Separable ODEs — technique, implicit solutions: Visual walkthrough

4.6.3 · D2 · Maths › Ordinary Differential Equations › Separable ODEs — technique, implicit solutions

Hum derive karne wale hain, bilkul scratch se, $\int \frac{d y}{h ( y )} = \int g (x) d x + C .$ Lekin abhi yeh line sirf kaagaz par marks hai. Aao har mark ko build karte hain.

Step 1 — $\frac{d y}{d x}$ asal mein hai kya? (ek picture par slope)

KYA. Hamare paas ek mystery curve $y (x)$ hai: horizontal axis par har input $x$ ke liye, yeh vertical axis par ek height $y$ deta hai. Symbol $\frac{d y}{d x}$ us curve ka slope hai — jab tum $x$ ko thoda sa right nudge karte ho toh height kitni tezi se chadti hai.

KYO. Ek ODE ek aisa rule hai jo tumhe har point par slope batata hai, aur tumse curve reconstruct karne ko kehta hai. Kuch bhi "solve" karne se pehle hume dekhna hoga ki $\frac{d y}{d x}$ ek geometric object hai: curve ko barely kissing karne wali line ki steepness.

PICTURE. Figure mein, black curve $y (x)$ hai. Uske kisi point ko pick karo. Lavender line bilkul usi point par touch karti hai — uski steepness (rise $d y$ over run $d x$ ) hi $\frac{d y}{d x}$ hai.

Figure — Separable ODEs — technique, implicit solutions

Step 2 — "Separable" kaisa dikhta hai: ek slope field jo factor hota hai

KYA. Ek separable ODE kehta hai ki point $(x, y)$ par slope ek product hai: $yahan slope \frac{d y}{d x} = sirf kitna right hai uspar depend g (x) \times sirf kitna upar hai uspar depend h (y) .$

KYO. Zyaadatar ODEs $x$ aur $y$ ko bekar tarike se mix kar dete hain (jaise $x + y$ — ek sum, jo split nahi hoga). Ek product $g (x) h (y)$ ka magic yeh hai ki baad mein hum har $x$ ko ek side aur har $y$ ko doosri side bhej sakte hain. Factoring hi exactly woh cheez hai jo separation possible banati hai.

PICTURE. Har point par hum ek chhota slope-tick draw karte hain. Ek separable field mein, tick ki steepness " $x$ -ingredient times $y$ -ingredient" hoti hai. Neeche wale mint ticks sabhi same height $y$ share karte hain, isliye unki steepness sirf $g (x)$ ke zariye different hoti hai; coral ticks sabhi same $x$ share karte hain, isliye woh sirf $h (y)$ ke zariye different hote hain.

Step 3 — $h (y)$ se divide karo: $y$ ko left ki taraf herding karo

KYA. Dono sides ko $h (y)$ se divide karo (hume assume karna hoga ki $h (y) \neq = 0$ — Step 7 ke liye yeh thought rakh lo): $pure- y weight \frac{1}{h ( y )} slope \frac{d y}{d x} = pure x g (x) .$

KYO. Hum chahte hain ki left side mein sirf $y$ -stuff ho (aur slope machinery), aur right side mein sirf $x$ -stuff ho. $h (y)$ se divide karna right side se last $y$ ko left side par sweep kar deta hai. Ab dono variables naddi ke opposite kinaaon par rehte hain.

PICTURE. Do bins imagine karo. Divide karne se pehle, $y$ -ingredient $h (y)$ right par baitha tha, $g (x)$ ke saath uljha hua. Divide karne ke baad, woh left par $1/ h (y)$ ki tarah flip ho jaata hai. Left bin: sirf $y$ . Right bin: sirf $x$ .

