Homework06 - Math 404 - Homework 6 Example Student...

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Unformatted text preview: Math 404 - Homework 6 Example Student Solutions February 15, 2011 S: 6.1.8 Plot Phase Portraits for the following system (the van der Pol oscilla- tor). x = y y =- x + y (1- x 2 ) Answer First, note that we will have fixed points when x = 0 and y = 0. This will happen when: 0 = y 0 =- x + 0 x = 0 Therefore, there is only one fixed point and it occurs at (0 , 0). We will thus focus our phase portrait graph around this point. Here we see that there is a closed orbit around the fixed point, and that within that closed orbit the fixed point acts like an unstable spiral. 1 Figure 1: Phase Portrait 2 S: 6.1.9 Plot Phase Portraits for the following system (the dipole fixed point). x = 2 xy y = y 2- x 2 Answer First note that we will have fixed points when x = 0 and y = 0. This will happen when: 0 = 2 xy and 0 = y 2- x 2 The second equation tells us that y = x and thus the first one implies that this only happens at (0 , 0). Therefore, there is only one fixed point, and we will focus our phase portrait around this point. Figure 2: Phase Portrait 3 S: 6.1.10 Plot Phase Portraits for the following system (the two-eyed mon- ster). x = y + y 2 y =- 1 2 x + 1 5 y- xy + 6 5 y 2 Answer First note that we will have fixed points when x = 0 and y = 0. 0 = y + y 2 factros to give us ( y + 1) y = 0 and finally y =- 1 and y = 0 When y = 0 and y = 0, we know that x = 0. Therefore, there is a fixed point at (0 , 0). When y =- 1 and y = 0: 0 =- 1 2 x- 1 5 + x + 6 5- 1 = 1 2 x x =- 2 Therefore, there is another fixed point at (- 2 ,- 1). We will thus focus the phase portrait around these two points....
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This note was uploaded on 03/13/2011 for the course MATH 404 taught by Professor Staff during the Spring '08 term at University of Michigan.

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Homework06 - Math 404 - Homework 6 Example Student...

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