Question

Let \( y(t) \) be a bacterial population whose growth is given by \[ \frac{dy}{dt} = \lambda (y + 2) \] where \( \lambda \) is the growth rate constant. If \( y(0) = 1 \) and \( y(1) = 4 \), then the value of \( \lambda \) is

• A. \( \ln 2 \)
• B. \( \ln 3 \)
• C. \( \ln 4 \)
• D. \( \ln 6 \)

Hint:

To solve separable differential equations, integrate both sides after separating the variables, and use the initial conditions to determine the constants.

Answer 1

The Correct Option is A

The differential equation describing the bacterial population is: \[ \frac{dy}{dt} = \lambda (y + 2). \] This is a separable differential equation. To solve, first separate the variables: \[ \frac{dy}{y + 2} = \lambda \, dt. \] Next, integrate both sides: \[ \int \frac{dy}{y + 2} = \lambda \int dt. \] The integral of \( \frac{1}{y + 2} \) is \( \ln |y + 2| \), so: \[ \ln |y + 2| = \lambda t + C, \] where \( C \) is a constant of integration. To solve for \( C \), use the initial condition \( y(0) = 1 \): \[ \ln |1 + 2| = C \quad \Rightarrow \quad \ln 3 = C. \] Thus, the equation becomes: \[ \ln |y + 2| = \lambda t + \ln 3. \] Now, apply the second condition \( y(1) = 4 \): \[ \ln |4 + 2| = \lambda(1) + \ln 3, \] \[ \ln 6 = \lambda + \ln 3. \] Solving for \( \lambda \): \[ \lambda = \ln 6 - \ln 3 = \ln \frac{6}{3} = \ln 2. \] Thus, the value of \( \lambda \) is \( \boxed{\ln 2} \).