Answer :
Answer:
[tex]n(t)=n_0e^{(1-e^{-t }-t)}[/tex]
Step-by-step explanation:
If n(t) represents the population size at time t, where n is measured in individuals and t is measured in years.
[tex]\frac{dn}{dt}=n(e^{-t }-1), n(0)=n_o[/tex]
[tex]\frac{dn}{n}=(e^{-t }-1)dt[/tex]
Taking the integral of both sides
[tex]\int\frac{dn}{n}=\int(e^{-t }-1)dt\\\int\frac{dn}{n}= \int e^{-t }dt-\int1dt[/tex]
ln |n| = [tex]-e^{-t }-t+C[/tex]
Where C is integration constant
Taking the exponential of both sides
[tex]n=e^{(-e^{-t }-t+C)}[/tex]
[tex]n=e^{(-e^{-t }-t)}e^C\\n=Ke^{(-e^{-t }-t)}[/tex] whee the exponential of a constant is a constant K.
When t=0, [tex]n(0)=n_o[/tex]
[tex]n_0=Ke^{-1[/tex]
Therefore:
[tex]n=n_0e^{1}e^{(-e^{-t }-t)}[/tex]
[tex]n(t)=n_0e^{(1-e^{-t }-t)}[/tex]