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

4.6.14 · D5 · Maths › Ordinary Differential Equations › Non-homogeneous — method of undetermined coefficients (annih

Questions se pehle, ek anchor jo aap har jagah reuse karenge:

Recall Wo ek idea kya hai jo poore method ko kaam karwata hai?

Har "nice" forcing term $g (x)$ — polynomial, exponential, sine/cosine, aur unke products — khud kisi constant-coefficient homogeneous ODE ka solution hota hai, isliye uska ek killer operator $A$ hota hai jiske liye $A [g] = 0$ . $g$ ko ulta padhkar uska characteristic root nikalo to $A$ milta hai; $A$ apply karne se $L [y] = g$ ek homogeneous problem $A L [y] = 0$ ban jaata hai jise aap pehle se solve karna jaante ho.

True ya false — justify karo

True ya false AUR reason bolo — reason hi asli cheez hai.

D - α

e^{α x}

ko annihilate karta hai kyunki

\frac{d}{d x} e^{α x} = α e^{α x}

, aur

α e^{α x} - α e^{α x} = 0

True — differentiate karne par function ka

α

guna wapas milta hai, isliye operator ke andar

α

guna function subtract karne par exactly cancel ho jaata hai.

D^{2} + 4

cos 2 x

ko annihilate karta hai lekin

sin 2 x

ko nahi.

False — dono

cos 2 x

aur

sin 2 x

same root pair

\pm 2 i

se aate hain, isliye

D^{2} + 4

dono ko kill karta hai; operator dono mein farq nahi kar sakta.

Kisi function ka annihilator unique hota hai.

False — annihilator ka koi bhi multiple (jaise

D (D - α)

) bhi

e^{α x}

ko annihilate karta hai; the annihilator matlab lowest-order wala, jo unique hota hai.

Agar

g (x) = 7

(ek constant) hai, to uska annihilator

D

hai.

True — constant

7 x^{0}

hai, jo root

r = 0

se ek baar generate hota hai, isliye

D

(yaani

D - 0

) ise zero kar deta hai kyunki

\frac{d}{d x} (7) = 0

g (x) = tan x

ka ek annihilator hai, isliye annihilator method apply hota hai.

False —

tan x

kisi bhi constant-coefficient homogeneous ODE ka solution nahi hai (uska koi finite-order killer nahi hai), isliye koi annihilator exist nahi karta; aapko Variation of parameters use karna hoga.

L [y] = g_{1} + g_{2}

ke liye aap

g_{1}

aur

g_{2}

ko alag-alag annihilate karke trial forms add kar sakte ho.

True — ye Superposition principle for linear ODEs hai:

L [y_{1}] = g_{1}

aur

L [y_{2}] = g_{2}

solve karo, phir

y_{p} = y_{p 1} + y_{p 2}

kyunki

L

linear hai.

g = x e^{2 x}

ke liye trial

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

hai.

False (generally) —

x e^{2 x}

root

2

doubled se aata hai, jiska annihilator

(D - 2)^{2}

hai; ye dono

e^{2 x}

aur

x e^{2 x}

produce karta hai, isliye trial

y_{p} = (A + B x) e^{2 x}

hai (koi overlap adjustment se pehle).

Agar

g

ke roots aur

L

ke roots mein se koi overlap nahi karta, to kabhi extra

x

ka factor nahi chahiye.

True — extra

x

tabhi aata hai jab annihilate karne se $L$ mein pehle se maujood root ki multiplicity badh jaaye; koi overlap nahi matlab koi multiplicity bump nahi, isliye koi

x

nahi.

D^{2}

har straight line

a x + b

ko annihilate karta hai.

True —

\frac{d}{d x} (a x + b) = a

aur ek baar aur differentiate karne par

0

milta hai; root

r = 0

do baar repeat hone par exactly

{1, x}

generate hota hai.

