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

4.6.14 · Maths › Ordinary Differential Equations

Annihilator kya hota hai?

KYU annihilators exist bhi karte hain? Kyunki "nice" $g (x)$ mein aane wali functions — polynomials, exponentials, sines, cosines, aur unke products — exactly constant-coefficient homogeneous ODEs ki solutions hoti hain. Har aisi solution kisi characteristic polynomial ke root se paida hoti hai, toh function ko ulta padhne se operator mil jaata hai.

Formula Standard annihilator table (derive karo, memorize mat karo)

Har row is baat se aati hai ki "kaunsa characteristic root yeh function produce karta hai?"

Function $g (x)$	Root jisse aata hai	Annihilator $A$
$x^{n}$ (yaani $1, x, \dots, x^{n}$ )	$r = 0$ repeated $n + 1$ baar	$D^{n + 1}$
$e^{α x}$	$r = α$	$D - α$
$x^{n} e^{α x}$	$r = α$ repeated $n + 1$ baar	$(D - α)^{n + 1}$
$cos β x, sin β x$	$r = \pm i β$	$D^{2} + β^{2}$
$x^{n} e^{α x} cos β x$ etc.	$r = α \pm i β$ rep. $n + 1$ baar	$[(D - α)^{2} + β^{2}]^{n + 1}$

Table ko ulta padhna (ek row ki derivation). Lo $g = e^{α x}$ . Root $r = α$ wali ODE hai $(r - α) = 0 \Rightarrow (D - α) y = 0$ , jiska solution $C e^{α x}$ hai. Toh $(D - α)$ , $e^{α x}$ ko annihilate karta hai: $(D - α) e^{α x} = α e^{α x} - α e^{α x} = 0. ✓$ $x^{n} e^{α x}$ ke liye root $α$ ko repeated $n + 1$ baar chahiye (repeated roots $x^{k} e^{α x}$ terms dete hain), isliye $(D - α)^{n + 1}$ .

Method, step by step

Figure — Non-homogeneous — method of undetermined coefficients (annihilator method)

Worked Example 1 — exponential, koi overlap nahin

Solve karo $y^{''} - 3 y^{'} + 2 y = 4 e^{3 x}$ , yaani $(D^{2} - 3 D + 2) y = 4 e^{3 x}$ .

Step 1 (Kyun): factor karo $D^{2} - 3 D + 2 = (D - 1) (D - 2)$ , roots $1, 2$ . Toh $y_{h} = c_{1} e^{x} + c_{2} e^{2 x} .$

Step 2 (Kyun): $g = 4 e^{3 x}$ root $3$ se aata hai, toh annihilator $A = (D - 3)$ . Apply karo: $(D - 3) (D - 1) (D - 2) y = 0.$

Step 3–4 (Yeh step kyun?): roots ab $1, 2, 3$ hain. General solution $c_{1} e^{x} + c_{2} e^{2 x} + c_{3} e^{3 x}$ . Pehle do $y_{h}$ hain, toh naya term form deta hai: $y_{p} = A e^{3 x} .$

Step 5 (Kyun): substitute karo. $y_{p}^{''} - 3 y_{p}^{'} + 2 y_{p} = (9 - 9 + 2) A e^{3 x} = 2 A e^{3 x} = 4 e^{3 x} \Rightarrow A = 2.$

Step 6: $y = c_{1} e^{x} + c_{2} e^{2 x} + 2 e^{3 x} .$

Worked Example 2 — overlap (resonance) ek $x$ force karta hai

Solve karo $y^{''} - 3 y^{'} + 2 y = 5 e^{2 x}$ .

Step 1: same $y_{h} = c_{1} e^{x} + c_{2} e^{2 x}$ .

Step 2 (Overlap kyun matter karta hai): $g = 5 e^{2 x}$ ka root $2$ hai, lekin $2$ $L$ ka already ek root hai. Annihilator $A = (D - 2)$ : $(D - 2) (D - 1) (D - 2) y = (D - 1) (D - 2)^{2} y = 0.$

Step 3–4 (Yeh step kyun?): roots $1, 2, 2$ (double). General solution $c_{1} e^{x} + c_{2} e^{2 x} + c_{3} x e^{2 x}$ . Pehle do $y_{h}$ hain; genuinely naya term $x e^{2 x}$ hai. Isliye $y_{p} = A x e^{2 x} .$ (Dekho — $x$ automatically aaya kyunki root double ho gaya.)

