Differential equations

Q: What is the standard form of a first-order linear differential equation, and what is its integrating factor?

The standard form is \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\), and the integrating factor is \(I(x) = \mathrm{e}^{\int P(x)\,\mathrm{d}x}\).

Q: What is the auxiliary equation for the second-order homogeneous differential equation \(a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = 0\)?

The auxiliary equation is \(am^2 + bm + c = 0\).

Q: What is the form of the Complementary Function (CF) when the auxiliary equation has two distinct real roots, \(m = \alpha\) and \(m = \beta\)?

The Complementary Function is \(y = A\mathrm{e}^{\alpha x} + B\mathrm{e}^{\beta x}\), where \(A\) and \(B\) are arbitrary constants.

Q: What is the form of the Complementary Function (CF) when the auxiliary equation has a repeated real root, \(m = \alpha\)?

The Complementary Function is \(y = (A + Bx)\mathrm{e}^{\alpha x}\).

Q: What is the form of the Complementary Function (CF) when the auxiliary equation has complex conjugate roots, \(m = p \pm \mathrm{i}q\)?

The Complementary Function is \(y = \mathrm{e}^{px}\big(A\cos(qx) + B\sin(qx)\big)\).

Q: How is the General Solution of a non-homogeneous second-order differential equation constructed?

It is the sum of the Complementary Function and the Particular Integral: \(y = y_{CF} + y_{PI}\).

Q: What trial form should be used for the Particular Integral (PI) if \(f(x) = k\cos(\omega x)\) or \(k\sin(\omega x)\)?

The trial form is \(y = \lambda\cos(\omega x) + \mu\sin(\omega x)\).

Q: What is the clash rule when selecting a trial form for the Particular Integral (PI)?

If the standard trial form for \(y_{PI}\) already appears in \(y_{CF}\), multiply the trial form by \(x\) (or \(x^2\) if it corresponds to a repeated root).

Q: When given a derivative boundary condition like \(\frac{\mathrm{d}y}{\mathrm{d}x} = v\) at \(x = x_0\), what function must be differentiated?

You must differentiate the complete general solution \(y = y_{CF} + y_{PI}\), not just the complementary function.

Q: How are non-standard differential equations solved when a substitution is given?

Differentiate the substitution (using the chain or product rule) to express derivatives in the new variable, substitute into the ODE to obtain a standard form, solve it, and convert back.

Further Mathematics 2330 · CCEA A Level · 9 min read · Updated: 31 Aug 2026

Differential Equations: Master Study Notes

Welcome to the chapter on Differential Equations for CCEA A2 Further Mathematics! Don't worry if this topic sounds intimidating at first. At its heart, solving a differential equation simply means "undoing" rates of change to find the original formula that describes a system.

Differential equations are used across physics, engineering, biology, and finance to model everything from vibrating guitar strings and electrical circuits to the spread of diseases. In this chapter, you will build upon the separation of variables you met in single A Level and learn powerful, systematic techniques to solve much broader families of first-order and second-order differential equations.

1. First-Order Linear Differential Equations: The Integrating Factor

What is a First-Order Linear Differential Equation?

A first-order linear differential equation involves \(x\), \(y\), and the first derivative \(\frac{\mathrm{d}y}{\mathrm{d}x}\). To solve it, we must first write it in standard form:

\(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\)

Notice that \(y\) and \(\frac{\mathrm{d}y}{\mathrm{d}x}\) both have a power of \(1\) and are not multiplied together.

The Idea: The Product Rule in Reverse

Think about differentiating a product using the product rule: \(\frac{\mathrm{d}}{\mathrm{d}x}(I(x)y) = I(x)\frac{\mathrm{d}y}{\mathrm{d}x} + I'(x)y\).
Our goal is to multiply the entire differential equation by a clever factor, called the Integrating Factor \(I(x)\), so that the left-hand side collapses neatly into the derivative of a single product.

Finding the Integrating Factor

The integrating factor is defined as:

\(I(x) = \mathrm{e}^{\int P(x)\,\mathrm{d}x}\)

Note: You do not need to add a constant of integration \(+C\) when calculating \(I(x)\). A factor of \(1\) is all we need!

Step-by-Step Method

Step 1: Rearrange the equation into standard form: \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\). Ensure the coefficient of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) is exactly \(1\).
Step 2: Identify \(P(x)\) and compute the integrating factor \(I(x) = \mathrm{e}^{\int P(x)\,\mathrm{d}x}\).
Step 3: Multiply every term of the standard form equation by \(I(x)\).
Step 4: Rewrite the left-hand side as \(\frac{\mathrm{d}}{\mathrm{d}x}\big(I(x)y\big)\).
Step 5: Integrate both sides with respect to \(x\): \(I(x)y = \int I(x)Q(x)\,\mathrm{d}x + C\).
Step 6: Divide by \(I(x)\) to make \(y\) the subject.

