Answer
a) $P(t)$ = $\frac{2000}{1+9e^{-0.3t}}$
b) $\approx$ $5.97$
Work Step by Step
a)
$P'(t)$ = $0.3P(t)(\frac{1-P(t)}{500})$
with general solution
$P(t)$ = $\frac{500}{1-\frac{e^{-0.3t}}{C}}$
the initial condition $P(0)$ = $50$ allow us to determine the value of C
$50$ = $\frac{2000}{1-\frac{1}{C}}$
$C$ = $-\frac{1}{9}$
then
$P(t)$ = $\frac{2000}{1+9e^{-0.3t}}$
b)
$P(t)$ = $200$
$200$ = $\frac{500}{1+9e^{-0.3t}}$
$t$ = $\frac{10}{3}\ln6$ $\approx$ $5.97$
