First order equations: separating variables

Separating the variables

If dy/dx = f(x)g(y), rearrange to ∫ 1/g(y) dy = ∫ f(x) dx, integrate both sides, and add one constant of integration.

With logarithms, write the constant as ln A so the answer tidies up: ln|y| = ½ln(x² + 1) + ln A gives y = A√(x² + 1).

General and particular solutions

The general solution contains an arbitrary constant, so it describes a family of curves. A particular solution uses a given condition, such as y = 3 when x = 0, to fix the constant.

Forgetting the constant of integration means you cannot find the particular solution, so it usually costs the later marks too.

Families of solution curves

To sketch a family, draw several members for different constants. For y = Ax the curves are straight lines through the origin; for y² - x² = c they are hyperbolas (with c = 0 giving the pair of lines y = ±x).

Check: where does each curve cross the axes, and what happens as x → ∞?

Forming an equation

Translate words into a derivative: "the rate of increase of N is proportional to N" means dN/dt = kN, with k > 0. Use the information given to find k and the constant, then answer the question in context.

Dividing by y loses the solution y = 0; mention it if the question asks for all solutions.

Worked example

Find the solution of dy/dx = xy/(x² + 1) with y = 3 when x = 0.

1. Separate: ∫ 1/y dy = ∫ x/(x² + 1) dx.
2. ln|y| = ½ln(x² + 1) + ln A, so y = A√(x² + 1).
3. x = 0, y = 3: A = 3.

Answer: y = 3√(x² + 1)

Check yourself

1. Solve dy/dx = 2xy with y = 5 when x = 0, and find y when x = 1, to 3 decimal places.
   Show answer

   13.591. ln|y| = x² + c, so y = Ae^x2. A = 5, so y(1) = 5e = 13.591.

2. What is the general solution of dy/dx = y/x, and what does the family look like?
   • y = x + c: parallel lines
   • y = Ax: straight lines through the origin
   • y = Ae^x: exponential curves
   • xy = c: hyperbolas
   Show answer

   y = Ax: straight lines through the origin. ∫ 1/y dy = ∫ 1/x dx gives ln|y| = ln|x| + ln A, so y = Ax.

3. Solve dy/dx = x/y with y = 2 when x = 0, and find y when x = 3, to 3 decimal places (take y > 0).
   Show answer

   3.606. ∫ y dy = ∫ x dx gives y²/2 = x²/2 + c, so y² = x² + 4. At x = 3: y = √13 = 3.606.

4. Which family of curves is described by the general solution y² - x² = c?
   • Hyperbolas, with the lines y = ±x when c = 0
   • Circles centred at the origin
   • Parabolas
   • Straight lines through the origin
   Show answer

   Hyperbolas, with the lines y = ±x when c = 0. y² - x² = c is a rectangular hyperbola for c ≠ 0; c = 0 gives (y - x)(y + x) = 0.

5. A population N satisfies dN/dt = kN. N = 200 when t = 0 and N = 300 when t = 5. Find N when t = 12, to 3 significant figures.
   Show answer

   529. N = 200e^kt. 300 = 200e^5k gives e^5k = 1.5. N(12) = 200 × 1.5^12/5 = 529.24..., so N = 529 to 3 significant figures.

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

   0.693. y = 4e^-2x. 1 = 4e^-2x gives e^-2x = 1/4, so x = ½ln 4 = ln 2 = 0.693.

7. When you divide dy/dx = 3x² y by y to separate the variables, which solution can be lost?
   • y = x³
   • y = e^x
   • y = 0
   • y = 1
   Show answer

   y = 0. Dividing by y assumes y ≠ 0, but y = 0 for all x also satisfies the equation.

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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

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