Integrating factors for non-exact equations — Hinglish notes

4.6.7 · Maths › Ordinary Differential Equations

Ek "fixer" jise tum ek stubborn ODE se multiply karte ho taaki woh exact ban jaye, aur phir tum use potential functions se solve kar sako.

Setup: "exact" ka matlab kya hai

Hum ek first-order ODE ko differential form mein likhte hain: $M (x, y) d x + N (x, y) d y = 0.$

Exactness test. Kyunki mixed partials commute karte hain ( $F_{x y} = F_{y x}$ smooth $F$ ke liye): $M_{y} = F_{x y} = F_{y x} = N_{x} .$ Isliye: $Exact ⟺ M_{y} = N_{x} .$

Problem: zyaadatar equations exact NAHI hoti

Agar $M_{y} \neq = N_{x}$ , toh koi $F$ exist nahi karta aur hum phanse hain... jab tak hum poori equation ko kisi function $μ (x, y)$ se multiply na karein, jise integrating factor kehte hain: $μ M d x + μ N d y = 0.$ Iske solution curves SAME hain (humne sirf kisi nonzero cheez se multiply kiya), lekin ab hum demand karte hain ki yeh exact ho.

$μ$ ke liye condition scratch se derive karna

Multiplied equation ki exactness ke liye zaroori hai $\frac{\partial}{\partial y} (μ M) = \frac{\partial}{\partial x} (μ N) .$ Dono sides ko product rule se expand karo: $μ_{y} M + μ M_{y} = μ_{x} N + μ N_{x} .$ Rearrange karo: $N μ_{x} - M μ_{y} = μ (M_{y} - N_{x})$

Yeh $μ$ ke liye ek PDE hai — generally original ODE jitna hi mushkil! Trick yeh hai ki guess karo ki $μ$ sirf ek hi variable par depend karta hai, jo ek term ko khatam kar deta hai.

Case 1: $μ = μ (x)$ only

Tab $μ_{y} = 0$ , $μ_{x} = μ^{'}$ . PDE ban jaati hai $N μ^{'} = μ (M_{y} - N_{x}) ⟹ \frac{μ ^{'}}{μ} = \frac{M _{y} - N _{x}}{N} .$

Case 2: $μ = μ (y)$ only

Symmetry se ( $μ_{x} = 0$ ): $- M μ^{'} = μ (M_{y} - N_{x}) ⟹ \frac{μ ^{'}}{μ} = \frac{N _{x} - M _{y}}{M} .$

Full procedure (the 80/20 core)

$M d x + N d y = 0$ ki tarah likho.
Test karo: kya $M_{y} = N_{x}$ hai? Agar haan, toh step 5 par jaao.
$\frac{M _{y} - N _{x}}{N}$ compute karo → $x$ ka function hai? $μ (x)$ use karo. Warna $\frac{N _{x} - M _{y}}{M}$ compute karo → $y$ ka function hai? $μ (y)$ use karo.
$μ$ se multiply karo. Ab exact hai.
$F$ dhundho: $F_{x} = M^{*}$ ko $x$ mein integrate karo, unknown $ϕ (y)$ add karo, $y$ mein differentiate karo, $N^{*}$ se match karo.
Solution: $F (x, y) = C$ .

Worked Example 1 — factor in $x$

Solve karo $(3 x y + y^{2}) d x + (x^{2} + x y) d y = 0.$

Yahaan $M = 3 x y + y^{2}$ , $N = x^{2} + x y$ .

$M_{y} = 3 x + 2 y$ , $N_{x} = 2 x + y$ . Kyun? $M$ ko $y$ ke saath differentiate karo, $N$ ko $x$ ke saath. Exact nahi kyunki $3 x + 2 y \neq = 2 x + y$ .

$\frac{M _{y} - N _{x}}{N} = \frac{( 3 x + 2 y ) - ( 2 x + y )}{x ^{2} + x y} = \frac{x + y}{x ( x + y )} = \frac{1}{x} .$ Yeh step kyun? Numerator factor aur cancel ho jaata hai $(x + y)$ se, aur sirf ek pure- $x$ function bachta hai — isliye Case 1 kaam karta hai.

