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

4.6.13 · D3 · Maths › Ordinary Differential Equations › Case 3 - complex conjugate roots — Euler's formula connecti

Case 3 — complex conjugate roots — Euler's formula connection ka child hai. Yahan hum har tarah ke complex-root problems dhundte hain jo exam mein aa sakte hain, aur har ek ko poori tarah solve karte hain. Agar parent note ne tumhe recipe sikhayi, toh yeh page wo drill hall hai jahan hum us recipe ko har sign, har degenerate input, aur har real-world dressing ke saath run karte hain.

Shuru karne se pehle: wo ek sentence yaad karo jo sab kuch power karta hai. $a y^{''} + b y^{'} + cy = 0$ ke liye, jisme characteristic roots $m = α \pm i β$ hain (ek complex conjugate pair, matlab $β \neq = 0$ ), real general solution hai: $y (x) = e^{α x} (C_{1} cos β x + C_{2} sin β x) .$ Yahan $α = - \frac{b}{2 a}$ envelope exponent hai (root ka real part) aur $β = \frac{4 a c - b ^{2}}{2 a} > 0$ angular frequency hai (imaginary part ki size). $m$ kahan se aata hai, yeh dekhne ke liye Characteristic Equation of Linear ODEs dekho.

Scenario matrix

Har Case-3 problem neeche ki exactly ek cell mein rehta hai. Hamara kaam hai inhe sab hit karna.

Cell	Kya cheez isse alag banati hai	$α$ ka sign	Example
A	Pure oscillation, koi first-derivative term nahi ( $b = 0$ )	$α = 0$	Ex 1
B	Decaying oscillation ( $α < 0$ )	$α < 0$	Ex 2
C	Growing oscillation ( $α > 0$ )	$α > 0$	Ex 3
D	IVP — data se $C_{1}, C_{2}$ pin karo	koi bhi	Ex 4
E	Non-monic leading coeff $a \neq = 1$ ( $α, β$ mein fractions)	$α < 0$	Ex 5
F	Degenerate boundary: $b^{2} = 4 a c$ almost — check karo ki hum Case 2 mein toh nahi gaye	$α < 0$	Ex 6
G	Real-world word problem (spring–mass with damping)	$α < 0$	Ex 7
H	Exam twist: jawab ko single-cosine amplitude–phase form mein convert karo	$α = 0$	Ex 8

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

Cell A — pure oscillation ( $b = 0$ , $α = 0$ )

Worked example Example 1 —

$y^{''} + 16 y = 0$ Forecast: padhne se pehle andaza lagao — kya frequency hogi, aur koi decay hogi?

Step 1. Characteristic equation likho: $m^{2} + 16 = 0$ . Yeh step kyun? $y^{''} \to m^{2}$ aur $y \to 1$ replace karo ( $y^{'}$ term absent hai isliye koi $m$ term nahi). Yeh ek differential equation ko ordinary algebra mein badal deta hai — dekho Characteristic Equation of Linear ODEs.

Step 2. Solve karo: $m^{2} = - 16 \Rightarrow m = \pm - 16 = \pm 4 i$ . Yeh step kyun? $- 16 = 16 - 1 = 4 i$ . Toh $α = 0$ , $β = 4$ .

Step 3. Solution likho: kyunki $α = 0$ , envelope $e^{0 \cdot x} = 1$ dikhna band ho jaata hai. $y = C_{1} cos 4 x + C_{2} sin 4 x .$ Yeh step kyun? Koi real part nahi matlab koi growth ya decay nahi — ek swing bina friction ke, hamesha oscillate karti rehti hai.

Verify: $y = cos 4 x$ lo. Toh $y^{''} = - 16 cos 4 x$ , isliye $y^{''} + 16 y = - 16 cos 4 x + 16 cos 4 x = 0$ . ✔ Frequency $β = 4$ , period $2 π /4 = π /2$ .

Cell B — decaying oscillation ( $α < 0$ )

Worked example Example 2 —

$y^{''} + 4 y^{'} + 13 y = 0$ Forecast: middle term $+ 4 y^{'}$ friction hai. Wiggle badhegi ya sikhudegi?

Step 1. Characteristic: $m^{2} + 4 m + 13 = 0$ . Yeh step kyun? $y^{''} \to m^{2}, y^{'} \to m, y \to 1$ .

