Variation of Parameters

The idea

Suppose you have already solved the homogeneous version of the ODE:

y'' + P(x) y' + Q(x) y = 0

and have two linearly independent solutions y₁(x), y₂(x). The homogeneous general solution is c₁ y₁ + c₂ y₂ with constants c₁, c₂. Now add a forcing term to get the non-homogeneous problem

y'' + P(x) y' + Q(x) y = f(x)

and look for a particular solution. Lagrange's idea (1774): assume y_p has the same shape as the homogeneous solution, but with the constants promoted to functions of x:

y_p(x) = c₁(x) y₁(x) + c₂(x) y₂(x)

The two functions c₁(x), c₂(x) are the new unknowns. One linear ODE for one unknown function y_p has been replaced by one ODE for two unknown functions, which is under-determined — we need one extra constraint. A clever choice that simplifies the algebra is:

c'₁ y₁ + c'₂ y₂ = 0   (constraint 1)

With that constraint, you can compute y'_p without second derivatives of c_i: y'_p = c₁ y'₁ + c₂ y'₂. Differentiate once more, plug into the original ODE, simplify using Ly_i = 0, and you get

c'₁ y'₁ + c'₂ y'₂ = f(x)   (constraint 2)

Now you have a 2×2 linear system for c'₁, c'₂:

[y₁   y₂ ] [c'₁]   [ 0 ]
[y'₁  y'₂] [c'₂] = [f(x)]

The determinant of the coefficient matrix is the Wronskian W = y₁ y'₂ − y'₁ y₂. By Cramer's rule,

c'₁ = −y₂ f / W,   c'₂ = y₁ f / W

Integrate, plug back into y_p, done.

The final formula

For a second-order linear ODE y'' + P(x) y' + Q(x) y = f(x) with homogeneous solutions y₁, y₂ and Wronskian W = y₁ y'₂ − y'₁ y₂,

y_p(x) = −y₁(x) ∫ (y₂(x') f(x') / W(x')) dx' + y₂(x) ∫ (y₁(x') f(x') / W(x')) dx'

The general solution is y = c₁ y₁ + c₂ y₂ + y_p, with c₁, c₂ now arbitrary constants determined by initial conditions.

Important. The formula assumes the coefficient of y'' has been normalised to 1. If your ODE is a(x) y'' + b(x) y' + c(x) y = g(x), divide through by a(x) first so that f(x) = g(x)/a(x) is the correct right-hand side.

Worked example — y'' + y = sec(x)

Step 1: solve the homogeneous equation. y'' + y = 0 has characteristic polynomial r² + 1 = 0 with roots r = ±i, so y₁ = cos x, y₂ = sin x.

Step 2: compute the Wronskian.

W = cos x · cos x − (−sin x) · sin x = cos²x + sin²x = 1

So W = 1 — the Wronskian is constant.

Step 3: assemble the formula. With f(x) = sec x = 1/cos x,

y_p = −cos x · ∫ (sin x · sec x / 1) dx + sin x · ∫ (cos x · sec x / 1) dx

Simplify the integrands:

• sin x · sec x = sin x / cos x = tan x
• cos x · sec x = 1

Step 4: integrate. ∫ tan x dx = −ln|cos x|. ∫ 1 dx = x.

Step 5: substitute.

y_p = −cos x · (−ln|cos x|) + sin x · x = cos x · ln|cos x| + x sin x

General solution.

y = c₁ cos x + c₂ sin x + cos x · ln|cos x| + x sin x

Why undetermined coefficients failed. sec(x) is not a polynomial, exponential, sine, or cosine — the trial-form catalogue has no entry. Variation of parameters does not care about the structure of f — it just needs you to compute two integrals.

