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lecture_35 - MA 36600 LECTURE NOTES MONDAY APRIL 20...

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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, APRIL 20 Homogeneous Systems with Constant Coefficients Example #2. We have already seen that the general solution to the homogeneous system d dt x =- 1 2 1- 1- 1 2 x is the function x ( t ) = c 1 cos t- sin t e- t/ 2 + c 2 sin t cos t e- t/ 2 . Figure 1 gives a plot of the direction field. We call this graph a spiral . Figure 1. Direction Field for d dt x =- 1 2 1- 1- 1 2 x-0.15-0.125-0.1-0.075-0.05-0.025 0.025 0.05 0.075 0.1 0.125 0.15-0.1-0.075-0.05-0.025 0.025 0.05 0.075 0.1 In general, let r = λ ± iμ be complex eigenvalues for a 2 × 2 matrix A with constant coefficients, and say that ξ = a ± i b are the corresponding eigenvectors. To draw the slope field, perform the following three steps: (1) Draw lines through the origin determined by a and b . (2) Along these lines, draw arrows going towards the origin if λ < 0, and away from the origin if λ > 0. (3) Using the relations A a = λ a- μ b A b = μ a + λ b spiral from b to a if μ/λ < 0, and from a to b if μ/λ > 0....
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This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue.

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lecture_35 - MA 36600 LECTURE NOTES MONDAY APRIL 20...

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