Case 2 - repeated real root — reduction of order: Exercises

4.6.12 · D4 · Maths › Ordinary Differential Equations › Case 2 - repeated real root — reduction of order

Yeh page tumhara self-test track hai Case 2: repeated real roots ke liye. Har problem ka poora worked solution ek collapsible callout mein chhupa hua hai — pehle khud try karo, phir reveal karo. Levels dheere-dheere recognise karne se lekar build karne tak jaate hain — poori solution families scratch se.

Level 1 — Recognition

Goal: sirf dekh ke bata do ki root repeated hai ya nahi, aur root kya hai — kuch mushkil solve karne ki zaroorat nahi.

Recall Solution 1.1

Discriminant $b^{2} - 4 a c$ sirf ek sawaal ka jawaab deta hai: kya dono roots equal hain? Repeated root ke liye yeh exactly $0$ hona chahiye.

(a) $a = 1, b = - 10, c = 25$ : $b^{2} - 4 a c = 100 - 100 = 0$ . Repeated — double root $r = - b /2 a = 10/2 = 5$ par. (b) $a = 1, b = 2, c = 5$ : $4 - 20 = - 16 < 0$ . Repeated nahi — complex roots (dekho Case 3 complex roots). (c) $a = 9, b = - 6, c = 1$ : $36 - 36 = 0$ . Repeated — double root $r = - b /2 a = 6/18 = \frac{1}{3}$ par.

Recall Solution 1.2

(C) correct hai.

(A) collapse hokar $(C_{1} + C_{2}) e^{- 2 x}$ ban jaata hai — ek single function, ek 1-dimensional family. 2nd-order ODE ke liye yeh bahut chhota hai.
(B) mein $C_{2}$ (ek bare constant) ko solution maan liya gaya hai, lekin constant tabhi ODE solve karta hai jab $c = 0$ ho. Yahan aisa nahi hai.
(C) genuinely naya partner $x e^{- 2 x}$ banane ke liye required factor $x$ attach karta hai.

Level 2 — Application

Goal: ek repeated-root ODE ko end to end solve karo, odd leading coefficients ke saath bhi.

Recall Solution 2.1

Characteristic: $r^{2} - 14 r + 49 = 0$ . Yeh ek perfect square hai: $(r - 7)^{2} = 0$ , isliye $r = 7$ (double). Check: discriminant $196 - 196 = 0$ . ✓
Solutions: $e^{7 x}$ aur $x e^{7 x}$ .
General solution: $y = (C_{1} + C_{2} x) e^{7 x}$ .

Recall Solution 2.2

$16 r^{2} + 8 r + 1 = 0$ . Perfect square ke roop mein: $(4 r + 1)^{2} = 0$ , isliye $4 r + 1 = 0 \Rightarrow r = - \frac{1}{4}$ (double). Cross-check $r = - b /2 a = - 8/32 = - \frac{1}{4}$ se. ✓
General solution: $y = (C_{1} + C_{2} x) e^{- x /4}$ .

Recall Solution 2.3

$r^{2} + 12 r + 3 = 0$ . Discriminant $b^{2} - 4 a c = 12 - 12 = 0$ — repeated. Root $r = - \frac{12}{2} = - \frac{2 3}{2} = - 3$ .
General solution: $y = (C_{1} + C_{2} x) e^{- 3 x}$ .

Level 3 — Analysis

Goal: initial-value problems, boundary problems, aur solution se ODE banana.

Recall Solution 3.1

$r^{2} - 2 r + 1 = (r - 1)^{2} = 0 \Rightarrow r = 1$ (double).
$y = (C_{1} + C_{2} x) e^{x}$ .
$y (0) = 2$ : $x = 0$ par, $e^{0} = 1$ aur $x$ -wala term zero ho jaata hai, isliye $C_{1} = 2$ .
Differentiate karo: $y^{'} = C_{2} e^{x} + (C_{1} + C_{2} x) e^{x} = (C_{1} + C_{2} + C_{2} x) e^{x}$ . $x = 0$ par: $y^{'} (0) = C_{1} + C_{2} = 5 \Rightarrow C_{2} = 5 - 2 = 3$ .
Answer: $y = (2 + 3 x) e^{x}$ .

Recall Solution 3.2

$r^{2} + 6 r + 9 = (r + 3)^{2} = 0 \Rightarrow r = - 3$ (double).
$y = (C_{1} + C_{2} x) e^{- 3 x}$ .
$y (0) = 0 \Rightarrow C_{1} = 0$ .
$y^{'} = C_{2} e^{- 3 x} - 3 (C_{1} + C_{2} x) e^{- 3 x} = (C_{2} - 3 C_{1} - 3 C_{2} x) e^{- 3 x}$ . $x = 0$ par: $y^{'} (0) = C_{2} - 3 C_{1} = 4 \Rightarrow C_{2} = 4$ .
Answer: $y = 4 x e^{- 3 x}$ .

Recall Solution 3.3

Solution form se pata chalta hai ki root $r = 5$ hai, repeated. Isliye characteristic equation $(r - 5)^{2} = 0$ hai, yaani $r^{2} - 10 r + 25 = 0$ . $a r^{2} + b r + c$ se match karte hue $a = 1$ ke saath: $b = - 10, c = 25.$ ODE: $y^{''} - 10 y^{'} + 25 y = 0$ .

