What is trial integral clashing with the complementary function?

If your trial y_p already appears in y_c, multiply it by x before substituting, or it cannot work.

Find the auxiliary equation of d^2ydx^2 + 2dydx + y = 0 and its root. [2 marks]

SQA AH style: complex roots-style practice: Find the general solution of d^2ydx^2 + 4y = 0.

Auxiliary equation: m^2 + 4 = 0, so m^2 = -4 and m = 2i (1 mark). Complex roots m = α β i with α = 0, β = 2 (1 mark). The general solution is y = e^α x(Acosβ x + Bsinβ x) (1 mark). With α = 0, β = 2: y = Acos 2x + Bsin 2x (1 mark). Markers reward the auxiliary equation, identifying complex roots, the correct form, and the final solution.

ScotlandMathsSyllabus dot point

How do you solve a linear second-order differential equation with constant coefficients?

Solve homogeneous and non-homogeneous second-order linear differential equations with constant coefficients using the auxiliary equation, the complementary function and a particular integral, covering distinct real, equal and complex roots.

A focused answer to the SQA Advanced Higher Mathematics second-order differential equations content, covering the auxiliary equation, the three cases of distinct real, equal and complex roots, the complementary function, finding a particular integral for non-homogeneous equations, and the general solution.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The auxiliary equation and the three cases
Non-homogeneous equations
Boundary conditions
Why the three cases look so different
Try this

What this dot point is asking

The SQA wants you to solve linear second-order differential equations with constant coefficients of the form $a\dfrac{d^2y}{dx^2} + b\dfrac{dy}{dx} + cy = f(x)$ . You build the solution from the auxiliary equation, handling distinct real, equal and complex roots, and add a particular integral when the right-hand side is non-zero.

The auxiliary equation and the three cases

For the homogeneous equation $a\dfrac{d^2y}{dx^2} + b\dfrac{dy}{dx} + cy = 0$ , trying $y = e^{mx}$ leads to $am^2 + bm + c = 0$ . The nature of the roots, decided by the discriminant, gives three forms for the solution.

Solve

\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0

For a homogeneous equation with constant coefficients, we try the solution $y = e^{mx}$ . Substituting this into the equation converts it into an algebraic equation in $m$ called the auxiliary equation, whose roots determine the form of the general solution.

Step 1: Form the auxiliary equation

Replace each derivative with the corresponding power of $m$ : $\dfrac{d^2y}{dx^2}$ becomes $m^2$ , $\dfrac{dy}{dx}$ becomes $m$ , and $y$ becomes $1$ .

m^2 - 5m + 6 = 0

Step 2: Solve for the roots

Factorise the quadratic to find the two values of $m$ .

(m - 2)(m - 3) = 0 \implies m = 2 \text{ or } m = 3

These are two distinct real roots, so we use the general solution form $y = Ae^{m_1 x} + Be^{m_2 x}$ .

Step 3: Write the general solution

Substitute $m_1 = 2$ and $m_2 = 3$ into the formula. The two arbitrary constants $A$ and $B$ are determined by boundary conditions.

Final answer: $y = Ae^{2x} + Be^{3x}$ .

Solve

\dfrac{d^2y}{dx^2} + 6\dfrac{dy}{dx} + 9y = 0

When the discriminant of the auxiliary equation is zero, both roots are equal. In this case $e^{mx}$ and $xe^{mx}$ are the two independent solutions, so the form of the general solution changes from the distinct-roots case.

Step 1: Write and factorise the auxiliary equation

Replace derivatives with powers of $m$ to form the quadratic.

m^2 + 6m + 9 = 0 \implies (m + 3)^2 = 0

Step 2: Identify the repeated root

The only solution is $m = -3$ , occurring twice. This is the equal-roots case, so we cannot simply write $Ae^{-3x} + Be^{-3x}$ (that collapses to a single constant). Instead, the second independent solution is $xe^{-3x}$ .

Step 3: Write the general solution using the repeated-root form

For an equal root $m$ , the general solution is $y = (A + Bx)e^{mx}$ .

Final answer: $y = (A + Bx)e^{-3x}$ .

Non-homogeneous equations

When the right-hand side $f(x)$ is not zero, the general solution is the complementary function plus a particular integral $y_p$ : a single function that satisfies the full equation. You find $y_p$ by trying a form that matches $f(x)$ (a polynomial for a polynomial, $pe^{kx}$ for an exponential, $p\cos kx + q\sin kx$ for a trigonometric term) and solving for its coefficients.

