}_0-j) u^{m-j} \int \frac{d v}{v^2/2-u} \, d v - (m-j) \int dv \right) \\&={{\tilde{G}}}_1(u) (v^2/2-u)^{j} + a_\ell ({{\tilde{k}}}_0-j) \sqrt{2} u^{m-j-1/2} (v^2/2-u)^j \text {arctanh} \, \left( \frac{v}{\sqrt{2} \sqrt{u}}\right) \\&\quad + a_\ell (m-j) (v^2/2-u)^j v u^{m-j-1}, \end{aligned} \end{aligned}$$

where ${{\tilde{G}}}_1(u)$ is a smooth function in the variable u. Taking into account that $G_1$ must be a polynomial of degree $2m-1$ we must have ${{\tilde{G}}}_1(u)=0$ and ${{\tilde{k}}}_0=j$. In short

$$\begin{aligned} G_1= a_\ell (m-j) Y^j X (X^2/2-Y)^{m-j-1}. \end{aligned}$$

The terms in (20) with $\mu ^2$ satisfy

$$\begin{aligned} L[G_2] + Y \frac{\partial G_1}{\partial Y} + {{\tilde{a}}}_{10} X \frac{\partial G_0}{\partial Y} = j X G_2 + j G_1. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} L[G_2]&=j X G_2 + a_\ell (m-j) (m-j-1) Y^{j+1} X (X^2/2-Y)^{m-j-2} \\&\quad -{{\tilde{a}}}_{10} a_\ell j Y^{j-1} X (X^2/2-Y)^{m-j} +{{\tilde{a}}}_{10} a_\ell (m-j) Y^{j} X (X^2/2-Y)^{m-j-1}. \end{aligned} \end{aligned}$$

Using the transformations in (22) and (23) and working in a similar way to solve ${\overline{G}}_1$ we get the ordinary differential equation

$$\begin{aligned} \begin{aligned} (v^2/2-u)\frac{d {\overline{G}}_2}{d v}&=j v {\overline{G}}_2 + a_\ell (m-j)(m-j-1) (v^2/2-u)^{j+1} v u^{m-j-2} \\&\quad -\,{{\tilde{a}}}_{10} a_\ell j (v^2/2-u)^{j-1} v u^{m-j} + {{\tilde{a}}}_{10} a_\ell (m-j) (v^2/2-u)^{j} v u^{m-j-1}. \end{aligned} \end{aligned}$$

Solving this linear differential equation we get

$$\begin{aligned} \begin{aligned} {\overline{G}}_2&={{\tilde{G}}}_2(u) (v^2/2-u)^{j} + a_\ell (v^2/2-u)^{j} \left( (m-j)(m-j-1) u^{m-j-2} \int v \, d v \right. \\&\quad \left. -\,{{\tilde{a}}}_{10} j u^{m-j} \int \frac{v}{(v^2/2-u)^{2}} \, d v + {{\tilde{a}}}_{10} (m-j) u^{m-j-1} \int \frac{v}{v^2/2-u} \, d v \right) \\&={{\tilde{G}}}_2(u) (v^2/2-u)^{j} +\frac{a_\ell }{2} (m-j) (m-j-1)(v^2/2-u)^{j} u^{m-j-2} v^2\\&\quad +\,{{\tilde{a}}}_{10} a_\ell j (v^2/2-u)^{j-1} u^{m-j} +{{\tilde{a}}}_{10} a_\ell (m-j)(v^2/2-u)^{j} u^{m-j-1} \log (v^2-2u), \end{aligned} \end{aligned}$$

where ${{\tilde{G}}}_2(u)$ is a smooth function in the variable u. Since $G_2$ must be a polynomial of degree $2m-2$ we must have $j=m$. Then $G_1=0$, $G_0=a_\ell Y^m$, ${{\tilde{k}}}_0=m$ and $G_2={{\tilde{a}}}_{10} a_\ell m Y^{m-1}$.

The terms in (20) with $\mu ^3$ satisfy

$$\begin{aligned} L[G_3] + Y \frac{\partial G_2}{\partial Y} + {{\tilde{a}}}_{10} X \frac{\partial G_1}{\partial Y} = m X G_3 + m G_2. \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned} L[G_3]&=m X G_3 + m^2 {{\tilde{a}}}_{10} a_\ell Y^{m-1} -{{\tilde{a}}}_{10} a_\ell m (m-1) Y^{m-1} = m X G_3 + {{\tilde{a}}}_{10} a_\ell Y^{m-1}. \end{aligned} \end{aligned}$$

Using the transformations in (22) and (23) and working in a similar way to solve ${\overline{G}}_2$ we get the ordinary differential equation

$$\begin{aligned} (v^2/2-u)\frac{d {\overline{G}}_3}{d v} =m v {\overline{G}}_3 + {{\tilde{a}}}_{10} a_\ell (v^2/2-u)^{m-1} \end{aligned}$$

Solving this linear differential equation we get

$$\begin{aligned} \begin{aligned} {\overline{G}}_3&={{\tilde{G}}}_3(u) (v^2/2-u)^{m} + {{\tilde{a}}}_{10} a_\ell (v^2/2-u)^{m} \int \frac{dv}{(v^2/2-u)^{2}} \\&={{\tilde{G}}}_3(u) (v^2/2-u)^{m} + {{\tilde{a}}}_{10} a_\ell (v^2/2-u)^{m} \left( \frac{1}{\sqrt{2}u^{3/2}} \text {arctanh} \, \left( \frac{v}{\sqrt{2} \sqrt{u}}\right) -\frac{v}{ u (v^2-2u)} \right) , \end{aligned} \end{aligned}$$

where ${{\tilde{G}}}_3(u)$ is a smooth function in the variable u. Since $G_3$ must be a polynomial of degree $2m-3$ and ${{\tilde{a}}}_{10} a_\ell \ne 0$ we get a contradiction.

