Visual walkthrough — First-order linear ODEs — integrating factor method (derivation)

4.6.4 · D2 · Maths › Ordinary Differential Equations › First-order linear ODEs — integrating factor method (derivat

Sab kuch ek akele zidd-bhar equation se shuru hota hai. Chalte hain usse milte hain.

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Step 1 — Equation se milo, aur dekho yeh "stuck" kyun hai

KYA HAI. Humare paas ek unknown function $y(x)$ hai — ek rahasya curve. Symbol $\frac{dy}{dx}$ ka matlab hai "the slope of that curve at each point"; yeh batata hai ki $x$ ke right move karne pe $y$ kitni tezi se badhti hai. $P(x)$ aur $Q(x)$ $x$ ke known functions hain (jo hum kisi bhi $x$ pe calculate kar sakte hain). Hum actual $y$ dhundna chahte hain.

YEH STUCK KYUN HAI. Left side do alag pieces ka addition hai: ek slope piece $\frac{dy}{dx}$ aur ek height piece $P(x)y$. Slope ko "undo" karne ke liye hum integrate karte hain — lekin slope ko cleanly tabhi integrate kar sakte ho jab poora left side ek hi cheez ka slope ho. Abhi yeh ek cheez ka slope hai plus kuch aur. Hum sum ko directly integrate nahi kar sakte.

PICTURE. Neeche, do pieces ko do alag arrows ke roop mein draw kiya gaya hai jo alag-alag directions mein khich rahe hain. Woh ek single motion mein combine hone se mana kar rahe hain — bilkul wahi "do bacche jo haath nahi pakdenge" wali baat jaise parent note mein thi.

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Step 2 — Wish: kya cheez ise un-stuck karti?

KYA HAI. Ek bilkul naya function $\mu (x)$ introduce karo — "mu of x" padho, bas ek naam ek helper multiplier ke liye jo hum choose kar sakte hain. Step 1 ki har term ko $\mu$ se multiply karo: $\underset{\text{slope piece}\times \mu }{\underbrace{\mu \frac{dy}{dx}}}+\underset{\text{height piece}\times \mu }{\underbrace{\mu Py}}=\underset{\text{push}\times \mu }{\underbrace{\mu Q}}.$

MULTIPLY KYUN KARTE HAIN? Multiply karne se equation ke solutions nahi badalte (jab tak $\mu \ne 0$): agar do cheezein barabar hain, toh $\mu$ ka unhe multiply karna fir bhi barabar rehta hai. Toh hum yeh free mein kar sakte hain, aur hum yeh freedom ek achhe left side ke liye khareedenge.

PICTURE. $\mu$ ko ek amber "coating" ki tarah draw kiya gaya hai jo dono pieces ke around wrap hai — sab pe same coating, taki woh finally merge ho sakein.

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Step 3 — Woh tool jo products merge karta hai: Product Rule

KYA HAI. Product Rule kehta hai: "$\mu$ times $y$" ka slope hai (first $\times$ slope of second) plus (slope of first $\times$ second). Yahan $\frac{d\mu }{dx}$ ka matlab hai "hamara helper $\mu$ khud kitni tezi se change hota hai".

LINE UP KYUN KARTE HAIN? Apna merged left side Product Rule ke right side ke saath side-by-side rakhke piece by piece compare karo — kyunki agar woh match karte hain, toh hamara left side simply $\frac{d}{dx}(\mu y)$ hi hai.

hamara left side (Step 2)	Product Rule (Step 3)	verdict
$\mu \frac{dy}{dx}$	$\mu \frac{dy}{dx}$	already identical ✓
$\mu Py$	$\frac{d\mu }{dx}y$	must be forced to match

PICTURE. Pehle terms perfectly overlap karte hain (green tick); doosre terms sirf shape mein same hain — dono "(kuch)$\times y$" — lekin "kuch" alag hai. Woh mismatch hi ek kaam bacha hai.

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Step 4 — Match force karo: $\mu$ pe ek condition ka janam

KYA HAI. Hum ab umeed nahi kar rahe — hum $\mu$ ko choose kar rahe hain taaki uski growth rate $\frac{d\mu }{dx}$ khud uske times $P$ ke barabar ho. Woh akeli equation poora admission price hai.

