A system of nonlinear differential equations We have the following system in $\mathbb{R}^{2}$
$$\dot{y}_1=2-y_1y_2-y_2^2$$
$$\dot{y}_2=2-y_1^2-y_1y_2$$
i) Calculate the equilibrium points en determine their stability.
ii) Draw the Phase Plot.
I know that I have to use the following to calculate the equilibrium points:
$$\dot{y}_1=0 , \dot{y}_2=0$$
That's all I know. So my question is how do I solve these equations, how do I determine the stability of the equilibrium points and how do I draw the phase plot of this system. 
An other question I have is what kind of solution you will get, I'm not able to understand what this system represents. So I would like to know how I should interpret such a system like this. Are we looking for something like $f(y_1,y_2)$ or do you need to find $y_1$ and $y_2$. And if so what would the solution mean?
 A: Critical Points:
$$2 - y_1 y_2 - y_2^2 = 0, 2 - y_1^2 - y_1 y_2 = 0 \implies (y_1, y_2) = (-1,-1),(1,1)$$
Jacobian:
$$J(y_1,y_2) = \begin{bmatrix} -y_2 & -2 y_1-y_2 \\ -y_1-2 y_2 & -y_1 \end{bmatrix}$$ 
Jacobian's eigenvalues  at each critical point:
$$J(-1,-1) = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}$$
The eigenvalues are $\lambda_1 = -2, \lambda_2 = 4$, which is a saddle point.
$$J(1,1) = \begin{bmatrix} -1 & -3 \\ -3 & -1 \end{bmatrix}$$
The eigenvalues are $\lambda_1 = -4, \lambda_2 = 2$,  which is a saddle point..
Phase portrait (you should learn to roughly draw by hand):

In an ideal world, we would like to find a closed-form solution for $y_1(t)$ and $y_2(t)$. For most nonlinear systems, this is not possible, so we resort to qualitative approaches like the above.
When you cannot find closed-form solutions, you can also resort to numerical methods to solve for $y_1(t)$ and $y_2(t)$. You can plot each of those solutions as functions of $t$ or parametrically plot $y_1(t) vs. y_2(t)$. You should try these both with a system you know the answers to (including drawing the phase portrait).
Now, what conclusions can be drawn from all of this?
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\left\lbrace\begin{array}{rcl}
     \dot{y}_{1} & = & 2 - y_{1}y_{2} - y_{2}^{2}
     \\
     \dot{y}_{2} & = & 2-y_{1}^{2} - y_{1}y_{2}
\end{array}\right.}$

$$
\dot{y}_{1} + \dot{y}_{2} = 4 - \pars{y_{1} + y_{2}}^{2}
\quad\imp\quad
\color{#00f}{\large y_{1} + y_{2} = 2\,{\expo{4t} - A \over \expo{4t} + A}}
\,,\quad A\ \mbox{is a constant}
$$

$$
\dot{y}_{1} - \dot{y}_{2} = \pars{y_{1} - y_{2}}\pars{y_{1} + y_{2}}
=\pars{y_{1} - y_{2}}\pars{2\,{\expo{4t} - A \over \expo{4t} + A}}
$$

$$
\ln\pars{y_{1} - y_{2}} = 2\bracks{-t + \half\,\ln\pars{A + \expo{4t}}}
+ \mbox{a constant}
$$

$$
\color{#00f}{\large y_{1} - y_{2} =
B\expo{-2t}\pars{A + \expo{4t}}}\,,\quad B\ \mbox{is a constant}
$$
