Visual walkthrough — Bernoulli equations — substitution

4.6.5 · D2 · Maths › Ordinary Differential Equations › Bernoulli equations — substitution

Yeh page kya hai: hum parent note ka Bernoulli trick lete hain aur use ek ek picture ke saath bilkul zero se rebuild karte hain. Last step tak aap dekh loge ki $y$ ke ek chunk ka naam badalna ek uljhe hue nonlinear equation ko ek smooth linear equation mein kyun badal deta hai — jo aap pehle se solve karna jaante ho.

Hum sirf yeh maan ke chalte hain ki aap $\frac{dy}{dx}$ ko "jitni tezi se $y$ badalta hai jab $x$ move karta hai" ke roop mein padh sakte ho. Baaki sab hum khud banate hain.

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Step 1 — Hum kya dekh rahe hain

KYA HAI. Hum ek first-order differential equation se milte hain jo is tarah dikhti hai:

$\underset{\text{curve ka slope}}{\underbrace{\frac{dy}{dx}}}+\underset{\text{jaana-maana }x\text{-only coefficient}}{\underbrace{P(x)}}\underset{\text{unknown function}}{\underbrace{y}}=\underset{\text{jaana-maana }x\text{-only term}}{\underbrace{Q(x)}}\underset{\text{asli villain}}{\underbrace{y^n}}$

Har symbol ka matlab:

• $\frac{dy}{dx}$ — woh rate jis par unknown curve $y(x)$ upar ya neeche jaati hai.
• $P(x)$ aur $Q(x)$ — functions jo sirf $x$ par depend karte hain (kabhi $y$ par nahi). Yeh "jaane-maane knobs" hain.
• $y^n$ — unknown $y$ ek fixed power $n$ par (ho sakta hai $2$, $\frac{1}{2}$, $-3$, ...).

KYUN zaroori hai. Agar right side sirf $Q(x)$ hoti (koi $y^n$ nahi), toh yeh ek linear equation hoti — woh friendly wali jise hum integrating factor se solve karte hain. Akela $y^n$ hi ek cheez hai jo hamare raaste mein khadi hai.

PICTURE. Left side par, linear world: $y$ pehli power par aata hai, ek seedha "ramp". Right side par, $y^n$ term us ramp ko ek curve mein mod deti hai — saari takleef ka source yehi hai.

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Step 2 — $n=0$ aur $n=1$ boring kyun hain (aur exclude kyun hain)

KYA HAI. Kuch bhi clever karne se pehle poochho: kya $n$ ki koi aisi values hain jahan koi takleef hi nahi?

• Agar $n=0$: $y^0=1$, toh equation pehle se hi $y^{\prime }+Py=Q$ hai. Pehle se linear.
• Agar $n=1$: right side $Qy$ hai, aur use left mein le jaane par $y^{\prime }+(P-Q)y=0$ milta hai. Pehle se linear.

KYUN. Koi substitution ki zaroorat nahi jab equation pehle se friendly kind ki ho. Isliye definition mein $n\ne 0,1$ likha hai. (Aage ki baat: hamara trick $1-n$ se divide karega; $n=1$ par woh factor $0$ hai, toh trick literally zero se divide karega — algebra khud mana kar deta hai.)

PICTURE. $n$ ki number line. Do dots $n=0$ aur $n=1$ par "pehle se aasaan — skip karo" likha hai. Baaki sab "Bernoulli territory" hai.

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Step 3 — Chhupi hui power expose karo: $y^n$ se divide karo

KYA HAI. Har term ko $y^n$ se divide karo (abhi ke liye maano $y\ne 0$):

$\underset{\text{slope ke saath }{y}^{-n}\text{ chipka hai}}{\underbrace{y^{-n}\frac{dy}{dx}}}+P(x)\underset{\text{ek saaf bachha hua power}}{\underbrace{y^{1-n}}}=Q(x)$

Term-by-term, division ne kya kiya:

• $\frac{dy}{dx}÷y^n=y^{-n}\frac{dy}{dx}$ — slope par ab $y^{-n}$ ka factor lag gaya.
• $Py÷y^n=P{y}^{1-n}$ — linear term ab ek power $y^{1-n}$ ban gaya.
• $Q{y}^n÷y^n=Q$ — villain chala gaya! Woh seedha $Q(x)$ ban gaya.

KYUN yahi move. Hum do cheezein chahte the: (1) right side ka ugly $y^n$ khatam karo, aur (2) sirf ek naya power, $y^{1-n}$, aur ek $y^{-n}{y}^{\prime }$ chunk bachne do. Jaise Step 4 dikhayega, woh chunk bilkul wohi hai jo chain rule swallow kar sakta hai.

