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

4.6.14 · D2 · Maths › Ordinary Differential Equations › Non-homogeneous — method of undetermined coefficients (annih

Step 0 — Shuru karne se pehle zaroori words

Koi bhi symbol aane se pehle, main un teen cheezon ke naam bata deta hoon jo hum baar baar draw karenge.

Step 1 — Equation ko "machine $y$ khaati hai, $g$ ugalti hai" ki tarah picture karo

KYA. Hum $L [y] = g (x)$ solve kar rahe hain jahan $L = D^{2} - 3 D + 2$ aur $g (x) = 5 e^{2 x}$ hai.

KYUN. Is chapter ka har method $L$ ko invert karne ka ek tarika hai: hum output $g$ jaante hain, hum input $y$ chahte hain. $L$ ko ek box ki tarah draw karna goal ko concrete banata hai — woh backward arrow dhundho.

PICTURE. Left mein, input $y$ (unknown, ek question mark). Box $L$ use transform karta hai. Right mein, known output $g = 5 e^{2 x}$ (amber curve). Humara kaam dashed backward arrow hai.

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

General solution split hoti hai $y = y_{h} + y_{p}$ ki tarah (yeh hai Superposition principle for linear ODEs): $y_{h}$ woh sab hai jise machine zero kar deti hai, $y_{p}$ ek aisa input hai jo target $g$ ko hit karta hai.

Step 2 — $y_{h}$ dhundho: woh inputs jo machine zero kar deti hai

KYA. $L [y] = 0$ solve karo, yaani $(D^{2} - 3 D + 2) y = 0$ . Factor karo: $D^{2} - 3 D + 2 = (D - 1) (D - 2)$ , toh roots hain $r = 1$ aur $r = 2$ .

KYUN. Homogeneous solutions machine ke "silent inputs" hain — inhe andar daalo aur kuch bhi nahi niklega. Inhe pehle jaanna zaroori hai, kyunki baad mein hum inhe copy karne se bachenge.

PICTURE. Root-number-line par do beej, $r = 1$ aur $r = 2$ par. Har beej upar ki taraf apne function mein ugta hai: $e^{x}$ aur $e^{2 x}$ .

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

$c_{1}, c_{2}$ free constants hain — knobs jo initial conditions baad mein set karenge. Abhi yeh bas yeh mark karte hain ki kaun se functions ko machine annihilate karti hai.

Step 3 — $g$ ko ulta padho aur uska annihilator $A$ banao

KYA. Hum sabse chhoti machine $A$ chahte hain jisme $A [g] = 0$ ho. Humara $g = 5 e^{2 x}$ single root $r = 2$ se uga hua hai. Ise ulta padhne par, beej $r = 2$ factor $(D - 2)$ se aata hai. Toh $A = D - 2$ .

KYUN. Yahi method ka dil hai. Agar $A$ $g$ ko zero kar deta hai, toh $A$ ko humare equation ke dono sides par apply karne se right-hand side gayab ho jaayega — ek hard non-homogeneous problem ko ek easy homogeneous mein badal dega.

PICTURE. Output curve $5 e^{2 x}$ chhoti machine $A = D - 2$ mein enter hoti hai aur zero par flat hokar nikalti hai. Aao picture mein aur symbols mein yeh crush verify karein:

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

Pehla term hai $D e^{2 x} = 2 e^{2 x}$ ; doosra hai $2$ times $e^{2 x}$ . Yeh exactly cancel ho jaate hain — wahi cancellation hai kyun $(D - 2)$ annihilator hai.

Step 4 — $A$ ko poori equation par apply karo: dono sides ab homogeneous

KYA. $L [y] = g$ ke har part ko $A$ se hit karo: $A L [y] = A [g] ⟹ (D - 2) (D - 1) (D - 2) y = (D - 2) 5 e^{2 x} = 0.$

KYUN. Right side ab zero hai — kyunki $A$ ko $g$ kill karne ke liye design kiya gaya tha. Humne ek price di hai: equation ab higher order hai. Lekin higher-order homogeneous with constant coefficients ek aisi cheez hai jo hum roots list karke hamesha solve kar sakte hain. Humne "hard, chhota" trade kiya "easy, bada" ke liye.

