A simple question about behavior of differential equation as it goes to infinity. Let $y^{\prime}=y+C$ where $C$ is any constant. 
The equilibrium value is $y=-C$
The behavior of y as $t\to \infty$ is what I am asking about. 
I think this is true: 


*

*If $t > -C $ then it goes to infinity,

*if $t < -C$ then it goes to negative infinity,

*lastly if $t=C$ it doesn't change.


If you have: $y^{\prime}=y^2-y-6$ which has two equilibrium solutions,$t=3,-2$
What exactly is the behavior as $t\to \infty$? 
is it: 


*

*If $t > 3$ goes to infity

*If $t = 3$ no change

*If $2<t<3$ goes to 2

*If $t = 2$ the no change

*If $t <2 $ goes to 2


Does this mean that 2 is more stable than 3? 
 A: In these types of problems, we are after the long term behavior of the system as a function of time. Also, we will often want to know if the behavior of the solution will depend on the value of the initial condition.
For the first example, if we solve $y' = y + C$, we get:
$$y(t) = A e^t - C$$
We find that the equilibrium is $-C$. 
Lets do an example with $C = 2$. If we take initial conditions around the equilibrium:


*

*$y(0) = -1 \rightarrow y(t) = e^t - 2$. So, what happens to $y(t)$ as $t \rightarrow \infty$?

*$y(0) = -2 \rightarrow y(t) = -2$. This is our equilibrium. So, what happens to $y(t)$ as $t \rightarrow \infty$?

*$y(0) = -3 \rightarrow y(t) = -e^t - 2$. So, what happens to $y(t)$ as $t \rightarrow \infty$?


If we plot the direction field for this example, we get:

If we choose any IC around the equilibrium, the behavior is qualitatively the same as above.
For the second example, we do a similar analysis. The solution is given by:
$$y(t) = \dfrac{-2e^{5(t+c)} - 3}{e^{5(t+c)}-1}$$
A direction field plot shows:

If you choose ICs around the equilibrium points, the solutions will be as this direction field plot shows (try this).  
I am not sure what you mean by $2$ is more stable than $3$? If we choose an IC around the equilibrium points, our solutions will either go to infinity or approach $y = -2$.  