YEH PROGRESS KYUN HAI. Dhyan se dekho: yeh ek ODE hai, lekin ek aasaan wala. Ismein sirf $\mu$ aur $x$ hain — $y$ bilkul nahi. Aur iska special shape hai "rate of change $=$ (kuch) $\times$ current value", jo ek separable equation hai. Humne ek mushkil problem (find $y$) ko ek aasaan problem (find $\mu$) se trade kiya.

PICTURE. Ek dial: requirement "second terms equal" turn karne pe boxed condition $\frac{d\mu }{dx}=\mu P$ mein collapse ho jaati hai. Dekho $y$ poori picture se gum ho gaya hai.

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Step 5 — Variables separate karke $\mu$ solve karo

KYA HAI. Har $\mu$ left mein, har $x$ right mein ikatha karo: $\underset{\text{all the }\mu \text{ stuff}}{\underbrace{\frac{d\mu }{\mu }}}=\underset{\text{all the }x\text{ stuff}}{\underbrace{P(x)dx}}.$ Dono sides integrate karo: $\int \frac{d\mu }{\mu }=\int P(x)dx⟹\underset{\text{from }\int d\mu /\mu }{\underbrace{\ln |\mu |}}=\int P(x)dx.$

LOGARITHM KYUN? $\frac{1}{\mu }$ ka integral $\ln |\mu |$ hai — wahi ek antiderivative hai jo ek reciprocal ko "undo" karta hai. $\mu$ ko logarithm ke andar se free karne ke liye hum exponential $e^{(\cdot )}$ apply karte hain, kyunki $e$ aur $\ln$ inverse operations hain: $|\mu |=e^{\int Pdx}.$

$+C$ kahan gaya? Ek integration constant $e^C\cdot e^{\int Pdx}$ ke roop mein saath aata — $\mu$ ka ek constant multiple. Lekin $\mu$ ko ek constant se multiply karna Step 2 ke har term ko usi constant se multiply karta hai, jo cancel ho jaata hai. Toh hum simplest choice $e^C=1$ rakhte hain.

PICTURE. Separation plane ko ek "$\mu$-column" aur ek "$x$-column" mein split karta hai; exponential curve accumulated area $\int Pdx$ se uthti hai.

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Step 6 — Dekho left side kaise collapse karta hai

KYA HAI. Do zidd-bhar pieces $\mu y$ ek single object ke slope mein fuse ho gaye hain. Problem ka messy hissa gaya.

YEH KYUN KAAM KIYA. Kyunki humne Step 4 mein $\frac{d\mu }{dx}=\mu P$ force kiya tha, Product Rule ke do terms ab hamare do terms se exactly equal hain — toh unka sum definition ke hisaab se $\frac{d}{dx}(\mu y)$ ke barabar hai, luck se nahi.

PICTURE. Step 1 ke do arrows, ab dono amber-coated, ek bold arrow mein snap ho jaate hain jiska label $\frac{d}{dx}(\mu y)$ hai.

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Step 7 — Dono sides ko un-slope karo aur $y$ solve karo

KYA HAI. $\frac{d}{dx}(\mu y)=\mu Q$ ko $x$ ke respect mein integrate karo. Ek slope ko integrate karna cheez khud return karta hai: $\underset{\text{recovered product}}{\underbrace{\mu y}}=\int \mu (x)Q(x)dx+\underset{\text{the constant, here at last}}{\underbrace{C}}.$

$+C$ ABHI KYUN? Yeh poori equation ka pehla genuine integration hai, toh yahi jagah hai jahan single family-labelling constant belong karta hai. (Ise pehle, $\mu$ ke andar daalna, cancel ho jaata — Step 5.)

Final divide. Hamare paas $\mu y$ hai, $y$ nahi. $\mu$ se divide karo (legal hai kyunki $\mu =e^{(\cdots )}$ kabhi zero nahi hota — ek exponential hamesha positive hota hai):

PICTURE. Integration single arrow ko $C$ ki value se stack hue curves ki ek family mein lift karta hai; $\mu$ se divide karna unhe actual solution curves $y(x)$ mein reshape karta hai.

