**Unformatted text preview: **Exercises 5.4
Notice that both of these equations will be satisﬁed if y = x. Thus, x = C and y = C , for any ﬁxed constant C , will be a solution to the given system of diﬀerential equations and one family of critical points is the line y = x. We also see that we have a critical point at the point (1, 1). (This critical point is, of course, also on the line y = x.) Next we will ﬁnd the integral curves. Therefore, we must solve the ﬁrst order diﬀerential equation given by dy dy/dt (x − y )(x − 1) = = dx dx/dt (y − x)(y − 1) we have (y − 1)dy = ⇒ ⇒ By completing the square, we obtain (x − 1)2 + (y − 1)2 = c, where c = 2C +2. Therefore, the integral curves are concentric circles with centers at the point (1, 1), including the critical point for the system of diﬀerential equations. The trajectories associated with the constants c = 1, 4, and 9, are sketched in Figure B.31 in the answers of the text. Finally we will determine the ﬂow along the trajectories. Notice that the variable t imparts a ﬂow to the trajectories of a solution to a system of diﬀerential equations in the same manner as the parameter t imparts a direction to a curve written in parametric form. We will ﬁnd this ﬂow by determining the regions in the xy -plane where x(t) is increasing (moving from left to right on each trajectory) and the regions where x(t) is decreasing (moving from right to left on each trajectory). Therefore, we will use four cases to study the equation dx/dt = (y − x)(y − 1), the ﬁrst equation in our system. 297 (1 − x)dx ⇒ dy 1−x = . dx y−1 We can solve this last diﬀerential equation by the method of separation of variables. Thus, y2 x2 −y = x− +C 2 2 x2 − 2x + y 2 − 2y = 2C. ...

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