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

4.6.13 · D5 · Maths › Ordinary Differential Equations › Case 3 - complex conjugate roots — Euler's formula connecti

Apne instincts ko sharpen karo. Neeche diya hua har item ek aisa jagah target karta hai jahan students sahi lagte hain lekin galat hote hain, ya ek boundary case jis par formula quietly depend karta hai. Pehle parent Case 3 - complex conjugate roots — Euler's formula connection padho, phir yahan khud ko test karo.

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

True or false — justify

A real-coefficient quadratic mein exactly ek complex root aur ek real root ho sakta hai.

False. Conjugation real coefficients fix karta hai, toh agar

α + i β

ek root hai toh

α - i β

bhi hona chahiye — complex roots hamesha pair mein aate hain, akele kabhi nahi.

Agar

Δ = b^{2} - 4 a c = 0

toh bhi tum Case 3 mein ho.

False.

Δ = 0

Case 2 hai (ek repeated real root, Case 2 — repeated real roots); Case 3 ko strictly

Δ < 0

chahiye, taaki

b^{2} - 4 a c = i 4 a c - b^{2}

genuinely imaginary ho aur

β \neq = 0

General solution

y = e^{α x} (C_{1} cos β x + C_{2} sin β x)

mein sirf do arbitrary constants hain haalaanki ODE mein "chaar" functions

e^{α x}, cos, sin

hain.

True. Envelope

e^{α x}

ek common factor hai, free choice nahi; ek second-order ODE mein exactly do free constants hote hain, jo

C_{1}, C_{2}

se match karte hain.

e^{i β x}

amplitude ko

x

badhne ke saath grow karta hai.

False.

∣ e^{i β x} ∣ = cos^{2} β x + sin^{2} β x = 1 = 1

sabhi

x

ke liye — yeh unit circle par ek pure rotation hai, isliye yeh amplitude kabhi scale nahi karta. Sirf

α

(real part) amplitude ko touch karta hai.

β > 0

ki jagah

β < 0

choose karne se genuinely alag solution set milta hai.

False.

cos

even hai aur

sin

odd hai, toh

β

ka sign flip karne se sirf

C_{2} \to - C_{2}

hota hai; kyunki

C_{2}

waise bhi saare reals pe range karta hai, solution space identical hai.

α = 0

ke liye solution periodic hai period

2 π / β

ke saath.

True. Bina envelope ke,

y = C_{1} cos β x + C_{2} sin β x

har baar repeat karta hai jab

β x

mein

2 π

ka advancement hota hai, yaani har

Δ x = 2 π / β

par.

Do complex solutions

e^{(α + i β) x}

aur

e^{(α - i β) x}

complex numbers ke upar linearly independent hain.

True. Ye distinct exponents wale distinct exponentials hain; unka Wronskian nonzero hai, toh ye independent hain — hum inhe realness ke liye recombine karte hain, is liye nahi ki ye dependent the.

Spot the error

(Yaad karo: $m$ ek characteristic root hai, yaani $a m^{2} + bm + c = 0$ ka solution, likha $m = α \pm i β$ .)

m = - 1 \pm 2 i

, toh envelope

e^{(- 1 + 2 i) x}

hai aur yeh single exponential se aur oscillate aur decay karta hai."

Envelope sirf

e^{α x} = e^{- x}

hai;

2 i

rehta hai

e^{2 i x} = cos 2 x + i sin 2 x

mein, jo oscillate karta hai lekin magnitude

1

hai.

e^{α x}

(size) ko

e^{i β x}

(rotation) se alag karna hi yahan Euler ka poora point hai.

"Roots hain

\pm 3 i

, toh

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

Double error:

α = 0

matlab envelope

e^{0} = 1

hai (aage

e^{3 i x}

nahi), aur tum real general solution mein kabhi

i

nahi chhodte. Jawab hai

y = C_{1} cos 3 x + C_{2} sin 3 x

y^{''} + 2 y^{'} + 5 y = 0

ke liye mujhe

m = - 1 \pm 2 i

milta hai, toh

y = C_{1} cos 2 x + C_{2} sin 2 x

e^{α x} = e^{- x}

envelope drop ho gaya. Kyunki

α = - 1 \neq = 0

, sahi jawab hai

y = e^{- x} (C_{1} cos 2 x + C_{2} sin 2 x)

; envelope bhoolne se decay bilkul kho jaata hai.

"Euler's formula ek lucky coincidence hai — koi reason nahi ki

e^{i θ}

cos + i sin

ke barabar hona chahiye."

Lucky nahi:

e^{i θ}

ko uski Maclaurin series se define karo aur

i

ki powers cycle karti hain

1, i, - 1, - i

, sum ko exactly cosine (even) aur

i sin

(odd) series mein split karti hain. Yeh forced hai, coincidental nahi.

