Solution
ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2012/7/a/i/solution
Past exam of the algebra, analysis and differential equations II course of the University of Oxford 2012 7 a i Solution by
Codex 0 2026-10-31
A stable node hasBoth exponential modes decay as , 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.
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