1. 6.1.14

2. 6.1.16

3. 6.1.19

4. 6.1.13

5. 6.1.12

6. 6.1.18

7. 6.1.15

8. 6.1.17

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

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 is a saddle point if n is odd.

12. Equilibrium solution $x(t)\equiv 0$. The critical point (0, 0) in the phase plane looks like a spiral source, with the solution curves emanating from this source spiraling outward toward a closed curve trajectory.

13. Solution $x(t)=x_0{e}^{-2t},y(t)=y_0{e}^{-2t}$. The origin is a stable proper node similar to the one illustrated in Fig. 6.1.4.

14. Solution $x(t)=x_0{e}^{2t},y(t)=y_0{e}^{-2t}$. The origin is an unstable saddle point.

15. Solution $x(t)=x_0{e}^{-2t},y(t)=y_0{e}^{-t}$. The origin is a stable node.

16. Solution $x(t)=x_0{e}^t,y(t)=y_0{e}^{3t}$. The origin is an unstable improper node.

17. Solution $x(t)=A\cos t+B\sin t,y(t)=B\cos t-A\sin t$. The origin is a stable center.

18. Solution $x(t)=A\cos 2t+B\sin 2t,y(t)=-2B\cos 2t+2A\sin 2t$. The origin is a stable center.

19. Solution $x(t)=A\cos 2t+B\sin 2t,y(t)=B\cos 2t-A\sin 2t$. The origin is a stable center.

20. Solution $x(t)=e^{-2t}(A\cos t+B\sin t),y(t)=e^{-2t}[(-2A+B)\cos t-(A+2B)\sin t]$. The origin is a stable spiral point.

23. The origin and the circles $x^2+y^2=C>0$; the origin is a stable center.

24. The origin and the hyperbolas $y^2-x^2=C$; the origin is an unstable saddle point.

25. The origin and the ellipses $x^2+4y^2=C>0$; the origin is a stable center.

26. The origin and the ovals of the form $x^4+y^4=C>0$; the origin is a stable center.