What is the auxiliary equation?

Trying a solution of the form y = e^λ x and substituting gives, after dividing by e^λ x, the auxiliary equation

What are wrong solution form for complex roots?

Complex roots α β i give e^α x(Acosβ x + Bsinβ x); do not leave the answer with e^(α + β i)x.

What is sign of the real part?

In e^α x, a negative α means decay; reading the sign wrongly inverts growth and decay.

Write the auxiliary equation of d^2 ydx^2 - dydx - 6y = 0. [1 mark]

SEAB Original-style practice: Solve d^2 ydx^2 + 4dydx + 13y = 0 given y = 0 and dydx = 6 when x = 0.

Auxiliary equation: λ^2 + 4λ + 13 = 0. The discriminant is 16 - 52 = -36 < 0, so the roots are complex: λ = -4 sqrt(-36)2 = -2 3i. For complex roots α β i, the general solution is y = e^α x(Acosβ x + Bsinβ x). Here α = -2, β = 3: $y = e^-2x(Acos 3x + Bsin 3x).$ Apply y = 0 at x = 0: 0 = 1·(A) A = 0. So y = Be^-2xsin 3x. Differentiate: dydx = B(-2e^-2xsin 3x + 3e^-2xcos 3x). At x = 0: y' = B(0 + 3) = 3B = 6, so B = 2. Hence y = 2e^-2xsin 3x. Markers reward the auxiliary equation, the complex roots -2 3i, the correct e^α x(Acosβ x + Bsinβ x) form, and applying both conditions to get y = 2e^-2xsin 3x.

SingaporeFurther MathsSyllabus dot point

How do we solve a homogeneous second-order linear differential equation with constant coefficients using the auxiliary equation?

Solve homogeneous second-order linear differential equations with constant coefficients using the auxiliary equation, covering real, repeated and complex roots

A focused answer to the H2 Further Mathematics outcome on homogeneous second-order linear ODEs. The auxiliary equation, the three cases of distinct real, repeated and complex roots, and applying two initial conditions to fix the constants.

Generated by Claude Opus 4.811 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

SEAB wants you to solve a homogeneous second-order linear differential equation with constant coefficients,

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

by forming the auxiliary (characteristic) equation, identifying which of three cases its roots fall into, writing the matching general solution, and using two initial conditions to fix the two constants.

The answer

The auxiliary equation

Trying a solution of the form $y = \mathrm{e}^{\lambda x}$ and substituting gives, after dividing by $\mathrm{e}^{\lambda x}$ , the auxiliary equation

a\lambda^2 + b\lambda + c = 0.

Its roots determine the form of the general solution, exactly as the characteristic equation does for second-order recurrence relations.

The three cases

Distinct real roots $\lambda_1 \neq \lambda_2$ : $\;y = A\mathrm{e}^{\lambda_1 x} + B\mathrm{e}^{\lambda_2 x}$ .
Repeated real root $\lambda$ : $\;y = (A + Bx)\mathrm{e}^{\lambda x}$ .
Complex roots $\lambda = \alpha \pm \beta i$ : $\;y = \mathrm{e}^{\alpha x}\big(A\cos\beta x + B\sin\beta x\big)$ .

In each case there are two arbitrary constants $A$ and $B$ .

Why complex roots give oscillation

When the discriminant $b^2 - 4ac < 0$ , the roots are $\alpha \pm \beta i$ . The real part $\alpha$ controls growth or decay through $\mathrm{e}^{\alpha x}$ , and the imaginary part $\beta$ sets the oscillation frequency through $\cos\beta x$ and $\sin\beta x$ . A negative $\alpha$ gives damped oscillation.

Applying initial conditions

A second-order equation needs two conditions, typically a value of $y$ and of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ at the same point. Differentiate the general solution, substitute both conditions, and solve the resulting simultaneous equations for $A$ and $B$ .

Worked example

Solve $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 6\dfrac{\mathrm{d}y}{\mathrm{d}x} + 9y = 0$ given $y = 1$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$ when $x = 0$ .

Step 1: Write the auxiliary equation

\lambda^2 + 6\lambda + 9 = 0.

Step 2: Solve and identify the case

$(\lambda + 3)^2 = 0$ , so $\lambda = -3$ is a repeated root.

Step 3: Write the matching general solution

For a repeated root,

y = (A + Bx)\mathrm{e}^{-3x}.

Step 4: Differentiate

\frac{\mathrm{d}y}{\mathrm{d}x} = B\mathrm{e}^{-3x} + (A + Bx)(-3)\mathrm{e}^{-3x} = \big(B - 3A - 3Bx\big)\mathrm{e}^{-3x}.

Step 5: Apply the two conditions

At $x = 0$ : $y = A = 1$ . And $y' = B - 3A = 0 \Rightarrow B = 3A = 3$ . So

y = (1 + 3x)\mathrm{e}^{-3x}.

