Cauchy-Euler (Equidimensional) equation: Worked examples

4.6.16 · D3 · Maths › Ordinary Differential Equations › Cauchy-Euler (Equidimensional) equation

Neeche har example batata hai ki woh matrix ki KAUN si cell fill karta hai. Pehle answer khud guess karo ("Forecast" line), phir worked steps se apne aap ko check karo.

The scenario matrix

Har Cauchy-Euler problem (a) roots ki nature aur (b) uske upar lagaa "twist" se pin down hoti hai. Yeh poora grid hai:

Cell	Root type	Twist / degeneracy	Example
A	Distinct real	plain homogeneous	Ex 1
B	Distinct real	initial-value problem (solve for $C_{1}, C_{2}$ )	Ex 2
C	Repeated	$ln x$ appears	Ex 3
D	Complex	oscillation in $ln x$	Ex 4
E	any	domain twist: $x < 0$ , use $	x
F	any	singular point $x = 0$ , limiting behaviour	Ex 6
G	Distinct real	non-homogeneous RHS (variation of parameters)	Ex 7
H	Distinct real	3rd-order (cubic indicial)	Ex 8
I	Complex	real-world word problem (radial physics)	Ex 9

Jin cheezon par hum rely karte hain: Constant-Coefficient Linear ODEs, the Characteristic / Auxiliary Equation, Euler's Formula, Reduction of Order, aur Variation of Parameters. Parent hai Cauchy-Euler (Equidimensional) equation.

Recall The master recipe (shuru karne se pehle yaad karo)

$a x^{2} y^{''} + b x y^{'} + cy = 0$ ke liye, $y = x^{m}$ guess karo aur indicial equation $am (m - 1) + bm + c = 0$ milegi. Phir:

distinct real $m_{1}, m_{2}$ : $y = C_{1} x^{m_{1}} + C_{2} x^{m_{2}}$
repeated $m$ : $y = (C_{1} + C_{2} ln ∣ x ∣) x^{m}$
complex $α \pm i β$ : $y = ∣ x ∣^{α} [C_{1} cos (β ln ∣ x ∣) + C_{2} sin (β ln ∣ x ∣)]$

Cell A — distinct real roots (plain)

Worked example Ex 1 ·

$x^{2} y^{''} + 2 x y^{'} - 6 y = 0$ Forecast: Do roots, do powers. Ek root positive hogi (growing curve), ek negative (ek $1/ x^{k}$ curve). Padhne se pehle guess karo ki kaun se powers honge.

Step 1. $y = x^{m}$ guess karo, toh $y^{'} = m x^{m - 1}$ , $y^{''} = m (m - 1) x^{m - 2}$ . Yeh step kyun? Ek equidimensional equation mein, $x^{k} \frac{d ^{k}}{d x ^{k}} x^{m}$ hamesha ek constant times $x^{m}$ hota hai, isliye pure power hi ek aisi shape hai jo survive kar sakti hai.

Step 2. Substitute karo: $x^{2} m (m - 1) x^{m - 2} + 2 x m x^{m - 1} - 6 x^{m} = 0$ , aur har $x$ -power collapse ho jaati hai $x^{m}$ mein: $[m (m - 1) + 2 m - 6] x^{m} = 0.$ Yeh step kyun? $x^{2} \cdot x^{m - 2} = x^{m}$ aur $x \cdot x^{m - 1} = x^{m}$ ; poori equation ban jaati hai (number) $\times x^{m}$ .

Step 3. $x \neq = 0$ ke liye bracket zero hona chahiye: $m^{2} + m - 6 = 0 \Rightarrow (m + 3) (m - 2) = 0 \Rightarrow m = 2, - 3$ . Yeh step kyun? Humne $m (m - 1) + 2 m = m^{2} + m$ collect kiya; factoring se do exponents milte hain.

Step 4. Distinct roots → independent powers → general solution $y = C_{1} x^{2} + C_{2} x^{- 3} .$ Yeh step kyun? Do powers jinke alag exponents hain linearly independent hain, isliye unka combination poori family hai.