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Step 8 — Edge & degenerate cases (jo log ghumba dete hain)

Case A — $P(x)=0$. Tab $\mu =e^{\int 0dx}=e^0=1$. Equation already $\frac{dy}{dx}=Q$ hai, toh tum seedha integrate karo. Method silently plain integration mein reduce ho jaata hai — kuch break nahi hota.

Case B — $P,Q$ dono constant. $P=k$ (ek number) ke saath, $\mu =e^{kx}$. Yeh constant-coefficient special case hai; same formula jaana-pehchana $y=\frac{Q}{k}+C{e}^{-kx}$ shape produce karta hai.

Case C — $\frac{dy}{dx}$ ka coefficient $1$ nahi hai. Agar tumhe $a(x)\frac{dy}{dx}+\dots$ mile, tum abhi standard form mein nahi ho. Pehle poori equation ko $a(x)$ se divide karo. Agar yeh skip kiya, toh jo term $P$ ke roop mein feed karte ho woh galat hai aur Step 6 ka collapse fail ho jaata hai.

PICTURE. Teen mini-panels: flat $\mu =1$, exponential $\mu =e^{kx}$, aur ek "STANDARDIZE FIRST" gate jisse tumhe guzarna zaroori hai.

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Ek-picture summary

Ek diagram saare aath steps compress karta hai: stuck equation → $\mu$ se coat karo → Product Rule se match karo → forced condition → separable solve → exponential $\mu$ → collapse → integrate → divide → solution.

Recall Feynman retelling — ek story ki tarah bolo

Humne ek aisi equation se shuru kiya jiska left side do unfriendly pieces ka sum tha: ek slope aur ek height. Aisa sum ek move mein integrate nahi kar sakte. Toh humne ek magic coating $\mu$ invent ki aur poori equation pe paint kar di. Ab humne demand ki ki coating ke baad, left side ek single lump $\mu y$ ka slope ban jaaye. Ek hi rule hai puri calculus mein jo "slope-of-a-product" ko do added terms mein turn karta hai — woh hai Product Rule — toh humne apni coated equation us se line up ki aur piece by piece match kiya. Pehle pieces free mein match ho gaye; doosre pieces sirf tabhi match hue jab $\mu$ rate $\mu P$ se bade. Woh chhoti si demand khud ek aasaan separable equation thi jiska jawab — kyunki "growth proportional to size" hamesha exponential ka matlab hota hai — $\mu =e^{\int P}$ hai. Us coating ke saath, do pieces fuse ho gaye, humne un-slope kiya (integrate kiya) taaki $\mu y$ mile, ek constant $C$ dala, aur kabhi-zero-nahi-hone-wale $\mu$ se divide karke $y$ reveal kiya. Har "special" case — $P=0$, constant coefficients, non-1 leading coefficient — bas yahi story hai jisme ek knob thoda ghuma hua hai.

Recall Rapid self-test

Hum sirf add karne ki jagah integrate kyun karte hain? ::: Left side ek slope hai; integration woh operation hai jo kisi function ko uske slope se recover karta hai. Poori construction ko kaun sa ek rule force karta hai? ::: Product Rule — yeh ek product ki derivative ka do summed terms mein ek hi expansion hai. $\mu$ se divide karna hamesha legal kyun hai? ::: $\mu =e^{\int P}$ ek exponential hai, jo strictly positive hai, isliye kabhi zero nahi hota. Agar $P=0$ ho toh kya hota hai? ::: $\mu =e^0=1$; method plain direct integration mein degenerate ho jaata hai.

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Connections

• Product Rule — Step 3 ka engine.
• Separable ODEs — Step 5 mein condition solve karta hai.
• Exact ODEs — jahan integrating factors generalise hote hain.
• Bernoulli Equations — substitution se is linear form mein reduce hoti hain.
• Linear Constant-Coefficient ODEs — Step 8 ka Case B.