Integrating Factor (AA HL)
Some first order differential equations can't be separated - the \(x\) and \(y\) terms are tangled together. When the equation is linear, multiplying through by a carefully chosen integrating factor turns the left-hand side into the derivative of a single product, which can then be integrated directly. This page covers the method, worked examples, and the mistakes that lose the most marks. It's part of the broader Differential Equations topic.
19 questions on this sub-topic.
The integrating factor method
Covered under IB syllabus reference AHL5.18. The technique only applies once the equation is arranged into standard form - do that first, then apply the formula mechanically.
Standard form
\(y' + P(x)y = Q(x)\)
Rearrange the equation so the coefficient of \(y'\) is exactly 1 before reading off \(P(x)\).
Integrating factor
\(I = e^{\int P(x)\,dx}\)
Multiplying both sides by \(I\) makes the left side equal to \(\dfrac{d}{dx}(Iy)\), so you can integrate both sides directly and solve for \(y\).
Need the full syllabus wording, Euler's method, and separable equations too? See Differential Equations.
Worked examples
Solve \(\dfrac{dy}{dx} + 3y = 6.\)
Worked solution
Integrating factor: \(I\) M1
\(= e^{3x}.\) A1
Multiply: \(\dfrac{d}{dx}(ye^{3x}) = 6e^{3x}.\) M1
Integrate: \(ye^{3x} = 2e^{3x} + C.\) A1
So \(y = 2 + Ce^{-3x}.\) A1
Solve \(\dfrac{dy}{dx} + \dfrac{1}{x}y = x\) for \(x > 0\), given \(y(1) = 2.\)
Worked solution
\(I = e^{\int \frac1x dx}\) M1 \(= x.\) A1
\(\dfrac{d}{dx}(xy) = x^2.\) M1 A1
\(xy = \tfrac{x^3}{3} + C \Rightarrow y\) M1 \(= \tfrac{x^2}{3} + \tfrac{C}{x}.\) A1
\(y(1)=2 \Rightarrow C = \tfrac53.\) \(y = \tfrac{x^2}{3} + \tfrac{5}{3x}.\) A1
On the GDC: once you have \(\tfrac13 + C = 2\), the equation solver on any of the three approved models will confirm \(C=\tfrac53\) instantly. See the parent topic's GDC guidance for calculator-specific steps.
Explain why \(\dfrac{d}{dx}(ye^{2x}) = e^{2x}\dfrac{dy}{dx} + 2ye^{2x}\) and hence write down the DE this is equivalent to.
Worked solution
By the product rule: \(\dfrac{d}{dx}(ye^{2x})\) M1
\(= e^{2x}\dfrac{dy}{dx} + y\cdot 2e^{2x}.\) A1
Dividing by \(e^{2x}\): \(\dfrac{dy}{dx} + 2y = e^{-2x}\dfrac{d}{dx}(ye^{2x}) = Q(x).\) A1
Common mistakes
- Forgetting to arrange the equation into standard form first. \(P(x)\) is only the coefficient of \(y\) once the coefficient of \(y'\) has been made exactly 1 - dividing through before reading off \(P(x)\) is essential.
- Multiplying only the left-hand side by the integrating factor. Every term in the equation, including \(Q(x)\) on the right, must be multiplied by \(I\) - a solution that skips this ends up with the wrong right-hand side to integrate.
- Dropping the constant of integration when applying an initial condition. A "solve given \(y(1)=2\)" question needs the general solution with \(C\) still present before you substitute the given point - skipping straight to a particular solution loses the method mark.
Ready to practise properly?
19 integrating-factor questions, marked instantly like the real exam.
Quick answers
What is the integrating factor for a first order linear differential equation?
For an equation written as \(y' + P(x)y = Q(x)\), the integrating factor is \(I = e^{\int P(x)\,dx}\).
Why does multiplying by the integrating factor work?
Multiplying both sides by \(I\) turns the left-hand side into the derivative of a single product, \(\dfrac{d}{dx}(Iy)\), which can then be integrated directly.