First-order linear ODEs — integrating factor method (derivation): Worked examples

4.6.4 · D3 · Maths › Ordinary Differential Equations › First-order linear ODEs — integrating factor method (derivat

Yeh page parent derivation ke method ka stress test hai. Hum sirf recipe repeat nahi karenge — hum deliberately har tarah ki situation tackle karenge jo ek first-order linear ODE de sakta hai: seedhe-saadhe wale, woh jahan signs flip hoti hain, woh jahan ek coefficient zero hai, woh jo secretly linear nahi lagte jab tak rearrange na karo, aur ek real-world word problem.

Ek bhi example touch karne se pehle, hum saare cases ka map neeche rakh rahe hain taaki tum har waqt exactly dekh sako ki hum territory ke kis corner mein khade hain.

The scenario matrix

Har row neeche ek class of problem hai. Is page ka poora point yeh hai ki koi bhi cell khaali nahi hai — har example us cell ke saath tagged hai jise woh fill karta hai.

Cell	Kya cheez isse alag banati hai	Kahan mushkil aati hai	Example
A. Constant $P, Q$	$P$ aur $Q$ plain numbers hain	trivial $\int P d x = P x$	Ex 1
B. Non-standard leading coefficient	equation $a (x) y^{'} + \dots$ se shuru hoti hai	pehle divide karna zaroori hai warna method fail ho jaata hai	Ex 2
C. Negative $P$	$P$ ki sign exponent ko flip karti hai	$\mu = e^{-\int	P
D. $P$ produces $ln$ / power $μ$	$P = n / x$ se $μ = x^{n}$ milta hai	absolute values, domain $x > 0$ vs $x < 0$	Ex 4
E. Degenerate: $P = 0$	koi $y$ term hi nahi	method kaam karta hai lekin plain integration mein reduce ho jaata hai	Ex 5
F. Initial-value problem (IVP)	ek condition $y (x_{0}) = y_{0}$ jo $C$ pin karti hai	IC sirf end mein apply karo	Ex 6
G. Real-world word problem	mixing / cooling — tumhe ODE banana padega	words ko $P, Q$ mein translate karna	Ex 7
H. Exam twist — looks non-linear	$y$ "galat tarah" appear karta hai; $x, y$ ke roles swap karo	$x$ ko unknown function ki tarah treat karo	Ex 8

Cell A — Constant coefficients

Worked example Example 1 —

$\frac{d y}{d x} + 3 y = 6$ Forecast: Aage padhne se pehle andaaza lagao: $x \to \infty$ par $y$ kaisa dikhega? (Hint: kya $y^{'} = 0$ koi steady value deta hai?)

Step 1 — $P, Q$ identify karo. Already standard hai: $P = 3$ , $Q = 6$ . Yeh step kyun? Har formula assume karta hai ki $y^{'}$ ka coefficient $1$ hai; yahan hai, toh hum directly $P, Q$ padh sakte hain.

Step 2 — Integrating factor. $\int 3 d x = 3 x$ , toh $μ = e^{3 x}$ . Yeh step kyun? $μ = e^{\int P d x}$ iss tarah bana hai ki $μ y^{'} + μ P y = \frac{d}{d x} (μ y)$ . Constant $P$ ke saath integral sirf $P x$ hota hai.

Step 3 — Collapse. $\frac{d}{d x} (e^{3 x} y) = 6 e^{3 x}$ . Yeh step kyun? Left side guaranteed ek single derivative hai jab hum sahi $μ$ use karte hain; right side $Q = 6$ ko $μ$ se multiply karo.

Step 4 — Integrate. $e^{3 x} y = \frac{6}{3} e^{3 x} + C = 2 e^{3 x} + C$ .

Step 5 — $μ$ se divide karo. $y = 2 + C e^{- 3 x}$ . Yeh step kyun? Hamare paas $μ y$ tha, $y$ nahi; poore right side ko $e^{3 x}$ se divide karo.

Verify: $x \to \infty$ par, $e^{- 3 x} \to 0$ toh $y \to 2$ — woh steady state jahan $y^{'} = 0 \Rightarrow 3 y = 6 \Rightarrow y = 2$ . ✓ Aur wapas plug karne par: $y^{'} = - 3 C e^{- 3 x}$ , toh $y^{'} + 3 y = - 3 C e^{- 3 x} + 6 + 3 C e^{- 3 x} = 6$ . ✓

Cell B — Non-standard leading coefficient

Worked example Example 2 —

$(x^{2}) \frac{d y}{d x} + 2 x y = 1$ Forecast: Left side $x^{2} y^{'} + 2 x y$ — kya yeh tumhe Product Rule se already jaani-pehchani kisi derivative ki yaad dilata hai? Aage padhne se pehle $\frac{d}{d x} (x^{2} y)$ guess karo.

