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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. G -∞0

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. G -∞0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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G Problems Problems 1 through 4 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 or in the ty-plane. unstable. Draw the phase line, and sketch several graphs of solutions G G 1. 2. 3. G 4. dy/dt = e- - 1,-00< Yo<∞ 5. 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 y A k dy/dt = ay+by², a > 0, b>0, -∞< yo<∞ dy/dt = y(y - 1)(y-2), yo 20 dy/dt = e' -1, -∞0<o<∞ 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. Gb. 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) = yo and confirm the conclusions reached in part b. o(t) = k t y k o(t) = k t 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. G 6. dy/dt = y²(y²-1), -∞0<yo <∞ 7. dy/dt = y(1- y²), -∞<yo <∞ -∞0<yo <∞ 8. dy/dt = y²(4- y²), 9. dy/dt = y²(1 - y)², -∞< y₁ <∞ (a) (b) FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt > 0. y. (11) of the logistic model by solving equation (10) for 10. Complete the derivation of the explicit formula for the solution equation (13). That is, solve the solution (11) for t. 11. In Example 1, complete the manipulations needed to arrive at 12. Complete the derivation of the location of the vertical asymptote in the solution (15) when yo > T. That is, derive formula (16) by finding the value of t when the denominator of the right-hand side of equation (15) is zero. 13. Complete the derivation of formula (18) for the locations of the inflection points of the solution of the logistic growth model with a threshold (17). Hint: Follow the steps outlined on p. 66. 14. Consider the equation dy/dt f(y) and suppose that У1 is a critical point that is, f(y₁) = 0. Show that the constant equilibrium if f'(y₁) > 0. solution (t) = y₁ is asymptotically stable if f'(y₁) < 0 and unstable = dy/dt = ry(1-(y/K)). 15. Suppose that a certain population obeys the logistic equation a. If yo = K/3, find the time at which the initial population has doubled. Find the value of 7 corresponding to r = 0.025 per year. ->> b. If yo/K = a, find the time T at which y(T)/K = 3, where 0 <a, 3 < 1. Observe that T→ ∞o as a → 0 or as B → 1. Find the value of T for r = 0.025 per year, a = 0.1, and 3 = 0.9. 15 G 16. Another equation that has been used to model population growth is the Gompertz¹5 equation dy K =ry ln dt where r and K are positive constants. a. Sketch the graph of f(y) versus y, find the critical points, and determine whether each is asymptotically stable or unstable. b. For 0 ≤ y ≤ K, determine where the graph of y versus t is concave up and where it is concave down. c. For each y in 0 ≤ y ≤ K, show that dy/dt as given by the Gompertz equation is never less than dy/dt as given by the logistic equation. 17. a. Solve the Gompertz equation dy ln dr = ry in (). dt subject to the initial condition y(0) = yo. Hint: You may wish to let u = ln(y/K). b. For the data given in Example 1 in the text (r = 0.71 per year, K = 80.5 x 106 kg, yo/K = 0.25), use the Gompertz model to find the predicted value of y(2). C. For the same data as in part b, use the Gompertz model to find the time at which y(T) = 0.75K. 15 Benjamin Gompertz (1779-1865) was an English actuary. He developed his model for population growth, published in 1825, in the course of constructing mortality tables for his insurance company.
Transcribed Image Text:G Problems Problems 1 through 4 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 or in the ty-plane. unstable. Draw the phase line, and sketch several graphs of solutions G G 1. 2. 3. G 4. dy/dt = e- - 1,-00< Yo<∞ 5. 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 y A k dy/dt = ay+by², a > 0, b>0, -∞< yo<∞ dy/dt = y(y - 1)(y-2), yo 20 dy/dt = e' -1, -∞0<o<∞ 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. Gb. 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) = yo and confirm the conclusions reached in part b. o(t) = k t y k o(t) = k t 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. G 6. dy/dt = y²(y²-1), -∞0<yo <∞ 7. dy/dt = y(1- y²), -∞<yo <∞ -∞0<yo <∞ 8. dy/dt = y²(4- y²), 9. dy/dt = y²(1 - y)², -∞< y₁ <∞ (a) (b) FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt > 0. y. (11) of the logistic model by solving equation (10) for 10. Complete the derivation of the explicit formula for the solution equation (13). That is, solve the solution (11) for t. 11. In Example 1, complete the manipulations needed to arrive at 12. Complete the derivation of the location of the vertical asymptote in the solution (15) when yo > T. That is, derive formula (16) by finding the value of t when the denominator of the right-hand side of equation (15) is zero. 13. Complete the derivation of formula (18) for the locations of the inflection points of the solution of the logistic growth model with a threshold (17). Hint: Follow the steps outlined on p. 66. 14. Consider the equation dy/dt f(y) and suppose that У1 is a critical point that is, f(y₁) = 0. Show that the constant equilibrium if f'(y₁) > 0. solution (t) = y₁ is asymptotically stable if f'(y₁) < 0 and unstable = dy/dt = ry(1-(y/K)). 15. Suppose that a certain population obeys the logistic equation a. If yo = K/3, find the time at which the initial population has doubled. Find the value of 7 corresponding to r = 0.025 per year. ->> b. If yo/K = a, find the time T at which y(T)/K = 3, where 0 <a, 3 < 1. Observe that T→ ∞o as a → 0 or as B → 1. Find the value of T for r = 0.025 per year, a = 0.1, and 3 = 0.9. 15 G 16. Another equation that has been used to model population growth is the Gompertz¹5 equation dy K =ry ln dt where r and K are positive constants. a. Sketch the graph of f(y) versus y, find the critical points, and determine whether each is asymptotically stable or unstable. b. For 0 ≤ y ≤ K, determine where the graph of y versus t is concave up and where it is concave down. c. For each y in 0 ≤ y ≤ K, show that dy/dt as given by the Gompertz equation is never less than dy/dt as given by the logistic equation. 17. a. Solve the Gompertz equation dy ln dr = ry in (). dt subject to the initial condition y(0) = yo. Hint: You may wish to let u = ln(y/K). b. For the data given in Example 1 in the text (r = 0.71 per year, K = 80.5 x 106 kg, yo/K = 0.25), use the Gompertz model to find the predicted value of y(2). C. For the same data as in part b, use the Gompertz model to find the time at which y(T) = 0.75K. 15 Benjamin Gompertz (1779-1865) was an English actuary. He developed his model for population growth, published in 1825, in the course of constructing mortality tables for his insurance company.
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