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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 6. dy/dt = y²(y² - 1), -∞

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), -∞

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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G G G (equilibrium) points, and classify each one as asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions in the ty-plane. G G 2. dy/dt = ay+by2, a > 0, b> 0, -∞< yo<∞ dy/dt = y(y - 1)(y-2), Yo ≥ 0 -∞<Yo <∞ -∞<yo <∞ 3. dy/dt = e) — 1, 4. dy/dt = e--1, 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 G G 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 away. Therefore, (t) = 1 is semistable. above it grow farther c. Solve equation (19) subject to the initial condition y(0) = Yo (cand confirm the conclusions reached in part b. y4 k 1. 9. 7. 8. o(t) = k t pily k o(t) = k t aus nouslugng ads 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 <∞ dy/dt = y(1- y²), -∞<yo <∞ dy/dt = y²(4- y²), -∞<yo <∞ dy/dt = y²(1-y)², -∞0</o <∞ TO 11. In Example 1, complete the manipulations needed to equation (13). That is, solve the solution (11) for t. jonge (a) (b) 103 Svo FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt ≥ 0. 12. Complete the derivation of the location of the vertical a in the solution (15) when yo > T. That is, derive formu finding the value of t when the denominator of the right-ha equation (15) is zero. mabigs 13. Complete the derivation of formula (18) for the loca inflection points of the solution of the logistic growth m threshold (17). Hint: Follow the steps outlined on p. 66. 14. Consider the equation dy/dt f(y) and suppos critical point- that is, f(y₁) = 0. Show that the constan solution (1) = y₁ is asymptotically stable if f'(y₁) < 0 if f'(y₁) > 0. boot lo 15. Suppose that a certain population obeys the log dy/dt = ry(1-(y/K)). a. If yo = K/3, find the time at which the ini has doubled. Find the value of T corresponding to our year. = a, find the time T at which b. If yo/K where 0 <a, ß < 1. Observe that To as 3- →1. Find the value of T for r = 0.025 pe and 3 = 0.9. Tato str in = In 15 G 16. Another equation that has been used to : growth is the Gompertz's equation (OS) dy dt = ry in (5). the ba where r and K are positive constants. a. Sketch the graph of f(y) versus y, find and determine whether each is asymptoticall b. For 0 ≤ y ≤ K, determine where the g concave up and where it is concave down. noc. For each y in 0 ≤ y ≤ K, show tha the Gompertz equation is never less than d logistic equation. 17. a. Solve the Gompertz equation dy dt =ry In K ( 5 y 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 year, K = = 80.5 × 106 kg, yo/K = 0 model to find the predicted value of y(2 c. For the same data as in part b, use the the time at which y(T) = 0.75K. 15 Benjamin Gompertz (1779-1865) was an Engl model for population growth, published in 1825. mortality tables for his insurance company.
Transcribed Image Text:G G G (equilibrium) points, and classify each one as asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions in the ty-plane. G G 2. dy/dt = ay+by2, a > 0, b> 0, -∞< yo<∞ dy/dt = y(y - 1)(y-2), Yo ≥ 0 -∞<Yo <∞ -∞<yo <∞ 3. dy/dt = e) — 1, 4. dy/dt = e--1, 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 G G 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 away. Therefore, (t) = 1 is semistable. above it grow farther c. Solve equation (19) subject to the initial condition y(0) = Yo (cand confirm the conclusions reached in part b. y4 k 1. 9. 7. 8. o(t) = k t pily k o(t) = k t aus nouslugng ads 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 <∞ dy/dt = y(1- y²), -∞<yo <∞ dy/dt = y²(4- y²), -∞<yo <∞ dy/dt = y²(1-y)², -∞0</o <∞ TO 11. In Example 1, complete the manipulations needed to equation (13). That is, solve the solution (11) for t. jonge (a) (b) 103 Svo FIGURE 2.5.9 In both cases the equilibrium solution (t) = k is semistable. (a) dy/dt ≤ 0; (b) dy/dt ≥ 0. 12. Complete the derivation of the location of the vertical a in the solution (15) when yo > T. That is, derive formu finding the value of t when the denominator of the right-ha equation (15) is zero. mabigs 13. Complete the derivation of formula (18) for the loca inflection points of the solution of the logistic growth m threshold (17). Hint: Follow the steps outlined on p. 66. 14. Consider the equation dy/dt f(y) and suppos critical point- that is, f(y₁) = 0. Show that the constan solution (1) = y₁ is asymptotically stable if f'(y₁) < 0 if f'(y₁) > 0. boot lo 15. Suppose that a certain population obeys the log dy/dt = ry(1-(y/K)). a. If yo = K/3, find the time at which the ini has doubled. Find the value of T corresponding to our year. = a, find the time T at which b. If yo/K where 0 <a, ß < 1. Observe that To as 3- →1. Find the value of T for r = 0.025 pe and 3 = 0.9. Tato str in = In 15 G 16. Another equation that has been used to : growth is the Gompertz's equation (OS) dy dt = ry in (5). the ba where r and K are positive constants. a. Sketch the graph of f(y) versus y, find and determine whether each is asymptoticall b. For 0 ≤ y ≤ K, determine where the g concave up and where it is concave down. noc. For each y in 0 ≤ y ≤ K, show tha the Gompertz equation is never less than d logistic equation. 17. a. Solve the Gompertz equation dy dt =ry In K ( 5 y 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 year, K = = 80.5 × 106 kg, yo/K = 0 model to find the predicted value of y(2 c. For the same data as in part b, use the the time at which y(T) = 0.75K. 15 Benjamin Gompertz (1779-1865) was an Engl model for population growth, published in 1825. mortality tables for his insurance company.
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