Second-Order Linear Differential Equations
Solve a·y″ + b·y′ + c·y = g(x) with constant coefficients, step by step: the characteristic equation, the homogeneous solution, a particular solution by undetermined coefficients, and the starting values. The model of every spring, pendulum and circuit that oscillates.
Why erx?
The derivative of erx is r·erx: the same function, times a number. Put y = erx into a·y″ + b·y′ + c·y = 0 and every term has the factor erx, which leaves an ordinary quadratic equation, a·r² + b·r + c = 0. Its two roots give two building blocks; every solution is a mix of them.
Complex roots α ± βi mean oscillation: eαx cos βx and eαx sin βx. For a mass on a spring, m·y″ + d·y′ + k·y = 0: little damping gives a dying wobble, a lot of damping a slow creep back, and just enough (critical damping) the fastest return without overshoot.
The Solution, Step by Step
rootsStep-by-Step Solution
The Three Cases
The second building block is x·erx, because erx alone would only give one.
A particular solution has the shape of g(x); if that shape already solves the homogeneous equation (resonance), multiply the guess by x.
Sources & Verification
- W. E. Boyce & R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems, 10th ed., Wiley, 2012 — Ch. 3 (Second-Order Linear Equations: 3.1, 3.3, 3.4 and 3.5 undetermined coefficients; 3.7 mechanical vibrations).
- E. Kreyszig, Advanced Engineering Mathematics, 10th ed., Wiley, 2011 — Ch. 2 (Second-Order Linear ODEs).
Frequently Asked Questions
How do you solve a·y″ + b·y′ + c·y = 0?
Try y = e^(rx): this gives the characteristic equation a·r² + b·r + c = 0. Its two roots give the solution. For y″ + 3y′ + 2y = 0 the roots are −1 and −2, so y = C₁·e^(−x) + C₂·e^(−2x). With y(0) = 1 and y′(0) = 0 the constants are C₁ = 2 and C₂ = −1.
What if the characteristic equation has complex roots?
Roots α ± βi give y = e^(αx)·(C₁·cos βx + C₂·sin βx): an oscillation that dies out (α < 0), keeps going (α = 0) or grows (α > 0). For a damped spring y″ + 0.4y′ + 4y = 0 the roots are −0.2 ± 1.99i: it wobbles with a period of about 3.16 and slowly comes to rest.
What if the characteristic equation has a double root?
Then e^(rx) gives only one solution, and the second is x·e^(rx). For y″ − 2y′ + y = 0 the double root is r = 1, so y = (C₁ + C₂·x)·e^x.
What is the method of undetermined coefficients?
A way to find one particular solution when the right side g(x) is a polynomial, an exponential or a sine/cosine: guess a function of the same shape with unknown coefficients, put it in the equation and compare. For y″ − y′ − 2y = 4x², guessing Ax² + Bx + C gives A = −2, B = 2, C = −3. The full solution is the homogeneous solution plus this particular one.
What is resonance in a differential equation?
When the right side has the same shape as a solution of the homogeneous equation, for example y″ + 4y = sin(2x), where sin(2x) already solves y″ + 4y = 0. The usual guess fails and must be multiplied by x: y_p = −x·cos(2x)/4. Without damping the swing then grows without limit, which is how a singer can break a glass.
What do underdamped, critically damped and overdamped mean?
For a mass on a spring with damping, m·y″ + d·y′ + k·y = 0. If d² < 4mk it is underdamped and oscillates while dying out; if d² = 4mk it is critically damped and returns fastest without overshooting; if d² > 4mk it is overdamped and creeps back slowly. Car shock absorbers are designed close to critical damping.