My question is about stability analysis and whether the equilibrium point is well-defined in example 2?. If they are not defined, how can one approach the stability analysis for those cases.

Example 1:

$ \frac{dx}{dt} = -x + c$

In this example, $ c$ is a constant and $ x=c$ is an asymptotically stable equilibrium point. If $ x<c$ , $ \frac{dx}{dt}$ will be positive, pushing $ x$ toward $ c$ and a similar statement holds for the negative case. In this case, the equilibrium point is well defined.

Example 2: We have a slightly modified equation:

$ \frac{dx}{dt} = -x + c(1-\frac{1}{M}\exp(-2t))$

and M is a very large positive constant. No matter where we start from, we’ll always approach $ c$ . Can we say $ x=c$ is also an equilibrium point in this case? The problem is when we plug this value in the equation $ \frac{dx}{dt}$ will not be zero at any time. My understanding is for an equilibrium point, we must be able to plug in the state at the equilibrium point and observe that it doesn’t change.