How Is Bacterial Population Growth Modeled by a Differential Equation?

Jin314159
The population growth of bacteria is proportional to the square of the population.
 
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dP/dt = kP^2
dP/P^2 = kdt
-1/P = kt + C

-1/Po = C

-1/P = kt - 1/Po = (Po*kt - 1)/Po

P = Po/(1 - Po*kt)

cookiemonster
 


To solve this differential equation, we can use the method of separation of variables. Let P(t) represent the population of bacteria at time t. The given information tells us that dP/dt is proportional to P^2, which can be written as:

dP/dt = kP^2

where k is a constant of proportionality. Now, we can separate the variables and integrate both sides:

1/P^2 dP = k dt

Integrating both sides gives us:

-1/P = kt + C

where C is the constant of integration. Solving for P, we get:

P(t) = -1/(kt + C)

However, we also know that the population cannot be negative, so we can discard the negative sign and rewrite the equation as:

P(t) = 1/(kt + C)

To find the specific solution, we can use the initial condition that the population at time t=0 is P0. This means that when t=0, P(t) = P0, so we can substitute these values into the equation:

P0 = 1/(0 + C)

Solving for C, we get:

C = 1/P0

Substituting this back into the equation, we get the final solution:

P(t) = 1/(kt + 1/P0)

This is the general solution to the given differential equation. To find the specific solution for a particular population growth scenario, we would need to know the value of the constant k and the initial population size P0.
 
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