4.6.14 · D5Ordinary Differential Equations
Question bank — Non-homogeneous — method of undetermined coefficients (annihilator method)
Before the questions, one anchor you will reuse everywhere:
Recall What is the single idea that makes the whole method work?
Every "nice" forcing term — polynomial, exponential, sine/cosine, and their products — is itself the solution of some constant-coefficient homogeneous ODE, so it has a killer operator with . Reading backwards to its characteristic root gives you ; applying turns into , a homogeneous problem you already know how to solve.
True or false — justify
Say true or false AND the reason — the reason is the whole point.
annihilates because , and .
True — differentiating gives back times the function, so subtracting times the function inside the operator cancels it exactly.
annihilates but not .
False — both and come from the same root pair , so kills both; the operator cannot tell them apart.
The annihilator of a function is unique.
False — any multiple of an annihilator (e.g. ) also annihilates ; the annihilator means the lowest-order one, which is unique.
If (a constant), its annihilator is .
True — a constant is , born from the root once, so (which is ) sends it to zero since .
has an annihilator, so the annihilator method applies.
False — is not a solution of any constant-coefficient homogeneous ODE (it has no finite-order killer), so no annihilator exists; you must use Variation of parameters instead.
For you may annihilate and separately and add the trial forms.
The trial for is .
False (in general) — comes from root doubled, whose annihilator is ; that produces both and , so the trial is (before any overlap adjustment).
If none of 's roots overlap the roots of , you never need an extra factor of .
True — the extra only appears when annihilating raises the multiplicity of a root already in ; no overlap means no multiplicity bump, so no .
annihilates every straight line .
True — and differentiating again gives ; the root repeated twice produces exactly .
Spot the error
Each line has a flawed claim — say what is wrong and the correct statement.
" is already in , so I'll still guess ."
Wrong — since solves , we get , which can never equal a nonzero ; multiply by (root now doubled) to get .
"For I only include because there's no sine on the right."
Wrong — the operator mixes sine and cosine when it differentiates , so you must include both: ; one may turn out zero, but you cannot assume it.
"The annihilator of is ."
Wrong — needs root repeated four times (to cover ), so the annihilator is ; the power is , not .
"After forming I keep the whole general solution as ."
Wrong — the roots of inside regenerate , which satisfies and contributes nothing to ; keep only the new terms from 's roots as the form of .
", so annihilator is ."
Wrong — you cannot add annihilators; the product comes from roots , giving annihilator , a single quadratic factor.
"Since makes the right side zero, means ."
Wrong — means lies in the kernel of (it's a solution of that homogeneous ODE), not that is the zero function; maps a nonzero to zero.
"For (resonance), the trial is ."
Wrong — has root , which is already in ; the overlap forces , the hallmark of Resonance in forced oscillations.
"I found , so the full answer is just ."
Wrong — the general solution is ; dropping discards the arbitrary constants needed to satisfy initial/boundary conditions.
Why questions
Answer with the mechanism, not the rule name.
Why do annihilators only exist for polynomials, exponentials, and sinusoids (and their products)?
Because these are precisely the functions that solve constant-coefficient homogeneous ODEs — each is generated by a characteristic root — so reading them backwards yields a finite-order operator; other functions have no such finite operator.
Why does an overlap between 's root and 's root force a factor of ?
The overlapping root gains multiplicity in , and a root of multiplicity produces the ladder ; the genuinely new term therefore carries an extra power of .
Why do we throw away the portion of the big solution when forming ?
Those terms come from the roots of , so sends them to zero — they cannot possibly produce the nonzero , hence only the annihilator's new roots can match .
Why must we plug back into the original rather than ?
The enlarged equation only tells us the form of ; the actual coefficients are fixed by matching the real right side , which only appears in the original equation.
Why does share its annihilator with ?
Both arise from the complex root pair , and factors as , so it annihilates every real combination of those two roots — sine and cosine alike.
Why is the annihilator method restricted to constant-coefficient ?
The trick needs the roots of and to combine cleanly into a single characteristic polynomial; with variable coefficients there is no characteristic equation, so no root-counting story and no guaranteed solution form.
Why does resonance (physical unbounded growth) correspond exactly to the algebraic overlap of roots?
When the forcing frequency matches a natural frequency, 's root coincides with 's root, the multiplicity rises, and the (or ) term grows without bound — see Resonance in forced oscillations.
Edge cases
The boundary situations the recipe quietly assumes away.
What is the annihilator of ?
The identity operator (order ) — since is already annihilated by any operator, the lowest-order one is trivially "do nothing"; practically and .
What happens if 's root overlaps a double root of ?
The root reaches multiplicity three in , so you multiply the trial by (the smallest power that steps past the two existing terms), e.g. .
Can ever need a special treatment beyond one term?
No — that sum is just , a single exponential with root ; you treat it as one term with annihilator (or if overlaps ).
For with , does the constant term of matter for whether we need an extra ?
Yes — the annihilator of is (root doubled); since is not a root of (roots ), there is no overlap, so with no extra factor.
If has a root and (constant), what is the trial?
The constant's root overlaps 's root , so multiply by : (or matching 's multiplicity of the root ).
What if is a sum whose pieces have overlapping annihilators, like ?
Combine into the single highest-order annihilator (covering both), then form one trial — do not double-count the shared root.
Is (complex exponential) allowed as forcing?
Yes — its annihilator is ; the complex-exponential shortcut solves and forcing at once by taking real/imaginary parts, avoiding the separate trial.