Answer
(a) $10lnP - 10ln(2000-P) + 10ln19 = t$
$P(t=20) = 557$
(b) $t \approx 33.5$
Work Step by Step
(a) $\frac{dP}{dt} = 0.1P(1- \frac{P}{2000})$
$\int \frac{dP}{0.1P(1- \frac{P}{2000})} = \int dt$....(1)
Now, $\frac{1}{0.1P(1- \frac{P}{2000})} = \frac{1- \frac{P}{2000}}{0.1P(1- \frac{P}{2000})} + \frac{\frac{P}{2000}}{0.1P(\frac{2000-P}{2000})}$
So, $\frac{1}{0.1P(1- \frac{P}{2000})} = \frac{10}{P} + \frac{10}{2000-P}$....(2)
Using (1) and (2):
$\int \frac{10}{P}dP + \int \frac{10}{2000-P} = \int dt$
$10lnP - 10ln(2000-P) + lnC = t$
Now, at t=0, P=100, so:
$10ln100 - 10ln1900 + lnC = 0$
$lnC = 10ln1900 - 10ln100$
$lnC = ln\frac{1900^{10}}{100^{10}}$
$C = 19^{10}$
Thus, $10lnP - 10ln(2000-P) + 10ln19 = t$
$ln(\frac{P^{10}19^{10}}{(2000-P)^{10}}) = t$.....(3)
For t=20:
$(\frac{19P}{2000-P})^{10} = e^{20}$
$\frac{19P}{2000-P} = e^2$
$\frac{19P}{2000-P} \approx 7.34$
$26.34P=14680$
$P(t=20) = 557$ {Since population $cannot$ be in decimal}
(b) $P=1200, t=?$
We know that: $10ln(\frac{19P}{2000-P}) = t$
$10ln(28.5) = t$
Thus, $t \approx 33.5$