Level 4 — Synthesis

Goal: reduction of order, Wronskian, aur long-run behaviour combine karo.

Recall Solution 4.1

$y = v (x) e^{- x}$ set karo. Tab $y^{'} = e^{- x} (v^{'} - v), y^{''} = e^{- x} (v^{''} - 2 v^{'} + v) .$ $y^{''} + 2 y^{'} + y = 0$ mein substitute karo aur common $e^{- x} \neq = 0$ drop karo: $(v^{''} - 2 v^{'} + v) + 2 (v^{'} - v) + v = 0.$ Group karo: $v^{''} + (- 2 + 2) v^{'} + (1 - 2 + 1) v = 0 \Rightarrow v^{''} = 0$ . Dono lower terms vanish ho jaate hain (repeated root ki pehchaan). Toh $v = C_{1} + C_{2} x$ , aur nayi piece $v = x$ hai, jisse $y_{2} = x e^{- x}$ milta hai. ✓

Recall Solution 4.2

Wronskian $W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ yeh poochhta hai: kya ek solution doosre ka constant multiple hai? Agar $W \neq = 0$ kahi bhi ho, toh nahi — woh independent hain.

$y_{1} = e^{- x}$ , $y_{1}^{'} = - e^{- x}$ .
$y_{2} = x e^{- x}$ , $y_{2}^{'} = e^{- x} - x e^{- x} = e^{- x} (1 - x)$ . $W = e^{- x} \cdot e^{- x} (1 - x) - (- e^{- x}) \cdot x e^{- x} = e^{- 2 x} [(1 - x) + x] = e^{- 2 x} .$ Kyunki $e^{- 2 x} > 0$ everywhere hai, woh independent hain. $x = 1$ par: $W (1) = e^{- 2} \approx 0.135$ .

Recall Solution 4.3

Limit: exponential decay $e^{- x}$ , linear growth $x$ ko beat kar deta hai, isliye $x e^{- x} \to 0$ jab $x \to + \infty$ (aur generally $(C_{1} + C_{2} x) e^{- x} \to 0$ ). Neeche figure dekho.

Maximum of $y = x e^{- x}$ : $y^{'} = e^{- x} (1 - x) = 0$ set karo $\Rightarrow x = 1$ . Tab $y (1) = 1 \cdot e^{- 1} = e^{- 1} \approx 0.368$ . Yeh ek maximum hai (ek hump), jo figure ki shape se match karta hai.

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

Level 5 — Mastery

Goal: pattern ko higher multiplicity aur non-standard forms tak extend karo.

Recall Solution 5.1

Characteristic: $r^{3} - 9 r^{2} + 27 r - 27 = 0$ . Binomial expansion pehchano: $(r - 3)^{3} = r^{3} - 9 r^{2} + 27 r - 27$ . Isliye $r = 3$ ki multiplicity 3 hai. Repeated roots ke pattern ke mutabik, multiplicity $m$ ka ek root $e^{r x}, x e^{r x}, \dots, x^{m - 1} e^{r x}$ contribute karta hai. Yahan $m = 3$ : $y = (C_{1} + C_{2} x + C_{3} x^{2}) e^{3 x} .$

Recall Solution 5.2

Repeated root ke liye discriminant $k^{2} - 4 (1) (9) = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6$ chahiye. $k > 0$ lete hain: $k = 6$ . Tab $r = - k /2 = - 3$ (double), aur $y = (C_{1} + C_{2} x) e^{- 3 x}$ .

Recall Solution 5.3

Characteristic: $r^{4} + 8 r^{2} + 16 = 0$ . $u = r^{2}$ lo: $u^{2} + 8 u + 16 = (u + 4)^{2} = 0 \Rightarrow u = - 4$ (double). Isliye $r^{2} = - 4 \Rightarrow r = \pm 2 i$ , har ek repeated (multiplicity 2). Ek repeated complex pair $\pm β i$ se solutions milte hain $cos β x, sin β x$ aur $x cos β x, x sin β x$ (Case 3 complex roots ke andar applied repeated-root rule). $β = 2$ ke saath: $y = (C_{1} + C_{2} x) cos 2 x + (C_{3} + C_{4} x) sin 2 x .$

Recall Poori ladder ka one-line summary

Spot the double root $r = - b /2 a$ ::: tab general solution $(C_{1} + C_{2} x) e^{r x}$ hai; multiplicity $m$ ke liye $x^{m - 1} e^{r x}$ tak jao.

Connections

Case 2 - repeated real root — reduction of order (index 4.6.12) — woh parent theory jise yeh drills exercise karti hain.
Characteristic equation of linear ODEs — har problem yahan se shuru hoti hai.
Case 1 distinct real roots / Case 3 complex roots — neighbouring cases (Ex 1.1b, 5.3).
Reduction of order (general method) — Exercise 4.1 mein yeh haath se kiya gaya hai.
Wronskian and linear independence — Exercise 4.2.
Higher-order repeated roots — Exercises 5.1 aur 5.3.