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309_pdfsam_math 54 differential equation solutions odd

309_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.4 (c) To find critical points for the system in (a), we solve v = 0 , x + 1 λ x = 0 v = 0 , x 2 λx + 1 = 0 v = 0 , x = λ ± λ 2 4 2 . For 0 < λ < 2, λ 2 4 < 0 and so both roots are complex numbers. However, for λ > 2 there are two distinct real solutions, x 1 = λ λ 2 4 2 and x 2 = λ + λ 2 4 2 , and the critical points are λ λ 2 4 2 , 0 and λ + λ 2 4 2 , 0 . (d) The phase plane diagrams for λ = 1 and λ = 3 are shown in Figures B.40 and B.41 in the answers section of the text. (e) From Figures B.40 we conclude that, for λ = 1, all solution curves approach the vertical line x = 1(= λ ). This means that the bar is attracted to the magnet. The case λ = 3 is more complicated. The behavior of the bar depends on the initial displacement x (0) and the initial velocity v (0) = x (0). From Figure B.41 we see that (with v (0) = 0) if
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