Answer
a) $P(t)$ = $\frac{2000}{1+3e^{-0.6t}}$
b) $\approx$ $1.83$
Work Step by Step
a)
$P'(t)$ = $0.6P(t)(\frac{1-P(t)}{2000})$
with general solution
$P(t)$ = $\frac{2000}{1-\frac{e^{-0.6t}}{C}}$
the initial condition $P(0)$ = $500$ allow us to determine the value of C
$500$ = $\frac{2000}{1-\frac{1}{C}}$
$C$ = $-\frac{1}{3}$
then
$P(t)$ = $\frac{2000}{1+3e^{-0.6t}}$
b)
$P(t)$ = $1000$
$1000$ = $\frac{2000}{1+3e^{-0.6t}}$
$t$ = $\frac{5}{3}\ln3$ $\approx$ $1.83$
