A single inspired substitution, assume y = erx, converts a second-order differential equation into a quadratic. The roots of that quadratic determine everything about the solution.
To solve ay″ + by′ + cy = 0, we need a function whose second derivative, first derivative, and the function itself all combine to zero. Exponentials are the only functions with that property.
Consider what happens when you differentiate erx. Each differentiation multiplies by r:
Since erx is never zero, we can divide through by it. What remains is a purely algebraic equation:
The Characteristic Equation
ar² + br + c = 0
This is the characteristic equation (also called the auxiliary equation). Its roots, found by the quadratic formula, tell us exactly what the solution looks like. This is the entire method. One substitution. One quadratic. Three possible cases depending on the discriminant b² − 4ac.
What the discriminant tells you
The discriminant Δ = b² − 4ac from the quadratic formula controls everything. If Δ > 0, two distinct real roots. If Δ = 0, one repeated real root. If Δ < 0, a pair of complex conjugate roots. Each case gives a qualitatively different family of solutions.
§ 02The Three Cases
Solving ar² + br + c = 0 gives roots r = (−b ± √(b²−4ac)) / 2a. What those roots look like determines the form of the general solution.
Case 1 — Δ > 0
Two Distinct Real Roots
b² − 4ac > 0 → r1 ≠ r2, both real
y = C1er₁x + C2er₂x
Two independent exponential solutions. If both roots are negative, the solution decays to zero (overdamped). The constants C₁ and C₂ are fixed by initial conditions.
Case 2 — Δ = 0
One Repeated Root
b² − 4ac = 0 → r1 = r2 = r = −b/2a
y = (C1 + C2x)erx
When the quadratic has a double root, the second independent solution picks up a factor of x. This is the borderline between oscillatory and non-oscillatory behaviour — critical damping.
Case 3 — Δ < 0
Complex Conjugate Roots
b² − 4ac < 0 → r = α ± βi
y = eαx(C1cos βx + C2sin βx)
Complex roots always come in conjugate pairs α ± βi. By Euler's formula, the complex exponentials combine into real sinusoids multiplied by eαx — oscillations that grow, decay, or stay constant depending on the sign of α.
Why Does the Repeated Root Case Need an Extra x?
When the characteristic equation has a double root r, we have only one exponential solution erx. A second-order ODE requires two linearly independent solutions. A method called reduction of order shows that the second solution is xerx: the factor of x is not a guess; it falls out of the calculation. You can verify it works by substituting back: if y = xerx and r = −b/2a, then ay″ + by′ + cy = 0 holds identically.
Where Euler's Formula Connects
In Case 3, the roots are r = α ± βi where α = −b/2a and β = √(4ac−b²)/2a. The two exponential solutions are e(α+βi)x and e(α−βi)x. By Euler's formula eiθ = cos θ + i sin θ, their sum and difference produce the real-valued pair:
From complex to real
e(α+βi)x = eαx(cos βx + i sin βx)
e(α−βi)x = eαx(cos βx − i sin βx)
Real basis: eαxcos βx, eαxsin βx
Physical meaning of α and β
In a mechanical or electrical oscillator, β is the natural oscillation frequency (in rad/s or rad per unit of the independent variable), while α controls the amplitude envelope. If α < 0 the oscillations die out exponentially — this is underdamping. If α = 0 (no damping, so b = 0) the system oscillates forever at constant amplitude.
§ 03Interactive Characteristic Equation Solver
Enter the coefficients of your ODE. The solver computes the discriminant, finds the roots, and writes out the general solution.
Solve ay″ + by′ + cy = 0
_y″ + _y′ + _y = 0
§ 04Worked Examples
Ten fully solved problems, two or three from each case, escalating in complexity.
Example 1 · Case 1Two distinct real roots
Solve: y″ − 5y′ + 6y = 0
Write the characteristic equation
r² − 5r + 6 = 0
Factorise
(r − 2)(r − 3) = 0 → r = 2, r = 3
Discriminant check: Δ = 25 − 24 = 1 > 0 ✓
General solution
y = C1e2x + C2e3x
y = C₁e²ˣ + C₂e³ˣ
Example 2 · Case 1Distinct real roots with initial conditions
⚠ Forgetting the x factor in the repeated root case
When Δ = 0, students often write y = C₁erx + C₂erx = (C₁+C₂)erx — which is really only one constant, not two. The two independent solutions must be erx and xerx. Always check whether you actually have two free constants.
⚠ Using the wrong sign when reading off α and β
If the roots are r = −3 ± 5i, then α = −3 and β = 5. A common error is to write e3x instead of e−3x. The real part of the root goes directly into the exponent.
⚠ Mixing up the characteristic equation with the ODE
The characteristic equation ar² + br + c = 0 is solved for r, not for y. Write it down explicitly before solving — do not try to read off roots directly from the ODE in one step.
⚠ Applying this method to non-constant-coefficient ODEs
The characteristic equation only works when a, b, c are constants. For variable-coefficient equations like y″ + (sin x)y′ + y = 0, you need a completely different approach. Do not apply ar² + br + c = 0 if the coefficients depend on x.
§ 06Practice Quiz — 10 Questions
Write the characteristic equation, find the roots, identify the case, and state the general solution.
Score: 0 / 0
Question 1 — Concept
For the ODE y″ − 5y′ + 6y = 0, what is the characteristic equation?
Question 2 — Roots
For y″ + 5y′ + 4y = 0, find the roots. Enter the larger root.
r² + 5r + 4 = 0. Try factoring: what two numbers multiply to 4 and add to 5?
r² + 5r + 4 = (r+1)(r+4) = 0
r = −1 or r = −4
Larger root: r = −1
Question 3 — Case identification
The characteristic equation ar² + br + c = 0 has discriminant Δ = 0. What is the form of the general solution?
Question 4 — Discriminant
For the ODE 2y″ + 3y′ + 2y = 0, compute the discriminant b² − 4ac.
a = 2, b = 3, c = 2. Δ = b² − 4ac = ?
a = 2, b = 3, c = 2
Δ = 9 − 16 = −7
Since Δ < 0, this gives complex conjugate roots → underdamped oscillation.
Question 5 — Solution form
For y″ + 16y = 0, the roots are ±4i. What is the general solution? Type it in the form shown, enter the value of β only (the number multiplying x inside sin and cos).
Solve the IVP y″ + 2y′ + 5y = 0, y(0) = 0, y′(0) = 4. The particular solution has the form y = e^(αx)·A·sin(βx). What is A?
Roots: r = −1 ± 2i. General solution: y = e⁻ˣ(C₁cos 2x + C₂sin 2x). Apply y(0) = 0 first.
r² + 2r + 5 = 0 → r = (−2 ± √(−16))/2 = −1 ± 2i
y = e⁻ˣ(C₁cos 2x + C₂sin 2x)
y(0) = C₁ = 0 → C₁ = 0
y = C₂e⁻ˣsin 2x
y′ = C₂e⁻ˣ(2cos 2x − sin 2x)
y′(0) = 2C₂ = 4 → C₂ = 2
Solution: y = 2e⁻ˣsin 2x
§ 07What to Study Next
The characteristic equation solves homogeneous second-order ODEs: equations where the right-hand side is zero. The natural next step is non-homogeneous equations, where a driving term g(x) is present:
Second-Order Linear ODEs, the full treatment of all three damping cases with physical applications.
Undetermined Coefficients, guessing a particular solution when g(x) is polynomial, exponential, or sinusoidal.