Cauchy-Euler (Equidimensional) equation: Question bank

4.6.16 · D5 · Maths › Ordinary Differential Equations › Cauchy-Euler (Equidimensional) equation

Jo cheez hum reason kar rahe hain usse yaad karo: equation $a x^{2} y^{''} + b x y^{'} + c y = 0$ , iska indicial equation $a m (m - 1) + b m + c = 0$ (equivalently $a m^{2} + (b - a) m + c = 0$ ), aur guess $y = x^{m}$ . Yahan $y^{''}$ ka matlab hai " $y$ ka second derivative", $ln x$ natural logarithm hai, aur $C_{1}, C_{2}$ arbitrary constants hain.

True or false — justify karo

Har second-order Cauchy-Euler equation mein kam se kam ek solution pure form

y = x^{m}

mein hota hai.

True — indicial quadratic

a m^{2} + (b - a) m + c = 0

ka hamesha kam se kam ek root

m

hota hai (real ya complex), aur us root se bana

x^{m}

equation ko hamesha solve karta hai; sirf doosre solution ko

ln x

cos / sin

ki zaroorat pad sakti hai.

a x^{2} y^{''} + b x y^{'} + cy = 0

ka indicial equation

a m^{2} + bm + c = 0

hota hai.

False —

x^{m}

ko do baar differentiate karne par

m (m - 1)

milta hai, isliye

x^{2} y^{''}

contribute karta hai

am (m - 1) = a m^{2} - am

; sahi equation hai

a m^{2} + (b - a) m + c = 0

, jisme ek extra

- am

aata hai.

Agar indicial roots

m_{1} = 2, m_{2} = 2

(repeated) hain, to general solution

y = (C_{1} + C_{2} x) x^{2}

hai.

False — repeated-root ka partner

ln x

ka factor hota hai,

x

ka nahi; answer hai

(C_{1} + C_{2} ln x) x^{2}

. "Extra

x

" wala idea constant-coefficient ODEs se illegally borrowed hai.

x^{i}

(jahan

i = - 1

) ek genuine complex-valued solution hai jab roots

m = \pm i

hon.

True —

x^{i} = e^{i l n x} = cos (ln x) + i sin (ln x)

by Euler's Formula; ye ODE ko solve karta hai, aur iske real aur imaginary parts do real solutions dete hain.

Kyunki equation "equidimensional" hai,

x \to λ x

scale karne par solution set unchanged rehta hai.

True — har term

x^{k} y^{(k)}

ka same dimension hota hai, isliye ODE scale-invariant hai; agar

y (x)

ise solve karta hai, to

y (λ x)

bhi karta hai, yahi wajah hai ki power functions (jo khud scale-covariant hain) natural solutions hain.

Cauchy-Euler equation disguise mein ek constant-coefficient ODE hai.

True — substitution

t = ln x

ise

a \overset{y}{¨} + (b - a) \overset{y}{˙} + cy = 0

mein convert kar deta hai jisme constant coefficients hain; "disguise" variable

x

hai, aur

t = ln x

use hata deta hai.

x^{2} y^{''} + b x y^{'} + cy = 0

mein origin

x = 0

ek ordinary point hai.

False —

x = 0

par leading coefficient

x^{2}

zero ho jaata hai, isliye

x = 0

ek singular point hai (actually ek regular singular point, dekho Frobenius Method & Regular Singular Points);

x^{- 1}

ln x

jaise solutions wahan blow up karte hain.

x < 0

ke liye solution

x^{m}

∣ x ∣^{m}

se aur

ln x

ln ∣ x ∣

se replace karna padta hai.

True —

ln x

aur non-integer powers negative

x

ke liye undefined hain; interval

x < 0

par hum

∣ x ∣

use karte hain taaki expressions real aur defined rahein.

Error dhundo

x^{2} y^{''} - 2 x y^{'} - 4 y = 0

se indicial

m^{2} - 2 m - 4 = 0

milta hai." — galti kahan hai?

x^{2} y^{''}

term se

m (m - 1) = m^{2} - m

milta hai,

m^{2}

nahi; sahi indicial hai

m^{2} - m - 2 m - 4 = m^{2} - 3 m - 4 = 0

. Likhne wala

m (m - 1)

se aane wala

- m

bhool gaya.