Worked Example

Problem: Find the general solution of \(x\frac{\mathrm{d}y}{\mathrm{d}x} + 3y = \frac{1}{x}\) for \(x > 0\).

Solution:
• Step 1: Divide throughout by \(x\) to get standard form:
\(\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{3}{x}y = \frac{1}{x^2}\)
Here, \(P(x) = \frac{3}{x}\) and \(Q(x) = \frac{1}{x^2}\).

• Step 2: Find the integrating factor:
\(\int P(x)\,\mathrm{d}x = \int \frac{3}{x}\,\mathrm{d}x = 3\ln(x) = \ln(x^3)\)
\(I(x) = \mathrm{e}^{\ln(x^3)} = x^3\)

• Step 3 & 4: Multiply the standard equation by \(x^3\):
\(x^3\frac{\mathrm{d}y}{\mathrm{d}x} + 3x^2y = x^3\left(\frac{1}{x^2}\right)\)
\(\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3 y\right) = x\)

• Step 5: Integrate both sides:
\(x^3 y = \int x\,\mathrm{d}x = \frac{1}{2}x^2 + C\)

• Step 6: Divide by \(x^3\):
\(y = \frac{1}{2x} + \frac{C}{x^3}\)

Common Mistakes to Avoid

• Forgetting to divide through so that the coefficient of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) is \(1\) before identifying \(P(x)\).
• Forgetting logarithm power rules: \(\mathrm{e}^{2\ln(x)} \neq 2x\), but rather \(\mathrm{e}^{\ln(x^2)} = x^2\).
• Forgetting to add the constant of integration \(+C\) before dividing by \(I(x)\). The constant must be divided by \(I(x)\) too!

Key Takeaway: First-order linear equations are solved by converting the equation into \(\frac{\mathrm{d}}{\mathrm{d}x}(Iy) = I Q\) using the factor \(I = \mathrm{e}^{\int P\,\mathrm{d}x}\).

2. Second-Order Homogeneous Linear Differential Equations

The General Form

A second-order linear differential equation with constant coefficients has the form:

\(a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = f(x)\)

When the right-hand side is zero, \(f(x) = 0\), the equation is called homogeneous:

\(a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = 0\)

The solution to this equation is called the Complementary Function (CF), denoted \(y_{CF}\).

The Auxiliary Equation

Because derivatives of \(\mathrm{e}^{mx}\) are proportional to \(\mathrm{e}^{mx}\), we try a solution of the form \(y = \mathrm{e}^{mx}\). Substituting this in leads directly to the Auxiliary Equation (AE):

\(am^2 + bm + c = 0\)

Solving this quadratic gives three possible cases depending on the discriminant \(\Delta = b^2 - 4ac\):

Case 1: Two Distinct Real Roots (\(m = \alpha\), \(m = \beta\))

When \(b^2 - 4ac > 0\), the auxiliary equation has two different real roots \(\alpha\) and \(\beta\).
The Complementary Function is:

\(y = A\mathrm{e}^{\alpha x} + B\mathrm{e}^{\beta x}\)

where \(A\) and \(B\) are arbitrary constants.

Case 2: One Repeated Real Root (\(m = \alpha\))

When \(b^2 - 4ac = 0\), there is a single repeated root \(\alpha\).
The Complementary Function is:

\(y = (A + Bx)\mathrm{e}^{\alpha x}\)

Did you know? We multiply the second term by \(x\) to ensure we still have two linearly independent parts for our general solution!

Case 3: Complex Conjugate Roots (\(m = p \pm \mathrm{i}q\))

When \(b^2 - 4ac < 0\), the roots are complex numbers: \(m = p \pm \mathrm{i}q\).
Using Euler's identity (\(\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta\)), the solution transforms into real trigonometric functions:

\(y = \mathrm{e}^{px}\big(A\cos(qx) + B\sin(qx)\big)\)

• The real part \(p\) controls the exponential growth or decay envelope.
• The imaginary part \(q\) gives the frequency of oscillation.

Quick Summary Table of CF Forms

• Distinct real roots \(\alpha, \beta\): \(y = A\mathrm{e}^{\alpha x} + B\mathrm{e}^{\beta x}\)
• Repeated real root \(\alpha\): \(y = (A + Bx)\mathrm{e}^{\alpha x}\)
• Complex roots \(p \pm \mathrm{i}q\): \(y = \mathrm{e}^{px}(A\cos(qx) + B\sin(qx))\)

Worked Example

Problem: Solve \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 6\frac{\mathrm{d}y}{\mathrm{d}x} + 13y = 0\).

