Differential Equations and Linear Algebra, 4th Edition

Section 9.1

1. 1. 6.1.14

2. 2. 6.1.16

3. 3. 6.1.19

4. 4. 6.1.13

5. 5. 6.1.12

6. 6. 6.1.18

7. 7. 6.1.15

8. 8. 6.1.17

9. 9. Equilibrium solutions $x(t)\equiv 0,\pm 2$. The critical point (0, 0) in the phase plane looks like a center, whereas the points $(\pm 2,0)$ look like saddle points.

   

10. 10. Equilibrium solution $x(t)\equiv 0$. The critical point (0, 0) in the phase plane looks like a spiral sink.

    

11. 11. Equilibrium solutions $x(t)\equiv \dots ,-2\pi ,-\pi ,0,\pi ,2\pi ,\dots$. The phase portrait shown in the solutions manual suggests that the critical point $(n\pi ,0)$ in the phase plane is a spiral sink if n is even, but ...