\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 121, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2011/121\hfil Computation of rational solutions]
{Computation of rational solutions for a
first-order nonlinear differential equation}
\author[D. Behloul, S. S. Cheng\hfil EJDE-2011/121\hfilneg]
{Djilali Behloul, Sui Sun Cheng} % in alphabetical order
\address{Djilali Behloul \newline
Facult\'{e} G\'{e}nie Electrique, D\'{e}partement informatique,
USTHB, BP32, El Alia, Bab Ezzouar, 16111, Algiers, Algeria}
\email{dbehloul@yahoo.fr}
\address{Sui Sun Cheng \newline
Department of Mathematics, Hua University, Hsinchu 30043, Taiwan}
\email{sscheng@math.nthu.edu.tw}
\thanks{Submitted August 23, 2011. Published September 19, 2011.}
\thanks{D. Behloul is supported by a national PNR (2011-2013) grant.}
\subjclass[2000]{34A05}
\keywords{Polynomial solution; rational solution;
nonlinear differential equation}
\begin{abstract}
In this article, we study differential equations of
the form $y'=\sum A_i(x)y^i/\sum B_i(x)y^i$ which
can be elliptic, hyperbolic, parabolic, Riccati, or quasi-linear.
We show how rational solutions can be computed in a systematic manner.
Such results are most likely to find applications in the theory of
limit cycles as indicated by Gin\'{e} et al \cite{g1}.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\section{Introduction}
When confronting an unfamiliar differential equation, it is natural
to try to find the simplest type of solutions such as polynomial
and rational solutions. Indeed, exact solutions
(such as polynomial and rational solutions) for the nonlinear differential
equation
\begin{equation}
\frac{dy}{dx}=P(x,y) \label{00}
\end{equation}
are of great interests, in particular understanding the whole set
of solutions and their dynamical properties.
In 1936, Rainville \cite{r1} determined all the Riccati differential
equations of the form
\[
\frac{dy}{dx}=y^{2}+A_1(x)y+A_0(x)
\]
with $A_0$ and $A_1$ polynomials, which have polynomial solutions.
In 1954, Campbell and Golomb \cite{g1} provided an algorithm for finding
all the polynomial solutions of the differential equation
\[
B_0(x)\frac{dy}{dx}=A_2(x)y^{2}+A_1(x)y+A_0(x)
\]
with $B_0$, $A_0$, $A_1$ and $A_2$ polynomials. In 2006, Behloul
and Cheng \cite{b1} (see also \cite{b2}) gave another algorithm
for looking for the rational solutions of the equation
\[
B_0(x)\frac{dy}{dx}=A_n(x)y^n+A_{n-1}(x)y^{n-1}+\dots +A_0(x)
\]
with $B_0$ and $A_i$ polynomials. In 2011, Gin\'{e} et al \cite{g1}
developed new results about `periodic' polynomial solutions for
\begin{equation}
\frac{dy}{dx}=A_n(x)y^n+A_{n-1}(x)y^{n-1}+\dots +A_0(x)
\label{0}
\end{equation}
with $A_i$ polynomials. Such results give rise to sharp
information on the number of polynomial limit cycles. These
conclusions are important since the theory of limit cycles is
an active and difficult research area. For a concise list of
references related to limit cycles of \eqref{0}, including the
works by Abel, Briskin, Gasull, Llibre, Neto, Lloyd, the
readers is referred to \cite{g1}.
Clearly, equations of the form \eqref{0} are among the easiest of
equations of the form \eqref{00}. The next level of difficulty
will come from studying the case when $P(x,y)$ is a rational
function. In this paper, we are concerned with the rational
solutions of the differential equation
\begin{equation}
y'=\frac{A_n(x)y^n+A_{n-1}(x)y^{n-1}+\dots +A_0(x)}{
B_m(x)y^{m}+B_{m-1}(x)y^{m-1}+\dots +B_0(x)}, \label{A}
\end{equation}
where $A_0,A_1,\dots,A_n$ and $B_0,B_1,\dots,B_m$ are (complex
valued) polynomials (of one independent complex variable) such
that $A_n$ and $B_m$ are not identically zero.