"Repeated root

m = 3

, isliye

y = (C_{1} + C_{2} x^{3} ln x)

." — ise fix karo.

Dono terms mein base power

x^{3}

honi chahiye; answer hai

y = (C_{1} + C_{2} ln x) x^{3}

ln x

doosri

x^{3}

copy ko multiply karta hai, ye pehli

x^{3}

ko replace nahi karta aur na hi us se alag float karta hai.

"Complex roots

m = α \pm i β

y = x^{α} (C_{1} cos β x + C_{2} sin β x)

milta hai." — kya galat hai?

Oscillation

ln x

mein hoti hai,

x

mein nahi: ye hona chahiye

cos (β ln x)

aur

sin (β ln x)

, kyunki

x^{i β} = e^{i β l n x}

cos β x

likhna constant-coefficient answer ko galti se import karna hai.

x^{2} y^{''} + x y^{'} + y = 0

solve karne ke liye main guess

y = e^{r x}

ke saath undetermined coefficients use karunga." — ye kyun fail hota hai?

e^{r x}

differentiate karne par doosra

e^{r x}

milta hai lekin

x^{2}

x

coefficients tab

x

-powers uncancelled chhod dete hain — terms collapse nahi hote. Equidimensional structure demand karta hai

y = x^{m}

taaki

x^{k} \frac{d ^{k}}{d x ^{k}} x^{m} \propto x^{m}

ho.

t = ln x

ke baad, equation

x^{2} y^{''} + b x y^{'} + cy = 0

a \overset{y}{¨} + b \overset{y}{˙} + cy = 0

ban jaata hai." — ise correct karo.

\overset{y}{˙}

ka coefficient

b - a

hota hai,

b

nahi, kyunki

x^{2} y^{''} = \overset{y}{¨} - \overset{y}{˙}

ek extra

- a \overset{y}{˙}

contribute karta hai. Transformed equation hai

a \overset{y}{¨} + (b - a) \overset{y}{˙} + cy = 0

Why questions

Hum yahan exponential

e^{r x}

(jo Constant-Coefficient Linear ODEs ke liye kaam karta hai) ki jagah power

x^{m}

kyun guess karte hain?

Kyunki

x^{m}

differentiate karne par power ek se kam ho jaati hai jabki coefficient

x^{k}

use

k

se wapas badhata hai, isliye

x^{k} y^{(k)}

x^{m}

ke proportional rehta hai aur har term (number)

\cdot x^{m}

mein collapse ho jaata hai. Exponentials ye trick sirf tab karte hain jab coefficients constant hon.

Repeated root

ln x

kyun force karta hai na ki ek second power?

t = ln x

ki duniya mein repeated root

e^{m t}

aur

t e^{m t}

deta hai (standard Reduction of Order result);

t = ln x

se wapas translate karne par

t

ln x

ban jaata hai, aur

x^{m}

aur

(ln x) x^{m}

milte hain.

t

mein transformed characteristic equation identical kyun hoti hai

x

mein indicial equation se?

Kyunki dono same collapse encode karte hain:

y = e^{m t}

a \overset{y}{¨} + (b - a) \overset{y}{˙} + cy = 0

mein substitute karne par

a m^{2} + (b - a) m + c = 0

milta hai, aur

e^{m t} = e^{m l n x} = x^{m}

exactly power guess hai. Dono methods ek hi calculation hain jo do variables mein dekhi jaati hai.

Naive quadratic-in-

m

ka middle coefficient

b

ki jagah

(b - a)

kyun hai?

Second-derivative term

m (m - 1) = m^{2} - m

produce karta hai; extra

- am

+ bm

mein merge hokar

(b - a) m

deta hai. Ye shift poore topic ka sabse common trap hai.

Euler identity

x

mein nahi,

ln x

mein oscillation kyun produce karta hai?

