Linear second-order DEs ay'' + by' + cy = 0. Solve the characteristic equation am² + bm + c = 0 — three cases by discriminant.
For y″ − 5y′ + 6y = 0, the characteristic equation m² − 5m + 6 = 0 has roots m = 2 and m = 3. What's the general solution?
The discriminant picks the case
Turn ay″ + by′ + cy = 0 into am² + bm + c = 0 and look at b² − 4ac: positive ⇒ two real roots (overdamped), zero ⇒ one repeated root (critically damped), negative ⇒ complex roots (oscillation, possibly decaying). The same b² − 4ac you met in the quadratic formula now controls how a spring or circuit behaves.
Solve y″ + 4y = 0.
m² + 4 = 0 ⇒ m = ±2i (α = 0, β = 2). General solution: y = C₁cos 2x + C₂sin 2x — pure undamped oscillation, period π.
stiffer spring or lighter mass ⇒ faster bounce: ω = √(k/m) = 3.46 rad/s, period T = 2π√(m/k) = 1.81 s. The red arrow is the restoring force F = −kx — always pointing back toward x = 0.
Classify the roots of y″ − 4y′ + 4y = 0 and write the solution.
Three cases
- Two real roots m₁, m₂ → y = C₁e^(m₁x) + C₂e^(m₂x).
- Repeated root m → y = (C₁ + C₂x)e^(mx).
- Complex roots α ± βi → y = e^(αx)(C₁ cos βx + C₂ sin βx).