What is trial clashing with y_c unnoticed?

If the trial appears in the complementary function, multiply by x; otherwise the substitution collapses to 0 = f(x).

What is wrong polynomial degree?

Match the trial polynomial's degree to the forcing; a constant forcing still needs a constant trial, but a clash with a zero root may force an extra x.

What trial particular integral suits forcing f(x) = 5e^3x (not in the complementary function)? [1 mark]

SEAB Original-style practice: Find the general solution of d^2 ydx^2 - 3dydx + 2y = 4x.

Complementary function: solve the homogeneous equation. Auxiliary equation λ^2 - 3λ + 2 = 0 gives (λ - 1)(λ - 2) = 0, so λ = 1, 2 and y_c = Ae^x + Be^2x. Particular integral: the forcing 4x is a polynomial of degree 1, so try y_p = ax + b. Then y_p' = a, y_p'' = 0. Substitute: $0 - 3a + 2(ax + b) = 4x 2ax + (2b - 3a) = 4x.$ Match coefficients: 2a = 4 a = 2; 2b - 3a = 0 2b = 6 b = 3. So y_p = 2x + 3. General solution: y = Ae^x + Be^2x + 2x + 3. Markers reward the complementary function, the polynomial trial form, matching coefficients to find a = 2, b = 3, and combining into the general solution.

SingaporeFurther MathsSyllabus dot point

How do we solve a non-homogeneous second-order linear differential equation using a complementary function and a particular integral?

Solve non-homogeneous second-order linear differential equations by finding the complementary function and a particular integral for standard forcing terms

A focused answer to the H2 Further Mathematics outcome on non-homogeneous second-order ODEs. The complementary function plus particular integral structure, trial forms for polynomial, exponential and trigonometric forcing, the breakdown case, and fitting initial conditions last.

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

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

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What this dot point is asking
The answer
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What this dot point is asking

SEAB wants you to solve a non-homogeneous second-order linear equation

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

by combining the complementary function (the general solution of the homogeneous equation) with a particular integral (any one solution of the full equation). You choose the particular integral by a trial form matched to the forcing term $\mathrm{f}(x)$ , and apply initial conditions only at the very end.

The answer

The structure of the general solution

The general solution is

y = y_c + y_p,

where $y_c$ is the complementary function (containing the two arbitrary constants) and $y_p$ is any particular integral. This works because the equation is linear: adding a solution of the homogeneous equation to a particular solution still satisfies the full equation.

Step one: the complementary function

Solve $a y'' + b y' + c y = 0$ via the auxiliary equation, using the three cases (distinct real, repeated, complex). This supplies $y_c$ and the constants $A$ and $B$ .

Step two: the particular integral by trial

Choose a trial $y_p$ matching the form of $\mathrm{f}(x)$ , then substitute and match coefficients:

polynomial of degree $n$ : try a general polynomial of degree $n$ ;
$\mathrm{e}^{kx}$ : try $a\mathrm{e}^{kx}$ ;
$\cos\omega x$ or $\sin\omega x$ : try $a\cos\omega x + b\sin\omega x$ (include both even if only one appears).

The breakdown (resonance) case

If the trial form already appears in the complementary function, substituting it gives $0 = \mathrm{f}(x)$ , which is impossible. Fix this by multiplying the trial by $x$ (or $x^2$ if it still clashes with a repeated root). Physically this is resonance: the forcing matches a natural mode, and the response grows like $x$ .

Step three: apply initial conditions last

Combine $y = y_c + y_p$ first, then differentiate and substitute the initial conditions to fix $A$ and $B$ . Fitting the constants to $y_c$ alone, before adding $y_p$ , is wrong.

Worked example

Find the general solution of $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} - 4y = \mathrm{e}^{x}$ .

Step 1: Complementary function

Auxiliary equation $\lambda^2 - 4 = 0$ gives $\lambda = \pm 2$ , so

y_c = A\mathrm{e}^{2x} + B\mathrm{e}^{-2x}.

Step 2: Choose the trial particular integral

The forcing is $\mathrm{e}^{x}$ , which is not in $y_c$ (the exponents there are $\pm 2$ , not $1$ ). Try $y_p = a\mathrm{e}^{x}$ .