Step 4 — Dono sides ko $x$ ke upar integrate karo: slopes ko sum karo

KYA. Dono sides par $\int (\dots) d x$ apply karo: $\int left bin \frac{1}{h ( y )} \frac{d y}{d x} d x = \int right bin g (x) d x .$

KYO. ODE har $x$ par slope fix karta hai. Integration woh tool hai jo infinitely many tiny slopes ko ek actual curve mein stitch karta hai — yeh bilkul "slope lo" ka exact inverse hai. Kyunki dono sides $x$ ki equal functions hain, $x$ ke upar unke integrals equal hain (ek constant tak). Yeh ek hi legal move hai jo slope information ko wapas height information mein badalta hai.

PICTURE. Right side literally $g (x)$ ke neeche shaded area hai: $x$ -axis ko thin strips mein kaato, har strip ka area $g (x) d x$ hai, aur $\int$ unhe add karta hai. Woh accumulated area hi curve ko pin down karta hai.

Step 5 — Chain rule, backwards chalao: left side ko collapse karo

KYA. $H$ ko $\frac{1}{h}$ ka ek antiderivative maan lo, yani $H^{'} (u) = \frac{1}{h ( u )}$ . Chain rule kehta hai $\frac{d}{d x} H (y (x)) = = 1/ h (y) H^{'} (y) \cdot inner slope \frac{d y}{d x} = \frac{1}{h ( y )} \frac{d y}{d x} .$ Toh messy left integrand $H (y (x))$ ka derivative hi hai aur kuch nahi. Isliye $\int \frac{1}{h ( y )} \frac{d y}{d x} d x = \int \frac{d y}{h ( y )} .$

KYO. Yahi honest reason hai ki " $d y / d x$ ek fraction hai" wala shortcut kaam karta hai. Humne actually fraction split nahi kiya; humne ek chain-rule pattern pehchana aur use undo kiya. Poora $d x$ -integral quietly ek $d y$ -integral ban jaata hai.

PICTURE. Chain rule ko ek do-gear machine ki tarah padho: outer gear $H$ rate $H^{'} (y) = 1/ h (y)$ par ghoomta hai, inner gear $y$ rate $d y / d x$ par ghoomta hai. Unka product hai ki jab $x$ move karta hai toh $H (y (x))$ kitni tezi se spin karta hai. Use reverse karna (integrate karna) tumhe $H (y)$ par land karaata hai, ek area $y$ -direction mein measure kiya hua.

Step 6 — Master formula, aur uski picture do areas ki tarah

KYA. Steps 4 aur 5 ko saath rakho aur ek constant attach karo: $\int \frac{d y}{h ( y )} = \int g (x) d x + C$

Term by term:

$\int \frac{d y}{h ( y )}$ ::: $y$ -weight $1/ h (y)$ ko accumulate karo jab $y$ change hoti hai — $y$ -world mein ek area, ise $H (y)$ kaho.
$\int g (x) d x$ ::: $g (x)$ ko accumulate karo jab $x$ change hota hai — $x$ -world mein ek area, ise $G (x)$ kaho.
$C$ ::: ek number jo dono worlds ke beech height offset set karta hai.

Sirf ek $C$ kyun? Har indefinite integral ek constant tak hi fix hota hai, lekin tum dono leftover constants ko ek single $C$ mein bundle kar sakte ho equals sign ke doosri taraf ek ko move karke. Do constants redundant hote.

PICTURE. Left area equals right area plus ek fixed gap $C$ . $C$ choose karna (starting point jaise $y (0) = 3$ ke zariye) sirf ek area ko upar ya neeche slide karna hai jab tak dono apne known point par balance na kar lein.

Step 7 — Forbidden move: $h (y) = 0$ aur kho gaaye horizontal solutions

KYA. Step 3 mein humne $h (y)$ se divide kiya — yeh jahan bhi $h (y) = 0$ ho wahan illegal hai. Agar kisi constant $y_{0}$ ke liye $h (y_{0}) = 0$ ho, toh flat line $y \equiv y_{0}$ ka slope $0$ hai, aur $\frac{d y}{d x} = 0 = g (x) \cdot h (y_{0}) = g (x) \cdot 0 = 0. ✓$ Toh $y \equiv y_{0}$ ek genuine solution hai — lekin division ne use erase kar diya.

