Determining equations for the compactified version of the system using $X$ and $Y$ as dependent variables Can someone please solve the problem? I am not sure how to approach it. Thank you for your time. 
Consider the system of equations $$x' = x+y-y^3$$ $$y' = -x+y+x^3.$$ I wish to determine equations for the compactified version of this system using $X$ and $Y$ as dependent variables. Desingularize the equation at infinity, i.e. for $Z=0,$ and how would the phase portrait look like?
 A: $\textbf{Solution:}$ Let $x = \frac{X}{Z}$ and $y= \frac{Y}{Z}$. So, $$X' = x'Z + xZ' = Z(x+y-y^3) + x\left(-\frac{(\frac{x(x+y-y^3) + y(-x+y+x^3)}{\sqrt{1+x^2+y^2}})}{(\sqrt{1+x^2+y^2})^2}\right)$$ $$=X+Y-\frac{Y^3}{Z^2} + \frac{X}{Z}\left(\frac{-\frac{X^2}{Z^2} - \frac{Y^2}{Z^2}+\frac{XY^3}{Z^4}-\frac{X^3Y}{Z^4}}{Z^{-3}}\right)$$ $$=X+Y - \frac{Y^3}{Z^2} + \frac{X}{Z}\left(-X^2Z-Y^2Z+ \frac{XY^3}{Z}-\frac{X^3Y}{Z}\right)$$ $$=X+Y - \frac{Y^3}{Z^2} - X^3 - XY^2+\frac{X^2Y^3}{Z^2}-\frac{X^4Y}{Z^2}$$ $$=\frac{1}{Z^2} (-Y^3 + X^2Y^3-X^4Y).$$ $$\text{Now, }Y'= y'Z + yZ' = Z(-x+y+x^2) + \frac{Y}{Z}(-X^2Z - Y^2Z + \frac{XY^2}{Z} - \frac{X^3Y}{Z})$$ $$=-X + Y + \frac{X^3}{Z^2} - X^2Y - Y^3 + \frac{XY^4 - X^3Y^2}{Z^2}$$ $$=\frac{1}{Z^2} (X^3 + XY^4 - Z^3Y^2).$$ Now, for $Z=0$ $$X=-Y^3 + X^2Y^3 - X^4Y = Y(-Y^2+X^2Y^2-X^4) \text{ and } Y= X^3 + XY^4 -X^3Y^2 = X(X^2+Y^4-X^2Y^2).$$ Observe $(0,\pm 1)$ is an equilibria point for $X,$ and $(\pm 1, 0)$ is an equilibria for $Y$. By comparing the right hand side equations with our given system, we arrive at $$Y^4 - X^4 -Y^2 + X^2= 0 \implies Y^4-X^4 = Y^2-X^2 \implies (X,Y) = (0,0).$$ However, by our established equilibrium points we have the following circle basically with points on the axis as $x=-1,1$ and at $y=-1,1$.