Step 3: Substitute

$y_p'' = a\mathrm{e}^{x}$ , so $y_p'' - 4y_p = a\mathrm{e}^{x} - 4a\mathrm{e}^{x} = -3a\mathrm{e}^{x}$ .

Step 4: Match coefficients

Set $-3a\mathrm{e}^{x} = \mathrm{e}^{x}$ , so $-3a = 1$ , giving $a = -\dfrac{1}{3}$ . Thus $y_p = -\dfrac{1}{3}\mathrm{e}^{x}$ .

Step 5: Combine

y = A\mathrm{e}^{2x} + B\mathrm{e}^{-2x} - \frac{1}{3}\mathrm{e}^{x}.

Examples in context

Example 1. Driven oscillator. A forced spring-mass system $\ddot{x} + \omega_0^2 x = F\cos\omega t$ has a sinusoidal particular integral giving the steady-state response, while resonance (when $\omega = \omega_0$ ) forces the $x\sin\omega t$ growth, the mathematical signature of a system driven at its natural frequency.

Example 2. Circuit with a source. An RLC circuit driven by a constant or sinusoidal voltage source is modelled by a non-homogeneous equation; the complementary function is the transient that decays away and the particular integral is the steady-state current that persists, the standard transient-plus-steady-state decomposition.

Try this

Q1. State the structure of the general solution of a non-homogeneous linear ODE. [1 mark]

Cue. $y = y_c + y_p$ : complementary function plus a particular integral.

Q2. What trial particular integral suits forcing $\mathrm{f}(x) = 5\mathrm{e}^{3x}$ (not in the complementary function)? [1 mark]

Cue. $y_p = a\mathrm{e}^{3x}$ .

Q3. Why must the trial be multiplied by $x$ when it matches the complementary function? [2 marks]

Cue. Otherwise it solves the homogeneous equation and substitution gives $0 = \mathrm{f}(x)$ ; the extra $x$ produces an independent form (resonance).

Exam-style practice questions

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

Original7 marksFind the general solution of

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

Show worked answer →

Complementary function: solve the homogeneous equation. Auxiliary equation $\lambda^2 - 3\lambda + 2 = 0$ gives $(\lambda - 1)(\lambda - 2) = 0$ , so $\lambda = 1, 2$ and $y_c = A\mathrm{e}^{x} + B\mathrm{e}^{2x}$ .

Particular integral: the forcing $4x$ is a polynomial of degree $1$ , so try $y_p = ax + b$ . Then $y_p' = a$ , $y_p'' = 0$ . Substitute:

0 - 3a + 2(ax + b) = 4x \Rightarrow 2ax + (2b - 3a) = 4x.

Match coefficients:

2a = 4 \Rightarrow a = 2

;

2b - 3a = 0 \Rightarrow 2b = 6 \Rightarrow b = 3

. So

y_p = 2x + 3

General solution: $y = A\mathrm{e}^{x} + B\mathrm{e}^{2x} + 2x + 3$ .

Markers reward the complementary function, the polynomial trial form, matching coefficients to find $a = 2$ , $b = 3$ , and combining into the general solution.

Original7 marksFind the general solution of

\dfrac{d^2 y}{dx^2} + y = 2\cos x

, noting the resonance between the forcing and the complementary function.

Show worked answer →

Complementary function: $\lambda^2 + 1 = 0$ gives $\lambda = \pm i$ , so $y_c = A\cos x + B\sin x$ .

The forcing $2\cos x$ already appears in the complementary function (resonance), so the usual trial $a\cos x + b\sin x$ would be absorbed and fail. Multiply the trial by $x$ : try $y_p = x(a\cos x + b\sin x)$ .

Differentiate twice. $y_p' = a\cos x + b\sin x + x(-a\sin x + b\cos x)$ , and

y_p'' = 2(-a\sin x + b\cos x) + x(-a\cos x - b\sin x).

Then

y_p'' + y_p = 2(-a\sin x + b\cos x)

(the

x

terms cancel). Set equal to

2\cos x

-2a = 0 \Rightarrow a = 0

and

2b = 2 \Rightarrow b = 1