Questions: Variation of Parameters Method

Question 1

Why does variation of parameters introduce functions u₁(x) and u₂(x) rather than directly guessing the form of the particular solution?

• A. Because u₁ and u₂ are always easier to compute than a direct guess
• B. To avoid using the Wronskian, which can be difficult to compute
• C. Because undetermined coefficients requires f(x) to have a specific form, but variation of parameters makes no assumption about f(x) — it works for any continuous right-hand side
• D. Because u₁ and u₂ represent the homogeneous solution, not the particular solution

Answer: (C)

The whole point of variation of parameters is generality. Undetermined coefficients works by guessing the form of the particular solution — which requires f(x) to be a polynomial, exponential, sine, cosine, or combination thereof. If f(x) = sec(x), ln(x), or any function that doesn't generate a finite family of derivatives, undetermined coefficients fails. Variation of parameters sidesteps this by assuming only that y_p = u₁y₁ + u₂y₂ and then solving for u₁ and u₂ through integration — no assumption about the form of f(x) is required.

Question 2

A student is solving y'' + y = sec(x). They try undetermined coefficients but get stuck. Which statement correctly explains why variation of parameters succeeds where undetermined coefficients fails?

• A. Variation of parameters only works for constant-coefficient equations, which is why it handles this case
• B. sec(x) and its derivatives form an infinite, non-repeating family (sec, sec·tan, sec³+sec·tan², ...) so no finite trial solution exists; variation of parameters integrates u₁' and u₂' directly without needing to guess the form
• C. sec(x) is not continuous, so undetermined coefficients fails on continuity grounds
• D. Undetermined coefficients requires a nonhomogeneous term, which sec(x) is not

Answer: (B)

Undetermined coefficients requires f(x) to generate a finite family of linearly independent derivatives so you can build a trial solution from them. sec(x) does not: each successive derivative introduces new functions (sec·tan, sec³, ...), so no finite trial solution exists. Variation of parameters avoids this entirely — it only requires integrating u₁' = −y₂f/W and u₂' = y₁f/W, and ∫sec(x)·cos(x)/W dx simplifies to something integrable. The method's lack of restriction on f(x) is precisely its advantage.

Question 3

The simplifying constraint u₁'y₁ + u₂'y₂ = 0 imposed in variation of parameters is an arbitrary choice that could be replaced by any other condition — it is just one of many equally valid approaches.

• T. True
• F. False

Answer: False

False — while the constraint is a choice (not forced by the algebra), it is far from arbitrary: it is the natural choice that simultaneously eliminates the highest-derivative cross-terms from the substitution and produces the cleanest possible 2×2 system in u₁' and u₂'. Without this constraint, the algebra does not reduce to a solvable system of two equations. Other constraints would produce more complex and less tractable systems. The constraint is conventional but deeply motivated.

Question 4

The Wronskian W = y₁y₂' − y₂y₁' must be nonzero for variation of parameters to succeed, because a zero Wronskian means y₁ and y₂ are linearly dependent and do not form a fundamental solution set.

• T. True
• F. False

Answer: True

True. The Wronskian appears in the denominator of the formulas u₁' = −y₂f/W and u₂' = y₁f/W. If W = 0, division by zero means the method breaks down. Crucially, W = 0 precisely when y₁ and y₂ are linearly dependent. Linearly dependent solutions do not span the solution space of the homogeneous equation, so the foundation of the method — using y₁ and y₂ as a basis — fails. This is why you need a fundamental solution set before starting.

Question 5

Explain the role that 'promoting constants to functions' plays in the logic of variation of parameters. Why does the method begin with the homogeneous solution's structure?

Model answer: The homogeneous solution y_h = c₁y₁ + c₂y₂ spans all solutions when f(x) = 0. Variation of parameters asks: what if the constants c₁ and c₂ were actually functions of x? This transforms the homogeneous structure into a flexible ansatz that can absorb the effect of f(x). By starting with y₁ and y₂ — functions already guaranteed to satisfy the homogeneous equation — the method reduces the problem to finding two scalar functions u₁ and u₂ via integration. The homogeneous solutions do the structural work; u₁ and u₂ provide the degrees of freedom needed to match f(x).

Instead of guessing a new function, you recycle the homogeneous solution's building blocks and 'vary the parameters' (the constants). Because y₁ and y₂ already satisfy the homogeneous ODE, substituting u₁y₁ + u₂y₂ into the full ODE with the simplifying constraint produces a system involving only u₁' and u₂' — no second derivatives of the unknown functions appear. This keeps the system algebraically tractable and reduces a second-order problem to two first-order integrations.