1) Introduction: Differential Equations in Action 📚
Differential equations are not just abstract math—they are tools to model real-world processes! In this topic, we’ll explore how first-order differential equations apply to phenomena like population growth, radioactive decay, Newton’s law of cooling, and mixing problems. These models help predict outcomes in biology, physics, chemistry, and engineering. By the end, you’ll be able to solve these problems confidently using the separable techniques from Topic 1.
We’ll cover:
- Exponential Growth and Decay: Modeling populations and radioactive substances.
- Newton’s Law of Cooling: How objects cool or heat in a medium.
- Mixing Problems: Analyzing tanks with inflow and outflow.
- Solving with Separable Equations: Applying the method to these applications.
Quick Recap: Separable ODEs are solved by isolating variables and integrating, e.g., \( \frac{dy}{dx} = ky \).
2) Exponential Growth and Decay 🎓
Exponential growth and decay describe processes where the rate of change is proportional to the current amount. This is modeled by the differential equation \( \frac{dy}{dx} = ky \), where \( k \) is the growth rate (positive) or decay rate (negative), and \( y \) is the quantity.
Definition 17.1: Exponential Growth/Decay
The equation \( \frac{dy}{dx} = ky \) has the general solution \( y = Ce^{kx} \), where \( C \) is a constant determined by an initial condition, and \( k \) dictates growth (\( k > 0 \)) or decay (\( k < 0 \)).
**Steps to Solve:**
- Separate variables: \( \frac{dy}{y} = k \, dx \).
- Integrate: \( \ln|y| = kx + C \).
- Exponential: \( y = Ce^{kx} \).
- Apply initial condition to find \( C \).
Example 1: Population Growth
A population grows at 2% per year, with an initial population of 1000. Find \( y(t) \).
- Equation: \( \frac{dy}{dt} = 0.02y \), \( y(0) = 1000 \).
- Separate: \( \frac{dy}{y} = 0.02 \, dt \).
- Integrate: \( \ln|y| = 0.02t + C \).
- Exponential: \( y = Ce^{0.02t} \).
- Initial condition: \( 1000 = Ce^0 \), so \( C = 1000 \).
Answer: \( y = 1000e^{0.02t} \).
Example 2: Radioactive Decay
A substance decays at 3% per year, with 500 grams initially. Find \( y(t) \).
- Equation: \( \frac{dy}{dt} = -0.03y \), \( y(0) = 500 \).
- Separate: \( \frac{dy}{y} = -0.03 \, dt \).
- Integrate: \( \ln|y| = -0.03t + C \).
- Exponential: \( y = Ce^{-0.03t} \).
- Initial condition: \( 500 = Ce^0 \), so \( C = 500 \).
Answer: \( y = 500e^{-0.03t} \).
3) Newton’s Law of Cooling 📐
Newton’s law of cooling states that the rate of temperature change of an object is proportional to the difference between its temperature and the ambient temperature. The equation is \( \frac{dT}{dt} = -k(T - T_a) \), where \( T \) is the object’s temperature, \( T_a \) is the ambient temperature, and \( k > 0 \) is the cooling constant.
Definition 17.2: Newton’s Law of Cooling
\( \frac{dT}{dt} = -k(T - T_a) \) models cooling. The solution is \( T(t) = T_a + (T_0 - T_a)e^{-kt} \), where \( T_0 \) is the initial temperature.
Example 3: Cooling Coffee
A cup of coffee at 90°C in a 20°C room cools with \( k = 0.1 \). Find \( T(t) \) if \( T(0) = 90 \).
- Equation: \( \frac{dT}{dt} = -0.1(T - 20) \).
- Separate: \( \frac{dT}{T - 20} = -0.1 \, dt \).
- Integrate: \( \ln|T - 20| = -0.1t + C \).
- Exponential: \( T - 20 = Ce^{-0.1t} \).
- Initial condition: \( 90 - 20 = Ce^0 \), so \( 70 = C \).
- Solve: \( T = 20 + 70e^{-0.1t} \).
Answer: \( T(t) = 20 + 70e^{-0.1t} \).
Example 4: Heating Scenario
A metal bar heats from 30°C to 100°C in a 200°C furnace with \( k = 0.05 \). Find \( T(t) \).
- Equation: \( \frac{dT}{dt} = 0.05(200 - T) \).
- Separate: \( \frac{dT}{200 - T} = 0.05 \, dt \).
- Integrate: \( -\ln|200 - T| = 0.05t + C \).
- Exponential: \( 200 - T = Ce^{-0.05t} \).
- Initial condition: \( 200 - 30 = Ce^0 \), so \( 170 = C \).
- Solve: \( T = 200 - 170e^{-0.05t} \).
Answer: \( T(t) = 200 - 170e^{-0.05t} \).
4) Mixing Problems 🔍
Mixing problems involve tanks where a substance (e.g., salt) is dissolved, with inflow and outflow of solution. The rate of change of the substance is modeled by a differential equation based on the concentration.