Solution:
• Write the auxiliary equation: \(m^2 - 6m + 13 = 0\)
• Use the quadratic formula:
\(m = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(13)}}{2} = \frac{6 \pm \sqrt{36 - 52}}{2} = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4\mathrm{i}}{2} = 3 \pm 2\mathrm{i}\)
• Here, \(p = 3\) and \(q = 2\). Therefore, the general solution is:
\(y = \mathrm{e}^{3x}\big(A\cos(2x) + B\sin(2x)\big)\)

Key Takeaway: The roots of the auxiliary quadratic equation dictate the structure of the homogeneous solution: real roots give exponentials, repeated roots introduce a factor of \(x\), and complex roots give decaying or growing sinusoidal waves.

3. Second-Order Non-Homogeneous Differential Equations

The Fundamental Principle

When the right-hand side is not zero (\(f(x) \neq 0\)):

\(a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = f(x)\)

The General Solution (GS) is the sum of two parts:

\(\mathbf{General\ Solution} = \mathbf{Complementary\ Function\ (CF)} + \mathbf{Particular\ Integral\ (PI)}\)

\(y = y_{CF} + y_{PI}\)

• \(y_{CF}\) solves the equation when RHS \(= 0\) (contains arbitrary constants \(A\) and \(B\)).
• \(y_{PI}\) is any single specific function that produces \(f(x)\) on the RHS (contains no arbitrary constants).

Choosing the Form for the Particular Integral (PI)

We make an educated guess (trial function) based on the mathematical form of \(f(x)\):

• Constant: \(f(x) = k \implies \text{Try } y = \lambda\)
• Linear: \(f(x) = kx + c \implies \text{Try } y = \lambda x + \mu\)
• Quadratic: \(f(x) = kx^2 + \dots \implies \text{Try } y = \lambda x^2 + \mu x + \nu\)
• Exponential: \(f(x) = k\mathrm{e}^{kx} \implies \text{Try } y = \lambda\mathrm{e}^{kx}\)
• Trigonometric: \(f(x) = k\cos(\omega x)\) or \(k\sin(\omega x) \implies \text{Try } y = \lambda\cos(\omega x) + \mu\sin(\omega x)\)

The "Clash" Rule (When the Trial Function is Part of the CF)

If your standard trial for \(y_{PI}\) already appears inside your \(y_{CF}\), it will simply give zero when substituted into the LHS. To fix this, multiply your trial function by \(x\) (or by \(x^2\) if it is a repeated root).

Worked Example: Full Non-Homogeneous Equation

Problem: Find the general solution of \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 3\frac{\mathrm{d}y}{\mathrm{d}x} + 2y = 4x + 2\).

Step 1: Find the Complementary Function (\(y_{CF}\))
Auxiliary equation: \(m^2 - 3m + 2 = 0 \implies (m - 1)(m - 2) = 0\)
Roots: \(m = 1, m = 2\)
\(y_{CF} = A\mathrm{e}^{x} + B\mathrm{e}^{2x}\)

Step 2: Find the Particular Integral (\(y_{PI}\))
Since \(f(x) = 4x + 2\) is linear, try \(y = \lambda x + \mu\).
Differentiate:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \lambda\)
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0\)
Substitute these into the original ODE:
\(0 - 3(\lambda) + 2(\lambda x + \mu) = 4x + 2\)
\(2\lambda x + (2\mu - 3\lambda) = 4x + 2\)

Equate coefficients:
• For \(x\): \(2\lambda = 4 \implies \lambda = 2\)
• Constants: \(2\mu - 3(2) = 2 \implies 2\mu - 6 = 2 \implies 2\mu = 8 \implies \mu = 4\)
So, \(y_{PI} = 2x + 4\).

Step 3: Combine to Form the General Solution
\(y = y_{CF} + y_{PI}\)
\(y = A\mathrm{e}^{x} + B\mathrm{e}^{2x} + 2x + 4\)

Key Takeaway: Always find \(y_{CF}\) first. Then pick an appropriate template for \(y_{PI}\), differentiate it, substitute it into the ODE, equate coefficients, and add both parts together.

4. Finding Particular Solutions: Initial and Boundary Conditions

The general solution has unknown constants (\(A\) and \(B\)). If additional conditions are given, we can find the exact numerical values of these constants to get a particular solution.

Types of Conditions

• Initial Conditions: Values given at a single starting point, usually \(x = 0\) (e.g., \(y(0) = 2\) and \(y'(0) = -1\)).
• Boundary Conditions: Values of \(y\) given at two different points (e.g., \(y(0) = 1\) and \(y(\pi) = 0\)).

