What is a wrong integrating factor?

I = e^int P\,dx, and you must put the equation in the standard form dydx + P(x)y = Q(x) first, so the coefficient of dydx is 1.

Find the integrating factor for dydx + 1xy = 1, x > 0. [2 marks]

SQA AH style: separable-style practice: Solve dydx = xy given that y = 2 when x = 0.

Separate the variables: 1y\,dy = x\,dx (1 mark). Integrate both sides: ln|y| = 12x^2 + C (1 mark). Apply the condition y = 2 at x = 0: ln 2 = C (1 mark). So ln y = 12x^2 + ln 2, giving y = 2e^x^2/2 (1 mark). Markers reward separation, integration of both sides, finding C, and the explicit particular solution.

SQA AH style: integrating factor-style practice: Solve dydx + 2xy = x, for x > 0.

This is linear with P(x) = 2x. Integrating factor I = e^int (2)/(x)\,dx = e^2ln x = x^2 (2 marks). Multiply through: x^2dydx + 2xy = x^3, so ddx(x^2 y) = x^3 (1 mark). Integrate: x^2 y = 14x^4 + C (1 mark). Divide by x^2: y = 14x^2 + Cx^2 (1 mark). Markers reward the integrating factor, the product-rule recognition, the integration, and solving for y.

ScotlandMathsSyllabus dot point

How do you solve a first-order differential equation that is separable or linear?

Solve first-order differential equations by separating the variables and by the integrating-factor method for linear equations, find particular solutions from initial conditions, and apply differential equations to growth and decay models.

A focused answer to the SQA Advanced Higher Mathematics first-order differential equations content, covering separable equations, the integrating-factor method for linear first-order equations, finding particular solutions from initial conditions, and applications to growth and decay.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-14

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

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What this dot point is asking
Separable equations
Linear equations and the integrating factor
Particular solutions and modelling
Choosing between the two methods
Try this

What this dot point is asking

The SQA wants you to solve first-order differential equations of two kinds: those you can solve by separating the variables, and linear equations you solve with an integrating factor. You also need to find a particular solution from an initial condition and apply the method to real growth and decay problems.

Separable equations

An equation is separable when you can get all the $y$ terms (with $dy$ ) on one side and all the $x$ terms (with $dx$ ) on the other. Once separated, you integrate each side independently and add a single constant.

Solve

\dfrac{dy}{dx} = \dfrac{x}{y}

The right-hand side is a product of a pure $x$ -function ( $x$ ) and a pure $y$ -function ( $\dfrac{1}{y}$ ), so the equation is separable. We rearrange so that all $y$ terms (with $dy$ ) sit on one side and all $x$ terms (with $dx$ ) on the other, then integrate each side independently.

Step 1: Separate the variables

Multiply both sides by $y$ and by $dx$ to move everything into the right position.

y\,dy = x\,dx

Step 2: Integrate both sides

Integrate the left side with respect to $y$ and the right side with respect to $x$ . Add only one constant of integration $C$ on one side; it absorbs any constants from both integrals.

\int y\,dy = \int x\,dx \implies \dfrac{1}{2}y^2 = \dfrac{1}{2}x^2 + C

Step 3: Write the general solution in a tidy form

Multiply through by $2$ and rename $2C$ as $A$ (since $A$ is still an arbitrary constant) to remove the fractions.

y^2 = x^2 + A

Final answer: $y = \pm\sqrt{x^2 + A}$ , where $A$ is an arbitrary constant.

Linear equations and the integrating factor

A first-order equation is linear when it is written as $\dfrac{dy}{dx} + P(x)y = Q(x)$ , with $y$ and its derivative appearing to the first power only. The integrating factor turns the left-hand side into a single derivative you can integrate.

Solve

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

This is a first-order linear equation in the standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$ , with $P(x) = 1$ and $Q(x) = e^{2x}$ . The integrating-factor method turns the left-hand side into a single recognisable derivative, making the equation integrable.

Step 1: Find the integrating factor

Compute $I = e^{\int P(x)\,dx}$ . With $P(x) = 1$ , the integral is simply $x$ , giving $I = e^{x}$ .