Derivation in detail

Start from the ansatz y_p = c₁ y₁ + c₂ y₂ with c_i functions. Differentiate:

y'_p = c'₁ y₁ + c'₂ y₂ + c₁ y'₁ + c₂ y'₂

The first two terms are the awkward ones. Impose the constraint c'₁ y₁ + c'₂ y₂ = 0 — a free choice we make to keep the algebra tidy. Then

y'_p = c₁ y'₁ + c₂ y'₂

Differentiate again:

y''_p = c'₁ y'₁ + c'₂ y'₂ + c₁ y''₁ + c₂ y''₂

Plug y_p, y'_p, y''_p into y'' + P y' + Q y = f:

[c'₁ y'₁ + c'₂ y'₂] + [c₁ (y''₁ + P y'₁ + Q y₁)] + [c₂ (y''₂ + P y'₂ + Q y₂)] = f

The bracketed expressions on the right of each c_i are zero because y_i satisfies the homogeneous equation. What survives is

c'₁ y'₁ + c'₂ y'₂ = f

which is constraint 2. Together with constraint 1 we have the 2×2 linear system. Determinant is W; Cramer's rule gives the c'_i. Integration produces c_i, and y_p follows.

Higher-order and matrix-form variants

• n-th order. For an n-th order linear ODE with homogeneous basis y₁, …, y_n and Wronskian W, the formula generalises: y_p = Σ y_i ∫ (W_i f / W) dx, where W_i is the n×n Wronskian determinant with the i-th column replaced by (0, …, 0, 1)^T.
• Variable-coefficient form. The technique requires P(x), Q(x) to be continuous and y_i to be known — it does not give you the homogeneous solutions; you must already have them by characteristic roots, series methods, or known special functions.
• Matrix / system form. For a first-order system y' = A(x) y + f(x) with fundamental matrix Φ(x), the particular solution is y_p = Φ(x) ∫ Φ⁻¹(x') f(x') dx'. The 2×2 Wronskian system is the scalar disguise of this matrix formula.
• Green's function. Variation of parameters can be repackaged as y_p(x) = ∫ G(x, x') f(x') dx', where G(x, x') is the Green's function of the operator L. Same formula, different bookkeeping.
• Duhamel's principle. The PDE analogue: for an inhomogeneous linear evolution equation, integrate the homogeneous propagator against the forcing — again, the same structural idea generalised to PDE.

Variation of parameters vs alternatives

Technique	Form of f(x) handled	Cost	Limitation
Undetermined coefficients	Polynomials, exponentials, sines, cosines, products thereof	Solve a small linear system in constants	Fails for f like sec(x), tan(x), 1/x, ln(x)
Variation of parameters	Any continuous f	Two extra integrals per particular solution	Need homogeneous basis y₁, y₂ first
Green's function	Any continuous f	Integrate against G(x, x')	Need G — once-and-done for a given operator
Laplace transform	Most f with simple transforms	Algebra in s-domain	Constant-coefficient ODEs only
Numerical (RK4)	Any f, including no closed form	O(N) steps for N grid points	No symbolic answer
Series solution	Analytic f	Tedious by hand	Convergence issues outside a disk

Common pitfalls

• Forgetting to normalise the leading coefficient. The formula assumes y'' + P y' + Q y = f. If your ODE has a(x) y'' on the left, divide through by a(x) first — otherwise f in the formula is wrong by a factor of a(x).
• Picking the wrong sign on c'₁. By Cramer's rule c'₁ = −y₂ f / W (with the minus sign) and c'₂ = +y₁ f / W. Mixing them up makes the algebra not just wrong but plausible-looking.
• Mis-computing the Wronskian. W = y₁ y'₂ − y'₁ y₂ (note the order). Sign-flipping gives the negative Wronskian, and your final answer flips sign.
• Using c_i(x) with the integration constant of integration. The constants of integration in ∫ c'_i dx are absorbed into the c₁, c₂ of the homogeneous part. You can drop them in the particular solution.
• Skipping linear independence of y₁, y₂. If the homogeneous basis is dependent, the Wronskian is zero and Cramer's rule fails. Always confirm W ≠ 0 on the interval of interest.
• Reaching for variation of parameters when undetermined coefficients works. The latter is faster because you avoid integrals. Reserve variation of parameters for forcings that defeat the trial-form catalogue.