Method of Undetermined Coefficients

Core Idea

To solve y'' + py' + qy = f(x), find the homogeneous solution y_h, then guess the form of a particular solution y_p based on f(x). For f polynomial, exponential, sine, or cosine, use corresponding y_p forms with unknown coefficients. Substitute into the equation and solve for these coefficients. The general solution is y = y_h + y_p. This method is efficient when applicable.

Explainer

You've already solved y″ + py′ + qy = 0 using the characteristic equation. The roots r₁, r₂ give the homogeneous solution y_h — for real distinct roots, y_h = C₁e^r₁x + C₂e^r₂x. Now add a forcing function f(x) on the right side. The equation is no longer asking "what decays to zero?" but "what produces exactly f(x) when differentiated and combined?" You need a particular solution y_p that satisfies the full equation, then combine: the general solution is y = y_h + y_p.

The insight behind undetermined coefficients is that differentiation preserves certain functional forms. Derivatives of polynomials are polynomials; derivatives of e^αx are multiples of e^αx; derivatives of sin(βx) and cos(βx) cycle back to sines and cosines. So if f(x) is built from these forms, a y_p of the same form has a chance of working. The strategy: guess the form of y_p, substitute into the ODE, and solve for the unknown coefficients by matching both sides.

The guessing rules: if f(x) = xⁿ (polynomial of degree n), try y_p = Aₙxⁿ + ··· + A₁x + A₀ (a full polynomial of degree n). If f(x) = e^αx, try y_p = Ae^αx. If f(x) = sin(βx) or cos(βx), always try y_p = A cos(βx) + B sin(βx) together — even if only sine or cosine appears in f, both terms are needed because differentiating introduces the other. Products combine: f(x) = x²e^3x calls for y_p = (Ax² + Bx + C)e^3x.

The critical exception is the modification rule (also called the resonance case). If your initial guess for y_p would duplicate a term already present in y_h, that guess will produce zero when substituted into the left side of the homogeneous part and can never match f(x). The fix: multiply the guess by x. If the duplication persists, multiply by x². For example, if y_h includes e^2x and f(x) = e^2x, the usual guess Ae^2x fails — use Axe^2x instead. This modification is analogous to the repeated-root adjustment in the characteristic equation method, and understanding why it's needed connects the algebra directly to the structure of the solution space.