Non-homogeneous — method of undetermined coefficients (annihilator method): Worked examples

Intuition What this page is for

The parent note gave you three clean examples. Real problems are messier: the right-hand side $g (x)$ can be a polynomial, a decaying exponential, a product of things, a sum of things, or a function that collides with the homogeneous solution. This page walks through one example for every kind of collision and shape, so you never meet a case you haven't seen.

Before we start, a reminder of the symbols we lean on, in plain words:

$D$ means "take one derivative": $D [y] = \frac{d y}{d x}$ . So $D^{2} [y] = y^{''}$ . This is the D-operator.
$L$ is our machine, a polynomial in $D$ with constant numbers, e.g. $L = D^{2} - 3 D + 2$ means $L [y] = y^{''} - 3 y^{'} + 2 y$ .
$g (x)$ is the forcing term — whatever sits on the right of $L [y] = g (x)$ .
$A$ (a curly capital A) is the annihilator of $g$ : the simplest polynomial-in- $D$ machine that turns $g$ into pure $0$ . We reserve the plain letters $A, B, C$ for the unknown coefficients in the trial solution, so the operator never wears the same badge as a number.

The full answer is always $y = y_{h} + y_{p}$ , where $y_{h}$ solves $L [y] = 0$ (the homogeneous piece) and $y_{p}$ is one particular solution.

The scenario matrix

Every problem this topic throws at you lands in exactly one of these cells. The examples below are labelled by cell number so you can see the whole space is covered.

#	Case class	What makes it tricky	Example
C1	Pure polynomial $g$	root $r = 0$ ; need $D^{n + 1}$ ; watch for $r = 0$ already in $L$	Ex 1
C2	Exponential, no overlap	copy the shape, simplest case	Ex 2
C3	Exponential, overlap (resonance)	root of $g$ already in $L$ → multiply by $x$	Ex 3
C4	Trig, no overlap	must include both $cos$ and $sin$	Ex 4
C5	Trig, overlap (pure resonance)	$\pm i β$ already in $L$ → multiply by $x$ ; grows forever	Ex 5
C6	Product $x^{n} e^{α x}$ (repeated-root annihilator)	need $(D - α)^{n + 1}$ , not just $(D - α)$	Ex 6
C7	Sum $g_{1} + g_{2}$ (superposition)	annihilate with $A_{1} A_{2}$ , add the trial forms	Ex 7
C8	Word problem (real-world, with initial conditions)	translate physics → ODE, then apply the method	Ex 8
C9	Exam twist: double overlap ( $x^{2}$ appears)	root already doubled in $L$ → multiply by $x^{2}$	Ex 9
C10	Polynomial × trig: $g = x sin x$	product needs $(D^{2} + β^{2})^{2}$ ; carry $x$ -times-both-trig	Ex 10
C11	$e^{α x} sin β x$ (damped oscillation drive)	complex roots $α \pm i β$ ; annihilator $(D - α)^{2} + β^{2}$	Ex 11

We also cover the degenerate / limiting corners inside these: $g$ a constant (Ex 1's simplest sub-case), $β \to$ the natural frequency (Ex 5, the resonance limit), and $α = 0$ reducing $x^{n} e^{α x}$ back to a polynomial (noted in Ex 6).

C1 — Pure polynomial forcing

Step 1 — homogeneous part. Why this step? We always peel off $y_{h}$ first, because the trial for $y_{p}$ must avoid whatever $y_{h}$ already covers. $D^{2} - 3 D + 2 = (D - 1) (D - 2)$ , roots $1, 2$ : $y_{h} = c_{1} e^{x} + c_{2} e^{2 x} .$

Step 2 — annihilate $g$ . Why this step? Here $n$ is the degree of the polynomial forcing term — for $g = 2 x^{2}$ that degree is $n = 2$ . A polynomial of degree $n$ is built from the root $r = 0$ repeated $n + 1$ times (the root $0$ gives $1$ ; doubled gives $x$ ; tripled gives $x^{2}$ ). The killer is $D^{n + 1}$ ; with $n = 2$ that is $A = D^{3}$ . Check: $D^{3} [2 x^{2}] = 0$ because differentiating a degree-2 polynomial three times wipes it out. ✓

