# Differential equations, stability of fixed points

Consider the differential equations:

$$\dot{x}=x^2-9$$

$$\dot{x}=x(x-1)(2-x)=-x^3+3x^2-2x$$

a. Find the stability type of each fixed point.

(I am not sure about the stability of the points. Do I determine the stability based on the attraction/repulsion of the points?)

$$\dot{x}=x^2-9$$ fixed points: $x=±3$

$f'(x)=2x$

$f'(3)=6$ >0 repelling

So if it is repelling, then it is unstable?

$f'(-3)=-6$ <0 attracting

So if it is attracting, then it is stable?

$$\dot{x}=x(x-1)(2-x)=-x^3+3x^2-2x$$

fixed points: $x=0$, $x=1$, $x=2$

$f'(x)=-3x^2+6x-2$

$f'(0)=-2 <0$ attracting, stable

$f'(1)=1 >0$ repelling, unstable

$f'(2)=-2 <0$ attracting, stable

b. Sketch the phase portrait on the line.  c. Sketch the graph of $x(t)=ϕ(t;x_0)$ for several representative initial conditions $x_0$.

I guess I am not really sure how to computer $x(t)=ϕ(t;x_0)$ or how to sketch the graph.

The idea for graphing is to let $t$ be your independent variable (that is, what is usually the "$x$-axis") and let $x$ be your dependent variable (that is, what is usually the "$y$-axis"), then graph as normal. Now, the key word in this is sketch. They aren't asking you (fortunately) to determine an explicit formula for $x(t)$ that applies for all $t\in\Bbb R$, using the ODE and your choice of initial condition. I recommend that you go ahead and draw the vector field on some grid, and use that to help you sketch what such a curve might look like. (Think back to your early multivariate calculus days for this one.) Fortunately, the vector field in each case will be well-behaved, and act precisely the same along any horizontal line. Don't forget to include the fixed points in the grid selection!