14.16: Section 4.4 Answers

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 $t>0$.