Verify: $y = x^{2}$ plug karo: $x^{2} (2) + 2 x (2 x) - 6 x^{2} = 2 x^{2} + 4 x^{2} - 6 x^{2} = 0$ . ✓ $y = x^{- 3}$ plug karo: $x^{2} (12 x^{- 5}) + 2 x (- 3 x^{- 4}) - 6 x^{- 3} = 12 x^{- 3} - 6 x^{- 3} - 6 x^{- 3} = 0$ . ✓

Cell B — distinct real + initial-value problem

Worked example Ex 2 ·

$x^{2} y^{''} - 2 x y^{'} - 4 y = 0$ , saath mein $y (1) = 3, y^{'} (1) = 0$ Forecast: Parent note se roots hain $m = 4, - 1$ . Initial data specific $C_{1}, C_{2}$ pick karenge. Kyunki $y^{'} (1) = 0$ (flat start) hai, expect karo ki do powers $x = 1$ par apni slopes balance karenge.

Step 1. General solution (parent Ex): $y = C_{1} x^{4} + C_{2} x^{- 1}$ . Yeh step kyun? Indicial $m^{2} - 3 m - 4 = (m - 4) (m + 1) = 0$ .

Step 2. $y (1) = 3$ apply karo: $C_{1} (1) + C_{2} (1) = 3 \Rightarrow C_{1} + C_{2} = 3$ . Yeh step kyun? $x = 1$ par dono powers $1$ ke barabar hain, jo ek simple sum deta hai.

Step 3. $y^{'} = 4 C_{1} x^{3} - C_{2} x^{- 2}$ ; $y^{'} (1) = 0$ apply karo: $4 C_{1} - C_{2} = 0 \Rightarrow C_{2} = 4 C_{1}$ . Yeh step kyun? Humhe ek doosri equation chahiye; differentiate karo phir evaluate karo.

Step 4. Solve karo: $C_{1} + 4 C_{1} = 3 \Rightarrow C_{1} = \frac{3}{5}, C_{2} = \frac{12}{5}$ . Yeh step kyun? Do linear equations, do unknowns. $y = \frac{3}{5} x^{4} + \frac{12}{5} x^{- 1} .$

Verify: $y (1) = \frac{3}{5} + \frac{12}{5} = \frac{15}{5} = 3$ ✓. $y^{'} (1) = 4 \cdot \frac{3}{5} - \frac{12}{5} = \frac{12}{5} - \frac{12}{5} = 0$ ✓.

Cell C — repeated root ( $ln x$ appears)

Worked example Ex 3 ·

$x^{2} y^{''} - 5 x y^{'} + 9 y = 0$ Forecast: Agar indicial quadratic ka discriminant zero ho, toh hume EK power milti hai aur partner ke liye $ln x$ add karna padta hai. Guess karo: kya doosra curve $x^{m}$ se faster ya slower grow karta hai? (Figure mein orange curve dekho — $ln x$ grow karta hai, toh yeh thoda faster grow karta hai.)

Step 1. Indicial: $m (m - 1) - 5 m + 9 = m^{2} - 6 m + 9 = (m - 3)^{2} = 0 \Rightarrow m = 3$ (double). Yeh step kyun? $b - a = - 5 - 1 = - 6$ ; perfect square ek repeated root signal karta hai.

Step 2. Ek solution hai $y_{1} = x^{3}$ . Doosra $x = e^{t}$ picture se aata hai: wahan ek repeated root $e^{3 t}$ AUR $t e^{3 t}$ deta hai; $t = ln x$ translate karne par $(ln x) x^{3}$ milta hai. Yeh step kyun? Yeh Reduction of Order disguise mein hai — $t$ -world automatically missing partner de deti hai.

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

Verify: $y = x^{3} ln x$ lo. Phir $y^{'} = 3 x^{2} ln x + x^{2}$ , $y^{''} = 6 x ln x + 5 x$ . Substitute karo: $x^{2} (6 x ln x + 5 x) - 5 x (3 x^{2} ln x + x^{2}) + 9 x^{3} ln x = 6 x^{3} ln x + 5 x^{3} - 15 x^{3} ln x - 5 x^{3} + 9 x^{3} ln x = 0$ ✓ (dono $ln$ terms aur plain terms cancel ho jaate hain).

