First order equations by substitution

How it works

1. From the substitution, write y (and dy/dx) in terms of the new variable.

2. Substitute into the equation and simplify until it is separable or linear.

3. Solve that equation.

4. Substitute back to get y in terms of x.

If part (a) says "show that the substitution transforms the equation into ...", every line of working must be there, because the answer is given.

Common substitutions

• y = vx (for equations like dy/dx = (x² + y²)/(xy)): dy/dx = v + x dv/dx, by the product rule.
• z = 1/y (for equations with a y² term): y = 1/z, so dy/dx = -(1/z²) dz/dx.
• z = y² or z = y^1-n for dy/dx + Py = Qyⁿ: these make the equation linear in z.

Exam points

Use the chain rule carefully, and include the constant before substituting back. When you take a square root at the end, consider ±; the June 2025 examiners' report notes a considerable number of students giving only one of the two square roots.

Worked example

Use z = 1/y to solve dy/dx + xy = xy².

1. y = 1/z gives dy/dx = -(1/z²) dz/dx.
2. Substitute: -(1/z²) dz/dx + x/z = x/z². Multiply by -z²: dz/dx - xz = -x.
3. IF = e^-x2/2: d/dx(ze^-x2/2) = -xe^-x2/2, so ze^-x2/2 = e^-x2/2 + c.
4. z = 1 + ce^x2/2, so y = 1/z.

Answer: y = 1/(1 + ce^x2/2)

Check yourself

1. If y = vx, where v is a function of x, what is dy/dx?
   • x dv/dx
   • v dv/dx
   • dv/dx
   • v + x dv/dx
   Show answer

   v + x dv/dx. Product rule: d/dx(vx) = v × 1 + x × dv/dx.

2. If z = 1/y, which expression equals dy/dx?
   • (1/z²) dz/dx
   • -(1/z²) dz/dx
   • -1/z²
   • -z² dz/dx
   Show answer

   -(1/z²) dz/dx. y = z^-1, so dy/dx = -z^-2 dz/dx by the chain rule.

3. Using y = vx, the equation dy/dx = (x² + y²)/(xy) becomes which equation?
   • x dv/dx = v + 1/v
   • dv/dx = 1/v
   • x dv/dx = v²
   • x dv/dx = 1/v
   Show answer

   x dv/dx = 1/v. (x² + v² x²)/(vx²) = (1 + v²)/v = 1/v + v. So v + x dv/dx = v + 1/v.

4. From x dv/dx = 1/v you get y² = 2x² ln x + cx². Given y = 2 when x = 1, find y when x = e, to 3 decimal places (y > 0).
   Show answer

   6.658. 4 = 0 + c, so c = 4. At x = e: y² = 2e² + 4e² = 6e², so y = e√6 = 6.658.

5. The solution of dy/dx + xy = xy² is y = 1/(1 + ce^x2/2). Given y = 1/2 when x = 0, find y when x = 1, to 4 decimal places.
   Show answer

   0.3775. 1/2 = 1/(1 + c), so c = 1. y(1) = 1/(1 + e^1/2) = 0.3775.

6. Why is a substitution used in these questions?
   • To change the equation into one that is separable or linear
   • To remove the constant of integration
   • To make the equation second order
   • To avoid using the condition given
   Show answer

   To change the equation into one that is separable or linear. The new equation can be solved with methods you know; then you substitute back.

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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.