Step 3 — bigger homogeneous ODE. Why this step? Applying $A$ to both sides turns the right side into $0$ , so now we only need to solve a homogeneous equation whose root list we can read off. $D^{3} (D - 1) (D - 2) y = 0, roots new 0, 0, 0, 1, 2.$

Step 4 — keep only the new terms. Why this step? Roots $1, 2$ just rebuild $y_{h}$ and contribute nothing to $g$ . The new root $0$ (tripled) contributes $1, x, x^{2}$ . So: $y_{p} = A x^{2} + B x + C .$ (This answers the forecast: you need all three, even though $g$ only shows $x^{2}$ — because $y^{''}$ and $y^{'}$ pull down lower powers.)

Step 5 — determine coefficients. Why this step? The form is fixed but the numbers $A, B, C$ are not; substituting into the original ODE pins them down. $y_{p}^{'} = 2 A x + B$ , $y_{p}^{''} = 2 A$ . Substitute into $y^{''} - 3 y^{'} + 2 y$ : $2 A - 3 (2 A x + B) + 2 (A x^{2} + B x + C) = 2 x^{2} .$ Group by power of $x$ : $2 A x^{2} + (- 6 A + 2 B) x + (2 A - 3 B + 2 C) = 2 x^{2} + 0 x + 0.$ Match:

$x^{2}$ : $2 A = 2 \Rightarrow A = 1$ .
$x^{1}$ : $- 6 A + 2 B = 0 \Rightarrow B = 3$ .
$x^{0}$ : $2 A - 3 B + 2 C = 0 \Rightarrow 2 - 9 + 2 C = 0 \Rightarrow C = \frac{7}{2}$ .

Step 6 — combine. $y = c_{1} e^{x} + c_{2} e^{2 x} + x^{2} + 3 x + \frac{7}{2} .$

Recall Verify: plug

$y_{p} = x^{2} + 3 x + \frac{7}{2}$ back. $y_{p}^{''} = 2$ , $y_{p}^{'} = 2 x + 3$ . Then $2 - 3 (2 x + 3) + 2 (x^{2} + 3 x + \frac{7}{2}) = 2 - 6 x - 9 + 2 x^{2} + 6 x + 7 = 2 x^{2}$ . ✓ matches $g$ .

C2 — Exponential, no overlap

Step 1 — homogeneous part. Why this step? Same reason as always: we must know $y_{h}$ before choosing the trial, to detect any overlap. $D^{2} + 4 = 0 \Rightarrow r = \pm 2 i$ (a pure imaginary pair, which produces oscillation), so $y_{h} = c_{1} cos 2 x + c_{2} sin 2 x$ .

Step 2 — annihilate $g$ . Why $(D + 1)$ ? $e^{- x}$ has root $- 1$ , so $A = D - (- 1) = D + 1$ annihilates it: $(D + 1) e^{- x} = - e^{- x} + e^{- x} = 0$ . ✓ And $- 1 \neq = \pm 2 i$ , so no overlap — no $x$ needed.

Step 3–4 — build and prune. Why this step? $A L [y] = 0$ has roots $- 1, \pm 2 i$ ; the pair $\pm 2 i$ merely rebuilds $y_{h}$ , so only the new root $- 1$ can produce $g$ : $y_{p} = C e^{- x} .$

Step 5 — determine the coefficient. Why this step? Substituting into the original ODE fixes $C$ . $y_{p}^{''} = C e^{- x}$ , so $y_{p}^{''} + 4 y_{p} = C e^{- x} + 4 C e^{- x} = 5 C e^{- x} = 3 e^{- x} \Rightarrow C = \frac{3}{5} .$

Step 6 — combine. $y = c_{1} cos 2 x + c_{2} sin 2 x + \frac{3}{5} e^{- x} .$

Recall Verify:

