Consider y'' - 4y = e^{2x}. The homogeneous solution is y_h = C₁e^{2x} + C₂e^{-2x}. A student guesses y_p = Ae^{2x} and substitutes into the equation. What will happen?
AThe substitution will work and give A = 1/4
BThe left side will reduce to 0, making it impossible to solve for A
CThe substitution gives A = -1/4, which is the correct particular solution
DThe method fails entirely because the right side is exponential
Since e^{2x} is already a term in y_h (the solution to the homogeneous equation), substituting Ae^{2x} into y'' - 4y gives A(4e^{2x}) - 4(Ae^{2x}) = 0 — the guess annihilates itself and can never equal e^{2x}. This is the resonance case: the standard guess is a solution to the homogeneous equation and always produces zero on the left side. The fix is the modification rule: use y_p = Axe^{2x} instead, which does not duplicate y_h.
Question 2 Multiple Choice
For the equation y'' + y = sin(x), a student correctly identifies that the forcing function is sin(x) and guesses y_p = A sin(x). What is wrong with this guess?
ANothing — A sin(x) is the correct form for this equation
BThe guess should be A cos(x) instead, since sin differentiates to cos
CThe guess should include both terms: A cos(x) + B sin(x), because differentiating introduces the other trig function
DSine forcing functions require a polynomial guess, not a trigonometric one
Whenever f(x) involves sin or cos, the correct guess always includes BOTH A cos(βx) and B sin(βx), even if only one appears in f(x). This is because differentiating sin introduces cos and vice versa: (A cos + B sin)'' = -A cos - B sin, which after substitution can match sin on the right side by choosing A and B independently. A guess of A sin(x) alone would require the cos term to be zero, but the equation has a cos component in its derivatives that forces a nonzero A cos term. Note also that for this particular equation, sin(x) is in y_h (since the characteristic roots are ±i), so the full guess Ax cos(x) + Bx sin(x) is needed.
Question 3 True / False
When f(x) in y'' + py' + qy = f(x) is a polynomial of degree n, the correct guess for y_p is a polynomial of degree n with all terms from degree 0 through n included.
TTrue
FFalse
Answer: True
The correct guess for a polynomial forcing function is a full polynomial of the same degree: y_p = Aₙxⁿ + Aₙ₋₁xⁿ⁻¹ + ... + A₁x + A₀. All terms must be included because after substitution, lower-degree terms from y_p'' and y_p' can generate contributions at any degree below n, and the coefficients at each degree must be matched. Truncating the guess (e.g., guessing only Axⁿ) would fail to account for these lower-degree contributions.
Question 4 True / False
The method of undetermined coefficients can be applied to any continuous function f(x), not just polynomials, exponentials, and sinusoids.
TTrue
FFalse
Answer: False
The method only works for functions whose derivatives eventually cycle back to the same family of functions: polynomials (derivatives are polynomials), exponentials (derivatives are multiples of the same exponential), and sines/cosines (derivatives cycle between them). For functions like f(x) = ln(x), tan(x), or 1/x, differentiation produces new functional forms that cannot be matched by a finite-term guess. For these forcing functions, variation of parameters is needed instead.
Question 5 Short Answer
Why does the modification rule (multiplying the initial guess by x) fix the resonance problem in the method of undetermined coefficients?
Think about your answer, then reveal below.
Model answer: When the initial guess y_p duplicates a term already in y_h, substituting it into the left side of the ODE produces zero (since y_h satisfies the homogeneous equation). Multiplying by x produces a new function — e.g., Axe^{2x} instead of Ae^{2x} — that is no longer in y_h, so it does not annihilate when substituted. Differentiation of xe^{αx} produces xe^{αx} and e^{αx} terms, the latter of which can now match the non-zero right side after collecting coefficients and solving for A.
The resonance situation is structurally identical to the repeated-root case in the homogeneous equation: when a root appears twice, the second solution is x times the first (e.g., xe^{rx} alongside e^{rx}). The modification rule imports this same logic to the particular solution — if the natural frequency of the forcing matches a natural frequency of the system, the response grows linearly in x (or x² for double resonance). This connection between algebraic structure and solution behavior is the deep insight linking the two methods.