527ex2sl

527ex2sl - 642:527 SOLUTIONS: EXAM 2 FALL 2007 1. (a) Find...

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Unformatted text preview: 642:527 SOLUTIONS: EXAM 2 FALL 2007 1. (a) Find the general solution of z = A z , where z = bracketleftbigg x y bracketrightbigg and A = bracketleftbigg 4 3 1 2 bracketrightbigg . (b) Give a careful drawing of the phase plane ( xy-plane) for this system, showing enough trajectories to indicate qualitatively the motion in each region of the plane, as well as any special (straight- line) trajectories. Your trajectories should be marked with arrowheads giving the direction in which the solution moves as t increases. Solution: (a) Since det( A I ) = 2 6 +5 = ( 5)( 1) the eigenvalues are 1 = 1, 2 = 5. The corresponding eigenvectors z ( i ) are found by solving ( A i ) z ( i ) = : 1 = 1 : bracketleftbigg 3 3 1 1 bracketrightbiggbracketleftbigg x y bracketrightbigg = = z (1) = bracketleftbigg 1 1 bracketrightbigg 2 = 5 : bracketleftbigg 1 3 1 3 bracketrightbiggbracketleftbigg x y bracketrightbigg = = z (2) = bracketleftbigg 3 1 bracketrightbigg The general solution is z ( t ) = c 1 e t z (1) + c 2 e 5 t z (2) . This is an unstable node . (b) Here is a solution plot, from Maple. I dont know how to get Maple to put arrow- heads on curves, but as is clear from the di- rection field or from the form of the solution, all trajectories are oriented away from the ori- gin. The straight line trajectories are parallel to the eigenvectors found in (a). The other trajectories are determined by the fact that as as t they are parallel to z (2) and as t to z (1) . x K 2 K 1 1 2 y K 2 K 1 1 2 2. (a) Consider the system x = 1 y, y = x 2 y. Determine its singular (equilibrium) points and classify each, insofar as possible, using linearization: for each equilibrium point, determine whether it is unstable, stable, or asymptotically stable, and if it is a focus, node, or saddle, if you have the information to do so. If one cannot determine the type from the linearization, say so. (b) Sketch the phase plane for this system, showing nullclines (curves on which the solution curves have horizontal or vertical tangents), direction of motion as the solutions cross the nullclines, and direction of motion in the regions separated by the nullclines. No trajectories need be sketched....
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527ex2sl - 642:527 SOLUTIONS: EXAM 2 FALL 2007 1. (a) Find...

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