Second-order linear ODEs with constant coefficients

Every second-order linear ODE with constant coefficients — the equation behind damped springs, RLC circuits and countless engineering models — reduces to a single algebraic step. Guess a solution of the form $y = e^{mx}$ , substitute, and the differential equation collapses into a quadratic in $m$ : the auxiliary equation. Whatever the physical system, once you know the roots of that quadratic you can write down the full family of solutions by pattern-matching to one of three cases. This article builds that pattern-matching machine from scratch. We derive why $y=e^{mx}$ is the right guess, work through the three root cases (real distinct, real repeated, complex conjugate), and show why a repeated root needs an extra factor of $x$ — a subtlety that trips up most students. A fully worked initial value problem ties the algebra to a concrete numerical answer, and a set of traps and quiz questions checks that the pattern has actually stuck.

The core insight

Substituting $y = e^{mx}$ into $ay'' + by' + cy = 0$ turns the ODE into the algebraic auxiliary equation $am^2 + bm + c = 0$ , and the nature of its two roots — real distinct, real repeated, or complex conjugate — fixes the exact form of the general solution.

The core idea

A second-order linear homogeneous ODE with constant coefficients has the form

a y'' + b y' + c y = 0,

where: $a, b, c$ are fixed real constants ( $a \neq 0$ ) and $y = y(x)$ is the unknown function. "Homogeneous" just means the right-hand side is zero; "linear" means $y$ , $y'$ , $y''$ appear only to the first power and are not multiplied together. Because the derivative of $e^{mx}$ is a constant multiple of itself ( $\frac{d}{dx}e^{mx} = me^{mx}$ ), trying $y = e^{mx}$ turns every derivative into a power of $m$ times $e^{mx}$ . Substituting and cancelling the common factor $e^{mx}$ (which is never zero) leaves the auxiliary equation (also called the characteristic equation):

am^2 + bm + c = 0.

where: $m$ is the unknown exponent we are solving for. This is an ordinary quadratic, solvable by factoring or the quadratic formula $m = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ . The discriminant $b^2 - 4ac$ sorts every second-order case into exactly three families:

Case 1 — real, distinct roots ( $b^2-4ac>0$ ): two different exponents $m_1 \neq m_2$ , giving $y_c = C_1e^{m_1x} + C_2e^{m_2x}$ .

Case 2 — real, repeated root ( $b^2-4ac=0$ ): one exponent $m$ used twice, giving $y_c = (C_1+C_2x)e^{mx}$ — note the extra factor of $x$ on the second term.

Case 3 — complex conjugate roots ( $b^2-4ac<0$ ): roots $m=\alpha\pm i\beta$ , giving $y_c = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$ , obtained from Euler's formula by combining the two complex exponentials into real sine and cosine terms. In every case $y_c$ , the complementary function, is the general solution of the homogeneous equation, carrying exactly two arbitrary constants ( $C_1, C_2$ ) because the ODE is second order. These constants are pinned down by two extra conditions — typically initial values $y(0)$ and $y'(0)$ , or boundary values at two points.

Figure

Three panels showing the auxiliary equation roots on the complex plane for real distinct roots, a real repeated root, and complex conjugate roots, each labeled with its complementary function

Derivation

Derivation AFrom the auxiliary equation to the general solution for real distinct roots

The goal is to show, starting only from the substitution $y = e^{mx}$ and ordinary algebra, exactly why a second-order homogeneous linear ODE with constant coefficients and a real-distinct-root auxiliary equation has general solution $y = C_1e^{m_1x} + C_2e^{m_2x}$ , and to confirm this solution genuinely satisfies the original ODE for any choice of the two constants.

Step 1. Substitute the trial solution. Let $y = e^{mx}$ in $ay'' + by' + cy = 0$ . Since $y' = me^{mx}$ and $y'' = m^2e^{mx}$ , the equation becomes $am^2e^{mx} + bme^{mx} + ce^{mx} = 0$ , or $e^{mx}(am^2+bm+c) = 0$ . Because $e^{mx}$ is never zero for any real $x$ or $m$ , the bracket must vanish: $am^2+bm+c=0$ , the auxiliary equation.

Step 2. Solve for the two roots. When $b^2-4ac>0$ , the quadratic formula gives two distinct real roots $m_1 = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $m_2 = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$ . Each root separately makes $y=e^{m_1x}$ and $y=e^{m_2x}$ individual solutions of the ODE, by Step 1 run in reverse.

Step 3. Combine by linearity. Because the ODE is linear and homogeneous, any linear combination of solutions is also a solution: if $y_1$ and $y_2$ both satisfy the equation, so does $y = C_1y_1 + C_2y_2$ for any constants $C_1, C_2$ — substitute and the equation splits into two copies of Step 1, each individually zero. So $y = C_1e^{m_1x} + C_2e^{m_2x}$ solves the ODE.

