A.4.4: Section 4.4 Answers

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1. \(\overline{y}=0\) 0 is a stable equilibrium; trajectories are \(v^{2}+\frac{y^{2}}{4}=c\)

2. \(\overline{y}=0\) 0 is an unstable equilibrium; trajectories are \(v^{2}+\frac{2y^{3}}{3}=c\)

3. \(\overline{y}=0\) 0 is a stable equilibrium; trajectories are \(v^{2}+\frac{2|y|^{3}}{3}=c\)

4. \(\overline{y}=0\) 0 is a stable equilibrium; trajectories are \(v^{2}-e^{-y}(y+1)=c\)

5. equilibria: \(0\) (stable) and \(−2, 2\) (unstable); trajectories: \(2v^{2} − y^{4} + 8y^{2} = c\); separatrix: \(2v^{2} − y^{4} + 8y^{2} = 16\)

6. equilibria: \(0\) (unstable) and \(−2, 2\) (stable); trajectories: \(2v^{2} + y^{4} − 8y^{2} = c\); separatrix: \(2v^{2} + y^{4} − 8y^{2} =0\)

7. equilibria: \(0, −2, 2\) (stable), \(−1, 1\) (unstable); trajectories: \(6v^{2} + y^{2}(2y^{4} − 15y^{2} + 24) = c\); separatrix: \(6v^{2} + y^{2} (2y^{4} − 15y^{2} + 24) = 11\)

8. equilibria: \(0, 2\) (stable) and \(−2, 1\) (unstable); trajectories: \(30v^{2} + y^{2}(12y^{3} − 15y^{2} − 80y + 120) = c\); separatrices: \(30v^{2} + y^{2} (12y^{3} − 15y^{2} − 80y + 120) = 496\) and \(30v^{2} + y^{2} (12y^{3} − 15y^{2} − 80y + 120) = 37\)

9. No equilibria if \(a < 0; 0\) is unstable if \(a = 0\); \(\sqrt{a}\) is stable and \(−\sqrt{a}\) is unstable if \(a > 0\).

10. \(0\) is a stable equilibrium if \(a ≤ 0\); \(−\sqrt{a}\) and \(\sqrt{a}\) are stable and \(0\) is unstable if \(a > 0\).

11. \(0\) is unstable if \(a ≤ 0\); \(−\sqrt{a}\) and \(\sqrt{a}\) are unstable and \(0\) is stable if \(a > 0\).

12. \(0\) is stable if \(a ≤ 0; 0\) is stable and \(−\sqrt{a}\) and \(\sqrt{a}\) are unstable if \(a ≤ 0\).

22. An equilibrium solution \(\overline{y}\) of \(y'' + p(y) = 0\) is unstable if there’s an \(€> 0\) such that, for every \(δ > 0\), there’s a solution of (A) with \(\sqrt{(y(0)-\overline{y})^{2}+v^{2}(0)}<\delta \), but \(\sqrt{(y(t)-\overline{y})^{2}+v^{2}(t)}\geq €\) for some \(t>0\).