Higher-order & linear differential equations
The idea
Second-order linear differential equations with constant coefficients — a y'' + b y' + c y = 0 — govern oscillating springs, RLC circuits, and damped vibrations, and they reward you with a complete, finite recipe. The key insight: exponential functions are eigenfunctions of differentiation, so guessing y = e^(rx) converts the differential equation into the characteristic polynomial a r² + b r + c = 0, an ordinary quadratic you have solved since high school.
The roots dictate the solution's personality. Two distinct real roots give a combination of two exponentials; a repeated root gives e^(rx) and x e^(rx); complex roots p ± qi give e^(px) times sines and cosines — oscillation with growing or decaying amplitude. Superposition then says the general solution is c₁y₁ + c₂y₂ for any two independent solutions, with two constants because two integrations are buried in a second derivative.
The matching misconception: one initial condition is not enough. A second-order equation needs both y(0) and y'(0) — position and velocity — to single out a trajectory, and the two constants are found by solving a small simultaneous system, not one at a time.
Worked example
Solve y'' − 5y' + 6y = 0 with initial conditions y(0) = 1 and y'(0) = 0.
- Substitute the exponential ansatz y = e^(rx): each derivative pulls down a factor of r, giving the characteristic equation r² − 5r + 6 = 0.
- Factor: (r − 2)(r − 3) = 0, so r = 2 and r = 3 — two distinct real roots, hence the general solution y = A e^(2x) + B e^(3x).
- Impose the first condition: y(0) = A + B = 1.
- Impose the second: y'(x) = 2A e^(2x) + 3B e^(3x), so y'(0) = 2A + 3B = 0.
- Solve the pair: from the first, A = 1 − B; substituting, 2(1 − B) + 3B = 2 + B = 0 gives B = −2 and A = 3.
- Verify: y = 3e^(2x) − 2e^(3x) has y(0) = 3 − 2 = 1 and y'(0) = 6 − 6 = 0, and plugging into the equation gives (12 − 30 + 18)e^(2x) + (−18 + 30 − 12)e^(3x) = 0 — everything cancels, as required.
Answer. The solution is y = 3e^(2x) − 2e^(3x).
Check your understanding
- Why does the guess y = e^(rx) turn a differential equation into an algebra problem?
- How do complex characteristic roots produce real oscillating solutions, and what controls the oscillation's decay?
- Why does a second-order equation need exactly two initial conditions, and what do they mean physically for a spring?
- What changes in the general solution when the characteristic equation has a repeated root, and why is a second independent solution needed at all?
Build the foundations first
Higher-order & linear differential equations builds on these concepts. If any feel shaky, start there.