Mathematics (9660) | Differential equations

Welcome to the Chapter on Differential Equations!

Hello there! This chapter might sound complicated, but it is one of the most powerful tools in Pure Mathematics. Differential equations (DEs) are simply equations that involve derivatives (rates of change).

If differentiation told you the rate of change of a quantity, then differential equations let you work backwards from that rate of change to find the original quantity or function!

Why is this important? Differential equations are the language of change. They are used to model almost everything that changes over time, including population growth, radioactive decay, spread of diseases, and motion in physics.

In this unit (P2), we will focus on solving the simplest, but most common, type of first-order DEs: those with separable variables.

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1. Understanding Differential Equations (DEs)

1.1 What is a Differential Equation?

A differential equation is an equation that relates an unknown function to its independent variable (like \(x\) or \(t\)) and one or more of its derivatives.

Example:
\(\frac{dy}{dx} = 3x^2\) (Simple, you can solve this just by integrating the right side!)
\(\frac{dy}{dx} = 5y\) (The derivative depends on the function itself—this is a DE!)

1.2 Order of a Differential Equation

The order of a differential equation is the order of the highest derivative present in the equation.

• First Order: Involves only the first derivative, \(\frac{dy}{dx}\) (or \(\frac{dx}{dt}\)). This is the type we focus on in P2.
• Example: \(\frac{dy}{dx} = 4y + \cos x\)
• Second Order: Involves the second derivative, \(\frac{d^2y}{dx^2}\). (Not required in P2/A2 Mathematics 9660 syllabus.)

Key Takeaway: For this chapter, remember that we are only required to solve First Order DEs where the variables can be separated.

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2. Forming Simple Differential Equations (Context)

Before solving, you must be able to translate a real-world statement about a rate of change into a mathematical DE. This often involves concepts of proportionality (the rate of change is proportional to something).

2.1 Rates of Change and Proportionality

In modelling problems, the rate of change is almost always taken with respect to time, \(t\). If \(X\) is the quantity (e.g., population, mass, temperature), its rate of change is \(\frac{dX}{dt}\).

Memory Aid for Translation:

• "The rate of increase of \(X\)..." means \(\frac{dX}{dt}\) is positive.
• "The rate of decrease of \(X\)..." means \(\frac{dX}{dt}\) is negative.
• "...is proportional to \(Y\)..." means \(= kY\), where \(k\) is a constant.

2.2 The Growth and Decay Model

The most common model in P2 is where the rate of change of a quantity is proportional to the quantity itself. This describes exponential growth or exponential decay (e.g., bacteria multiplying, radioactive material decaying).

Statement: The rate of increase of a quantity \(P\) is proportional to the value of \(P\) at that time.

DE Formation:

\(\frac{dP}{dt} \propto P\)
\(\frac{dP}{dt} = kP\)

If \(k\) is positive, it's growth. If \(k\) is negative, it's decay.

Did you know? This DE, \(\frac{dP}{dt} = kP\), leads directly to the standard exponential formula \(P = Ae^{kt}\), where \(A\) is the initial amount.

Key Takeaway: Look for "rate of change" (\(\frac{d}{dt}\)) and "proportional to" (\(k\)) to form your equation.

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3. Analytical Solution: Separable Variables

Don't worry if this seems tricky at first—solving DEs with separable variables is a straightforward, repetitive process once you master the steps!

A first-order differential equation is separable if it can be written in the form:

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

This means you can separate the \(x\) terms (and \(dx\)) from the \(y\) terms (and \(dy\)).

3.1 Step-by-Step Method for Solving Separable DEs

Let's use a clear example: Solve \(\frac{dy}{dx} = \frac{x^2}{y}\).

Step 1: Separate the Variables

Move all \(y\) terms (including \(dy\)) to the left side and all \(x\) terms (including \(dx\)) to the right side. We treat \(dy\) and \(dx\) as separate components for separation purposes.

Original: \(\frac{dy}{dx} = \frac{x^2}{y}\)
Separate: \(y \, dy = x^2 \, dx\)

Step 2: Integrate Both Sides

Now, integrate both sides of the separated equation:

\(\int y \, dy = \int x^2 \, dx\)

Crucial Rule: You only need one constant of integration, \(+C\), and you should always add it to the side containing the independent variable (often \(x\) or \(t\)).

