What is not in standard form before μ?

The integrating factor uses the coefficient of y when the coefficient of dydx is 1; divide through first.

State the integrating factor for dydx + 3y = x. [1 mark]

Separate the variables in dydx = xy. [1 mark]

SEAB Original-style practice: Solve dydx + 2y = e^-x given y = 0 when x = 0, using an integrating factor.

The equation is linear in the form dydx + P(x)y = Q(x) with P = 2. The integrating factor is μ = e^int P\,dx = e^2x. Multiply through: e^2xdydx + 2e^2xy = e^2xe^-x = e^x. The left side is ddx(e^2xy), so $(d)/(dx)(e^2xy) = e^x.$ Integrate: e^2xy = e^x + C, so y = e^-x + Ce^-2x. Apply y = 0 at x = 0: 0 = 1 + C, so C = -1. Hence y = e^-x - e^-2x. Markers reward identifying the linear form, the integrating factor e^2x, recognising the left side as a derivative of a product, integrating, and applying the condition to get y = e^-x - e^-2x.

SingaporeFurther MathsSyllabus dot point

How do we solve first-order differential equations by separating variables and by the integrating factor method?

Solve first-order differential equations by separation of variables and by the integrating factor method, applying initial conditions

A focused answer to the H2 Further Mathematics outcome on first-order differential equations. Separation of variables, the integrating factor method for linear equations, general and particular solutions, and applying initial conditions.

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
The answer
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Try this

What this dot point is asking

SEAB wants you to solve first-order differential equations by two methods: separation of variables, for equations where the variables can be split onto opposite sides, and the integrating factor method, for linear equations of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$ . You must produce a general solution and then a particular solution by applying an initial condition.

The answer

Separation of variables

If a first-order equation can be written so that all the $y$ (and $\mathrm{d}y$ ) terms are on one side and all the $x$ (and $\mathrm{d}x$ ) terms on the other,

\mathrm{g}(y)\,\mathrm{d}y = \mathrm{h}(x)\,\mathrm{d}x,

then integrate both sides. The single constant of integration (combine the two into one) is fixed later by the initial condition.

Recognising a separable equation

An equation is separable when $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ equals a product (or quotient) of a function of $x$ and a function of $y$ . If $x$ and $y$ are tangled additively (for example $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x + y$ ), separation fails and the integrating factor method is the tool.

The integrating factor method

For a linear first-order equation

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

multiply through by the integrating factor

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

The left side then collapses to the derivative of a product: $\dfrac{\mathrm{d}}{\mathrm{d}x}\big(\mu y\big) = \mu Q$ . Integrate both sides and divide by $\mu$ to find $y$ .

General and particular solutions

Both methods give a general solution containing one arbitrary constant. Substituting the initial condition (a value of $y$ at a given $x$ ) fixes the constant, producing the particular solution that the question asks for.

Worked example

Solve $x\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = x^2$ for $x > 0$ , given $y = 2$ when $x = 1$ .

Step 1: Put it in standard linear form

Divide by $x$ : $\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{1}{x}y = x$ . So $P = \dfrac{1}{x}$ and $Q = x$ .

Step 2: Find the integrating factor

\mu = \mathrm{e}^{\int (1/x)\,\mathrm{d}x} = \mathrm{e}^{\ln x} = x.

Step 3: Multiply through and recognise the product derivative

x\frac{\mathrm{d}y}{\mathrm{d}x} + y = x^2 \;\Rightarrow\; \frac{\mathrm{d}}{\mathrm{d}x}(xy) = x^2.

Step 4: Integrate

xy = \int x^2\,\mathrm{d}x = \frac{x^3}{3} + C \;\Rightarrow\; y = \frac{x^2}{3} + \frac{C}{x}.

Step 5: Apply the initial condition

$y = 2$ at $x = 1$ : $2 = \dfrac{1}{3} + C$ , so $C = \dfrac{5}{3}$ . The particular solution is

y = \frac{x^2}{3} + \frac{5}{3x}.

