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(a) For $x' = 0$, we need $x = 0$ or $y=1$. For $y' = 0$, we need $y = 0$ or $x=1$. Thus the critical points are $(0,0)$ and $(1,1)$. The Jacobian matrix is $$\pmatrix{-1+y & x\cr -y & 1-x\cr}$$. At $(0,0)$ this is $$\pmatrix{-1 & 0\cr 0 & 1\cr}$$, which has eigenvalues $-1$ and 1. Therefore $(0,0)$ is a saddle. At $(1,1)$ the Jacobian matrix is $$\pmatrix{0 & 1\cr -1 & 0\cr}$$, which has eigenvalues $i$ and $-i$. This is a centre in the linearization, but we don't know yet whether it is a centre in the nonlinear system. However, we may be able to find out by doing part (c).

(c) By the Chain Rule we have

$$\frac{dy}{dx} = \frac{y'}{x'} = \frac{y (1-x)}{x(-1+y)}$$

This is separable, and we can solve it:

$$\int \frac{-1+y}{y} \, dy = \int \frac{1-x}{x} \, dx$$

$$-\ln \vert y\vert + y = \ln \vert x\vert - x + c$$

Thus the trajectories (apart from those on $x = 0$ and $y = 0$) have implicit equations $-\ln \vert y\vert + y - \ln \vert x\vert + x =$constant.

(a continued) Since the function $-\ln \vert y\vert + y - \ln \vert x\vert + x$ is smooth near $(1,1)$, its level curves there can be closed curves but not spirals. So $(1,1)$ really is a centre.

(b) Note that the trajectories entering and leaving the saddle $(0,0)$ are on the $x$ and $y$ axes respectively.