# Math Help - semistable equilibrium solutions

1. ## semistable equilibrium solutions

Hi, I need help on this problem

Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable

Consider the equations
dy/dt = k(1 - y)^2 where k is a positive constant. Show that y=1 is the only critical point, with the corresponding equilibrium solution theta(t) = 1.

2. Originally Posted by Recklessid
Hi, I need help on this problem

Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable

Consider the equations
dy/dt = k(1 - y)^2 where k is a positive constant. Show that y=1 is the only critical point, with the corresponding equilibrium solution theta(t) = 1.

i'm not sure about that "corresponding equilibrium solution theta(t) = 1" thing, but we can see that y = 1 is the only critical point since critical points occur when the derivative is 0 (or undefined, but this is a polynomial and hence defined everywhere, so we don't have to worry about that).

so for critical points, dy/dt = 0
=> k(1 - y)^2 = 0
=> k = 0 or (1 - y)^2 = 0
k cannot be zero, since it is a positive constant, therefore, (1 - y)^2 = 0
=> 1 - y = 0 ..............squared both sides
=> y = 1
thus this is the only critical point since it is the only solution to dy/dt = 0

3. Originally Posted by Recklessid
Hi, I need help on this problem

Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable

Consider the equations
dy/dt = k(1 - y)^2 where k is a positive constant. Show that y=1 is the only critical point, with the corresponding equilibrium solution theta(t) = 1.