Case 3 - complex conjugate roots — Euler's formula connection: Exercises

4.6.13 · D4 · Maths › Ordinary Differential Equations › Case 3 - complex conjugate roots — Euler's formula connecti

Ye exercises Case 3 ki poori machine ko drill karti hain: negative discriminant pehchanna, $α$ aur $β$ padhna, aur real solution $y = e^{α x} (C_{1} cos β x + C_{2} sin β x)$ likhna. Har problem ke saath ek collapsible full solution hai — pehle khud try karo, phir reveal karo. Difficulty L1 (pehchaan) se L5 (mastery) tak badhti hai.

Shuru karne se pehle, machine ki ek shared reminder, taaki koi bhi symbol unexplained na rahe.

Neeche wala figure har answer ke peeche ka mental picture hai: ek envelope ek wiggle ko squeeze kar rahi hai. (Description: ek blue curve left-to-right oscillate karti hai; do dashed yellow curves — ek upar, ek neeche mirror mein — ek shrinking funnel banate hain jo blue wiggle ko trap karta hai, dikhata hai ki amplitude fade ho rahi hai jab $x$ badhta hai kyunki $α < 0$ hai.)

Figure — Case 3 - complex conjugate roots — Euler's formula connection

Level 1 — Recognition

Goal: ek quadratic diya ho, case decide karo aur $α, β$ padho. Abhi solve nahi karna.

L1.1 $y^{''} + 16 y = 0$ ke liye, roots $m$ nikalo aur $α, β$ batao.

Recall Solution L1.1

Step 1 (WHAT + WHY). $y^{''} \to m^{2}$ , $y \to 1$ replace karke characteristic equation banao: $m^{2} + 16 = 0$ . Ye replacement kyun: trial solution $y = e^{m x}$ substitute karne par har derivative $m$ ki power ban jaati hai ( $y^{''} \to m^{2} e^{m x}$ , $y \to e^{m x}$ ), aur kabhi-zero-na-hone-wale factor $e^{m x}$ ko divide karne par exactly $m^{2} + 16 = 0$ milta hai. Step 2 (WHY complex). Yahan $a = 1, b = 0, c = 16$ , toh $Δ = b^{2} - 4 a c = 0 - 64 = - 64 < 0$ → negative discriminant ka matlab roots complex hain (Case 3). Step 3. $m^{2} = - 16 \Rightarrow m = \pm - 16 = \pm 4 i$ . Toh $α = 0$ , $β = 4$ . Answer: $m = \pm 4 i$ , $α = 0$ , $β = 4$ .

L1.2 $y^{''} + 6 y^{'} + 13 y = 0$ ke liye, $Δ$ compute karo aur confirm karo ki ye Case 3 hai; phir $α, β$ padho.

Recall Solution L1.2

Step 1. $a = 1, b = 6, c = 13$ . $Δ = 6^{2} - 4 (1) (13) = 36 - 52 = - 16 < 0$ ✔ Case 3. Step 2 (WHY the formula). Quadratic formula split hoti hai $α = - \frac{b}{2 a} = - \frac{6}{2} = - 3$ (real part) mein. Step 3. $β = \frac{4 a c - b ^{2}}{2 a} = \frac{52 - 36}{2} = \frac{16}{2} = \frac{4}{2} = 2$ . Answer: $α = - 3$ , $β = 2$ , toh $m = - 3 \pm 2 i$ .

L1.3 Inme se kaun sa Case 3 (complex roots) hai? (a) $m^{2} - 5 m + 6 = 0$ (b) $m^{2} + 4 = 0$ (c) $m^{2} + 2 m + 1 = 0$ .

Recall Solution L1.3

Har discriminant $Δ = b^{2} - 4 a c$ compute karo:

(a) $Δ = 25 - 24 = 1 > 0$ → real distinct, Case 1.
(b) $Δ = 0 - 16 = - 16 < 0$ → complex, Case 3 ✔
(c) $Δ = 4 - 4 = 0$ → repeated real, Case 2. Answer: sirf (b) Case 3 hai.

Level 2 — Application

Goal: quadratic se seedha real general solution tak jaana.

L2.1 $y^{''} + 25 y = 0$ solve karo.

Recall Solution L2.1

Step 1. $m^{2} + 25 = 0 \Rightarrow m = \pm 5 i$ . Toh $α = 0, β = 5$ . Step 2 (WHY the general solution form). Complex roots $α \pm i β$ se real independent solutions $e^{α x} cos β x$ aur $e^{α x} sin β x$ milte hain (Euler + superposition, parent note mein built hai); yahan $α = 0$ toh $e^{α x} = e^{0} = 1$ aur envelope flat hai — undamped. Answer: $y = C_{1} cos 5 x + C_{2} sin 5 x$ .

