All subjects · Further Mathematics · WFM02 Further Pure Mathematics 2

Second order equations by substitution

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

Equations such as x² d^2y/dx² + ax dy/dx + by = 0 do not have constant coefficients. A given substitution, usually x = e^u, turns them into the constant coefficient type you can already solve.

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The substitution x = e^u

With x = e^u, dx/du = e^u = x. By the chain rule dy/du = (dy/dx)(dx/du) = x dy/dx.

Differentiating again with the product rule: d^2y/du² = x dy/dx + x² d^2y/dx², so

x dy/dx = dy/du and x² d^2y/dx² = d^2y/du² - dy/du.

Method

1. Derive (or use, if given) the expressions for x dy/dx and x² d^2y/dx².

2. Substitute to get an equation in y and u with constant coefficients.

3. Solve it as usual (auxiliary equation, CF, PI).

4. Replace u by ln x, using e^ku = x^k.

The scheme of work notes that problems arise in applying the chain rule within the product rule to find the second derivative: lay out each derivative on its own line.

Other substitutions

A question may instead give y = xv or z = dy/dx. The approach is the same: find dy/dx and d^2y/dx² in the new variables with the product and chain rules, substitute, and simplify to the required form. Show every step in a "show that".

Worked example

Use x = e^u to solve x² d^2y/dx² - 2x dy/dx + 2y = 0.

Substitute: (d^2y/du² - dy/du) - 2dy/du + 2y = 0, so d^2y/du² - 3dy/du + 2y = 0.
Auxiliary equation m² - 3m + 2 = 0, m = 1 or 2: y = Ae^u + Be^2u.
e^u = x and e^2u = x².

Answer: y = Ax + Bx²

Key idea: With x = e^u: x dy/dx = dy/du and x² d^2y/dx² = d^2y/du² - dy/du. Solve in u, then put e^u = x.

Check yourself

If x = e^u, which expression equals x dy/dx?
- dy/du
- d^2y/du²
- e^u dy/du
- dy/du - y
Show answer
dy/du. dy/du = (dy/dx)(dx/du) = (dy/dx)e^u = x dy/dx.
If x = e^u, which expression equals x² d^2y/dx²?
- d^2y/du²
- d^2y/du² + dy/du
- d^2y/du² - dy/du
- (dy/du)²
Show answer
d^2y/du² - dy/du. Differentiate dy/du = x dy/dx with respect to u: d^2y/du² = x dy/dx + x² d^2y/dx² (using d/du = x d/dx). So x² y'' = d^2y/du² - dy/du.
Use x = e^u to find the general solution of x² y'' + xy' - 4y = 0.
- y = Ax² + B/x²
- y = Ae^2x + Be^-2x
- y = Ax² + Bx^-1
- y = A cos(2 ln x) + B sin(2 ln x)
Show answer
y = Ax² + B/x². (y_uu - y_u) + y_u - 4y = y_uu - 4y = 0, so y = Ae^2u + Be^-2u = Ax² + Bx^-2.
For x² y'' + xy' - 4y = 0, with y = 2 and dy/dx = 0 when x = 1, find y when x = 2.
Show answer
4.25. y = Ax² + Bx^-2, y' = 2Ax - 2Bx^-3. A + B = 2 and 2A - 2B = 0, so A = B = 1. y(2) = 4 + 1/4 = 4.25.
Use x = e^u to find the general solution of x² y'' - 3xy' + 4y = 0.
- y = (A + Bx)e^2x
- y = (A + B ln x)x²
- y = Ax² + Bx^-2
- y = (A + Bx)x²
Show answer
y = (A + B ln x)x². y_uu - y_u - 3y_u + 4y = y_uu - 4y_u + 4y = 0: repeated root m = 2. y = (A + Bu)e^2u = (A + B ln x)x².
Use x = e^u to find the general solution of x² y'' + 3xy' + 5y = 0.
- y = e^-x(A cos 2x + B sin 2x)
- y = x(A cos(2 ln x) + B sin(2 ln x))
- y = (1/x)(A cos 2x + B sin 2x)
- y = (1/x)(A cos(2 ln x) + B sin(2 ln x))
Show answer
y = (1/x)(A cos(2 ln x) + B sin(2 ln x)). y_uu - y_u + 3y_u + 5y = y_uu + 2y_u + 5y = 0, m = -1 ± 2i. y = e^-u(A cos 2u + B sin 2u), and e^-u = 1/x, u = ln x.

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