Visual walkthrough — Cauchy-Euler (Equidimensional) equation

4.6.16 · D2 · Maths › Ordinary Differential Equations › Cauchy-Euler (Equidimensional) equation

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Step 1 — Symbols ka matlab kya hai?

KYA. Kuch bhi solve karne se pehle, equation ko ek sentence ki tarah padhte hain. Teen characters hain:

• $y$ — $x$ ka ek unknown function. Socho ek curve hai jiska height position $x$ par abhi pata nahi.
• $y^{\prime }$ — us curve ki slope (height kitni tezi se change hoti hai jab hum right step karte hain). Prime mark ${}^{\prime }$ ka matlab hai "$x$ ke saath rate of change".
• $y^{\prime \prime }$ — slope ki slope (curve kitna bend karta hai).

Poori equation hai $\underset{\text{number}}{\underbrace{a}}\underset{\text{coeff.}}{\underbrace{x^2}}\underset{\text{bend}}{\underbrace{y^{\prime \prime }}}+\underset{\text{number}}{\underbrace{b}}\underset{\text{coeff.}}{\underbrace{x}}\underset{\text{slope}}{\underbrace{y^{\prime }}}+\underset{\text{number}}{\underbrace{c}}\underset{\text{height}}{\underbrace{y}}=0.$

KYUN. Coefficients mein pattern dekho: bend term $x^2$ pahnta hai, slope term $x^1$ pahnta hai, height term $x^0=1$ pahnta hai. $x$ ki power hamesha primes ki sankhya ke barabar hoti hai. Yahi "equidimensional" balance hai — yeh ek hi fact hai jise poora page exploit karta hai.

PICTURE. Neeche, ek curve $y(x)$ jisme ek point par height, slope arrow (tangent), aur bend mark kiya gaya hai.

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Step 2 — Power $x^m$ kyun natural guess hai

KYA. Hum trial shape $y=x^m$ propose karte hain, jahan $m$ ek aisa number hai jo abhi pata nahi (yeh $2$ ho sakta hai, ya $-1$, ya imaginary bhi — hum pata kar lenge). Yahan $x^m$ ka matlab bas "$x$ ko power $m$ tak raise karna" hai.

KYUN yeh shape, koi wave nahi? Ordinary constant-coefficient ODE mein (dekho Constant-Coefficient Linear ODEs) magic shape $e^{rx}$ hoti hai, kyunki $e^{rx}$ ko differentiate karne par sirf $r$ se multiply hota hai — shape survive karti hai. Yahan coefficients powers of $x$ hain, constants nahi, isliye surviving shape ko $x$ se multiply karne ke saath nicely interact karna hoga. Power exactly yahi karti hai: $x^m$ ko $x$ se multiply karne par $x^{m+1}$ milta hai (power ek upar), $x^m$ ko differentiate karne par $m{x}^{m-1}$ milta hai (power ek neeche). Yeh do moves inverse hain. Toh $x^2{y}^{\prime \prime }$ power ko do baar neeche push karta hai phir do baar upar — yeh $x^m$ par wapas aata hai.

PICTURE. Ek "power ladder" par do operations arrows ke roop mein: differentiation ek rung neeche jaata hai aur scale karta hai, multiply-by-$x$ wapas upar jaata hai.

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Step 3 — Guess ko differentiate karo, powers ko slide hote dekho

KYA. Differentiation rule $\frac{d}{dx}{x}^p=p{x}^{p-1}$ do baar apply karo:

$y=x^m,y^{\prime }=\underset{\text{scale}}{\underbrace{m}}{x}^{\stackrel{\text{power dropped 1}}{\overbrace{m-1}}},y^{\prime \prime }=\underset{\text{scale}}{\underbrace{m(m-1)}}{x}^{\stackrel{\text{power dropped 2}}{\overbrace{m-2}}}.$

KYUN. Har derivative exponent se ek chheenta hai aur purana exponent multiplying number ke roop mein peeche chod jaata hai. Do derivatives ke baad power $2$ se drop ho gayi aur hum ne aage product $m(m-1)$ collect kar liya. Yahi product aane waale algebra ka star hoga.

PICTURE. Exponent value ko plot karte hain jaise woh $m\to m-1\to m-2$ slide karta hai, saath mein accumulated scale factor likha hua hai.

