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
dPdt=0.2P(t)(1−P(t)200),P(0)=150a:Autonomous since the RHS does not depend on tb:1000dPP(P−200)=dt5∫(1P−200−1P)dP=dt5ln∣P−200P∣=t+C′P−200P=Cet/5P(t)=2001−Cet/5P(0)=150⇒2001−C=150⇒C=−13P(t)=2001+13et/5\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}}dtdP=0.2P(t)(1−200P(t)),P(0)=150a:AutonomoussincetheRHSdoesnotdependontb:P(P−200)1000dP=dt5∫(P−2001−P1)dP=dt5ln∣∣PP−200∣∣=t+C′PP−200=Cet/5P(t)=1−Cet/5200P(0)=150⇒1−C200=150⇒C=−31P(t)=1+31et/5200
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments