Differential equations - Mathematics (9709) - Cambridge A-Level

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Differential Equations (Pure Mathematics 3: Topic 3.8)

Welcome to the chapter on Differential Equations! Don't worry if this sounds intimidating—it’s actually one of the most powerful and fascinating tools in mathematics. Differential equations are simply equations that involve derivatives ($dy/dx$).

What you will learn: You will learn how to translate real-world concepts involving rates of change (like population growth, cooling tea, or dissolving chemicals) into mathematical equations, and then how to solve those equations to predict the future behavior of the system.

Why it matters: Differential equations are the language of physics, engineering, economics, and biology. Mastering this topic allows you to mathematically model and understand how systems change over time.

1. Introduction to Differential Equations (DEs)

What is a Differential Equation?

A Differential Equation (DE) is any equation that contains at least one derivative of an unknown function. We use DEs to model situations where the rate of change of a quantity depends on the quantity itself or on time.

Example: $ \frac{dy}{dx} = 2x $ is a simple DE.
The Solution: The solution to a DE is not a number, but a function $ y = f(x) $. If we solve the example above by integration, we get $ y = x^2 + C $.

The First-Order Separable DE (Our Focus)

In the P3 syllabus, we focus exclusively on first-order DEs where the variables are separable.

First-Order: This means the highest derivative present is the first derivative ($dy/dx$ or $dx/dt$).
Separable: This means the equation can be rearranged so that all terms involving the dependent variable (usually $y$) and $dy$ are on one side, and all terms involving the independent variable (usually $x$ or $t$) and $dx$ (or $dt$) are on the other side.

Did you know? The motion of a swinging pendulum, the spread of a virus, and the path of a rocket are all described using differential equations.

2. Formulating Differential Equations from Context

This section involves translating simple statements about rates of change into the standard mathematical notation.

Key Translation Phrases

Always identify the variable and the rate of change requested:

"The rate of change of quantity $X$" $\longrightarrow \frac{dX}{dt} $
"is proportional to" $\longrightarrow = k \times (\text{something}) $ (where $k$ is the constant of proportionality)
"is inversely proportional to" $\longrightarrow = \frac{k}{(\text{something})} $
"varies as the square of $y$" $\longrightarrow \propto y^2 $
"decreases" or "decays" $\longrightarrow $ means the rate must be negative (usually achieved by introducing a negative sign, e.g., $ \frac{dy}{dt} = -ky $)

Step-by-Step Formulation Example

Imagine a chemical reaction where the rate of increase of the concentration, $C$, is proportional to the concentration itself.

Step 1: Identify the Rate of Change:
Rate of increase of $C$ means $ \frac{dC}{dt} $.

Step 2: Identify the Relationship:
"is proportional to the concentration $C$" means $ \propto C $.

Step 3: Combine and Introduce Constant:
$$ \frac{dC}{dt} = kC $$

(If it had been "rate of decrease," we would write $ \frac{dC}{dt} = -kC $.)

Key Takeaway (Formulation)

The structure is always: Rate (LHS) = Relationship (RHS). Remember to always include the constant of proportionality, $k$, whenever you see the word "proportional."

3. Solving Separable Differential Equations

Once you have formulated the DE, the next job is to solve it using the technique of Separation of Variables and then Integration.

3.1 The Method of Separation of Variables

This method works only if you can isolate the variables. We will use the general form: $ \frac{dy}{dx} = f(x)g(y) $.

Step 1: Separate the Variables
Gather all terms involving $y$ (and $dy$) on the left, and all terms involving $x$ (and $dx$) on the right.

$$ \frac{dy}{dx} = f(x)g(y) \quad \longrightarrow \quad \frac{1}{g(y)} dy = f(x) dx $$

Step 2: Integrate Both Sides
Integrate the LHS with respect to $y$ and the RHS with respect to $x$.

$$ \int \frac{1}{g(y)} dy = \int f(x) dx $$

Step 3: Introduce the Constant of Integration
You must introduce the constant $C$ on one side only (conventionally the side with the independent variable, usually $x$ or $t$).

$$ \int \frac{1}{g(y)} dy = \int f(x) dx + C $$

Step 4: Express $y$ in terms of $x$ (if possible)
Rearrange the resulting equation to make $y$ the subject. This may involve using logarithms or exponentials.