Case 2: $a_{20} \ne 0$, $a_{11}=0$and$a_{01} \ne 0$. In this case system (2) becomes

$$\begin{aligned} x' =y, \quad y'= a_{00} +a_{10} x +a_{20} x^2 +a_{01} y. \end{aligned}$$

(24)

We introduce the change of variables and parametrization of time

$$\begin{aligned} x_1= \frac{a_{20}}{a_{01}^2} x, \quad y_1= \frac{a_{20}}{a_{01}^3}y, \quad t =\frac{1}{a_{01}} s \end{aligned}$$

(25)

if $a_{01} \ne 0$, and the change of variables

$$\begin{aligned} x_1= a_{20} x, \quad y_1= a_{20} y, \quad t = s \end{aligned}$$

if $a_{01}=0$. With these changes of variables, system (24) becomes

$$\begin{aligned} x_1 ' = y_1, \quad y_1' = {{\tilde{a}}}_{00} + {{\tilde{a}}}_{10} x_1 +x_1^2 + {{\tilde{a}}}_{01} y_1, \end{aligned}$$

(26)

where

$$\begin{aligned} {{\tilde{a}}}_{01}={\left\{ \begin{array}{ll} 1, &{} \text {if } a_{01} \ne 0, \\ 0,&{} \text {if } a_{01} = 0, \end{array}\right. } \end{aligned}$$

${{\tilde{a}}}_{00}=a_{00} a_{20}/a_{01}^4$, ${{\tilde{a}}}_{10}=a_{10}/a_{01}^2$ and the prime means derivative with respect to the new time s.

Let $g=g(x_1,y_1)=f(x,y)=0$ be an invariant algebraic curve of system (26). First we show that the cofactor $K(x_1,y_1)$ must be constant. Indeed, since system (26) has degree two, the cofactor has degree at most one, i.e., $K(x_1,y_1)={\hat{k}}_0 +{\hat{k}}_1 x_1 +{\hat{k}}_2 y_1$ and then E q. (5) writes as

$$\begin{aligned} y_1 \frac{\partial g}{\partial x_1} + ({{\tilde{a}}}_{00} + {{\tilde{a}}}_{10} x_1 +x_1^2 + {{\tilde{a}}}_{01} y_1)\frac{\partial g}{\partial y_1} = ({\hat{k}}_0 + {\hat{k}}_1 x_1 + {\hat{k}}_2 y_1) g. \end{aligned}$$

(27)

Now we write g as sum of its homogeneous parts in the form

$$\begin{aligned} g=\sum _{j=0}^n g_j (x_1,y_1), \quad \text {where } g_j\in \mathbb {R}[x_1,y_1] \hbox { is } \hbox {homogeneous of degree } j. \end{aligned}$$

Imposing that the higher order term in expression (27) vanishes que get the partial differential equation

$$\begin{aligned} x_1^2 \frac{\partial g_n}{\partial y_1} = ({\hat{k}}_1 x_1 +{\hat{k}}_2 y_1) g_n. \end{aligned}$$

Solving it we get

$$\begin{aligned} g_n = {{\tilde{g}}}_n (x_1) e^{\frac{{\hat{k}}_1}{x_1} y_1 + \frac{{\hat{k}}_2}{2 x_1^2} y_1^2}, \end{aligned}$$

where ${{\tilde{g}}}_n$ is a smooth function in $x_1$. Since $g_n$ has to be a polynomial we get that ${\hat{k}}_2={\hat{k}}_1=0$ and so $K(x_1,y_1)={\hat{k}}_0 \in \mathbb {C}$.

We introduce the weight-change of variables of the form

$$\begin{aligned} x_1 =\mu ^{-2}X, \quad y_1 = \mu ^{-3} Y, \quad s=\mu \tau . \end{aligned}$$

In this form, system (26) becomes

$$\begin{aligned} \begin{aligned} X'&= Y, \\ Y'&=X^2 + {{\tilde{a}}}_{01} \mu Y +{{\tilde{a}}}_{10} \mu ^2 X +{{\tilde{a}}}_{00} \mu ^4, \end{aligned} \end{aligned}$$

(28)

where the prime denotes derivative in $\tau$. Now we set

$$\begin{aligned} G(X,Y)=\mu ^\ell g(\mu ^{-2} X, \mu ^{-3} Y) \quad \text {and} \quad K=\mu {\hat{k}}_0 \end{aligned}$$

where $\ell$ is the highest weight degree in the weight homogeneous components of g in $x_1$, $y_1$ with weight (2, 3).

