Separable ODEs — technique, implicit solutions: Worked examples

4.6.3 · D3 · Maths › Ordinary Differential Equations › Separable ODEs — technique, implicit solutions

Agar neeche koi word unfamiliar lage, to wo parent note the parent topic mein build kiya gaya tha. Hum wahi vocabulary reuse karte hain, kabhi contradict nahi karte.

The scenario matrix

Har row ek case class hai — ek genuinely alag situation jo technique tumhare saamne la sakti hai. Last column batata hai kaun sa worked example ise cover karta hai.

#	Case class	Isme kya alag hai	Covered by
C1	Clean explicit + IVP	Separates, integrates, $y$ isolate karo, $C$ fix karo	Ex 1
C2	Implicit-only answer	$y$ cleanly isolate nahi ho sakta — $F (x, y) = C$ chhod do	Ex 2
C3	Partial-fractions needed	$h (y)$ ek product hai $\Rightarrow$ integrate karne se pehle split karo	Ex 3
C4	Equilibrium / lost solution	$h (y_{0}) = 0$ ek constant solution delete kar deta hai	Ex 3 & Ex 4
C5	Sign / absolute-value branch	$ln ∣ y ∣$ aur IVP kaun sa branch select karta hai	Ex 4
C6	Degenerate: $g (x) = 0$	RHS $0$ hai $\Rightarrow$ $y$ constant hai	Ex 5
C7	Limiting / long-run behaviour	$x \to \infty$ par ya blow-up ke paas kya hota hai	Ex 6
C8	Real-world word problem	Words ko translate karo → ODE → units check	Ex 7 (cooling)
C9	Exam twist: disguised separable	Sum/exponent jaisa lagta hai lekin factor hota hai	Ex 8

Hum inhe aik aisa order mein work karte hain jo build up karta jaata hai.

Ex 1 — Case C1: clean explicit answer with an initial value

Worked example Statement

$\frac{d y}{d x} = - 2 x y$ ko solve karo, $y (0) = 5$ ke saath.

Forecast: shape guess karo. Jab $x > 0$ aur $y > 0$ ho to rate $d y / d x$ negative hai, aur iska size $x$ ke saath badhta hai — to $y (0) = 5$ se hum expect karte hain ki $y$ tezi se giregi, ek bell-jaisi decay. Wo picture yaad rakho.

1. Separate karo. $\frac{d y}{y} = - 2 x d x$ . Ye step kyun? Right-hand side already $g (x) h (y) = (- 2 x) (y)$ ke roop mein factor hoti hai, isliye saara $y$ left jaata hai, saara $x$ right — method ki defining move.

2. Dono sides integrate karo. $ln ∣ y ∣ = - x^{2} + C$ . Ye step kyun? $\int \frac{d y}{y} = ln ∣ y ∣$ aur $\int - 2 x d x = - x^{2}$ . Ek constant $C$ (parent: constants merge karo).

3. Exponentiate karo. $∣ y ∣ = e^{C} e^{- x^{2}}$ , isliye $y = A e^{- x^{2}}$ jahan $A = \pm e^{C}$ . Ye step kyun? $ln$ undo karo. Absolute value ka $\pm$ ek naye nonzero constant $A$ mein fold ho jaata hai.

4. $y (0) = 5$ se $C$ fix karo. $5 = A e^{0} = A$ , isliye $A = 5$ . Ye step kyun? Ek initial value ek free constant ko pin down karti hai. $y = 5 e^{- x^{2}}$

Verify karo: differentiate karo: $y^{'} = 5 \cdot (- 2 x) e^{- x^{2}} = - 2 x (5 e^{- x^{2}}) = - 2 x y$ . ✓ Aur $y (0) = 5$ . ✓ Ye wahi decaying bell hai jo humne forecast ki thi (baad mein Gaussian curve figure reuse hoti hai).

Ex 2 — Case C2: answer explicit banne se mana karta hai

Worked example Statement

$\frac{d y}{d x} = \frac{3 x ^{2}}{2 y - cos y}$ solve karo.

Forecast: denominator $2 y - cos y$ ka koi tidy inverse operation nahi hai, isliye shuru karne se pehle hi bet lagao ki $y$ isolate nahi hogi — honest answer implicit hogi.