PICTURE. Equation collapse ho jaati hai. $A$ right par $g$ ko absorb kar leta hai; left par operators stack up ho jaate hain. Repeated factor ko group karo: $(D - 1) same factor twice (D - 2) (D - 2) y = (D - 1) (D - 2)^{2} y = 0.$

Notice karo amber alarm bell: factor $(D - 2)$ ab do baar appear hota hai. Root $2$ pehle se $L$ mein tha (from $y_{h}$ ), aur $A$ ne ek aur copy add ki. Woh doubling hi poori kahani hai — yeh thought Step 6 ke liye pakad ke rakho.

Step 5 — Badi equation ke har root ki list banao

KYA. Bada operator $(D - 1) (D - 2)^{2}$ ke roots hain: $r = 1$ (ek baar), aur $r = 2$ (do baar — ek repeated root).

KYUN. Hume badi homogeneous ODE ka general solution chahiye, aur uski shape poori tarah inhi roots se tay hoti hai. Simple root $r$ deta hai $e^{r x}$ ; repeated root of multiplicity $2$ deta hai $e^{r x}$ aur $x e^{r x}$ — yeh rule hai Characteristic equation and repeated roots se.

PICTURE. Number line par teen beej: ek $r = 1$ par, aur $r = 2$ par ek stacked pair. Stacked pair do functions mein ugta hai, $e^{2 x}$ (chhota) aur $x e^{2 x}$ (lamba — $x$ ise climb karta banata hai).

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

Pehle do terms exactly $y_{h}$ hain Step 2 se. Sirf aakhri term, $x e^{2 x}$ , genuinely naya hai.

Step 6 — Sirf naya term rakho: wahi $y_{p}$ ki shape HAI

KYA. $c_{1} e^{x}$ aur $c_{2} e^{2 x}$ discard karo (yeh $y_{h}$ hain). Jo bachta hai woh hai trial form: $y_{p} = A x e^{2 x} (A ek unknown coefficient dhundhna hai) .$

KYUN — is page ka sabse gehra "kyun". $L$ ke roots waale terms $L [y] = 0$ satisfy karte hain; inhe machine mein daalo aur tum paoge zero, kabhi $g$ nahi. Toh yeh $g$ produce nahi kar sakte. Sirf woh roots jo $A$ ne nayi contribute ki hain woh $g$ ko hit kar sakti hain. Yahan $A$ ka contribution root $2$ ki doosri copy tha, aur doosri copy hamesha ek extra $x$ ki tarah dikhti hai. Isliye overlap factor $x$ ko force karta hai — koi memorized rule se nahi, balki isliye ki number line par literally do beej ek hi jagah stack ho gaye.

PICTURE. Terms ke do pile. Grey pile ( $e^{x}, e^{2 x}$ ) labeled hai "pehle se $y_{h}$ — fenko." Amber pile ( $x e^{2 x}$ ) labeled hai "NAYA — yahi $y_{p}$ hai."

Step 7 — Real machine $L$ mein dial $A$ tune karo

KYA. $y_{p} = A x e^{2 x}$ ko original $L [y] = 5 e^{2 x}$ mein substitute karo aur $A$ solve karo.

KYUN. Annihilator ne humein shape diya lekin size nahi. Size fix karne ke liye hum sachi machine par wapas aate hain aur demand karte hain ki output exactly $5 e^{2 x}$ ke barabar ho.

PICTURE. $A$ marked ek dial; hum ise tab tak rotate karte hain jab tak output curve $L [y_{p}]$ (cyan) exactly target $5 e^{2 x}$ (amber) par land na kare. Derivatives term by term compute karo:

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

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

$x$ -terms zero par cancel ho jaate hain (yeh hona hi chahiye — woh piece ek silent input tha), ek clean constant bachta hai: $A e^{2 x} = 5 e^{2 x} ⟹ A = 5.$

Toh $y_{p} = 5 x e^{2 x}$ , aur full answer: $y = c_{1} e^{x} + c_{2} e^{2 x} + 5 x e^{2 x}$

Step 8 — Degenerate aur edge cases, ek number line par

KYA. Kya hota agar koi overlap nahi hota? Kya hota agar root $2$ $L$ mein pehle se double hota? Number-line picture dono ka jawab ek hi nazar mein deti hai.

