Bernoulli equations — substitution: Foundations — Hinglish deep dive

4.6.5 · D1 · Maths › Ordinary Differential Equations › Bernoulli equations — substitution

Is page par assume kiya gaya hai ki tumne is notation mein se kuch bhi nahi dekha. Hum har ek symbol ko bilkul shuruaat se banate hain, uss order mein jis order mein unhe ek doosre ki zaroorat hai, phir tumhe parent Bernoulli topic par wapas bhejte hain.

0 · $y$ kya hai, aur $x$ kya hai?

Kisi bhi equation se pehle, actors par agree karo.

Topic ko yeh kyun chahiye: ek differential equation ek ऐसी curve ke baare mein ek statement hai jo hum abhi nahi jaante. Neeche sab kuch uss curve ki talash hai.

1 · Derivative $\frac{d y}{d x}$ — steepness

Do naam kyun? $\frac{d y}{d x}$ yaad dilata hai ki kya change ho raha hai ( $y$ mein ek choti change aur $x$ mein ek choti change ka ratio); $y^{'}$ sirf shorthand hai jab meaning clear ho. Yeh dono ek hi cheez hain.

Topic ko yeh kyun chahiye: ek Bernoulli equation mein $\frac{d y}{d x}$ hoti hai. Yeh humein har point par slope batati hai — uss position ke hisaab se — ek recipe jo curve ko maanni padti hai.

2 · Ek ODE actually kya kehta hai

3 · $y$ ki powers: $y^{n}$ , $y^{- n}$ , $y^{1 - n}$

Bernoulli mein jo bhi khaas hai woh sab $y$ ki power mein hai. Toh powers theek se samjho.

Topic ko specifically $y^{1 - n}$ kyun chahiye: substitution $v = y^{1 - n}$ naam deta hai. $1 - n$ aata kahan se hai? Arithmetic dekho: equation se $y^{n}$ ki power cancel karne ke liye hum usse divide karte hain, $y$ ko $y^{1 - n} = y^{1} / y^{n}$ mein badal deta hai (division rule use karke upar se). Woh ek subtraction $1 - n$ hi poore trick ki pehchaan hai.

\frac{d y}{d x}

kya measure karta hai, ek word mein?

Curve ka slope (steepness) ek point par.

y^{- 2}

aur

y^{1/2}

ko exponents ke bina likhna.

\frac{1}{y ^{2}}

aur

y

y^{0}

kya hai, aur yeh

n = 1

case kyun break karta hai?

y^{0} = 1

;

n = 1

ke saath,

v = y^{1 - 1} = y^{0} = 1

constant hai — useless.

Kya

P (x)

aur

Q (x)

kabhi

y

par depend karte hain?

Nahi — yeh sirf

x

par depend karte hain.

\frac{d v}{d x}

ke liye chain rule batao jab

v

y

par depend kare aur

y

x

par.

\frac{d v}{d x} = \frac{d v}{d y} \cdot \frac{d y}{d x}

v = y^{1 - n}

x

ke respect mein differentiate karo.

\frac{d v}{d x} = (1 - n) y^{- n} \frac{d y}{d x}

Ek ODE "

y

mein linear" kab hoti hai?

y

aur

y^{'}

sirf first power par aate hain; koi

y^{n}

nahi,

y

aur

y^{'}

ke koi products nahi.

\int \dots d x

kya karta hai?

Antiderivative dhoondhta hai — woh function jiska slope integrand hai (area accumulation).

Integrate karne ke baad arbitrary constant

C

kyun aata hai?

Kaafi curves ek slope field share karte hain;

C

choose karta hai kaun sa.

Connections

Bernoulli equations — substitution (index 4.6.5) (parent — yahan aage jao)
Linear First-Order ODEs — Integrating Factor (woh machine jisme hum reduce karte hain)
Separable Equations
Exact Equations and Integrating Factors
Substitution Methods in ODEs
Riccati Equations

Foundations — Bernoulli equations — substitution

0 · $y$ kya hai, aur $x$ kya hai?

1 · Derivative $\frac{d y}{d x}$ — steepness

2 · Ek ODE actually kya kehta hai

3 · $y$ ki powers: $y^{n}$ , $y^{- n}$ , $y^{1 - n}$

4 · Sirf $x$ ki functions: $P (x)$ aur $Q (x)$

5 · " $y$ mein linear" — woh equation jo hum chahte hain

6 · Chain rule — woh engine jo $y^{n}$ ko gayab karta hai

7 · Integrating factor $μ (x) = e^{\int (1 - n) P d x}$ — ek jhaanki

8 · Arbitrary constant $C$

Prerequisite map

Equipment checklist

Connections