 Solution

Intersect the nullclines $y=x(2-x)$ and $y=kx$. Their intersection equation is $x(2-k-x)=0$, giving the origin and $(2-k,k(2-k))$. The Jacobian matrix isAt the origin, $\mathrm{tr}M=3$, $\det M=2-k$ andFor $k>2$, only the origin is in the nonnegative quadrant, and it is a saddle because its determinant is negative. For $0<k<2$, the origin is an unstable node, with two distinct positive eigenvalues, and there is a second equilibrium in the open quadrant. At that second equilibrium, $\det M=k-2<0$, so it is a saddle. These signs give hyperbolic equilibria, so the nonlinear classifications follow from the linearization stability theorem. The boundary value $k=2$ is excluded from the two stated cases.