Answer:
[tex]P(t)=\frac{38125e^{kt}}{7(1+\frac{5}{56} e^{kt})}[/tex]
Step-by-step explanation:
The size of fish population satisfies the logistic equation:
[tex]\frac{dP}{dt} =kP(1-\frac{P}{K})[/tex]
K=6100 is the carry capacity. Substitute to get:
[tex]\frac{dP}{dt} =kP(1-\frac{P}{6100})[/tex]
We separate variables to obtain:
[tex]\frac{6100dP}{P(6100-P)} =k\cdot dt[/tex]
Use partial fraction decomposition to get:
[tex]\frac{dP}{P}+\frac{dP}{6100-P} =k\cdot dt[/tex]
Integrate to obtain:
[tex]\ln \frac{|P|}{|6100-P}=kt+C_1[/tex]
[tex]\implies \frac{P}{6100-P} =Ce^{kt}------(1)[/tex]
We apply the ICs to get, [tex]C=\frac{5}{56}[/tex]
The expression for population size is
[tex]P(t)=\frac{6100*\frac{5}{6} e^{kt}}{1+\frac{5}{6} e^{kt}}[/tex]
[tex]P(t)=\frac{38125e^{kt}}{7(1+\frac{5}{56} e^{kt})}[/tex]
b)
Substitute P=3050 into [tex]\frac{P}{6100-P} =Ce^{kt}[/tex]
This implies that:
[tex]\implies \frac{3050}{6100-3050} =\frac{5}{56} e^{kt}[/tex]
Solve for t to get:
[tex]t=\frac{\ln(\frac{56}{5} )}{k} ,k\ne 0[/tex]
