Solve.
-A local pond can only hold up to 39 geese. Six geese are introduced into the pond. Assume that the rate of grow of the population is $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 39 - \mathrm { P } ) \mathrm { P }$ where $t$ is time in weeks. Find a formula for the goose population in terms of $\mathrm { t }$.
A) $\mathrm { P } ( \mathrm { t } ) = \frac { 39 } { 1 + 5.5 \mathrm { e } ^ { - 0.051 \mathrm { t } } }$
B) $P ( t ) = \frac { 39 } { 1 + 33 e ^ { - 39 t } }$
C) $\mathrm { P } ( \mathrm { t } ) = \frac { 39 } { 1 + 5.5 \mathrm { e } ^ { - 39 \mathrm { t } } }$
D) $\mathrm { P } ( \mathrm { t } ) = \frac { 39 } { 1 + 33 \mathrm { e } ^ { - 0.051 \mathrm { t } } }$

Question

Quizplus · Accepted Answer

The given differential equation is a logistic growth model. To solve it, we separate variables and integrate. The initial condition is $P(0) = 6$, which helps determine the constant after integration. The correct formula, considering the logistic growth model and initial condition, is $P(t) = \frac{39}{1 + 5.5e^{-0.051t}}$, which matches choice A.

Solve -A Local Pond Can Only Hold Up to 39 dPdt=(0.0013)(39−P)P\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 39 - \mathrm { P } ) \mathrm { P }dtdP​=(0.0013)(39−P)P

Solve
-A Local Pond Can Only Hold Up to 39 $\frac { \mathrm { dP } } { \mathrm { dt } } = ( 0.0013 ) ( 39 - \mathrm { P } ) \mathrm { P }$