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Series solutions of differential equations

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

Many differential equations cannot be solved exactly. A Taylor series gives an approximate solution near the point where the conditions are given, using the equation itself to supply the derivatives.

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The method

1. Use the given values of y (and y') at x = a.

2. Use the differential equation to find the next derivative at x = a.

3. Differentiate the equation (implicitly, with respect to x) to get higher derivatives, and evaluate each at x = a.

4. Substitute into y = y(a) + y'(a)(x - a) + y''(a)(x - a)2/2! + ....

The answer must start "y = ..."; the scheme of work says writing f(x) = ... or nothing at all, instead of y = ..., is penalised.

Differentiating implicitly

Every term containing y must be differentiated with the chain rule: d/dx(y2) = 2y dy/dx, d/dx(xy) = y + x dy/dx.

The June 2025 examiners' report notes some students leaving out dy/dx when differentiating a y2 term.

Second order equations

For y'' + xy' + y = 0 with y(0) = 1, y'(0) = 0: the equation gives y''(0) = -1. Differentiating gives y''' + 2y' + xy'' = 0, so y'''(0) = 0, and once more y'''' + 3y'' + xy''' = 0, so y''''(0) = 3.

So y = 1 - x2/2 + x4/8 + ....

Exam advice

Do not try to find the general solution: the question wants the series. Keep the derivative values in a neat list, and remember the factorials.

Worked example

Find the series solution of dy/dx = x + y2, with y = 1 at x = 0, up to the term in x4.

  1. y'(0) = 0 + 1 = 1.
  2. y'' = 1 + 2yy', so y''(0) = 1 + 2 = 3.
  3. y''' = 2(y')2 + 2yy'', so y'''(0) = 2 + 6 = 8.
  4. y'''' = 6y'y'' + 2yy''', so y''''(0) = 18 + 16 = 34.
  5. y = 1 + x + 3x2/2! + 8x3/3! + 34x4/4!.

Answer: y = 1 + x + (3/2)x2 + (4/3)x3 + (17/12)x4 + ...

Key idea: Use the equation to get y'', differentiate it implicitly for higher derivatives, evaluate at the given point, then write y = Taylor series.

Check yourself

  1. dy/dx = xy + 1, with y = 2 when x = 0. Find the value of d2y/dx2 at x = 0.
    Show answer

    2. d2y/dx2 = y + x dy/dx (product rule). At x = 0: 2 + 0 = 2.

  2. For the same equation dy/dx = xy + 1, y(0) = 2, find the coefficient of x3 in the series solution. Give your answer as a fraction in its simplest form.
    Show answer

    1/3. y''' = 2dy/dx + x d2y/dx2. y'(0) = 1, so y'''(0) = 2. Coefficient = 2/3! = 1/3.

  3. y'' + xy' + y = 0, with y = 1 and y' = 0 at x = 0. Find the coefficient of x2 in the series solution. Give your answer as a fraction in its simplest form.
    Show answer

    -1/2. y''(0) = -0 × 0 - 1 = -1, so the coefficient is -1/2! = -1/2.

  4. For the same equation, y''''(0) = 3. Find the coefficient of x4. Give your answer as a fraction in its simplest form.
    Show answer

    1/8. 3/4! = 3/24 = 1/8.

  5. Use y = 1 + x + (3/2)x2 + (4/3)x3 + (17/12)x4 to estimate y when x = 0.1, to 4 decimal places.
    Show answer

    1.1165. 1 + 0.1 + 0.015 + 0.001333... + 0.000141... = 1.1165.

  6. When differentiating dy/dx = x + y2 with respect to x, what is the derivative of y2?
    • 2y
    • 2x
    • 2y dy/dx
    • y2 dy/dx
    Show answer

    2y dy/dx. y is a function of x, so the chain rule gives 2y × dy/dx.

  7. dy/dx = y2 - x, with y = 2 when x = 1. Find the coefficient of (x - 1)2 in the Taylor series solution about x = 1. Give your answer as a fraction in its simplest form.
    Show answer

    11/2. y'(1) = 4 - 1 = 3. y'' = 2yy' - 1, so y''(1) = 12 - 1 = 11. Coefficient = 11/2!.

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