(b) Assume that a model describing two competing species in a closed environment is given by the following system of two differential equations: \[ \frac{d x}{d t}=x-x^{2}-2 x y, \] \( \frac{d y}{d t}=y-\left(2 x y+y^{2}\right) \), where \( x \) and \( y \) represent two interacting species. (i) Determine the \( x \)-isoclines and \( y \)-isoclines; and the signs of \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \). (ii) Determine the equilibrium points of the system. (iii) Draw the phase diagram for the system. Clearly label the equilibrium points, isoclines and the (8) trajectories. Please note that only hand-drawn phase diagrams will be marked. Thus, should you use other drawing tools (like online-plotters, chatbots, AI-assisted plotters or Miscrosoft paint etc.), your diagram will be awarded a mark of zero. Hint: You should consider only the 1st quadrant of the xy-plane.

To solve the problem step by step, we will follow the outlined parts (i), (ii), and (iii). ### (i) Determine the \( x \)-isoclines and \( y \)-isoclines; and the signs of \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \). **1. Finding the \( x \)-isoclines:** The \( x \)-isoclines are found by setting \( \frac{d x}{d t} = 0 \): \[ x - x^2 - 2xy = 0 \] Factoring out \( x \): \[ x(1 - x - 2y) = 0 \] This gives us two cases: - \( x = 0 \) (the x-axis) - \( 1 - x - 2y = 0 \) or \( y = \frac{1 - x}{2} \) **2. Finding the \( y \)-isoclines:** The \( y \)-isoclines are found by setting \( \frac{d y}{d t} = 0 \): \[ y - (2xy + y^2) = 0 \] Factoring out \( y \): \[ y(1 - 2x - y) = 0 \] This gives us two cases: - \( y = 0 \) (the y-axis) - \( 1 - 2x - y = 0 \) or \( y = 1 - 2x \) **3. Signs of \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \):** - For \( \frac{d x}{d t} \): - If \( x(1 - x - 2y) > 0 \), then \( \frac{d x}{d t} > 0 \) (population of species \( x \) increases). - If \( x(1 - x - 2y) < 0 \), then \( \frac{d x}{d t} < 0 \) (population of species \( x \) decreases). - For \( \frac{d y}{d t} \): - If \( y(1 - 2x - y) > 0 \), then \( \frac{d y}{d t} > 0 \) (population of species \( y \) increases). - If \( y(1 - 2x - y) < 0 \), then \( \frac{d y}{d t} < 0 \) (population of species \( y \) decreases). ### (ii) Determine the equilibrium points of the system. Equilibrium points occur when both \( \frac{d x}{d t} = 0 \) and \( \frac{d y}{d t} = 0 \). From the isoclines: 1. \( x = 0 \) gives \( y \) can be any value (but we consider only the first quadrant). 2. \( y = 0 \) gives \( x \) can be any value (but we consider only the first quadrant). 3. Setting \( 1 - x - 2y = 0 \) and \( 1 - 2x - y = 0 \): From \( y = \frac{1 - x}{2} \) and substituting into \( y = 1 - 2x \): \[ \frac{1 - x}{2} = 1 - 2x \] Multiplying through by 2: \[ 1 - x = 2 - 4x \] Rearranging gives: \[ 3x = 1 \implies x = \frac{1}{3} \] Substituting \( x = \frac{1}{3} \) back into \( y = 1 - 2x \): \[ y = 1 - 2 \cdot \frac{1}{3} = 1 - \frac{2}{3} = \frac{1}{3} \] Thus, the equilibrium point is: \[ \left( \frac{1}{3}, \frac{1}{3} \right) \] ### (iii) Draw the phase diagram for the system. To draw the phase diagram, we will sketch the isoclines and indicate the equilibrium point. 1. **Isoclines:** - The \( x \)-isocline \( y = \frac{1 - x}{2} \) is a line with a y-intercept of \( \frac{1}{2} \) and a slope of \( -\frac{1}{2} \). - The \( y \)-isocline \( y = 1 - 2x \) is a line with a y-intercept of 1 and a slope of \( -2 \). 2. **Equilibrium Point:** - Mark the point \( \left( \frac{1}{3}, \frac{1}{3} \right) \) on the diagram. 3. **Direction of Trajectories:** - Use the signs of \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \) to indicate the direction of trajectories in the first quadrant. ### Conclusion You can now sketch the phase diagram by hand, labeling the isoclines, equilibrium point, and indicating the direction of trajectories based on the signs of the derivatives.