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« Last post by **Victor Ivrii ** on* April 18, 2018, 06:48:47 AM* »
**a.** Solving $x(x-y+1)=0$, $y(x-2)=0$ we get cases

\begin{align*}

&x=y=0 &&\implies A_1=(0,0),\\

&x=x-2=0 &&\implies \text{impossible}\\

&y=x-y+1=0 &&\implies A_2=(-1,0),\\

&x-y+1=x-2=0 &&\implies A_3=(2,3).

\end{align*}

**b. **Linearizations at these points have matrices

\begin{align*}

&

\begin{pmatrix}

1 &\ \ 0\\

0 &-2

\end{pmatrix}

&&

\begin{pmatrix}

-1 &\ \ 1\\

0 &-3

\end{pmatrix}

&&

\begin{pmatrix}

2 &-2\\

3 &0

\end{pmatrix}

\\[5pt]

\text{with eigenvalues }&\{1,-2\} &&\{-1,-3\} && \{1-\sqrt{5}i,1+\sqrt{5}i \}

\end{align*}

and therefore

* $A_1$ is a saddle,

* $A_2$ is a stable node, and

* $A_3$ is unstable focal point and since left bottom number is $3>0$ it is counterclockwise oriented.

**c.** Axis are:

in $A_1$: $\mathbf{e}_1=(1,0)^T$ unstable ($\lambda_1=1$), $\mathbf{e}_2=(0,1)^T$ stable ($\lambda_2=-2$).

in $A_2$: $\mathbf{f}_1=(1,0)^T$ ($\lambda_1=-1$), $\mathbf{f}_2=(1,-2)^T$ ($\lambda_1=-3$). Since $\lambda_1 >\lambda_2$, all trajectories have an entry directions $\pm \mathbf{f}_1$ (except two, which have entry directions $\pm \mathbf{f}_2$). Then we draw trajectories near critical points (See attachment P5-loc.png).

**d**. One should observe that either $x=0$ in every point of the trajectory, or in no point; and that $y=0$ in every point of the trajectory, or in no point. It allows us to make a "skeleton'' of the phase portrait (see attachment), impose local pictures on it and finally draw a global portrait