Higher-Order Linear ODEs with Constant Coefficients

A higher-order linear ODE with constant coefficients looks like ay″ + by′ + cy = 0, or with more derivatives, where a, b, c are just numbers. The constant coefficients make these equations completely solvable: every solution is built from exponentials, sines, cosines, and polynomials, dictated by a single algebraic equation.

These equations matter because they model every vibrating spring, every RLC circuit, and every small oscillation in engineering. The method is algorithmic: write the characteristic equation, find its roots, and translate each root into solution pieces.

Key Concepts

• Characteristic equation. For ay″ + by′ + cy = 0, it is ar² + br + c = 0; for order n, an r^n + … + a0 = 0.
• Distinct real roots. Each root r gives a solution e^{rx}; distinct roots give independent pieces.
• Repeated roots. A root r of multiplicity m contributes e^{rx}, xe^{rx}, …, x^{m−1}e^{rx}.
• Complex roots. A pair α ± iβ gives e^{αx}cos(βx) and e^{αx}sin(βx).
• Homogeneous vs nonhomogeneous. This article covers the homogeneous case (right side zero); nonhomogeneous equations add a particular solution on top.

Formulas

• Trial solution: assume y = e^{rx}; then y^{(k)} = r^k e^{rx}, turning the ODE into the characteristic polynomial equation.
• General solution: y = C1y1 + … + Cnyn, a linear combination of n independent pieces, one per root counting multiplicity.
• Complex pair: from r = α ± iβ, use Euler's formula to get the real pair e^{αx}cos βx, e^{αx}sin βx.
• Assumptions and limits: coefficients must be constant and the leading coefficient nonzero; the method as stated solves homogeneous equations only.

Worked Example

Solve y‴ − 3y″ + 3y′ − y = 0.

1. Assume y = e^{rx} and substitute: r³e^{rx} − 3r²e^{rx} + 3re^{rx} − e^{rx} = 0, so the characteristic equation is r³ − 3r² + 3r − 1 = 0.
2. Recognize the pattern: r³ − 3r² + 3r − 1 = (r − 1)³, the binomial expansion of (r − 1)³.
3. Root: r = 1 with multiplicity 3 (a triple root).
4. Translate the triple root: the three independent pieces are e^x, xe^x, and x²e^x.
5. General solution: y = C1e^x + C2xe^x + C3x²e^x = (C1 + C2x + C3x²)e^x.
6. Spot-check the first piece: for y = e^x, y′ = y″ = y‴ = e^x, so y‴ − 3y″ + 3y′ − y = e^x − 3e^x + 3e^x − e^x = 0. Correct.

Common Misconceptions

• Using e^{rx} once per repeated root. A root of multiplicity m needs the polynomial factors 1, x, …, x^{m−1} times e^{rx}; without them you have fewer than n independent solutions.
• Leaving complex exponentials in the answer. For real problems, convert α ± iβ into the real pair e^{αx}cos βx, e^{αx}sin βx using Euler's formula.
• Applying the method to variable coefficients. If the coefficients depend on x, the e^{rx} trial fails; equations like Cauchy–Euler need their own method.

Real-World Applications

• Spring–mass–damper systems: my″ + cy′ + ky = 0 classifies motion as overdamped, critically damped, or underdamped from the characteristic roots.
• RLC circuits: the charge on the capacitor obeys LQ″ + RQ′ + Q/C = 0, with the same root analysis.
• Control systems: the poles of a system (characteristic roots) determine its stability: negative real parts mean decay to equilibrium.

Constant-coefficient homogeneous ODEs are solved by the characteristic equation: each root r contributes e^{rx}, repeated roots bring polynomial factors, and complex pairs become damped oscillations. The triple root (r − 1)³ giving (C1 + C2x + C3x²)e^x shows the whole method in one line.