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

Before we begin, one reminder of the four symbols we reuse everywhere:

Level 1 — Recognition

Goal: look at a function and instantly write the operator that kills it. No solving yet — just reading the function backwards to its characteristic root.

L1.1 State the annihilator of $g (x) = 7 e^{- 4 x}$ .

Recall Solution L1.1

An exponential $e^{α x}$ is born from the single characteristic root $r = α$ . Here $α = - 4$ , so the root is $r = - 4$ , and the operator with that root is $(D - (- 4)) = D + 4$ . Check: $(D + 4) 7 e^{- 4 x} = 7 (- 4) e^{- 4 x} + 4 \cdot 7 e^{- 4 x} = 0. ✓$ The constant $7$ is irrelevant — annihilators kill the shape, not the size. $A = D + 4.$

L1.2 State the annihilator of $g (x) = x^{2}$ .

Recall Solution L1.2

Powers $1, x, x^{2}, \dots, x^{n}$ all come from the root $r = 0$ repeated $n + 1$ times, because a root $0$ of multiplicity $k$ produces $e^{0 x}, x e^{0 x}, \dots = 1, x, \dots, x^{k - 1}$ . For $x^{2}$ we have $n = 2$ , so we need $r = 0$ repeated $3$ times: $A = D^{3}$ . Check: differentiate $x^{2}$ three times: $2 x \to 2 \to 0. ✓$ $A = D^{3} .$

L1.3 State the annihilator of $g (x) = sin 5 x$ .

Recall Solution L1.3

A pure sine or cosine of frequency $β$ comes from the pair of complex roots $r = \pm i β$ (see Characteristic equation and repeated roots). Building a polynomial with roots $+ i β$ and $- i β$ : $(r - i β) (r + i β) = r^{2} + β^{2} .$ With $β = 5$ that is $A = D^{2} + 25$ . Check: $(D^{2} + 25) sin 5 x = - 25 sin 5 x + 25 sin 5 x = 0. ✓$ $A = D^{2} + 25.$

L1.4 State the annihilator of $g (x) = x e^{2 x}$ .

Recall Solution L1.4

The extra factor $x$ means the root $α = 2$ must be repeated once more. A root of multiplicity $2$ gives $e^{2 x}$ and $x e^{2 x}$ . For $x^{n} e^{α x}$ the rule is multiplicity $n + 1$ ; here $n = 1$ , so multiplicity $2$ : $A = (D - 2)^{2} .$ Trap avoided: $(D - 2)$ alone kills $e^{2 x}$ but NOT $x e^{2 x}$ — you need the square.

Level 2 — Application

Goal: run the full six-step algorithm on clean cases with NO overlap between $g$ and $y_{h}$ .

L2.1 Solve $y^{''} - y^{'} - 2 y = 6 e^{4 x}$ .

Recall Solution L2.1

Step 1 — homogeneous. Characteristic equation $r^{2} - r - 2 = (r - 2) (r + 1) = 0$ , roots $2, - 1$ . So $y_{h} = c_{1} e^{2 x} + c_{2} e^{- x} .$ Step 2 — annihilate. $g = 6 e^{4 x}$ has root $4$ . It is NOT $2$ or $- 1$ , so no overlap. $A = D - 4$ . Step 3–4 — bigger homogeneous. Roots become $2, - 1, 4$ . The genuinely new root is $4$ , giving the trial $y_{p} = A e^{4 x} .$ Step 5 — determine. $y_{p}^{'} = 4 A e^{4 x}$ , $y_{p}^{''} = 16 A e^{4 x}$ . Substitute: $16 A - 4 A - 2 A = 10 A, 10 A e^{4 x} = 6 e^{4 x} \Rightarrow A = \frac{3}{5} .$ Step 6. $y = c_{1} e^{2 x} + c_{2} e^{- x} + \frac{3}{5} e^{4 x} .$

L2.2 Solve $y^{''} + 4 y = 8 x + 3$ .

