Finding critical points of $dX/dt = k(\alpha - x)(\beta -x)$ I need to find the critical points and the assess the stability of the following equation and describe its behavior as t approaches infinity if X(0) = 0.
$\frac{dX}{dt} = k(\alpha - x)(\beta -x)$
I was able to separate the variables and solve the differential equation with the initial condition. I got 
$\frac{ (\alpha - x)}{(\beta -x)} = \frac {\alpha}{\beta} e^{k(\beta - \alpha)t} $
I am now having trouble getting the critical points from this and sketching a phase diagram. Can someone please help?
 A: Part 1: Looks like you understand setting $x' = 0$ and how to find the two critical points:
$$x = \alpha, \beta$$
Part 2: Plot the direction field for $k \gt 0 = 1$, with $\alpha = \beta > 0 = 1$ (try those with other values and convince yourself that it doesn't really change things qualitatively, it just moves the single critical point):
$~~~~~~~~~~~~$
Do you notice the single critical point that is equal to one in the direction field? What happens if you choose an initial point above the critical point? What happens if you choose an initial point below the critical point?
Case 1: Using the values of Part 2, choose an $x(0) \lt \alpha = -1$ (since $\alpha = 1$) and describe the behavior of $x(t)$ as $t \rightarrow \infty$. We have:
$$x' = (1 - x)(1 -x), x(0) = -1$$
Solving this (it is separable), we have
$$x(t) = \dfrac{2 t-1}{2 t+1}$$
A plot of $x(t)$ shows:
$~~~~~~~~~~~~$
Do you see what happens as $t \rightarrow \infty$? What is the limit? 
If we plot the direction field using the initial condition, we have:

Case 2: Using the values of Part 2, choose an $x(0) \gt \alpha = 2$ (since $\alpha = 1$) and describe the behavior of $x(t)$ as $t \rightarrow \infty$. We have:
$$x' = (1 - x)(1 -x), x(0) = 2$$
Solving this (it is separable), we have
$$x(t) = \dfrac{t-2}{t-1}$$
A plot of $x(t)$ shows:
$~~~~~~~~~~~~$
Do you see what happens as $t \rightarrow \infty$? What is the limit? 
If we plot the direction field using the initial condition, we have:

