First-order linear ODEs — integrating factor method (derivation): Question bank

4.6.4 · D5 · Maths › Ordinary Differential Equations › First-order linear ODEs — integrating factor method (derivat

True ya false — justify karo

Har woh equation jo

\frac{d y}{d x}

aur

y

contain kare, ek linear ODE hoti hai.

False. Linearity ke liye zaroori hai ki

y

aur

\frac{d y}{d x}

dono sirf first power par aayein, kabhi aapas mein multiply na hon ya kisi function mein feed na hon; jaise

y \frac{d y}{d x}

sin y

ise tod deta hai.

\frac{d y}{d x} + x y^{2} = x

ko integrating factor method se jaise-taise solve kiya ja sakta hai.

False.

y^{2}

term ise nonlinear banata hai, isliye product-rule collapse fail ho jaata hai. Yeh ek Bernoulli equation hai (

n = 2

) aur pehle substitution

v = y^{1 - n}

chahiye.

Integrating factor

μ = e^{\int P d x}

hamesha positive hota hai.

True. Yeh ek exponential hai, aur

e^{(koi bhi real number)} > 0

; isliye hum final equation ko

μ

se freely divide kar sakte hain bina zero se divide karne ki chinta ke.

Har baar integrate karte waqt

+ C

likhna zaroori hai,

μ

dhundhte waqt bhi.

False.

μ

dhundhte waqt hume sirf ek working factor chahiye; extra constant ek multiplicative

e^{C}

ban jaata hai jo upar-neeche cancel ho jaata hai.

+ C

sirf final integration mein rakho.

Agar

P (x) = 0

ho, toh bhi integrating factor method apply hoti hai.

True, aur yeh gracefully degenerate ho jaati hai:

μ = e^{\int 0 d x} = e^{0} = 1

, toh equation already

\frac{d}{d x} (y) = Q

ban jaati hai — sirf seedha integration, jo separable baseline hai.

Method ke liye zaroori hai ki

Q (x)

closed form mein integrable ho.

False. General solution

y = \frac{1}{μ} [\int μ Q d x + C]

hamesha valid hai; agar

\int μ Q d x

ka koi elementary form na ho, toh hum use simply integral ki tarah chhod dete hain. Method kabhi fail nahi hoti, antiderivative bas simplify nahi ho sakta.

μ

se multiply karna ODE ka solution set badal deta hai.

False (jab tak

μ \neq = 0

hai, jo guaranteed hai). Dono sides ko ek never-zero function se multiply karne par ek equivalent equation milti hai, isliye koi solution na banta hai na khoota hai.

Ek first-order linear ODE ka hamesha exactly ek solution hota hai.

False. General solution mein ek free constant

C

hota hai, jo solutions ki ek family deta hai; tumhe ek akela solution tab milta hai jab koi initial condition jaise

y (x_{0}) = y_{0}

impose karo.

Error dhundho

x \frac{d y}{d x} + y = cos x

ke liye main

P = 1

leta hoon isliye

μ = e^{x}

Error:

\frac{d y}{d x}

ka coefficient

x

hai,

1

nahi. Pehle standardize karna zaroori hai

x

se divide karke, jisse

P = \frac{1}{x}

milta hai aur

μ = x

, na ki

e^{x}

μ y = \int μ Q d x + C

par pahunchne ke baad, mera answer

y = \int μ Q d x + C

hai."

Error: tum $μ$ se divide karna bhool gaye. Left side

μ y

hai, isliye

y = \frac{1}{μ} [\int μ Q d x + C]

— aur

C

ko bhi

μ

se divide karna padega, isliye yeh

y

mein ek bare additive constant nahi hai.

μ = e^{\int \frac{1}{x} d x} = e^{l n ∣ x ∣ + C}

, isliye main

+ C

rakhta hoon."

Error: constant drop karna hi poora point hai.

e^{l n ∣ x ∣ + C} = e^{C} ∣ x ∣

;

e^{C}

factor cancel ho jaata hai jab yeh dono sides ko multiply karta hai, isliye hum sabse simple

μ = ∣ x ∣

lete hain (working domain par aksar

x

μ = e^{\int 3 d x} = e^{3}

kyunki

\int 3 d x = 3

Error:

\int 3 d x = 3 x

hai, $3$ nahi. Integrating factor

μ = e^{3 x}

hai;

x

bhoolne se ek function ek bekar constant ban jaata hai jo left side ko collapse nahi karega.

"Left side

\frac{d}{d x} (μ y)

ban jaati hai, isliye main ise

μ \frac{d y}{d x}

bhi likh sakta hoon."

Error:

\frac{d}{d x} (μ y) = μ \frac{d y}{d x} + \frac{d μ}{d x} y

Product Rule ke anusaar — ismein do terms hain.

\frac{d μ}{d x} y

piece drop karna exactly woh term kho deta hai jisne collapse ko kaam karaya.

"Maine initial condition

μ y = \int μ Q d x + C

par

y

ke liye solve karne se pehle apply ki."

Zaroorat nahi ki galat ho, lekin risky aur premature hai. Sabse clean hai

y (x_{0}) = y_{0}

ko final

y (x)

par apply karna; jaldi apply karne se

μ (x_{0})

sahi evaluate karna padta hai aur arithmetic slips ka khatra rehta hai.