Step 2. Quadratic formula: $m = \frac{- 4 \pm 16 - 52}{2} = \frac{- 4 \pm - 36}{2}$ . Yeh step kyun? Discriminant $Δ = 16 - 52 = - 36 < 0$ , yeh Case 3 confirm karta hai (complex roots).

Step 3. $- 36 = 6 i$ , toh $m = \frac{- 4 \pm 6 i}{2} = - 2 \pm 3 i$ . Isliye $α = - 2, β = 3$ . Yeh step kyun? Dono parts ko 2 se divide karo. Real part $- 2$ envelope exponent hai; imaginary part $3$ frequency hai.

Step 4. $y = e^{- 2 x} (C_{1} cos 3 x + C_{2} sin 3 x) .$ Yeh step kyun? $α = - 2 < 0$ → amplitude $e^{- 2 x}$ ki tarah sikhudti hai (friction wali swing). Damped Harmonic Oscillator (RLC / spring-mass) se connect hota hai.

Verify: $y_{1} = e^{- 2 x} cos 3 x$ ke saath, $y_{1}^{'} = e^{- 2 x} (- 2 cos 3 x - 3 sin 3 x)$ aur $y_{1}^{''} = e^{- 2 x} (- 5 cos 3 x + 12 sin 3 x)$ compute karo. Toh $y_{1}^{''} + 4 y_{1}^{'} + 13 y_{1} = e^{- 2 x} [(- 5 - 8 + 13) cos 3 x + (12 - 12) sin 3 x] = 0$ . ✔

Cell C — growing oscillation ( $α > 0$ )

Cell D — initial value problem (constants pin karo)

Worked example Example 4 —

$y^{''} + 6 y^{'} + 13 y = 0$ , $y (0) = 1$ , $y^{'} (0) = - 3$ Forecast: do conditions → do equations → unique $C_{1}, C_{2}$ . Andaza lagao ki yeh decay karega ya nahi.

Step 1. $m^{2} + 6 m + 13 = 0 \Rightarrow m = \frac{- 6 \pm 36 - 52}{2} = \frac{- 6 \pm 4 i}{2} = - 3 \pm 2 i$ . Yeh step kyun? $Δ = - 16 < 0$ ; $- 16 = 4 i$ . Toh $α = - 3, β = 2$ .

Step 2. General solution $y = e^{- 3 x} (C_{1} cos 2 x + C_{2} sin 2 x)$ . Yeh step kyun? Case-3 template apply karo.

Step 3. $y (0) = 1$ use karo: $x = 0$ par, $e^{0} = 1$ , $cos 0 = 1$ , $sin 0 = 0$ , toh $y (0) = C_{1} = 1$ . Yeh step kyun? $x = 0$ substitute karne se har trig term apni value par collapse ho jaata hai, $C_{1}$ isolate ho jaata hai.

Step 4. Differentiate karo (product rule): $y^{'} = e^{- 3 x} [(- 3 C_{1} + 2 C_{2}) cos 2 x + (- 3 C_{2} - 2 C_{1}) sin 2 x] .$ $x = 0$ par: $y^{'} (0) = - 3 C_{1} + 2 C_{2}$ . $- 3$ ke barabar set karo: $- 3 (1) + 2 C_{2} = - 3 \Rightarrow C_{2} = 0$ . Yeh step kyun? Derivative ko start mein diya hua slope match karna chahiye; sirf $cos$ term $x = 0$ par survive karta hai.

Step 5. $y = e^{- 3 x} cos 2 x .$

Verify: $y (0) = e^{0} cos 0 = 1$ ✔. $y^{'} = e^{- 3 x} (- 3 cos 2 x - 2 sin 2 x)$ , toh $y^{'} (0) = - 3$ ✔. Aur $y^{''} + 6 y^{'} + 13 y$ : $y^{''} = e^{- 3 x} (5 cos 2 x + 12 sin 2 x)$ ke saath, sum $= e^{- 3 x} [(5 - 18 + 13) cos 2 x + (12 - 12) sin 2 x] = 0$ ✔.

