Question

The number N(t) of supermarkets throughout the country that are using a computerized checkout system is described by the initial-value problem dN dt = N(1 − 0.0004N), N(0) = 1.

Answer 1

${N(t)(1-0.0004N(t)=0.9996e^t$

Step-by-step explanation:

$\frac{dN}{dt}=N(1- 0.0004N)$

$\Leftrightarrow\frac{dN}{N(1-0.0004N)}=dt$

$\Leftrightarrow (\frac{1}{N} +\frac{0.0004}{(1-0.0004N)} )dN = dt$

Integrating both sides

$log N-log(1-0.0004N)=t +C$            [ where C integrating constant]

$\Leftrightarrow logN(1-0.0004N) =t +C$

Given condition N=1 when t=0

$log1(1-0.0004)=0+C$

$\therefore C = log(0.9996)$

Therefore  $logN(1-0.0004N)-log 0.9996=t$

$\Leftrightarrow {N(t)(1-0.0004N(t)=0.9996e^t$