Question

A bacteria sample of certain number of bacteria is observed to grow exponentially in a given amount of time. Using 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 \( t \). bf{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 given differential equation is: \[ \frac{dP}{dt} = kP. \] Separate the variables \( P \) and \( t \): \[ \frac{1}{P} \, dP = k \, dt. \] Integrate both sides: \[ \int \frac{1}{P} \, dP = \int k \, dt. \] Solve the integrals: \[ \ln |P| = kt + C, \] where \( C \) is the constant of integration. Rewrite in exponential form: \[ P = e^{kt + C} = e^C \cdot e^{kt}. \] Let \( e^C = P_0 \), where \( P_0 \) is the initial population. Then: \[ P = P_0 e^{kt}. \] Final Answer (i): The general solution is: \[ P = P_0 e^{kt}. \]