Respuesta :
Solution:
Given :
[tex]$\frac{dP}{dt}= aP-bP^2$[/tex] .............(1)
where, B = aP = birth rate
D = [tex]$bP^2$[/tex] = death rate
Now initial population at t = 0, we have
[tex]$P_0$[/tex] = 220 , [tex]$B_0$[/tex] = 9 , [tex]$D_0$[/tex] = 15
Now equation (1) can be written as :
[tex]$ \frac{dP}{dt}=P(a-bP)$[/tex]
[tex]$\frac{dP}{dt}=bP(\frac{a}{b}-P)$[/tex] .................(2)
Now this equation is similar to the logistic differential equation which is ,
[tex]$\frac{dP}{dt}=kP(M-P)$[/tex]
where M = limiting population / carrying capacity
This gives us M = a/b
Now we can find the value of a and b at t=0 and substitute for M
[tex]$a_0=\frac{B_0}{P_0}$[/tex] and [tex]$b_0=\frac{D_0}{P_0^2}$[/tex]
So, [tex]$M=\frac{B_0P_0}{D_0}$[/tex]
= [tex]$\frac{9 \times 220}{15}$[/tex]
= 132
Now from equation (2), we get the constants
k = b = [tex]$\frac{D_0}{P_0^2} = \frac{15}{220^2}$[/tex]
= [tex]$\frac{3}{9680}$[/tex]
The population P(t) from logistic equation is calculated by :
[tex]$P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$[/tex]
[tex]$P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$[/tex]
[tex]$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$[/tex]
As per question, P(t) = 110% of M
[tex]$\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$[/tex]
[tex]$ 220-88e^{\frac{-99}{2420} t}=200$[/tex]
[tex]$ e^{\frac{-99}{2420} t}=\frac{5}{22}$[/tex]
Now taking natural logs on both the sides we get
t = 36.216
Number of months = 36.216