$y_{p} = \frac{3}{5} e^{- x}$ . $y_{p}^{''} = \frac{3}{5} e^{- x}$ , so $y_{p}^{''} + 4 y_{p} = \frac{3}{5} e^{- x} + \frac{12}{5} e^{- x} = \frac{15}{5} e^{- x} = 3 e^{- x}$ . ✓

C3 — Exponential overlap (single resonance)

Step 1 — homogeneous part. Why this step? We need $y_{h}$ up front precisely because $g = e^{2 x}$ looks suspiciously like it might already live there. $D^{2} - 4 D + 4 = (D - 2)^{2}$ , root $2$ doubled: $y_{h} = c_{1} e^{2 x} + c_{2} x e^{2 x} .$

Step 2 — annihilate $g$ . Why the collision matters: $g = e^{2 x}$ has root $2$ , already a double root of $L$ . Annihilator $A = (D - 2)$ : $(D - 2) (D - 2)^{2} y = (D - 2)^{3} y = 0.$

Step 3–4 — build and prune. Why this step? Root $2$ now has multiplicity three, giving $e^{2 x}, x e^{2 x}, x^{2} e^{2 x}$ . The first two are $y_{h}$ ; the genuinely new one is $x^{2} e^{2 x}$ : $y_{p} = A x^{2} e^{2 x} .$ (The double overlap forced $x^{2}$ — look at Ex 9 for the same effect on the exam.)

Step 5 — determine the coefficient. Why this step? Substitute the trial into the original ODE to pin down $A$ . With $y_{p} = A x^{2} e^{2 x}$ : $y_{p}^{'} = A e^{2 x} (2 x + 2 x^{2})$ , $y_{p}^{''} = A e^{2 x} (2 + 8 x + 4 x^{2})$ . Then $y_{p}^{''} - 4 y_{p}^{'} + 4 y_{p} = A e^{2 x} [(2 + 8 x + 4 x^{2}) - 4 (2 x + 2 x^{2}) + 4 x^{2}] = A e^{2 x} \cdot 2 = e^{2 x} .$ So $A = \frac{1}{2}$ .

Step 6 — combine. $y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{2} x^{2} e^{2 x} .$

This is resonance in its exponential form: the forcing matches the system's own decaying mode, so the response is a bigger multiple of the same mode.

Recall Verify:

$y_{p} = \frac{1}{2} x^{2} e^{2 x}$ . Coefficient bracket $= 2 A = 2 \cdot \frac{1}{2} = 1$ , giving exactly $e^{2 x}$ . ✓

C4 — Trig, no overlap

Step 1 — homogeneous part. Why this step? We list $y_{h}$ first to check whether the drive frequency $3$ clashes with the natural frequency hidden in $y_{h}$ . $D^{2} + 1 = 0 \Rightarrow r = \pm i$ , $y_{h} = c_{1} cos x + c_{2} sin x$ (natural frequency $1$ ).

Step 2 — annihilate $g$ . Why $D^{2} + 9$ ? $sin 3 x$ comes from roots $\pm 3 i$ ; the annihilator for $cos β x, sin β x$ is $D^{2} + β^{2}$ with $β = 3$ . Since $\pm 3 i \neq = \pm i$ , no overlap.

Step 3–4 — build and prune. Why this step? $(D^{2} + 9) (D^{2} + 1) y = 0$ has roots $\pm i, \pm 3 i$ ; the $\pm i$ pair rebuilds $y_{h}$ , so only the new $\pm 3 i$ pair matters: $y_{p} = A cos 3 x + B sin 3 x .$

Step 5 — determine coefficients. Why this step? Substitute into the original ODE and match the $cos$ and $sin$ parts separately. $y_{p}^{''} = - 9 A cos 3 x - 9 B sin 3 x$ , so $y_{p}^{''} + y_{p} = - 8 A cos 3 x - 8 B sin 3 x = sin 3 x .$ Match: $- 8 A = 0 \Rightarrow A = 0$ ; $- 8 B = 1 \Rightarrow B = - \frac{1}{8}$ .

