Case 2 - repeated real root — reduction of order

4.6.12 · Maths › Ordinary Differential Equations

Constant-coefficient ODE $a y^{''} + b y^{'} + cy = 0$ ke liye, jab characteristic equation ka repeated real root $r$ ho (yaani discriminant $b^{2} - 4 a c = 0$ ), tab do independent solutions hote hain $e^{r x}$ aur $== x e^{r x} ==$ .

The Setup

$x e^{r x}$ kyun? — Do derivations scratch se

Derivation A: Reduction of Order (honest method)

$a y^{''} + b y^{'} + cy = 0$ se shuru karo aur $y = v e^{r x}$ lo.

Derivatives compute karo (product rule): $y^{'} = e^{r x} (v^{'} + r v), y^{''} = e^{r x} (v^{''} + 2 r v^{'} + r^{2} v) .$

Yeh step kyun? Hume $y^{'}, y^{''}$ ko $v$ ke terms mein chahiye taaki substitute kar sakein.

Substitute karo aur $e^{r x} \neq = 0$ factor out karo: $a (v^{''} + 2 r v^{'} + r^{2} v) + b (v^{'} + r v) + c v = 0.$

$v^{''}, v^{'}, v$ ke hisaab se group karo: $a v^{''} + (2 a r + b) v^{'} + (a r^{2} + b r + c) v = 0.$

Yeh step kyun? Coefficients alag karte hain taaki $r$ ke baare mein jo jaante hain woh use kar sakein.

Ab is case ki do facts use karo:

$r$ ek root hai: $== a r^{2} + b r + c = 0 ==$ → $v$ -term vanish ho jaata hai.
$r$ repeated root $r = - b /2 a$ hai, isliye $2 a r + b = 2 a (- b /2 a) + b === 0 ==$ → $v^{'}$ -term vanish ho jaata hai.

Sirf yeh bachta hai $a v^{''} = 0 \Rightarrow v^{''} = 0.$

Derivation B: Limit / coalescing roots (intuition booster)

Independence check (taaki yeh valid second solution ho)

$y_{1} = e^{r x}, y_{2} = x e^{r x}$ ke saath: $y_{2}^{'} = e^{r x} + r x e^{r x} = e^{r x} (1 + r x) .$ $W = e^{r x} \cdot e^{r x} (1 + r x) - r e^{r x} \cdot x e^{r x} = e^{2 r x} [(1 + r x) - r x] === e^{2 r x} == \neq = 0.$

Theek hai — har jagah independent. ✓

Figure — Case 2 - repeated real root — reduction of order

Worked Examples

Common Mistakes

Flashcards

Constant-coefficient ODE

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

ka repeated root kab aata hai?

Jab discriminant

b^{2} - 4 a c = 0

ho; single root hai

r = - b /2 a

Repeated real root

r

ke liye do independent solutions kya hain?

e^{r x}

aur

x e^{r x}

Reduction of order mein hum 2nd solution ke liye kya form assume karte hain?

y_{2} = v (x) y_{1} = v (x) e^{r x}

, jahan

v

unknown ho.

y = v e^{r x}

ko repeated-root ODE mein substitute karne ke baad

v

ke liye kya equation bachti hai?

a v^{''} = 0

, yaani

v^{''} = 0

, jisse

v = C_{1} + C_{2} x

milta hai.

Repeated-root reduction mein

v

aur

v^{'}

dono terms kyun vanish ho jaate hain?

v

-term:

a r^{2} + b r + c = 0

(root).

v^{'}

-term:

2 a r + b = 0

kyunki

r = - b /2 a

(repeated root).

e^{r x}

aur

x e^{r x}

ka Wronskian kya hai?

W = e^{2 r x} \neq = 0

, toh yeh linearly independent hain.

y^{''} - 6 y^{'} + 9 y = 0

ka general solution?

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

C_{1} e^{r x} + C_{2} e^{r x}

valid general solution kyun nahi hai?

Yeh ek function

(C_{1} + C_{2}) e^{r x}

mein collapse ho jaata hai — 1D space, jo 2nd-order ODE ke liye bahut chhota hai.

x e^{r x}

ki limit interpretation kya hai?

lim_{ε \to 0} \frac{e ^{(r + ε) x} - e ^{r x}}{ε} = \frac{\partial}{\partial r} e^{r x} = x e^{r x}

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho tumhe koi bhi tower (solution) banane ke liye do alag building blocks chahiye. Normally do roots do blocks dete hain. Lekin kabhi kabhi do roots same number nikalta hai — toh sirf ek block milta hai, aur sab kuch nahi bana sakte. Trick: apna ek block $e^{r x}$ lo aur usmein $x$ chipka do taaki doosra, genuinely alag block $x e^{r x}$ bane. Ab phir se do blocks hain aur koi bhi tower bana sakte ho. $x$ kyun kaam karta hai: socho do roots almost-but-not-quite equal hain aur unhe slide karke milao, unke do solutions ka difference exactly $x e^{r x}$ ban jaata hai.

Connections

Characteristic equation of linear ODEs — jahan se root $r$ aata hai.
Case 1 distinct real roots — contrast: do roots, do clean exponentials.
Case 3 complex roots — jab discriminant $< 0$ ho.
Reduction of order (general method) — yahan use ki gayi technique, kisi bhi known $y_{1}$ ke liye kaam karti hai.
Wronskian and linear independence — prove karta hai ki $e^{r x}, x e^{r x}$ independent hain.
Higher-order repeated roots — multiplicity $m$ ka root $e^{r x}, x e^{r x}, \dots, x^{m - 1} e^{r x}$ deta hai.