*This preview shows
page 1. Sign up to
view the full content.*

**Unformatted text preview: **Chapter 5
To ﬁnd the critical points, we solve the system of equations given by v = 0 and −y (1 + y 4 ) = 0. This system is satisﬁed only when v = 0 and y = 0. Thus, the only critical point is the point (0, 0). To ﬁnd the integral curves, we solve the ﬁrst order equation given by dv dv/dt −y − y 5 = = . dy dy/dt v This is a separable equation and can be written as v dv = −y − y 5 dy ⇒ ⇒ v2 y2 y6 =− − +C 2 2 6 (c = 6 C ), 3v 2 + 3y 2 + y 6 = c where we have integrated to obtain the second equation above. Therefore, the integral curves for this system are given by the equations 3v 2 + 3y 2 + y 6 = c for each positive constant c. To determine the ﬂow along the trajectories, we will examine the equation dy/dt = v . Thus, we see that dy dy > 0 when v > 0, and < 0 when v < 0. dt dt Therefore, y will be increasing when v > 0 and decreasing when v < 0. Hence, above the y -axis the ﬂow will be from left to right and below the x-axis the ﬂow will be from right to left. Thus, the ﬂow on these trajectories will be clockwise (Figure B.35 in the answers of the text). Thus (0, 0) is a center (stable). 23. With v = y , v = y , the equation transforms to the system y = v, v +y−y = 0
4 ⇒ y = v, v = y 4 − y. (5.26) Therefore, f (y, v ) = v and g (y, v ) = y 4 − y = y (y 3 − 1). We ﬁnd critical points by solving v = 0, y (y 3 − 1) = 0 ⇒ v = 0, y = 0 or y = 1. Hence, system (5.26) has two critical points, (0, 0) and (1, 0). In the upper half plane, y = v > 0 and so solutions ﬂow to the right; similarly, solutions ﬂow to the left in the lower half-plane. See Figure B.36 in the text for the direction ﬁeld. This 300 ...

View Full
Document