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# lecture_35 (dragged) - steps(1 Draw lines through the...

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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 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 ± i μ be complex eigenvalues for a 2 × 2 matrix A with constant coe cients, and say that ξ = a ± i b are the corresponding eigenvectors. To draw the slope field, perform the following three
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Unformatted text preview: steps: (1) Draw lines through the origin determined by a and b . (2) Along these lines, draw arrows going towards the origin if Î» < 0, and away from the origin if Î» > 0. (3) Using the relations A a = Î» a âˆ’ Î¼ b A b = Î¼ a + Î» b spiral from b to a if Î¼/Î» < 0, and from a to b if Î¼/Î» > 0. If Î» is positive, we say that the direction Â±eld is a nodal spiral . If Î» is negative, we say that the direction Â±eld is a spiral sink . 1...
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