Solve

\dfrac{d^2y}{dx^2} - y = 3e^{2x}

Because the right-hand side is non-zero, the general solution is the complementary function (solving the homogeneous equation) plus a particular integral (a single solution of the full equation). We find each part in turn.

Step 1: Find the complementary function

Solve the associated homogeneous equation $\dfrac{d^2y}{dx^2} - y = 0$ by forming the auxiliary equation and finding its roots.

m^2 - 1 = 0 \implies (m-1)(m+1) = 0 \implies m = 1 \text{ or } m = -1

Two distinct real roots give $y_c = Ae^{x} + Be^{-x}$ .

Step 2: Choose and substitute a trial particular integral

The right-hand side $3e^{2x}$ is an exponential with exponent $2x$ , which does not appear in $y_c$ , so we try $y_p = pe^{2x}$ . Differentiate twice: $y_p' = 2pe^{2x}$ and $y_p'' = 4pe^{2x}$ .

Substitute into the original equation to find $p$ :

4pe^{2x} - pe^{2x} = 3e^{2x} \implies 3p = 3 \implies p = 1

So $y_p = e^{2x}$ .

Step 3: Combine to form the general solution

The general solution is the sum of the complementary function and the particular integral.

Final answer: $y = Ae^{x} + Be^{-x} + e^{2x}$ .

Boundary conditions

The general solution of a second-order equation contains two arbitrary constants, so you need two pieces of information to find a particular solution: typically the value of $y$ and of $\dfrac{dy}{dx}$ at a point, or the value of $y$ at two points. Apply the conditions to the full general solution $y = y_c + y_p$ , not just to the complementary function.

Why the three cases look so different

The three forms all come from the same idea, that solutions are built from exponentials $e^{mx}$ . With two distinct real roots you simply have two independent exponentials. With a repeated root there is only one exponential, so a second independent solution is manufactured by multiplying by $x$ , giving the $(A + Bx)e^{mx}$ shape. With complex roots $\alpha \pm \beta i$ , the exponentials $e^{(\alpha \pm \beta i)x}$ are rewritten with Euler's relation into the real combination $e^{\alpha x}(A\cos\beta x + B\sin\beta x)$ , where $\alpha$ controls growth or decay and $\beta$ controls the oscillation. This is why oscillating physical systems, such as a mass on a spring with light damping, lead naturally to the complex-root case.

Try this

Q1. Write the general solution form if the auxiliary equation has roots $m = -1 \pm 3i$ . [2 marks]

Cue. $\alpha = -1$ , $\beta = 3$ , so $y = e^{-x}(A\cos 3x + B\sin 3x)$ .

Q2. Find the auxiliary equation of $\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} + y = 0$ and its root. [2 marks]

Cue. $m^2 + 2m + 1 = (m + 1)^2 = 0$ , equal root $m = -1$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: complex roots4 marksFind the general solution of

\dfrac{d^2y}{dx^2} + 4y = 0

Show worked answer →

Auxiliary equation: $m^2 + 4 = 0$ , so $m^2 = -4$ and $m = \pm 2i$ (1 mark).

Complex roots $m = \alpha \pm \beta i$ with $\alpha = 0$ , $\beta = 2$ (1 mark).

The general solution is $y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)$ (1 mark).

With $\alpha = 0$ , $\beta = 2$ : $y = A\cos 2x + B\sin 2x$ (1 mark). Markers reward the auxiliary equation, identifying complex roots, the correct form, and the final solution.

AH style: non-homogeneous6 marksFind the general solution of

\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + 2y = 4x

Show worked answer →

Auxiliary equation: $m^2 - 3m + 2 = 0$ , so $(m - 1)(m - 2) = 0$ and $m = 1, 2$ (1 mark). Complementary function $y_c = Ae^{x} + Be^{2x}$ (1 mark).

Try particular integral $y_p = ax + b$ , so $y_p' = a$ , $y_p'' = 0$ (1 mark).

Substitute: $0 - 3a + 2(ax + b) = 4x$ , giving $2a = 4$ so $a = 2$ , and $-3a + 2b = 0$ so $b = 3$ (2 marks).

General solution $y = Ae^{x} + Be^{2x} + 2x + 3$ (1 mark). Markers reward the complementary function, the trial form, solving for the coefficients, and combining into the general solution.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)