= Solution
At an <equilibrium point>, $x(1-x-y)=0$ and $2-2x-y=0$. The first equation gives $x=0$ or $y=1-x$. Substituting in the second gives, respectively,
$$
\boxed{(x,y)=(0,2)\quad\text{or}\quad(1,0).}
$$
These are all the <equilibrium points>. The <Jacobian matrix> of the <vector field> is
$$
J(x,y)=\begin{pmatrix}1-2x-y&-x\\-2&-1\end{pmatrix}.
$$
At $(0,2)$,
$$
J(0,2)=\begin{pmatrix}-1&0\\-2&-1\end{pmatrix}.
$$
The <eigenvalue> $-1$ has algebraic multiplicity two and <eigenspace> $\operatorname{span}\{(0,1)^T\}$. For the linearized coordinates $u=x$, $v=y-2$,
$$
u(t)=Ae^{-t},\qquad v(t)=(B-2At)e^{-t}.
$$
Both components tend to zero. The repeated <eigenvalue> with one independent <eigenvector> makes this a \b[stable improper <degenerate node>], a <sink equilibrium>. The term $t e^{-t}$ is the signature of the nontrivial <Jordan block>. Since it is a <hyperbolic equilibrium point>, the <linearization stability theorem> gives local <asymptotic stability> of the nonlinear system. Approaching trajectories are tangent to the vertical <eigendirection>.
At $(1,0)$,
$$
J(1,0)=\begin{pmatrix}-1&-1\\-2&-1\end{pmatrix},\qquad
\lambda_+=-1+\sqrt2,\quad v_+=\begin{pmatrix}1\\-\sqrt2\end{pmatrix},\qquad
\lambda_-=-1-\sqrt2,\quad v_-=\begin{pmatrix}1\\\sqrt2\end{pmatrix}.
$$
One <eigenvalue> is positive and one negative, so this point is a \b[<saddle equilibrium>]. Its <stable manifold> is tangent to $y=\sqrt2(x-1)$, and its <unstable manifold> is tangent to $y=-\sqrt2(x-1)$.
For the <phase portrait>, the <nullclines> are
$$
\dot x=0:\quad x=0\text{ or }y=1-x,\qquad
\dot y=0:\quad y=2-2x.
$$
Above the second line the vertical flow is downward, and below it upward. For $x>0$, horizontal flow is to the right below $y=1-x$ and to the left above it; for $x<0$ the directions reverse. The line $x=0$ is invariant, with $\dot y=2-y$, so it flows toward $(0,2)$. Together with the saddle's two invariant directions, these signs determine the sketch.
\Image[/past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2014-quadratic-phase-portrait.png]
{title=<Phase portrait> with the stable improper <degenerate node> at $(0,2)$, the <saddle equilibrium> at $(1,0)$, and their <nullclines>}
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