Step 6 — combine. $y = c_{1} cos x + c_{2} sin x - \frac{1}{8} sin 3 x .$

Recall Verify:

$y_{p} = - \frac{1}{8} sin 3 x$ . $y_{p}^{''} = \frac{9}{8} sin 3 x$ ; $y_{p}^{''} + y_{p} = \frac{9}{8} sin 3 x - \frac{1}{8} sin 3 x = sin 3 x$ . ✓

C5 — Trig overlap (pure resonance) with a figure

Step 1 — homogeneous part. Why this step? We need $y_{h}$ 's frequency to see the collision: the drive $cos 2 x$ is at frequency $2$ , exactly the natural one. $D^{2} + 4 = 0 \Rightarrow r = \pm 2 i$ , $y_{h} = c_{1} cos 2 x + c_{2} sin 2 x$ .

Step 2 — annihilate $g$ . Why the collision matters: $cos 2 x$ has roots $\pm 2 i$ — exactly the roots of $L$ . Annihilator $A = D^{2} + 4$ : $(D^{2} + 4) (D^{2} + 4) y = (D^{2} + 4)^{2} y = 0.$

Step 3–4 — build and prune. Why this step? Roots $\pm 2 i$ now have multiplicity two, giving $cos 2 x, sin 2 x, x cos 2 x, x sin 2 x$ . The first two are $y_{h}$ ; the new ones carry the extra $x$ : $y_{p} = x (A cos 2 x + B sin 2 x) .$

Step 5 — determine coefficients. Why this step? Substituting into the original ODE fixes $A, B$ ; watch the un-shifted $x$ -terms cancel — that cancellation is exactly why the extra $x$ was needed. Differentiate $y_{p} = A x cos 2 x + B x sin 2 x$ : $y_{p}^{'} = A (cos 2 x - 2 x sin 2 x) + B (sin 2 x + 2 x cos 2 x),$ $y_{p}^{''} = A (- 4 sin 2 x - 4 x cos 2 x) + B (4 cos 2 x - 4 x sin 2 x) .$ Then $y_{p}^{''} + 4 y_{p}$ : the $x cos 2 x$ and $x sin 2 x$ terms cancel, leaving $- 4 A sin 2 x + 4 B cos 2 x = cos 2 x .$ Match: $- 4 A = 0 \Rightarrow A = 0$ ; $4 B = 1 \Rightarrow B = \frac{1}{4}$ .

Step 6 — combine. $y = c_{1} cos 2 x + c_{2} sin 2 x + \frac{1}{4} x sin 2 x .$

The figure below plots this particular solution $y_{p} = \frac{1}{4} x sin 2 x$ . The horizontal axis is $x$ (time), the vertical axis is the displacement $y$ . In chalk blue is the wiggling solution itself; the two pink dashed lines are the envelope $\pm \frac{1}{4} x$ — the straight-line ceiling and floor the oscillation keeps bumping against. Because that envelope is a line through the origin with slope $\frac{1}{4}$ , the amplitude grows without bound as $x$ increases: each swing is a little taller than the last. That linear-growth envelope is the visual signature of resonance — the limiting behaviour when the drive frequency equals the natural frequency exactly.

Recall Verify:

$y_{p} = \frac{1}{4} x sin 2 x$ . With $A = 0, B = \frac{1}{4}$ : $y_{p}^{''} + 4 y_{p} = 4 B cos 2 x = cos 2 x$ . ✓

C6 — Product $x e^{α x}$ (repeated-root annihilator)

Step 1 — homogeneous part. Why this step? We must know $y_{h}$ to detect that the drive's root $1$ already lives there. $D^{2} - 1 = (D - 1) (D + 1)$ , roots $\pm 1$ : $y_{h} = c_{1} e^{x} + c_{2} e^{- x} .$

Step 2 — annihilate $g$ . Why $(D - 1)^{2}$ ? The factor $x^{n} e^{α x}$ needs root $α$ repeated $n + 1$ times; here $n = 1, α = 1$ , so $A = (D - 1)^{2}$ . (If $α$ were $0$ , this would collapse to the polynomial case C1 — that's the degenerate corner.) Because root $1$ is already in $L$ once, applying $A$ gives it total multiplicity $1 + 2 = 3$ : $(D - 1)^{2} (D - 1) (D + 1) y = (D - 1)^{3} (D + 1) y = 0.$

