If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$ If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$
My Attempt:
$$p^3-p(x^2+xy+y^3)+x^2y+xy^2=0$$
$$p^3-p(x+y)^2+pxy+xy(x+y)=0$$
$$p(p^2-(x+y)^2)+xy(p+(x+y))=0$$
$$p\cdot (p+(x+y))\cdot (p-(x+y))+xy(p+(x+y))=0$$
$$(p+(x+y))\cdot (p(p-(x+y))+xy)=0$$
I don't get what to do after this.

We get three equations: $p=x,y-x-y$
We get three solutions: $$y=x^2/2+C, y=Ce^x, y=1-x+Ce^{-x}$$
The total solution of this first order ODE is
$$(x^2/2-C)(ye^{-x}-C)[(y-1+x)e^{x}-C]=0,$$
with only one contant of integration: $C$.