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Step 4 — Powers cancel ho jaate hain aur equation collapse ho jaati hai

KYA. $y,y^{\prime },y^{\prime \prime }$ ko $a{x}^2{y}^{\prime \prime }+bx{y}^{\prime }+cy=0$ mein substitute karo:

$a{x}^2\cdot m(m-1)x^{m-2}+bx\cdot m{x}^{m-1}+c{x}^m=0.$

Ab $x$ ki powers ko term by term combine karo:

$\underset{\text{first term}}{\underbrace{x^2\cdot x^{m-2}=x^m}},\underset{\text{second term}}{\underbrace{x\cdot x^{m-1}=x^m}},\underset{\text{third term}}{\underbrace{x^0\cdot x^m=x^m}}.$

Har ek term ek number times $x^m$ ban gayi:

$[am(m-1)+bm+c]x^m=0.$

KYUN. Yahi Step 1 ke equidimensional balance ka poora point hai. $x^2$ ne woh do powers cancel kar diye jo second derivative ne hataaye the; $x^1$ ne woh ek cancel kar diya jo first derivative ne hataya. Toh $x^m$ cleanly factor out ho jaata hai aur hum ek pure number puzzle bracket ke andar chod jaate hain.

PICTURE. Teen coloured tiles, har ek common height $x^m$ tak collapse ho rahi hain, phir ek bracket mein merge ho rahi hain.

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Step 5 — Indicial equation: ek quadratic sab decide karta hai

KYA. Kisi bhi $x\ne 0$ ke liye factor $x^m$ zero nahi hai, isliye bracket zero hona chahiye:

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

$m(m-1)=m^2-m$ expand karo aur gather karo:

$a{m}^2+\underset{\text{isko dhyan se dekho!}}{\underbrace{(b-a)}}m+c=0.$

Yeh indicial (auxiliary) equation hai — Characteristic / Auxiliary Equation jaisa hi idea lekin exponentials ki jagah powers ke liye.

KYUN $(b-a)$ aur sirf $b$ nahi? Bend term $x^2{y}^{\prime \prime }$ ne $m(m-1)=m^2-m$ produce kiya, aur woh chhupa hua $-m$ linear coefficient se ek $a$ chura leta hai. Yeh sabse common galti hai — linear coefficient $b-a$ hai, $b$ nahi.

PICTURE. Quadratic $f(m)=a{m}^2+(b-a)m+c$ ek parabola ki tarah draw ki gayi; horizontal axis ke crossings woh roots $m$ hain jo hum dhundh rahe hain.

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Step 6 — Case A: do alag real roots

KYA. Agar parabola axis ko do alag real points $m_1\ne m_2$ par cross kare, to hume do power solutions milte hain aur hum unhe freely combine karte hain:

$y=C_1{x}^{m_1}+C_2{x}^{m_2}.$

Yahan $C_1,C_2$ arbitrary constants hain (ek 2nd-order ODE mein 2-parameter family of answers hoti hai).

KYUN. $x^{m_1}$ aur $x^{m_2}$ genuinely alag shapes hain (independent), isliye ek ka koi combination doosre ko fake nahi kar sakta — saath milkar woh har solution ko span karte hain.

PICTURE. Alag exponents ki do power curves (jaise $x^4$ growing, $x^{-1}$ decaying) $x>0$ ke liye plot ki gayi hain.

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Step 7 — Case B: repeated root, aur $\ln x$ kahan chhupta hai

KYA. Agar parabola sirf axis ko chhue (uska lowest point axis par ho), to do roots ek value $m$ mein merge ho jaate hain. Ab hamare paas sirf EK power $x^m$ hai. Doosra solution hai $y=(C_1+C_2\ln x)x^m,$ aur mystery yeh hai: $\ln x$ kyun?

KYUN. Equation ko $x=e^t$, yaani $t=\ln x$ se disguise hatao (yeh Reduction of Order / substitution trick hai jo parent prove karta hai). $t$-world mein equation constant-coefficient ban jaati hai, aur wahan repeated root pair $e^{mt}$ aur $t{e}^{mt}$ deta hai. Wapas translate karo: $e^{mt}=x^m$ aur $t{e}^{mt}=(\ln x)x^m$. Extra partner ek factor of $t=\ln x$ hai, $x$ ka factor NAHI. $x\cdot x^m$ likhna galat world ki copy karna hoga.

PICTURE. Left: axis ko tangent parabola (double root). Right: $x^2$ aur uska partner $(\ln x)x^2$ alag hote hue taaki dekh sako ki woh independent hain.