Proceeding as in Sect. 2 Case 2, we have that $G=0$ is an invariant algebraic curve of system (28) with cofactor K. Assume that $G=\sum _{i=0}^j G_i$ where $G_i$ is a weight homogeneous polynomial in X, Y with weight degree $\ell -i$ for $i=0,\ldots ,j$ and $\ell \ge j$. From the definition of invariant algebraic curve we have

$$\begin{aligned} \begin{aligned} Y \sum _{i=0}^j \mu ^i \frac{\partial G_i}{\partial X} +\big (X^2 + {{\tilde{a}}}_{01} \mu Y +{{\tilde{a}}}_{10} \mu ^2 X +{{\tilde{a}}}_{00} \mu ^4\big ) \sum _{i=0}^j \frac{\partial G_i}{\partial Y} = \mu {\hat{k}}_0 \sum _{i=0}^j \mu ^i G_i. \end{aligned} \end{aligned}$$

Equating the terms with $\mu ^0$ and $\mu ^1$ we get

$$\begin{aligned} L[G_0]=0, \quad L[G_1]=-\,{{\tilde{a}}}_{01} Y \frac{\partial G_0}{\partial Y} + {\hat{k}}_0 G_0 \end{aligned}$$

(29)

where

$$\begin{aligned} L=Y \frac{\partial }{\partial X} + X^2 \frac{\partial }{\partial Y}. \end{aligned}$$

The characteristic equations associated with the first linear partial differential equation of system (29) are

$$\begin{aligned} \frac{d X}{d Y} = \frac{Y}{X^2}. \end{aligned}$$

This system has the general solution $u=Y^2/2 - X^3/3 =\kappa$, where $\kappa$ is a constant.

According with the method of characteristics we make the change of variables

$$\begin{aligned} u=Y^2/2 - X^3/3, \quad v= X. \end{aligned}$$

(30)

Its inverse transformation is

$$\begin{aligned} Y=\pm \sqrt{2 u+2 v^3/3}, \quad X=v. \end{aligned}$$

(31)

In the following for simplicity we only consider the case $Y= \sqrt{2 u+2 v^3/3}$. Under changes (30) and (31), the first equation in (29) becomes the ordinary differential equation (for fixed u)

$$\begin{aligned} \sqrt{2 u+2 v^3/3} \frac{d {\overline{G}}_0}{d v} =0 \end{aligned}$$

where ${\overline{G}}_0$ is $G_0$ written in the variables u, v. The above equation has the general solution

$$\begin{aligned} {\overline{G}}_0={\overline{F}}_0 (u), \end{aligned}$$

where ${\overline{F}}_0$ is an arbitrary smooth function in the variable u. In order that $G_0$ be a weight homogenous polynomial with weight degree m, since X and Y have weight degrees 2 and 3 respectively we get that $G_0$ should be of weight degree $\ell =6 m$ for some convenient $m \in \mathbb {N}$. So, $G_0$ has the form

$$\begin{aligned} G_0= a_\ell \left( \frac{Y^2}{2} - \frac{X^3}{3}\right) ^m, \quad a_\ell \in \mathbb {C}\setminus \{0\}. \end{aligned}$$

Note that by assumptions we have that $a_{01} \ne 0$. So, ${{\tilde{a}}}_{01}=1$ and

$$\begin{aligned} L[G_1]=-\, Y \frac{\partial G_0}{\partial X} + {\hat{k}}_0 G_0. \end{aligned}$$

So,

$$\begin{aligned} \begin{aligned} L[G_1]&= -\, a_\ell m Y^2 \left( \frac{Y^2}{2} - \frac{X^3}{3}\right) ^{m-1}+{\hat{k}}_0 a_\ell \left( \frac{Y^2}{2} - \frac{X^3}{3}\right) ^{m} \\&= -\, a_\ell \ell \left( 2 \left( \frac{Y^2}{2} - \frac{X^3}{3}\right) +\frac{2}{3} X^3 \right) \left( \frac{Y^2}{2} - \frac{X^3}{3}\right) ^{m-1} +{\hat{k}}_0 a_\ell \left( \frac{Y^2}{2} - \frac{X^3}{3}\right) ^{m} \\&=a_\ell ({\hat{k}}_0 -2 m)\left( \frac{Y^2}{2} - \frac{X^3}{3}\right) ^{m} -\frac{2}{3} a_\ell m X^3 \left( \frac{Y^2}{2} - \frac{X^3}{3}\right) ^{m-1}. \end{aligned} \end{aligned}$$

Using the transformations in (30) and (31) and working in a similar way to solve ${\overline{G}}_0$ we get the ordinary differential equation

$$\begin{aligned} \sqrt{2 u-2 \gamma v^3/3} \frac{d {\overline{G}}_1}{d v} =a_\ell ({\hat{k}}_0 - 2 m)u^{m} +\frac{2}{3} a_\ell m \gamma v^3 u^{m-1}. \end{aligned}$$