Step 3–4 — build and prune. Why this step? Root $1$ triple → $e^{x}, x e^{x}, x^{2} e^{x}$ ; root $- 1$ → $e^{- x}$ . Of these, $e^{x}$ and $e^{- x}$ are $y_{h}$ . The new terms are $x e^{x}, x^{2} e^{x}$ : $y_{p} = x (A x + B) e^{x} = (A x^{2} + B x) e^{x} .$

Step 5 — determine coefficients. Why this step? Substituting into the original ODE fixes $A, B$ . To differentiate a product $p (x) e^{x}$ (where $p$ is any polynomial), the product rule gives $D [p (x) e^{x}] = p^{'} (x) e^{x} + p (x) e^{x} = (p^{'} (x) + p (x)) e^{x},$ because $e^{x}$ differentiates to itself. So each derivative just replaces the polynomial $p$ by $p^{'} + p$ . Here $p = A x^{2} + B x$ , so $p^{'} = 2 A x + B$ and: $y_{p}^{'} = (p^{'} + p) e^{x} = (A x^{2} + (B + 2 A) x + B) e^{x}$ , $y_{p}^{''} = ((p^{'} + p)^{'} + (p^{'} + p)) e^{x} = (A x^{2} + (B + 4 A) x + (2 B + 2 A)) e^{x}$ . Then $y_{p}^{''} - y_{p} = [(B + 4 A) x + (2 B + 2 A) - B x] e^{x} = (4 A x + 2 A + 2 B) e^{x} = x e^{x} .$ Match: $4 A = 1 \Rightarrow A = \frac{1}{4}$ ; $2 A + 2 B = 0 \Rightarrow B = - \frac{1}{4}$ .

Step 6 — combine. $y = c_{1} e^{x} + c_{2} e^{- x} + (\frac{1}{4} x^{2} - \frac{1}{4} x) e^{x} .$

Recall Verify:

$y_{p} = (\frac{1}{4} x^{2} - \frac{1}{4} x) e^{x}$ . $4 A x + 2 A + 2 B = 4 (\frac{1}{4}) x + 2 (\frac{1}{4}) + 2 (- \frac{1}{4}) = x + 0 = x$ , so $y_{p}^{''} - y_{p} = x e^{x}$ . ✓

C7 — Sum of forcings (superposition)

Step 1 — homogeneous part. Why this step? We list $y_{h}$ first so we can tell which of the two forcing pieces collides with it. $D^{2} + 1$ , roots $\pm i$ : $y_{h} = c_{1} cos x + c_{2} sin x$ .

Step 2 — annihilate $g$ using superposition. Why this step? Linearity lets us treat each summand's root independently, then multiply the annihilators. Split $g = g_{1} + g_{2}$ with $g_{1} = e^{x}$ (root $1$ , annihilator $D - 1$ , no overlap) and $g_{2} = 2 cos x$ (roots $\pm i$ , annihilator $D^{2} + 1$ , overlaps $y_{h}$ ). Combined annihilator $A = (D - 1) (D^{2} + 1)$ , applied: $(D - 1) (D^{2} + 1) (D^{2} + 1) y = (D - 1) (D^{2} + 1)^{2} y = 0.$

Step 3–4 — build and prune. Why this step? Roots: $1$ (new), $\pm i$ doubled. New terms: $e^{x}$ from root $1$ ; and from the doubled $\pm i$ the new pieces $x cos x, x sin x$ . So: $y_{p} = for g_{1} C e^{x} + for g_{2} x (A cos x + B sin x) .$

Step 5 — determine coefficients. Why this step? Substitute; linearity lets us solve the two pieces separately. Piece 1: $C e^{x}$ : $y^{''} + y = C e^{x} + C e^{x} = 2 C e^{x} = e^{x} \Rightarrow C = \frac{1}{2} .$ Piece 2: $x (A cos x + B sin x)$ . Exactly as in C5 with $β = 1$ , the surviving terms are $- 2 A sin x + 2 B cos x = 2 cos x \Rightarrow A = 0, B = 1.$

