Integrating factors for non-exact equations: Worked examples

4.6.7 · D3 · Maths › Ordinary Differential Equations › Non-exact equations ke liye Integrating Factors

Yeh page Integrating factors for non-exact equations ki "shooting range" hai. Hum tumhare upar har tarah ki problem chalate hain — clean $x$ -factors, clean $y$ -factors, woh trap jahan tumhe beech mein switch karna padta hai, ek already-exact equation (degenerate case), ek limiting/undefined case, ek word problem, aur ek exam twist — aur hum har ek ko end tak solve karte hain, phir verify bhi karte hain.

Shuru karne se pehle: yahan sab kuch parent note ke ek test aur do formulas par tika hua hai. Main unhe seedhe shabdon mein dobara bolunga taaki koi symbol bina samjhe naa aaye.

Recall Teen tools jinpar hum rely karte hain

Exactness test. Equation ko $M d x + N d y = 0$ ki tarah likho. Yahan $M$ hai "woh cheez jo $d x$ step ko right mein multiply karti hai" aur $N$ hai "woh cheez jo $d y$ step ko upar multiply karti hai". Equation exact hoti hai jab $M_{y} = N_{x}$ — yaani "M ka $y$ -slope, N ke $x$ -slope ke barabar hai". ( $M_{y}$ ka matlab: $x$ ko freeze karo, dekho $M$ kitni tez $y$ ke badhne se change hota hai.) $x$ -factor test: agar $\frac{M _{y} - N _{x}}{N}$ mein koi $y$ nahi bachta, toh ise $g (x)$ kaho aur $μ = exp \int g d x$ rakho. $y$ -factor test: agar $\frac{N _{x} - M _{y}}{M}$ mein koi $x$ nahi bachta, toh ise $h (y)$ kaho aur $μ = exp \int h d y$ rakho. Pehle figure mein aane wali hai ek picture ki " $μ$ kya fix karta hai".

Scenario matrix

Is topic ki har problem in cells mein se kisi ek mein aati hai. Neeche ke worked examples un cells ke saath tagged hain jo woh hit karte hain, aur milke yeh sabhi ko cover karte hain.

#	Cell (scenario class)	Kya khaas baat hai	Example
A	Already exact ( $M_{y} = N_{x}$ )	degenerate case: $μ = 1$ , koi fixing nahi chahiye	Ex 1
B	Clean $x$ -factor	ratio apna $y$ kho deta hai	Ex 2
C	Clean $y$ -factor	pehla ratio abhi bhi $x$ rakhta hai; switch karo aur jeet lo	Ex 3
D	Sign trap in the $y$ -numerator	$N_{x} - M_{y}$ vs $M_{y} - N_{x}$ — purana sign carry nahi karna	Ex 4
E	Neither pure- $x$ nor pure- $y$ (limiting/failure)	dono ratios dono letters rakhte hain — kya karein	Ex 5
F	Word problem (real units)	model → $M, N$ identify karo → solve → units check karo	Ex 6
G	Linear ODE in disguise	$μ = e^{\int P d x}$ ek special $x$ -factor hai	Ex 7
H	Exam twist: negative/zero regions	$μ$ ka sign, domain jahan $μ$ undefined hai	Ex 8

Figure — Integrating factors for non-exact equations

Cell A — already exact (degenerate case)

Worked example Ex 1.

$(2 x y + 1) d x + (x^{2}) d y = 0$ solve karo. Forecast: padhne se pehle guess karo — kya tumhe integrating factor ki zaroorat padegi? Pehle apne dimag mein dono slopes compute karo.

Step 1 — $M, N$ padhna. $M = 2 x y + 1$ , $N = x^{2}$ . Yeh step kyun? Baad ka sab kuch $M$ aur $N$ ke named hone par depend karta hai.

Step 2 — exactness test. $M_{y} = 2 x$ ( $x$ freeze karo, $y$ mein differentiate karo), $N_{x} = 2 x$ ( $y$ freeze karo, $x$ mein differentiate karo). Equal! Yeh step kyun? Agar $M_{y} = N_{x}$ toh hum Cell A mein hain: $μ = 1$ , saari factor machinery skip karo.

