For the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = x (5-2x-3y)\, , \\
&y'_t = y (5-3x-2y)
\end{aligned}\right.
\end{equation*}
(a) describe the locations of all critical points,
(b) classify their types (including whatever relevant: stability, orientation, etc.),
(c) sketch the phase portraits near the critical points,
(d) sketch the phase portrait of this system of ODEs.
Solution(a) Solving $x (5-2x-3y=0$, $y (5-3x-2y )$ we have 4 cases $x=y=0$, $x=5-3x-2y=0$, $y=5-2x-3y=0$ and $5-3x-2y =5-2x-3=0$ giving us 4 points $(0,0)$, $(0,\frac{5}{2})$, $(\frac{5}{2},0)$ and $(1,1)$.
(b) Let $f= x (5-2x-3y)=5x-2x^2-3xy$, $g=y (5-3x-2y)=5y-2y^2-3xy$. Then $f_x=5-4x-3y$, $f_y=-3x$, $g_x=-3y$, $g_y=5-4y-3x$.
- $(0,0)$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals $\begin{pmatrix} 5 & 0 \\ 0 &5\end{pmatrix}$ with eigenvalues $r_1=r_2=5$; and eigenvectors $(1, 0)^T$ and $(0,1)^T$; unstable node;
- $(0,\frac{5}{2})$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals $\begin{pmatrix} -\frac{5}{2} & 0 \\ -\frac{15}{2} &-5\end{pmatrix}$ with eigenvalues $r_1=-\frac{5}{2},\ r_2=-5$ and eigenvectors $(1,-3)^T$ and $(1,0)^T$ respectively; stable node;
- $(\frac{5}{2},0)$; the same as in (2);
- $(1,1)$; matrix $\begin{pmatrix} f_x & f_y \\ g_x &g_y\end{pmatrix}$ at this point equals $\begin{pmatrix} -2 & -3 \\ -3 &-2\end{pmatrix}$ with eigenvalues $r_1=-5$ and $r_2=1$ and eigenvectors $(1,1)^T$ and $(1,-1)^T$ respectively; saddle.
(c-d) Plotting
Remark This is ``two competing species'' system.