Why questions

Hum

μ

se multiply kyun karte hain, kuch add ya substitute kyun nahi karte?

Kyunki multiplication hi hai jo Product Rule ko fire karne deti hai:

\frac{d}{d x} (μ y)

ek product ki derivative hai, isliye sirf ek multiplicative factor hi

μ \frac{d y}{d x} + μ P y

ko ek single derivative mein reshape kar sakta hai.

μ = e^{\int P d x}

kyun hai, koi aur function kyun nahi?

Kyunki collapse ke liye zaroori hai

\frac{d μ}{d x} = μ P

; yeh condition ek separable ODE hai jiska solution forced hai

μ = e^{\int P d x}

hone ke liye. Yeh guess nahi hai — yeh woh unique factor hai jo condition allow karti hai.

\frac{d y}{d x}

ka coefficient shuru karne se pehle exactly

1

kyun hona chahiye?

Kyunki jis term ko

\frac{d μ}{d x} y

ke barabar hona hai woh

μ P y

hai, aur

P

is assumption par read hota hai ki

\frac{d y}{d x}

ka coefficient

1

hai. Agar woh

a (x)

ho, toh sahi

P

hai

\frac{( coeff of y )}{a ( x )}

, isliye normalization skip karne par galat

P

use hota hai.

μ

dhundhna ek separable problem kyun ban jaata hai?

Kyunki collapse condition

\frac{d μ}{d x} = μ P (x)

mein rearrange karne par

\frac{d μ}{μ} = P d x

milta hai — ek separable equation ki textbook shape, jisme

μ

ek side aur

x

doosri side par hai.

Hum

e^{l n ∣ x ∣} = ∣ x ∣

mein absolute-value subtleties ko ignore karke sirf

x

kyun use kar sakte hain?

Kyunki hum ek aisi interval par kaam karte hain jahan

x

ek sign rakhta hai (solution ka domain); us interval par

∣ x ∣ = \pm x

hai, aur sign us multiplicative constant mein absorb ho jaata hai jise hum pehle se drop karne par raazi ho gaye hain.

Wahi

C

jo integration mein ek baar appear hota hai, answer mein

μ

se divide kyun ho jaata hai?

Kyunki

C

bracket ke andar

μ y = \int μ Q d x + C

par introduce hota hai, aur bilkul agla step poori right side ko

μ

se divide karta hai, isliye

C

saath mein

\frac{C}{μ ( x )}

ki tarah aata hai — jo aksar

x

ka ek genuine function hota hai, constant nahi.

Constant-coefficient case (Linear Constant-Coefficient ODEs) sirf ek special instance kyun hai?

Jab

P

ek constant

k

ho,

\int P d x = k x

aur

μ = e^{k x}

— method bilkul waise hi chalti hai; derivation mein kahi bhi assume nahi kiya gaya ki

P

vary karta hai.

Edge cases

Agar

P (x)

mein singularity ho, jaise

P = \frac{1}{x}

x = 0

par, toh method ka kya hoga?

Integrating factor

μ = x

x = 0

par vanish karta hai, isliye

μ

se divide karna wahan invalid hai. Solution sirf ek aisi interval par guaranteed hai jo singularity contain nahi karti (jaise

x > 0

Agar

Q (x) \equiv 0

ho, toh kya method phir bhi kaam karti hai, aur kya milta hai?

Haan.

Q = 0

ke saath equation homogeneous hai;

\frac{d}{d x} (μ y) = 0

μ y = C

milta hai, isliye

y = \frac{C}{μ} = C e^{- \int P d x}

— ek pure decay/growth family, koi particular part nahi.

Kya integrating factor kabhi zero ho sakta hai aur final division tod sakta hai?

Nahi. Kyunki

μ = e^{(\dots)}

ek exponential hai, yeh strictly positive hota hai jahan bhi exponent finite ho;

μ

se division definition ki interval par hamesha valid hai.

Kya

\frac{d y}{d x} + P (x) y = Q (x)

jab

P, Q

discontinuous hon, ek piece mein solve ho sakta hai?

Generally nahi. Agar

P

Q

jump kare, toh solution sirf har us interval par continuous guaranteed hai jahan woh continuous hain; tum piece-by-piece solve karte ho aur joins par values match karte ho.

Kya ho agar equation multiply karne se pehle already exact derivative ho, yaani

μ

1

nikle?

Tab

P = 0

hai aur left side already

\frac{d}{d x} (y)

hai; integrating factor

μ = 1

hai aur method direct integration mein reduce ho jaati hai — Exact ODEs viewpoint ke saath consistent, jahan equation shuru se exact thi.

Kya kabhi roles swap karna —

x

unknown,

y

variable — help karta hai?

Haan — agar koi equation

y

mein nonlinear ho lekin

\frac{d x}{d y} + P (y) x = Q (y)

ke roop mein dekhen toh linear ho, tum identical method

x

aur

y

interchange karke apply karte ho; linearity is baat ki property hai ki unknown kaise appear karta hai, uske naam ki nahi.

Connections

Separable ODEs — woh sub-problem jo $μ$ deta hai.
Product Rule — upar har collapse claim ke peeche ka mechanism.
Exact ODEs — $μ = 1$ / already-exact edge case.
Bernoulli Equations — " $y^{2}$ linear nahi hai" wale traps yahan point karte hain.
Linear Constant-Coefficient ODEs — constant- $P$ special case.