Step 1 — Standardize. $x^{2}$ se divide karo: $\frac{d y}{d x} + \frac{2}{x} y = \frac{1}{x ^{2}}$ . Toh $P = \frac{2}{x}$ , $Q = \frac{1}{x ^{2}}$ . Yeh step kyun? Agar hum yeh skip karein, toh $y^{'}$ ka coefficient $x^{2}$ hoga, $1$ nahi, aur $μ P y$ match karne wala term galat ho jaayega — collapse fail ho jaayega.

Step 2 — Integrating factor. $\int \frac{2}{x} d x = 2 ln ∣ x ∣ = ln x^{2}$ , toh $μ = e^{l n x^{2}} = x^{2}$ . Yeh step kyun? $\int \frac{n}{x} d x = n ln ∣ x ∣$ , aur exponentiate karne se log undo ho jaata hai.

Step 3 — Collapse. $\frac{d}{d x} (x^{2} y) = x^{2} \cdot \frac{1}{x ^{2}} = 1$ . Yeh step kyun? Dhyaan do is cancellation par — yeh ek accha sign hai ki standardization sahi tha. Aur $μ = x^{2}$ tumhare forecast ko confirm karta hai: original left side already $\frac{d}{d x} (x^{2} y)$ tha.

Step 4 — Integrate. $x^{2} y = x + C$ .

Step 5 — Divide. $y = \frac{1}{x} + \frac{C}{x ^{2}}$ ( $x \neq = 0$ ke liye).

Verify: $y^{'} = - \frac{1}{x ^{2}} - \frac{2 C}{x ^{3}}$ . Phir $x^{2} y^{'} + 2 x y = - 1 - \frac{2 C}{x} + 2 x (\frac{1}{x} + \frac{C}{x ^{2}}) = - 1 - \frac{2 C}{x} + 2 + \frac{2 C}{x} = 1$ . ✓

Cell C — Negative $P$ (sign growth ko flip karti hai)

Cell D — $P = n / x$ : power-law integrating factors aur domain question

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

Worked example Example 4 —

$\frac{d y}{d x} - \frac{3}{x} y = x^{3}$ Forecast: $P = - 3/ x$ . Kya $μ$ hoga $x^{3}$ , $x^{- 3}$ , ya kuch absolute value ke saath? Pehle exponent ki sign guess karo.

Step 1 — Identify. Already standard: $P = - \frac{3}{x}$ , $Q = x^{3}$ .

Step 2 — Integrating factor. $\int - \frac{3}{x} d x = - 3 ln ∣ x ∣ = ln ∣ x ∣^{- 3}$ , toh $μ = ∣ x ∣^{- 3}$ . Domain $x > 0$ par hum $μ = x^{- 3}$ likhte hain. Absolute value kyun? $\int \frac{1}{x} d x = ln ∣ x ∣$ dono signs of $x$ ke liye valid hai, lekin negative number ka $ln$ real nahi hota. Kyunki ek solution curve $x = 0$ ke ek taraf rehti hai (ODE wahan singular hai — figure mein red dashed line dekho), hum branch choose karte hain aur bars hata dete hain.

Step 3 — Collapse. $\frac{d}{d x} (x^{- 3} y) = x^{- 3} \cdot x^{3} = 1$ . Yeh step kyun? Phir se clean cancellation confirm karta hai ki $μ$ sahi hai.

Step 4 — Integrate. $x^{- 3} y = x + C$ .

Step 5 — Divide. $y = x^{4} + C x^{3}$ ( $x > 0$ ke liye; mirror branch $x < 0$ ka algebra same hai).