Integrating this equation with respect to v we get

$$\begin{aligned} \begin{aligned} {\overline{G}}_1&={{\tilde{G}}}_1(u) - \frac{2 \ell u^{m-1}}{(5) \sqrt{3}} X \sqrt{2 u + v^{3}/(3)} \\&\quad + \frac{1}{\sqrt{2}(5)} u^{m-1/2} v (5 {\hat{k}}_0-6 m ) \,_2F_1\left( \frac{1}{2}, \frac{1}{3}, \frac{4}{3}, \frac{- X^{3}}{6v}\right) , \end{aligned} \end{aligned}$$

where ${{\tilde{G}}}_1(u)$ is a smooth function in the variable u and

$$\begin{aligned} \,_2F_1(a,b,c,y)=\sum _{k=0}^\infty \frac{a(a+1) \cdots (a+k-1)}{b(b+1) \cdots (b+k-1)c (c+1) \cdots (c+k-1)} \frac{x^k}{k!} \end{aligned}$$

is the hypergeometric function (see, for instance [1] for its definition and properties) that is well defined if b, c are not negative integers. In particular, it is a polynomial if and only if a is a negative integer. Note that in our case $a=1/2$, $b=1/3$ and $c=4/3$. Consequently, $\,_2F_1$ is never a polynomial. Since $G_1(X,Y)={{\tilde{G}}}_1(u) = {{\tilde{G}}}_1(Y^2/2 - X^3/3)$ in order that ${\overline{G}}_1$ is a weight homogeneous polynomial of weight degree $6m-1$ we must have $\tilde{G}_1=0$ and

$$\begin{aligned} 5 {\hat{k}}_0 - 6 m =0, \quad \text {that is } {\hat{k}}_0= \frac{6 m }{5}. \end{aligned}$$

Now we apply Theorem 4. We recall that since ${\hat{k}}_0$ is a constant ${\hat{k}}_0 \ne 0$, in view of Theorem 4, $g=g(x_1,y_1)$ must vanish in the critical points of system (26), which are

$$\begin{aligned} u_1=\left( \frac{-{{\tilde{a}}}_{10} + \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}}{2},0 \right) \quad \text {and} \quad u_2=\left( \frac{-{{\tilde{a}}}_{10} - \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}}{2},0 \right) . \end{aligned}$$

Moreover, the critical point $u_1$ has the eigenvalues

$$\begin{aligned} \lambda _1^+ = \frac{1}{2} + \frac{\sqrt{1 + 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}}}{2} \quad \text {and} \quad \lambda _1^- =\frac{1}{2} - \frac{\sqrt{1 + 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}}}{2} \end{aligned}$$

and the critical point $u_2$ has the eigenvalues

$$\begin{aligned} \lambda _2^+ = \frac{1}{2} + \frac{\sqrt{1 - 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}}}{2} \quad \text {and} \quad \lambda _2^- =\frac{1}{2} - \frac{\sqrt{1 - 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}}}{2}. \end{aligned}$$

Now we consider different cases:

Case 1: ${{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}=0$. In this case, $u_1=u_2$ and ${\lambda _1}^+ ={\lambda _2}^+=1$ while $\lambda _1^-={\lambda _2}^-=0$. In view of Theorem 4 and since ${\hat{k}}_0 \ne 0$, we must have ${\hat{k}}_0 =1$. But then

$$\begin{aligned} \frac{6 m }{5}=1 \quad \text {that is} \quad m=\frac{5}{6} \end{aligned}$$

which is not possible because $m \in \mathbb {N}$.

Case 2: ${{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00} < 0$. In this case, $\lambda _1^+$ and $\lambda _1^-$ are rationally independent as well as $\lambda _2^+$ and $\lambda _2^-$. So, in view of Theorem 4 and since ${\hat{k}}_0 \ne 0$ we must have

$$\begin{aligned} {\hat{k}}_0 \in \{\lambda _1^+, \lambda _1^-, \lambda _1^++ \lambda _1^-\} = \{\lambda _1^+, \lambda _1^-, 1\} \end{aligned}$$

and

$$\begin{aligned} {\hat{k}}_0 \in \{\lambda _2^+, \lambda _2^-, \lambda _2^++ \lambda _2^-\} = \{\lambda _2^+, \lambda _2^-, 1\}. \end{aligned}$$

Proceeding as above we note that ${\hat{k}}_0 \ne 1$ and since $\lambda _1^+ \not \in \{\lambda _1^-,\lambda _2^+,\lambda _2^-\}$ we reach to a contradiction and this case is not possible.

Case 3: ${{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}>0$. In this case $u_1$ is a saddle. In view of Theorem 4 and since ${\hat{k}}_0 \not =1 =\lambda _1^++\lambda _1^-$ we must have that ${\hat{k}}_0 \in \{\lambda _1^+,\lambda _1^-\}$. So, either ${\hat{k}}_0=\lambda _1^+$ or ${\hat{k}}_0=\lambda _1^-$.

