306_pdfsam_math 54 differential equation solutions odd

306_pdfsam_math 54 differential equation solutions odd - ,...

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Chapter 5 in the left half-plane. Solutions starting on the y -axis stay on it ( x 0 = 0); they ﬂow up if the initial point is in the upper half-plane (because y 0 = y> 0) and ﬂow down if the initial point in the lower half-plane. This matches the Fgure for unstable node. (b) Solving the phase plane equation for this system, we get dy dx = 4 x y ydy = 4 xdx y 2 +4 x 2 = C. Thus the integral curves are ellipses. (Also, notice that the solutions ﬂow along these ellipses in clockwise direction because x increases in the upper half-plane and decreases in the lower half-plane.) Therefore, here we have a center (stable). (c) Solving 5 x +2 y> 0and x 4 y> 0 we Fnd that x increases in the half-plane y> 5 x and decreases in the half-plane y< 5 x ,and y increases in the half-plane y<x/ 4and decreases in the half-plane y>x/ 4. This leads to the scheme . % - for the solution’s ﬂows. Thus all solutions approach the critical point (0 , 0), as t →∞ , which corresponds to a stable node. (d) An analysis, similar to that in (c), shows that all the solutions ﬂow away from (0
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Unformatted text preview: , 0). Among pictures shown in ±igure 5.7, only the unstable node and the unstable spiral have this feature. Since the unstable node is the answer to (a), we have the unstable spiral in this case. (e) The phase plane equation dy dx = 4 x − 3 y 5 x − 3 y , has two linear solutions, y = 2 x and y = 2 x/ 3. (One can Fnd them by substituting y = ax into the above phase plane equation and solving for a .) Solutions starting from a point on y = 2 x in the Frst quadrant, have x = 5 x − 3(2 x ) = − x < 0 and so ﬂow toward (0 , 0); similarly, solutions, starting from a point on this line in the third quadrant, have x = − x > 0 and, again, ﬂow to (0 , 0). On the other line, y = 2 x/ 3, the picture is opposite: in the Frst quadrant, x = 5 x − 3(2 x/ 3) = 3 x > 0, and x < 0 in the third quadrant. Therefore, there are two lines, passing through the critical point (0 , 0), such 302...
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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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