By providing a systematic scheme for computing all the rational
solutions of \eqref{A}, we hope that our results lead to
estimates of the number of `rational limit cycles', more general
than those in Gin\'{e} et al \cite{g1}, and to qualitative results
for nonlinear equations of the form \eqref{A} but with an
additional nonlinear perturbations.
As another motivation for our study, we quote a result by
Malmquist \cite{m1} which states: If the differential equation
\eqref{A} is not one of the two forms
$$
B_0(x)\frac{dy}{dx}=A_1(x)y+A_0(x)$$
or
$$
B_0(x) \frac{dy}{dx}=A_2(x)y^{2}+A_1(x)y+A_0(x)
$$
then all its one-valued solutions must be \emph{rational}. For
example the equation $\frac{dy}{dx} =y$ admits $e^{x}$ as a
one-valued solution which is not rational and the equation
$\frac{dy}{dx}=1+y^{2}$ admits $\tan x$ as a one-valued solution
which is not rational.
Clearly, equation \eqref{A} is only defined at places where the
denominator does not vanish. However, a root of the denominator
may also be a root of the numerator and \eqref{A} may
still be meaningful by assigning proper values to the rational
function on the right-hand side. To avoid such technical details,
we will define a polynomial solution to be a polynomial function
$y=y(x)$ such that
\begin{equation}
\begin{split}
&y'(x)\{ B_m(x)y^{m}+B_{m-1}(x)y^{m-1}+\dots +B_0(x)\} \\
&\equiv A_n(x)y^n+A_{n-1}(x)y^{n-1}+\dots +A_0(x);
\end{split} \label{AP}
\end{equation}
and a rational solution to be a pair of polynomials
$(U(x),V(x))$ such that the degree of $V$ is greater than or
equal to $1$ and
\begin{equation}
\begin{split}
&(V(x)U'(x)-U(x)V'(x)) \{ B_m(x)y^{m}+\dots
+B_0(x)t\} \\
&\equiv V^{2}(x)\{ A_n(x)y^n+\dots
+A_0(x)\} .
\end{split} \label{AR}
\end{equation}
Since the right-hand sides and the left-hand sides are polynomials,
singularities are thus avoided.
To motivate what follows, let us consider the specific example
\begin{equation}
y'=\frac{y^3+2x}{2x^{2}y+x}. \label{01}
\end{equation}
Suppose we try to find a constant (polynomial) solution of the form
$ y(x)=\lambda $. Then substituting it into the above equation, we
see that
\[
0=0\cdot \{ 2x^{2}\lambda +x\} \equiv \lambda ^3+2x
\]
for all $x$, which is impossible. Next, we try polynomial
solutions with degree $1$. Then $y''(x)\equiv 0$, such that
\[
0=y''=\big(\frac{y^3+2x}{2x^{2}y+x}\big) '
=\frac{y'(-4x^{2}+4xy^3+3y^{2}) }{x(
2xy+1) ^{2}}-\frac{1}{x^{2}}\frac{y(
4x^{2}+4xy^3+y^{2}) }{(2xy+1) ^{2}}.
\]
Replacing $y'$ by $\frac{y^3+2x}{2x^{2}y+x}$ in the above
equation and rearranging term,
\begin{equation}
\begin{split}
&[ (-4x) y^{5}+(8x^{2}-3) y^{4}+6xy^3+(
1-4x^{2}) y^{2}+(8x^3-6x) y+(4x^{2}) ]y\\
&\equiv -8x^3.
\end{split} \label{AL}
\end{equation}
Thus $y(x)$ is a factor of the polynomial $x^3$. Hence
$y=\lambda x$ for a nonzero number $\lambda $. Then from
\eqref{01},
\[
\lambda (2x^{2}\lambda +1) \equiv \lambda ^3x^{2}+2,
\]
so that $\lambda =2$. We may easily check that $y(x)=2x$ is indeed
a solution of \eqref{01}.
Next we may try polynomials with higher degrees of course.
But we should stop for a while and consider the existence and
uniqueness of all polynomial and
rational solutions as well as schematic methods for computing them.
