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Suppose that the equation dx dt = kx(M − x) − Ex models the population of fish in a lake, where harvesting occurs at the rate of Ex fish per month (E is a positive constant, called the harvesting effort). Suppose that 0 < E < kM, (a) Find the solution of this ODE. (b) What is the limiting population?

User Fakrul
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Final answer:

The given differential equation dx/dt = kx(M - x) - Ex represents the fish population in a lake considering both natural growth and harvesting efforts. The solution to this ODE reveals how the fish population changes over time, while the limiting population is found by setting the population's rate of change to zero and solving for the equilibrium points.

Step-by-step explanation:

The differential equation dx/dt = kx(M - x) - Ex models the population of fish in a lake where harvesting occurs. The parameter k is the intrinsic growth rate, M is the carrying capacity of the lake, and E is the harvesting effort. This equation is a form of the logistic growth model that has been adjusted to include harvesting. Solving this ordinary differential equation (ODE) typically involves separating variables and integrating both sides to find the general solution, which will give us the population of fish over time, x(t).

The limiting population is the population the system settles at as time goes to infinity. Mathematically, we find this by setting dx/dt to 0 and solving for x to find the equilibrium solutions. Given the condition 0 < E < kM, there will generally be two equilibrium points: one at x = 0 (extinction) and another at some positive value where x equals some function of M, k, and E. The non-zero equilibrium is the limiting population when the fish population stabilizes with the presence of harvesting.

User Laurent LAPORTE
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