Step 3 — $F$ banana. $F = \int M d x = \int (2 x y + 1) d x = x^{2} y + x + ϕ (y) .$ $ϕ (y)$ kyun? $x$ mein integrate karne se $y$ freeze ho jaata hai, isliye "constant" secretly $y$ par depend kar sakta hai.

Step 4 — $N$ match karo. $F_{y} = x^{2} + ϕ^{'} (y)$ ko $N = x^{2}$ ke barabar hona chahiye, toh $ϕ^{'} (y) = 0$ , $ϕ = const$ .

$x^{2} y + x = C .$

Verify: answer ko implicitly differentiate karo: $d (x^{2} y + x) = (2 x y + 1) d x + x^{2} d y = 0$ . Bilkul original left side. ✓

Cell B — clean $x$ -factor

Worked example Ex 2.

$(3 x y + y^{2}) d x + (x^{2} + x y) d y = 0$ solve karo. Forecast: tumhara bet kaun sa single-variable ratio apna extra letter khoega — $x$ -wala ya $y$ -wala?

Step 1 — naam lo aur test karo. $M = 3 x y + y^{2}$ , $N = x^{2} + x y$ . $M_{y} = 3 x + 2 y$ , $N_{x} = 2 x + y$ . Equal nahi → exact nahi. Yeh step kyun? Confirm karta hai ki $F$ banana waste karne se pehle factor chahiye.

Step 2 — $x$ -test try karo. $\frac{M _{y} - N _{x}}{N} = \frac{( 3 x + 2 y ) - ( 2 x + y )}{x ^{2} + x y} = \frac{x + y}{x ( x + y )} = \frac{1}{x} .$ $(x + y)$ cancel ho jaata hai aur hum pure- $x$ function ke saath reh jaate hain — toh $x$ -factor exist karta hai. Yeh clean Cell B hai. Yeh step kyun? Cell B ka poora point yahi hai: $x$ -ratio sach mein apna $y$ kho deta hai.

Step 3 — $μ$ banao. $μ = exp \int \frac{1}{x} d x = e^{l n x} = x .$

Step 4 — multiply karo aur re-test karo. $M_{⋆} = μ M$ , $N_{⋆} = μ N$ use karke: $M_{⋆} = x (3 x y + y^{2}) = 3 x^{2} y + x y^{2}$ , $N_{⋆} = x (x^{2} + x y) = x^{3} + x^{2} y$ . $(M_{⋆})_{y} = 3 x^{2} + 2 x y$ , $(N_{⋆})_{x} = 3 x^{2} + 2 x y$ . Equal ✓ Re-test kyun? Parent note ki #1 galti: factor ko re-check kiye bina kabhi trust mat karo.

Step 5 — $F$ banao. $F = \int M_{⋆} d x = \int (3 x^{2} y + x y^{2}) d x = x^{3} y + \frac{1}{2} x^{2} y^{2} + ϕ (y) .$ $F_{y} = x^{3} + x^{2} y + ϕ^{'} (y) = N_{⋆} = x^{3} + x^{2} y \Rightarrow ϕ^{'} = 0.$

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

Verify: $d (x^{3} y + \frac{1}{2} x^{2} y^{2}) = (3 x^{2} y + x y^{2}) d x + (x^{3} + x^{2} y) d y = μ (M d x + N d y)$ jahan $μ = x$ . ✓

Cell C — pehla ratio $x$ rakhta hai, switch karo aur jeet lo

Worked example Ex 3.

$y d x + (2 x - y e^{y}) d y = 0$ solve karo. (parent ka Ex 2, yahan re-checked) Forecast: kaun sa test jeetega, aur $ϕ (y)$ ke liye kya chahiye hoga (plain integral ya by-parts)?

