Understanding the application of differential equations in real-world scenarios is a critical aspect of mathematics, especially in fields like environmental science, engineering, and finance. This section aims to provide a comprehensive understanding of how to interpret the solutions of differential equations in various contexts.

Introduction to Differential Equations in Real-World Scenarios

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are essential in modelling the dynamics of systems where change is continuous and dependent on the current state of the system.

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The Role of Differential Equations in Modelling

Differential equations are used to model a wide range of phenomena, from the growth of populations to the decay of radioactive materials. They allow us to predict future states of a system based on its current state and the rate of change.

Example: Environmental Application

Consider a scenario involving environmental management:

"A lake initially contains 1 million gallons of water with a pollution level of 5 ppm (parts per million). Clean water flows into the lake at a rate of 100,000 gallons per day, and the mixed water flows out at the same rate. Determine the pollution level in the lake over time."

Differential Equation Formation:

• Let $P(t)$ be the pollution level at time $t$ in days.
• Equation: $\frac{dP}{dt} = \text{rateIn} - \text{rateOut}$.
• Since incoming water is clean, $\text{rateIn} = 0$.
• Outflow rate of pollution: $\text{rateOut} = \frac{P(t) \times 100,000}{1,000,000}$.
• This yields $\frac{dP}{dt} = -0.1P(t)$.

Solving the Differential Equation:

1. Separation of Variables:

• Rearrange to $\frac{dP}{P} = -0.1 dt$.

2. Integrate Both Sides:

• Integrate $\int \frac{dP}{P} = \int -0.1 dt$.
• This results in $\ln|P| = -0.1t + C$, where $C$ is the integration constant.

3. Solve for $P(t)$:

• Exponentiate both sides to remove the natural log, giving $P = e^{-0.1t + C}$.
• Rewrite as $P(t) = Ce^{-0.1t}$, where $C = e^C$ is a new constant.

4. Apply Initial Condition (y(0) = 5 ppm):

• Substitute $t = 0$ and $P(0) = 5$ into $P(t) = Ce^{-0.1t}$.
• $5 = Ce^0$ implies $C = 5$.

5. Final Solution:

• $P(t) = 5e^{-0.1t}$.

Conclusion:

• The solution $P(t) = 5e^{-0.1t}$ shows the pollution level decreases exponentially over time due to the dilution by the inflow of clean water.

Real-World Implications

Understanding the dynamics of pollution in natural water bodies is critical. This model allows for the prediction of pollution concentration over time, which is essential for:

• Planning and implementing clean-up operations.
• Informing the public and local businesses about water quality.
• Making legislative decisions regarding environmental protection.

Example: Pollution Level Prediction

With the formula $P(t) = 5e^{-0.001t}$ ppm, determine the pollution level in the lake after a duration of 1000 days.

Solution:

1. Substitute $t = 1000$:

• $P(1000) = 5e^{-0.001 \times 1000}$.

2. Calculate Numerical Value:

• Evaluate $5e^{-1}$.
• $P(1000) \approx 5 \times \frac{1}{e}$.
• Approximate with $e \approx 2.71828$.
• Resulting in $P(1000) \approx 1.84$ ppm.

Conclusion: After 1000 days, the pollution level is approximately 1.84 ppm.