Welcome to Constructing Differential Equations!
In this chapter, we are going to learn how to translate real-world descriptions of change into mathematical equations. This is one of the most powerful tools in a mathematician's toolkit. Instead of just looking at where something is right now, we look at how fast it is changing. Whether it’s a population of rabbits growing or the price of a new smartphone dropping, differential equations allow us to model the "speed" of these changes.
Don't worry if this seems a bit abstract at first! By the end of these notes, you’ll see that constructing these equations is really just like translating a sentence from English into Mathematics.
1. What is a Differential Equation?
A differential equation is simply any equation that contains a derivative (like \( \frac{dy}{dx} \), \( \frac{dV}{dt} \), or \( \frac{dh}{dt} \)).
In most GCSE and early A Level problems, you might be used to equations like \( y = x^2 + 5 \). In this chapter, we look at equations that describe the rate of change. For example:
\( \frac{dy}{dx} = 3x \)
This tells us that the "gradient" or "rate of change of \( y \)" depends on the value of \( x \).
Analogy: The Speedometer vs. The Odometer
Think of a car. Your odometer shows your total distance (that's like \( y \)). Your speedometer shows your rate of change of distance (that's like \( \frac{dy}{dt} \)). A differential equation is like a rule that explains how your speedometer reading relates to other things, like how hard you are pressing the gas pedal!
Quick Review:
- \( \frac{dy}{dx} \) means: "The rate of change of \( y \) with respect to \( x \)."
- \( \frac{dP}{dt} \) means: "The rate of change of Population (\( P \)) over Time (\( t \))."
2. The Language of "Proportionality"
Most exam questions won't give you the equation directly. Instead, they will use specific keywords. The most important term to look for is proportional to.
If a question says a rate is proportional to something, we use the symbol \( \propto \). To turn this into an equation, we replace the \( \propto \) with \( = k \), where \( k \) is a constant of proportionality.
- "The rate of increase of \( x \) is proportional to \( x \)."
Translates to: \( \frac{dx}{dt} = kx \) - "The rate of decrease of volume \( V \) is inversely proportional to the time \( t \)."
Translates to: \( \frac{dV}{dt} = -\frac{k}{t} \)
Memory Aid: The "k" Key
Whenever you see "proportional," you must include the letter \( k \). It is the "key" that unlocks the equation! Also, remember: if something is decreasing, your derivative should usually be negative.
Key Takeaway: "Rate of change" always goes on the left as a fraction (the derivative), and the "proportional" part goes on the right with a constant \( k \).
3. Context 1: Population Growth
The OCR syllabus specifically mentions population growth as a key context. In the real world, the more people (or bacteria, or animals) there are, the more "births" occur, so the population grows faster.
The Scenario: "A population of bacteria, \( P \), grows at a rate that is proportional to the number of bacteria present at time \( t \)."
Step-by-Step Construction:
1. Identify the rate of change: \( \frac{dP}{dt} \)
2. Identify what it is proportional to: \( P \)
3. Use the constant \( k \): \( \frac{dP}{dt} = kP \)
Did you know? This simple equation leads to what we call "Exponential Growth." It’s why a small number of bacteria can become a massive colony in just a few hours!
4. Context 2: Kinematics (Motion)
In Mechanics and Pure Math, we often model how objects move. You should already know these two definitions, but they are vital for constructing equations:
- Velocity (\( v \)) is the rate of change of displacement (\( s \)): \( v = \frac{ds}{dt} \)
- Acceleration (\( a \)) is the rate of change of velocity (\( v \)): \( a = \frac{dv}{dt} \)
Example: "A car accelerates at a rate proportional to the square root of its velocity."
Equation: \( \frac{dv}{dt} = k\sqrt{v} \)
Common Mistake to Avoid: Don't confuse the variable with its rate. If the question says "acceleration is proportional to...", do not write \( a \propto \dots \). Instead, write \( \frac{dv}{dt} \propto \dots \) because the goal of a differential equation is to show the derivative!
5. Context 3: Price and Demand
The syllabus also mentions the relationship between price (\( P \)) and demand (\( D \)). Generally, as the price of an item increases, the demand for it decreases.
The Scenario: "The rate at which the demand for a product changes with respect to its price is inversely proportional to the square of the price."
Step-by-Step Construction:
1. The rate of change of demand with respect to price: \( \frac{dD}{dP} \)
2. Inversely proportional to the square of price: \( \frac{1}{P^2} \)
3. Assemble with \( k \): \( \frac{dD}{dP} = \frac{k}{P^2} \)
Key Takeaway: Always read carefully to see what is changing "with respect to" what. In this case, it was \( P \), not time \( t \).
6. Summary Checklist for Success
When you are faced with a "Constructing Differential Equations" question, follow these steps:
- Identify the variables: What letters represent the quantities? (e.g., \( V \) for Volume, \( r \) for radius).
- Identify the independent variable: Usually time (\( t \)), but could be something else like price (\( P \)) or distance (\( x \)).
- Look for the "Rate" word: This becomes your \( \frac{d(something)}{d(something)} \).
- Spot the "Proportionality": Is it direct (\( kx \)) or inverse (\( \frac{k}{x} \))?
- Check the sign: Is the quantity increasing (positive \( k \)) or decreasing (negative \( k \))?
Key Takeaway: You aren't being asked to solve the equation here—just to build it. Think of it as drawing the blueprint before building the house!
Quick Review Box
Direct Proportionality: \( \frac{dy}{dt} = ky \)
Inverse Proportionality: \( \frac{dy}{dt} = \frac{k}{y} \)
Proportional to the square: \( \frac{dy}{dt} = ky^2 \)
Rate of decrease: \( \frac{dy}{dt} = -k \dots \)