To this end, we first settle on a convenient notation. We
will let $\mathbb{N}$ be the set of nonnegative integers,
$\mathbb{N}^{*}$ the set of positive integers, and $\mathbb{C}$
the set of complex numbers. When $G=G(x)$ is a nontrivial
polynomial, its degree is denoted by $\deg G(x)$, and when it is
the zero polynomial, its degree is defined to be $-\infty $. When
$ H=H(x,y)$ is a bivariate polynomial of the form
\[
H(x,y)=h_n(x)y^n+h_{n-1}(x)y^{n-1}+\dots +h_0(x)
\]
where $h_0,\dots,h_n$ are polynomials with $h_n$ not
identically zero, then $\deg _yH(x,y)$ is taken to be $n$
(e.g., if $H(x,y)=3xy^{2}+y$ then $\deg _yH(x,y)=2$, although
$H(0,y)=y$). We will set $a_i=\deg A_i(x)$ for $i=0,1,\dots,n$
and $b_i=\deg B_i(x)$ for $i=0,1,2,\dots,m$,
\begin{gather}
P(x,y)=A_n(x)y^n+A_{n-1}(x)y^{n-1}+\dots +A_0(x), \label{P}
\\
Q(x,y)=B_m(x)y^{m}+B_{m-1}(x)y^{m-1}+\dots +B_0(x). \label{Q}
\end{gather}
Let us also write $A_n$ and $B_m$ in the form
\begin{gather*}
A_n(x)=Ax^{a_n}+\dots, \\
B_m(x)=Bx^{b_m}+\dots
\end{gather*}
where $A,B\neq 0$.
The derivative of a function $g(x)$ of one variable is denoted by
$g'(x)$ or $g^{(1)}(x)$ and the higher order derivatives by $
g^{(2)}(x),g^{(3)}(x),\dots$ as usual and partial derivatives of a
function $ H(x,y)$ of two variables are denoted respectively by
$H_x'(x,y)$ and $H_y'(x,y)$.
Let $G=G(x)$ be a polynomial. We recall that the multiplicity of
$G$ at $ \alpha $ is defined to be $0$ if $\alpha $ is not a root
of $G$, and be the positive integer $s$ if $\alpha $ is a root of
$G$ with multiplicity $s$. Let $H=H(x)$ be another polynomial
which is not identically zero. For the rational function
$F(x)=G(x)/H(x)$, if $F$ is not identically zero, its valuation
$v_{\alpha }(F)$ at $\alpha $ is the difference of the
multiplicity of $F$ at $\alpha $ and the multiplicity of $G$ at
$\alpha $; otherwise, its valuation is $+\infty $. For example, if
$F(x)=x(x+1)/( x^3-2x^{2})$, then $v_0(F)=1-2=-1$,
$v_{-1}(F)=1-0=1$, $v_2(F)=0-1=-1$ and $v_{\alpha }(F)=0$ if
$\alpha \notin \{-1,0,2\}$.
In the rest of our discussions, we will assume that
$P$ and $Q$ are coprime; i.e., $\gcd (P,Q)=1$.
Since $n,m\in \mathbb{N}$, we may classify \eqref{A} into five
mutually distinct and exhaustive cases:
Case I: If $n>m+2$, then \eqref{A} is said to be
\textit{elliptic}.
Case II: If $nm+1$. Then for
all sufficiently large $d$, $nd+a_n>md+b_m+d-1$ and
\begin{align*}
&f(d)\\
&= \max\big\{
a_0,a_1+d,\dots,a_n+nd;b_0+d-1,b_1+d+d-1,\dots, b_m+md+d-1\big\} \\
&= \max \{ a_n+nd,b_m+md-1\} \\
&= \max \{ a_n+nd\} \\
&= f_n(d) \\
&>\max \{
f_0(d),f_1(d),\dots,f_{n-1}(d);f_{n+1}(d),\dots,f_{n+m+1}(d)\}.
