14.16: Section 4.4 Answers
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1. ¯y=0 0 is a stable equilibrium; trajectories are v2+y24=c
2. ¯y=0 0 is an unstable equilibrium; trajectories are v2+2y33=c
3. ¯y=0 0 is a stable equilibrium; trajectories are v2+2|y|33=c
4. ¯y=0 0 is a stable equilibrium; trajectories are v2−e−y(y+1)=c
5. equilibria: 0 (stable) and −2,2 (unstable); trajectories: 2v2−y4+8y2=c; separatrix: 2v2−y4+8y2=16
6. equilibria: 0 (unstable) and −2,2 (stable); trajectories: 2v2+y4−8y2=c; separatrix: 2v2+y4−8y2=0
7. equilibria: 0,−2,2 (stable), −1,1 (unstable); trajectories: 6v2+y2(2y4−15y2+24)=c; separatrix: 6v2+y2(2y4−15y2+24)=11
8. equilibria: 0,2 (stable) and −2,1 (unstable); trajectories: 30v2+y2(12y3−15y2−80y+120)=c; separatrices: 30v2+y2(12y3−15y2−80y+120)=496 and 30v2+y2(12y3−15y2−80y+120)=37
9. No equilibria if a<0;0 is unstable if a=0; √a is stable and −√a is unstable if a>0.
10. 0 is a stable equilibrium if a≤0; −√a and √a are stable and 0 is unstable if a>0.
11. 0 is unstable if a≤0; −√a and √a are unstable and 0 is stable if a>0.
12. 0 is stable if a≤0;0 is stable and −√a and √a are unstable if a≤0.
22. An equilibrium solution ¯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 √(y(0)−¯y)2+v2(0)<δ, but √(y(t)−¯y)2+v2(t)≥€ for some t>0.