The initial population is P(0) = 600.
The carrying capacity for the pond is K=1500.
The rate of growth is r=1.4 (or 140%).
As this model is described by the logistic equation, we can write:
[tex]\frac{dP}{dt}=rP(1-\frac{P}{K})=1.4P(1-\frac{P}{1500})[/tex]We can solve this differential equation as:
[tex]\begin{gathered} dP=rP(1-\frac{P}{K})dt \\ \frac{dP}{P(1-\frac{P}{K})}=r\cdot dt \\ \int \frac{dP}{P(1-\frac{P}{K})}=\int r\cdot dt \\ -K\int \frac{dP}{P(P-K)}=\int r\cdot dt \\ -K\int \frac{dP}{(1-\frac{K}{P})P^2}=\int r\cdot dt \end{gathered}[/tex]We can substitute the variables as:
[tex]\begin{gathered} u=1-\frac{K}{P} \\ \frac{du}{dP}=\frac{K}{P^2} \\ dP=\frac{P^2}{K}du \end{gathered}[/tex]Replacing in the integral:
[tex]\begin{gathered} -K\int (\frac{1}{(1-\frac{K}{P})P^2}dP=-K\int \frac{1}{u\cdot P^2}\frac{P^2}{K}du=-\int \frac{1}{u}du \\ -\int \frac{1}{u}du=-\ln |u|=-\ln |1-\frac{K}{P}|=-\ln (\frac{K}{P}-1) \\ -\ln (\frac{K}{P}-1)=-\ln (\frac{K-P}{P})=\ln (\frac{P}{K-P}) \end{gathered}[/tex]The other integral is solved as:
[tex]\int r\cdot dt=rt+C[/tex]Then, we can write:
[tex]\begin{gathered} \ln (\frac{P}{K-P})=rt+C_1 \\ \frac{P}{K-P}=C\cdot e^{rt} \\ P=(K-P)Ce^{rt}_{} \\ P=K\cdot Ce^{rt}-P\cdot Ce^{rt} \\ P+P\cdot Ce^{rt}=K\cdot Ce^{rt} \\ P(1+Ce^{rt})=K\cdot Ce^{rt} \\ P=\frac{KCe^{rt}}{1+Ce^{rt}} \end{gathered}[/tex]We can find the value of the constant C using the information of the initial condition:
[tex]\begin{gathered} t=0\Rightarrow P(0)=600 \\ \frac{P}{K-P}=Ce^{rt} \\ \frac{600}{1500-600}=C\cdot e^{1.4\cdot0}=C\cdot e^0=C\cdot1=C \\ C=\frac{600}{900}=\frac{2}{3} \end{gathered}[/tex]Then, the model becomes:
[tex]P(t)=\frac{1500\cdot\frac{2}{3}\cdot e^{1.4t}}{1+\frac{2}{3}e^{1.4t}}=\frac{1000e^{1.4t}}{1+\frac{2}{3}e^{1.4t}}[/tex]For the first season (t=1), the population will be:
[tex]P(1)=\frac{1000e^{1.4\cdot1}}{1+\frac{2}{3}e^{1.4\cdot1}}\approx\frac{1000\cdot4.055}{1+0.667\cdot4.055}\approx\frac{4055}{3.70}\approx1096[/tex]After the second season (t=2), the population will be:
[tex]P(2)=\frac{1000e^{1.4\cdot2}}{1+\frac{2}{3}e^{1.4\cdot2}}\approx\frac{1000\cdot16.444}{1+0.667\cdot16.444}\approx\frac{16444}{11.963}\approx1375[/tex]Answer:
Population after the first season P(1) = 1096 fish.
Population after the second season P(1) = 1375 fish.