Moreover, if $1 - 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}<0$ then $\lambda _2^+$ and $\lambda _2^-$ would be rationally independent and in view of Theorem 4 we should have that ${\hat{k}}_0 \in \{\lambda _1^+,\lambda _1^-\}$ and ${\hat{k}}_0 \in \{\lambda _2^+,\lambda _2^-\}$ which as we saw in Subcase 2.2 this is not possible. So, $1 - 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}\ge 0$. Note that $k_0 \in \{\lambda _1^+,\lambda _1^-\}$ can be written as

$$\begin{aligned} \frac{12 m-5}{5} = \pm \sqrt{1 + 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}}. \end{aligned}$$

So

$$\begin{aligned} 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}=\left( \frac{12m -5}{5} \right) ^2 -1 \end{aligned}$$

Imposing that

$$\begin{aligned} 0 \le 1 - 4 \sqrt{{{\tilde{a}}}_{10}^2 -4 {{\tilde{a}}}_{00}}=2-\left( \frac{12m -5}{5} \right) ^2 \end{aligned}$$

we get

$$\begin{aligned} \left( \frac{12m -5}{5} \right) ^2 \le 2. \end{aligned}$$

Taking into account that m is an integer we must have $m=1$. Therefore,

$$\begin{aligned} {\hat{k}}_0=\frac{6 }{5}, \quad {{\tilde{a}}}_{00} =\frac{625 {{\tilde{a}}}_{10}^2 -36}{2500} , \quad G_0= a_1 \left( \frac{Y^2}{2} - \frac{ X^3}{3}\right) . \end{aligned}$$

This implies that the highest order terms in the invariant algebraic curve $g(x_1,y_1)$ in the variable $x_1$ is $- x_1^3/3$ and in the variable $y_1$ is $y_1^2/2$. Hence

$$\begin{aligned} g(x_1,y_1)=-\,\frac{1}{3} x_1^3 + \frac{1}{2} y_1^2 + g_1(x_1,y_1) \end{aligned}$$

where

$$\begin{aligned} g_1(x_1,y_1)=b_1 x_1^2 +b_2 x_1 y_1 + b_3 x_1 + b_4 y_1 + b_5, \quad b_i \in \mathbb {C}, \ i=1,\ldots , 5. \end{aligned}$$

Imposing that g satisfies (27) with ${\hat{k}}_0= \frac{6 }{5}$ we readily get that the unique invariant algebraic curve is $g(x_1,y_1)=0$ with

$$\begin{aligned} \begin{aligned} g(x_1,y_1)&=-\,\frac{(6+25 {{\tilde{a}}}_{10})(625 {{\tilde{a}}}_{10}^2-36)}{375000} -\frac{(6+25 {{\tilde{a}}}_{10})(25 {{\tilde{a}}}_{10}-2)}{2500} x_1 -\frac{6 + 25 {{\tilde{a}}}_{10}}{125} y_1 \\&\quad -\, \frac{2 + 25 {{\tilde{a}}}_{10}}{50} x_1^2 -\frac{2}{5} x_1 y_1 +\frac{1}{2} y_1^2- \frac{1}{3} x_1^3. \end{aligned} \end{aligned}$$

Under the change of variables (25) we get that system (2) has an invariant algebraic curve if and only if

$$\begin{aligned} a_{00}= -\,\frac{(6 a_{01}^2 - 25 a_{10}) (6 a_{01}^2 + 25 a_{10})}{2500 a_{20}} \end{aligned}$$

and the invariant algebraic curve is

$$\begin{aligned} \begin{aligned}&\frac{ (6 a_{01}^2 - 25 a_{10}) (6 a_{01}^2 + 25 a_{10})^2 }{375000 a_{01}a_{20}} + \frac{12 a_{01}^4-100 a_{01}^2 a_{10}-625 a_{10}^2}{2500 a_{01}} x -\frac{6 a_{01}^2 + 25 a_{10}}{125 } y \\&\quad -\,\frac{(2 a_{01}^2 + 25 a_{10}) a_{20} }{50 a_{01}} x^2 -\frac{2 a_{20}}{5 } x y + \frac{a_{20}}{2 a_{01}} y^2 -\frac{a_{20}^2}{3a_{01}} x^3 =0. \end{aligned} \end{aligned}$$

This concludes the proof of the theorem.

5 Proof of Theorem 2

First note that if $a_{20} \ne 0$ and $a_{11}=a_{01}=0$, system (2) has the polynomial first integral $I=a_{00} x +\frac{a_{10}}{2} x^2 -\frac{1}{2} y^2+ \frac{a_{20}}{3} x^3$ and so it is Liouvillian integrable. On the other hand, if $a_{20}=0$, $a_{11}a_{10} \ne 0$ and $a_{01} a_{10}=a_{00}a_{11}$, system (2) has the first integral

$$\begin{aligned} (a_{10}+a_{11}y) \exp \left( \frac{a_{11}^2 \big (\frac{a_{00}}{a_{10}}+x \big )^2 -2 (a_{11}y +a_{10})}{2a_{10}}\right) . \end{aligned}$$

and so it is Liouvillian integrable.

On the other direction, we recall that in view of Theorem 1 we can assume that we are under the assumptions of Theorem 3, i.e., $|a_{20}|^2 +|a_{11}|^2 \ne 0$, $|a_{10}|^2 +|a_{20}|^2 \ne 0$ and hypothesis (H1) holds. Hence, system (2) has an equilibrium point. Moreover, under the assumptions $a_{20} \ne 0$ and $a_{11}=a_{01}=0$ or $a_{20}=0$, $a_{11}a_{10} \ne 0$ and $a_{01} a_{10}=a_{00}a_{11}$, system (2) is Liouvillian integrable, so in order to conclude the proof of the Theorem we will show that in all the other cases system (2) does not have a Liouvillian first integral. We separate into different cases.