\end{align*}
Thus we may let $d_0$ be the first positive integer such that the
above chain of inequalities hold for all $d\geq d_0$. If $t$ is
feasible, then by Lemma \ref{lem1}, $t\max \{
f_0(d),f_1(d),\dots,f_{n+m}(d)\} \label{c1}
\end{equation}
for sufficiently large $d$. Let $d_0$ be the first positive
integer such that the above chain of inequalities hold for all
$d\geq d_0$. If $t$ is feasible, then by Lemma \ref{lem1}, $tb_m-1$. Then $nd+a_n>md+b_m+d-1$
for all $ d, $ and for all sufficiently large $d$,
\[
f(d)=f_n(d)>\max \{
f_0(d),f_1(d),\dots,f_{n-1}(d);f_{n+1}(d),\dots,f_{n+m+1}(d)\}.
\]
As in the first case, we let $d_0$ be the first positive integer
such that the above chain of inequalities hold for all $d\geq
d_0$. If $t$ is feasible, then by Lemma \ref{lem1}, $t1$, we see further that $\Omega =\{1\}$.
Let $y$ be a polynomial solution of degree $1$. Then as we have
already seen in the Introduction, \eqref{AL} must hold, and $y=2x$
is a polynomial solution and hence is also the unique polynomial
solution of \eqref{A1}.
\end{example}
\section{Rational solutions}
We now turn to rational solutions of \eqref{A}.
\begin{theorem} \label{thm2}
If \eqref{A} is elliptic,
then any rational solution of \eqref{A} is of the form
$y=u/A_n$ where $ u $ is a polynomial; and
if \eqref{A} is hyperbolic or $ (n,m)=(1,0)$, then there
exists $\varrho \in \mathbb{N}$ (which can be
determined) such that any rational solution of \eqref{A} is of the
form $y=u/B_m^{\varrho }$, where $u$ is a
polynomial.
\end{theorem}
Before we turn to the proof, recall from Taylor's expansion that
\begin{gather*}
A_n(x)=A_n(x_0)+\dots +A_n^{(k)}(x_0)\frac{(x-x_0)^{k}}{k!}
+\dots +A_n^{(a_n)}(x_0)\frac{(x-x_0)^{a_n}}{a_n!},
\\
B_m(x)=B_m(x_0)+\dots +B_m^{(k)}(x_0)\frac{(x-x_0)^{k}}{k!}
+\dots +B_m^{(b_m)}(x_0)\frac{(x-x_0)^{b_m}}{b_m!}.
\end{gather*}
Furthermore, if $x_0$ is a root of $A_n$ or $B_m$, then we can
write
\begin{gather*}
A_n(x)=A_n^{(\alpha )}(x_0)\frac{(x-x_0)^{\alpha }}{\alpha !}
+\dots
+A_n^{(a_n)}(x_0)\frac{(x-x_0)^{a_n}}{a_n!},\;\alpha
=v_{x_0}(A_n),
\\
B_m(x)=B_m^{(\beta )}(x_0)\frac{(x-x_0)^{\beta }}{\beta
!}+\dots
+B_m^{(b_m)}(x_0)\frac{(x-x_0)^{b_m}}{b_m!},\;\beta
=v_{x_0}(B_m).
\end{gather*}
\begin{proof}[Proof of Theorem \ref{thm2}]
First note that if $u$ is a rational
solution of an elliptic equation \eqref{A}, then a pole of $u$ is
a root of $A_n$. Indeed, let $\alpha $ be a pole of $u$ with
order $k>0$. If $A_n(\alpha )$ is not null, then the valuation
of\ $P(x,u)$ (as a function of $x$) at $ \alpha $ is exactly $-nk$
and the valuation of $Q(x,y)y'$ (as a function of $x$) at $\alpha
$ is at least $-mk-k-1$. Since $n>m+2$, the equality
$Q(x,u)u'=P(x,u)$ is then impossible.
Now let $y$ be a rational solution of \eqref{A}. Then it can be
written as $ u/A_n$ where $u$ is rational. From \eqref{A}, we
have
\[
\big(B_m(\frac{u}{A_n}) ^{m}+\dots +B_0\big)
\big(\frac{u'A_n-uA_n'}{A_n^{2}}\big)
=A_n(\frac{u}{A_n}) ^n+\dots +A_1\frac{u}{A_n}+A_0.
