Phase plane for guerrilla vs. conventional warfare For a class, I'm modeling a battle between a conventional force, $C(t)$, and a guerrilla force, $G(t)$. I already know that the time derivative of $C(t)=-gG$, where $g$ is the effectiveness of the guerrilla army, and the time derivative of $G(t)=-cCG$.
Then I take the ratio to get $dC/dG = g/cC$. Separating these and integrating, I have $gG = 0.5cC^2 +K$. I'm not sure what $K$ is supposed to be, but I include it since I integrated.
Now I must plot the phase plane, and I'm quite lost on how to do this. Does it have to do with plotting the slope field of $dC/dG=1/C$? I also must derive a condition that will tell who wins and loses.
 A: Please forgive my changing variable names because having a $c$ and $C$ is very confusing.
We have the system
$$\dfrac{dx}{dt} = -~c x y \\ \dfrac{dy}{dt} = -~ d x$$
Solving for $\dfrac{dy}{dx}$
$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{- d x }{-c x y} = \dfrac{d}{cy}$$
This is a separable equation, so we can seperate variables and integrate each side
$$\displaystyle c\int_{y_0}^y y ~ dy = d \int_{x_0}^x dx \implies \dfrac{c}{2}(y^2 - y_0^2) = d(x-x_0)$$
Simplifying
$$c~ y^2 - 2~ d~ x = c~y_0^2 - 2~ d~ x_0 = M$$
These are parabolas. Using the equation for a parabola above, the phase portrait (WLOG, let $c = d = 1$, but you can play around with other values, but qualitatively, basically the same as shown).

Observations: If $M > 0, y$ wins, if $M = 0$, tie, if $M < 0, x$ wins. You can certainly add three different colored initial points on the phase portrait as shown above.
It is worth noting that you could just plot the phase portrait without solving using details of the system, but you have likely learned these methods or can search them out.
