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

Step 0 — The words we need before we start

Before a single symbol appears, let me name the three things we will draw over and over.

Step 1 — Picture the equation as "machine eats $y$ , spits out $g$ "

WHAT. We are solving $L [y] = g (x)$ with $L = D^{2} - 3 D + 2$ and $g (x) = 5 e^{2 x}$ .

WHY. Every method in this chapter is a way of inverting $L$ : we know the output $g$ , we want the input $y$ . Drawing $L$ as a box makes the goal concrete — find the arrow that goes backwards.

PICTURE. On the left, the input $y$ (unknown, a question mark). The box $L$ transforms it. On the right, the known output $g = 5 e^{2 x}$ (the amber curve). Our job is the dashed backward arrow.

$the machine D^{2} - 3 D + 2 L [unknown input y] = known output g 5 e^{2 x}$

The general solution splits as $y = y_{h} + y_{p}$ (this is the Superposition principle for linear ODEs): $y_{h}$ is everything the machine crushes to zero, $y_{p}$ is one input that hits the target $g$ .

Step 2 — Find $y_{h}$ : the inputs the machine crushes to zero

WHAT. Solve $L [y] = 0$ , i.e. $(D^{2} - 3 D + 2) y = 0$ . Factor: $D^{2} - 3 D + 2 = (D - 1) (D - 2)$ , so the roots are $r = 1$ and $r = 2$ .

WHY. The homogeneous solutions are the machine's "silent inputs" — feed them in and nothing comes out. We must know them first, because later we will have to avoid copying them.

PICTURE. Two seeds on the root-number-line, at $r = 1$ and $r = 2$ . Each seed grows upward into its function: $e^{x}$ and $e^{2 x}$ .

$y_{h} = free dial c_{1} root 1 e^{x} + free dial c_{2} root 2 e^{2 x}$

The $c_{1}, c_{2}$ are free constants — knobs the initial conditions will later set. Right now they just mark which functions the machine annihilates.

Step 3 — Read $g$ backwards to build its annihilator $A$

WHAT. We want the smallest machine $A$ with $A [g] = 0$ . Our $g = 5 e^{2 x}$ is grown from the single root $r = 2$ . Reading that backwards, the seed $r = 2$ comes from the factor $(D - 2)$ . So $A = D - 2$ .

WHY. This is the heart of the method. If $A$ crushes $g$ to zero, then applying $A$ to both sides of our equation will make the right-hand side vanish — turning a hard non-homogeneous problem into an easy homogeneous one.

PICTURE. The output curve $5 e^{2 x}$ enters the small machine $A = D - 2$ and comes out flat at zero. Let us verify the crush on the picture and in symbols:

$(D - 2) e^{2 x} = derivative 2 e^{2 x} - times 2 2 e^{2 x} = 0 ✓$

The first term is $D e^{2 x} = 2 e^{2 x}$ ; the second is $2$ times $e^{2 x}$ . They cancel exactly — that cancellation is why $(D - 2)$ is the annihilator.

Step 4 — Apply $A$ to the whole equation: both sides now homogeneous

WHAT. Hit every part of $L [y] = g$ with $A$ : $A L [y] = A [g] ⟹ (D - 2) (D - 1) (D - 2) y = (D - 2) 5 e^{2 x} = 0.$

WHY. The right side is now zero — because $A$ was designed to kill $g$ . We have paid a price: the equation is now higher order. But higher-order homogeneous with constant coefficients is something we can always solve by listing roots. We traded "hard, small" for "easy, big."

PICTURE. The equation collapses. $A$ absorbs $g$ on the right; on the left the operators stack up. Group the repeated factor: $(D - 1) same factor twice (D - 2) (D - 2) y = (D - 1) (D - 2)^{2} y = 0.$

Notice the amber alarm bell: the factor $(D - 2)$ now appears twice. The root $2$ was already in $L$ (from $y_{h}$ ), and $A$ added another copy. That doubling is the whole story — hold that thought for Step 6.

Step 5 — List every root of the big equation

WHAT. The big operator $(D - 1) (D - 2)^{2}$ has roots: $r = 1$ (once), and $r = 2$ (twice — a repeated root).

WHY. We need the general solution of the big homogeneous ODE, and its shape is dictated entirely by these roots. A simple root $r$ gives $e^{r x}$ ; a repeated root of multiplicity $2$ gives $e^{r x}$ and $x e^{r x}$ — this is the rule from Characteristic equation and repeated roots.

PICTURE. Three seeds on the number line: one at $r = 1$ , and a stacked pair at $r = 2$ . The stacked pair grows into two functions, $e^{2 x}$ (short) and $x e^{2 x}$ (taller — the $x$ makes it climb).

$y_{big} = root 1 c_{1} e^{x} + root 2, copy 1 c_{2} e^{2 x} + root 2, copy 2 c_{3} x e^{2 x}$

The first two terms are exactly $y_{h}$ from Step 2. Only the last term, $x e^{2 x}$ , is genuinely new.