PICTURE. Division ko "har exponent ko $n$ se neeche khiskaana" socho. Right-side ka $y^n$ khisak ke $y^0=1$ ban jaata hai; linear $y^1$ khisak ke $y^{1-n}$ ban jaata hai; slope ka chhupta $y^0$ khisak ke $y^{-n}$ ban jaata hai.

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Step 4 — Clean power ka naam rakho: $v=y^{1-n}$ lo

KYA HAI. Ek bilkul naya variable define karo:

$\boxed{v=y^{1-n}}$

KYUN yahi naam. Step 3 mein do matching leftovers dikhte hain: power $y^{1-n}$ aur chunk $y^{-n}{y}^{\prime }$. Hum umeed karte hain ki yeh ek hi cheez ke do chehere hain — yaani $v$ aur uska derivative $v^{\prime }$. Step 5 yeh confirm karta hai. $v=y^{1-n}$ chunna luck nahi hai: yeh isliye engineer kiya gaya hai taaki ise differentiate karne par $y^{-n}$ ka factor wapas nikle.

PICTURE. Ek "relabel" box: messy expression $y^{1-n}$ andar jaata hai, ek saaf akela letter $v$ bahar aata hai. Same object, cleaner naam.

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Step 5 — Engine: chain rule se $v$ differentiate karo

KYA HAI. $v=y^{1-n}$ ko $x$ ke saath differentiate karo. Kyunki $y$ khud $x$ par depend karta hai, hum chain rule use karte hain (power neeche laao, ek ghatao, andar ki derivative se multiply karo):

$\frac{dv}{dx}=\underset{\text{power neeche aata hai}}{\underbrace{(1-n)}}\underset{\text{exponent ek kam ho jaata hai}}{\underbrace{y^{-n}}}\underset{\text{andar wale }y\text{ ki derivative}}{\underbrace{\frac{dy}{dx}}}$

KYUN chain rule aur kuch nahi. Hum "ek function ki power" differentiate kar rahe hain, $(y(x){)}^{1-n}$. Yeh chain rule ka textbook trigger hai: outer operation = "power $1-n$ tak raise karo", inner function = $y(x)$. Koi product rule nahi (kuch multiply nahi ho raha), koi quotient rule nahi (kuch divide nahi ho raha) — chain rule exactly sahi tool hai.

Ab Step 3 mein dikha chunk isolate karne ke liye rearrange karo:

$y^{-n}\frac{dy}{dx}=\frac{1}{1-n}\frac{dv}{dx}$

Right-to-left padho: awkward $y^{-n}{y}^{\prime }$ barabar hai ek plain $v^{\prime }$ jo $\frac{1}{1-n}$ se scale hua ho. Puri trick ek line mein yehi hai.

PICTURE. Chain rule ek do-gear machine ki tarah: outer "power" gear aur inner "$y$" gear dono saath ghoomne par $(1-n)y^{-n}{y}^{\prime }$ output milta hai.

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Step 6 — Substitute karo aur dekho kaise ek line mein collapse ho jaata hai

KYA HAI. Step 3 lo aur uske do matching pieces ko Step 5 use karke replace karo:

• $y^{-n}{y}^{\prime }\to \frac{1}{1-n}\frac{dv}{dx}$
• $y^{1-n}\to v$

$\frac{1}{1-n}\frac{dv}{dx}+P(x)v=Q(x)$

Fraction hatane ke liye $(1-n)$ se multiply karo:

$\boxed{\frac{dv}{dx}+\underset{\text{naya coefficient}}{\underbrace{(1-n)P(x)}}v=\underset{\text{naya right side}}{\underbrace{(1-n)Q(x)}}}$

Term-by-term, kya badla:

• Unknown ab $v$ hai, sirf pehli power par — yeh linear in $v$ hai.
• Chain rule ka factor $(1-n)$ dono $P$ aur $Q$ par chipak jaata hai. Yeh sabse zyada bhoolne wali detail hai: $v^{\prime }+Pv=Q$ likhna aur $(1-n)$ bhool jaana aasaan lagta hai, kyunki substitution $v=y^{1-n}$ aapko original ki clean copy ki umeed dilata hai. $(1-n)$ ko dono sides par hamesha rakho.

KYUN yeh jeet hai. Humne ek anjaan nonlinear equation ko exactly us shape mein convert kar diya jo integrating-factor method naashte mein khaata hai.

PICTURE. Bumpy $y^n$ curve (Step 1) $v$ mein ek seedhe ramp mein badal jaati hai — nonlinearity "naam badal ke hataa di gayi" hai.