KYO. Zero se divide karna undefined hai, isliye algebra ne quietly un flat solutions ko overboard phek diya. Master formula sirf curved family ko capture karta hai; flat equilibrium solutions ko haath se wapas add karna padta hai.

PICTURE. Logistic $h (y) = y (1 - y)$ ke liye, roots $y = 0$ aur $y = 1$ hain: do flat coral lines. Formula se aane wale lavender S-curves unke beech flow karte hain lekin kabhi unhe include nahi karte. Tumhe flat lines alag se report karni padengi.

Step 8 — Do endings: explicit vs implicit (kyun tum kabhi kabhi jaldi rok dete ho)

KYA. Integrate karne ke baad tumhare paas ek relation $H (y) = G (x) + C$ hota hai.

Agar tum algebraically $y$ isolate kar sakte ho, tumhe ek explicit solution milta hai $y = (stuff in x)$ .
Agar $y$ isolate karne ke liye $\pm$ ya mess chahiye, ise implicit solution $F (x, y) = C$ chhod do — bilkul valid hai.

KYO. $y$ ke liye solve karna ek bonus hai, requirement nahi. Relation pehle se hi curve define kar deta hai; kya tum use rearrange kar sakte ho yeh ek alag algebra question hai. Isolation force karna sign errors introduce kar sakta hai.

PICTURE. Do curves side by side: ek jise ek rearrangement clean $y = f (x)$ (explicit) mein flatten karta hai, aur ek — jaise $y^{2} + y - x^{2} = C$ — jo ek knotted contour ki tarah rehta hai jise tum level set ki tarah padhte ho (implicit).

Ek-picture summary

KYA. Sab kuch ek saath: ek product slope-field (Step 2) → $y$ ko left herd karo (Step 3) → har bank ko integrate karo (Steps 4–5) → $C$ se offset equal areas (Step 6), flat equilibrium lines (Step 7) upar drawn, aur ek label ki ending explicit hai ya implicit (Step 8).

Recall Feynman retelling — poora walkthrough plain words mein

Ek ODE ek aisa rule hai jo ek chupi hui curve ki steepness har jagah whisper karta hai. Jab woh steepness hoti hai "ek $x$ -flavour times ek $y$ -flavour," hum mess sort kar sakte hain: har $y$ -ingredient ko left dhakelo, har $x$ -ingredient ko right. Ab har side tiny steepnesses ka ek pile hai. Integration woh glue hai jo un tiny steps ko wapas real heights mein stack karta hai — aur chain rule ko reverse mein run karne ki wajah se, left pile quietly $y$ -direction mein area measure karta hai jabki right pile $x$ -direction mein area measure karta hai. Dono areas ko equal set karo, vertical offset ke liye ek fudge number $C$ allow karo, aur tum ho gaye — almost. $y$ -flavour se divide karne ne secretly koi bhi flat line delete kar di jahan woh flavour zero tha, isliye hum un roots ko dhoondh ke flat solutions wapas staple karte hain. Aakhir mein, agar bacha hua equation " $y =$ kuch" mein suljhata hai, great — explicit; nahi toh hum use $F (x, y) = C$ ki tarah knotted chhod dete hain — implicit, aur utna hi sach.

Connections

Separable ODEs — Technique & Implicit Solutions (parent)
First-Order ODEs — Overview
Initial Value Problems ( $C$ choose karna = area gap slide karna)
Partial Fractions (integrate karne se pehle $1/ h (y)$ split karna)
The Logistic Equation (Step 7 mein equilibrium picture)
Exact ODEs (implicit solutions, generalized)
Substitution Method — Homogeneous ODEs
Integrating Factor & Linear ODEs