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Draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y asf→ If this behavior depends on the initial value of y at = 0, describe this dependency. y' = y(y - 5)² Where a = 5. Equilibrium solution: y(t) = 5. Behavior of y(t) as → is independent of initial value y(t): y(fo) → 5 for all y(fo). Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(m): y() > 5:y(1) diverges from y = 5. 0 < y(t) < 5:y(t) → 5. y(to) < 0:y(t) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(1) ast → y(to) > 5:y(1) diverges from y = 5. 0 < y(t) < 5:y(t) → 0. y(to) < 0: y(t) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(to) > 0: y(t) → 5. y(to) < 0: y(t) diverges from y = 0. a depends on initial value y():

Draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y asf→ If this behavior depends on the initial value of y at = 0, describe this dependency. y' = y(y - 5)² Where a = 5. Equilibrium solution: y(t) = 5. Behavior of y(t) as → is independent of initial value y(t): y(fo) → 5 for all y(fo). Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(m): y() > 5:y(1) diverges from y = 5. 0 < y(t) < 5:y(t) → 5. y(to) < 0:y(t) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(1) ast → y(to) > 5:y(1) diverges from y = 5. 0 < y(t) < 5:y(t) → 0. y(to) < 0: y(t) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(to) > 0: y(t) → 5. y(to) < 0: y(t) diverges from y = 0. a depends on initial value y():

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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Draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y ast. If this behavior depends on the initial value of y at t = 0, describe this dependency. y' = y(y - 5)² Where a = 5. Equilibrium solution: y(t) = 5. Behavior of y(t) as t → is independent of initial value y(t): y(to) → 5 for all y(to). U Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(t) > 5:y(t) diverges from y = 5. 0 < y(tg) < 5:y(t) → 5. y(to) < 0:y(t) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(to) > 5:y(1) diverges from y = 5. 0 < y(t) < 5:y(t) → 0. y(to) < 0:y(1) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(to) > 0: y(t)→ 5. y(to) < 0:y(t) diverges from y = 0. O a 0 a 0.
Transcribed Image Text:Draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y ast. If this behavior depends on the initial value of y at t = 0, describe this dependency. y' = y(y - 5)² Where a = 5. Equilibrium solution: y(t) = 5. Behavior of y(t) as t → is independent of initial value y(t): y(to) → 5 for all y(to). U Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(t) > 5:y(t) diverges from y = 5. 0 < y(tg) < 5:y(t) → 5. y(to) < 0:y(t) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(to) > 5:y(1) diverges from y = 5. 0 < y(t) < 5:y(t) → 0. y(to) < 0:y(1) diverges from y = 0. Where a = 5. Equilibrium solutions: y(t) = 0 and y(t) = 5. Behavior of y(t) ast → depends on initial value y(t): y(to) > 0: y(t)→ 5. y(to) < 0:y(t) diverges from y = 0. O a 0 a 0.
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