\]
But $n-1\geq m+2$, thus
\[
(A_n^{n-m-3}B_mu^{m}+\dots
+A_n^{n-3}B_0)(u'A_n-uA_n')=u^n+\dots
+A_n^nA_1\frac{u}{A_n} +A_n^{n-1}A_0,
\]
and
\begin{align*}
&(A_n^{n-m-2}B_mu^{m}+\dots +A_n^{n-2}B_0)u'\\
&= u^n+\dots
+(A_n^{n-m-3}B_mu^{m+1}+\dots
+A_n^{n-3}B_0u)A_n'+\dots +A_n^{n-1}A_0,
\end{align*}
so that \eqref{A} becomes the so called ``reduced
equation''
\begin{equation}
(\tilde{B}_mu^{m}+\dots +\tilde{B}_0)u'=u^n+\tilde{A}
_{n-1}u^{n-1}+\dots +\tilde{A}_0 \label{R}
\end{equation}
where $\tilde{B}_i$, $\tilde{A}_i$ are polynomials and
$\tilde{B}_m$ is not identically zero. Note that \eqref{R} is
also elliptic. Thus by what we have discussed above, a pole
$\alpha $ of $u$ as a solution of \eqref{R} must be a root of the
leading coefficient of the right hand side. But since this
coefficient is $1$, $u$ cannot have any poles. We conclude that
$u$ is a polynomial.
Suppose \eqref{A} is hyperbolic. Let $y$ be a rational function
and $\alpha $ a pole of order $k$ $>0\ $of $y$. If $B_m(\alpha
)$ is not null, then the valuation of\ $Q(x,y)y'$ (as a function
of $x$) at $\alpha $ is exactly $-mk-k-1$ and the valuation of
$P(x,y)$ (as a function of $x$) at $ \alpha $ is at least $-nk$.
Since $n\leq m+1$, the equality $Q(x,y)y'=P(x,y)$ is then
impossible, unless $B_m(\alpha )=0$. We may conclude that any
rational solution of \eqref{A} is of the form $u/B_m^{r} $where
$ u$ is a polynomial and $r\in \mathbb{N}$.
Let $x_0$ a root of $B_m$ of order $v_{x_0}(B_m)\in
\mathbb{N}^{*}$ ,\ and $y$ a rational solution with the pole
$x_0$:
\[
y=\frac{c}{(x-x_0)^{-v_{x_0}(y)}}+R,
\]
where $c\in \mathbf{C\backslash }\{0\}$, $R$ is rational and $
v_{x_0}(R)>v_{x_0}(y)$.
Let us show that there exists $k_{x_0}'\in \mathbb{N}$ (which can
be determined) such that
\[
-v_{x_0}(y)\leq k_{x_0}'.
\]
First there exists a least integer $k_{x_0}\in \mathbb{N}^{\ast }$
which can easily be determined, such that for any integer $k\geq
k_{x_0}$, we have
\[
nk-v_{x_0}(A_n)>ik-v_{x_0}(A_i)
\]
for $i=0,\dots,n-1$, and
\[
mk-v_{x_0}(B_m)+k+1>ik-v_{x_0}(B_i)+k+1
\]
for $i=0,\dots,m-1$. (In practice one uses $
mk-v_{x_0}(B_m)>ik-v_{x_0}(B_i).$)
Next, if $m+1>n$, then $mk-v_{x_0}(B_m)+k+1>nk-v_{x_0}(A_n)$
so that $(m+1-n)k+1>v_{x_0}(B_m)-v_{x_0}(A_n)$ for
sufficiently large $k$.
If $nv_0(y)$.
Let us find $k_0\in \mathbb{N}$ such that for any integer $k\geq
k_0$, we have
\[
3k-v_0(A_{3})>ik-v_0(A_i)
\]
for $i=0,1,2$ and
\[
2k-v_0(B_2)>ik-v_0(B_i)
\]
for $i=0,1$.
Since $A_2=A_1=B_1\equiv 0$, we see that $
v_0(A_2)=v_0(A_1)=v_0(B_1)=+\infty $, and $v_0(A_{3})=0=\alpha $,
$v_0(A_0)=0$, $v_0(B_2)=1=\beta $, $v_0(B_0)=0$. Therefore, it is
clear that $k_0=1$.
