I'm already familiar with bifurcations of differential equations with only $1$ parameter $\lambda$, like: $$ x'=-\lambda x - x^4 $$ but what if we're given a differential equation with $2$ parameters: $$ x'=\lambda-\mu x^2 +4x^4 \equiv F(x,\lambda,\mu) $$ My attempt:
The global minimum of $f(x)$ is the solution $x^*$ to the equation: $$ \frac{\partial F(x,\lambda,\mu)}{\partial x}=0, \quad\frac{\partial^2 F(x^*,\lambda,\mu)}{\partial x^2}>0 $$ which gives: $$ x=\pm \sqrt{\frac{\mu}{8}} $$ Plugging this solution to $F$'s formula, we yield: $$ a(\lambda,\mu)=\lambda-\frac{\mu^2}{8}+\frac{4\mu^4}{8^4} $$ which is the function of $F$'s mimima. Therefore, we have a bifurcation when we cross the curve: $$ \lambda=\frac{\mu^2}{8}+\frac{4\mu^4}{8^4} $$ in the $(\lambda, \mu)$ plane.
From here, how can one determine the number of equilibria and their stability for the different values of $\lambda, \mu$?