What is algebra slips in the substitution?

Differentiating and substituting is error-prone; keep terms organised and multiply out carefully to reach the clean second-order equation.

The system x = -y, y = x reduces to which second-order equation in x? [1 mark]

SEAB Original-style practice: Solve the system dxdt = x + 2y, dydt = 3x + 2y, given x = 1 and y = 0 when t = 0.

Eliminate y. From the first equation, 2y = dxdt - x, so y = 12(dxdt - x). Differentiate: dydt = 12(d^2 xdt^2 - dxdt). Substitute this and y into the second equation dydt = 3x + 2y: $12(x - x) = 3x + 2·12(x - x) = 3x + x - x = 2x + x.$ Multiply by 2: x - x = 4x + 2x, so x - 3x - 4x = 0. Auxiliary equation λ^2 - 3λ - 4 = 0 (λ - 4)(λ + 1) = 0, so λ = 4, -1 and x = Ae^4t + Be^-t. Recover y = 12(x - x) = 12(4Ae^4t - Be^-t - Ae^4t - Be^-t) = 12(3Ae^4t - 2Be^-t). Apply x(0) = 1: A + B = 1. Apply y(0) = 0: 12(3A - 2B) = 0 3A = 2B. Solving: A = 25, B = 35. So x = 25e^4t + 35e^-t and y = 35e^4t - 35e^-t. Markers reward eliminating y, the second-order equation in x, its solution, recovering y, and applying both initial conditions.

SingaporeFurther MathsSyllabus dot point

How do we solve a coupled system of first-order linear differential equations by reducing it to a single second-order equation?

Solve coupled systems of first-order linear differential equations by reduction to a single second-order equation

A focused answer to the H2 Further Mathematics outcome on coupled linear systems. Eliminating one variable to obtain a single second-order equation, solving it, recovering the second variable, and applying initial conditions to both.

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
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Try this

What this dot point is asking

SEAB wants you to solve a coupled system of two first-order linear differential equations, in which the rates of change of $x$ and $y$ each depend on both variables. The standard method is to eliminate one variable to obtain a single second-order equation in the other, solve that with the auxiliary-equation method, then recover the eliminated variable and apply the initial conditions to both.

The answer

What a coupled system looks like

A linear system has the form

\frac{\mathrm{d}x}{\mathrm{d}t} = ax + by, \qquad \frac{\mathrm{d}y}{\mathrm{d}t} = cx + dy,

where each derivative depends on both $x$ and $y$ . You cannot solve either equation alone because they are intertwined.

Step one: express one variable from the other equation

From the first equation, solve for the variable you want to eliminate. For example, if $b \neq 0$ , rearrange $\dfrac{\mathrm{d}x}{\mathrm{d}t} = ax + by$ to give $y = \dfrac{1}{b}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t} - ax\right)$ .

Step two: differentiate and substitute

Differentiate that expression for $y$ to get $\dfrac{\mathrm{d}y}{\mathrm{d}t}$ in terms of $x$ , $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ and $\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2}$ . Substitute both $y$ and $\dfrac{\mathrm{d}y}{\mathrm{d}t}$ into the second equation. The result is a single second-order linear equation in $x$ alone.

Step three: solve and recover

Solve the second-order equation for $x$ using the auxiliary equation. Then substitute $x$ back into the expression for $y$ from step one to recover $y$ , without introducing new constants.

Step four: apply the initial conditions

Use the given initial values $x(0)$ and $y(0)$ to fix the two constants. Note that $y(0)$ is applied through the recovered expression for $y$ , which is why no separate constants appear there.

Worked example

Solve $\dfrac{\mathrm{d}x}{\mathrm{d}t} = 3x - 4y$ , $\dfrac{\mathrm{d}y}{\mathrm{d}t} = x - y$ , with $x(0) = 2$ , $y(0) = 1$ .

Step 1: Express y from the first equation

$4y = 3x - \dot{x}$ , so $y = \tfrac{1}{4}(3x - \dot{x})$ .

Step 2: Differentiate and substitute

$\dot{y} = \tfrac{1}{4}(3\dot{x} - \ddot{x})$ . Substitute $y$ and $\dot{y}$ into $\dot{y} = x - y$ :

\tfrac{1}{4}(3\dot{x} - \ddot{x}) = x - \tfrac{1}{4}(3x - \dot{x}).

Multiply by $4$ : $3\dot{x} - \ddot{x} = 4x - 3x + \dot{x} = x + \dot{x}$ , so $-\ddot{x} + 2\dot{x} - x = 0$ , that is $\ddot{x} - 2\dot{x} + x = 0$ .

Step 3: Solve the second-order equation

Auxiliary equation $\lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0$ , a repeated root $\lambda = 1$ , so

x = (A + Bt)\mathrm{e}^{t}.