Here $n=m+1=3$ and $v_0(A_{3})=v_0(B_2)-1=0$, then put $
v_0(y)=\gamma $, so that replacing $y$ by $(cx^{\gamma }+R) $in
\eqref{A} , as in proof of Theorem \ref{thm2}, we obtain
\[
\gamma =(\alpha +1)\frac{A_n^{(\alpha )}(x_0)}{B_m^{(\alpha
+1)}(x_0) }=1
\]
since $\alpha =0,\ n=3,\ m=2,\ x_0=0,\ A_{3}(0)=1$ and
$B_2^{(1)}(0)=1$. Since $\gamma >0$, one takes $\varrho =k_0=1$.
The reduced equation satisfied by the polynomial $u$ is
\begin{equation}
u'=\frac{2u^3-xu-x^3}{xu^{2}-x^{2}}. \label{16}
\end{equation}
Here $\Omega =\{1,2\}$. Then $u^{(3)}=0$. By differentiating both
sides of ( \ref{16}), we obtain
\[
u''=\frac{(
2u^{4}-5u^{2}x+2ux^3+x^{2}) u'}{x(x-u^{2})
^{2}}-\frac{(2u^{5}-4u^3x+2u^{2}x^3+ux^{2}-x^{4})
}{x^{2}(x-u^{2}) ^{2}}.
\]
Replacing $u'\ $by $\frac{2u^3-xu-x^3}{xu^{2}-x^{2}}$, we see
that
\[
u''=\frac{2(
-u^{7}+3u^{5}x-u^3x^{2}-3u^{2}x^{4}+ux^{6}+x^{5})
}{x^{2}(x-u^{2}) ^3}.
\]
By differentiating both sides again, we obtain
\begin{align*}
u^{(3)} &= \frac{(
u^{8}-4u^{6}x+12u^{4}x^{2}-12u^3x^{4}+5u^{2}x^{6}-3u^{2}x^3+x^{7})
u'}{x^{2}(x-u^{2}) ^{4}} \\
&\quad +\frac{2u(
-2u^{8}+8u^{6}x-12u^{4}x^{2}+6u^3x^{4}-4u^{2}x^{6}+3u^{2}x^3+x^{7}
) }{x^3(x-u^{2}) ^{4}}.
\end{align*}
If we replace $u'\ $by $\frac{2u^3-xu-x^3}{xu^{2}-x^{2}}$ and
$u^{(3)}$ by $0$ in the above equation, we see that
\begin{equation}
\begin{split}
0 &= 2u^{11}-11xu^{9}+x^3u^{8}+12x^{2}u^{7}+8x^{4}u^{6}-(
2x^{6}+12x^3) u^{5} \notag \\
&\quad +12x^{5}u^{4}+(3x^{4}-19x^{7}) u^3+(
5x^{9}-3x^{6}) u^{2}+3x^{8}u+x^{10}.
\end{split} \label{AL''}
\end{equation}
We may conclude that $u$ divides $x^{10}$. Thus $y$ is a constant
function or $u=\lambda x$ or $u=\lambda x^{2}$, where
$\lambda \in \mathbb{C} \backslash \{0\}$. It is clear that \eqref{16}
has non-constant solutions only, because replacing $y$ by a constant
$\lambda $ in \eqref{16}, we obtain for all $x\in \mathbb{C}$
that $2\lambda ^3-x\lambda -x^3=0$ which is impossible. If
$u=\lambda x$ then
\[
\lambda =\frac{2(\lambda x)^3-x(\lambda x)-x^3}{x(\lambda
x)^{2}-x^{2}};
\]
i.e., $\lambda ^3=1$. One concludes that $u=x,xe^{\frac{2i\pi
}{3}},xe^{ \frac{4i\pi }{3}}$. If $u=\lambda x^{2}$ then
\[
2\lambda x=\frac{2(\lambda x^{2})^3-x(\lambda
x^{2})-x^3}{x(\lambda x^{2})^{2}-x^{2}};
\]
i.e., $\lambda =1$. One concludes that $u=x^{2}$.