Verify: $y^{'} = 4 x^{3} + 3 C x^{2}$ . Phir $y^{'} - \frac{3}{x} y = 4 x^{3} + 3 C x^{2} - \frac{3}{x} (x^{4} + C x^{3}) = 4 x^{3} + 3 C x^{2} - 3 x^{3} - 3 C x^{2} = x^{3}$ . ✓

Cell E — Degenerate case: $P = 0$

Cell F — Initial-value problem (andar integration by parts)

Worked example Example 6 — IVP:

$\frac{d y}{d x} + y = x$ , $y (0) = 3$ Forecast: General solution mein $C e^{- x}$ hoga. Guess karo: $y (0) = 3$ apply karne ke baad, kya $C$ positive hoga ya negative?

Step 1 — Identify. $P = 1$ , $Q = x$ .

Step 2 — Integrating factor. $\int 1 d x = x$ , toh $μ = e^{x}$ .

Step 3 — Collapse. $\frac{d}{d x} (e^{x} y) = x e^{x}$ .

Step 4 — Integration by parts se integrate karo. $\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + C$ . Parts kyun? Integrand $x$ (algebraic) aur $e^{x}$ (exponential) ka product hai; parts woh tool hai jo polynomial factor ko ek derivative ek time par uthata hai. Toh $e^{x} y = x e^{x} - e^{x} + C$ .

Step 5 — Divide. $y = x - 1 + C e^{- x}$ .

Step 6 — Initial condition last mein apply karo. $y (0) = 0 - 1 + C = 3 \Rightarrow C = 4$ . Last mein kyun? $C$ tabhi meaningful hai jab general solution exist kare; isse pehle pin karna integration ko corrupt kar dega.

Answer: $y = x - 1 + 4 e^{- x}$ .

Verify: $y^{'} = 1 - 4 e^{- x}$ . Phir $y^{'} + y = 1 - 4 e^{- x} + x - 1 + 4 e^{- x} = x$ . ✓ Aur $y (0) = 0 - 1 + 4 = 3$ . ✓

Cell G — Real-world word problem (mixing tank)

Worked example Example 7 — Salt mixing

Ek tank mein $100$ litres pure water hai. Brine jisme $2$ grams salt per litre hai, $5$ L/min ki speed se andar aata hai, aur acchi tarah se hila hua mixture same $5$ L/min par bahar jaata hai. $y (t)$ ko tank mein salt ke grams maano. $y (t)$ nikalo.

Forecast: Inflow ek constant rate par salt add karta hai; outflow salt ko proportional remove karta hai jitni wahan hai. Solve karne se pehle long-run amount of salt (steady state) guess karo. (Tank volume $= 100$ L, incoming concentration $2$ g/L…)

Step 1 — ODE banao. Rate in $= (2 g/L) (5 L/min) = 10$ g/min. Rate out $= (\frac{y}{100} g/L) (5 L/min) = \frac{y}{20}$ g/min. Toh $\frac{d y}{d t} = 10 - \frac{y}{20} .$ Yeh step kyun? "Rate of change = in − out" physical law hai; tank mein concentration $y /100$ hai kyunki volume $100$ L rehta hai (equal flow rates).

Step 2 — Standardize. $\frac{d y}{d t} + \frac{1}{20} y = 10$ . Toh $P = \frac{1}{20}$ , $Q = 10$ .

Step 3 — Integrating factor. $\int \frac{1}{20} d t = \frac{t}{20}$ , toh $μ = e^{t /20}$ .

Step 4 — Collapse & integrate. $\frac{d}{d t} (e^{t /20} y) = 10 e^{t /20}$ , toh $e^{t /20} y = 10 \cdot 20 e^{t /20} + C = 200 e^{t /20} + C .$ $\times 20$ kyun? $\int e^{t /20} d t = 20 e^{t /20}$ (inner coefficient $1/20$ se divide karo).

Step 5 — Divide. $y = 200 + C e^{- t /20}$ .

Step 6 — Initial condition. $t = 0$ par pure water matlab $y (0) = 0$ : $0 = 200 + C \Rightarrow C = - 200$ . Answer: $y (t) = 200 (1 - e^{- t /20})$ grams.

Verify (steady state + units): $t \to \infty$ par, $y \to 200$ g. Physically check karo: steady state par $y^{'} = 0 \Rightarrow y /20 = 10 \Rightarrow y = 200$ g. Yeh $200/100 = 2$ g/L hai — bilkul inflow concentration ke barabar, exactly jaisa intuition demand karta hai. ✓ Units: $[10] =$ g/min, $[y /20] =$ g/min ✓.

