Complementary Function of Linear Differential Equations with Constant Coefficients

The key idea. The complementary function (CF) is the general solution of the homogeneous equation $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0.$ Because differentiating $e^{mx}$ just multiplies it by $m$, an exponential is the natural trial solution.

Substitute $y = e^{mx}$. Then $\dfrac{dy}{dx} = m\,e^{mx}$ and $\dfrac{d^2y}{dx^2} = m^2 e^{mx}$. The equation becomes $am^2 e^{mx} + bm\,e^{mx} + c\,e^{mx} = 0.$

Divide by $e^{mx}$ (never zero) to leave a pure quadratic in $m$: $am^2 + bm + c = 0.$

This is the auxiliary equation (also called the characteristic equation). To form it you simply replace $\frac{d^2y}{dx^2}\to m^2, \qquad \frac{dy}{dx}\to m, \qquad y\to 1.$

Solving this quadratic gives the values of $m$ for which $e^{mx}$ works. Because a second-order linear ODE has a two-parameter general solution, we combine two independent building blocks with arbitrary constants $A$ and $B$.

Cambridge tip. The trial $y = e^{mx}$ does NOT need to be quoted in the exam — go straight to "the auxiliary equation is $am^2 + bm + c = 0$". But understanding the substitution is what stops you mixing up the three cases below.

Once the auxiliary equation $am^2 + bm + c = 0$ is solved, the discriminant $\Delta = b^2 - 4ac$ tells you which of exactly three forms the complementary function takes.

Case 1 — Distinct real roots ($\Delta > 0$). Two different real values $m_1 \neq m_2$. Each gives an independent exponential: $y = Ae^{m_1 x} + Be^{m_2 x}.$

Case 2 — Repeated real root ($\Delta = 0$). One value $m$ (a double root). A single $e^{mx}$ only gives ONE building block, so the second is $x\,e^{mx}$: $y = (A + Bx)e^{mx}.$ The $x$ factor is not optional — without it you have only one arbitrary constant and an incomplete general solution.

Case 3 — Complex conjugate roots ($\Delta < 0$). The roots are $m = p \pm qi$. Using $e^{(p+qi)x} = e^{px}(\cos qx + i\sin qx)$ and combining the conjugate pair gives the real form: $y = e^{px}\bigl(A\cos qx + B\sin qx\bigr).$

Cambridge tip. Always state which case you are in ("the auxiliary equation has complex roots $\ldots$") before writing the CF — examiners credit the recognition, and it stops you reaching for the wrong template under pressure.

A particularly important sub-case of the complex roots is when $b = 0$, so the auxiliary equation is $am^2 + c = 0$ with $a, c$ the same sign. Then $m^2 = -\frac{c}{a} < 0 \quad\Rightarrow\quad m = \pm qi, \qquad q = \sqrt{\tfrac{c}{a}}.$ The real part $p = 0$, so the exponential factor $e^{0\cdot x} = 1$ disappears and the CF is pure oscillation: $y = A\cos qx + B\sin qx.$

This is exactly the simple harmonic motion solution of $\dfrac{d^2y}{dx^2} + q^2 y = 0$ — undamped, constant-amplitude. It appears constantly in mechanics-flavoured 9231 questions.

Verify it satisfies the ODE. Take $y = A\cos qx + B\sin qx$. Then $y' = -Aq\sin qx + Bq\cos qx$ and $y'' = -Aq^2\cos qx - Bq^2\sin qx = -q^2 y$. Hence $y'' + q^2 y = 0$ — confirmed.

Sign matters. If instead $a$ and $c$ have opposite signs, $m^2 = +\frac{c}{a} > 0$ gives real roots $m = \pm q$ and you are back in Case 1 with $y = Ae^{qx} + Be^{-qx}$ (which can be rewritten with $\cosh$ and $\sinh$). Do not confuse the two.

Cambridge tip. Spotting "no $\frac{dy}{dx}$ term" should immediately flag either pure oscillation (same signs) or pure exponential/hyperbolic growth (opposite signs). Check the sign of $-\frac{c}{a}$ first.

A complementary function is only correct if it (i) satisfies the homogeneous ODE and (ii) contains two arbitrary constants (for a second-order equation). Both are quick to check and worth the seconds in an exam.

Two-constant check. Each of the three templates carries exactly $A$ and $B$. The classic error is the repeated-root case: writing $y = Ae^{mx}$ (one constant) instead of $y = (A + Bx)e^{mx}$ (two). If your CF has fewer than two constants, it is incomplete.

Substitution check (repeated root). Verify $y = (A + Bx)e^{mx}$ solves $y'' - 2my' + m^2 y = 0$ (whose auxiliary equation $(\,\cdot - m)^2 = 0$ has the double root $m$): $y' = Be^{mx} + m(A + Bx)e^{mx} = \bigl(mA + B + mBx\bigr)e^{mx},$ $y'' = \bigl(m^2 A + 2mB + m^2 Bx\bigr)e^{mx}.$ Then $y'' - 2my' + m^2 y = \bigl[(m^2A + 2mB + m^2Bx) - 2m(mA + B + mBx) + m^2(A + Bx)\bigr]e^{mx}.$ The bracket collapses term-by-term to $0$, confirming $(A+Bx)e^{mx}$ is a genuine solution.

Cambridge tip. When a full ODE has a particular integral too, the CF is only half the answer — but the CF must still be written with both constants, because those constants are fixed later by the boundary/initial conditions, not by the particular integral.