Phase plane diagram for system of non-linear odes Hello i was wondering if anyone could tell me if the phase diagram i have plotted is correct for the system below 
$\frac{dx}{dt}=-2x-y+2$ and $\frac{dy}{dt}=xy$
I calculated the equilibrium points to be $(\bar x,\bar y)=(0,2)$ and $(\bar x,\bar y)=(1,0)$
For the first equilibrium point i calculated that the eigenvalues to be $\lambda=-1\pm 2i$ which implies that this is a stable spiral?
For the second equilibrium point the eigenvalues are $\lambda_1=-2,\lambda_2=-1$ which implies that this equilibrium point is a stable node?
Now for the maple diagram my code was
     DEplot(sys, [x(t), y(t)], t = 10 .. -10, x = -3 .. 3, y = -3 .. 3, [[x(0) = 0, y(0) = 2]], [[x(0) = 0, y(0) = 1]], stepsize = .1, linecolor = blue, thickness = 2, arrows = medium);

where 
           sys := {diff(x(t), t) = -2*x(t)-y(t)+2, diff(y(t), t) = x(t)*y(t)}

This gave the following plot 
if this is incorrect can anyone tell me where i'm going wrong? it doesnt look right to me but i cant think of any other way plotting it using maple.
 A: The Jacobian is given by 
$$J[x, y] =\begin{bmatrix} \dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y} \\ \dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y} \end{bmatrix}=\begin{bmatrix} -2 & - 1 \\ y & x \end{bmatrix}$$
When we evaluate this at the first critical (equilibrium) point, $(x, y) = (2, 0)$, we have
$$J[x, y] = \begin{bmatrix} -2 & - 1 \\ 0 & 2 \end{bmatrix} \implies \lambda_{1,2} = -1 \pm  i \implies \mbox{Stable Spiral}.$$
When we evaluate this at the second critical point, $(x, y) = (0,1)$, we have
$$J[x, y] = \begin{bmatrix} -2 & - 1 \\ 0 & 1 \end{bmatrix} \implies \lambda_{1,2} = -2, 1 \implies \mbox{Unstable Saddle}.$$
As noted in comments, a saddle is always an unbstable equilibrium.
In this second matrix, we find the roots of characteristic polynomial using $|A-\lambda I| = 0 \implies \lambda ^2+\lambda -2 = (\lambda -1) (\lambda +2) = 0 \implies \lambda_1 = -2, \lambda_2 = 1$.
The phase portrait is correct. Here is another variant

Update
In the first matrix, the characteristic polynomial is given by:
$$ = \lambda ^2+2 \lambda +2 = (\lambda + (1 - i)) (\lambda + (1 + i)) \implies \lambda_{1,2} = -1 \pm i$$