Cell E — non-monic leading coefficient (fractions aate hain)

Cell F — degenerate boundary (sure karo ki yeh Case 3 hai, Case 2 nahi)

Worked example Example 6 —

$y^{''} + 2 y^{'} + 2 y = 0$ vs. uska dangerous neighbour Forecast: agar $c$ thoda chhota hota aur $Δ$ exactly zero hit karta? Kya hamara formula tab bhi apply hota?

Step 1. $m^{2} + 2 m + 2 = 0$ , $Δ = 4 - 8 = - 4 < 0$ . Yeh step kyun? Humein pehle $Δ$ ka sign check karna chahiye. Sirf $Δ < 0$ Case 3 hai; $Δ = 0$ Case 2 — repeated real roots hota aur $Δ > 0$ Case 1 — real distinct roots hota.

Step 2. $m = \frac{- 2 \pm - 4}{2} = \frac{- 2 \pm 2 i}{2} = - 1 \pm i$ . Toh $α = - 1, β = 1$ . Yeh step kyun? $- 4 = 2 i$ ; $2$ se divide karo. Note karo $β = 1$ — chhota hai par nonzero, toh genuinely oscillatory hai.

Step 3. $y = e^{- x} (C_{1} cos x + C_{2} sin x) .$ Yeh step kyun? Envelope $e^{- x}$ , sabse slow possible whole oscillation.

Step 4 (boundary lesson). Agar equation $y^{''} + 2 y^{'} + 1 y = 0$ hoti, toh $Δ = 4 - 4 = 0$ : Case 3 nahi — repeated root $m = - 1$ , jisse $y = (C_{1} + C_{2} x) e^{- x}$ milta koi wiggle nahi. Jis moment $β \to 0$ , cosine flat hokar $1$ ban jaata hai aur Case 3, Case 2 ko hand over kar deta hai. Yeh step kyun? Yeh Case 3 ka limiting behaviour hai: jab $β \to 0$ , oscillation period $2 π / β \to \infty$ , matlab oscillate karna band ho jaata hai.

Verify: $y = e^{- x} cos x$ ke liye: $y^{'} = e^{- x} (- cos x - sin x)$ , $y^{''} = e^{- x} (2 sin x)$ . Toh $y^{''} + 2 y^{'} + 2 y = e^{- x} [(0 - 2 + 2) cos x + (2 - 2 + 0) sin x] = 0$ ✔.

Cell G — real-world word problem (damped spring)

Worked example Example 7 — spring–mass with damping

Ek mass $m = 1 kg$ ek spring of stiffness $k = 25 N/m$ par lata hai, damper coefficient $c = 6 N\cdots/m$ ke saath. Newton's law deta hai $x^{''} + 6 x^{'} + 25 x = 0$ , jahan $x (t)$ metres mein displacement hai. $x (0) = 0.1 m$ se shuru karo, rest se release $x^{'} (0) = 0$ . $x (t)$ nikalo.

Forecast: underdamped (wiggle karte hue marna chahiye) ya overdamped (bina wiggle ke wapas aana)?

Step 1. Characteristic: $m^{2} + 6 m + 25 = 0$ , $Δ = 36 - 100 = - 64 < 0$ . Yeh step kyun? $Δ < 0$ matlab underdamped — Case 3 ka physical naam. Ek Damped Harmonic Oscillator (RLC / spring-mass) ki real damped motion.

Step 2. $m = \frac{- 6 \pm - 64}{2} = \frac{- 6 \pm 8 i}{2} = - 3 \pm 4 i$ . Toh $α = - 3 s^{- 1}, β = 4 rad/s$ . Yeh step kyun? $α$ ke units inverse seconds hain (ek decay rate); $β$ angular frequency hai rad/s mein. Units confirm karte hain ki humne sahi set up kiya.

Step 3. $x (t) = e^{- 3 t} (C_{1} cos 4 t + C_{2} sin 4 t)$ .

Step 4. $x (0) = C_{1} = 0.1$ . Kyun? trig $t = 0$ par collapse ho jaata hai.

Step 5. $x^{'} = e^{- 3 t} [(- 3 C_{1} + 4 C_{2}) cos 4 t + (- 3 C_{2} - 4 C_{1}) sin 4 t]$ ; $x^{'} (0) = - 3 C_{1} + 4 C_{2} = 0 \Rightarrow C_{2} = \frac{3 C _{1}}{4} = 0.075$ . Kyun? "released from rest" matlab zero starting velocity.

