301_pdfsam_math 54 differential equation solutions odd

301_pdfsam_math 54 differential equation solutions odd -...

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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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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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