Finally $y=1,e^{\frac{2i\pi}{3}},e^{\frac{4i\pi }{3}}$ or $x$.
\end{example}
\section{The parabolic case}
\begin{theorem} \label{thm3}
Let us consider the differential equation
\begin{equation}
y'=\frac{A_{m+2}y^{m+2}+\dots +A_0}{B_my^{m}+\dots +B_0},
\label{B}
\end{equation}
where $m\in \mathbb{N}^{*}$,
$A_i$, $B_i$ are polynomials such that
$A_{m+2}$ and $B_m$ are not identically
zero, and $A_{m+2}y^{m+2}+\dots +A_0$ and
$B_my^{m}+\dots +B_0$ are coprime.
Then \eqref{B} admits a finite number of rational solutions.
\end{theorem}
\begin{proof} There are two cases.
\textbf{Case 1:}
Suppose $A_0=0$, (i.e. $y=0$ is a solution). Then
$ B_0\neq 0$. Let $z=1/y$, equation \eqref{B} becomes
\begin{equation}
z'=-\frac{A_{m+2}+\dots +A_1z^{m+1}}{B_m+\dots +B_0z^{m}},
\label{B1}
\end{equation}
which is hyperbolic. Thus equation \eqref{B} admits a finite
number of rational solutions.
\textbf{Case 2:} Suppose $A_0\neq 0$. If \eqref{B} admits a
rational solution $f$. If $z=y-f$, equation \eqref{B} becomes
\begin{equation}
z'=\frac{C_{m+2}z^{m+2}+\dots +C_1z}{D_mz^{m}+\dots +D_0},
\label{B2}
\end{equation}
where $C_i$, $D_i$ are polynomials, $C_0=0\ $and $D_0$ is not
identically zero. This is exactly the first case; i.e., $z=0$ is a
solution. Let $\varphi =\frac{1}{z}$ then \eqref{B}
becomes hyperbolic which has a finite number of rational solutions
$\varphi $. But $\varphi = \frac{1}{z}=\frac{1}{y-f}$, thus
$y=\frac{1}{\varphi }+f$.
\end{proof}
As a corollary, we can compute all the rational solutions of
\eqref{B} if we have at least one particular rational solution of
\eqref{B}.
\begin{example} \label{examp5} \rm
Consider the equation
\begin{equation}
y'=\frac{y^{4}-y}{-y^{2}+x}. \label{A1''}
\end{equation}
This equation is parabolic, we can compute all its polynomial
solutions. Furthermore, if we find a polynomial solution of
\eqref{A1''}, we can compute all its rational solutions. Since
$\Omega =\emptyset$, we only look for constant solutions. This
leads us to $y'=0$, and $ y^{4}-y=0$. Thus
$y=0,1,e^{2i\pi/3},e^{4i\pi/3}$. We have four constant solutions of
\eqref{A1''}. Let $z=1/y$ (as in the proof of
Theorem \ref{thm3}, $z=1/(y-f)$ with $f=0$). From \eqref{A1''}, we have
\begin{equation}
z'=\frac{z^3-1}{xz^{2}-1}, \label{B1'}
\end{equation}
which is the same equation in Example \ref{examp4}. In conclusion,
$0,1,e^{2i\pi /3},e^{4i\pi/3}$ and $1/x$ are the
rational solutions of \eqref{A1''}.
\end{example}
\section{The quasi-linear and Riccati cases}
Suppose first that \eqref{A} is quasi-linear. It suffices to
consider the equation
\begin{equation}
B_0y'=A_1y+A_0. \label{QL}
\end{equation}
We may first determine $\delta $ (the upper bound of $\deg y$)
by Lemma \ref{lem1}. Replacing $y$ by
$y_{\delta }x^{\delta }+\dots +y_0$ in \eqref{QL}, we obtain
\[
B_0(\delta y_{\delta }x^{\delta -1}+\dots +y_1)=A_1(y_{\delta
}x^{\delta }+\dots +y_0)+A_0.
