Answer:
Solution of differential equation is
[tex]p(t)=\frac{QA}{(A+(Q-A)e^{-Qkt})}[/tex]
Step-by-step explanation:
As complete question is not given so considering complete statement as attached below.
Growth of population = p(t)
As growth is unchecked, for constant k it is given as:
[tex]p'=kp\\\\\frac{dp}{dt}=kp\\\\\frac{1}{p}dp=k.dt[/tex]
Integrating both sides
[tex]\frac{1}{p}dp=k.dt\\\\ln|p|=kt+c\\\\p=Ae^{kt}[/tex]
Model differential equation is:
p′=kp(Q−p)--(1)
Initial condition p(0)=A.
From (1) by separating variables
[tex]\frac{dp}{dt}=kp(Q-p)\\\\\frac{1}{p(Q-p)}dp=k.dt\\\\\int\limits^t_0 {\frac{1}{p(Q-p)}} \, dp= \int\limits^t_0 {k} \, dt---(2)\\[/tex]
[tex]{\frac{1}{p(Q-p)}[/tex] can be factorize as:
[tex]\frac{1}{P(Q-p)}=\frac{A}{p}+\frac{B}{Q-p}\\\\1=A(Q-p)+Bp[/tex]
Equating coefficients of P
-A+B=0
A=B
Equating constants
AQ=1
A=1/Q
Then (2) becomes
[tex]\int {\frac{1}{Qp}+\frac{1}{Q(Q-p)}} \, dp= \int{k} \, dt\\\\\int {\frac{1}{p}+\frac{1}{(Q-p)}} \, dp= \int{Qk} \, dt\\\\\int {(\frac{1}{p}+\frac{1}{(Q-p)})} \, dp= \int {Qk} \, dt\\\\ln|p|-ln|Q-p|=Qkt+c\\\\ln|\frac{p}{Q-p}|=Qkt+c\\\\\frac{p}{Q-p}=e^{Qkt+c}\\\\p=(Q-p)(e^{Qkt+c})\\\\p=Q(e^{Qkt+c})-p(e^{Qkt+c})\\\\p+p(e^{Qkt+c})=Q(e^{Qkt+c})\\\\p(1+e^{Qkt+c})=Qe^{Qkt+c}\\\\p=\frac{Qe^{Qkt+c}}{(1+e^{Qkt+c})}\\\\p=\frac{Q}{(1+e^{-Qkt-c})}\\\\e^{-c}=D\\\\p(t)=\frac{Q}{(1+De^{-Qkt})}\\[/tex]
at t=0 p(0) = A
then
[tex]p(t)=\frac{Q}{(1+De^{-Qkt})}\\\\p(0)=\frac{Q}{(1+De^{0})}\\\\A=\frac{Q}{(1+D)}\\1+D=\frac{Q}{A}\\\\D=\frac{Q-A}{A}[/tex]
[tex]p(t)=\frac{Q}{(1+\frac{(Q-A)}{A}e^{-Qkt})}\\\\p(t)=\frac{QA}{(A+(Q-A)e^{-Qkt})}[/tex]
The solution to the separable differential equation is:
[tex]p = \frac{Q}{exp(-k*Q*t)*(Q/A - 1) + 1}[/tex]
Here we need to solve the differential equation:
p' = kp(Q - p)
This is a separable differential equation, we can write:
[tex]\frac{dp}{dt} = kp*(Q - p)\\\\\frac{dp}{p*(Q - p)} = k*dt\\[/tex]
Now we just integrate in both sides, so we will get:
[tex]\int\limits\frac{dp}{p*(Q - p)} = \int\limits k*dt\\\\\\-\frac{ln(Q/p -1)}{Q} = k*t + C[/tex]
Where C is a constant of integration.
Solving for p we get:
[tex]{ln(Q/p -1) = -Q*k*t -Q*C\\\\[/tex]
[tex]Q/p - 1 = exp(-Q*k*t - Q*C)\\\\Q/p = exp(-Q*k*t - Q*C) + 1\\\\p = \frac{Q}{exp(-Q*k*t - Q*C) + 1}[/tex]
Now, we know that for t = 0, p = A, replacing that we get:
[tex]p = \frac{Q}{exp(-Q*C) + 1} = A\\\\p = \frac{Q}{a + 1} = A\\\\a = Q/A - 1[/tex]
Then we can write:
[tex]p = \frac{Q}{exp(-k*Q*t)*(Q/A - 1) + 1}[/tex]
If you want to learn more about differential equations, you can read:
https://brainly.com/question/18760518
