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Semistable Equilibrium Solutions. Sometimes a constant equilibrium solution has the
Chapter 2, Problem 5(choose chapter or problem)
Semistable Equilibrium Solutions. Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it (see Figure 2.5.9). In this case the equilibrium solution is said to be semistable. a. Consider the equation dy/dt = k(1 y)2, (19) where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution (t) = 1. G b. Sketch f ( y) versus y. Show that y is increasing as a function of t for y < 1 and also for y > 1. The phase line has upward-pointing arrows both below and above y = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore, (t) = 1 is semistable. c. Solve equation (19) subject to the initial condition y(0) = y0 and confirm the conclusions reached in part b.
Questions & Answers
QUESTION:
Semistable Equilibrium Solutions. Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it (see Figure 2.5.9). In this case the equilibrium solution is said to be semistable. a. Consider the equation dy/dt = k(1 y)2, (19) where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution (t) = 1. G b. Sketch f ( y) versus y. Show that y is increasing as a function of t for y < 1 and also for y > 1. The phase line has upward-pointing arrows both below and above y = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore, (t) = 1 is semistable. c. Solve equation (19) subject to the initial condition y(0) = y0 and confirm the conclusions reached in part b.
ANSWER:Step 1 of 6
a) Consider the equation .
The equation is of the form is an autonomous equation.
To determine the critical (or equilibrium) points, equate the function to 0.
The critical points are the zeros of .
Solve the equation for .
Therefore, there is only one critical point, , corresponding to the equilibrium solutions .