Case 1: Assume that system (2) has no invariant algebraic curves. Then it can only have an exponential factor of the form $E=\exp (g)$ with $g \in \mathbb {C}[x,y] \setminus \mathbb {C}$. The cofactor must be of degree one and so $L=\beta _0+\beta _1 x +\beta _2 y$ with $\beta _i \in \mathbb {C}$. Note that setting $h=g-\beta _2 x$ then $E=\exp (h)$ is an exponential factor with cofactor $L=\beta _0+\beta _1 x$ and so, in view of (13), it satisfies

$$\begin{aligned} y \frac{\partial h}{\partial x} + \big (a_{00}+a_{10} x +a_{20} x^2 + y(a_{01} +a_{11}x )\big ) \frac{\partial h}{\partial y} =\beta _0+\beta _1 x . \end{aligned}$$

(32)

Now we consider different subcases.

Case 1.1: $a_{20} \ne 0$. Consider first the case in which $a_{00}+a_{10} x +a_{20} x^2=0$ has to complex roots, that is, $a_{00}+a_{10} x +a_{20} x^2=a_{20} (x-x_1)(x-x_2)$ with $x_1 \ne x_2$. In this case, evaluating (32) on $y=0$, $x=x_1$ and on $y=0$, $x=x_2$, we get, respectively,

$$\begin{aligned} \beta _0+\beta _1 x_1=0 \quad \text {and} \quad \beta _0+\beta _1 x_2=0 \end{aligned}$$

which yields $\beta _0=\beta _1=0$. But then h must be a first integral, which is not possible.

Consider now that $a_{00}+a_{10} x +a_{20} x^2=0$ has a multiple root, that is, $a_{00}+a_{10} x +a_{20} x^2=a_{20} (x-x_1)^2$. In this case, evaluating (32) on $y=0$, $x=x_1$ we get $\beta _0=-\,\beta _1 x_1$ and then (32) becomes

$$\begin{aligned} y \frac{\partial h}{\partial x} + \big (a_{20} (x-x_1)^2 + y(a_{01} +a_{11}x )\big ) \frac{\partial h}{\partial y} =\beta _1(x-x_1). \end{aligned}$$

(33)

Evaluating (33) on $y=0$ we get

$$\begin{aligned} (x-x_1) \left( a_{20} \frac{\partial h}{\partial y}|_{y=0} -\beta _1\right) =0, \end{aligned}$$

which yields $\beta _1=0$. But then $\beta _0=\beta _1=0$ and so h must be a first integral, which is not possible.

Case 1.2: $a_{20}=0$. Then $a_{10}\ne 0$ (we recall that $|a_{10}|^2 +|a_{20}|^2 \ne 0$). In this case, note that

$$\begin{aligned} E=\exp \left( g-\beta _2 x-\frac{\beta _1}{a_{10}} \left( y-a_{01} x -a_{11} \frac{x^2}{2}\right) \right) =\exp (h) \end{aligned}$$

is another exponential factor with cofactor $L={\tilde{\beta }}_0$, where ${\tilde{\beta }}_0= \beta _0 -\beta _1 a_{00}/a_{10}$. Then in view of Theorem 8 the cofactor $L=L(x,y)$ of the exponential factor $\exp (h)$ must satisfy

$$\begin{aligned} d {\tilde{\beta }}_0 = -\,a_{01} -a_{11}x, \quad d \in \mathbb {C}. \end{aligned}$$

Since $a_{11} \ne 0$ this is not possible and we are done.

Case 2: Assume that system (2) has invariant algebraic curves. In view of Theorem 3 we will distinguish between the cases in statements (b), (c) and (e).

Case 2.1: $a_{20} a_{11} \ne 0$and$a_{00} =\frac{(a_{01}a_{11}+a_{20})(a_{10} a_{11}^2 -a_{01} a_{11}a_{20}-a_{20}^2)}{a_{11}^4}$. In this case system (2) can have an exponential factor of the form $E=\exp \big (f/g^n\big )$ with $n > 0$,

$$\begin{aligned} g=-\,\frac{a_{01} a_{11}a_{20} +a_{20}^2-a_{10} a_{11}^2}{a_{11}^3} +\frac{a_{20}}{a_{11}} x+y \end{aligned}$$

and f coprime with g. We will show that this is not possible. Assume that it is so, then f satisfies, after simplifying the exponential term E,

$$\begin{aligned} \begin{aligned}&y \frac{\partial f}{\partial x} + \big (a_{00}+a_{10} x +a_{20} x^2 + y(a_{01} +a_{11}x )\big ) \frac{\partial f}{\partial y} - \frac{n(a_{01} a_{11} + a_{20} + a_{11}^2 x)}{a_{11}} f\\&\quad =(\beta _0+\beta _1 x +\beta _2 y) g^n. \end{aligned} \end{aligned}$$

(34)

