Consider the system $$ \dot{x}=x+ay-y^3,\quad \dot{y}=b-2y+x. $$ Analyse the existence of equilibria and determine their bidurcation.
The equilibria can be determined by setting $$ 0=x+ay-y^3,\quad 0=b-2y+x. $$ This yields $x=2y-b$ and $y^3-(a+2)y+b=0$.
Now, in order to have fold bifurcation that can yield to born equilibria, we need to have simple eigenvalue $\lambda=0$. Hence, I consider the linearization matrix which is $$ \begin{pmatrix}1 & a-3y^2\\1 & -2\end{pmatrix}. $$ Looking at the char. polynomial and equating it to 0, I get $$ \mu^2+\mu-2-a+3y^2=0 $$ so that this has simple eigenvalue $0$ iff $$ a+2=3y^2. $$ Hence, I get from the last equilibrium equation from above that $$ y^3-3y^3+b=0\Leftrightarrow b=2y^3. $$ Moreover, I get $$ y=\pm\left(\frac{a+2}{3}\right)^{1/2} $$ hence $b=\pm 2\left(\frac{a+2}{3}\right)^{3/2}$. So when I draw the $(a,b)$-diagram, i.e. a on the $x$-axis and $b$ on the $y$-axis, I get the bifurcation curve $L$ which looks as
where the blue line is $b=2\left(\frac{a+2}{3}\right)^{3/2}$ and the red line is $b=-2\left(\frac{a+2}{3}\right)^{3/2}$.
How can I now see where exactly I have how many equilibria? Don't know whether for parameter values $b$ larger or smaller as the values on the bifurcation curve two equiliria are born resp. vanish.