1. Separate karo. $(2 y - cos y) d y = 3 x^{2} d x$ . Ye step kyun? $y$ -piece RHS denominator ka reciprocal hai; cross multiply karke $y$ ko $d y$ ke saath aur $x$ ko $d x$ ke saath gather karo.

2. Dono sides integrate karo. $y^{2} - sin y = x^{3} + C$ . Ye step kyun? $\int (2 y - cos y) d y = y^{2} - sin y$ aur $\int 3 x^{2} d x = x^{3}$ .

3. Ruko — $y$ ke liye solve nahi kar sakte. Koi algebra $y$ ko $y^{2} - sin y$ se isolate nahi kar sakti. Ye step kyun? Ek implicit relation $F (x, y) = C$ ek fully valid solution hai (parent definition). $y^{2} - sin y - x^{3} = C$

Verify karo: implicitly differentiate karo: $2 y y^{'} - cos y y^{'} - 3 x^{2} = 0 \Rightarrow y^{'} (2 y - cos y) = 3 x^{2} \Rightarrow y^{'} = \frac{3 x ^{2}}{2 y - cos y}$ . ✓ Original se match karta hai — trust karo ise.

Ex 3 — Cases C3 + C4: partial fractions aur do hidden equilibria

Worked example Statement

$\frac{d y}{d x} = y (y - 2)$ solve karo, phir sabhi solutions list karo.

Forecast: RHS $y = 0$ aur $y = 2$ par zero hai — do flat equilibrium lines expect karo, aur moving solutions unse door ya unki taraf push hoti hain. Guess karo ki kya $y = 2$ ek "wall" hai jisse solutions door bhaagti hain.

1. Pehle $h (y) = 0$ check karo. $h (y) = y (y - 2) = 0$ at $y = 0$ aur $y = 2$ , isliye $y \equiv 0$ aur $y \equiv 2$ equilibrium solutions hain. Ye step kyun? Parent ka SISC rule: constants ko pehle dhundho divide karne se pehle, kyunki divide karna unhe erase kar deta hai.

2. Separate karo ( $y \neq = 0, 2$ ke liye). $\frac{d y}{y ( y - 2 )} = d x$ .

3. Partial fractions. $\frac{1}{y ( y - 2 )} = \frac{- 1/2}{y} + \frac{1/2}{y - 2}$ . Ye step kyun? Har simple piece ek log mein integrate hoti hai — dekho Partial Fractions. (Cover-up: $y = 0$ par, $\frac{1}{0 - 2} = - \frac{1}{2}$ ; $y = 2$ par, $\frac{1}{2} = \frac{1}{2}$ .)

4. Integrate karo. $- \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 2 ∣ = x + C$ , yaani $\frac{1}{2} ln \frac{y - 2}{y} = x + C$ .

5. $y$ ke liye solve karo. 2 se multiply karo, exponentiate karo: $\frac{y - 2}{y} = B e^{2 x}$ ( $B = \pm e^{2 C}$ ). Rearrange karne par, $y = \frac{2}{1 - B e ^{2 x}} plus y \equiv 0, y \equiv 2.$ Equilibria report kyun karo? Formula kabhi $0$ ya $2$ output nahi kar sakta, isliye step 1 ke bina wo lost ho jaate.

Verify karo: $B = 0$ lo: formula $y = 2$ deta hai — equilibrium ke saath consistent hai. General form differentiate karo (VERIFY mein kiya gaya) taaki $y^{'} = y (y - 2)$ recover ho. ✓ Ye The Logistic Equation se logistic-flavour split hai.

Ex 4 — Cases C4 + C5: initial value dwara choose ki gayi sign branch

Worked example Statement

$\frac{d y}{d x} = \frac{x}{y}$ solve karo (a) $y (0) = 3$ aur (b) $y (0) = - 3$ ke saath.

Forecast: $y^{2} - x^{2}$ constant hoga — solution curves hyperbolas hain. $y (0)$ ka sign upper ya lower branch decide karta hai. Aage padhne se pehle guess karo kaun sa sign kidhar jaata hai.

1. Danger line note karo. $h (y) = y$ hai, lekin yahan $h$ denominator mein hai, isliye $y = 0$ excluded hai (rate blow up hoti hai), equilibrium nahi. Ye step kyun? $1/ h (y)$ vs $h (y)$ zeros par alag behave karte hain — hamesha dekho.