Step 4: Recover y

$\dot{x} = (B)\mathrm{e}^t + (A + Bt)\mathrm{e}^t = (A + B + Bt)\mathrm{e}^t$ . Then

y = \tfrac{1}{4}(3x - \dot{x}) = \tfrac{1}{4}\big(3(A + Bt) - (A + B + Bt)\big)\mathrm{e}^t = \tfrac{1}{4}(2A - B + 2Bt)\mathrm{e}^t.

Step 5: Apply the initial conditions

$x(0) = A = 2$ . $y(0) = \tfrac{1}{4}(2A - B) = 1 \Rightarrow 2(2) - B = 4 \Rightarrow B = 0$ . So

x = 2\mathrm{e}^{t}, \qquad y = \mathrm{e}^{t}.

Examples in context

Example 1. Predator-prey dynamics. A linearised predator-prey model couples the rates of change of the two populations; eliminating one gives a second-order equation whose oscillatory solutions (complex auxiliary roots) reproduce the cyclic rise and fall of predator and prey numbers.

Example 2. Two coupled tanks. Brine flowing between two connected tanks gives a coupled system for the two salt masses. Reducing to a single second-order equation yields how each concentration evolves and the common steady state they approach, a standard compartmental model.

Try this

Q1. What is the first step in solving a coupled linear system by reduction? [1 mark]

Cue. Express one variable from one equation, then differentiate and substitute to eliminate it, leaving a single second-order equation.

Q2. When you recover the second variable, why are no new constants introduced? [2 marks]

Cue. It is found algebraically from the already-solved first variable, so the two constants are shared, not added to.

Q3. The system $\dot{x} = -y$ , $\dot{y} = x$ reduces to which second-order equation in $x$ ? [1 mark]

Cue. $\ddot{x} + x = 0$ (simple harmonic motion).

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.

Original8 marksSolve the system

\dfrac{dx}{dt} = x + 2y

\dfrac{dy}{dt} = 3x + 2y

, given

x = 1

and

y = 0

when

t = 0

Show worked answer →

Eliminate $y$ . From the first equation, $2y = \dfrac{dx}{dt} - x$ , so $y = \tfrac{1}{2}\left(\dfrac{dx}{dt} - x\right)$ .

Differentiate: $\dfrac{dy}{dt} = \tfrac{1}{2}\left(\dfrac{d^2 x}{dt^2} - \dfrac{dx}{dt}\right)$ . Substitute this and $y$ into the second equation $\dfrac{dy}{dt} = 3x + 2y$ :

\tfrac{1}{2}\left(\ddot{x} - \dot{x}\right) = 3x + 2\cdot\tfrac{1}{2}(\dot{x} - x) = 3x + \dot{x} - x = 2x + \dot{x}.

Multiply by

2

\ddot{x} - \dot{x} = 4x + 2\dot{x}

, so

\ddot{x} - 3\dot{x} - 4x = 0

Auxiliary equation $\lambda^2 - 3\lambda - 4 = 0 \Rightarrow (\lambda - 4)(\lambda + 1) = 0$ , so $\lambda = 4, -1$ and $x = A\mathrm{e}^{4t} + B\mathrm{e}^{-t}$ .

Recover $y = \tfrac{1}{2}(\dot{x} - x) = \tfrac{1}{2}\left(4A\mathrm{e}^{4t} - B\mathrm{e}^{-t} - A\mathrm{e}^{4t} - B\mathrm{e}^{-t}\right) = \tfrac{1}{2}(3A\mathrm{e}^{4t} - 2B\mathrm{e}^{-t})$ .

Apply $x(0) = 1$ : $A + B = 1$ . Apply $y(0) = 0$ : $\tfrac{1}{2}(3A - 2B) = 0 \Rightarrow 3A = 2B$ . Solving: $A = \tfrac{2}{5}$ , $B = \tfrac{3}{5}$ .

So $x = \tfrac{2}{5}\mathrm{e}^{4t} + \tfrac{3}{5}\mathrm{e}^{-t}$ and $y = \tfrac{3}{5}\mathrm{e}^{4t} - \tfrac{3}{5}\mathrm{e}^{-t}$ .

Markers reward eliminating $y$ , the second-order equation in $x$ , its solution, recovering $y$ , and applying both initial conditions.

Original6 marksBy forming a single second-order equation, find the general solution for

x

in the system

\dfrac{dx}{dt} = -y

\dfrac{dy}{dt} = x

Show worked answer →

Differentiate the first equation: $\dfrac{d^2 x}{dt^2} = -\dfrac{dy}{dt}$ . From the second equation $\dfrac{dy}{dt} = x$ , so

\frac{d^2 x}{dt^2} = -x, \quad \text{that is}\quad \ddot{x} + x = 0.

Auxiliary equation

\lambda^2 + 1 = 0

gives

\lambda = \pm i

, so the general solution is

x = A\cos t + B\sin t.

(The system describes circular motion in the $(x, y)$ plane; eliminating $y$ produces simple harmonic motion in $x$ .)

Markers reward differentiating to eliminate $y$ , the equation $\ddot{x} + x = 0$ , and the oscillatory general solution $x = A\cos t + B\sin t$ .