Step 1 — naam lo aur test karo. $M = y$ , $N = 2 x - y e^{y}$ . $M_{y} = 1$ , $N_{x} = 2$ . Exact nahi.

Step 2 — $x$ -test. $\frac{M _{y} - N _{x}}{N} = \frac{- 1}{2 x - y e ^{y}}$ — abhi bhi $x$ hai. ✗ $y$ -test. $\frac{N _{x} - M _{y}}{M} = \frac{2 - 1}{y} = \frac{1}{y}$ — pure $y$ . ✓ (Cell C.)

Step 3 — $μ$ . $μ = exp \int \frac{1}{y} d y = y .$

Step 4 — multiply karo, re-test karo. $M_{⋆} = y^{2}$ , $N_{⋆} = 2 x y - y^{2} e^{y}$ . $(M_{⋆})_{y} = 2 y = (N_{⋆})_{x}$ . ✓

Step 5 — $F$ . $F = \int y^{2} d x = x y^{2} + ϕ (y)$ . Phir $F_{y} = 2 x y + ϕ^{'} (y) = 2 x y - y^{2} e^{y} \Rightarrow ϕ^{'} = - y^{2} e^{y}$ . Do baar integration by parts: $\int y^{2} e^{y} d y = e^{y} (y^{2} - 2 y + 2)$ , toh $ϕ = - e^{y} (y^{2} - 2 y + 2)$ .

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

Verify: $ϕ^{'} (y) = \frac{d}{d y} [- e^{y} (y^{2} - 2 y + 2)] = - e^{y} (y^{2} - 2 y + 2) - e^{y} (2 y - 2) = - e^{y} y^{2}$ . ✓

Cell D — sign trap

Worked example Ex 4.

$(x y - 1) d x + (x^{2} - x y) d y = 0$ solve karo. Forecast: yahan danger yeh hai ki $M_{y} - N_{x}$ ko $y$ -formula mein copy kar lo. Upar kaun sa sign aana chahiye?

Step 1 — naam lo aur test karo. $M = x y - 1$ , $N = x^{2} - x y$ . $M_{y} = x$ , $N_{x} = 2 x - y$ . Exact nahi.

Step 2 — $x$ -test. $\frac{M _{y} - N _{x}}{N} = \frac{x - ( 2 x - y )}{x ^{2} - x y} = \frac{- x + y}{x ( x - y )} = \frac{- ( x - y )}{x ( x - y )} = \frac{- 1}{x}$ — pure $x$ . ✓ Yeh clean kyun hua? Numerator $y - x$ exactly $- (x - y)$ hai, jo $N$ mein chupi $(x - y)$ ko cancel karta hai.

Step 3 — $μ$ . $μ = exp \int \frac{- 1}{x} d x = e^{- l n x} = x^{- 1} .$

Step 4 — multiply karo, re-test karo. $M_{⋆} = \frac{x y - 1}{x} = y - \frac{1}{x}$ , $N_{⋆} = \frac{x ^{2} - x y}{x} = x - y$ . $(M_{⋆})_{y} = 1$ , $(N_{⋆})_{x} = 1$ . ✓

Step 5 — $F$ . $F = \int (y - \frac{1}{x}) d x = x y - ln ∣ x ∣ + ϕ (y)$ . $F_{y} = x + ϕ^{'} (y) = x - y \Rightarrow ϕ^{'} = - y \Rightarrow ϕ = - \frac{1}{2} y^{2} .$

$x y - ln ∣ x ∣ - \frac{1}{2} y^{2} = C .$

Verify: $d (answer) = (y - \frac{1}{x}) d x + (x - y) d y = \frac{1}{x} (M d x + N d y)$ . ✓

Cell E — na pure- $x$ na pure- $y$ (limiting/failure case)

Worked example Ex 5.

$(y) d x + (x^{2} + y^{2}) d y = 0$ par single-variable factor se koshish karo. Forecast: kya tumhe $μ (x)$ ya $μ (y)$ exist karne ki umeed hai? "Neither" predict karo aur dekho kyun.

