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Second order equations: the complementary function

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

A second order linear equation with constant coefficients, a d^2y/dx² + b dy/dx + cy = f(x), is solved in two parts. This lesson finds the first part, the complementary function, which solves the equation with f(x) = 0.

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The auxiliary equation

Try y = e^mx in a y'' + by' + cy = 0: e^mx(am² + bm + c) = 0. So e^mx is a solution when

am² + bm + c = 0, the auxiliary equation.

The type of roots decides the form of the solution.

Three cases

Real distinct roots α and β: y = Ae^αx + Be^βx.
Equal roots α: y = (A + Bx)e^αx.
Complex roots p ± qi: y = e^px(A cos qx + B sin qx).

If the roots are ±ωi (no y' term, for example y'' + 9y = 0): y = A cos ωx + B sin ωx, simple harmonic motion.

Roots of am² + bm + c = 0	Complementary function
α, β real and different	Ae^αx + Be^βx
α repeated	(A + Bx)e^αx
p ± qi	e^px(A cos qx + B sin qx)
±ωi	A cos ωx + B sin ωx

Complementary functions

Why the complex case works

With m = p ± qi, e^((p + qi)x) = e^px(cos qx + i sin qx) by Euler's relation. Combining the two complex solutions with suitable constants gives the real form e^px(A cos qx + B sin qx).

Common slips

The scheme of work warns that the wrong sign on the power of e is common: if the auxiliary equation is m² - 2m + 5 = 0, the roots are 1 ± 2i, so the factor is e^+x. Keep the variable consistent: if the equation is in t, the solution is in t.

Worked example

Find the general solution of d^2y/dx² - 5dy/dx + 6y = 0.

Auxiliary equation: m² - 5m + 6 = 0.
(m - 2)(m - 3) = 0, so m = 2 or m = 3.
Real distinct roots.

Answer: y = Ae^2x + Be^3x

Key idea: Solve am² + bm + c = 0. Distinct real roots: two exponentials. Repeated root: (A + Bx)e^αx. Complex p ± qi: e^px(A cos qx + B sin qx).

Check yourself

Find the general solution of y'' + 4y' + 4y = 0.
- y = Ae^-2x + Be^-2x
- y = (A + Bx)e^2x
- y = (A + Bx)e^-2x
- y = A cos 2x + B sin 2x
Show answer
y = (A + Bx)e^-2x. m² + 4m + 4 = (m + 2)² = 0, a repeated root m = -2.
Find the general solution of y'' + 2y' + 5y = 0.
- y = e^-x(A cos 2x + B sin 2x)
- y = e^x(A cos 2x + B sin 2x)
- y = e^-2x(A cos x + B sin x)
- y = Ae^-x + Be^2x
Show answer
y = e^-x(A cos 2x + B sin 2x). m = (-2 ± √(4 - 20))/2 = -1 ± 2i. So p = -1 and q = 2.
Find the general solution of y'' + 9y = 0.
- y = Ae^3x + Be^-3x
- y = A cos 9x + B sin 9x
- y = A cos 3x + B sin 3x
- y = (A + Bx)e^3x
Show answer
y = A cos 3x + B sin 3x. m² + 9 = 0 gives m = ±3i, so p = 0 and q = 3.
Find the roots of the auxiliary equation for 2y'' + y' - 3y = 0.
Show answer
m = 1 or m = -1.5. 2m² + m - 3 = (2m + 3)(m - 1) = 0.
Find the general solution of y'' - 4y = 0.
- y = A cos 2x + B sin 2x
- y = (A + Bx)e^2x
- y = Ae^4x + B
- y = Ae^2x + Be^-2x
Show answer
y = Ae^2x + Be^-2x. m² - 4 = 0 gives m = ±2, real and distinct.
A student solves y'' - 2y' + 5y = 0 and writes y = e^-x(A cos 2x + B sin 2x). What is the error?
- The trig functions should be cos x and sin x
- The roots are 1 ± 2i, so the factor should be e^x
- There should be no trig functions
- Nothing: the answer is correct
Show answer
The roots are 1 ± 2i, so the factor should be e^x. m = (2 ± √(4 - 20))/2 = 1 ± 2i. The real part is +1.
y'' - 3y' + 2y = 0 has general solution y = Ae^x + Be^2x. Given y = 1 and dy/dx = 0 when x = 0, find y when x = 1, to 3 decimal places.
Show answer
-1.952. A + B = 1 and A + 2B = 0, so B = -1, A = 2. y(1) = 2e - e² = -1.952.

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