$by$ the differential equation

$\frac{dP}{dt}=\frac{1}{6}P(1-\frac{P}{800})$

This $is$ a separable differential equation that can $be$ solved $by$ integrating

$800{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{P(800-P)}dP=\frac{1}{6}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dt$

The solution $to$ this equation $is$ given $by$

$P(t)=\frac{800}{1+A{e}^{-\frac{t}{6}}}$

where $Ais$ a constant determined $by$ the initial conditions. $Ifwe$ let $tbe$ the number $of$ years since $1950,$ then our

initial condition becomes $P(0)=600.$ From this $we$ determine that

$A=$

Furthermore $if$ the environmental conditions were $to$ remain unchanged, then $(to$ the nearest million rabbits$)$ the rabbit

population $in1980$ would $be$

million.

Interestingly the rabbit population will approach the maximum sustainable population $of$

$\underset{t}{lim}P(t)=,$ million $by$ the differential equation

$\frac{dP}{dt}=\frac{1}{6}P(1-\frac{P}{800})$

This $is$ a separable differential equation that can $be$ solved $by$ integrating

$800{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{P(800-P)}dP=\frac{1}{6}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dt$

The solution $to$ this equation $is$ given $by$

$P(t)=\frac{800}{1+A{e}^{-\frac{t}{6}}}$

where $Ais$ a constant determined $by$ the initial conditions. $Ifwe$ let $tbe$ the number $of$ years since $1950,$ then our

initial condition becomes $P(0)=600.$ From this $we$ determine that

$A=$

Furthermore $if$ the environmental conditions were $to$ remain unchanged, then $(to$ the nearest million rabbits$)$ the rabbit

population $in1980$ would $be$

million.

Interestingly the rabbit population will approach the maximum sustainable population $of$

$\underset{t}{lim}P(t)=,$ million