Cell H — Exam twist: $y$ mein non-linear lagta hai, roles swap karo

Worked example Example 8 —

$\frac{d y}{d x} = \frac{1}{x + y ^{2}}$ Forecast: Jaisa likha hai, right side mein $y^{2}$ hai — yeh non-linear lagta hai, aur koi bhi $μ$ formula fit nahi hota. Lekin agar hum fraction ko flip karein aur $x$ ko $y$ ka function sochen? Guess karo $\frac{d x}{d y}$ kya banega.

Step 1 — Derivative invert karo. Kyunki $\frac{d y}{d x} = \frac{1}{d x / d y}$ , flip karne par $\frac{d x}{d y} = x + y^{2} .$ Yeh step kyun? Derivative ko reciprocate karna legal hai jahan bhi $d y / d x \neq = 0$ ho. Yeh key move hai: equation $y$ mein non-linear hai lekin $x$ mein linear hai.

Step 2 — $x (y)$ mein linear ki tarah likhao. $\frac{d x}{d y} - x = y^{2}$ . Ab unknown function $x$ hai, variable $y$ hai, aur $P (y) = - 1$ , $Q (y) = y^{2}$ . Yeh step kyun? Standard form $\frac{d x}{d y} + P (y) x = Q (y)$ se compare karo — yeh match karta hai jahan $x$ woh role play karta hai jo usually $y$ karta hai.

Step 3 — Integrating factor ( $y$ mein). $μ = e^{\int - 1 d y} = e^{- y}$ .

Step 4 — Collapse. $\frac{d}{d y} (e^{- y} x) = y^{2} e^{- y}$ .

Step 5 — Integration by parts (do baar). $\int y^{2} e^{- y} d y = - y^{2} e^{- y} - 2 y e^{- y} - 2 e^{- y} + C = - (y^{2} + 2 y + 2) e^{- y} + C .$ Parts do baar kyun? $y^{2}$ ko vanish hone ke liye do derivatives chahiye; har application polynomial degree ek se kam karta hai. Toh $e^{- y} x = - (y^{2} + 2 y + 2) e^{- y} + C$ .

Step 6 — $μ$ se divide karo ( $e^{y}$ se multiply karo): $x = - (y^{2} + 2 y + 2) + C e^{y}$ .

Answer: $x = C e^{y} - y^{2} - 2 y - 2$ ( $y$ ke terms mein $x$ ka explicit formula).

Verify: $\frac{d x}{d y} = C e^{y} - 2 y - 2$ . Phir $\frac{d x}{d y} - x = (C e^{y} - 2 y - 2) - (C e^{y} - y^{2} - 2 y - 2) = y^{2}$ . ✓

Active Recall

Recall Kis cell ko exponent mein sign rakhna pada, aur kyun?

Cell C ( $P = - 2$ ): integrating factor hai $e^{\int P d x} = e^{- 2 x}$ . Minus hataane se ek alag ODE solve hota hai. ::: $P$ ki sign $\int P d x$ ke andar rakho.

Recall Jab

$P = 0$ ho toh $μ$ kya hota hai, aur method kis cheez mein reduce ho jaata hai? $μ = e^{0} = 1$ ; method ordinary direct integration mein degrade ho jaata hai (Cell E). :::

Recall Mixing problem mein tank concentration

$y /100$ kyun thi? Inflow aur outflow rates equal thein ( $5$ L/min), toh volume $100$ L par constant raha. :::

Recall Exam twist: kya cheez batati hai ki

$y (x)$ ki jagah $x (y)$ solve karo? Equation $y$ mein non-linear hai lekin $\frac{d x}{d y}$ lene ke liye reciprocate karne par linear ho jaati hai. :::

Connections

Separable ODEs — Cell E iske saath collapse hota hai; $μ$ khud solve karne mein bhi use hota hai.
Product Rule — collapse $μ y^{'} + μ P y = \frac{d}{d x} (μ y)$ directly Cell B mein dikh jaata hai ( $\frac{d}{d x} (x^{2} y)$ ).
Exact ODEs — Cell H mein "roles swap" karna spirit mein uss integrating factor dhundhne ke kareeb hai jo equations ko exact banata hai.
Bernoulli Equations — "non-linear lagta hai, linear ban jaata hai" ka aagla level $v = y^{1 - n}$ ke zariye.
Linear Constant-Coefficient ODEs — Cell A exactly yeh special case hai.

Worked examples — First-order linear ODEs — integrating factor method (derivation)