Step 6 — combine. $y = c_{1} cos x + c_{2} sin x + \frac{1}{2} e^{x} + x sin x .$

Recall Verify: substitute

$y_{p} = \frac{1}{2} e^{x} + x sin x$ symbolically. Piece 1: for $C e^{x}$ , $y^{''} + y = 2 C e^{x}$ ; with $C = \frac{1}{2}$ this is $e^{x}$ . ✓ Piece 2: for $x sin x$ , differentiate: $(x sin x)^{'} = sin x + x cos x$ , $(x sin x)^{''} = 2 cos x - x sin x$ . So $(x sin x)^{''} + x sin x = 2 cos x - x sin x + x sin x = 2 cos x$ . ✓ Adding the two pieces gives $y_{p}^{''} + y_{p} = e^{x} + 2 cos x = g$ . ✓

C8 — Word problem (mass–spring, real world)

Step 1 — homogeneous part. Why this step? $y_{h}$ encodes the spring's natural oscillation; comparing its frequency to the drive frequency tells us whether we hit resonance. $D^{2} + 9$ , roots $\pm 3 i$ : $y_{h} = c_{1} cos 3 t + c_{2} sin 3 t$ . (Natural angular frequency $3 rad/s$ .)

Step 2 — annihilate $g$ . Why no $x$ : drive $cos t$ has roots $\pm i \neq = \pm 3 i$ — no overlap, no resonance. Annihilator $A = D^{2} + 1$ .

Step 3–4 — build and prune. Why this step? $A L [y] = 0$ has roots $\pm 3 i$ (rebuild $y_{h}$ ) and the new $\pm i$ ; only the new pair produces $g$ : $y_{p} = A cos t + B sin t .$

Step 5 — determine coefficients. Why this step? Substitute into the original ODE to fix $A, B$ . $y_{p}^{''} = - A cos t - B sin t$ : $y_{p}^{''} + 9 y_{p} = 8 A cos t + 8 B sin t = 6 cos t \Rightarrow A = \frac{3}{4}, B = 0.$ So $y_{p} = \frac{3}{4} cos t$ .

Step 6 — apply initial conditions. Why this step? A word problem has one specific motion; the two constants $c_{1}, c_{2}$ are fixed by the given start state. $y = c_{1} cos 3 t + c_{2} sin 3 t + \frac{3}{4} cos t$ .

$y (0) = 0$ : $c_{1} + \frac{3}{4} = 0 \Rightarrow c_{1} = - \frac{3}{4} .$
$y^{'} (t) = - 3 c_{1} sin 3 t + 3 c_{2} cos 3 t - \frac{3}{4} sin t$ ; $y^{'} (0) = 3 c_{2} = 0 \Rightarrow c_{2} = 0.$

$y (t) = - \frac{3}{4} cos 3 t + \frac{3}{4} cos t .$

Verify (units + sanity): $y (0) = - \frac{3}{4} + \frac{3}{4} = 0$ m ✓; $y^{'} (0) = 0$ ✓. The response is a bounded beat between the two frequencies — no runaway growth because we're off resonance. Units: $y^{''}$ has $m/s^{2}$ , $9 y$ has (rad/s) $^{2} \cdot$ m $=$ m/s $^{2}$ , $6 cos t$ in m/s $^{2}$ — consistent.

Recall Verify at

$t = π$ . $y (π) = - \frac{3}{4} cos 3 π + \frac{3}{4} cos π = - \frac{3}{4} (- 1) + \frac{3}{4} (- 1) = \frac{3}{4} - \frac{3}{4} = 0$ . ✓

C9 — Exam twist: double overlap forces $x^{2}$

Step 1 — homogeneous part. Why this step? We must count the multiplicity of the root that matches the drive — that count controls the power of $x$ . $D^{2} - 2 D + 1 = (D - 1)^{2}$ , root $1$ doubled: $y_{h} = c_{1} e^{x} + c_{2} x e^{x}$ .

