Algebraic solutions of a differential equation. Given a differential equation $y' = (1/x)(y^2 + y^3)$. 
My question is how does one go about finding the solutions of this differential equation which are algebraic over the field $\Bbb{C}(x)$,if any.  
Notation- $\Bbb{C}(x)$ is the quotient field of $\Bbb{C}[x]$.
 A: If you have an algebraic solution then you have an irreducible polynomial $P(x,y) \in \Bbb C[x,y]$ such that $P(x,y) = 0$.
Differentiating with respect to $x$ gives you $x (dP/dx) + (y^2+y^3) (dP/dy) = 0$ whenever $P(x,y)=0$, and so $x (dP/dx) + (y^2+y^3) (dP/dy) = QP$ for some polynomial $Q$.
Now if you look at the degrees, $Q$ has to have degree $0$ in $x$ and has degree at most $2$ in $y$, so $Q = a+by+cy^2$
Evaluating at $x=0$ you get $(y^2+y^3) dP/dy (0,y) = Q(y)P(0,y)$ which forces $a=0$ (or $P(0,y) = 0$, which forces $P(x,y) = \lambda x$ which is not an interesting solution).
Evaluating at $y=0$ you get $xdP/dx(x,0) = 0$, so either $P(x,y)=\lambda y$ or $P(x,0)$ is a nonzero constant. 
In that case, going back to the $x=0$ equation we get that $c$ has to be a positive integer, $P(0,y) = \lambda (1+y)^c$ and $Q(y) = cy^2$.
Let $P_k(y)$ be the coefficient of $x^k$. Then each $P_k$ has to satisfy the differential equation $(y^2+y^3)P_k' = (cy^2-k)P_k$, which has no nonzero polynomial solution for $k>0$ (just look at the nonzero coefficient of smallest degree). Therefore $P = \lambda (1+y)^c$. Since $P$ is irreducible we must have $c=0$ or $c=1$.
So the irreducible polynomial solutions of that equation are the multiples of $1,X,Y$, and $1+Y$. Hence the only algebraic solutions are the two constant solutions.
