Question

A population grows exponentially according to the differential equation dP, dt equals k times P , where P is the population, t is time, and k is a positive constant. If P(0) = A, what is the time for the population to quadruple its initial value?

Answer 1

Answer:

$\frac{\ln 4}{k}$

Step-by-step explanation:

Given equation that shows the change in population with respect to time,

$\frac{dP}{dt}=kP$

$\frac{dP}{P}=kdt$

On integrating,

$\ln P = kt + C$

$\implies P = e^{kt+C}$

$P=e^C e^{kt}$

Put $e^C=P_0$

$P = P_0 e^{kt}$

According to the question,

If P = A if t = 0,

$A = P_0 e^0\implies A = P_0$

Thus, the required function,

$P = A e^{kt}$

If the final population = 4A,

$4A = A e^{kt}$

$4 = e^{kt}$

Taking natural log both sides,

$\ln 4 = \ln e^{kt}$

$\ln 4 = kt\ln e$

$\ln 4 = kt$

$\implies t = \frac{\ln 4}{k}$

Hence, after $\frac{\ln 4}{k}$ time, the population will be quadruple its initial value.