Step 2 — annihilate $g$ . Why $x^{2}$ is coming: $g = e^{x}$ has root $1$ , already multiplicity $2$ in $L$ . Annihilator $A = (D - 1)$ pushes total multiplicity to $3$ : $(D - 1) (D - 1)^{2} y = (D - 1)^{3} y = 0.$

Step 3–4 — build and prune. Why this step? Root $1$ triple → $e^{x}, x e^{x}, x^{2} e^{x}$ . First two are $y_{h}$ ; new term is $x^{2} e^{x}$ : $y_{p} = A x^{2} e^{x} .$

Step 5 — determine the coefficient. Why this step? Substitute into the original ODE to fix $A$ ; use the product-rule shortcut $D [p (x) e^{x}] = (p^{'} + p) e^{x}$ from C6. $y_{p} = A x^{2} e^{x}$ , with $p = A x^{2}$ so $p^{'} = 2 A x$ : $y_{p}^{'} = A (x^{2} + 2 x) e^{x}$ , $y_{p}^{''} = A (x^{2} + 4 x + 2) e^{x}$ . Then $y_{p}^{''} - 2 y_{p}^{'} + y_{p} = A [(x^{2} + 4 x + 2) - 2 (x^{2} + 2 x) + x^{2}] e^{x} = A \cdot 2 e^{x} = e^{x} .$ So $A = \frac{1}{2}$ .

Step 6 — combine. $y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{2} x^{2} e^{x} .$

Notice the ladder: single overlap gave $x$ (Ex 3 for exponential / Ex 5 for trig), double overlap gives $x^{2}$ . The rule "multiplicity $m$ in $L$ + your annihilator ⟹ multiply trial by $x^{m}$ " is fully exposed here.

Recall Verify:

$y_{p} = \frac{1}{2} x^{2} e^{x}$ . Bracket $= 2 A = 2 \cdot \frac{1}{2} = 1$ , giving $e^{x}$ . ✓

C10 — Polynomial × trigonometric forcing

Step 1 — homogeneous part. Why this step? We check whether the trig frequency in $g$ (here $β = 1$ ) matches $y_{h}$ 's frequency — it does, so overlap is coming. $D^{2} + 1$ , roots $\pm i$ : $y_{h} = c_{1} cos x + c_{2} sin x$ .

Step 2 — annihilate $g$ . Why $(D^{2} + 1)^{2}$ ? The pattern $x^{n} e^{α x} cos β x$ (with $α = 0, β = 1, n = 1$ ) comes from roots $\pm i$ repeated $n + 1 = 2$ times. The annihilator is $[(D - 0)^{2} + 1]^{2} = (D^{2} + 1)^{2}$ . Check it: $(D^{2} + 1)$ once kills $cos x, sin x$ ; a second application is needed to also kill the $x cos x, x sin x$ produced by the extra factor of $x$ .

Step 3–4 — build and prune. Why this step? Apply $A = (D^{2} + 1)^{2}$ to the ODE, whose $L$ already contributes one factor $(D^{2} + 1)$ : $(D^{2} + 1)^{2} (D^{2} + 1) y = (D^{2} + 1)^{3} y = 0.$ Roots $\pm i$ now have multiplicity three, giving $cos x, sin x, x cos x, x sin x, x^{2} cos x, x^{2} sin x$ . The first pair is $y_{h}$ ; the four new terms form the trial. To keep the algebra clean we write it as $x$ times a full first-degree-polynomial-in- $x$ combination of both trig functions: $y_{p} = x [(A x + B) cos x + (C x + E) sin x] .$ (We use $E$ for the fourth coefficient to keep $D$ free as the derivative operator.)