$μ = exp (\int \frac{1}{x} d x) = e^{l n x} = x .$

Multiply karo: $M^{*} = 3 x^{2} y + x y^{2}$ , $N^{*} = x^{3} + x^{2} y$ . Exact check karo: $M_{y}^{*} = 3 x^{2} + 2 x y$ , $N_{x}^{*} = 3 x^{2} + 2 x y$ . ✓

$F$ dhundho: $F = \int (3 x^{2} y + x y^{2}) d x = x^{3} y + \frac{1}{2} x^{2} y^{2} + ϕ (y) .$ $ϕ (y)$ kyun? $x$ ke saath integration ka "constant" $y$ par depend kar sakta hai. $F_{y} = x^{3} + x^{2} y + ϕ^{'} (y)$ ko $N^{*} = x^{3} + x^{2} y$ ke barabar hona chahiye, isliye $ϕ^{'} (y) = 0$ .

$x^{3} y + \frac{1}{2} x^{2} y^{2} = C .$

Worked Example 2 — factor in $y$

Solve karo $y d x + (2 x - y e^{y}) d y = 0.$

$M = y$ , $N = 2 x - y e^{y}$ . $M_{y} = 1$ , $N_{x} = 2$ . Exact nahi.

Case 1 try karo: $\frac{M _{y} - N _{x}}{N} = \frac{- 1}{2 x - y e ^{y}}$ — abhi bhi $x$ hai. ✗ Case 2 try karo: $\frac{N _{x} - M _{y}}{M} = \frac{2 - 1}{y} = \frac{1}{y}$ — pure $y$ . ✓

$μ = exp (\int \frac{1}{y} d y) = y .$

Multiply karo: $M^{*} = y^{2}$ , $N^{*} = 2 x y - y^{2} e^{y}$ . Check karo: $M_{y}^{*} = 2 y$ , $N_{x}^{*} = 2 y$ . ✓

$F = \int y^{2} d x = x y^{2} + ϕ (y)$ . Phir $F_{y} = 2 x y + ϕ^{'} (y) = 2 x y - y^{2} e^{y}$ , isliye $ϕ^{'} (y) = - y^{2} e^{y}$ . Integration by parts se: $ϕ = - e^{y} (y^{2} - 2 y + 2)$ .

$x y^{2} - e^{y} (y^{2} - 2 y + 2) = C .$

Common mistakes

Recall Feynman: explain to a 12-year-old

Ek pahadi socho jahan height $F$ hai. Aise chalna ki teri height kabhi na badle matlab contour line par chalna — yahi solution hai. Kuch "chalne ke rules" ( $M d x + N d y$ ) kisi real pahadi ki contour ko bilkul trace nahi karte. Integrating factor $μ$ aise magic glasses pehanne jaisa hai jo map ko rescale karte hain jab tak rule ek real pahadi se match na kare. Jab match ho jaata hai, tum bas contour padh lo: "height $C$ par raho."

Flashcards

M d x + N d y = 0

exact kab hoti hai?

Jab

M_{y} = N_{x}

ho (tab ek potential

F

exist karta hai jisme

F_{x} = M, F_{y} = N

ho).

Exactness solution

F = C

kyun deti hai?

Kyunki

M d x + N d y = d F

hai, aur

d F = 0

ka matlab hai

F

solutions ke saath constant hai.

PDE jo integrating factor

μ

ko satisfy karni chahiye?

N μ_{x} - M μ_{y} = μ (M_{y} - N_{x})

x

-only integrating factor ke liye condition kya hai?

\frac{M _{y} - N _{x}}{N}

sirf

x

ka function ho.

x

-only factor ka formula kya hai?

μ (x) = exp \int \frac{M _{y} - N _{x}}{N} d x

y

-only factor ke liye condition aur formula kya hai?

\frac{N _{x} - M _{y}}{M}

sirf

y

ka function ho;

μ (y) = exp \int \frac{N _{x} - M _{y}}{M} d y

μ

se multiply karna legitimate kyun hai?

Yeh woh jagah nahi badlaata jahan expression zero hai (same solution curves), sirf form badlata hai.

μ

se multiply karne ke baad kya karna zaroori hai?

M_{y}^{*} = N_{x}^{*}

re-verify karo, phir partial integration se

F

dhundho.

Connections

Exact differential equations
Total differentials and potential functions
Linear first-order ODEs — factor $e^{\int P d x}$ ek special integrating factor hai
Separable equations
Mixed partial derivatives (Clairaut's theorem)
Conservative vector fields