Examples in context

Example 1. Newton's law of cooling. The temperature of a cooling body obeys $\dfrac{\mathrm{d}T}{\mathrm{d}t} = -k(T - T_s)$ , a separable equation whose solution $T = T_s + (T_0 - T_s)\mathrm{e}^{-kt}$ is the exponential approach to the surroundings, the workhorse of cooling and heating models.

Example 2. A mixing tank. The mass of dissolved substance in a tank with inflow and outflow satisfies a linear equation $\dfrac{\mathrm{d}m}{\mathrm{d}t} + \dfrac{r}{V}m = c$ , solved by an integrating factor, which is why the integrating factor method appears throughout chemical and environmental modelling.

Try this

Q1. State the integrating factor for $\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y = x$ . [1 mark]

Cue. $\mu = \mathrm{e}^{\int 3\,\mathrm{d}x} = \mathrm{e}^{3x}$ .

Q2. Separate the variables in $\dfrac{\mathrm{d}y}{\mathrm{d}x} = xy$ . [1 mark]

Cue. $\dfrac{1}{y}\,\mathrm{d}y = x\,\mathrm{d}x$ , integrating to $\ln|y| = \tfrac{1}{2}x^2 + C$ .

Q3. Why does the integrating factor method work after multiplying through? [2 marks]

Cue. The left side becomes $\dfrac{\mathrm{d}}{\mathrm{d}x}(\mu y)$ by the product rule, so the equation can be integrated directly.

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 the differential equation

\dfrac{dy}{dx} = \dfrac{x(1 + y^2)}{y}

, given that

y = 1

when

x = 0

Show worked answer →

This is separable. Gather $y$ on the left and $x$ on the right:

\frac{y}{1 + y^2}\,dy = x\,dx.

Integrate both sides. On the left, the numerator is half the derivative of

1 + y^2

\int \frac{y}{1 + y^2}\,dy = \tfrac{1}{2}\ln(1 + y^2), \qquad \int x\,dx = \tfrac{1}{2}x^2.

\tfrac{1}{2}\ln(1 + y^2) = \tfrac{1}{2}x^2 + C_1

, that is

\ln(1 + y^2) = x^2 + C

Apply $y = 1$ at $x = 0$ : $\ln 2 = 0 + C$ , so $C = \ln 2$ . Hence $\ln(1 + y^2) = x^2 + \ln 2$ , giving $1 + y^2 = 2\mathrm{e}^{x^2}$ and $y = \sqrt{2\mathrm{e}^{x^2} - 1}$ (positive root, since $y = 1 > 0$ ).

Markers reward separating correctly, both integrals (recognising the $\ln(1 + y^2)$ form), applying the initial condition, and an explicit $y$ .

Original7 marksSolve

\dfrac{dy}{dx} + 2y = \mathrm{e}^{-x}

given

y = 0

when

x = 0

, using an integrating factor.

Show worked answer →

The equation is linear in the form $\dfrac{dy}{dx} + P(x)y = Q(x)$ with $P = 2$ . The integrating factor is $\mu = \mathrm{e}^{\int P\,dx} = \mathrm{e}^{2x}$ .

Multiply through: $\mathrm{e}^{2x}\dfrac{dy}{dx} + 2\mathrm{e}^{2x}y = \mathrm{e}^{2x}\mathrm{e}^{-x} = \mathrm{e}^{x}$ . The left side is $\dfrac{d}{dx}\left(\mathrm{e}^{2x}y\right)$ , so

\frac{d}{dx}\left(\mathrm{e}^{2x}y\right) = \mathrm{e}^{x}.

Integrate:

\mathrm{e}^{2x}y = \mathrm{e}^{x} + C

, so

y = \mathrm{e}^{-x} + C\mathrm{e}^{-2x}

Apply $y = 0$ at $x = 0$ : $0 = 1 + C$ , so $C = -1$ . Hence $y = \mathrm{e}^{-x} - \mathrm{e}^{-2x}$ .

Markers reward identifying the linear form, the integrating factor $\mathrm{e}^{2x}$ , recognising the left side as a derivative of a product, integrating, and applying the condition to get $y = \mathrm{e}^{-x} - \mathrm{e}^{-2x}$ .