Recall Solution L2.2

Step 1. $r^{2} + 4 = 0 \Rightarrow r = \pm 2 i$ , so $y_{h} = c_{1} cos 2 x + c_{2} sin 2 x$ . Step 2. $g = 8 x + 3$ is a degree-1 polynomial → root $0$ repeated twice → $A = D^{2}$ . (No overlap: $0$ is not $\pm 2 i$ .) Step 3–4. New roots $0, 0$ give terms $1$ and $x$ , so $y_{p} = A x + B .$ Step 5. $y_{p}^{'} = A$ , $y_{p}^{''} = 0$ . Then $y_{p}^{''} + 4 y_{p} = 4 A x + 4 B = 8 x + 3$ . Match: $4 A = 8 \Rightarrow A = 2$ ; $4 B = 3 \Rightarrow B = \frac{3}{4}$ . Step 6. $y = c_{1} cos 2 x + c_{2} sin 2 x + 2 x + \frac{3}{4} .$

L2.3 Solve $y^{''} - 2 y^{'} + 5 y = sin x$ .

Recall Solution L2.3

Step 1. $r^{2} - 2 r + 5 = 0 \Rightarrow r = \frac{2 \pm 4 - 20}{2} = 1 \pm 2 i$ . So $y_{h} = e^{x} (c_{1} cos 2 x + c_{2} sin 2 x) .$ Step 2. $g = sin x$ has roots $\pm i$ (that is $β = 1$ ). Annihilator $A = D^{2} + 1$ . No overlap ( $\pm i \neq = 1 \pm 2 i$ ). Step 3–4. New terms $cos x, sin x$ , so we MUST include both (differentiation mixes them): $y_{p} = A cos x + B sin x .$ Step 5. $y_{p}^{'} = - A sin x + B cos x$ , $y_{p}^{''} = - A cos x - B sin x$ . Substitute into $y_{p}^{''} - 2 y_{p}^{'} + 5 y_{p}$ : $cos x : (- A - 2 B + 5 A) = 4 A - 2 B$ $sin x : (- B + 2 A + 5 B) = 2 A + 4 B$ Set equal to $sin x$ (so $cos$ -coeff $= 0$ , $sin$ -coeff $= 1$ ): $4 A - 2 B = 0, 2 A + 4 B = 1.$ From the first, $B = 2 A$ . Sub: $2 A + 8 A = 1 \Rightarrow A = \frac{1}{10}$ , $B = \frac{1}{5}$ . Step 6. $y = e^{x} (c_{1} cos 2 x + c_{2} sin 2 x) + \frac{1}{10} cos x + \frac{1}{5} sin x .$

Level 3 — Analysis (overlap / resonance / sums)

Goal: detect when $g$ shares a root with $L$ (forcing an extra $x$ ), and handle sums by superposition.

L3.1 Solve $y^{''} - 4 y^{'} + 4 y = e^{2 x}$ .

Recall Solution L3.1

Step 1. $r^{2} - 4 r + 4 = (r - 2)^{2} = 0$ , a double root $r = 2$ . So $y_{h} = c_{1} e^{2 x} + c_{2} x e^{2 x}$ . Step 2 — spot the overlap. $g = e^{2 x}$ has root $2$ , but $2$ is already a root of $L$ with multiplicity $2$ . Annihilator $A = D - 2$ . Applied: $(D - 2) (D - 2)^{2} y = (D - 2)^{3} y = 0.$ Step 3–4. Root $2$ now has multiplicity $3$ , giving $e^{2 x}, x e^{2 x}, x^{2} e^{2 x}$ . The first two are exactly $y_{h}$ ; the genuinely new term is $x^{2} e^{2 x}$ : $y_{p} = A x^{2} e^{2 x} .$ (The multiplicity jumped from 2 to 3, so the new term carries $x^{2}$ — that is why we multiply by $x^{m}$ where $m$ is the old multiplicity, here $m = 2$ .) Step 5. With $u = x^{2}$ : $y_{p} = A x^{2} e^{2 x}$ . $y_{p}^{'} = A e^{2 x} (2 x + 2 x^{2})$ , $y_{p}^{''} = A e^{2 x} (2 + 8 x + 4 x^{2})$ . Then $y_{p}^{''} - 4 y_{p}^{'} + 4 y_{p} = A e^{2 x} [(2 + 8 x + 4 x^{2}) - 4 (2 x + 2 x^{2}) + 4 x^{2}] = A e^{2 x} \cdot 2.$ So $2 A e^{2 x} = e^{2 x} \Rightarrow A = \frac{1}{2}$ . Step 6. $y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{2} x^{2} e^{2 x} .$ This is resonance in exponential clothing: forcing at a system's own root pumps energy in, and the response grows like $x^{2}$ .

