The population of Aedes mosquitoes which carry the Dengue virus can be modeled by a differential equation which describes the rate of growth of the population. The population growth rate dP/dt is given by dP/dt = rP(1-P/k) , where r is a positive constant and k is the carrying capacity.
 
The population sizes of the mosquitoes in a certain area at different times, in days, are
given in the table below.

time

0

35

63

91

105

126

140

182

203

224

number of mosquitoes

49

77

125

196

240

316

371

534

603

658

time

245

252

322

371

392

406

441

504

539

567

number of mosquitoes

701

712

776

791

794

796

798

799

800

800

 

 

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Inspection of the derivative function shows that it's a parabola​. When P=0 or k, dP/dt=0. And we can rewrite the function: P'=dP/dt=(rP/k)(k-P)=(r/k)(Pk-P^2)=(r/k)(k^2/4-P^2+Pk-k^2/4)=(r/k)((k/2)^2-(k-P)^2). The vertex is at (k/2,rk/2). So the distance between the point where the curve cuts the P (horizontal) axis gives us a measure for k; and, using this value for k we get r by inspecting the vertex height above the P axis and dividing by k/2.

The table shows P against t, but we need P' against P; but we can roughly calculate dt and dP by subtracting pairs of numbers. When P=800 in the table we clearly have two values for t and the gradient P'=0. Therefore, k=800, implying that k/2=400. When P=400 in the table we can see that t is between 140 and 182 and P is between 371 and  534. So the gradient is roughly (534-371)/(182-140)=163/42=3.9. Therefore rk/2=3.9 so r=3.9/400=0.01.

dP/P(1-P/k)=rdt. So int(dP/P(1-P/k))=rt+C, where C is to be determined from the table.

Int(dP/P)-int(dP/(P-k))=rt+C; ln(P)-ln|P-k|=rt+C; ln(P/|P-k|)=rt+C; P/|P-k|=ae^rt where a replaces constant C (a=e^C).

P=|P-k|ae^rt, so P=ake^rt/(1+ae^rt).

From the table, when t=0, P=49, so 49=ak/(1+a); 49+49a=ak and a(k-49)=49, a=49/(k-49). Using the approximate value of k=800, a is approximately 49/(800-49)=0.065. dP/dt=rP-rP^2/k. Since r and k are constants the DE can be solved by separation of variables:

So putting in our approximate values P=0.065*800*e^0.01t(1+0.065e^0.01t)⇒

P=52e^0.01t(1+0.065e^0.01t).

The values in the table can be plotted on a graph of P against t, but to plot dP/dt against P would require us to know r and k, for which we have only the approximate values 0.01 and 800.

 

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