Let's solve the problem step by step as requested. --- ## Step 1: **Set up and simplify the differential equation (part a)** ### **Information Given:** - Initial population: \( P_ = 634 \) - Birth rate per inhabitant: \( b = .35 \) - Death rate per inhabitant: \( d = .5 \) - Net migration out: \( M = 241 \) inhabitants/year ### **Population Change Equation:** \[ \frac{dP}{dt} = \text{(births)} - \text{(deaths)} - \text{(migration out)} \] \[ \frac{dP}{dt} = (b - d)P - M \] Substitute the values: \[ \frac{dP}{dt} = (.35 - .5)P - 241 \] \[ \frac{dP}{dt} = -.15P - 241 \] --- **Explanation:** The rate of change of the population is the sum of natural growth (births minus deaths) and net migration (here, negative because people leave). The equation reflects that. --- ## Step 2: **Determine the population model (part b)** This is a **linear first-order differential equation**: \[ \frac{dP}{dt} + .15P = -241 \] ### **Solving the differential equation:** The integrating factor \( \mu(t) = e^{.15t} \): Multiply both sides by \( e^{.15t} \): \[ e^{.15t}\frac{dP}{dt} + .15e^{.15t}P = -241e^{.15t} \] Left side is derivative of \( e^{.15t}P \): \[ \frac{d}{dt}(e^{.15t}P) = -241e^{.15t} \] Integrate both sides: \[ e^{.15t}P = \int -241e^{.15t} dt + C \] \[ e^{.15t}P = -241 \cdot \frac{e^{.15t}}{.15} + C \] \[ e^{.15t}P = -\frac{241}{.15}e^{.15t} + C \] \[ P = -\frac{241}{.15} + Ce^{-.15t} \] Now use the initial condition \( P() = 634 \): \[ 634 = -\frac{241}{.15} + C \] \[ 634 = -1606.67 + C \] \[ C = 634 + 1606.67 = 7946.67 \] So, the **population model** is: \[ \boxed{P(t) = -1606.67 + 7946.67e^{-.15t}} \] --- **Explanation:** We solved the differential equation using integrating factors, then solved for the constant using the initial population. The final formula gives the population at any time \( t \) in years. --- ## **Final Answer (Step 1 & 2):** ### **Step 1:** **Differential equation:** \[ \frac{dP}{dt} = -.15P - 241 \] ### **Step 2:** **Population model:** \[ \boxed{P(t) = -1606.67 + 7946.67e^{-.15t}} \] --- *Let me know if you'd like to proceed with the next parts!*

## Step 3: **Construct a table showing the population over the next ten years (part c)** ### **Calculating Population for Each Year:** We’ll use the population model derived in the previous step: \[ P(t) = -1606.67 + 7946.67e^{-0.15t} \] ### **Calculating for \( t = 0 \) to \( t = 10 \)** | Year \( t \) | Population \( P(t) \) | |--------------|------------------------| | 0 | \( 634 \) | | 1 | \( -1606.67 + 7946.67e^{-0.15 \cdot 1} \approx 5664.24 \) | | 2 | \( -1606.67 + 7946.67e^{-0.15 \cdot 2} \approx 5114.40 \) | | 3 | \( -1606.67 + 7946.67e^{-0.15 \cdot 3} \approx 4590.75 \) | | 4 | \( -1606.67 + 7946.67e^{-0.15 \cdot 4} \approx 4090.87 \) | | 5 | \( -1606.67 + 7946.67e^{-0.15 \cdot 5} \approx 3613.22 \) | | 6 | \( -1606.67 + 7946.67e^{-0.15 \cdot 6} \approx 3155.96 \) | | 7 | \( -1606.67 + 7946.67e^{-0.15 \cdot 7} \approx 2717.31 \) | | 8 | \( -1606.67 + 7946.67e^{-0.15 \cdot 8} \approx 2297.43 \) | | 9 | \( -1606.67 + 7946.67e^{-0.15 \cdot 9} \approx 1895.57 \) | | 10 | \( -1606.67 + 7946.67e^{-0.15 \cdot 10} \approx 1509.75 \) | ### **Final Table:** | \( t \) | \( P \) | |---------|-----------------| | 0 | \( 634 \) | | 1 | \( 5664.24 \) | | 2 | \( 5114.40 \) | | 3 | \( 4590.75 \) | | 4 | \( 4090.87 \) | | 5 | \( 3613.22 \) | | 6 | \( 3155.96 \) | | 7 | \( 2717.31 \) | | 8 | \( 2297.43 \) | | 9 | \( 1895.57 \) | | 10 | \( 1509.75 \) | --- **Explanation:** For each year from 0 to 10, we calculated the population using the derived model. Each entry shows how the population changes over the years. --- ## Step 4: **Determine when the population drops to zero (part d)** ### **Setting the Population Model to Zero:** To find when the population drops to zero, we set the population equation to zero: \[ 0 = -1606.67 + 7946.67e^{-0.15t} \] ### **Solving for \( t \):** \[ 7946.67 e^{-0.15t} = 1606.67 \] \[ e^{-0.15t} = \frac{1606.67}{7946.67} \] Taking the natural logarithm on both sides: \[ -0.15t = \ln\left(\frac{1606.67}{7946.67}\right) \] \[ t = -\frac{1}{0.15} \ln\left(\frac{1606.67}{7946.67}\right) \] Calculating \( t \): \[ t \approx -\frac{1}{0.15} \ln(0.202) \] \[ t \approx -\frac{1}{0.15} \cdot (-1.589) \approx 10.59 \] ### **Final Answer:** The population will drop to zero at approximately: \[ \boxed{10.59 \text{ years}} \] --- **Explanation:** We solved for \( t \) when the population function equals zero, finding that it occurs slightly over 10.5 years from the initial population.