1. **Problem 2: Population Growth** The population grows at a rate proportional to its current population. Initial population $P(0) = 10000$ at year 2020, and $P(5) = 15000$ at year 2025. 2. **Formulating the differential equation:** The rate of change of population $P(t)$ is proportional to $P(t)$: $$\frac{dP}{dt} = kP$$ where $k$ is the proportionality constant. 3. **General solution:** This is a separable differential equation with solution: $$P(t) = P_0 e^{kt}$$ where $P_0 = 10000$ is the initial population. 4. **Find $k$ using $P(5) = 15000$:** $$15000 = 10000 e^{5k} \implies e^{5k} = 1.5 \implies 5k = \ln(1.5) \implies k = \frac{\ln(1.5)}{5}$$ 5. **Population in 2030 ($t=10$):** $$P(10) = 10000 e^{10k} = 10000 e^{10 \times \frac{\ln(1.5)}{5}} = 10000 e^{2 \ln(1.5)} = 10000 (1.5)^2 = 22500$$ 6. **Time to reach 50000:** Set $P(t) = 50000$: $$50000 = 10000 e^{kt} \implies 5 = e^{kt} \implies kt = \ln(5) \implies t = \frac{\ln(5)}{k} = \frac{\ln(5)}{\frac{\ln(1.5)}{5}} = 5 \frac{\ln(5)}{\ln(1.5)} \approx 5 \times 4.965 = 24.825$$ So population reaches 50000 approximately 25 years after 2020, i.e., around year 2045. --- 7. **Problem 3: Salt in Tank** Initial salt: 40 pounds in 600 gallons. Inflow: 3 pounds/gallon at 4 gallons/minute. Outflow: 5 gallons/minute. Let $y(t)$ be the amount of salt in pounds at time $t$ minutes. 8. **Differential equation:** Rate of salt in: $3 \times 4 = 12$ pounds/min. Rate of salt out: concentration in tank $= \frac{y}{600 - t}$ pounds/gallon (since volume decreases by 1 gallon/min), outflow rate 5 gallons/min, so outflow salt rate: $$5 \times \frac{y}{600 - t} = \frac{5y}{600 - t}$$ Thus, $$\frac{dy}{dt} = 12 - \frac{5y}{600 - t}$$ 9. **General solution:** Rewrite: $$\frac{dy}{dt} + \frac{5}{600 - t} y = 12$$ Use integrating factor: $$\mu(t) = e^{\int \frac{5}{600 - t} dt} = e^{-5 \ln|600 - t|} = (600 - t)^{-5}$$ Multiply both sides: $$(600 - t)^{-5} \frac{dy}{dt} + \frac{5}{600 - t} (600 - t)^{-5} y = 12 (600 - t)^{-5}$$ Left side is derivative: $$\frac{d}{dt} \left[y (600 - t)^{-5} \right] = 12 (600 - t)^{-5}$$ Integrate both sides: $$y (600 - t)^{-5} = \int 12 (600 - t)^{-5} dt + C$$ Substitute $u = 600 - t$, $du = -dt$: $$\int 12 u^{-5} dt = -12 \int u^{-5} du = -12 \times \frac{u^{-4}}{-4} = 3 u^{-4} = 3 (600 - t)^{-4}$$ So, $$y (600 - t)^{-5} = 3 (600 - t)^{-4} + C$$ Multiply both sides by $(600 - t)^5$: $$y = 3 (600 - t) + C (600 - t)^5$$ 10. **Use initial condition $y(0) = 40$:** $$40 = 3 \times 600 + C \times 600^5 \implies 40 = 1800 + C \times 600^5 \implies C = \frac{40 - 1800}{600^5} = \frac{-1760}{600^5}$$ 11. **Amount of salt after 12 minutes:** $$y(12) = 3 (600 - 12) + C (600 - 12)^5 = 3 \times 588 + \frac{-1760}{600^5} \times 588^5$$ Calculate: $$3 \times 588 = 1764$$ $$\frac{588^5}{600^5} = \left(\frac{588}{600}\right)^5 = (0.98)^5 \approx 0.9039$$ So, $$y(12) \approx 1764 - 1760 \times 0.9039 = 1764 - 1590.9 = 173.1$$ Approximately 173.1 pounds of salt after 12 minutes.