Crucial Tip for Finding Derivative Values

When given a condition for \(\frac{\mathrm{d}y}{\mathrm{d}x}\), you must differentiate the entire general solution (\(y_{CF} + y_{PI}\)), not just the complementary function!

Mini-Example

Using our previous general solution \(y = A\mathrm{e}^x + B\mathrm{e}^{2x} + 2x + 4\), suppose \(y(0) = 5\) and \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3\) when \(x = 0\):

1. Substitute \(x = 0, y = 5\):
\(5 = A\mathrm{e}^0 + B\mathrm{e}^0 + 0 + 4 \implies 5 = A + B + 4 \implies A + B = 1\)

2. Differentiate the general solution:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = A\mathrm{e}^x + 2B\mathrm{e}^{2x} + 2\)

3. Substitute \(x = 0, \frac{\mathrm{d}y}{\mathrm{d}x} = 3\):
\(3 = A(1) + 2B(1) + 2 \implies A + 2B = 1\)

4. Solve simultaneously:
Subtracting equations gives \(B = 0 \implies A = 1\).
Hence, the particular solution is \(y = \mathrm{e}^x + 2x + 4\).

5. Differential Equations Reducible by Substitution

Sometimes you will meet an equation that does not look linear or second-order with constant coefficients. In the exam, you are often guided to use a given substitution (such as \(y = vx\), \(z = y^2\), or \(z = y - x\)) to transform the equation into a solvable standard form.

Strategy for Substitutions

Step 1: Carefully differentiate the substitution using the chain rule or product rule to express derivatives like \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) in terms of the new variable.
Step 2: Substitute these expressions into the original differential equation.
Step 3: Simplify the new equation—it will reduce to a standard form (either integrating factor or constant coefficient 2nd order).
Step 4: Solve for the new variable.
Step 5: Transform back into the original variables (\(x\) and \(y\)).

Key Takeaway: Be meticulous with the Product Rule and Chain Rule when substituting derivatives. Once converted, follow the standard methods.

Quick Revision Checklist

Before sitting your exam, make sure you can confidently:

• Recognize and solve \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\) using \(I = \mathrm{e}^{\int P\,\mathrm{d}x}\).
• Form and solve the auxiliary equation \(am^2 + bm + c = 0\).
• Write down the complementary function for real distinct, real repeated, and complex conjugate roots.
• Select the correct trial function for the particular integral and resolve clashes by multiplying by \(x\).
• Apply initial and boundary conditions to determine arbitrary constants.
• Carry out transformations using given substitutions.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the standard form of a first-order linear differential equation, and what is its integrating factor?

The standard form is \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\), and the integrating factor is \(I(x) = \mathrm{e}^{\int P(x)\,\mathrm{d}x}\).

What is the auxiliary equation for the second-order homogeneous differential equation \(a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = 0\)?

The auxiliary equation is \(am^2 + bm + c = 0\).

What is the form of the Complementary Function (CF) when the auxiliary equation has two distinct real roots, \(m = \alpha\) and \(m = \beta\)?

The Complementary Function is \(y = A\mathrm{e}^{\alpha x} + B\mathrm{e}^{\beta x}\), where \(A\) and \(B\) are arbitrary constants.

What is the form of the Complementary Function (CF) when the auxiliary equation has a repeated real root, \(m = \alpha\)?

The Complementary Function is \(y = (A + Bx)\mathrm{e}^{\alpha x}\).

What is the form of the Complementary Function (CF) when the auxiliary equation has complex conjugate roots, \(m = p \pm \mathrm{i}q\)?

The Complementary Function is \(y = \mathrm{e}^{px}\big(A\cos(qx) + B\sin(qx)\big)\).

How is the General Solution of a non-homogeneous second-order differential equation constructed?

It is the sum of the Complementary Function and the Particular Integral: \(y = y_{CF} + y_{PI}\).

What trial form should be used for the Particular Integral (PI) if \(f(x) = k\cos(\omega x)\) or \(k\sin(\omega x)\)?

The trial form is \(y = \lambda\cos(\omega x) + \mu\sin(\omega x)\).

What is the clash rule when selecting a trial form for the Particular Integral (PI)?

If the standard trial form for \(y_{PI}\) already appears in \(y_{CF}\), multiply the trial form by \(x\) (or \(x^2\) if it corresponds to a repeated root).

When given a derivative boundary condition like \(\frac{\mathrm{d}y}{\mathrm{d}x} = v\) at \(x = x_0\), what function must be differentiated?

You must differentiate the complete general solution \(y = y_{CF} + y_{PI}\), not just the complementary function.

How are non-standard differential equations solved when a substitution is given?

Differentiate the substitution (using the chain or product rule) to express derivatives in the new variable, substitute into the ODE to obtain a standard form, solve it, and convert back.

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