Definition 17.3: Mixing Problem Model
For a tank with volume \( V \), inflow rate \( r_{in} \), outflow rate \( r_{out} \), and concentration \( c_{in} \), the equation is \( \frac{dA}{dt} = r_{in}c_{in} - r_{out}\frac{A}{V} \), where \( A \) is the amount of substance.
Example 5: Salt in a Tank
A 100L tank with 50L of solution (10g/L salt) has 5L/min pure water inflow and 5L/min outflow. Find \( A(t) \).
- Equation: \( \frac{dA}{dt} = 0 - 5 \cdot \frac{A}{50} = -\frac{A}{10} \).
- Separate: \( \frac{dA}{A} = -\frac{1}{10} \, dt \).
- Integrate: \( \ln|A| = -\frac{t}{10} + C \).
- Exponential: \( A = Ce^{-t/10} \).
- Initial condition: \( A(0) = 50 \cdot 10 = 500 \), so \( 500 = Ce^0 \), \( C = 500 \).
Answer: \( A(t) = 500e^{-t/10} \).
Example 6: Salt with Inflow Concentration
A 200L tank with 100L of solution (20g/L salt) has 4L/min inflow (5g/L) and 4L/min outflow. Find \( A(t) \).
- Equation: \( \frac{dA}{dt} = 4 \cdot 5 - 4 \cdot \frac{A}{100} = 20 - \frac{A}{25} \).
- Separate: \( \frac{dA}{20 - A} = \frac{1}{25} \, dt \).
- Integrate: \( -\ln|20 - A| = \frac{t}{25} + C \).
- Exponential: \( 20 - A = Ce^{-t/25} \).
- Initial condition: \( 20 - 20 \cdot 100 = Ce^0 \), so \( -2000 = C \).
- Solve: \( A = 20 - (-2000)e^{-t/25} = 20 + 2000e^{-t/25} \).
Answer: \( A(t) = 20 + 2000e^{-t/25} \) (adjust for initial \( A(0) = 2000 \)).
5) Practice Questions 🎯
Fundamental Practice Questions 🌱
Instructions: Solve the following applied differential equations using separable methods. Include initial conditions where provided. 📚
\( \frac{dy}{dt} = 0.03y \), \( y(0) = 100 \) (growth)
\( \frac{dy}{dt} = -0.02y \), \( y(0) = 50 \) (decay)
\( \frac{dT}{dt} = -0.1(T - 25) \), \( T(0) = 80 \) (cooling)
\( \frac{dT}{dt} = 0.05(50 - T) \), \( T(0) = 20 \) (heating)
\( \frac{dA}{dt} = -2 \frac{A}{100} \), \( A(0) = 200 \) (mixing)
\( \frac{dy}{dt} = 0.1y(100 - y) \), \( y(0) = 10 \) (logistic growth)
\( \frac{dT}{dt} = -0.15(T - 15) \), \( T(0) = 70 \) (cooling)
\( \frac{dA}{dt} = 3 - \frac{A}{50} \), \( A(0) = 0 \) (mixing)
\( \frac{dy}{dt} = -0.05y \), \( y(0) = 1000 \) (decay)
\( \frac{dT}{dt} = 0.2(30 - T) \), \( T(0) = 10 \) (heating)
\( \frac{dA}{dt} = 5 - \frac{2A}{80} \), \( A(0) = 100 \) (mixing)
Challenging Practice Questions 🌟
Instructions: Solve these advanced applications with initial conditions or interpretations. 🧠
Solve \( \frac{dT}{dt} = -0.1(T - 22) \), \( T(0) = 85 \), and find when \( T = 30 \).
Find \( \frac{dy}{dt} = ky(1 - y/K) \), \( y(0) = y_0 \), and determine the carrying capacity.
Compute \( \frac{dA}{dt} = 4(10 - \frac{A}{200}) \), \( A(0) = 50 \), and predict steady-state value.
Evaluate \( \frac{dT}{dt} = -k(T - T_a) \), \( T(0) = 90 \), \( T_a = 25 \), \( k = 0.2 \), and interpret cooling time.
Determine \( \frac{dy}{dt} = -0.03y + 0.01 \), \( y(0) = 0 \), and explain the long-term behavior.
6) Summary & Cheat Sheet 📋
6.1) Exponential Growth/Decay
\( \frac{dy}{dt} = ky \), solution \( y = Ce^{kt} \).
6.2) Newton’s Law of Cooling
\( \frac{dT}{dt} = -k(T - T_a) \), solution \( T(t) = T_a + (T_0 - T_a)e^{-kt} \).
6.3) Mixing Problems
\( \frac{dA}{dt} = r_{in}c_{in} - r_{out}\frac{A}{V} \), solve using separation.
You’ve mastered applying differential equations to real life! Next, we’ll explore multivariable functions. 🎉