L3.2 Solve $y^{''} + 9 y = cos 3 x$ .

Recall Solution L3.2

Step 1. $r^{2} + 9 = 0 \Rightarrow r = \pm 3 i$ , so $y_{h} = c_{1} cos 3 x + c_{2} sin 3 x$ . Step 2 — overlap. $g = cos 3 x$ has roots $\pm 3 i$ , the same as $L$ ! Annihilator $A = D^{2} + 9$ . Applied: $(D^{2} + 9) (D^{2} + 9) y = (D^{2} + 9)^{2} y = 0.$ Step 3–4. Roots $\pm 3 i$ now have multiplicity $2$ , giving $cos 3 x, sin 3 x, x cos 3 x, x sin 3 x$ . Drop the first two ( $y_{h}$ ); the new pair carries the $x$ : $y_{p} = x (A cos 3 x + B sin 3 x) .$ Step 5. Compute $y_{p}^{''}$ . Let $y_{p} = A x cos 3 x + B x sin 3 x$ . $y_{p}^{''} = A (- 6 sin 3 x - 9 x cos 3 x) + B (6 cos 3 x - 9 x sin 3 x)$ . Then $y_{p}^{''} + 9 y_{p} = - 6 A sin 3 x + 6 B cos 3 x$ (the $x$ -terms cancel — that is the point of resonance). Match to $cos 3 x$ : $6 B = 1 \Rightarrow B = \frac{1}{6}$ ; $- 6 A = 0 \Rightarrow A = 0$ . Step 6. $y = c_{1} cos 3 x + c_{2} sin 3 x + \frac{1}{6} x sin 3 x .$ The $x sin 3 x$ term is a growing oscillation — textbook resonance.

L3.3 Solve $y^{''} - y = 2 e^{x} + 4$ (a sum — use superposition).

Recall Solution L3.3

Step 1. $r^{2} - 1 = (r - 1) (r + 1) = 0$ , roots $\pm 1$ . So $y_{h} = c_{1} e^{x} + c_{2} e^{- x}$ . Step 2 — split by superposition. Solve $L [y] = 2 e^{x}$ and $L [y] = 4$ separately, then add.

Piece 1: $g_{1} = 2 e^{x}$ , root $1$ — overlaps $y_{h}$ 's root $1$ (simple). So multiply by $x$ : $y_{p 1} = A x e^{x}$ . $y_{p 1}^{'} = A e^{x} (1 + x)$ , $y_{p 1}^{''} = A e^{x} (2 + x)$ . Then $y_{p 1}^{''} - y_{p 1} = A e^{x} (2 + x) - A x e^{x} = 2 A e^{x}$ . Set $2 A e^{x} = 2 e^{x} \Rightarrow A = 1$ . So $y_{p 1} = x e^{x}$ .
Piece 2: $g_{2} = 4$ (a constant → root $0$ , no overlap). Trial $y_{p 2} = C$ . Then $- C = 4 \Rightarrow C = - 4$ . Step 6. $y = c_{1} e^{x} + c_{2} e^{- x} + x e^{x} - 4.$

Level 4 — Synthesis

Goal: assemble everything — build annihilators for products, mix overlap with sums, and choose forms carefully.

L4.1 Give the trial form (unknown coefficients only, do NOT solve) for $y^{''} - 2 y^{'} + y = x^{2} e^{x} + sin x,$ given that $L = (D - 1)^{2}$ .