Error dhundho

Har line mein ek galat claim hai — kya galat hai aur sahi statement kya hai batao.

g = e^{2 x}

pehle se

y_{h}

mein hai, isliye main phir bhi

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

guess karunga."

Galat — kyunki

e^{2 x}

L [y] = 0

solve karta hai,

L [A e^{2 x}] = 0

milega, jo kabhi nonzero

g

ke barabar nahi ho sakta;

x

se multiply karo (root ab doubled ho gaya) to

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

milega.

g = cos 2 x

ke liye main sirf

A cos 2 x

include karunga kyunki right side mein koi sine nahi hai."

Galat — operator

L

cos

differentiate karte waqt sine aur cosine mix karta hai, isliye dono include karne chahiye:

y_{p} = A cos 2 x + B sin 2 x

; ek zero nikle, lekin aap pehle se assume nahi kar sakte.

x^{3}

ka annihilator

D^{3}

hai."

Galat —

x^{3}

ke liye root

0

ko chaar baar repeat karna padega (taaki

1, x, x^{2}, x^{3}

cover ho), isliye annihilator

D^{4} = D^{3 + 1}

hai; power

n + 1

hoti hai,

n

nahi.

A L [y] = 0

form karne ke baad main poori general solution ko

y_{p}

rakh leta hoon."

Galat —

L

ke roots jo

A L

mein hain wo

y_{h}

regenerate karte hain, jo

L [y] = 0

satisfy karta hai aur

g

mein kuch contribute nahi karta; sirf

A

ke roots se aane wale naye terms ko

y_{p}

ki form mein rakho.

g = e^{2 x} cos 3 x

, isliye annihilator

(D - 2) + (D^{2} + 9)

hai."

Galat — annihilators ko add nahi kar sakte;

e^{2 x} cos 3 x

roots

2 \pm 3 i

se aata hai, annihilator

(D - 2)^{2} + 9

deta hai, jo ek single quadratic factor hai.

"Kyunki

A

right side ko zero kar deta hai,

A [g] = 0

ka matlab

g = 0

hai."

Galat —

A [g] = 0

ka matlab hai

g

A

ke kernel mein hai (wo us homogeneous ODE ka solution hai), ye nahi ki

g

zero function hai;

A

ek nonzero

g

ko zero pe map karta hai.

y^{''} + y = cos x

(resonance) ke liye trial

y_{p} = A cos x + B sin x

hai."

Galat —

cos x

ka root

\pm i

hai, jo pehle se

y_{h}

mein hai; is overlap ki wajah se

y_{p} = x (A cos x + B sin x)

hoga, jo Resonance in forced oscillations ki pehchaan hai.

"Maine

y_{p}

find kar liya, to poora answer sirf

y_{p}

hai."

Galat — general solution

y = y_{h} + y_{p}

hai;

y_{h}

drop karne se wo arbitrary constants discard ho jaate hain jo initial/boundary conditions satisfy karne ke liye chahiye.

Why questions

Mechanism se jawab do, rule ka naam mat batao.

Annihilators sirf polynomials, exponentials, aur sinusoids (aur unke products) ke liye hi kyun exist karte hain?

Kyunki ye precisely wahi functions hain jo constant-coefficient homogeneous ODEs solve karte hain — har ek kisi characteristic root se generate hota hai — isliye unhe ulta padhne par finite-order operator milta hai; baaki functions ka aisa koi finite operator nahi hota.

g

ke root aur

L

ke root ka overlap kyun ek

x

ka factor force karta hai?

Overlapping root

A L [y] = 0

mein multiplicity gain karta hai, aur multiplicity

k

wala root

e^{α x}, x e^{α x}, \dots, x^{k - 1} e^{α x}

ka ladder produce karta hai; genuinely naya term isliye extra power of

x

carry karta hai.

Bade solution ka

y_{h}

portion

y_{p}

banate waqt kyun throw away karte hain?

