Differential equations provide powerful tools for describing the behavior of dyn

Question

Differential equations provide powerful tools for describing the behavior of dynamically changing processes. Here we look at the fish population in one of the Great Lakes and explore what harvesting rates maintain both the population and the fishing industry at acceptable long-term levels. We will use a differential equation to describe the population changes over time given birth, death, and harvesting rates. The clue that a differential equation may describe what is going on lies in the words birth, death, and harvesting rates. The key word here is rates. Rates are derivatives with respect to time. In this project we will consider the population of two species of fish in the great lakes. The New York State Department of Natural Resources has asked you to provide a report about hem- the two species interact with one another. You are given the birth rates but are not sure about the parameters associated with how the species interact. Here we will use a general linear model. d/dt[x(t) y(t)] = [0.2 a b 0.35] [x(t) y(t)], where x(t) is the number of fish of the first species. y(t) is the number of fish of the second species, and the constants a and b determine how do they interact. You are asked to determine the ranges of values of the constants a and b that will result in different behaviors of the solutions to the linear model, equation (??). For example, in a predator-prey relationship where the x species eats the y species then at first glance we might expect that a will be positive and b will be negative. You need to determine the full range of possible relationships between a and b, as to whether one or both species benefit or do not benefit with respect to the other population. You need to also determine if this model is reasonable and justify your conclusions. Based on your results you should provide a conclusion if this approach is adequate or they should start over and come up with a new model.

Explanation / Answer

x'(t) = 0.2x(t) + a y(t)

y'(t) = b x(t) + 0.35 y(t)

differntitate first equation

x''(t) = 0.2 x'(t) + a y'(t)

= 0.2 [0.2x(t) + a y(t)] + a [ b x(t) + 0.35 y(t)]

= (ab +0.04)x(t) + (0.55a)y(t)

= (ab +0.04)x(t) + (0.55)[x'(t) -0.2x(t)]

=(ab +0.04 -1.1)x(t) + 0.55x'(t)

= (ab -1.06)x(t) + 0.55x'(t)

x''(t) = Ax(t) +Bx'(t) where A =ab -1.06 and B = 0.55

So

Z2 -(ab -1.06)Z - 0.55 =0

for non imaginery solution

(ab -1.06)2 -4 (-0.55) >=0

(ab)2 - 2.12ab + 3.3236 >=0

(ab -3.169)(ab +1.049) >= 0

[ab > 3.169 or ab < -1.49 ] range for a and b

I think this model is good and adequate

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