Recall Solution L4.1

$y_{h}$ : $(r - 1)^{2} = 0$ , double root $1$ , so $y_{h} = c_{1} e^{x} + c_{2} x e^{x}$ . Piece $x^{2} e^{x}$ : annihilator $(D - 1)^{3}$ . But root $1$ already has multiplicity $2$ in $L$ , so applying it makes multiplicity $2 + 3 = 5$ , giving $e^{x}, x e^{x}, x^{2} e^{x}, x^{3} e^{x}, x^{4} e^{x}$ . Drop the first two ( $y_{h}$ ). New terms: $x^{2} e^{x}, x^{3} e^{x}, x^{4} e^{x}$ . Trial: $y_{p 1} = (A x^{2} + B x^{3} + C x^{4}) e^{x} .$ (Equivalently: base form $(A + B x + C x^{2}) e^{x}$ times $x^{2}$ because of the double overlap.) Piece $sin x$ : roots $\pm i$ , annihilator $D^{2} + 1$ . No overlap with root $1$ . Trial: $y_{p 2} = E cos x + F sin x .$ Combined trial: $y_{p} = (A x^{2} + B x^{3} + C x^{4}) e^{x} + E cos x + F sin x .$ (Six unknowns $A, B, C, E, F$ — wait, five. Solving them is a separate exercise; here we only needed the correct shape.)

L4.2 Solve $y^{''} + y = 4 cos x$ (overlap at the resonant frequency) and interpret.

Recall Solution L4.2

Step 1. $r^{2} + 1 = 0 \Rightarrow r = \pm i$ , $y_{h} = c_{1} cos x + c_{2} sin x$ . Step 2 — overlap. $g = 4 cos x$ has roots $\pm i$ , identical to $L$ . $A = D^{2} + 1$ , so $(D^{2} + 1)^{2} y = 0$ , multiplicity 2. Step 3–4. New terms: $x cos x, x sin x$ . So $y_{p} = x (A cos x + B sin x)$ . Step 5. With $y_{p} = A x cos x + B x sin x$ : $y_{p}^{''} = A (- 2 sin x - x cos x) + B (2 cos x - x sin x)$ . $y_{p}^{''} + y_{p} = - 2 A sin x + 2 B cos x$ . Match $4 cos x$ : $2 B = 4 \Rightarrow B = 2$ ; $- 2 A = 0 \Rightarrow A = 0$ . Step 6. $y = c_{1} cos x + c_{2} sin x + 2 x sin x .$ Interpretation. The amplitude $2 x$ grows without bound — pure resonance. See Resonance in forced oscillations. Physically, pushing a swing at exactly its natural frequency makes it swing ever higher.

L4.3 Solve $y^{'''} - y^{''} = 6$ (a repeated root at $0$ overlapping the constant forcing).

Recall Solution L4.3

Step 1. Characteristic: $r^{3} - r^{2} = r^{2} (r - 1) = 0$ . Roots $0$ (double), $1$ . So $y_{h} = c_{1} + c_{2} x + c_{3} e^{x} .$ Step 2 — overlap. $g = 6$ is a constant → root $0$ . But $0$ already has multiplicity $2$ in $L$ ! Annihilator $A = D$ . Applied: $D \cdot r^{2} (r - 1) \to r^{3} (r - 1)$ , so root $0$ has multiplicity $3$ . Step 3–4. Terms from root $0$ : $1, x, x^{2}$ . First two are $y_{h}$ ; the new one is $x^{2}$ . Trial: $y_{p} = A x^{2} .$ Step 5. $y_{p}^{''} = 2 A$ , $y_{p}^{'''} = 0$ . Then $y_{p}^{'''} - y_{p}^{''} = - 2 A = 6 \Rightarrow A = - 3$ . Step 6. $y = c_{1} + c_{2} x + c_{3} e^{x} - 3 x^{2} .$

Level 5 — Mastery

Goal: reason about the method itself — general statements, degenerate inputs, and why the machinery works.

L5.1 Without solving, state the order (degree of the operator) of the smallest homogeneous equation $A L [y] = 0$ you would build for $L (D - 1) (D - 2) y = x e^{x} + e^{2 x} .$ Then state how many unknown coefficients the trial $y_{p}$ has.

Recall Solution L5.1

$L$ has order $2$ (roots $1, 2$ ).

