Suppose that a certain population satisfies the initial value problem dy/dt = r

Question

Suppose that a certain population satisfies the initial value problem dy/dt = r (t )y - k y(0 ) = y0 where the growth rate r (t ) is given by r (t ) = (1 + sin t )/5, and k represents the rate of predation. (a) Suppose that k = 1/5. Plot y versus t for several values of y0 between1/2 and1. (b) Estimate the critical initial population yc below which the population will become extinct. (c) Choose other values of k and find the corresponding yc for each one. (d) Use the data you have found in parts (b) and(c) to plot yc versus k (this is problem 15 from sections 2.2 of Differential Equations with Boundary Value Problems)

Explanation / Answer

follow this a. Solving the initial value problem, I will start by separating the variables. Some algebra yields: dy/(y(1-y)) = k dt The L.H.S. can be further broken down by rewriting the expression as: dy/(y(1 - y)) = dy/y - dy/(1 - y) (I leave it to you to show how the rational expression was split into the two parts shown...) Rewriting our differential equation, we then have: dy/y - dy/(1 - y) = k dt Now integrate both sides: ln y - ln (1 - y) = kt + C Combine the natural log terms: ln (y/(1 - y)) = kt + C And then rewrite the equation using exponential notation: y/(1 - y) = Ce^(kt) At this point, let's solve for our constant of integration "C" using the initial condition: y(0)/(1 - y(0)) = Ce^(k*0) ==> C = y(0)/(1 - y(0)). Now, let's solve the equation for y: ==> y = Ce^(kt) (1 - y) ==> y(1 + Ce^(kt)) = Ce^(kt) ==> y = Ce^(kt)/(1 + Ce^(kt)) (I assume you can fill in the missing algebraic steps...) Let's rewrite our equation for y by dividing the top and bottom by e^(kt), giving us: y = C/(e^(-kt) + C), where again C = y(0)/(1 - y(0)) from above. This completes our solution. Now, note that as t -> infinity, e^(-kt) goes to zero. So: limit (t -> infinity) y = C/C, OR just 1.

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