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. !enter image description here !enter image description here 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.

Excellent work on parts a and b! You're spot on, all the way.

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!