Question

A bacteria sample of certain number of bacteria is observed to grow exponentially in a given amount of time. Using the exponential growth model, the rate of growth of this sample of bacteria is calculated.

The differential equation representing the growth of bacteria is given as:

\[ \frac{dP}{dt} = kP, \] where \( P \) is the population of bacteria at any time \( t \). 
Based on the above information, answer the following questions:
(i) Obtain the general solution of the given differential equation and express it as an exponential function of \( t \).
(ii) If the population of bacteria is 1000 at \( t = 0 \), and 2000 at \( t = 1 \), find the value of \( k \).

Hint:

For exponential growth or decay, always separate variables, integrate, and rewrite in exponential form.

Answer 1

(i) General solution of the differential equation: The differential equation is: \[ \frac{dP}{dt} = kP. \] Separating variables \( P \) and \( t \): \[ \frac{1}{P} \, dP = k \, dt. \] Integrating both sides: \[ \int \frac{1}{P} \, dP = \int k \, dt. \] Solving the integrals: \[ \ln |P| = kt + C, \] where \( C \) is the constant of integration. Rewriting in exponential form: \[ P = e^{kt + C} = e^C \cdot e^{kt}. \] Letting \( e^C = P_0 \), where \( P_0 \) is the initial population, we get: \[ P = P_0 e^{kt}. \] Final Answer (i): The general solution is: \[ P = P_0 e^{kt}. \]