Bernoulli equations — substitution: Question bank

4.6.5 · D5 · Maths › Ordinary Differential Equations › Bernoulli equations — substitution

Shuru karne se pehle, ek shared vocabulary reminder taaki koi symbol ajnabi na lage:

Error dhundho

Student likhta hai: "

v = y^{- 1}

, isliye

v^{'} = y^{- 2} y^{'}

Error:

y^{- 1}

ko differentiate karne par

v^{'} = - y^{- 2} y^{'}

milta hai (power rule se minus sign). Us sign ko drop karna baad ke har term ka sign palat deta hai.

Student

y^{'} + P y = Q y^{n}

y^{n}

se divide karta hai aur likhta hai

y^{- n} y^{'} + P y^{1 - n} = Q y^{- n}

Error: middle term

P y

P y^{1 - n}

ban jaata hai (sahi hai), lekin right-hand side

Q y^{n}

y^{n}

se divide karne par sirf

Q

milta hai, na ki

Q y^{- n}

. Divide karne ka poora reason hi power ko right side se hataana hai.

Student

v = y^{1 - n}

substitute karta hai aur linear ODE

v^{'} + P v = Q

likhta hai.

Error: chain rule se

(1 - n)

dono sides ko multiply karta hai. Sahi form hai

v^{'} + (1 - n) P v = (1 - n) Q

Student ek Bernoulli equation

n = 3

ke saath solve karta hai aur general solution

v = \dots

ek arbitrary constant ke saath report karta hai, phir kehta hai "

y = v

Do errors: constant theek hai, lekin final step

v = y^{1 - n} = y^{- 2}

hai, isliye

y = \pm v^{- 1/2}

— power ko invert karna hoga aur jahan relevant ho

\pm

rakhna hoga.

Student claim karta hai: "Dividing by

y^{n}

par koi comment ki zaroorat nahi kyunki

y

kabhi zero nahi hota."

Error:

y

zero ho sakta hai, aur agar

n > 0

hai toh

y \equiv 0

ek valid solution hai. Use divide karne se pehle alag rakhna hoga aur alag report karna hoga.

Student integrating factor banate waqt

P

y

ka function maanta hai.

Error: Bernoulli equation mein

P

aur

Q

sirf

x

ke functions hote hain. Agar

P

sach mein

y

par depend karta, toh poori linear-ODE machinery apply nahi hoti.

Why questions

Hum

y^{n}

se divide pehle kyun karte hain substitute karne se, rather than directly substitute karne ke?

Divide karne se exact pattern

y^{- n} y^{'}

expose hota hai jo

\frac{d}{d x} y^{1 - n}

produce karta hai; iske bina chain rule align nahi hota aur

y

-powers cancel nahi hote.

v = y^{1 - n}

"sahi" choice kyun hai aur, say,

v = y^{n}

nahi?

Kyunki

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

precisely wahi

y^{- n} y^{'}

term manufacture karta hai jo divide karne ke baad bachi rehti hai — yeh

y^{n}

ki problem ka tailor-made antidote hai.

Factor

(1 - n)

dono

P

aur

Q

par kyun aata hai?

Chain rule deta hai

y^{- n} y^{'} = \frac{1}{1 - n} v^{'}

; us fraction ko clear karne ke liye poori equation ko

(1 - n)

se multiply karna har term ko scale karta hai, isliye

P

aur

Q

dono

(1 - n)

pick up karte hain.

Negative

n

(jaise

n = - 2

) ko bhi same tarike se kyun handle kar sakte hain?

Derivation ne kabhi assume nahi kiya

n > 0

;

v = y^{1 - n} = y^{3}

theek kaam karta hai. Sirf "lost

y \equiv 0

" caveat badalta hai, kyunki

n < 0

ke liye

y^{n}

term

y = 0

par vanish hone ki jagah blow up karta hai.

Bernoulli equation ek linear ODE mein reduce kyun hoti hai na ki separable mein?

Transformed equation mein phir bhi ek explicit

P (x) v

term hota hai jo

v

aur

x

ko couple karta hai, isliye solution route integrating factor hai — separability sirf special case

P \equiv 0

mein aati hai.

Riccati equation Bernoulli se zyada mushkil kyun hai, jabki dono nonlinear hain?

Riccati equation mein

y^{2}

term aur ek stand-alone term bina

y

ke hota hai; woh extra term clean

y^{- n}

cancellation ko block karta hai, isliye pehle ek known particular solution chahiye hoti hai use Bernoulli form mein laane ke liye.

Edge cases

n \to 1

ke saath substitution ka kya hota hai?

Factor

\frac{1}{1 - n}

blow up karta hai — yeh signal hai ki method degenerate ho raha hai.

n = 1

par equation

y^{'} + (P - Q) y = 0

pehle se hi linear hai, isliye koi substitution ki zaroorat nahi.

n = 0

par kya hota hai?

y^{'} + P y = Q

milta hai, jo standard linear form hai — Bernoulli ki machinery sirf

1

se multiply karegi, kuch useful nahi karega.

n = \frac{1}{2}

ke liye, kya

y \equiv 0

abhi bhi ek solution hai jis par dhyan dena ho?

Haan:

n = \frac{1}{2} > 0

, isliye

y \equiv 0

y^{'} - y = y^{1/2}

ko satisfy karta hai (dono sides zero) aur note karna zaroori hai;

y^{1/2}

se divide karne par isse aasaani se khoya ja sakta hai.

n = - 1

ke liye, kya

y^{n}

se divide karna

y \equiv 0

lose karne ka risk hai?

Nahi: yahan

y^{n} = y^{- 1}

hai, aur

y \equiv 0

, equation ko solve nahi karta (

Q y^{- 1}

term

y = 0

par undefined hai), isliye koi trivial solution preserve nahi karni hai.

Agar dono

P \equiv 0

aur

n = 2

hon, equation kya ban jaati hai aur kaise solve karte hain?

Yeh

y^{'} = Q y^{2}

mein reduce hoti hai, jo directly separable hai (

y^{- 2} d y = Q d x

) — Bernoulli substitution abhi bhi kaam karta hai lekin overkill hai.

Fractional

n

jaise

n = \frac{1}{3}

ke liye,

y

ka domain state karte waqt kyun careful rehna chahiye?

v = y^{2/3}

aur

y = v^{3/2}

recover karna sign ya domain restrictions force kar sakta hai (negatives ke roots), isliye solution sirf wahan valid ho sakta hai jahan

y \geq 0

ho — hamesha check karo ki recovered

y

real aur consistent hai.

Agar recovered relation implicit ho, jaise

y^{- 3} = 2 x^{6} + C x^{3}

, kya yeh acceptable final answer hai?

Haan:

y

mein implicit solution ODEs ke liye complete aur standard hai; explicitly

y

ke liye solve karna (yahan

y = (2 x^{6} + C x^{3})^{- 1/3}

) optional hai jab tak problem demand na kare.

Recall Ek-sentence summary

Bernoulli ka method exactly chaar jagah fragile hai — $(1 - n)$ factor, $v^{'}$ mein sign, lost $y \equiv 0$ , aur back-substitution — aur is page ka har trap inhi chaar mein se ek hai ek disguise mein.

Connections

Bernoulli equations — substitution (index 4.6.5) (parent)
Linear First-Order ODEs — Integrating Factor
Separable Equations
Substitution Methods in ODEs
Riccati Equations

Question bank — Bernoulli equations — substitution

True or false — justify karo