Step 6 — Keep only the new term: that IS the shape of $y_{p}$

WHAT. Discard $c_{1} e^{x}$ and $c_{2} e^{2 x}$ (they are $y_{h}$ ). The survivor is the trial form: $y_{p} = A x e^{2 x} (A an unknown coefficient to find) .$

WHY — the deepest "why" on this page. The terms from the roots of $L$ satisfy $L [y] = 0$ ; feed them to the machine and you get zero, never $g$ . So they cannot produce the target. Only the roots that $A$ newly contributed can hit $g$ . Here $A$ 's contribution was the second copy of root $2$ , and a second copy always shows up as an extra $x$ . That is why overlap forces the factor $x$ — not by a memorized rule, but because the number line literally stacked two seeds at the same spot.

PICTURE. Two piles of terms. The grey pile ( $e^{x}, e^{2 x}$ ) is labelled "already $y_{h}$ — throw away." The amber pile ( $x e^{2 x}$ ) is labelled "NEW — this is $y_{p}$ ."

Step 7 — Tune the dial $A$ inside the real machine $L$

WHAT. Substitute $y_{p} = A x e^{2 x}$ into the original $L [y] = 5 e^{2 x}$ and solve for $A$ .

WHY. The annihilator gave us the shape but not the size. To fix the size we return to the true machine and demand the output equal $5 e^{2 x}$ exactly.

PICTURE. A dial marked $A$ ; we rotate it until the output curve $L [y_{p}]$ (cyan) lands exactly on the target $5 e^{2 x}$ (amber). Compute the derivatives term by term:

$y_{p}^{'} = D hits x \to 1 A e^{2 x} + D hits e^{2 x} 2 A x e^{2 x} = A e^{2 x} (1 + 2 x)$ $y_{p}^{''} = A e^{2 x} (4 + 4 x) (differentiate again, same product rule)$

Now assemble $L [y_{p}] = y_{p}^{''} - 3 y_{p}^{'} + 2 y_{p}$ : $A e^{2 x} [y_{p}^{''} (4 + 4 x) - - 3 y_{p}^{'} 3 (1 + 2 x) + + 2 y_{p} 2 x] = A e^{2 x} [4 x - 6 x + 2 x + 4 - 3] = A e^{2 x} \cdot 1$

The $x$ -terms cancel to zero (they must — that piece was a silent input), leaving a clean constant: $A e^{2 x} = 5 e^{2 x} ⟹ A = 5.$

So $y_{p} = 5 x e^{2 x}$ , and the full answer: $y = c_{1} e^{x} + c_{2} e^{2 x} + 5 x e^{2 x}$

Step 8 — The degenerate & edge cases, on one number line

WHAT. What if there were no overlap? What if root $2$ were already double in $L$ ? The number-line picture answers both at a glance.

WHY. A method you only trust in one case is not a method. Every scenario is just "how many seeds already sit at the target root, and how many does $A$ add."

PICTURE. Three mini number-lines, side by side:

No overlap (e.g. $g = 5 e^{3 x}$ , root $3$ not in $L$ ): $A$ plants a fresh seed at $3$ , multiplicity $1$ → trial $A e^{3 x}$ , no $x$ .
Simple overlap (our example, root $2$ once in $L$ ): stack becomes $2$ → trial $A x e^{2 x}$ , one $x$ .
Double overlap (if $L$ had root $2$ twice already): $A$ makes it $3$ → trial $A x^{2} e^{2 x}$ , $x^{2}$ .

The one-picture summary

Everything above, compressed: the equation flows from hard non-homogeneous (top) through the annihilator into an easy homogeneous problem (middle), whose root-line is read off (bottom), and only the new seed survives as $y_{p}$ .

Recall Feynman retelling — explain the whole walkthrough to a friend

You have a machine $L$ that eats a function and spits out $g = 5 e^{2 x}$ . You want the input. First, list the machine's silent inputs — the ones it turns into pure zero: here $e^{x}$ and $e^{2 x}$ . That's $y_{h}$ . Now look at the output $g$ and ask: "what tiny machine $A$ would grind $g$ down to nothing?" Since $5 e^{2 x}$ grows from the seed $r = 2$ , that tiny machine is $D - 2$ . Run the whole equation through $A$ . The right side becomes zero — the problem is now homogeneous, just longer. List its seeds: $1$ , and $2$ twice, because the seed $2$ was already there and $A$ dropped another on top. Two stacked seeds always sprout an extra $x$ , so the new plant is $x e^{2 x}$ . The old plants are just $y_{h}$ — throw them out. The new plant $x e^{2 x}$ is the shape of your answer. Feed $A x e^{2 x}$ into the real machine and turn the dial $A$ until the output hits $5 e^{2 x}$ exactly — it lands at $A = 5$ . Full answer: $y = c_{1} e^{x} + c_{2} e^{2 x} + 5 x e^{2 x}$ . The extra $x$ was never a rule to memorize; it was two seeds standing in the same place.

Active recall

Recall Why does the right-hand side become zero after applying

$A$ ? Because $A$ is built to annihilate $g$ : $A [g] = 0$ by construction, so $A L [y] = A [g] = 0$ .

Where does the factor

x

come from geometrically?

The overlapping root gets a second stacked seed on the root-line; a doubled root of multiplicity 2 sprouts

e^{r x}

and

x e^{r x}

, and the new term carries the

x

Which terms of the big homogeneous solution do we discard?

The ones matching the roots of

L

— they are

y_{h}

, satisfy

L [y] = 0

, and cannot produce

g

g = 5 e^{3 x}

and

3

is not a root of

L

, what is the trial?

y_{p} = A e^{3 x}

(no

x

, because the seed at 3 is fresh, multiplicity 1).