# Bifurcation value and description

Find the bifurcation of $a$ and describe the bifurcation that take place at each value

$\displaystyle dy/dt=e^{-y^2}+a$

I let $\displaystyle e^{-y^2}+a=0$ then solve for y. I got $y^2=-\ln(a)$ What do I do next to find $a$?

You started well to let $e^{-y^2} + a =0$ and then solve for $y$. This will allow us to determine the equilibrium points. You will arrive at $$y^2 = -\ln(-a).$$ (I believe you misplaced a negative sign.) From this we see that first of all, we must have $a < 0$. In order to find bifurcation points, we need to consider what values of $a$ will yield a change in the nature of the equilibrium points. That is, what value of $a$ will cause a change in the number or behavior of the equilibrium points? Is there a value of $a$ that will cause no solution to the equation?
• Thank you for fixing my mistake, I did forget the negative sign infron ogf a. I try some value of $a$ and notice that If $a<-1$, no equilibrium point . There is one equilibrium point when $a=-1$. If $-1<a<0$, there are 2 equilibrias. If $a>0$, no equilibrium point. Am I correct? May 14, 2013 at 19:42
• That is what I have found. Therefore we say that the bifurcation occurs when $a=-1$ since this is the value of $a$ that changes the nature of the equilibrium solutions. The next step is to also describe the behavior of the equilibrium solutions. That is, are they saddles, sources or sinks? May 14, 2013 at 20:08