Kyunki

x^{i β} = e^{(i β) l n x}

, isliye

cos

sin

mein feed hone wala angle

θ = β ln x

hai; argument woh hota hai jo exponent mein

i

ko multiply karta hai, aur woh

ln x

hai.

Non-homogeneous Cauchy-Euler equation ko pehle

t = ln x

substitute karke kyun solve kar sakte hain?

Substitution coefficients ko constant bana deta hai, isliye hum phir

t

mein freely undetermined coefficients ya Variation of Parameters use kar sakte hain, aur end mein sirf

t = ln x

wapas translate karte hain — equidimensional structure guarantee karta hai ki ye conversion exact hai.

Equation "equidimensional" kyun kahlaati hai aur ye power-law solutions ko kaise predict karta hai?

Har term

x^{k} y^{(k)}

ek physical dimension share karta hai, isliye equation mein koi built-in length scale nahi hai; ek scale-free equation naturally scale-covariant functions se solve hoti hai, aur pure powers

x^{m}

exactly woh functions hain jo cleanly scale karte hain.

Edge cases

Agar dono indicial roots zero hain (

m_{1} = m_{2} = 0

), to general solution kya hai?

y = (C_{1} + C_{2} ln x) x^{0} = C_{1} + C_{2} ln x

— ek constant plus

ln x

; ye

x^{2} y^{''} + x y^{'} = 0

ke liye hota hai jahan

c = 0

aur

b = a

ho.

Complex case

m = α \pm i β

mein

β \to 0

hone par solution structure ka kya hoga?

Roots ek real repeated root

m = α

mein merge ho jaate hain; correspondingly

cos (β ln x) \to 1

aur

\frac{s i n ( β l n x )}{β} \to ln x

, smoothly repeated-root form

(C_{1} + C_{2} ln x) x^{α}

recover ho jaata hai.

Kya

x = 0

kabhi Cauchy-Euler solution ke domain mein ek legal point ho sakta hai?

x^{- 1}

ln x

, ya non-integer powers jaise solutions ke liye nahi — ye

0

par singular hain; sirf jab dono roots non-negative integers hon tabhi ek solution wahan finite reh sakta hai, lekin general solution phir bhi

x = 0

ko exclude karta hai kyunki independence typically singular partner demand karti hai.

x < 0

ke liye, irrational

m

wala

x^{m}

problematic kyun hai, aur remedy kya hai?

Negative base ko irrational power pe uthane par real value nahi milti, isliye

x^{m}

undefined hota hai; remedy hai

∣ x ∣^{m}

, jo same ODE ko

x < 0

par solve karta hai aur real rehta hai.

Agar first-order Cauchy-Euler equation

x y^{'} + cy = 0

diya gaya ho, to kya "

ln x

" kabhi appear hoga?

First order mein koi repeated-root

ln x

appear nahi ho sakta — indicial linear hai (

m + c / a

-style) ek single root ke saath;

ln x

sirf tab aata hai jab repeated root ek doosra independent solution demand kare, jiske liye order

\geq 2

chahiye.

Kya substitution

t = ln x

verbatim interval

x < 0

par use ki ja sakti hai?

Nahi —

ln x

wahan undefined hai; tumhe

t = ln ∣ x ∣

use karna hoga, jiske baad constant-coefficient machinery negative branch par identically kaam karti hai.

x^{2} y^{''} + b x y^{'} + cy = 0

x^{2}

se divide karne par kya equation singular kahan hai ye change ho jaata hai?

Nahi — divide karne par

y^{''} + \frac{b}{x} y^{'} + \frac{c}{x ^{2}} y = 0

milta hai; coefficients phir bhi

x = 0

par blow up karte hain, confirming karta hai ki

x = 0

ek genuine (regular) singular point hai, ye sirf likhne ke tarike ka artefact nahi hai.

Recall One-line self-test

Ek saanp mein teen traps: middle coefficient == $(b - a)$ not $b$ hai, repeated-root partner $ln x$ not $x$ hai, aur oscillation $ln x$ not $x$ == mein hoti hai.