Solving the Bacterial Growth Differential Equation

The problem asks for the growth rate constant $ \lambda $ given the differential equation $ \frac{dy}{dt} = \lambda(y + 2) $ and the conditions $ y(0) = 1 $ and $ y(1) = 4 $. We need to solve this differential equation.

Differential Equation Solution Steps

1. Separate the variables:

   $ \frac{dy}{y + 2} = \lambda dt $

2. Integrate both sides:

   $ \int \frac{dy}{y + 2} = \int \lambda dt $ $ \ln|y + 2| = \lambda t + C_1 $ where $ C_1 $ is the constant of integration.

3. Exponentiate to remove the logarithm:

   $ |y + 2| = e^{\lambda t + C_1} = e^{C_1} e^{\lambda t} $ Let $ C = \pm e^{C_1} $. Since population $ y $ is typically non-negative and $ y+2 $ should be positive in this context, we can write: $ y + 2 = C e^{\lambda t} $ $ y(t) = C e^{\lambda t} - 2 $

4. Use the initial condition $ y(0) = 1 $ to find $ C $:

   $ 1 = C e^{\lambda(0)} - 2 $ $ 1 = C(1) - 2 $ $ C = 3 $ The equation becomes $ y(t) = 3 e^{\lambda t} - 2 $.

5. Use the second condition $ y(1) = 4 $ to find $ \lambda $:

   $ 4 = 3 e^{\lambda(1)} - 2 $ $ 4 + 2 = 3 e^{\lambda} $ $ 6 = 3 e^{\lambda} $ $ e^{\lambda} = \frac{6}{3} $ $ e^{\lambda} = 2 $

6. Solve for $ \lambda $ by taking the natural logarithm:

   $ \lambda = \ln(2) $

The value of $ \lambda $ is $ \ln(2) $. This corresponds to the first option.