2. Separate karo. $y d y = x d x$ .

3. Integrate karo. $\frac{y ^{2}}{2} = \frac{x ^{2}}{2} + C_{0} \Rightarrow y^{2} - x^{2} = C$ ( $C = 2 C_{0}$ set karo).

4a. $y (0) = 3$ apply karo. $9 - 0 = C \Rightarrow C = 9$ , isliye $y^{2} = x^{2} + 9$ . Kyunki $y (0) = 3 > 0$ hai, positive root choose karo: $y = + x^{2} + 9$ . Ye step kyun? $\cdot$ ke do branches hain; $y (0) > 0$ upper sheet select karta hai — IVP sign resolve karta hai.

4b. $y (0) = - 3$ apply karo. Phir se $9 = C$ , lekin ab $y (0) = - 3 < 0$ hai isliye $y = - x^{2} + 9$ select hoti hai. $y = + x^{2} + 9 (case a), y = - x^{2} + 9 (case b) .$

Verify karo: (a) ke liye, $y^{'} = \frac{x}{x ^{2} + 9} = \frac{x}{y}$ . ✓ $x = 0$ par, $y = 3$ . ✓ (b) ke liye overall minus ke saath bhi $y^{'} = x / y$ rehta hai. ✓ Figure dekho: dono branches kabhi $y = 0$ ko touch nahi karti, exactly woh excluded line.

Ex 5 — Case C6: degenerate RHS, $g (x) = 0$

Ex 6 — Case C7: limiting / long-run behaviour aur ek finite-time blow-up

Worked example Statement

$\frac{d y}{d x} = y^{2}$ solve karo $y (0) = 1$ ke saath, phir $x$ badhne par behaviour describe karo.

Forecast: rate $y$ ka square hai — ek feedback jo khud ko accelerate karta hai. Guess karo: kya $y$ hamesha ke liye finite rehta hai, ya kisi finite $x$ par explode ho jaata hai?

1. Equilibrium check karo. $h (y) = y^{2} = 0$ at $y = 0$ , isliye $y \equiv 0$ ek equilibrium solution hai. Ye step kyun? SISC — lost constant record karo.

2. Separate & integrate karo ( $y \neq = 0$ ke liye). $\frac{d y}{y ^{2}} = d x \Rightarrow - \frac{1}{y} = x + C$ .

3. $y$ ke liye solve karo aur $C$ fix karo. $y = \frac{- 1}{x + C}$ . $x = 0$ par, $y = 1 \Rightarrow - 1/ C = 1 \Rightarrow C = - 1$ . $y = \frac{1}{1 - x} .$

4. Limiting behaviour. Jab $x \to 1^{-}$ , denominator $\to 0^{+}$ , isliye $y \to + \infty$ : $x = 1$ par ek finite-time blow-up. Solution sirf $x < 1$ par jeeti hai. Ye step kyun? Squared feedback growth super-exponential banata hai — ye finite $x$ mein infinity escape kar jaata hai, unlike $e^{x}$ .

Verify karo: $y^{'} = \frac{d}{d x} (1 - x)^{- 1} = (1 - x)^{- 2} = y^{2}$ . ✓ $y (0) = 1$ . ✓ Figure mein $x = 1$ par vertical asymptote (plum dashed) dikhta hai.

Ex 7 — Case C8: real-world word problem (Newton's cooling), with units

Worked example Statement

Ek coffee $9 0^{\circ} C$ par hai aur $2 0^{\circ} C$ ke room mein rakhi hai. Uska temperature $T (t)$ ( $^{\circ} C$ mein, $t$ minutes mein) Newton's law follow karta hai: cooling rate us se proportional hai jitna wo room temperature se upar hai, $\frac{d T}{d t} = - k (T - 20), k > 0.$ Given ki wo $10$ minutes baad $7 0^{\circ} C$ tak cool ho jaati hai, $T (t)$ dhundho aur $t = 20$ par temperature batao.

Forecast: $90$ se $20$ ki taraf girna chahiye aur level off hona chahiye — room temperature ki taraf exponential approach. Guess karo ki $t = 20$ par $\frac{1}{2} (90 + 20) = 55$ se upar hoga ya neeche.

1. Separability expose karne ke liye substitute karo. $u = T - 20$ lo, isliye $\frac{d u}{d t} = \frac{d T}{d t} = - k u$ . Ye step kyun? $u$ "room ke upar excess" hai — RHS clean factor $(- k) (u)$ ban jaata hai.