Step 1 — test. $M = y$ , $N = x^{2} + y^{2}$ . $M_{y} = 1$ , $N_{x} = 2 x$ . Exact nahi.

Step 2 — $x$ -test. $\frac{M _{y} - N _{x}}{N} = \frac{1 - 2 x}{x ^{2} + y ^{2}}$ — DONO letters hain. ✗ $y$ -test. $\frac{N _{x} - M _{y}}{M} = \frac{2 x - 1}{y}$ — $x$ bach jaata hai. ✗ Yeh kyun hota hai? Koi bhi ratio ek variable mein collapse nahi hota, isliye koi single-variable $μ$ exist nahi karta. Yeh Cell E hai, honest failure case.

Step 3 — iske bajaaye kya karein. Tumhe ya toh $μ (x^{a} y^{b})$ ki form ka factor guess karna hoga ya koi aur shape, ya phir bilkul alag method par jaana hoga. Yeh mention kyun? Taaki tum kabhi aise formula par forever grind naa karo jo apply hi nahi ho sakta. Jaldi failure recognize karna ek skill hai.

Escape route (bonus). $\frac{d x}{d y} = - \frac{x ^{2} + y ^{2}}{y}$ ki tarah rewrite karo — yeh hamare topic ka nahi hai, lekin homogeneous-type substitution $x = v y$ handle kar leta hai. Lesson wahi rehta hai: Cell E ka matlab parent ke dono formulas genuinely kuch nahi dete.

Verify (ki dono ratios sachchi mein dono letters rakhte hain): $(x, y) = (1, 1)$ par $x$ -ratio hai $\frac{1 - 2}{2} = - \frac{1}{2}$ lekin $(1, 2)$ par yeh hai $\frac{1 - 2}{5} = - \frac{1}{5}$ — value fixed $x$ par $y$ ke saath badli, prove karta hai ki yeh akele $x$ ka function nahi. ✓

Cell F — real units wala word problem

Worked example Ex 6. Ek conserved-quantity model deta hai

$2 x y d x + (x^{2} + 3 y^{2}) d y = 0$ , jahan $x$ (metres) ek position coordinate hai aur $y$ (metres) ek equipotential trace karne wale particle ka doosra position coordinate hai. Curves ki family dhundho aur dimensional consistency check karo. Forecast: exact hai ya nahi? Dono slopes peek karo.

Step 1 — naam lo aur test karo. $M = 2 x y$ , $N = x^{2} + 3 y^{2}$ . $M_{y} = 2 x$ , $N_{x} = 2 x$ . Already exact (word problem ke andar Cell A pattern). Yeh step kyun? Real models often purpose se exact banaye jaate hain; $μ$ dhundne se pehle hamesha test karo.

Step 2 — $F$ banao. $F = \int 2 x y d x = x^{2} y + ϕ (y)$ . $F_{y} = x^{2} + ϕ^{'} (y) = x^{2} + 3 y^{2} \Rightarrow ϕ^{'} = 3 y^{2} \Rightarrow ϕ = y^{3} .$

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

Step 3 — dimensional check (sahi tarike se kiya). Dono $x$ aur $y$ metres mein hone par, term $x^{2} y$ ki units hain $m^{2} \cdot m = m^{3}$ aur $y^{3}$ ki units hain $m^{3}$ — yeh match karte hain, toh sum $x^{2} y + y^{3} = C$ dimensionally sound hai jahan $C$ in $m^{3}$ . Yeh kyun matter karta hai: agar $F$ ke dono terms ki alag units hoti toh tumhe pata chal jaata ki algebra mein koi galti hui — ek free error-catcher. (Isliye maine dono variables metres mein chunhe instead of seconds aur degrees mix karne ke, jo artificial rescaling force karta.)

