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**Unformatted text preview: **Exercises 5.4
ﬁgure indicates that (0, 0) is a stable critical point (center) whereas (1, 0) is a saddle point (unstable). 25. This system has two critical points, (0, 0) and (1, 0), which are solutions to the system y = 0, −x + x3 = 0. The direction ﬁeld for this system is depicted in Figure B.37. From this ﬁgure we conclude that (a) the solution passing through the point (0.25, 0.25) ﬂows around (0, 0) and thus is periodic; (b) for the solution (x(t), y (t)) passing through the point (2, 2), y (t) → ∞ as t → ∞, and so this solution is not periodic; (c) the solution passing through the critical point (1, 0) is a constant (equilibrium) solution and so is periodic. 27. The direction ﬁeld for given system is shown in Figure B.38 in the answers of the text. From the starting point, (1, 1), following the direction arrows the solution ﬂows down and to the left, crosses the x-axis, has a turning point in the fourth quadrant, and then does to the left and up toward the critical point (0, 0). Thus we predict that, as t → ∞, the solution (x(t), y (t)) approaches (0, 0). 29. (a) The phase plane equation for this system is 3y dy = . dx x It is separable. Separating variables and integrating, we get 3dx dy = y x ⇒ ln |y | = 3 ln |x| + C ⇒ y = cx3 . So, integral curves are cubic curves. Since in the right half-plane x = x > 0, in the left half-plane x < 0, the solutions ﬂow to the right in the right half-plane and to the left 301 ...

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