Cell D — complex roots (oscillation in $ln x$ )

Worked example Ex 4 ·

$x^{2} y^{''} + 3 x y^{'} + 5 y = 0$ Forecast: Negative discriminant matlab roots complex plane mein spin kar jaati hain. Solution ek plain power nahi hogi — yeh oscillate karegi, lekin $ln x$ mein, isliye wiggles $x$ badhne par stretch out hote hain (figure mein compressing-then-stretching wave dekho). Real part $α$ set karta hai ki envelope grow karegi ya decay karegi.

Step 1. Indicial: $m (m - 1) + 3 m + 5 = m^{2} + 2 m + 5 = 0$ . Yeh step kyun? $b - a = 3 - 1 = 2$ , $c = 5$ .

Step 2. Quadratic formula: $m = \frac{- 2 \pm 4 - 20}{2} = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i$ . Toh $α = - 1, β = 2$ . Yeh step kyun? $- 16 = 4 i$ ; hum real part $α$ aur imaginary part $β$ mein split karte hain.

Step 3. $x^{α \pm i β} = x^{α} e^{\pm i β l n x}$ use karo aur Euler's Formula $e^{i θ} = cos θ + i sin θ$ with $θ = β ln x$ : $y = x^{- 1} [C_{1} cos (2 ln x) + C_{2} sin (2 ln x)] .$ Yeh step kyun? Euler complex power ko real oscillations mein convert karta hai; $x^{- 1}$ ek decaying envelope hai isliye wiggles $x \to \infty$ par shrink hote hain.

Verify: $y = x^{- 1} cos (2 ln x)$ lo. Numerically $x = e$ par (toh $ln x = 1$ ): $y, y^{'}, y^{''}$ ko ODE mein directly substitute karne par $0$ aata hai — machine check mein confirm hua.

Cell E — domain twist: $x < 0$ , $∣ x ∣$ use karo

Worked example Ex 5 ·

$x^{2} y^{''} - 2 x y^{'} - 4 y = 0$ on $x < 0$ Forecast: Parent Ex 1 jaisi same equation (roots $m = 4, - 1$ ) lekin ab $x$ negative hai. Negative number ko $- 1$ power karna theek hai, lekin general theory $∣ x ∣$ chahti hai. Guess karo: answer identical dikhega ya koi sign aa jaayega?

Step 1. Roots abhi bhi $m = 4, - 1$ hain (indicial equation ne $x$ ka sign kabhi nahi dekha). Yeh step kyun? Indicial equation pure algebra in $m$ hai.

Step 2. $x < 0$ ke liye har power $x^{m}$ ko $∣ x ∣^{m}$ se replace karo: $y = C_{1} ∣ x ∣^{4} + C_{2} ∣ x ∣^{- 1} .$ Yeh step kyun? Jab $m$ non-integer ho, $x^{m}$ $x < 0$ ke liye real nahi hota; $∣ x ∣^{m}$ use karne se solutions poori negative ray par real rehte hain. (Yahan $∣ x ∣^{4} = x^{4}$ lekin $∣ x ∣^{- 1} = - x^{- 1}$ for $x < 0$ , toh constant sign absorb kar leta hai.)

Step 3. Kyunki $ln x$ $x < 0$ ke liye undefined hai, kisi bhi repeated/complex case mein $ln x$ ki jagah $ln ∣ x ∣$ use hoga. Yahan yeh arise nahi hota, lekin rule yaad rakho: positive ray se door hamesha $∣ x ∣$ use karo.

Verify: $x = - 2$ lo, $y = ∣ x ∣^{- 1} = \frac{1}{2}$ . Phir $y = - x^{- 1}$ , $y^{'} = x^{- 2}$ , $y^{''} = - 2 x^{- 3}$ . General $x < 0$ par substitute karo: $x^{2} (- 2 x^{- 3}) - 2 x (x^{- 2}) - 4 (- x^{- 1}) = - 2 x^{- 1} - 2 x^{- 1} + 4 x^{- 1} = 0$ ✓.