L2.2 $y^{''} + 4 y^{'} + 8 y = 0$ solve karo.

Recall Solution L2.2

Step 1 (WHAT + WHY). Characteristic equation $m^{2} + 4 m + 8 = 0$ (trial $y = e^{m x}$ , $e^{m x}$ divide karo). Quadratic formula apply karo kyunki ye kisi bhi $a m^{2} + bm + c = 0$ ko exactly solve karta hai: $m = \frac{- 4 \pm 16 - 32}{2} = \frac{- 4 \pm - 16}{2} = \frac{- 4 \pm 4 i}{2} = - 2 \pm 2 i$ . Step 2. $α = - 2, β = 2$ . Step 3 (WHY this final form). Do real independent solutions $e^{- 2 x} cos 2 x$ , $e^{- 2 x} sin 2 x$ superposition se general solution mein combine hote hain. Answer: $y = e^{- 2 x} (C_{1} cos 2 x + C_{2} sin 2 x)$ . $e^{- 2 x}$ decaying envelope hai (damped oscillation).

L2.3 $4 y^{''} + y = 0$ solve karo (dhyan do — $a \neq = 1$ ).

Recall Solution L2.3

Step 1. $4 m^{2} + 1 = 0 \Rightarrow m^{2} = - \frac{1}{4} \Rightarrow m = \pm \frac{1}{2} i$ . Step 2. $α = 0, β = \frac{1}{2}$ . Answer: $y = C_{1} cos \frac{x}{2} + C_{2} sin \frac{x}{2}$ . Note karo $β = \frac{1}{2}$ small hai → slow oscillation, period $2 π / β = 4 π$ .

Level 3 — Analysis

Goal: initial value problems aur answer se physical meaning padhna.

L3.1 IVP $y^{''} + 16 y = 0$ , $y (0) = 3$ , $y^{'} (0) = 8$ solve karo.

Recall Solution L3.1

Step 1 (WHAT + WHY). L1.1 se roots $m = \pm 4 i$ hain, toh Case 3 form se general solution hai $y = C_{1} cos 4 x + C_{2} sin 4 x$ (envelope $e^{0 x} = 1$ ). Do constants kyun: second-order ODE ko do independent solutions chahiye, aur do initial data $y (0), y^{'} (0)$ $C_{1}, C_{2}$ pin down karenge. Step 2 ( $y (0)$ apply karo). $y (0) = C_{1} cos 0 + C_{2} sin 0 = C_{1} = 3$ . Kyun: $cos 0 = 1, sin 0 = 0$ . Step 3 (differentiate phir $y^{'} (0)$ apply karo). $y^{'} = - 4 C_{1} sin 4 x + 4 C_{2} cos 4 x$ , toh $y^{'} (0) = 4 C_{2} = 8 \Rightarrow C_{2} = 2$ . Answer: $y = 3 cos 4 x + 2 sin 4 x$ .

L3.2 IVP $y^{''} + 2 y^{'} + 2 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 0$ solve karo.

Recall Solution L3.2

Step 1. $m = \frac{- 2 \pm 4 - 8}{2} = \frac{- 2 \pm 2 i}{2} = - 1 \pm i$ , toh $α = - 1, β = 1$ . Step 2. $y = e^{- x} (C_{1} cos x + C_{2} sin x)$ . Step 3 ( $y (0)$ ). $y (0) = e^{0} (C_{1} \cdot 1 + C_{2} \cdot 0) = C_{1} = 1$ . Step 4 ( $y^{'}$ ). Product rule use karo: $y^{'} = - e^{- x} (C_{1} cos x + C_{2} sin x) + e^{- x} (- C_{1} sin x + C_{2} cos x)$ . $x = 0$ par: $y^{'} (0) = - (C_{1}) + (C_{2}) = - C_{1} + C_{2} = - 1 + C_{2}$ . $= 0$ set karo $\Rightarrow C_{2} = 1$ . Answer: $y = e^{- x} (cos x + sin x)$ .

L3.3 Ek damped oscillator se milta hai $y = e^{- 0.5 x} (C_{1} cos 3 x + C_{2} sin 3 x)$ . Kya hai (a) oscillation period, (b) wo factor jitna envelope ek period mein shrink karta hai?