Wo terms

L

ke roots se aati hain, isliye

L

unhe zero bhej deta hai — wo nonzero

g

produce kar hi nahi sakti, isliye sirf annihilator ke naye roots hi

g

ko match kar sakte hain.

y_{p}

A L [y] = 0

mein nahi balki original

L [y] = g

mein plug kyun karna padta hai?

Enlarged equation

A L [y] = 0

sirf

y_{p}

ki form batati hai; actual coefficients real right side

g

se match karke fix hote hain, jo sirf original equation mein appear karta hai.

cos β x

apna annihilator

D^{2} + β^{2}

sin β x

ke saath kyun share karta hai?

Dono complex root pair

\pm i β

se arise karte hain, aur

D^{2} + β^{2}

factor hota hai

(D - i β) (D + i β)

mein, isliye ye un dono roots ke har real combination ko annihilate karta hai — sine aur cosine dono ko.

Annihilator method constant-coefficient

L

tak hi kyun restricted hai?

Is trick ke liye

L

aur

A

ke roots ko ek single characteristic polynomial mein cleanly combine hona chahiye; variable coefficients ke saath koi characteristic equation nahi hoti, isliye koi root-counting story nahi aur koi guaranteed solution form nahi.

Resonance (physically unbounded growth) algebraic root overlap ke saath exactly kyun correspond karta hai?

Jab forcing frequency natural frequency se match karti hai,

g

ka root

L

ke root se coincide karta hai, multiplicity badhti hai, aur

x e^{α x}

(ya

x cos β x

) term bina bound ke badhta hai — dekho Resonance in forced oscillations.

Edge cases

Wo boundary situations jinhe recipe chupke se assume kar leti hai.

g (x) = 0

ka annihilator kya hai?

Identity operator (order

0

) — kyunki

0

pehle se hi kisi bhi operator se annihilate ho jaata hai, sabse lowest-order wala trivially "kuch mat karo" hai; practically

y_{p} = 0

aur

y = y_{h}

Agar

g

ka root

L

ke double root se overlap kare to kya hoga?

Root

A L

mein multiplicity teen tak pahunch jaata hai, isliye trial ko

x^{2}

se multiply karo (sabse chhoti power jo do existing

y_{h}

terms se aage step kare), jaise

y_{p} = A x^{2} e^{α x}

Kya

g = e^{2 x} + e^{2 x}

ko kabhi ek

e^{2 x}

term se aage special treatment chahiye?

Nahi — wo sum sirf

2 e^{2 x}

hai, ek single exponential with root

2

; ise ek term samjho jiska annihilator

(D - 2)

hai (ya

(D - 2)^{2}

agar

2

L

se overlap kare).

g = x

aur

L = D^{2} - 3 D + 2

ke liye, kya

L

ka constant term matter karta hai is baat ke liye ki extra

x

chahiye ya nahi?

Haan —

x

ka annihilator

D^{2}

hai (root

0

doubled); kyunki

r = 0

D^{2} - 3 D + 2

ka root nahi hai (roots

1, 2

hain), koi overlap nahi, isliye

y_{p} = A x + B

bina extra factor ke.

Agar

L

ka root

0

hai aur

g = 5

(constant) hai, to trial kya hai?

Constant ka root

0

L

ke root

0

se overlap karta hai, isliye

x

se multiply karo:

y_{p} = A x

(ya

A x^{k}

L

ki root

0

ki multiplicity se match kare).

Agar

g

ek aisa sum hai jiske pieces ke annihilators overlap karte hain, jaise

g = e^{2 x} + x e^{2 x}

Ek single highest-order annihilator

(D - 2)^{2}

mein combine karo (jo dono cover kare), phir ek trial

y_{p} = (A + B x) e^{2 x}

banao — shared root double-count mat karo.

Kya

g = e^{i x}

(complex exponential) forcing ke roop mein allowed hai?

Haan — uska annihilator

(D - i)

hai; complex-exponential shortcut

cos

aur

sin

forcing dono ek saath real/imaginary parts lete hue solve karta hai, alag

A cos + B sin

trial se bachata hai.