Integration:
\(\frac{1}{2}y^2 = \frac{1}{3}x^3 + C\)

This result is the General Solution.

Step 3: Rearrange to find the General Solution (if required)

Often, you will be asked to express \(y\) in terms of \(x\). You may need to manipulate the constant \(C\).

Rearrange:
\(y^2 = \frac{2}{3}x^3 + 2C\)

Since \(2C\) is just another arbitrary constant, we can simplify notation by letting \(A = 2C\).

General Solution: \(y^2 = \frac{2}{3}x^3 + A\)

3.2 Handling Exponential and Logarithmic DEs

When solving DEs related to growth/decay (\(\frac{dy}{dx} = ky\)), the integration often involves \(\ln\) and \(e\). Be careful with the algebraic manipulation of the constant \(C\).

Example: Solve \(\frac{dy}{dt} = 5y\)

Step 1: Separate
\(\frac{1}{y} \, dy = 5 \, dt\)

Step 2: Integrate
\(\int \frac{1}{y} \, dy = \int 5 \, dt\)
\(\ln |y| = 5t + C\)

Step 3: Rearrange (Exponentiate)
\(|y| = e^{5t + C}\)
\(|y| = e^{5t} \cdot e^C\)

Since \(e^C\) is a positive constant, we replace it with \(A\): \(A = e^C\). We also drop the modulus sign (since \(A\) covers the sign choice).

General Solution: \(y = Ae^{5t}\)

Key Takeaway: When you integrate \(\frac{1}{y}\) to get \(\ln|y| = f(x) + C\), the final algebraic form will often be \(y = Ae^{f(x)}\), where \(A\) is a new constant combining \(e^C\) and the modulus sign.

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4. Finding the Particular Solution

The General Solution contains the arbitrary constant \(C\) (or \(A\)). It describes a whole family of curves that satisfy the differential equation.

The Particular Solution is a single, unique curve from that family. To find it, you need an initial condition or a boundary condition—a specific point \((x_0, y_0)\) that the solution must pass through.

4.1 Process for Finding C

Using the example from Section 3.1: \(y^2 = \frac{2}{3}x^3 + C\). Let's say the initial condition is that when \(x=0\), \(y=2\).

Step 1: Substitute the initial condition into the General Solution.

Substitute \(x=0\) and \(y=2\) into the equation:

\(2^2 = \frac{2}{3}(0)^3 + C\)
\(4 = 0 + C\)
\(C = 4\)

Step 2: Write the Particular Solution.

Substitute the value of \(C\) back into the General Solution.

\(y^2 = \frac{2}{3}x^3 + 4\)

If required, solve for \(y\): \(y = \sqrt{\frac{2}{3}x^3 + 4}\)

Key Takeaway: A problem asking for "the solution" will require a particular solution, meaning you must find \(C\) using the given initial conditions.

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5. Applications to Practical Problems

The ability to solve separable DEs allows us to model real-world scenarios, typically involving exponential growth or decay in contexts like population, concentration, or finance.

5.1 Population Growth Example

A simplified model for population (\(P\)) growth is often \(\frac{dP}{dt} = kP\).

If a town's population increases by 10% per year, the rate of increase is proportional to the current population (\(k=0.1\)).

If the starting population (at \(t=0\)) is 5000, we found the General Solution is \(P = Ae^{kt}\).
When \(t=0\), \(P=5000\):
\(5000 = Ae^{k(0)} \implies 5000 = A(1)\)
So, \(A=5000\).

Particular Solution: \(P = 5000e^{0.1t}\)

We can now use this equation to predict the population at any future time \(t\).

5.2 Decay Example (Physics/Chemistry)

The rate of decay of radioactive material (\(M\)) is proportional to the mass remaining. Since it's decay, the constant \(k\) is negative.

DE: \(\frac{dM}{dt} = -kM\)

Solving this yields \(M = M_0e^{-kt}\), where \(M_0\) is the initial mass. Problems often ask you to find \(k\) using a given half-life (the time taken for the mass to halve).

Key Takeaway: In applications, the arbitrary constant \(A\) often represents the initial value of the quantity being modelled (i.e., the value when time \(t=0\)).