4a 4.- The population of zebra in an area varies around an equilibrium level X.. The population of lions in an areavaries 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 bydx(2)= -aydtwhere a is a positive parameter.The change in the population of zebras also depends on fluctuations in the food supply. We can model this bychanging equation (2) to:dx= -ay + Asin(3)dtWhere 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 initialvalues:x(0) = 1500 and y(0) = 300 and a = B = 1, A = 4 yearsi and T, = 6 yearsWe are going in (d) and (e) look at the same problem, but write equations (2) and (4)_as a system of two firstorder 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 2x2matrix.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 theequilibrium of this system:dx= X(11dtdY= Y(-1+6001(7)dt3000 *)Look at the equation system (7) and find equilibrium point for the system, also examine the stabilityof the equilibrium points. With what period does the zebra/lion population vary?

We will answer to the question 4a only as it is mentioned and demanded.

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?