Recall Solution L3.3

(a) Period. $β = 3$ , toh period $T = \frac{2 π}{β} = \frac{2 π}{3} \approx 2.094$ . (b) Ek period mein envelope. Envelope hai $e^{- 0.5 x}$ . $Δ x = T = \frac{2 π}{3}$ par ye $e^{- 0.5 \cdot 2 π /3} = e^{- π /3} \approx 0.351$ se multiply hota hai. Answer: period $\frac{2 π}{3} \approx 2.094$ ; amplitude har period mein $\approx 0.351$ (roughly $35%$ ) tak drop hoti hai.

Level 4 — Synthesis

Goal: machine ko ulta chalao, ya kisi twist ke saath.

L4.1 Ek real second-order ODE construct karo jiska general solution $y = e^{2 x} (C_{1} cos 5 x + C_{2} sin 5 x)$ ho.

Recall Solution L4.1

Step 1 (roots padho + WHY ye machine reverse karta hai). Form $e^{α x} (C_{1} cos β x + C_{2} sin β x)$ sirf roots $m = α \pm i β$ se banta hai; toh diye gaye solution se match karne par $α = 2, β = 5$ milta hai, matlab $m = 2 \pm 5 i$ . Step 2 (roots se quadratic banao). Roots $m_{1}, m_{2}$ wala monic quadratic hai $(m - m_{1}) (m - m_{2}) = m^{2} - (m_{1} + m_{2}) m + m_{1} m_{2}$ .

Sum $m_{1} + m_{2} = (2 + 5 i) + (2 - 5 i) = 4$ .
Product $m_{1} m_{2} = (2 + 5 i) (2 - 5 i) = 4 - 25 i^{2} = 4 + 25 = 29$ . Step 3. Characteristic equation $m^{2} - 4 m + 29 = 0$ , toh ODE hai $y^{''} - 4 y^{'} + 29 y = 0$ . Check: $Δ = 16 - 116 = - 100 < 0$ ✔, $α = - b /2 a = 4/2 = 2$ ✔, $β = 100 /2 = 5$ ✔. Answer: $y^{''} - 4 y^{'} + 29 y = 0$ .

L4.2 $a, b, c$ nikalo (jahan $a = 1$ ) taaki $a y^{''} + b y^{'} + cy = 0$ ke solutions period $π$ se oscillate karein aur amplitude har period mein half ho jaaye. (Envelope $e^{α x}$ lo; "har period mein half" matlab $e^{α T} = \frac{1}{2}$ .)

Recall Solution L4.2

Step 1 (period se frequency). Period $T = \frac{2 π}{β} = π \Rightarrow β = 2$ . Step 2 (envelope condition). $e^{α T} = \frac{1}{2}$ jahan $T = π$ : $α π = ln \frac{1}{2} = - ln 2 \Rightarrow α = - \frac{ln 2}{π} \approx - 0.2206$ . Step 3 (coefficients banao). $a = 1$ ke saath: $b = - 2 α = \frac{2 ln 2}{π} \approx 0.4413$ , aur $c = α^{2} + β^{2}$ (kyunki $α^{2} + β^{2} = ∣ m ∣^{2} =$ conjugate roots ka product $= c / a$ ). $c \approx 0.220 6^{2} + 4 \approx 4.0487$ . Answer: $y^{''} + \frac{2 ln 2}{π} y^{'} + (\frac{ln ^{2} 2}{π ^{2}} + 4) y = 0$ , numerically $y^{''} + 0.4413 y^{'} + 4.0487 y = 0$ .

Level 5 — Mastery

Goal: sab kuch ek saath — conjugate structure, Euler, independence, limits.

L5.1 Seedha dikhaao (using Euler's formula) ki $y_{1} = e^{(α + i β) x}$ aur $y_{2} = e^{(α - i β) x}$ ko $u = \frac{1}{2} (y_{1} + y_{2})$ se recombine karne par ek real function milta hai, aur ise compute karo.

Recall Solution L5.1

Step 1 (har ek par Euler). $y_{1} = e^{α x} (cos β x + i sin β x)$ , $y_{2} = e^{α x} (cos β x - i sin β x)$ . Step 2 (add karo). $y_{1} + y_{2} = e^{α x} [(cos β x + cos β x) + i (sin β x - sin β x)] = e^{α x} \cdot 2 cos β x$ . $i$ -parts cancel ho jaate hain — conjugates pair karne ka yehi poora point hai. Step 3 (half karo). $u = \frac{1}{2} (y_{1} + y_{2}) = e^{α x} cos β x$ — purely real ✔. Superposition se, $u$ phir se ek solution hai. (Isi tarah $v = \frac{1}{2 i} (y_{1} - y_{2}) = e^{α x} sin β x$ .)