Set now

$$\begin{aligned} {\bar{f}}=f|_{g=0}=f|_{y=\frac{a_{01} a_{11}a_{20} +a_{20}^2-a_{10} a_{11}^2}{a_{11}^3} -\frac{a_{20}}{a_{11}} x}. \end{aligned}$$

Then ${\bar{f}} \ne 0$ (otherwise f would not be coprime with g) and it satisfies Eq. (34) restricted to $y=\frac{a_{01} a_{11}a_{20} +a_{20}^2-a_{10} a_{11}^2}{a_{11}^3} -\frac{a_{20}}{a_{11} x}$, i.e.,

$$\begin{aligned} \left( \frac{a_{01} a_{11}a_{20} +a_{20}^2-a_{10} a_{11}^2}{a_{11}^3} -\frac{a_{20}}{a_{11} x}\right) \frac{d {\bar{f}}}{d x} =n \frac{a_{01} a_{11} + a_{20} + a_{11}^2 x}{a_{11}} {\bar{f}}. \end{aligned}$$

Solving this linear differential equation we conclude that

$$\begin{aligned} {\bar{f}} = \frac{\alpha }{(3a_{20} +a_{11}(a_{01}+a_{11}x))^2} \exp \left( \frac{n a_{11}^2 x}{2a_{20}}\right) , \quad \alpha \in \mathbb {C}\setminus \{0\}. \end{aligned}$$

Since ${\bar{f}}$ must be a polynomial, and $n \alpha a_{11} \ne 0$ we get a contradiction. So, this case is not possible. Summarizing, the unique possible exponential factors must be of the form $E=\exp (g)$ with $g \in \mathbb {C}[x,y]$. Proceeding as in case 1 taking into account that in this case there are no polynomial first integrals we get to a contradiction.

Case 2.2: $a_{20}a_{11} \ne 0$and$a_{10} =\frac{2(a_{01}a_{11}a_{20}+2 a_{20}^2)}{a_{11}^2}$. In this case system (2) can have an exponential factor of the form $E=\exp \big (f/g^n\big )$ with $n > 0$,

$$\begin{aligned} g= \frac{a_{00}a_{11}}{2a_{20}} -\frac{a_{01} a_{11} +2 a_{20}}{a_{11}} x+y-\frac{a_{11}}{2} x^2 \end{aligned}$$

and f coprime with g. We will show that this is not possible. Assume that it is so, then f satisfies, after simplifying the exponential term E,

$$\begin{aligned} y \frac{\partial f}{\partial x} + \big (a_{00}+a_{10} x +a_{20} x^2 + y(a_{01} +a_{11}x )\big ) \frac{\partial f}{\partial y} +\frac{2 n a_{20}}{a_{11}} f =(\beta _0+\beta _1 x +\beta _2 y) g^n.\nonumber \\ \end{aligned}$$

(35)

Set now

$$\begin{aligned} {\bar{f}}=f|_{g=0}=f|_{y= \frac{a_{00}a_{11}}{2a_{20}} +\frac{a_{01} a_{11} -2 a_{20}}{a_{11}} x+\frac{a_{11}}{2} x^2 +\frac{a_{20}}{a_{11}} x}. \end{aligned}$$

Then ${\bar{f}} \ne 0$ (otherwise f would not be coprime with g) and it satisfies Eq. (35) restricted to $y=\frac{a_{00}a_{11}}{2a_{20}} +\frac{a_{01} a_{11} -2 a_{20}}{a_{11}} x+\frac{a_{11}}{2} x^2 +\frac{a_{20}}{a_{11}} x$, i.e.,

$$\begin{aligned} \left( \frac{a_{00}a_{11}}{2a_{20}} +\frac{a_{01} a_{11} -2 a_{20}}{a_{11}} x+\frac{a_{11}}{2} x^2 +\frac{a_{20}}{a_{11}} x\right) \frac{d {\bar{f}}}{d x} =-\,\frac{2 n a_{20}}{a_{11}}{\bar{f}}. \end{aligned}$$

Solving this linear differential equation we conclude that

$$\begin{aligned} {\bar{f}} =\alpha \exp \left( -\frac{4 n a_{20}^{3/2} \arctan \left( \frac{\sqrt{a_{20}} (a_{01} a_{11} + 2 a_{20} + a_{11}^2 x)}{\sqrt{a_{00} a_{11}^4 - a_{20} (a_{01} a_{11} + 2 a_{20})^2}}\right) }{\sqrt{a_{00} a_{11}^4 - a_{20} (a_{01} a_{11} + 2 a_{20})^2}}\right) , \quad \alpha \in \mathbb {C}\setminus \{0\}. \end{aligned}$$

Note that ${\bar{f}}$ can be written as

$$\begin{aligned} \alpha \left( \frac{\sqrt{a_{00} a_{11}^4 - a_{20} (a_{01} a_{11} + 2 a_{20})^2} -i \sqrt{a_{20}}(2a_{20}+a_{11}(a_{01}+a_{11}x))}{\sqrt{a_{00} a_{11}^4 - a_{20} (a_{01} a_{11} + 2 a_{20})^2} +i \sqrt{a_{20}}(2a_{20}+a_{11}(a_{01}+a_{11}x))} \right) ^{\frac{2 i n a_{20}^{3/2}}{\sqrt{a_{00} a_{11}^4 - a_{20} (a_{01} a_{11} + 2 a_{20})^2}}}. \end{aligned}$$

Since ${\bar{f}}$ must be a polynomial, and $n \alpha a_{20} \ne 0$ we get a contradiction. So, this case is not possible. Hence, the unique possible exponential factors must be of the form $E=\exp (g)$ with $g \in \mathbb {C}[x,y]$. Proceeding as in case 1 taking into account that in this case there are no polynomial first integrals we get to a contradiction.

