Question

The growth of a bacterial population \(y(t)\) is given by the differential equation \(\frac{dy}{dt} = k(y+3)\): where \(k\) is the growth rate constant. If \(y(0) = 1\) and \(y(1) = 5\), then the value of \(k\) is _______

• A. \(\ln 2\)
• B. \(\ln 4\)
• C. \(\ln 8\)
• D. \(\ln 12\)

Hint:

For any differential equation of the form \(y' = k(y + a)\), the solution is always \(y(t) + a = (y(0) + a)e^{kt}\).
Using this shortcut directly: \(y(1) + 3 = (y(0) + 3)e^{k \cdot 1} \implies 8 = 4 e^k \implies e^k = 2 \implies k = \ln 2\). This takes under 15 seconds!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question describes a bacterial population growth model governed by a first-order separable ordinary differential equation.
We are given the initial population at \(t = 0\) and the population at \(t = 1\), and we need to determine the growth rate constant \(k\).

Step 2: Key Formula or Approach:
The given differential equation is:
\[ \frac{dy}{dt} = k(y+3) \] We can solve this by separating the variables \(y\) and \(t\) and integrating both sides:
\[ \int \frac{dy}{y+3} = \int k \, dt \]

Step 3: Detailed Explanation:
Let us execute the integration and apply the boundary conditions systematically:
1. Integrate both sides:
\[ \ln(y + 3) = kt + C \] where \(C\) is the constant of integration.
2. Convert the logarithmic equation to exponential form:
\[ y(t) + 3 = e^{kt + C} = e^{C} \cdot e^{kt} \] Let us define \(C' = e^{C}\) as a new constant:
\[ y(t) + 3 = C' e^{kt} \] \[ y(t) = C' e^{kt} - 3 \] 3. Apply the first boundary condition, \(y(0) = 1\), to find \(C'\):
\[ 1 = C' e^{k(0)} - 3 \] \[ 1 = C'(1) - 3 \implies C' = 4 \] So the population equation becomes:
\[ y(t) = 4 e^{kt} - 3 \] 4. Apply the second boundary condition, \(y(1) = 5\), to find the growth constant \(k\):
\[ 5 = 4 e^{k(1)} - 3 \] Add 3 to both sides:
\[ 8 = 4 e^{k} \] Divide both sides by 4:
\[ e^{k} = 2 \] Take the natural logarithm of both sides:
\[ k = \ln 2 \] This matches Option (A) perfectly.

Step 4: Final Answer
The value of the growth rate constant \(k\) is \(\ln 2\), which corresponds to option (A).