Posted by **Erica** on Thursday, February 7, 2013 at 9:47pm.

Locate the bifurcation values of a for the one-parameter family and describe the bifurcation that takes place at each such value.

dy/dt=e^(-y^2)+a

I got that the equilibrium points will be at -sqrt(ln(a)). I don't know if that is fine. I took the partial derivative with respect to y and got -2ye^(-y^2) and that is zero when a is 0 if I plug in -sqrt(ln(a)) as y.

## Answer This Question

## Related Questions

- Differential Equations - Locate the bifurcation values of a for the one-...
- differential equations - dS/dt = f(S) = kS(1-S/N)(S/M-1) of a fox squirrel ...
- differential equations - dS/dt = f(S) = kS(1-S/N)(S/M-1) of a fox squirrel ...
- Calculus - Second Order Differential Equations - Posted by COFFEE on Monday, ...
- physiology - Need help with this one, can't really locate the answer. Consider ...
- math - How do I know that x^2-7=y is a function? thanks you know its a function ...
- diffeq - how do i find the bifurcation for this: d^x/dt^2 = -.3 x - (x - 1)/((x...
- Differential Equations - a) Sketch the phase line for the differential equation ...
- Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...
- Math - A two population model is satisfied by a system of differential equations...

More Related Questions