Step 5 (Kyun): $y_{p} = A x e^{2 x}$ ke saath, $y_{p}^{'} = A e^{2 x} (1 + 2 x)$ , $y_{p}^{''} = A e^{2 x} (4 + 4 x)$ . Tab $y_{p}^{''} - 3 y_{p}^{'} + 2 y_{p} = A e^{2 x} [(4 + 4 x) - 3 (1 + 2 x) + 2 x] = A e^{2 x} (1) = 5 e^{2 x} \Rightarrow A = 5.$

Step 6: $y = c_{1} e^{x} + c_{2} e^{2 x} + 5 x e^{2 x} .$

Worked Example 3 — trig with the $D^{2} + β^{2}$ annihilator

Solve karo $y^{''} + y = cos 2 x$ .

Step 1: $D^{2} + 1 = 0 \Rightarrow r = \pm i$ , toh $y_{h} = c_{1} cos x + c_{2} sin x$ .

Step 2 (Kyun): $cos 2 x$ roots $\pm 2 i$ se aata hai, annihilator $A = D^{2} + 4$ . Koi overlap nahin (kyunki $i \neq = 2 i$ ). $(D^{2} + 4) (D^{2} + 1) y = 0.$

Step 3–4: roots $\pm i, \pm 2 i$ . Naye terms: $cos 2 x, sin 2 x$ . Toh $y_{p} = A cos 2 x + B sin 2 x .$

Step 5 (Kyun): $y_{p}^{''} = - 4 A cos 2 x - 4 B sin 2 x$ , toh $y_{p}^{''} + y_{p} = - 3 A cos 2 x - 3 B sin 2 x = cos 2 x \Rightarrow A = - \frac{1}{3}, B = 0.$

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

Common mistakes (Steel-man + fix)

Active recall

Recall Ek sentence mein reason kya hai ki

$y_{p}$ sirf " $A$ -roots" rakhta hai? Kyunki $L$ -roots $y_{h}$ wapas generate karte hain, jo $L [y] = 0$ satisfy karta hai aur isliye $g$ mein kuch contribute nahin karta; sirf annihilator ke introduce kiye roots hi $g$ produce kar sakte hain.

Recall Overlap

$x$ ka factor kyun force karta hai? Overlapping root $A L$ mein multiplicity gain karta hai, aur multiplicity- $k$ root $e^{α x}, x e^{α x}, \dots$ terms deta hai, toh naya term extra $x$ carry karta hai.

Recall Feynman: ek 12-saal ke bachche ko explain karo

Tumhare paas ek machine $L$ hai jo ek function khaati hai aur $g$ ugalti hai. Function guess karne ke liye, $g$ dekho aur poochho "kaunsi simple machine $A$ , $g$ ko zero kar deti hai?" $g$ ko $A$ se guzaaro — woh gayab ho jaata hai. Ab dono sides zero hain, aur "zero-banane-waale" list karne aasaan hain. Answer-function naye zero-banane-waalon mein se ek hai jo $A$ ne add kiya. Use real machine $L$ mein try karo aur dial (coefficient) tune karo jab tak output $g$ se match na kare.

Flashcards

Annihilator of

e^{α x}

D - α

Annihilator of

x^{n} e^{α x}

(D - α)^{n + 1}

Annihilator of

cos β x

sin β x

D^{2} + β^{2}

Annihilator of

x^{n}

(and lower powers)

D^{n + 1}

Annihilator of

e^{α x} cos β x

(D - α)^{2} + β^{2}

A L [y] = 0

apply karne ke baad, kaun se terms

y_{p}

bante hain?

A

ke naye roots se aane wale terms (

y_{h}

terms drop kar do)

g

aur

y_{h}

overlap karne par trial ko

x

se multiply kyun karte hain?

Overlap

A L

mein root ki multiplicity badhaata hai, toh naya solution term extra

x

carry karta hai

L [y] = g

ke solution ki general structure

y = y_{h} + y_{p}

g = 5 e^{2 x}

ke liye trial jab

2

L

ka simple root ho

y_{p} = A x e^{2 x}

y_{p}

ki form milne ke baad aagla step

Original

L [y] = g

mein substitute karo aur undetermined coefficients solve karo

Connections

Linear differential operators and the D-operator
Homogeneous linear ODEs with constant coefficients
Characteristic equation and repeated roots
Variation of parameters (tab use hota hai jab $g$ annihilator-friendly nahin ho, jaise $sec x$ )
Superposition principle for linear ODEs
Resonance in forced oscillations ( $x$ -factor ka physical meaning)