KYUN. Ek method jis par tum sirf ek case mein trust karo woh koi method nahi hai. Har scenario sirf yeh hai: "target root par pehle se kitne beej baithe hain, aur $A$ kitne add kar raha hai."

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

No overlap (e.g. $g = 5 e^{3 x}$ , root $3$ $L$ mein nahi): $A$ $3$ par ek fresh beej lagata hai, multiplicity $1$ → trial $A e^{3 x}$ , koi $x$ nahi.
Simple overlap (humara example, root $2$ $L$ mein ek baar): stack $2$ ho jaata hai → trial $A x e^{2 x}$ , ek $x$ .
Double overlap ( $L$ mein root $2$ pehle se do baar hota): $A$ ise $3$ kar deta hai → trial $A x^{2} e^{2 x}$ , $x^{2}$ .

Ek-picture summary

Upar sab kuch, compressed: equation hard non-homogeneous (top) se annihilator ke through ek easy homogeneous problem (middle) mein flow karti hai, jiska root-line padhna hota hai (bottom), aur sirf naya beej $y_{p}$ ke roop mein bachta hai.

Recall Feynman retelling — poora walkthrough ek dost ko explain karo

Tumhare paas ek machine $L$ hai jo ek function khaati hai aur $g = 5 e^{2 x}$ ugalti hai. Tum input chahte ho. Pehle, machine ke silent inputs list karo — woh jo ise pure zero mein badal deti hai: yahan $e^{x}$ aur $e^{2 x}$ . Yeh hai $y_{h}$ . Ab output $g$ dekho aur pucho: "kaun si chhoti machine $A$ $g$ ko kuch nahi mein pees degi?" Kyunki $5 e^{2 x}$ beej $r = 2$ se ugta hai, woh chhoti machine hai $D - 2$ . Poori equation ko $A$ se chalao. Right side zero ho jaati hai — problem ab homogeneous hai, bas lambi. Uske beej list karo: $1$ , aur $2$ do baar, kyunki beej $2$ pehle se tha aur $A$ ne uske upar ek aur daal diya. Do stacked seeds hamesha ek extra $x$ ugaate hain, toh naya paudha hai $x e^{2 x}$ . Purane paudhe sirf $y_{h}$ hain — unhe fenko. Naya paudha $x e^{2 x}$ tumhare answer ki shape hai. $A x e^{2 x}$ ko real machine mein daalo aur dial $A$ tab tak ghoomaao jab tak output $5 e^{2 x}$ ko exactly hit na kare — yeh $A = 5$ par land karta hai. Full answer: $y = c_{1} e^{x} + c_{2} e^{2 x} + 5 x e^{2 x}$ . Extra $x$ koi rule yaad karne ke liye nahi tha; woh do beej ek hi jagah khade the.

Active recall

Recall

$A$ apply karne ke baad right-hand side zero kyun ho jaata hai? Kyunki $A$ ko $g$ annihilate karne ke liye banaya gaya hai: $A [g] = 0$ by construction, toh $A L [y] = A [g] = 0$ .

Geometrically factor

x

kahan se aata hai?

Overlapping root ko root-line par ek doosra stacked seed milta hai; multiplicity 2 ka doubled root

e^{r x}

aur

x e^{r x}

ugaata hai, aur naya term

x

carry karta hai.

Which terms of the big homogeneous solution do we discard?

L

ke roots se match karne waale terms — yeh

y_{h}

hain,

L [y] = 0

satisfy karte hain, aur

g

produce nahi kar sakte.

Agar

g = 5 e^{3 x}

aur

3

L

ka root nahi hai, toh trial kya hai?

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

(koi

x

nahi, kyunki

3

par beej fresh hai, multiplicity 1).