L5.2 Verify karo ki $u = e^{α x} cos β x$ aur $v = e^{α x} sin β x$ linearly independent hain unka Wronskian compute karke, aur batao ye kahan fail ho sakta hai.

Recall Solution L5.2

Step 0 (WHAT Wronskian hai). Do functions $u, v$ ke liye, Wronskian determinant hai $W = u u^{'} v v^{'} = u v^{'} - v u^{'} .$ Ye kyun matter karta hai: agar $W \neq = 0$ ek bhi point par, toh $u$ aur $v$ ek-doosre ke constant multiples nahi ho sakte → wo independent hain → wo solution space ka valid basis banate hain. Step 1 (derivatives, product rule). $u^{'} = e^{α x} (α cos β x - β sin β x)$ , $v^{'} = e^{α x} (α sin β x + β cos β x)$ . Step 2 (determinant). $W = u v^{'} - v u^{'}$ . $e^{2 α x}$ factor out karo: $W = e^{2 α x} [cos β x (α sin β x + β cos β x) - sin β x (α cos β x - β sin β x)] .$ $α$ -terms cancel ho jaate hain; $β$ -terms dete hain $β (cos^{2} β x + sin^{2} β x) = β$ . Step 3. $W = β e^{2 α x}$ . Kab fail ho sakta hai? $W = 0$ sirf tab agar $β = 0$ . Lekin Case 3 ke liye $Δ < 0 \Rightarrow β > 0$ zaroori hai, aur $e^{2 α x} > 0$ hamesha. Toh $W \neq = 0$ har jagah → independent ✔.

L5.3 Jab $c$ apni boundary value tak jaata hai, Case 3 Case 2 mein degenerate ho jaata hai. $y^{''} + 2 y^{'} + cy = 0$ ke liye, wo value of $c$ nikalo jahan roots complex hona band kar dete hain, aur describe karo ki $β$ aur solution shape ka kya hota hai jab $c$ upar se us value ke paas aata hai.

Recall Solution L5.3

Step 1 (discriminant). $a = 1, b = 2$ : $Δ = 4 - 4 c$ . Complex ke liye $Δ < 0 \Rightarrow c > 1$ chahiye. Step 2 (boundary). $c = 1$ par, $Δ = 0$ → repeated real root Case 2, $m = - 1$ (double). Step 3 ( $β$ ka limit). $β = \frac{4 c - 4}{2} = c - 1$ . Jab $c \to 1^{+}$ , $β \to 0^{+}$ : oscillation infinitely long period tak slow ho jaati hai ( $T = 2 π / β \to \infty$ ). Wiggle seedha hokar Case 2 ka non-oscillating $e^{- x} (C_{1} + C_{2} x)$ ban jaata hai. Answer: boundary $c = 1$ par; jab $c \to 1^{+}$ , $β \to 0$ aur oscillation gayab ho jaati hai (period $\to \infty$ ), continuously Case 2 se match karta hua.

Neeche wala figure L5.3 ke limit ko visible banata hai. (Description: teen coloured curves dikhate hain $e^{α x} sin β x / β$ shrinking $β$ = 1.5 (red), 0.6 (yellow), 0.15 (green) ke liye. Jab $β$ girta hai wiggle stretch hoti hai aur uska pehla hump choda ho jaata hai; dashed blue curve $x e^{α x}$ wo $β \to 0$ limit hai jise teeno approach karte hain — Case 2 solution vanishing oscillation se emerge hoti hui.)

Picture ko algebra ke saath padho: kyunki $\frac{sin β x}{β} \to x$ jab $β \to 0$ , green ( $β = 0.15$ ) curve already dashed blue $x e^{α x}$ se chipki hui hai — oscillation almost poori tarah Case 2 factor $x e^{α x}$ mein badal gayi hai. Ye Case 3 aur Case 2 ke beech ka continuous bridge hai jiske baare mein L5 mistake box warn karta hai.

Recall One-line self-test recap

Case 3 teen moves mein ::: (1) $Δ < 0$ ; (2) $α = - b /2 a$ , $β = 4 a c - b^{2} /2 a$ padho; (3) $y = e^{α x} (C_{1} cos β x + C_{2} sin β x)$ likho. Roots $α \pm i β$ se ODE reverse-engineer karna ::: sum $= 2 α$ , product $= α^{2} + β^{2}$ , toh $m^{2} - 2 α m + (α^{2} + β^{2}) = 0$ .