Answer
a) $y'(t)$ = $ky(1-y)$
b) $y(t)$ = $\frac{1}{1+9{e^{-0.896t}}}$
c) $3.67839$ days
Work Step by Step
a)
$y'(t)$ = $ky(1-y)$
where $k$ is a constant of proportionality
b)
$y(t)$ = $\frac{1}{1-\frac{e^{-kt}}{C}}$
the initial condition $y(0)$ = $\frac{1}{10}$ allow us to determine the value of C
$\frac{1}{10}$ = $\frac{1}{1-\frac{1}{C}}$
$C$ = $-\frac{1}{9}$
the condition $y(2)$ = $\frac{2}{5}$ allow us to determine the value of $k$
$\frac{2}{5}$ = $\frac{1}{1+9\frac{e^{-2k}}{C}}$
$k$ = $\frac{1}{2}\ln6$ $\approx$ $0.896$ $days^{-1}$
the particular solution of the differential equation for y is then
$y(t)$ = $\frac{1}{1+9{e^{-0.896t}}}$
c)
$\frac{3}{4}$ = $\frac{1}{1+9{e^{-0.896t}}}$
$t$ = $\frac{\ln27}{0.896}$ $\approx$ $3.67839$ days
