I need help with this problem. It is a differential equation.

Question

The following differential equation is used to model population growth in an environment with limited resources.

y' = 0.2y (1- y/85)

Find the values of y at which the population is in equilibrium.

Use a technology tool, like the direction field explorer to plot a direction field for this differential equation. Be sure to adjust the axes so the vertical axis goes beyond the equilibria you found, and adjust the horizontal axis to go out far enough so you can the long term trend. Include a screenshot of the direction field in your answer.

If the population starts out small, describe what will happen to the population over time.

Answer 1

I would explain this as follows:

This is a variables separable ODE: 85 dy/[y(85-y)]=.2 dt

Partial fractions gives ln[85/(85-y)]=.2t+C

If you do some algebraic manipulation you get:

y=85[1-(1/(e^.2t+C+1)]

y approaches 85 asymptotically very quickly irrespective of C

Answer 2

The equation y' = .2y(1 - y/85) is in the form of y' = f(y) also known as a first order autonomous differential equation. To find equilibrium values, find y values that make f(y) = 0. In this example the values y = 0 and y = 85 make f(y) and hence y' = 0 and so are equilbrium values. To find out whether these represent stable or unstable equilibria, plug numbers into f(y) that are a little greater and a little less than a given equilibrium value. if y' > 0 above y_eq (positive slope) and y' < 0 below y_eq (negative slope), then y_eq is unstable, so solutions diverge. If y' < 0 above and y' > 0 below then y_eq is stable. If y' above and below have the same sign (+ or -) then y_eq is like a saddle point and can be regarded as unstable.

Answer 3

dy/[y(1 - y/85)]= 0.2dt; ∫ (1/y + (1/85)/(1 - y/85)) = ∫0.2dt; lny + 1/85ln(1 - y/85) = 0.2t + lnC;

y(1-y/85) = Ce^0.2t and so on