Question

Let P(t) be the population of a certain animal species. Assume the

P(t) satisfies the logistic growth equation,

dP

dt = 0.2P(t) (1 −

P(t)

200) , y(0) = 150

a) Is the above equation autonomous? if yes (explain your answer with

proper reasons), if no (justify you answer).

b) Solve the above initial value problem, and find the value of solution at

time t = 0.5 using separation of variables

Answer 1

$\frac{dP}{dt}=0.2P\left( t \right) \left( 1-\frac{P\left( t \right)}{200} \right) ,P\left( 0 \right) =150\\a: Autonomous\,\,\sin ce\,\,the\,\,RHS\,\,does\,\,not\,\,depend\,\,on\,\,t\\b: \frac{1000dP}{P\left( P-200 \right)}=dt\\5\int{\left( \frac{1}{P-200}-\frac{1}{P} \right) dP}=dt\\5\ln \left| \frac{P-200}{P} \right|=t+C'\\\frac{P-200}{P}=Ce^{t/5}\\P\left( t \right) =\frac{200}{1-Ce^{t/5}}\\P\left( 0 \right) =150\Rightarrow \frac{200}{1-C}=150\Rightarrow C=-\frac{1}{3}\\P\left( t \right) =\frac{200}{1+\frac{1}{3}e^{t/5}}$