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Step 7 — Line solve karo, phir wapas translate karo

KYA HAI. Linear equation ko integrating factor se solve karo

$\mu (x)=e^{\int (1-n)P(x)dx},$

jo left side ko ek perfect derivative $(\mu v{)}^{\prime }=\mu (1-n)Q$ bana deta hai. Ek baar integrate karo, phir $y$ recover karne ke liye back-substitute karo $v=y^{1-n}$.

KYUN $\mu$. $\mu$ se multiply karna woh standard move hai jo $v^{\prime }+(\text{coeff})v$ ko $(\mu v{)}^{\prime }$ mein package karta hai, toh ek integration kaam khatam kar deta hai. (Poori derivation Linear First-Order ODEs — Integrating Factor mein hai.)

Concrete run (parent ka Example 1): $y^{\prime }+\frac{y}{x}=x{y}^2$ mein $P=\frac{1}{x},Q=x,n=2$, toh $1-n=-1$ aur $v=y^{-1}$:

$v^{\prime }-\frac{1}{x}v=-x\stackrel{\mu =1/x}{\to }(\frac{v}{x}{)}^{\prime }=-1\Rightarrow \frac{v}{x}=-x+C\Rightarrow \boxed{\frac{1}{y}=Cx-x^2}.$

PICTURE. Solution curves $\frac{1}{y}=Cx-x^2$ kai $C$ values ke liye, jo method se milne wali family dikhata hai.

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Step 8 — Edge case: $y\equiv 0$ mat bhoolna

KYA HAI. Step 3 mein humne $y^n$ se divide kiya aur quietly maana $y\ne 0$.

KYUN yeh khatarnaak hai. Agar $n>0$ ho, toh constant function $y\equiv 0$ original equation ke dono sides ko zero banata hai — yeh ek solution hai. Lekin $y^n$ se divide karne ne ise mita diya, kyunki zero se divide nahi ho sakta. Yeh $v=y^{1-n}$ ke zariye kabhi wapas nahi aata.

Fix. Solve karne ke baad, jab bhi $n>0$ ho $y\equiv 0$ manually wapas add karo. Yeh ek singular solution hai — $C$-family ka hissa nahi, lekin genuine hai. (Aur Step 3 ki zyada strict warning yaad karo: agar $n$ non-integer hai, toh humne pehle se $y>0$ restrict kar liya tha, toh $y=0$ allowed domain ki boundary par hai, andar nahi.)

PICTURE. Curves ki $C$-family aur ek flat red line $y=0$ alag se hover karti hui, "division mein kho gayi, haath se restore ki gayi" label ke saath.

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Ek-picture summary

Upar ki saari baatein, ek pipeline mein compress karke: nonlinear $y^n$ → $y^n$ se divide karo → $v=y^{1-n}$ rename karo → chain rule engine chalata hai → $v$ mein linear ODE → integrating factor → back-substitute → $y$ (aur $n>0$ hone par singular $y=0$ bhi).

Recall Feynman retelling — poora walkthrough plain words mein

Humne ek aisi equation se shuru kiya jo almost easy kind thi, bas ek piece mein $y$ ek weird power $n$ par tha. Woh power hi akela villain hai. Pehle humne sab kuch $y^n$ se divide kiya; isse right side ka villain ek plain "$1$" mein gayab ho gaya, aur peeche exactly do matching crumbs bache: ek akela power $y^{1-n}$ aur ek slope-chunk $y^{-n}{y}^{\prime }$. (Agar woh power ek fraction hai, toh hum quietly agree kar lete hain ki zero ke ek side par rehenge — maano $y>0$ — taaki $\sqrt{y}$ aur aise terms sense banayein.) Phir humne akele power ko ek nickname diya, $v=y^{1-n}$. Jab humne us nickname ko chain rule se differentiate kiya, slope-chunk bahar aaya (ek constant $1-n$ ke saath) — yeh prove karta hai ki dono crumbs asliyat mein $v$ aur $v^{\prime }$ the disguise mein. Unhe swap karne par, poori equation $v$ mein ek seedhi, linear equation mein flat ho gayi, jise hum pehle se jaante integrating-factor trick se solve karte hain. $v$ ke liye solve karo, nickname ko wapas $y$ mein translate karo, aur — agar power positive thi — $y=0$ wala flat solution wapas dena yaad rakho jo division ne nigal liya tha. Ho gaya.

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Connections

• Bernoulli equations — substitution (parent — poore worked examples wahan hain)
• Linear First-Order ODEs — Integrating Factor (woh engine jise Step 7 handoff karta hai)
• Separable Equations (jab $P$ ya $Q$ vanish ho jaaye tab kya bachta hai)
• Exact Equations and Integrating Factors
• Substitution Methods in ODEs (wahi "rename to simplify" spirit)
• Riccati Equations (nonlinearity mein ek step upar — Bernoulli par reduce hota hai)