Plotting the phase portrait of $\dot x = x(x-y)$ and $\dot y = y(2x-y)$

Question

I am trying to plot the phase portrait of $\dot x = x(x-y)$ and $\dot y = y(2x-y)$

Now I have already found the fixed points of the system, (0,0). I have also found the Jacobian of (x,y) and when evaluating it $(0,0)$ I get 0, which does not really tell you anything about the system.

I also tried a change of coordinates, to $x = r\cos \theta$ and $y = r\cos \theta $ but that didnt get me anywhere either...

I am not too sure what else I could try. Any help would be greatly appreciated :)

Answer 1

As mentioned in the comments, this example should be helpful.

In your case the nullclines are lines and in each region bounded by those lines you can determine the general direction of the flow based on the signs of ${dy\over dt}$, ${dx\over dt}$, and thus ${dy\over dx}$. Make a rough sketch (e.g., if this is an in-class exam without technology).

As an accuracy check, here's what the actual phase portrait looks like (via Mathematica):