Differential Equations · step 3 of 14

First-order linear equations & the integrating factor

Any first-order linear ODE can be written y′ + P(x)y = Q(x). Multiplying by μ = e^(∫P dx) makes the left side an exact derivative (μy)′, so integrating once solves the equation. The essential prerequisite is getting the equation into standard form with a coefficient of 1 on y′ — skipping that gives the wrong P and everything after it fails.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Put it in standard formDivide through so that y′ has coefficient 1. Then read off P(x) and Q(x).
Compute μ = e^(∫P dx)No constant of integration is needed here — any one integrating factor works.
Multiply through and recognise the product ruleThe left side becomes (μy)′ = μQ.
Integrate and divide by μμy = ∫μQ dx + C, so y = (1/μ)(∫μQ dx + C). The constant must be inside the bracket.

Worked example

Solve y′ + 2y = e^x.

Already in standard form with P = 2 and Q = e^x.
μ = e^(∫2dx) = e^(2x).
Multiply: e^(2x)y′ + 2e^(2x)y = e^(3x), and the left side is (e^(2x)y)′.
Integrate: e^(2x)y = e^(3x)/3 + C.

Answer. y = e^x/3 + Ce^(−2x). Check: y′ + 2y = (e^x/3 − 2Ce^(−2x)) + (2e^x/3 + 2Ce^(−2x)) = e^x ✓.

Where marks get dropped

These are the specific errors that cost credit on first-order linear equations & the integrating factor questions — QED's rubric penalises each of them separately.

Not dividing to standard form first. With 2y′ + 4y = x, P is 2, not 4.
Dividing only the integral by μ and leaving C outside. The constant arises inside the integration, so y = (∫μQ dx + C)/μ.
Adding a constant when computing μ. It cancels, so including it just adds work.

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First-order linear equations & the integrating factor — frequently asked questions

Why does the integrating factor work?

Because (μy)′ = μy′ + μ′y, and choosing μ′ = Pμ makes this exactly μ(y′ + Py). That condition is a separable equation whose solution is e^(∫P dx).

Can I use this on a separable equation?

Yes, if it is also linear — dy/dx = xy is both, and either method works. Separation is usually quicker there.

What if Q(x) = 0?

The equation is homogeneous and also separable, with solution y = Ce^(−∫P dx). That is the complementary function of the general case.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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