# System of ODE's

The question asks me to consider the system of differential equations:

$\frac{dx}{dt} = 1 - 2x + x^2y$
$\frac{dy}{dt} = x-x^2y$

It asks me to find the fixed point(s), and determine their stability, also to draw the phase plane.

So to find the fixed points, I set both equations equal to zero:
$1 - 2x + x^2y = 0 \Rightarrow x^2y = 2x-1$
$x-x^2y = 0 \Rightarrow x^2y = x$

Setting them equal to each other:

$2x-1 = x \Rightarrow x=1$ and from this we can deduce $y=1$. So the fixed point is (1,1).

Now this is where I'm not really sure on what to do and get confused by my lecture notes. I believe I have to use some Jacobian to work out the stability yet I have no clue why i'm doing this (I know eigenvalues/eigenvectors are involved somehow). Setting $$f(x,y)= 1 - 2x + x^2y$$ and $$g(x,y)= x-x^2y$$ I worked out the Jacobian at (1,1) to be:

\left|
\begin{array}{cc}
0 & 1 \\
-1 & -1 \end{array}
\right|

(Sorry about the lack of detail about this part but I'm new to latex and struggled with doing matrices, the jacobian is fx and fy on the first row and gx and gy on the second row.

Following lecture notes, I work out the 'trace' (T) of this matrix which is simply (a11+a22) which is -1 in this case, and also work out the determinant which is 1.

Since $T^2 - 4D = -3 < 0$, we have complex eigenvalues and I believe this means the phase plane will be a spiral, and since T < 0 this means the general trajactory will be moving towards the fixed point (stable). But do I need to work out the eigenvectors to see exactly what this spiral will look like? I'm not sure if I've worked out enough to complete the question. Also any help on actually explaining why I'm working out the Jacobian stuff would be great.

Thanks