Answer
a) $P(60)=289$
$P(100)= 233$
b) $t=128$
Work Step by Step
a) Solve the differential equation:
$$\frac{dP}{P}=kP$$
$$\frac{1}{P}\frac{dP}{P}=k$$
Intergrate:
$$\int \frac{1}{P}dP=\int k dt$$
$$\ln (P)=kt + c_1$$
$$P=e^{kt+c_1}$$
Take power of $e$ in both side:
$$P(t)=c_1e^{kt}$$
Since $P_0$ is the population at $t=0$ then $C_0=P_0$
$$P(t)=P_0e^{kt}$$
We are given $P_0=400,P(30)=340$
$$340=400e^{30k}$$
$$\rightarrow k=-0.0052$$
After 60 days:
$$P(60)=400e^{-0.0052\times60}=289$$
After 100 days:
$$P(100)=400e^{-0.0052\times100} \approx 233$$
b) Find $t$ time required for the population of swans to be cut in half:
$$200=400e^{-0.0052t}$$
$\rightarrow t=-\frac{1}{0.0052}\ln (\frac{1}{2}) \approx 128$