Step 6. $x (t) = e^{- 3 t} (0.1 cos 4 t + 0.075 sin 4 t) m .$ Yeh step kyun? Mass $4$ rad/s par oscillate karta hai shrinking envelope $e^{- 3 t}$ ke andar: thodi baar hilta hai aur settle ho jaata hai. Damped trace ke liye figure dekho.

Verify: $x (0) = 0.1$ ✔; $x^{'} (0) = - 3 (0.1) + 4 (0.075) = - 0.3 + 0.3 = 0$ ✔ (rest). Units: poore mein metres ✔.

Cell H — exam twist: amplitude–phase form

Worked example Example 8 —

$y = 3 cos 2 x + 4 sin 2 x$ ko ek cosine mein rewrite karo Yeh $y^{''} + 4 y = 0$ ka answer hai jisme $y (0) = 3, y^{'} (0) = 8$ . Exams aksar single amplitude–phase form $R cos (2 x - φ)$ maangti hai.

Forecast: peak height $R$ kya hai, aur kya phase $φ$ quadrant I, II, III ya IV mein padega?

Step 1. Chahiye $R cos (2 x - φ) = R cos φ cos 2 x + R sin φ sin 2 x$ . Yeh step kyun? Cosine subtraction identity bridge hai — yeh "do waves" ko "ek shifted wave" mein badal deta hai. Coefficients ko $3 cos 2 x + 4 sin 2 x$ se match karo.

Step 2. Match karo: $R cos φ = 3$ , $R sin φ = 4$ . Yeh step kyun? Do equations, do unknowns $R, φ$ .

Step 3. $R = 3^{2} + 4^{2} = 25 = 5$ . Yeh step kyun? Square aur add karo: $R^{2} (cos^{2} φ + sin^{2} φ) = 9 + 16$ , aur $cos^{2} + sin^{2} = 1$ (unit-circle identity from Euler's Formula and the Unit Circle).

Step 4. $φ = arctan (\frac{R s i n φ}{R c o s φ}) = arctan \frac{4}{3} \approx 0.9273 rad$ . Yeh step kyun? $R cos φ = 3 > 0$ aur $R sin φ = 4 > 0$ dono, toh $φ$ quadrant I mein hai — plain $arctan$ safe hai yahan (koi quadrant fix nahi chahiye). Agar koi negative hota toh $\pm π$ se adjust karte.

Step 5. $y = 5 cos (2 x - 0.9273) .$ Yeh step kyun? Ek clean wave: amplitude $5$ , phase shift $0.9273$ rad.

Verify: $x = 0$ par: $5 cos (- 0.9273) = 5 cos (0.9273) = 5 \cdot \frac{3}{5} = 3 = y (0)$ ✔. Aur $y^{'} = - 10 sin (2 x - 0.9273)$ ; $y^{'} (0) = - 10 sin (- 0.9273) = 10 sin (0.9273) = 10 \cdot \frac{4}{5} = 8 = y^{'} (0)$ ✔.

Recall checkpoint

Recall Main kis cell mein hun? (self-test)

$3 y^{''} + 0 \cdot y^{'} + 12 y = 0$ diya, kaun sa cell? ::: Cell A — $b = 0$ toh $α = 0$ , pure oscillation ( $m = \pm 2 i$ ). Roots $m = 2 \pm i$ diye, kya amplitude badhegi ya decay karegi? ::: Badhegi — $α = + 2 > 0$ , envelope $e^{2 x}$ (Cell C). Jab $β \to 0$ , Case 3 kaun se case mein badal jaata hai? ::: Case 2 — repeated real roots — oscillation period blow up ho jaata hai aur wiggle gayab ho jaati hai. Case 1 vs 2 vs 3 decide karne wala single number kya hai? ::: Discriminant $Δ = b^{2} - 4 a c$ : $> 0, = 0, < 0$ respectively. Spring problem mein, Case 3 ka physical naam kya hai? ::: Underdamped oscillation.

Wapas Case 3 — complex conjugate roots — Euler's formula connection · foundations in Superposition Principle for Linear ODEs aur Wronskian and Linear Independence.