1. **Stating the problem:** We have the system of ordinary differential equations (ODEs): $$X' = X(1 - Y)$$ $$Y' = \frac{1}{3} Y \left(X - \frac{3}{2}\right)$$ We want to analyze the system and draw its phase portrait. 2. **Find equilibrium points:** Equilibrium points occur where $X' = 0$ and $Y' = 0$ simultaneously. From $X' = X(1 - Y) = 0$, either: - $X = 0$ or - $Y = 1$ From $Y' = \frac{1}{3} Y \left(X - \frac{3}{2}\right) = 0$, either: - $Y = 0$ or - $X = \frac{3}{2}$ Combining these: - If $X=0$, then $Y=0$ or $Y=1$ gives equilibria at $(0,0)$ and $(0,1)$. - If $Y=0$, then $X=0$ or $X=\frac{3}{2}$ gives equilibria at $(0,0)$ and $(\frac{3}{2},0)$. - If $Y=1$, then $X=0$ or $X=\frac{3}{2}$ gives equilibria at $(0,1)$ and $(\frac{3}{2},1)$. So the equilibrium points are: $$ (0,0), (0,1), \left(\frac{3}{2},0\right), \left(\frac{3}{2},1\right) $$ 3. **Linearize the system at each equilibrium:** Compute the Jacobian matrix $$J = \begin{bmatrix} \frac{\partial X'}{\partial X} & \frac{\partial X'}{\partial Y} \\ \frac{\partial Y'}{\partial X} & \frac{\partial Y'}{\partial Y} \end{bmatrix}$$ Calculate partial derivatives: - $\frac{\partial X'}{\partial X} = 1 - Y$ - $\frac{\partial X'}{\partial Y} = -X$ - $\frac{\partial Y'}{\partial X} = \frac{1}{3} Y$ - $\frac{\partial Y'}{\partial Y} = \frac{1}{3} (X - \frac{3}{2})$ Evaluate $J$ at each equilibrium: - At $(0,0)$: $$J = \begin{bmatrix} 1 - 0 & -0 \\ \frac{1}{3} \cdot 0 & \frac{1}{3} (0 - \frac{3}{2}) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{2} \end{bmatrix}$$ Eigenvalues: $1$, $-\frac{1}{2}$ (saddle point). - At $(0,1)$: $$J = \begin{bmatrix} 1 - 1 & -0 \\ \frac{1}{3} \cdot 1 & \frac{1}{3} (0 - \frac{3}{2}) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ \frac{1}{3} & -\frac{1}{2} \end{bmatrix}$$ Eigenvalues: Solve $\lambda^2 + \frac{1}{2} \lambda = 0$ gives $\lambda = 0, -\frac{1}{2}$ (non-hyperbolic). - At $(\frac{3}{2},0)$: $$J = \begin{bmatrix} 1 - 0 & -\frac{3}{2} \\ \frac{1}{3} \cdot 0 & \frac{1}{3} (\frac{3}{2} - \frac{3}{2}) \end{bmatrix} = \begin{bmatrix} 1 & -\frac{3}{2} \\ 0 & 0 \end{bmatrix}$$ Eigenvalues: $1$, $0$ (non-hyperbolic). - At $(\frac{3}{2},1)$: $$J = \begin{bmatrix} 1 - 1 & -\frac{3}{2} \\ \frac{1}{3} \cdot 1 & \frac{1}{3} (\frac{3}{2} - \frac{3}{2}) \end{bmatrix} = \begin{bmatrix} 0 & -\frac{3}{2} \\ \frac{1}{3} & 0 \end{bmatrix}$$ Eigenvalues satisfy $\lambda^2 + \frac{1}{2} = 0$, so $\lambda = \pm i \sqrt{\frac{1}{2}}$ (center, purely imaginary eigenvalues). 4. **Interpretation:** - $(0,0)$ is a saddle point. - $(0,1)$ and $(\frac{3}{2},0)$ are non-hyperbolic; stability requires nonlinear analysis. - $(\frac{3}{2},1)$ is a center with closed orbits nearby. 5. **Phase portrait:** - Trajectories near $(0,0)$ move away along unstable manifold and towards along stable manifold. - Near $(\frac{3}{2},1)$, trajectories form closed orbits. - The nullclines are: - $X$-nullcline: $X=0$ or $Y=1$ - $Y$-nullcline: $Y=0$ or $X=\frac{3}{2}$ These divide the plane into regions with different vector field directions. 6. **Summary:** The system has four equilibria with different stability types, and the phase portrait shows saddle behavior, centers, and nonlinear dynamics. This completes the analysis and phase portrait description.