Question

The population \( p(t) \) at time \( t \) of a certain mouse species satisfies the differential equation:
\[ \frac{d p(t)}{dt} = 0.5p(t) - 450. \] If \( p(0) = 850 \), then the time at which the population becomes zero is:

• A. \( 2 \ln 18 \)
• B. \( \ln 9 \)
• C. \( \frac{1}{2} \ln 18 \)
• D. \( \ln 18 \)

Hint:

For solving first-order linear differential equations, use separation of variables and then integrate to find the general solution.

Answer 1

The Correct Option is A

The given differential equation is: \[ \frac{d p(t)}{dt} = \frac{1}{2} p(t) - 450. \] Rewriting the equation: \[ \frac{d p(t)}{dt} = \frac{p(t) - 900}{2}. \] Next, integrate both sides: \[ 2 \int \frac{d p(t)}{p(t) - 900} = \int - dt. \] This results in: \[ 2 \ln|p(t) - 900| = -t + C. \] Using the initial condition \( p(0) = 850 \): \[ 2 \ln(50) = C. \] Now, solving for \( p(t) \) when \( p(t) = 0 \): \[ p(t) = 900 - 50e^{-t/2}. \] Set \( p(t) = 0 \), solving for \( t \) gives: \[ t = 2 \ln 18. \]

Answer 2

Step 1: Write the given differential equation
We are given the differential equation for the population \( p(t) \) of a mouse species: \[ \frac{d p(t)}{dt} = 0.5 p(t) - 450. \]
Step 2: Rearrange the equation
Rearrange the equation to separate the variables: \[ \frac{d p(t)}{dt} = 0.5 p(t) - 450 \quad \Rightarrow \quad \frac{d p(t)}{dt} = 0.5 \left(p(t) - 900\right). \] Now, separate the variables \( p(t) \) and \( t \): \[ \frac{1}{p(t) - 900} \, dp(t) = 0.5 \, dt. \]
Step 3: Integrate both sides
Integrating both sides: \[ \int \frac{1}{p(t) - 900} \, dp(t) = \int 0.5 \, dt. \] The integral on the left side is \( \ln |p(t) - 900| \), and the integral on the right side is \( 0.5t + C \): \[ \ln |p(t) - 900| = 0.5 t + C. \]
Step 4: Apply the initial condition
We are given the initial condition \( p(0) = 850 \). Substitute \( t = 0 \) and \( p(0) = 850 \) into the equation: \[ \ln |850 - 900| = 0.5(0) + C \quad \Rightarrow \quad \ln 50 = C. \] Thus, the equation becomes: \[ \ln |p(t) - 900| = 0.5 t + \ln 50. \]
Step 5: Solve for \( p(t) \)
Exponentiate both sides to eliminate the logarithm: \[ |p(t) - 900| = 50 e^{0.5t}. \] Since \( p(t) - 900 \) is positive (population cannot be negative), we have: \[ p(t) - 900 = 50 e^{0.5t}. \] Thus, the population as a function of time is: \[ p(t) = 900 + 50 e^{0.5t}. \]
Step 6: Find the time when the population becomes zero
We are asked to find the time when the population \( p(t) = 0 \). Set \( p(t) = 0 \) and solve for \( t \): \[ 0 = 900 + 50 e^{0.5t} \quad \Rightarrow \quad 50 e^{0.5t} = -900. \] This simplifies to: \[ e^{0.5t} = -18 \quad \Rightarrow \quad 0.5 t = \ln 18 \quad \Rightarrow \quad t = 2 \ln 18. \] Thus, the time at which the population becomes zero is: \[ \boxed{2 \ln 18}. \]