I = e^{\int 1\,dx} = e^{x}

Step 2: Multiply through by $I$ and recognise the exact derivative

Multiplying both sides by $e^x$ turns the left side into the derivative of the product $e^x y$ , by the product rule in reverse. This is the whole point of the integrating factor.

e^{x}\dfrac{dy}{dx} + e^{x}y = e^{x}\cdot e^{2x} = e^{3x}

\dfrac{d}{dx}(e^{x}y) = e^{3x}

Step 3: Integrate both sides and solve for $y$

Integrate the right side with respect to $x$ , include the constant $C$ , then divide through by $e^x$ to isolate $y$ .

e^{x}y = \dfrac{1}{3}e^{3x} + C

Final answer: $y = \dfrac{1}{3}e^{2x} + Ce^{-x}$ , where $C$ is an arbitrary constant.

Particular solutions and modelling

The general solution carries an arbitrary constant; an initial or boundary condition fixes it, giving the particular solution that fits the situation. Growth and decay problems are the classic application: the statement "the rate of change is proportional to the amount present" becomes $\dfrac{dN}{dt} = kN$ , a separable equation whose solution is $N = N_0 e^{kt}$ .

A model: radioactive decay

Radioactive decay is a classic application of separable differential equations. The physical fact that the decay rate is proportional to the amount present translates directly into a first-order separable equation, and solving it recovers the familiar exponential decay formula.

Step 1: Write the differential equation

"The rate of change is proportional to the amount present" means $\dfrac{dm}{dt} \propto m$ . Because mass decreases, we use a negative constant $-k$ with $k > 0$ .

\dfrac{dm}{dt} = -km

Step 2: Separate the variables and integrate

Move all $m$ terms to the left and all $t$ terms to the right, then integrate each side.

\dfrac{1}{m}\,dm = -k\,dt

\ln m = -kt + C

m = e^{C}e^{-kt} = Ae^{-kt}

where $A = e^C$ is an arbitrary positive constant.

Step 3: Apply the initial condition to find the particular solution

At time $t = 0$ the mass is $m_0$ , so we substitute $t = 0$ and $m = m_0$ to pin down $A$ .

m_0 = Ae^{0} = A

Final answer: $m = m_0 e^{-kt}$ . The mass falls exponentially, halving over each fixed interval of length $\dfrac{\ln 2}{k}$ .

Choosing between the two methods

Before solving, decide which method fits. If you can split the equation so that every $y$ sits with $dy$ and every $x$ with $dx$ , it is separable and direct integration finishes it. If the $y$ term cannot be detached from an $x$ term but the equation is linear in $y$ , write it in the standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$ and use the integrating factor. Some equations are both, and either route works; many are only one. Always rearrange into the standard linear form first, because the coefficient of $\dfrac{dy}{dx}$ must be $1$ before you read off $P(x)$ .

A useful check on any solution is to differentiate it and substitute back: a correct general solution must satisfy the original equation for every value of the arbitrary constant. This catches sign slips and missing factors, and it is worth the few seconds in an exam where method marks ride on a clean final line.

Try this

Q1. Solve $\dfrac{dy}{dx} = 3y$ . [2 marks]

Cue. Separable: $\dfrac{1}{y}\,dy = 3\,dx$ , so $\ln y = 3x + C$ and $y = Ae^{3x}$ .

Q2. Find the integrating factor for $\dfrac{dy}{dx} + \dfrac{1}{x}y = 1$ , $x > 0$ . [2 marks]

Cue. $I = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x$ .

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: separable4 marksSolve

\dfrac{dy}{dx} = xy

given that

y = 2

when

x = 0

Show worked answer →

Separate the variables: $\dfrac{1}{y}\,dy = x\,dx$ (1 mark).

Integrate both sides: $\ln|y| = \dfrac{1}{2}x^2 + C$ (1 mark).

Apply the condition $y = 2$ at $x = 0$ : $\ln 2 = C$ (1 mark).

So $\ln y = \dfrac{1}{2}x^2 + \ln 2$ , giving $y = 2e^{x^2/2}$ (1 mark). Markers reward separation, integration of both sides, finding $C$ , and the explicit particular solution.

AH style: integrating factor5 marksSolve

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

, for

x > 0

Show worked answer →

This is linear with $P(x) = \dfrac{2}{x}$ . Integrating factor $I = e^{\int \frac{2}{x}\,dx} = e^{2\ln x} = x^2$ (2 marks).

Multiply through: $x^2\dfrac{dy}{dx} + 2xy = x^3$ , so $\dfrac{d}{dx}(x^2 y) = x^3$ (1 mark).

Integrate: $x^2 y = \dfrac{1}{4}x^4 + C$ (1 mark).

Divide by $x^2$ : $y = \dfrac{1}{4}x^2 + \dfrac{C}{x^2}$ (1 mark). Markers reward the integrating factor, the product-rule recognition, the integration, and solving for $y$ .

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)