Cell F — the singular point $x = 0$ (limiting behaviour)

Worked example Ex 6 ·

$x^{2} y^{''} + 2 x y^{'} - 6 y = 0$ ka $x \to 0^{+}$ par behaviour Forecast: Ex 1 se solution hai $C_{1} x^{2} + C_{2} x^{- 3}$ . $x = 0$ par $y^{''}$ ka coefficient (yaani $x^{2}$ ) vanish ho jaata hai — yeh ek regular singular point hai (Frobenius Method & Regular Singular Points). Guess karo kaun sa piece explode karta hai aur kaun sa bounded rehta hai.

Step 1. $y = C_{1} x^{2} + C_{2} x^{- 3}$ likho aur $x \to 0^{+}$ lo. Yeh step kyun? Limits reveal karte hain ki singular point ke paas kaun sa basis solution "physical" hai.

Step 2. $x^{2} \to 0$ (bounded, actually vanishing) lekin $x^{- 3} \to + \infty$ . Yeh step kyun? Positive exponent origin mein decay karta hai; negative exponent blow up karta hai. $∣ m ∣$ jitna bada, blow-up utna zyada violent.

Step 3. Origin par bounded rehne ke liye hume $C_{2} = 0$ set karna hoga, $y = C_{1} x^{2}$ bachta hai. Yeh step kyun? Ek disk par physical problems (centre par temperature, koi source nahi) infinities reject karti hain, regular branch select karti hain.

Verify: $x = 0.1$ par: $x^{2} = 0.01$ (tiny), $x^{- 3} = 1000$ (huge). Ratio $1 0^{5}$ confirm karta hai ki $x^{- 3}$ near $0$ dominate karta hai. Jaise $x \to 0^{+}$ , $x^{- 3} \to \infty$ ✓.

Cell G — non-homogeneous (variation of parameters)

Worked example Ex 7 ·

$x^{2} y^{''} - 2 x y^{'} - 4 y = x^{3}$ on $x > 0$ Forecast: Homogeneous part Ex 1 hai ( $C_{1} x^{4} + C_{2} x^{- 3}$ ). Forcing $x^{3}$ khud ek power hai, toh expect karo particular solution $\propto x^{3}$ — jab tak $x^{3}$ kisi homogeneous power se collide na kare (nahi karta; $3 \neq = 4, - 3$ ).

Step 1. $y_{h} = C_{1} x^{4} + C_{2} x^{- 3}$ ; try $y_{p} = A x^{3}$ . Yeh step kyun? RHS ek power hai, aur koi bhi homogeneous root $3$ ke barabar nahi hai, isliye undetermined-coefficient power kaam karta hai. (Colliding power ke liye $ln x$ se multiply karoge; general method ke liye Variation of Parameters dekho.)

Step 2. $y_{p}^{'} = 3 A x^{2}$ , $y_{p}^{''} = 6 A x$ . Substitute karo: $x^{2} (6 A x) - 2 x (3 A x^{2}) - 4 A x^{3} = 6 A x^{3} - 6 A x^{3} - 4 A x^{3} = - 4 A x^{3} .$ Yeh step kyun? Har term $x^{3}$ ka multiple hai; collect karo.

Step 3. RHS $x^{3}$ se match karo: $- 4 A = 1 \Rightarrow A = - \frac{1}{4}$ . Toh $y_{p} = - \frac{1}{4} x^{3}$ . Yeh step kyun? Left side par $x^{3}$ ka coefficient $1$ ke barabar hona chahiye.

Step 4. $y = C_{1} x^{4} + C_{2} x^{- 3} - \frac{1}{4} x^{3} .$

Verify: $y_{p} = - \frac{1}{4} x^{3}$ substitute karo: LHS $= - 4 (- \frac{1}{4}) x^{3} = x^{3} =$ RHS ✓.

Cell H — third order (cubic indicial)

Worked example Ex 8 ·

$x^{3} y^{'''} + 2 x^{2} y^{''} - 4 x y^{'} + 4 y = 0$ Forecast: General $n$ -th order Cauchy-Euler abhi bhi $m$ mein ek polynomial deta hai — is baar ek cubic. Teen exponents expect karo. Guess karo ki koi equal hain ya nahi.

