Differential Equations · step 9 of 14

Variation of parameters

Variation of parameters handles ANY continuous forcing term, including those undetermined coefficients cannot touch. Given homogeneous solutions y₁ and y₂, the particular solution is y_p = −y₁∫(y₂g/W)dx + y₂∫(y₁g/W)dx, where W is the Wronskian y₁y₂′ − y₂y₁′ and g is the forcing in standard form.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Get the homogeneous solutions y₁ and y₂Usually via the characteristic equation.
Put the equation in standard formCoefficient 1 on y″. The forcing g(x) is what remains on the right — a step that is easy to skip and fatal to omit.
Compute the WronskianW = y₁y₂′ − y₂y₁′. It is never zero for independent solutions.
Apply the formula and integratey_p = −y₁∫(y₂g/W)dx + y₂∫(y₁g/W)dx, dropping the constants (they belong to the complementary function).

Worked example

Find a particular solution of y″ + y = sec x.

Homogeneous solutions: y₁ = cos x, y₂ = sin x. Wronskian W = cos²x + sin²x = 1.
g = sec x, already in standard form.
First integral: ∫(y₂g/W)dx = ∫sin x sec x dx = ∫tan x dx = −ln|cos x|.
Second: ∫(y₁g/W)dx = ∫cos x sec x dx = ∫1 dx = x.

Answer. y_p = cos x·ln|cos x| + x sin x — a case undetermined coefficients cannot handle at all.

Where marks get dropped

These are the specific errors that cost credit on variation of parameters questions — QED's rubric penalises each of them separately.

Forgetting to normalise so y″ has coefficient 1. Using the raw right side as g gives an answer wrong by that factor.
Dropping the minus sign on the first term of the formula.
Including constants of integration in the two integrals, which just re-adds the complementary function.

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Variation of parameters — frequently asked questions

When should I use this over undetermined coefficients?

Whenever the forcing is not a polynomial, exponential or sinusoid — sec x, tan x, ln x, 1/x. For the nice cases undetermined coefficients is faster.

What does the Wronskian tell me?

Whether the two solutions are independent: W ≠ 0 confirms it. It also appears in the denominator of the formula, which is why independence is required.

Does it extend to higher order?

Yes, with a determinant formulation using n homogeneous solutions. The algebra grows quickly but the structure is identical.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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