Question 7.4.3: Finding Equilibrium Solutions of a System of Equations Find ...

If (x, y) is an equilibrium solution, then the constant functions x(t) = x and
y(t) = y satisfy the system of equations with x′(t) = 0 and y′(t) = 0. Substituting into the equations, we have

0=0.3 y-0.2 y^2-0.1 x y.

Notice that both equations factor, to give

0=0.1 y(3-2 y-x).

The equations are equally complicated, so we work with both equations simultaneously. From the top equation, either x = 0 or x + 4y = 4. From the bottom equation, either y = 0 or x + 2y = 3. Summarizing, we have

and  y=0 \quad \text { or } \quad x+2 y=3.

Taking x = 0 from the top line and y = 0 from the bottom gives us the equilibrium solution (0, 0). Taking x = 0 from the top and substituting into x + 2y = 3 on the bottom, we get y=\frac{3}{2} \text { so that }\left(0, \frac{3}{2}\right) is a second equilibrium solution. Note that (0, 0) corresponds to the case where neither species exists and (0, 3\2) corresponds to the case where species Y exists but species X does not.
Substituting y = 0 from the bottom line into x + 4y = 4, we get x = 4 so that (4, 0) is an equilibrium solution, corresponding to the case where species X exists but species Y does not. The fourth and last possibility has x + 4y = 4 and x + 2y = 3. Subtracting the equations gives 2 y=1 or y=\frac{1}{2} With y=\frac{1}{2}, x+2 y=3 gives us x = 2. The final equilibrium solution is then (2, 12). In this case, both species exist, with 4 times as many of species X.
Since we have now considered all combinations that make both equations true, we have found all equilibrium solutions of the system: \text (0,0),\left(0, \frac{3}{2}\right),(4,0) \text { and }\left(2, \frac{1}{2}\right).