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Step 8 — Case C: complex roots, ek curve jo badhte hue spin karta hai

KYA. Agar parabola axis ko kabhi nahi chhuti, to roots complex hain: $m=\alpha \pm i\beta$, jahan $\alpha$ real part hai aur $\beta$ imaginary part, aur $i=\sqrt{-1}$. Solution hai $y=x^{\alpha }[C_1\cos (\beta \ln x)+C_2\sin (\beta \ln x)].$

KYUN. $x^{\alpha +i\beta }=x^{\alpha }\cdot x^{i\beta }$ split karo. Ajeeb piece $x^{i\beta }=e^{i\beta \ln x}$ hai, aur Euler's Formula se $e^{i\theta }=\cos \theta +i\sin \theta$ jahan $\theta =\beta \ln x$ hai. Toh imaginary exponent $\ln x$ variable mein ek real oscillation ban jaata hai. $x^{\alpha }$ overall growth/decay envelope set karta hai; cosine/sine curve ko wobble karaate hain jaise $x$ badhta hai — ek spiral in disguise.

PICTURE. Ek oscillation jiska horizontal axis $\ln x$ hai (toh wiggles $x=0$ ke paas bunch up hote hain aur large $x$ ke liye stretch hote hain), ek $x^{\alpha }$ envelope par sawaar.

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Step 9 — Degenerate edge: $x=0$ par aur $x<0$ ke liye kya hota hai

KYA. $x=0$ par $y^{\prime \prime }$ ka coefficient $a{x}^2$ zero ho jaata hai, isliye equation apna leading term kho deta hai — $x=0$ ek singular point hai (specifically ek regular singular point; dekho Frobenius Method & Regular Singular Points). $x<0$ ke liye raw $\ln x$ undefined hai.

KYUN / fix. $x^{-1}$ ya $\ln x$ jaisi solutions $x=0$ ke paas blow up ya undefined ho sakti hain, isliye hum hamesha ek side tak restrict karte hain. $x<0$ ke liye $x^m\to |x{|}^m$ aur $\ln x\to \ln |x|$ replace karo — Steps 3–5 ka algebra untouched rehta hai kyunki usse sign ki kabhi parwaah nahi thi, sirf power arithmetic ki.

PICTURE. $x$-axis split: ek green working region $x>0$, $x=0$ par ek red forbidden point, aur $x<0$ ka mirrored region $|x|$ se handle kiya gaya.

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Ek-picture summary

Upar sab kuch compress karke: ek power guess karo → derivatives exponent slide karte hain → $x$-powers cancel ho jaate hain → ek quadratic nikalta hai → uske roots (real/repeated/complex) teen answer-shapes mein se ek choose karte hain.

Recall Feynman: poora walkthrough plain words mein

Hamare paas ek equation hai jo care nahi karti ki hum $x$ ko kitna bada draw karte hain — zoom in ya out karo, yeh same dikhti hai. Jo shapes zooming mein survive karti hain woh power curves hain, $x$-to-some-power, toh hum ek guess karte hain. Jab hum power differentiate karte hain, exponent neeche slide karta hai aur aage ek number drop karta hai; har derivative ke saath $x$'s stuck woh wapas upar slide kar dete hain. Us perfect cancellation ki wajah se, poora differential equation ek chote se quadratic mein shrink ho jaata hai jo poochhta hai "kaun si powers fit hoti hain?" Quadratic solve karo. Agar do clean answers hain, to do power curves milti hain aur hum unhe add karte hain. Agar double answer hai toh hum ek solution short hain, isliye hum disguised $x=e^t$ world mein jhankate hain jahan missing partner ek "extra $t$" hai — ise wapas translate karo aur woh $t$ $\ln x$ ban jaata hai. Agar quadratic ke answers imaginary hain, Euler's formula unhe ek aisi curve mein convert karta hai jo badhte hue gently spin karti hai, $\ln x$ ke cosines aur sines deta hai. Aur hum hamesha yaad rakhte hain: $x=0$ par leading term mar jaata hai, isliye hum ek side rehte hain, $|x|$ use karte hain agar left jaate hain.

Recall

Cauchy-Euler equation ke liye trial solution ::: $y=x^m$ Substitution ke baad, har term ban jaata hai ek number times ::: $x^m$ $a{x}^2{y}^{\prime \prime }+bx{y}^{\prime }+cy=0$ ke liye indicial equation ::: $am(m-1)+bm+c=0$, yaani $a{m}^2+(b-a)m+c=0$ Repeated root $m$ extra factor deta hai ::: $\ln x$ ($x$ nahi) Complex roots $\alpha \pm i\beta$ dete hain ::: $x^{\alpha }[C_1\cos (\beta \ln x)+C_2\sin (\beta \ln x)]$ $x=0$ special kyun hai ::: wahan $y^{\prime \prime }$ ka coefficient zero ho jaata hai — ek singular point