First order linear equations: the integrating factor

Standard form

Rearrange to dy/dx + P(x)y = Q(x). The coefficient of dy/dx must be 1: if the equation is x dy/dx + 2y = 4x², divide by x first to get dy/dx + (2/x)y = 4x.

Getting the standard form right comes first, because P is read from it.

The integrating factor

IF = e^{∫P dx}. Multiplying the equation by the IF makes the left side the derivative of y × IF (the product rule in reverse):

d/dx (y × IF) = Q × IF.

You may quote the IF without proof. Simplify it: e^{2ln x} = x², e^{∫tan x dx} = e^{ln sec x} = sec x, e^{-∫cot x dx} = cosec x.

The five steps

1. Rearrange to standard form.

2. Find the IF (no constant needed here).

3. Multiply through and write the left side as d/dx(y × IF).

4. Integrate both sides, including + c.

5. Rearrange to y = ... and use any condition.

Behaviour of solutions

The general solution often has a part that depends on c and a part that does not. For y = 3 - 2e^-2x, the exponential term dies away and y → 3 whatever the starting value. Questions may ask which value of c keeps a solution finite.

Worked example

Solve x dy/dx + 2y = 4x², for x > 0.

1. Standard form: dy/dx + (2/x)y = 4x.
2. IF = e^{∫2/x dx} = e^{2ln x} = x².
3. d/dx(x² y) = 4x³, so x² y = x⁴ + c.

Answer: y = x² + c/x²

Check yourself

1. Find the integrating factor for dy/dx + 3y = e^x.
   • e³
   • 3e^x
   • e^x3
   • e^3x
   Show answer

   e^3x. P = 3, so ∫P dx = 3x and IF = e^3x.

2. Find the integrating factor for dy/dx + y tan x = sec x.
   • cos x
   • sec x
   • e^{tan x}
   • ln(sec x)
   Show answer

   sec x. ∫tan x dx = ln|sec x|, so IF = e^{ln sec x} = sec x.

3. Find the integrating factor for dy/dx - y cot x = sin x.
   • sin x
   • e^{-cot x}
   • -sin x
   • cosec x
   Show answer

   cosec x. P = -cot x. ∫-cot x dx = -ln|sin x|, so IF = e^{-ln sin x} = 1/sin x = cosec x. The June 2025 examiners' report notes some students giving sin x instead of cosec x.

4. Solve dy/dx + 2y = 6 with y = 1 when x = 0. Find y when x = 1, to 3 decimal places.
   Show answer

   2.729. IF = e^2x. d/dx(ye^2x) = 6e^2x, so ye^2x = 3e^2x + c, y = 3 + ce^-2x. c = -2. y(1) = 3 - 2e^-2 = 2.729.

5. For x dy/dx + 2y = 4x², the general solution is y = x² + c/x². Given y = 3 when x = 1, find y when x = 2.
   Show answer

   4.5. 3 = 1 + c, so c = 2. y(2) = 4 + 2/4 = 4.5.

6. Solve dy/dx - y/x = x for x > 0, with y = 2 when x = 1. Find y when x = 3.
   Show answer

   12. IF = e^{-ln x} = 1/x. d/dx(y/x) = 1, so y/x = x + c and y = x² + cx. c = 1, so y(3) = 9 + 3 = 12.

7. The solution of a differential equation is y = 3 - 2e^-2x. What happens to y as x → ∞?
   • y → 1
   • y → ∞
   • y → 3
   • y → 0
   Show answer

   y → 3. e^-2x → 0, so y → 3.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit F2
• Pearson Edexcel IAL Further Pure Mathematics 2 scheme of work
• Pearson Edexcel IAL Further Pure Mathematics F2 (WFM02) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.