2. Separate & integrate karo. $\frac{d u}{u} = - k d t \Rightarrow ln ∣ u ∣ = - k t + C \Rightarrow u = A e^{- k t}$ . $∣ u ∣$ kyun? Pehle absolute value; $\pm$ $A$ mein fold hota hai. Yahan $u > 0$ hai (coffee room se hotter hai), isliye $A > 0$ .

3. Back-substitute karo & $T (0) = 90$ use karo. $T = 20 + A e^{- k t}$ ; $t = 0$ par, $90 = 20 + A \Rightarrow A = 70$ .

4. $T (10) = 70$ se $k$ dhundho. $70 = 20 + 70 e^{- 10 k} \Rightarrow e^{- 10 k} = \frac{50}{70} = \frac{5}{7} \Rightarrow k = \frac{1}{10} ln \frac{7}{5}$ . $T (t) = 20 + 70 (\frac{5}{7})^{t /10} .$

5. $t = 20$ evaluate karo. $T (20) = 20 + 70 (\frac{5}{7})^{2} = 20 + 70 \cdot \frac{25}{49} = 20 + \frac{250}{7} \approx 55.7 1^{\circ} C$ . Ye step kyun? Puche gaye value ka answer deta hai.

Verify karo (units + sanity): $[k] = min^{- 1}$ , $[k t]$ dimensionless — exponent unitless hai. ✓ $T (0) = 90$ , $T (10) = 70$ , aur $T \to 20$ jab $t \to \infty$ (room temp par level off). ✓ $55.71 > 55$ , halfway mark se thoda upar — slow-tail decay se match karta hai. ✓

Ex 8 — Case C9: exam twist — ek disguised separable ODE

Worked example Statement

$\frac{d y}{d x} = e^{2 x - y}$ solve karo $y (0) = 0$ ke saath.

Forecast: ek exponent ke andar subtraction non-separable lagta hai — lekin exponents sums ko products mein turn karte hain. Guess karo ki ye split ho jaayega.

1. Exponent split karo. $e^{2 x - y} = e^{2 x} e^{- y}$ , isliye $\frac{d y}{d x} = e^{2 x} e^{- y} = g (x) h (y)$ . Ye step kyun? $e^{a - b} = e^{a} e^{- b}$ hidden product reveal karta hai — trap ye sochna hai ki sum separate nahi ho sakta.

2. Separate karo. $e^{y} d y = e^{2 x} d x$ . Ye step kyun? $1/ h (y) = 1/ e^{- y} = e^{y}$ ; ise left move karo.

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

4. $y (0) = 0$ se $C$ fix karo. $e^{0} = \frac{1}{2} e^{0} + C \Rightarrow 1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}$ .

5. $y$ ke liye solve karo. $e^{y} = \frac{1}{2} e^{2 x} + \frac{1}{2} = \frac{1}{2} (e^{2 x} + 1)$ , isliye $y = ln (\frac{e ^{2 x} + 1}{2}) .$

Verify karo: $y^{'} = \frac{1}{( e ^{2 x} + 1 ) /2} \cdot \frac{2 e ^{2 x}}{2} = \frac{2 e ^{2 x}}{e ^{2 x} + 1}$ . Aur $e^{2 x - y} = e^{2 x} e^{- y} = e^{2 x} \cdot \frac{2}{e ^{2 x} + 1} = \frac{2 e ^{2 x}}{e ^{2 x} + 1}$ . ✓ Match karte hain, aur $y (0) = ln 1 = 0$ . ✓

Kya humne har cell fill ki?

Recall Har matrix row ko uske example se map karo

C1→Ex1, C2→Ex2, C3→Ex3, C4→Ex3 & Ex4, C5→Ex4, C6→Ex5, C7→Ex6, C8→Ex7, C9→Ex8.

Connections

Separable ODEs — Technique & Implicit Solutions (parent)
Initial Value Problems (Ex 1, 4, 5, 6, 7, 8 mein $C$ fix karna)
Partial Fractions (Ex 3)
The Logistic Equation (Ex 3 ka cousin)
Exact ODEs (implicit solutions, Ex 2)
Substitution Method — Homogeneous ODEs (Ex 7 mein $u = T - 20$ trick ek baby substitution hai)
First-Order ODEs — Overview
Integrating Factor & Linear ODEs (jab RHS factor nahi hogi)