Memory Aid for Integration:

You Do Integration, Clearly! (Y's and DY, Integrate, add Constant).

3.2 The Constant of Integration and Solution Types

When solving DEs, you always obtain a family of curves (because of $C$).

General Solution: The solution containing the arbitrary constant $C$. This describes the general behaviour of the system.
Particular Solution: The specific solution found by using an initial condition (a given point or starting value) to calculate the value of $C$.

Common Mistake Alert! Students often forget to include the constant $C$. If you forget $C$, you cannot find the particular solution, and you lose marks!

3.3 Crucial Integration Techniques (P3 Links)

Solving DEs often requires the advanced integration techniques covered in Topic 3.5. Be prepared to use:

(a) Integration leading to $\ln|x|$
You must be able to recognise the form $ \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C $.
This is essential when separating variables leads to $ \int \frac{1}{y} dy \text{ or } \int \frac{1}{ay+b} dy $.

(b) Partial Fractions
If the $y$-side is complicated (e.g., $ \frac{1}{y(y-1)} $), you must decompose it first before integration.

Example: $ \int \frac{1}{y(y-1)} dy = \int \left( \frac{1}{y-1} - \frac{1}{y} \right) dy = \ln|y-1| - \ln|y| + C $

(c) Substitution or Integration by Parts
The $x$-side (or $t$-side) may require integration by parts (e.g., $\int x \ln x dx$) or a specific substitution (if provided). Check Topic 3.5 carefully.

Example: Solving a Separable DE

Find the general solution of $ \frac{dy}{dx} = x^2 y $.

Step 1: Separate
$$ \frac{1}{y} dy = x^2 dx $$

Step 2: Integrate
$$ \int \frac{1}{y} dy = \int x^2 dx $$ $$ \ln|y| = \frac{1}{3}x^3 + C $$

Step 3: Rearrange (Find $y$)
We use the definition $ e^{\ln A} = A $:
$$ |y| = e^{(\frac{1}{3}x^3 + C)} $$ $$ y = A e^{\frac{1}{3}x^3 } \quad \text{where } A = \pm e^C \text{ or } A=0 $$ (Note: In A-Level, $C$ is often immediately replaced by $A$ or another constant when dealing with exponentials, as $ e^C $ is just another constant.)

Quick Review: Exponential Form

If you get $ \ln y = f(x) + C $, your general solution should look like $ y = A e^{f(x)} $, where $ A $ is the resulting constant derived from $ e^C $.

4. Real-Life Applications and Interpretation

The final step is usually applying the solution back to the context of the problem.

4.1 Using Initial Conditions (Finding the Particular Solution)

An initial condition is a specific point that the function must pass through (e.g., $y=5$ when $t=0$).

Step-by-Step: Finding C

Solve the DE to find the general solution (involving $C$).
Substitute the numerical values from the initial condition (e.g., $x=0, y=5$) into the general solution.
Solve the resulting numerical equation to find the value of $C$.
Substitute the calculated value of $C$ back into the general solution to get the particular solution.

It is often simplest to calculate $C$ immediately after the integration step, before trying to isolate $y$.

4.2 Interpreting the Solution

Once you have a particular solution $ y = f(t) $, you can answer prediction questions:

Example Context: Population $P$ growing over time $t$.
Question: Find the population after 10 years.
Solution: Substitute $ t=10 $ into your particular solution $ P = f(10) $ and calculate the result.

Crucial Syllabus Note: You will not be expected to know specialized facts about the context (e.g., advanced physics constants). All necessary information, including rates and initial conditions, will be provided in the question.

Analogy: Population Growth

The simplest growth model is based on the idea that the rate of growth is proportional to the current population: $ \frac{dP}{dt} = kP $.
This leads to an exponential solution: $ P = A e^{kt} $. This model shows that the more people you have, the faster the population grows—a classic example of modeling with a separable DE.

Key Takeaway (Differential Equations)

Differential equations in P3 require three key skills: Formulation (writing the DE), Separation and Integration (solving the DE, often requiring advanced P3 integration techniques), and Application (using initial conditions to find $C$ and interpreting the result).