Verify: $d (x^{2} y + y^{3}) = 2 x y d x + (x^{2} + 3 y^{2}) d y = 0$ . ✓

Cell G — linear ODE disguise mein

Worked example Ex 7. Linear ODE

$\frac{d y}{d x} + \frac{2}{x} y = x$ ko integrating factor use karke solve karo, aur ise ek special $x$ -factor ki tarah dekho. Forecast: ek first-order linear ODE $y^{'} + P y = Q$ ka hamesha $μ = e^{\int P d x}$ hota hai. Compute karne se pehle $μ$ guess karo.

Step 1 — differential form mein rakho. Sab kuch ek side move karo: $(\frac{2}{x} y - x) d x + d y = 0$ . Toh $M = \frac{2}{x} y - x$ , $N = 1$ . Yeh step kyun? $M, N$ machinery use karne ke liye humein $d x, d y$ layout chahiye.

Step 2 — $x$ -test. $M_{y} = \frac{2}{x}$ , $N_{x} = 0$ . $\frac{M _{y} - N _{x}}{N} = \frac{2/ x}{1} = \frac{2}{x}$ — pure $x$ . ✓ Yeh guaranteed kyun hai: linear $y^{'} + P y = Q$ ke liye yeh ratio hamesha exactly $P (x)$ hota hai, isliye $x$ -factor formula $μ = e^{\int P d x}$ reproduce karta hai. Dekho Linear first-order ODEs.

Step 3 — $μ$ . $μ = exp \int \frac{2}{x} d x = e^{2 l n x} = x^{2} .$

Step 4 — multiply karo, re-test karo. $M_{⋆} = 2 x y - x^{3}$ , $N_{⋆} = x^{2}$ . $(M_{⋆})_{y} = 2 x = (N_{⋆})_{x}$ . ✓

Step 5 — $F$ . $F = \int (2 x y - x^{3}) d x = x^{2} y - \frac{1}{4} x^{4} + ϕ (y)$ . $F_{y} = x^{2} + ϕ^{'} = x^{2} \Rightarrow ϕ^{'} = 0.$

$x^{2} y - \frac{1}{4} x^{4} = C ⟺ y = \frac{C}{x ^{2}} + \frac{x ^{2}}{4} .$

Verify: $y = \frac{C}{x ^{2}} + \frac{x ^{2}}{4}$ ke saath, $y^{'} = - \frac{2 C}{x ^{3}} + \frac{x}{2}$ ; phir $y^{'} + \frac{2}{x} y = - \frac{2 C}{x ^{3}} + \frac{x}{2} + \frac{2 C}{x ^{3}} + \frac{x}{2} = x$ . ✓ ( $Q = x$ se match karta hai)

Cell H — exam twist: monomial factor, signs, zeros

Worked example Ex 8.

$(3 y + 4 x y^{2}) d x + (2 x + 3 x^{2} y) d y = 0$ solve karo, phir batao ki integrating factor kahan undefined ya zero hai. Forecast: dono single-variable tests fail honge — toh hum monomial $μ = x^{a} y^{b}$ dhundhte hain. Padhne se pehle $a, b$ guess karo.

Step 1 — test. $M = 3 y + 4 x y^{2}$ , $N = 2 x + 3 x^{2} y$ . $M_{y} = 3 + 8 x y$ , $N_{x} = 2 + 6 x y$ . Exact nahi.

Step 2 — single-variable tests dono fail karte hain. $\frac{M _{y} - N _{x}}{N} = \frac{1 + 2 x y}{2 x + 3 x ^{2} y}$ ( $y$ hai); $\frac{N _{x} - M _{y}}{M} = \frac{- 1 - 2 x y}{3 y + 4 x y ^{2}}$ ( $x$ hai). Dono fail → $μ = x^{a} y^{b}$ try karo. Yeh step kyun? Jab dono single-variable tests fail hon (Cell E symptom), monomial guess often ek exam problem rescue kar leta hai.

