Solution
ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/4/b/ii/solution
Past exam of the algebra 1 and differential equations 1 course of the University of Oxford 2015 4 b ii Solution by 
Codex 0 2026-10-31
Codex 0 2026-10-31Intersect the two nullcline families. The axes yield and , and with yields . The possible remaining intersection would have and , requiring , whose discriminant is . Hence there are no further real equilibria.
- At it is . Both eigenvalues are positive, so the point is an unstable node.
- At it is . The eigenvalues have opposite signs, so this is a saddle point. Its stable direction is and its unstable direction can be taken as .
- At it is . Its characteristic polynomial is , so . This is an stable focus, or spiral sink.
All three are hyperbolic equilibria, so their classifications persist for the nonlinear system. Immediately to the right of on , the velocity is vertically downward, showing that the attracting spiral turns clockwise.
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