Variation of Parameters

Undetermined coefficients is fast, but it lives on a short list of forcing terms — polynomials, exponentials, sines and cosines, and their products. Feed it something off the menu — \sec x, \tan x, \ln x, 1/x — and there is simply no form to guess. The method has nothing to substitute.

Variation of parameters is the method that never runs out. Given the two homogeneous solutions y_1 and y_2, it writes down a particular solution of

y'' + p(x)\,y' + q(x)\,y = g(x)

for any continuous g(x) — no guessing, no lucky list. The price is a couple of integrals, which may or may not be pretty; the guarantee is that the recipe always gives you those integrals.

The idea: let the constants vary

The homogeneous solution is y_h = C_1 y_1 + C_2 y_2, with C_1, C_2 constant. Lagrange's trick — the one that names the method — is to promote the constants to functions and look for a particular solution of the same shape:

y_p = u_1(x)\,y_1 + u_2(x)\,y_2.

We have two unknown functions but only one equation to satisfy, so we get to impose one extra condition of our own choosing. Differentiate once:

y_p' = \big(u_1' y_1 + u_2' y_2\big) + \big(u_1 y_1' + u_2 y_2'\big).

The clever choice is to kill the first bracket — the imposed condition:

Require u_1' y_1 + u_2' y_2 = 0. This keeps y_p' free of the second derivatives of u_1, u_2, so the algebra stays linear and first-order in the unknowns.

With that bracket gone, y_p' = u_1 y_1' + u_2 y_2'. Differentiate again and substitute y_p, y_p', y_p'' into the ODE. Every term containing only u_1, u_2 collapses, because y_1 and y_2 each solve the homogeneous equation. What survives is a second, clean equation:

u_1' y_1' + u_2' y_2' = g(x).

Now stack the two conditions as a linear system in u_1' and u_2':

\begin{cases} u_1' y_1 + u_2' y_2 = 0, \\[2pt] u_1' y_1' + u_2' y_2' = g(x). \end{cases}

The Wronskian solves it

The coefficient matrix of that system is exactly the pair y_1, y_2 and their derivatives, and its determinant is the Wronskian of the homogeneous solutions:

W = W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1'.

Because y_1, y_2 are linearly independent solutions, W \neq 0 everywhere, so Cramer's rule cracks the system straight open:

\displaystyle u_1' = \frac{-\,y_2\, g}{W}, so \displaystyle u_1 = -\int \frac{y_2\, g}{W}\,dx.
\displaystyle u_2' = \frac{\;y_1\, g}{W}, so \displaystyle u_2 = \int \frac{y_1\, g}{W}\,dx.
The particular solution is y_p = u_1 y_1 + u_2 y_2, and the general solution is y = C_1 y_1 + C_2 y_2 + y_p.

That is the whole method. Notice the shape of it: you need nothing about the forcing g except that you can integrate y_2 g / W and y_1 g / W. No form to match, no resonance special-case — a term that overlaps y_h just gets absorbed into C_1, C_2 automatically.

Undetermined coefficients is a pattern-matcher: it only works because a polynomial forcing has a polynomial y_p, an exponential has an exponential, and so on — the derivative keeps you inside a small, closed family. But \dfrac{d}{dx}\sec x = \sec x \tan x, and differentiating again spawns \sec^3 x: the family never closes, so there is no finite trial form to write down. Variation of parameters sidesteps the whole question — it does not care what family g belongs to, because it never differentiates g. It integrates it. That is why it is the general method and undetermined coefficients is the shortcut.

Worked example 1: y'' + y = \sec x

Here g(x) = \sec x is nowhere on the undetermined-coefficients menu, so variation of parameters is the only game in town.

Step 1 — homogeneous solutions. The characteristic equation r^2 + 1 = 0 gives r = \pm i, so

y_1 = \cos x, \qquad y_2 = \sin x.

Step 2 — the Wronskian.

W = y_1 y_2' - y_2 y_1' = \cos x\cos x - \sin x(-\sin x) = \cos^2 x + \sin^2 x = 1.

A Wronskian of 1 — the integrals will be as clean as they get.

Step 3 — the two derivatives. With g = \sec x = 1/\cos x,

u_1' = \frac{-y_2\,g}{W} = -\sin x \cdot \frac{1}{\cos x} = -\tan x, \qquad u_2' = \frac{y_1\,g}{W} = \cos x \cdot \frac{1}{\cos x} = 1.

Step 4 — integrate.

u_1 = -\int \tan x\,dx = \ln|\cos x|, \qquad u_2 = \int 1\,dx = x.

(We drop the integration constants — they only re-create C_1 y_1 + C_2 y_2.)

Step 5 — assemble y_p = u_1 y_1 + u_2 y_2.

y_p = \cos x\,\ln|\cos x| + x\,\sin x.

So the general solution is

y = C_1 \cos x + C_2 \sin x + \cos x\,\ln|\cos x| + x\,\sin x.

Try producing that with a lucky guess — the \ln|\cos x| could never come from a finite trial form.

Worked example 2: a sanity check on y'' - y = e^{x}

This forcing is on the undetermined-coefficients menu, so we can grade our machine against a method we trust. (Watch the resonance: e^x already solves the homogeneous equation.)

Step 1 — homogeneous solutions. r^2 - 1 = 0 \Rightarrow r = \pm 1, so y_1 = e^{x}, y_2 = e^{-x}.

Step 2 — Wronskian.

W = e^{x}(-e^{-x}) - e^{-x}(e^{x}) = -1 - 1 = -2.

Step 3 — derivatives with g = e^{x}:

u_1' = \frac{-y_2 g}{W} = \frac{-e^{-x}e^{x}}{-2} = \tfrac12, \qquad u_2' = \frac{y_1 g}{W} = \frac{e^{x}e^{x}}{-2} = -\tfrac12 e^{2x}.

Step 4 — integrate.

u_1 = \tfrac12 x, \qquad u_2 = -\tfrac14 e^{2x}.

Step 5 — assemble.

y_p = \tfrac12 x\,e^{x} - \tfrac14 e^{2x}\,e^{-x} = \tfrac12 x\,e^{x} - \tfrac14 e^{x}.

The -\tfrac14 e^{x} is just a multiple of y_1 — absorb it into C_1 — leaving the essential particular solution y_p = \tfrac12 x\,e^{x}. That is exactly the x\,e^{x} resonance form undetermined coefficients would force by hand. The two methods agree, and variation of parameters found the x-factor on its own.

Three traps sink most attempts:

Standard form first. The formulas assume the coefficient of y'' is 1. For 3y'' + \dots = h(x) you must divide through, so the g in u_i' = \pm y_j g / W is h(x)/3, not h(x). Read g off the standardised equation.
Wronskian of the homogeneous solutions. W = y_1 y_2' - y_2 y_1' uses y_1, y_2 — the solutions of the homogeneous equation, not anything involving g.
Don't forget the imposed condition. The whole derivation rests on choosing u_1' y_1 + u_2' y_2 = 0. Skip it and you have one equation for two unknowns and the second-derivative terms never cancel — the tidy formulas simply do not follow.

See the answer take shape

This is the full solution of y'' + y = \sec x on (-\tfrac{\pi}{2}, \tfrac{\pi}{2}), where \cos x > 0. The faint curve is the particular part y_p = \cos x\,\ln(\cos x) + x\sin x alone; the bold curve is the general solution y = C_1\cos x + C_2\sin x + y_p. Slide C_1 and C_2 — the homogeneous constants slide the answer around, while the particular part carries the response to the forcing. Notice y_p diving toward -\infty near the edges: that is \ln(\cos x) as \cos x \to 0.