Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 7 / a / i / Solution

A stable node hasBoth exponential modes decay as $t\to \infty$, so the equilibrium is asymptotically stable. Generic linear orbits approach tangent to the eigenvector for the less negative eigenvalue; the eigenvector directions themselves are invariant straight lines. There is no rotation because both eigenvalues are real.