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FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt ≥ 0. Problems 6 through 9 involve equations of the form dy/dt = f(y). n each problem sketch the graph of f(y) versus y, determine the ritical (equilibrium) points, and classify each one as asymptotically table, unstable, or semistable (see Problem 5). Draw the phase line, nd sketch several graphs of solutions in the ty-plane. G 6. dy/dt = y²(y² - 1), -∞

FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt ≥ 0. Problems 6 through 9 involve equations of the form dy/dt = f(y). n each problem sketch the graph of f(y) versus y, determine the ritical (equilibrium) points, and classify each one as asymptotically table, unstable, or semistable (see Problem 5). Draw the phase line, nd sketch several graphs of solutions in the ty-plane. G 6. dy/dt = y²(y² - 1), -∞

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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14 as ed ly gh ot to er ly G G (19) subject to the initial condition y(0) = yo and confirm the conclusions reached in part b. YA k G G o(t) = k t y A Problems 6 through 9 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 5). Draw the phase line, and sketch several graphs of solutions in the ty-plane. 6. dy/dt = y²(y2 - 1), -∞<yo <∞ 7. dy/dt = y(1- y²), -∞<yo <∞ 8. dy/dt = y²(4- y²), -∞0<yo <∞ 9. dy/dt = y²(1 - y)², -∞<yo<∞ k o(t) = k (a) (b) FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt > 0. G 16 growth where a. a b t 1 17. a 15 Be mod mor
Transcribed Image Text:14 as ed ly gh ot to er ly G G (19) subject to the initial condition y(0) = yo and confirm the conclusions reached in part b. YA k G G o(t) = k t y A Problems 6 through 9 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 5). Draw the phase line, and sketch several graphs of solutions in the ty-plane. 6. dy/dt = y²(y2 - 1), -∞<yo <∞ 7. dy/dt = y(1- y²), -∞<yo <∞ 8. dy/dt = y²(4- y²), -∞0<yo <∞ 9. dy/dt = y²(1 - y)², -∞<yo<∞ k o(t) = k (a) (b) FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt > 0. G 16 growth where a. a b t 1 17. a 15 Be mod mor
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