Step 1. $y = x^{m}$ ke liye: $y^{'} = m x^{m - 1}$ , $y^{''} = m (m - 1) x^{m - 2}$ , $y^{'''} = m (m - 1) (m - 2) x^{m - 3}$ . Yeh step kyun? Same equidimensional trick; $x^{3} \cdot x^{m - 3} = x^{m}$ , etc.

Step 2. Indicial: $m (m - 1) (m - 2) + 2 m (m - 1) - 4 m + 4 = 0.$ Expand karo: $m (m - 1) (m - 2) = m^{3} - 3 m^{2} + 2 m$ aur $2 m (m - 1) = 2 m^{2} - 2 m$ , toh $m^{3} - 3 m^{2} + 2 m + 2 m^{2} - 2 m - 4 m + 4 = m^{3} - m^{2} - 4 m + 4 = 0.$ Yeh step kyun? $m$ ki saari powers ek cubic mein collect karo.

Step 3. Factor karo: $m = 1$ test karo: $1 - 1 - 4 + 4 = 0$ ✓. $(m - 1)$ factor out karo: $m^{3} - m^{2} - 4 m + 4 = (m - 1) (m^{2} - 4) = (m - 1) (m - 2) (m + 2)$ . Roots $m = 1, 2, - 2$ . Yeh step kyun? Rational-root guess phir remaining quadratic factor karo.

Step 4. Teen distinct real roots → teen independent powers: $y = C_{1} x + C_{2} x^{2} + C_{3} x^{- 2} .$

Verify: $y = x^{2}$ plug karo: $x^{3} (0) + 2 x^{2} (2) - 4 x (2 x) + 4 x^{2} = 0 + 4 x^{2} - 8 x^{2} + 4 x^{2} = 0$ ✓. $y = x^{- 2}$ plug karo: $x^{3} (- 24 x^{- 5}) + 2 x^{2} (6 x^{- 4}) - 4 x (- 2 x^{- 3}) + 4 x^{- 2} = - 24 x^{- 2} + 12 x^{- 2} + 8 x^{- 2} + 4 x^{- 2} = 0$ ✓.

Cell I — real-world word problem (radial physics)

Worked example Ex 9 · Hollow disk mein steady temperature

Ek얇e annular plate mein steady-state temperature $u (r)$ (angular variation ke saath) satisfy kar sakta hai $r^{2} u^{''} + r u^{'} - 4 u = 0, 1 \leq r \leq e, u (1) = 0, u (e) = 10.$ Forecast: Laplace's operator se radial equations hamesha Cauchy-Euler hoti hain ( $r^{2}$ , $r$ , constant pattern geometry mein baked in hai). Yahan $c = - 4 < 0$ distinct real roots $\pm 2$ deta hai: ek growing aur ek decaying radial mode.

Step 1. Indicial: $m (m - 1) + m - 4 = m^{2} - 4 = 0 \Rightarrow m = \pm 2$ . Yeh step kyun? $b - a = 1 - 1 = 0$ , toh linear $m$ term nahi; bas $m^{2} - 4$ .

Step 2. General radial solution: $u = C_{1} r^{2} + C_{2} r^{- 2}$ . Yeh step kyun? Distinct real roots → do power modes.

Step 3. $u (1) = 0$ apply karo: $C_{1} + C_{2} = 0 \Rightarrow C_{2} = - C_{1}$ . $u (e) = 10$ apply karo: $C_{1} e^{2} + C_{2} e^{- 2} = 10 \Rightarrow C_{1} (e^{2} - e^{- 2}) = 10$ . Yeh step kyun? Boundary conditions constants pin karte hain.

Step 4. $C_{1} = \frac{10}{e ^{2} - e ^{- 2}}$ , $C_{2} = - C_{1}$ . Numerically $e^{2} - e^{- 2} \approx 7.2537$ , toh $C_{1} \approx 1.3786$ , $C_{2} \approx - 1.3786$ . $u (r) = 1.3786 (r^{2} - r^{- 2}) (approx) .$

Verify: $u (1) = 1.3786 (1 - 1) = 0$ ✓. $u (e) = 1.3786 (e^{2} - e^{- 2}) = 1.3786 \times 7.2537 \approx 10$ ✓. Units: $u$ °C mein, $r$ dimensionless (scaled radius), har term $r^{\pm 2}$ dimensionless × °C — consistent.