Step 5 — determine coefficients. Why this step? Substitute into the original ODE $y^{''} + y = x sin x$ and match the four independent shapes $x cos x, cos x, x sin x, sin x$ . Carrying out the differentiation (product rule twice) and simplifying $y_{p}^{''} + y_{p}$ gives $y_{p}^{''} + y_{p} = 4 C x cos x + (2 A + 2 E) cos x - 4 A x sin x + (2 C - 2 B) sin x .$ Match against $x sin x = 0 \cdot x cos x + 0 \cdot cos x + 1 \cdot x sin x + 0 \cdot sin x$ :

$x cos x$ : $4 C = 0 \Rightarrow C = 0$ .
$x sin x$ : $- 4 A = 1 \Rightarrow A = - \frac{1}{4}$ .
$cos x$ : $2 A + 2 E = 0 \Rightarrow E = \frac{1}{4}$ .
$sin x$ : $2 C - 2 B = 0 \Rightarrow B = 0$ .

Step 6 — combine. $y = c_{1} cos x + c_{2} sin x - \frac{1}{4} x^{2} cos x + \frac{1}{4} x sin x .$

Recall Verify:

$y_{p} = - \frac{1}{4} x^{2} cos x + \frac{1}{4} x sin x$ . Plugging into $y^{''} + y$ (see VERIFY block) returns exactly $x sin x$ . ✓ Note both the $x^{2}$ (from the overlap raising multiplicity) and the mixing of $cos$ into a $sin$ -driven problem.

C11 — Damped-oscillation forcing $e^{α x} sin β x$

Step 1 — homogeneous part. Why this step? We list $y_{h}$ 's roots ( $\pm i$ ) to compare against the drive's root $1 \pm i$ and check for overlap. $D^{2} + 1$ , roots $\pm i$ : $y_{h} = c_{1} cos x + c_{2} sin x$ .

Step 2 — annihilate $g$ . Why $(D - 1)^{2} + 1$ ? The pattern $e^{α x} cos β x$ or $e^{α x} sin β x$ comes from the complex roots $α \pm i β$ ; the annihilator is $(D - α)^{2} + β^{2}$ . Here $α = 1, β = 1$ , giving $A = (D - 1)^{2} + 1 = D^{2} - 2 D + 2$ . Its roots $1 \pm i$ are different from $\pm i$ (the real part $1 \neq = 0$ ), so no overlap — no extra $x$ .

Step 3–4 — build and prune. Why this step? $A L [y] = 0$ has roots $\pm i$ (rebuild $y_{h}$ ) and the new pair $1 \pm i$ . The new pair produces $e^{x} cos x, e^{x} sin x$ : $y_{p} = e^{x} (A cos x + B sin x) .$

Step 5 — determine coefficients. Why this step? Substitute into the original ODE $y^{''} + y = e^{x} sin x$ . Differentiating $y_{p} = e^{x} (A cos x + B sin x)$ : $y_{p}^{'} = e^{x} [(A + B) cos x + (B - A) sin x],$ $y_{p}^{''} = e^{x} [2 B cos x - 2 A sin x] .$ Then $y_{p}^{''} + y_{p} = e^{x} [(A + 2 B) cos x + (B - 2 A) sin x] = e^{x} sin x .$ Match: $A + 2 B = 0$ and $B - 2 A = 1$ . From the first, $A = - 2 B$ ; substitute: $B - 2 (- 2 B) = 5 B = 1 \Rightarrow B = \frac{1}{5}$ , $A = - \frac{2}{5}$ .

Step 6 — combine. $y = c_{1} cos x + c_{2} sin x + e^{x} (- \frac{2}{5} cos x + \frac{1}{5} sin x) .$

Recall Verify:

$y_{p} = e^{x} (- \frac{2}{5} cos x + \frac{1}{5} sin x)$ . $A + 2 B = - \frac{2}{5} + \frac{2}{5} = 0$ (no $cos$ ) and $B - 2 A = \frac{1}{5} + \frac{4}{5} = 1$ (matches $sin$ ). So $y_{p}^{''} + y_{p} = e^{x} sin x$ . ✓

Wrap-up: the deciding questions

Recall Which cell forces an

$x^{2}$ in $y_{p}$ , and why? C9: the forcing root already has multiplicity $2$ in $L$ ; the annihilator raises it to $3$ , so the two lowest terms are $y_{h}$ and