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4.- The population of zebra in an area varies around an equilibrium level X.. The population of lions in an area varies around an equilibrium level Y,. We let x(t) be the deviation from the equilibrium level X, for the zebra. We let y(t) be the deviation from the equilibrium level Y, for the zebra. The time t is given in years. The zebra population decreases when y is positive and the diferential equation for x(t) is given by dx (2) = -ay dt where a is a positive parameter. The change in the population of zebras also depends on fluctuations in the food supply. We can model this by changing equation (2) to: dx = -ay + Asin (3) dt Where A is a real positive amplitude and T, is the period of the fluctuations in the food supply. The lion population increases when x(t) is positive and the differential equation for y(t) is given by: dy (4) dt = Bx, where B is a positive parameter. a) From equations (2) and (3) we can make a second order differential equation for x in the form: d?x + ax = f(t) (5) dt?

4.- The population of zebra in an area varies around an equilibrium level X.. The population of lions in an area varies around an equilibrium level Y,. We let x(t) be the deviation from the equilibrium level X, for the zebra. We let y(t) be the deviation from the equilibrium level Y, for the zebra. The time t is given in years. The zebra population decreases when y is positive and the diferential equation for x(t) is given by dx (2) = -ay dt where a is a positive parameter. The change in the population of zebras also depends on fluctuations in the food supply. We can model this by changing equation (2) to: dx = -ay + Asin (3) dt Where A is a real positive amplitude and T, is the period of the fluctuations in the food supply. The lion population increases when x(t) is positive and the differential equation for y(t) is given by: dy (4) dt = Bx, where B is a positive parameter. a) From equations (2) and (3) we can make a second order differential equation for x in the form: d?x + ax = f(t) (5) dt?

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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4.- The population of zebra in an area varies around an equilibrium level X.. The population of lions in an area varies around an equilibrium level Y, . We let x(t) be the deviation from the equilibrium level X, for the zebra. We let y(t) be the deviation from the equilibrium level Y, for the zebra. The time t is given in years. The zebra population decreases when y is positive and the diferential equation for x(t) is given by dx (2) = -ay dt where a is a positive parameter. The change in the population of zebras also depends on fluctuations in the food supply. We can model this by changing equation (2) to: dx = -ay + Asin (3) dt Where A is a real positive amplitude and T, is the period of the fluctuations in the food supply. The lion population increases when x(t) is positive and the differential equation for y(t) is given by: dy (4) dt = Bx, where B is a positive parameter. a) From equations (2) and (3) we can make a second order differential equation for x in the form: d?x (5) dr + ax = f(t) Show how this is done and find a and fit) in the above equation. b) Find the general solution for x(t) by using the equation you found in (a). c) Find an expression for y(t)_from (4) and find the exact solution for y(t) and x(t) given the initial values: x(0) = 1500 and y(0) = 300 and a = B = 1, A = 4 yearsi and T, = 6 years We are going in (d) and (e) look at the same problem, but write equations (2) and (4)_as a system of two first order differential equation given in the form: (6) G) - ; C) d) Find the general solution to the equation (6) expressed by the eigenvalues and eigenvectors of a 2x2 matrix. e) Find the general solution of equation (6) expressed with real functions. f) We revise the model for zebra/lion population. The total population of zebras is given by X(t) = X, + x(t) and the total population of lions is given by Y(t) = Y + y(t). We will now look at the equilibrium of this system: dx = X(1 1 dt dY = Y(-1+ 600 1 (7) dt 3000 *) Look at the equation system (7) and find equilibrium point for the system, also examine the stability of the equilibrium points. With what period does the zebra/lion population vary?
Transcribed Image Text:4.- The population of zebra in an area varies around an equilibrium level X.. The population of lions in an area varies around an equilibrium level Y, . We let x(t) be the deviation from the equilibrium level X, for the zebra. We let y(t) be the deviation from the equilibrium level Y, for the zebra. The time t is given in years. The zebra population decreases when y is positive and the diferential equation for x(t) is given by dx (2) = -ay dt where a is a positive parameter. The change in the population of zebras also depends on fluctuations in the food supply. We can model this by changing equation (2) to: dx = -ay + Asin (3) dt Where A is a real positive amplitude and T, is the period of the fluctuations in the food supply. The lion population increases when x(t) is positive and the differential equation for y(t) is given by: dy (4) dt = Bx, where B is a positive parameter. a) From equations (2) and (3) we can make a second order differential equation for x in the form: d?x (5) dr + ax = f(t) Show how this is done and find a and fit) in the above equation. b) Find the general solution for x(t) by using the equation you found in (a). c) Find an expression for y(t)_from (4) and find the exact solution for y(t) and x(t) given the initial values: x(0) = 1500 and y(0) = 300 and a = B = 1, A = 4 yearsi and T, = 6 years We are going in (d) and (e) look at the same problem, but write equations (2) and (4)_as a system of two first order differential equation given in the form: (6) G) - ; C) d) Find the general solution to the equation (6) expressed by the eigenvalues and eigenvectors of a 2x2 matrix. e) Find the general solution of equation (6) expressed with real functions. f) We revise the model for zebra/lion population. The total population of zebras is given by X(t) = X, + x(t) and the total population of lions is given by Y(t) = Y + y(t). We will now look at the equilibrium of this system: dx = X(1 1 dt dY = Y(-1+ 600 1 (7) dt 3000 *) Look at the equation system (7) and find equilibrium point for the system, also examine the stability of the equilibrium points. With what period does the zebra/lion population vary?
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