$g_{1} = x e^{x}$ : annihilator $(D - 1)^{2}$ , order $2$ .
$g_{2} = e^{2 x}$ : annihilator $(D - 2)$ , order $1$ . Combined annihilator $A = (D - 1)^{2} (D - 2)$ , order $3$ . So $A L$ has order $2 + 3 = 5$ . Overlap accounting for the trial:
Root $1$ : originally multiplicity $1$ in $L$ , forcing needs $(D - 1)^{2}$ → in $A L$ multiplicity $1 + 2 = 3$ . Terms $e^{x}, x e^{x}, x^{2} e^{x}$ ; drop $e^{x}$ ( $y_{h}$ ), new: $x e^{x}, x^{2} e^{x}$ → coefficients from base $(A + B x) e^{x} \cdot x$ .
Root $2$ : originally multiplicity $1$ , forcing $(D - 2)$ → multiplicity $2$ . Terms $e^{2 x}, x e^{2 x}$ ; drop $e^{2 x}$ , new: $x e^{2 x}$ . So $y_{p} = (A x + B x^{2}) e^{x} + C x e^{2 x}$ : 3 unknown coefficients $A, B, C$ .

L5.2 Degenerate input. What is the annihilator of $g (x) = 0$ ? What does the method then produce, and is that consistent?

Recall Solution L5.2

The zero function is annihilated by the identity operator... trivially by anything, and its lowest-order annihilator is the "do nothing to reach zero" case: literally $A = 1$ (the identity times a constant), since $1 \cdot 0 = 0$ and no lower order exists. Applying $A = 1$ to $L [y] = 0$ gives $L [y] = 0$ unchanged — no new roots. So $y_{p}$ has no new terms: $y_{p} = 0$ . Consistency: the equation $L [y] = 0$ IS homogeneous; its full solution is just $y_{h}$ . The method correctly returns $y = y_{h} + 0$ . Everything agrees. ✓

L5.3 Explain the mechanism. In one paragraph, argue why the terms coming from the roots of $L$ (inside the big equation $A L [y] = 0$ ) can never help build $y_{p}$ .

Recall Solution L5.3

Any term $ϕ$ generated by a root of $L$ satisfies $L [ϕ] = 0$ by definition — that is what "root of $L$ " means (see Characteristic equation and repeated roots). So if we tried to use such a $ϕ$ inside $y_{p}$ , plugging it into the original equation gives $L [ϕ] = 0$ , which contributes nothing to the required right-hand side $g$ . Only the new roots that the annihilator $A$ adds produce terms $ψ$ with $L [ψ] \neq = 0$ , and those are the only ones capable of matching $g$ . Hence $y_{p}$ is built purely from the " $A$ -part," and the " $L$ -part" is discarded because it merely reconstructs $y_{h}$ . This is also why overlap forces an $x$ : a shared root gains multiplicity in $A L$ , and the extra multiplicity's term carries the higher power of $x$ that is genuinely new (not already in $y_{h}$ ).

L5.4 Solve completely with the initial conditions $y (0) = 0, y^{'} (0) = 1$ : $y^{''} - 3 y^{'} + 2 y = 4 e^{3 x} .$ (This is Worked Example 1 from the parent, now with an IVP — the mastery step is pinning the constants.)

Recall Solution L5.4

From the parent note, the general solution is $y = c_{1} e^{x} + c_{2} e^{2 x} + 2 e^{3 x} .$ Apply $y (0) = 0$ : $c_{1} + c_{2} + 2 = 0$ . $y^{'} = c_{1} e^{x} + 2 c_{2} e^{2 x} + 6 e^{3 x}$ ; apply $y^{'} (0) = 1$ : $c_{1} + 2 c_{2} + 6 = 1$ . Subtract the first equation from the second: $c_{2} + 4 = 1 \Rightarrow c_{2} = - 3$ . Then $c_{1} = - 2 - c_{2} = - 2 + 3 = 1$ . $y = e^{x} - 3 e^{2 x} + 2 e^{3 x} .$ Verify: $y (0) = 1 - 3 + 2 = 0$ ✓; $y^{'} (0) = 1 - 6 + 6 = 1$ ✓.

Wrap-up

Recall One-line summary of the whole page

Read $g$ backwards to its roots → write $A$ → build $A L [y] = 0$ → keep only the roots $A$ added (multiply by $x^{m}$ for any shared root of multiplicity $m$ ) → solve coefficients in the real equation → add $y_{h}$ → then (if given) apply initial conditions to the total.