Cauchy-Euler (Equidimensional) equation — Hinglish notes

4.6.16 · Maths › Ordinary Differential Equations

WHAT is it?

Defining feature: degree of $x$ = order of derivative. Yahi balance hai jo substitution $y = x^{m}$ ko kaam karwata hai.

HOW we solve it — derive from scratch

Method 1: the power guess

Try karo $y = x^{m}$ . Toh $y^{'} = m x^{m - 1}, y^{''} = m (m - 1) x^{m - 2} .$

Substitute karo $a x^{2} y^{''} + b x y^{'} + cy = 0$ mein: $a x^{2} \cdot m (m - 1) x^{m - 2} + b x \cdot m x^{m - 1} + c x^{m} = 0$

Yeh step kyun? Har $x$ -power cancel ho jaati hai: $x^{2} \cdot x^{m - 2} = x^{m}$ aur $x \cdot x^{m - 1} = x^{m}$ . Toh: $[a m (m - 1) + b m + c] x^{m} = 0.$

$x \neq = 0$ ke liye hume bracket $= 0$ chahiye:

The three cases (aur WHY har form)

Case 1 — distinct real roots $m_{1} \neq = m_{2}$ : $y = C_{1} x^{m_{1}} + C_{2} x^{m_{2}} .$ Kyun: do independent power solutions hain, linear combination general hai.

Case 2 — repeated root $m_{1} = m_{2} = m$ : Humein sirf ek solution $x^{m}$ milta hai. Doosra kahaan se aata hai?

$y = (C_{1} + C_{2} ln x) x^{m} .$

Case 3 — complex roots $m = α \pm i β$ : $x^{α \pm i β} = x^{α} x^{\pm i β} = x^{α} e^{\pm i β l n x}$ . Euler's formula $e^{i θ} = cos θ + i sin θ$ use karte hue $θ = β ln x$ ke saath: $y = x^{α} [C_{1} cos (β ln x) + C_{2} sin (β ln x)] .$

Figure — Cauchy-Euler (Equidimensional) equation

Method 2: substitution $x = e^{t}$ (Method 1 ko prove karta hai)

Maano $t = ln x$ , toh $\frac{d t}{d x} = \frac{1}{x}$ . Chain rule se: $\frac{d y}{d x} = \frac{1}{x} \frac{d y}{d t} \Rightarrow x y^{'} = \frac{d y}{d t} .$ Phir se differentiate karo (product + chain rule): $y^{''} = \frac{1}{x ^{2}} (\frac{d ^{2} y}{d t ^{2}} - \frac{d y}{d t}) \Rightarrow x^{2} y^{''} = \frac{d ^{2} y}{d t ^{2}} - \frac{d y}{d t} .$

Yeh step kyun? Inhe $a x^{2} y^{''} + b x y^{'} + cy = 0$ mein substitute karne se har $x$ khatam ho jaata hai: $a (\overset{y}{¨} - \overset{y}{˙}) + b \overset{y}{˙} + c y = 0 ⟹ a \overset{y}{¨} + (b - a) \overset{y}{˙} + c y = 0,$ yeh $t$ mein ek constant-coefficient ODE hai jis ka characteristic equation hai $a m^{2} + (b - a) m + c = 0$ — indicial equation se identical. $t$ mein solve karo, phir $t = ln x$ replace karo. Isliye repeated roots produce karte hain $t e^{m t} \to (ln x) x^{m}$ .

Worked Examples

Steel-manned Mistakes

Recall Feynman: explain to a 12-year-old

Ek special spring puzzle imagine karo jahaan rule same rehta hai chahe aap usse kitna bada ya chota draw karo — isko koi favourite size nahi hai (yahi "equidimensional" hai). Jo springs size ki parwah nahi karte, unki natural shapes bendy waves nahi hoti balki stretchy power curves hoti hain jaise $x$ , $x^{2}$ , $\frac{1}{x}$ . Toh wiggles guess karne ki jagah, hum guess karte hain " $x$ to some power", plug in karte hain, aur poori equation ek choti number-puzzle (ek quadratic) mein simat jaati hai jo batati hai ki kaun si powers fit hoti hain. Agar do powers equal nikalein, toh hum ek $ln x$ chupa ke doosra hidden answer dhundhte hain. Agar imaginary nikalein, toh curve grow hote hue gently spin karta hai — yahi se $cos (ln x)$ aur $sin (ln x)$ aate hain.

Flashcards

Kaun sa substitution Cauchy-Euler equation ko constant-coefficient mein convert karta hai?

x = e^{t}

, yaani

t = ln x

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

ke liye, indicial equation kya hai?

am (m - 1) + bm + c = 0

, yaani

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

y = x^{m}

guess karna kyun kaam karta hai?

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

hamesha ek constant times

x^{m}

hota hai, toh har term collapse hokar (number)

\cdot x^{m}

ban jaata hai.

Distinct real roots

m_{1}, m_{2}

ke liye general solution?

y = C_{1} x^{m_{1}} + C_{2} x^{m_{2}}

Repeated root

m

ke liye general solution?

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

Complex roots

α \pm i β

ke liye general solution?

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

Repeated-root case mein

ln x

kahaan se aata hai?

t

-world mein

t e^{m t}

doosre solution se;

t = ln x

Isse "equidimensional" kyun kehte hain?

Har term

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

ke same physical dimensions hote hain, toh equation scale-invariant hai.

x^{2} y^{''}

t = ln x

derivatives ke terms mein kya likhte hain?

\overset{y}{¨} - \overset{y}{˙}

(aur

x y^{'} = \overset{y}{˙}

Kya

x = 0

ek regular point hai?

Nahi, yeh ek singular point hai — leading coefficient

a x^{2}

wahan vanish ho jaata hai.

Connections

Constant-Coefficient Linear ODEs — Cauchy-Euler $x = e^{t}$ ke zariye ek ban jaata hai.
Characteristic / Auxiliary Equation — same role, yahan yeh indicial equation hai.
Euler's Formula — $x^{i β}$ ko real $cos (ln x), sin (ln x)$ mein convert karta hai.
Reduction of Order — $ln x$ doosre solution tak pahunchne ka alternate route.
Frobenius Method & Regular Singular Points — Cauchy-Euler sabse simple regular singular point hai; indicial equation yahan generalize hoti hai.
Variation of Parameters — non-homogeneous case ke liye.