\]
Then rearranging terms in the resulting equation, we obtain
\[
K_{l}x^{l}+K_{l-1}x^{l-1}+\dots +K_0=0,
\]
where each $K_i$ may depend on $y_0,y_1,\dots,y_{\delta }$,
which is equivalent to following linear system:
$K_{l}=K_{l-1}=\dots =K_0=0$. After solving this system, we obtain
$y_i$.
Next let us compute rational solutions of \eqref{QL}. If
$A_1\equiv 0$, by means of the (classical) partial fraction
decomposition, we see that
\[
y'=\frac{A_0}{B_0}=p(x)+\sum_{d,\alpha }\frac{c(d,\alpha )}{
(x-\alpha )^{d}},
\]
where $p(x)$ is a polynomial, $c(d,\alpha )\in \mathbb{C}$ and the
sum is over the set of roots $\alpha $ of $B_0$ with multiplicity
$d$. Using direct integration, we see that a solution $y$ is
rational if and only if $ c(1,\alpha )=0$ for all $\alpha $.
If $A_1\neq 0$, by Theorem \ref{thm2}, there exits
$\varrho \in \mathbb{N}$
such that $y=u/B_0^{\varrho }$, where $u$ is a
polynomial. Replacing $y$ by $u/B_0^{\varrho }$ in
\eqref{QL}, we obtain
\[
B_0u'=(A_1+\varrho B_0')u+B_0^{\varrho }A_0.
\]
We may then determine $u$.
Note that the number of polynomial or rational solutions of
\eqref{QL} may not be finite. As an example, the equation
\[
xy'=y+x^{2},
\]
has polynomial solutions of the form $x^{2}+\lambda x$ where
$\lambda $ is an arbitrary complex number. As another example,
the equation
\[
y'=\frac{1}{x^{2}}
\]
has rational solutions of the form $-\frac{1}{x}+\lambda $ where
$\lambda $ is an arbitrary complex number.
We now suppose \eqref{A} is Riccati. It suffices to consider the
equation
\begin{equation}
B_0y'=A_2y^{2}+A_1y+A_0 \label{Ri}
\end{equation}
By Theorem \ref{thm1}, we can compute all its polynomial solutions
(finite number). If we have a rational solution $f$ of \eqref{Ri},
then by letting $z=1/(y-f)$, we obtain
\[
-B_0z'=(2fA_2+A_1)z+A_2,
\]
which is a quasi-linear equation. Again, we remark that the number
of rational solutions of \eqref{Ri} may not be finite. For
example, all solutions of the Riccati equation $y'=-y^{2}$ are of
the form
\[
y=\frac{1}{x+\lambda }
\]
where $\lambda $ is an arbitrary complex number.
As our final remark. we can find in \cite[Chapter I]{i1}
elementary methods of integration of classical ODE which can be
used to find the desired solutions in this section.
\begin{thebibliography}{0}
\bibitem{b1} D. Behloul and S. S. Cheng;
\emph{Computation of all polynomial solutions of a class of
nonlinear differential equations}, Computing,
77(2006), 163--177.
\bibitem{b2} D. Behloul and S. S. Cheng;
\emph{Polynomial solutions of a class of
algebraic differential equations with quadratic nonlinearities},
Southeast Asian Bulletin of Mathematics, 33(2009), 1029--1040.
\bibitem{c1} J. G. Campbell and M. Golomb;
\emph{On the polynomial solutions of a Riccati equation},
American Mathematical Monthly, 61(1954), 402--404.
\bibitem{g1} J. Gin\'{e}, M. Grau and J. Llibre;
\emph{On the polynomial limit cycles of polynomial differential
equations}, Israel Journal of Math,
181(2011), 461--475.
\bibitem{i1} E. L. Ince;
\emph{Ordinary Differential Equations}, Dover Publications,
New York, 1956.
\bibitem{m1} J. Malmquist;
\emph{Sur les fonctions a un nombre fini de branches
d\'{e}finies par les \'{e}quations diff\'{e}rentielles du premier
ordre}, Acta Math., 36(1) (1913), 297--343.
\bibitem{r1} E. D. Rainville;
\emph{Necessary conditions for polynomial solutions of
certain Riccati equation}, American Mathematical Monthly,
43 (1936), 473--476.
\end{thebibliography}
\end{document}