Step 4. Confirm this is the general solution, not just a solution. Two functions $e^{m_1x}$ and $e^{m_2x}$ are linearly independent whenever $m_1 \neq m_2$ (their Wronskian $W = (m_2-m_1)e^{(m_1+m_2)x}$ is never zero). A second-order linear ODE has exactly a two-dimensional solution space, so two independent solutions span it completely — every solution of the ODE is some combination $C_1e^{m_1x}+C_2e^{m_2x}$ , and the two arbitrary constants are exactly what is needed to match two initial or boundary conditions.

Worked example

Solve the initial value problem $y'' - 5y' + 6y = 0$ with $y(0) = 1$ , $y'(0) = 0$ .

Step 1. Write the auxiliary equation: $m^2 - 5m + 6 = 0$ .

Step 2. Factor: $(m-2)(m-3) = 0 \implies m = 2,\ 3$ . These are real and distinct, so Case 1 applies and the general solution is $y = C_1e^{2x} + C_2e^{3x}$ .

Step 3. Apply $y(0) = 1$ : $y(0) = C_1e^0 + C_2e^0 = C_1 + C_2 = 1$ .

Step 4. Differentiate and apply $y'(0) = 0$ . Since $y' = 2C_1e^{2x} + 3C_2e^{3x}$ , we get $y'(0) = 2C_1 + 3C_2 = 0$ .

Step 5. Solve the pair of equations. From Step 3, $C_1 = 1 - C_2$ . Substituting into Step 4: $2(1-C_2) + 3C_2 = 0 \implies 2 - 2C_2 + 3C_2 = 0 \implies C_2 = -2$ , and then $C_1 = 1 - (-2) = 3$ .

Step 6. Write the particular solution and check it. $y = 3e^{2x} - 2e^{3x}$ . Check: $y(0) = 3 - 2 = 1$ ✓; $y' = 6e^{2x} - 6e^{3x}$ , so $y'(0) = 6 - 6 = 0$ ✓. Substituting back into the ODE, $y'' - 5y' + 6y = 12e^{2x} - 18e^{3x} - 5(6e^{2x}-6e^{3x}) + 6(3e^{2x}-2e^{3x}) = (12-30+18)e^{2x} + (-18+30-12)e^{3x} = 0$ ✓.

The particular solution is $y = 3e^{2x} - 2e^{3x}$ .

Common traps

Forgetting the extra factor of $x$ for a repeated root — writing $y_c = C_1e^{mx}+C_2e^{mx}$ instead of $(C_1+C_2x)e^{mx}$ silently collapses two constants into one.
Mixing up $\alpha$ and $\beta$ from complex roots $\alpha \pm i\beta$ : $\alpha$ multiplies the exponential envelope $e^{\alpha x}$ , $\beta$ multiplies $x$ inside $\cos\beta x$ and $\sin\beta x$ .
Losing a sign under the square root when $b^2-4ac<0$ , writing a real-root form for what is actually a complex-root case.
Applying initial conditions to $y_c$ before differentiating, or dropping the product rule when differentiating $e^{\alpha x}(\cos\beta x,\sin\beta x)$ terms.

Ask the tutor: Want to work through a higher-order auxiliary equation with mixed root types, or get your complementary function checked step by step?

Why does the guess y = e^(mx) work for every constant-coefficient linear ODE?

Because differentiating $e^{mx}$ just multiplies it by $m$ each time, so every derivative in the ODE becomes a power of $m$ times the same $e^{mx}$ factor. Substituting turns a differential equation into a plain polynomial equation in $m$ , since $e^{mx}$ cancels out (it is never zero). This trick works for any order and any constant coefficients — the auxiliary equation is always the same polynomial with $y^{(k)}$ replaced by $m^k$ .

Why does a repeated root need an extra factor of x instead of just doubling the constant?

A single root $m$ used twice, as in $y = C_1e^{mx} + C_2e^{mx}$ , is really just $(C_1+C_2)e^{mx}$ — one function, one effective constant, not enough to match the two conditions a second-order ODE requires. Direct substitution shows $y = xe^{mx}$ also solves the ODE exactly when $m$ is a repeated root, and it is linearly independent of $e^{mx}$ (their Wronskian is nonzero). So the correct pair is $e^{mx}$ and $xe^{mx}$ , giving $y_c = (C_1+C_2x)e^{mx}$ .

How do I know which of the three root cases applies without fully solving the quadratic?

Compute the discriminant $b^2-4ac$ first. If it is positive, expect two real distinct roots (Case 1); if it is exactly zero, expect one repeated real root (Case 2); if it is negative, expect a complex conjugate pair (Case 3). Checking the sign of the discriminant before grinding through the quadratic formula saves time and avoids the "lost sign under the square root" trap that silently turns a complex case into a wrong real-root answer.

Second-order linear ODEs with constant coefficients

The core idea

Derivation

Worked example

Common traps

Before this

Where this leads