Step 3 — $a, b$ ke liye systematic matching. $μ = x^{a} y^{b}$ ke saath, $M_{⋆} = μ M$ , $N_{⋆} = μ N$ likho aur exactness $(M_{⋆})_{y} = (N_{⋆})_{x}$ impose karo. Compute karo: $M_{⋆} = 3 x^{a} y^{b + 1} + 4 x^{a + 1} y^{b + 2}, N_{⋆} = 2 x^{a + 1} y^{b} + 3 x^{a + 2} y^{b + 1} .$ $(M_{⋆})_{y} = 3 (b + 1) x^{a} y^{b} + 4 (b + 2) x^{a + 1} y^{b + 1},$ $(N_{⋆})_{x} = 2 (a + 1) x^{a} y^{b} + 3 (a + 2) x^{a + 1} y^{b + 1} .$ Ab like power-terms match karo (equation ka parser): $x^{a} y^{b}$ terms aur $x^{a + 1} y^{b + 1}$ terms ko alag-alag agree karna chahiye. $3 (b + 1) = 2 (a + 1) (first term),$ $4 (b + 2) = 3 (a + 2) (second term) .$ "Like terms match" kyun? Do polynomials tabhi equal hote hain jab har matching monomial ka coefficient equal ho — yeh ek exactness equation ko ek chhoti linear system mein convert kar deta hai.

Step 4 — system solve karo. Pehle se: $3 b + 3 = 2 a + 2 \Rightarrow 2 a - 3 b = 1$ . Doosre se: $4 b + 8 = 3 a + 6 \Rightarrow 3 a - 4 b = 2$ . Solve karo: pehle ko $3$ se multiply karo, doosre ko $2$ se: $6 a - 9 b = 3$ aur $6 a - 8 b = 4$ ; subtract karo: $- b = - 1 \Rightarrow b = 1$ , phir $2 a - 3 = 1 \Rightarrow a = 2$ . Toh $μ = x^{2} y$ — koi guessing nahi, equations ne force kiya.

Step 5 — multiply karo, re-test karo. $M_{⋆} = 3 x^{2} y^{2} + 4 x^{3} y^{3}$ , $N_{⋆} = 2 x^{3} y + 3 x^{4} y^{2}$ . $(M_{⋆})_{y} = 6 x^{2} y + 12 x^{3} y^{2}$ , $(N_{⋆})_{x} = 6 x^{2} y + 12 x^{3} y^{2}$ . Equal ✓

Step 6 — $F$ banao. $F = \int (3 x^{2} y^{2} + 4 x^{3} y^{3}) d x = x^{3} y^{2} + x^{4} y^{3} + ϕ (y)$ . $F_{y} = 2 x^{3} y + 3 x^{4} y^{2} + ϕ^{'} = N_{⋆} = 2 x^{3} y + 3 x^{4} y^{2} \Rightarrow ϕ^{'} = 0.$

$x^{3} y^{2} + x^{4} y^{3} = C .$

Step 7 — $μ$ kahan khatam hota hai. $μ = x^{2} y$ axes $x = 0$ aur $y = 0$ par zero hai. Un lines par $μ$ se multiply karna information destroy kar deta hai (hum zero se multiply kar rahe the), isliye koi bhi solution branch jo kisi axis par lie kare use alag se check karna padega — exam ke hidden marks.

Verify: $d (x^{3} y^{2} + x^{4} y^{3}) = (3 x^{2} y^{2} + 4 x^{3} y^{3}) d x + (2 x^{3} y + 3 x^{4} y^{2}) d y = x^{2} y (M d x + N d y)$ . ✓

Coverage check

Scenario matrix ke sabhi eight cells mein ab fully worked, verified examples hain. ✓

Connections

Integrating factors for non-exact equations — parent method jise yeh examples drill karte hain
Exact differential equations — Cell A yahi hai, unchanged
Total differentials and potential functions — har " $F$ banana" step
Linear first-order ODEs — Ex 7 ka $μ = e^{\int P d x}$
Separable equations — Cell E mein fallback
Mixed partial derivatives (Clairaut's theorem) — exactness test kyun kaam karta hai
Conservative vector fields — $M d x + N d y = d F$ conservative-field ki picture hai

Worked examples — Integrating factors for non-exact equations