Common traps

Missing the $x$ factor for a repeated root: A repeated root gives $(A + Bx)\mathrm{e}^{\lambda x}$ , not $A\mathrm{e}^{\lambda x}$ ; without $Bx$ you have only one constant.
Wrong solution form for complex roots: Complex roots $\alpha \pm \beta i$ give $\mathrm{e}^{\alpha x}(A\cos\beta x + B\sin\beta x)$ ; do not leave the answer with $\mathrm{e}^{(\alpha + \beta i)x}$ .
Using one condition: Two constants need two conditions; one value cannot determine both $A$ and $B$ .
Differentiating carelessly: Applying the second condition requires the derivative of the full general solution, including the product rule for the repeated-root and complex cases.
Sign of the real part: In $\mathrm{e}^{\alpha x}$ , a negative $\alpha$ means decay; reading the sign wrongly inverts growth and decay.

Examples in context

Example 1. Damped mechanical oscillation. A mass on a spring with damping obeys $m\ddot{x} + c\dot{x} + kx = 0$ . The auxiliary equation's roots distinguish overdamped (distinct real), critically damped (repeated) and underdamped (complex, oscillatory) motion, the central classification of vibration analysis.

Example 2. An LC and RLC circuit. The charge in an electrical circuit with inductance, resistance and capacitance satisfies the same form of equation. Complex auxiliary roots give the oscillating current of an underdamped RLC circuit, directly paralleling the mechanical case.

Try this

Q1. Write the auxiliary equation of $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} - \dfrac{\mathrm{d}y}{\mathrm{d}x} - 6y = 0$ . [1 mark]

Cue. $\lambda^2 - \lambda - 6 = 0$ , with roots $\lambda = 3$ and $\lambda = -2$ .

Q2. State the general solution form for complex auxiliary roots $-1 \pm 2i$ . [2 marks]

Cue. $y = \mathrm{e}^{-x}(A\cos 2x + B\sin 2x)$ .

Q3. What solution form corresponds to a repeated auxiliary root $\lambda = 4$ ? [1 mark]

Cue. $y = (A + Bx)\mathrm{e}^{4x}$ .

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.

Original6 marksSolve

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

given

y = 2

and

\dfrac{dy}{dx} = 5

when

x = 0

Show worked answer →

The auxiliary equation (try $y = \mathrm{e}^{\lambda x}$ ) is $\lambda^2 - 5\lambda + 6 = 0$ , factoring as $(\lambda - 2)(\lambda - 3) = 0$ , so $\lambda = 2$ or $\lambda = 3$ (distinct real roots).

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

Then $\dfrac{dy}{dx} = 2A\mathrm{e}^{2x} + 3B\mathrm{e}^{3x}$ .

Apply the conditions at $x = 0$ : $y = A + B = 2$ and $y' = 2A + 3B = 5$ . Subtracting twice the first from the second: $B = 1$ , hence $A = 1$ .

So $y = \mathrm{e}^{2x} + \mathrm{e}^{3x}$ .

Markers reward the auxiliary equation, its roots, the general solution form, and solving the simultaneous equations for $A = 1$ , $B = 1$ .

Original7 marksSolve

\dfrac{d^2 y}{dx^2} + 4\dfrac{dy}{dx} + 13y = 0

given

y = 0

and

\dfrac{dy}{dx} = 6

when

x = 0

Show worked answer →

Auxiliary equation: $\lambda^2 + 4\lambda + 13 = 0$ . The discriminant is $16 - 52 = -36 < 0$ , so the roots are complex: $\lambda = \dfrac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i$ .

For complex roots $\alpha \pm \beta i$ , the general solution is $y = \mathrm{e}^{\alpha x}(A\cos\beta x + B\sin\beta x)$ . Here $\alpha = -2$ , $\beta = 3$ :

y = \mathrm{e}^{-2x}(A\cos 3x + B\sin 3x).

Apply

y = 0

x = 0

0 = 1\cdot(A) \Rightarrow A = 0

. So

y = B\mathrm{e}^{-2x}\sin 3x

Differentiate: $\dfrac{dy}{dx} = B\left(-2\mathrm{e}^{-2x}\sin 3x + 3\mathrm{e}^{-2x}\cos 3x\right)$ . At $x = 0$ : $y' = B(0 + 3) = 3B = 6$ , so $B = 2$ .

Hence $y = 2\mathrm{e}^{-2x}\sin 3x$ .

Markers reward the auxiliary equation, the complex roots $-2 \pm 3i$ , the correct $\mathrm{e}^{\alpha x}(A\cos\beta x + B\sin\beta x)$ form, and applying both conditions to get $y = 2\mathrm{e}^{-2x}\sin 3x$ .