Case 2.3: $a_{20} a_{01} \ne 0$, $a_{11}=0$and$a_{00}= -\,\frac{(6 a_{01}^2 - 25 a_{10}) (6 a_{01}^2 + 25 a_{10})}{2500 a_{20}}$. In this case system (2) can have an exponential factor of the form $E=\exp \big (f/g^n\big )$ with $n > 0$,

$$\begin{aligned} \begin{aligned} g&=\frac{ (6 a_{01}^2 - 25 a_{10}) (6 a_{01}^2 + 25 a_{10})^2 }{375000 a_{01}a_{20}} + \frac{12 a_{01}^4-100 a_{01}^2 a_{10}-625 a_{10}^2}{2500 a_{01}} x -\frac{6 a_{01}^2 + 25 a_{10}}{125 } y \\&\quad -\,\frac{(2 a_{01}^2 + 25 a_{10}) a_{20} }{50 a_{01}} x^2 -\frac{2 a_{20}}{5 } x y + \frac{a_{20}}{2 a_{01}} y^2 -\frac{a_{20}^2}{3a_{01}} x^3 \end{aligned} \end{aligned}$$

and f coprime with g. We will show that this is not possible. Assume that it is so, then f satisfies, after simplifying the exponential term E,

$$\begin{aligned} y \frac{\partial f}{\partial x} + \big (a_{00}+a_{10} x +a_{20} x^2 + y(a_{01} +a_{11}x )\big ) \frac{\partial f}{\partial y} -\frac{6 n}{5} f =(\beta _0+\beta _1 x +\beta _2 y) g^n.\nonumber \\ \end{aligned}$$

(36)

We introduce the change of variables $X=6 a_{01}^2 + 25 (a_{10} + 2 a_{20} x), Y=y$. In these new variables we have

$$\begin{aligned} g(X,Y)= \frac{a_{01}}{31250 a_{20}} X^2 -\frac{1}{375000 a_{01} a_{20}} X^3 - \frac{1}{125} X Y + \frac{a_{20}}{2 a_{01}} Y^2. \end{aligned}$$

Let $F(X,Y)=f(x,y)$. Note that the two branches of $g(X,Y)=0$ are

$$\begin{aligned} Y=\frac{6 a_{01} X - \sqrt{3} X^{3/2}}{750 a_{20}} \quad \text {and} \quad Y=\frac{6 a_{01} X + \sqrt{3} X^{3/2}}{750 a_{20}}. \end{aligned}$$

(37)

We consider only the first condition in (37) since the second one is done in a similar manner. Let

$$\begin{aligned} {\bar{F}}=F(X,Y)|_{Y=\frac{6 a_{01} X - \sqrt{3} X^{3/2}}{750 a_{20}}} \quad \text {and set} \quad W=\sqrt{X}. \end{aligned}$$

Then if we set ${{\tilde{F}}}(W)={\bar{F}}$, taking into account that since $x' =y$, we get

$$\begin{aligned} Y= 50 a_{20} \frac{d X}{dt}, \end{aligned}$$

the first condition in (37) rewrites as

$$\begin{aligned} W'=\frac{d W}{dt } =-\,\frac{X'}{2 \sqrt{X}}=\frac{6 a_{01} W + \sqrt{3} W^2}{30}. \end{aligned}$$

Hence, (36) in these new variables becomes

$$\begin{aligned} \Bigl (-\frac{X'}{2 \sqrt{X}}=\frac{6 a_{01} W + \sqrt{3} W^2}{30}\Bigr ) \frac{d {{\tilde{F}}}}{d W} = \frac{6 n}{5} {{\tilde{F}}}. \end{aligned}$$

Solving this linear differential equation we get

$$\begin{aligned} {{\tilde{F}}}=\alpha \left( \frac{W}{6 a_{01} + \sqrt{3} W}\right) ^{\frac{ 6 n}{a_{01}}} \end{aligned}$$

that is

$$\begin{aligned} {\bar{F}}= \alpha \left( \frac{X}{6 a_{01} + \sqrt{3} X^2}\right) ^{\frac{12 n}{a_{01}}}. \end{aligned}$$

Since ${\bar{F}}$ must be a polynomial, and $n \alpha a_{01} \ne 0$ we get a contradiction. So, this case is not possible. Proceeding exactly in the same manner for the second condition in (37) we get also to a contradiction. So, the unique possible exponential factors must be of the form $E=\exp (g)$ with $g \in \mathbb {C}[x,y]$. Proceeding as in case 1 taking into account that in this case there are no polynomial first integrals we get to a contradiction. This concludes the proof of the theorem.