"Real solutions paane ke liye main

y_{1} + y_{2}

add karta hoon; yahi mera ek real solution hai."

Tumhe do real solutions milte hain:

u = \frac{1}{2} (y_{1} + y_{2}) = e^{α x} cos β x

aur

v = \frac{1}{2 i} (y_{1} - y_{2}) = e^{α x} sin β x

. Dono ek two-dimensional solution space ke liye zaroori hain.

"Kyunki complex form

C_{1} e^{(α + i β) x} + C_{2} e^{(α - i β) x}

mein

i

hai, yeh galat hai."

Yeh actually ek valid general solution hai — lekin sirf tab agar tum

C_{1}, C_{2}

ko complex conjugates lene do taaki imaginary parts cancel ho jayein. Real form prefer ki jaati hai kyunki woh realness ko constants par constraints ke bina clearly dikhati hai.

Why questions

Hum

y_{1}, y_{2}

ko recombine karne se pehle superposition principle kyun appeal karte hain?

Kyunki ODE linear aur homogeneous hai, solutions ka koi bhi linear combination phir se ek solution hai — woh licence humein

u

aur

v

banane deta hai, jo combinations hain aur jo specifically imaginary parts kill karne ke liye choose ki gayi hain.

v = \frac{1}{2 i} (y_{1} - y_{2})

banate waqt

2

se nahi,

2 i

se kyun divide karte hain?

Difference

y_{1} - y_{2}

equal hai

2 i e^{α x} sin β x

, toh

2 i

se divide karne par

i

aur factor

2

strip ho jaata hai, clean real function

e^{α x} sin β x

bach jaata hai.

Sirf

u

aur

v

ko dekhne ki jagah Wronskian kyun check karna zaroori hai?

Is liye ki guarantee ho ki do solutions poore solution space ko span karte hain;

W = β e^{2 α x}

nonzero hai (kyunki

β \neq = 0

e^{2 α x} > 0

), jo independence ko appearance se nahi balki rigorously prove karta hai.

α = - \frac{b}{2 a}

growth versus decay kyun decide karta hai?

α

real envelope

e^{α x}

ka exponent hai:

α < 0

amplitude ko shrink karta hai (damped),

α > 0

ise grow karta hai,

α = 0

ise constant rakhta hai — yeh ek damped oscillator mein friction term hai.

Case 3 ka method (Euler) Case 1 ke do real exponentials wale method se kyun replace nahi ho sakta?

Kyunki

Δ < 0

ke saath "exponents"

α \pm i β

real nahi hain, toh

e^{(α \pm i β) x}

complex-valued hain — Case 1 — real distinct roots ko kabhi Euler ki zaroorat nahi padti kyunki uske exponents real rehte hain.

e^{i β x}

ko unit circle par ek rotating arrow kyun best picture kiya jaata hai?

Kyunki

cos β x

aur

sin β x

uske horizontal aur vertical shadows hain, aur unke squares ka sum

1

hota hai; jaise

x

badhta hai arrow rate

β

se spin karta hai, oscillation deta hai bina kabhi length change kiye.

Negative discriminant physically oscillation kyun force karta hai?

Negative

Δ

matlab restoring force damping par dominate karta hai, toh system overshoot karta hai aur baar baar wapas swing karta hai — imaginary part

β

exactly us swing ki angular frequency hai.

Edge cases

Jab

β \to 0^{+}

(roots ek real repeated root ki taraf merge ho rahe hain) toh solution form ka kya hota hai?

Oscillation frequency vanish ho jaati hai aur

cos β x \to 1

\frac{s i n β x}{β} \to x

, toh basis degenerate ho jaata hai

{e^{α x}, x e^{α x}}

mein — smoothly Case 2 ke repeated-root solution se match karta hai.

Agar

b = 0

(toh

α = - \frac{b}{2 a} = 0

) aur

c > 0

, toh yeh physical system kya hai?

Pure undamped oscillation:

a y^{''} + cy = 0

deta hai

m = \pm i c / a

, yaani simple harmonic motion bina envelope ke, jaise ek frictionless spring-mass.

Kya

β = 0

kabhi ho sakta hai jabki "Case 3" mein bhi ho?

Nahi —

β = \frac{4 a c - b ^{2}}{2 a}

;

β = 0

matlab

4 a c - b^{2} = 0

, yaani

Δ = 0

, jo Case 2 hai, Case 3 nahi. Case 3 ko strictly

β > 0

chahiye.

x \to + \infty

limit mein envelope ka behaviour

α

ke har sign ke liye kya hai?

α < 0

: envelope

\to 0

(oscillation khatam ho jaata hai);

α = 0

: envelope

1

rehta hai (steady oscillation);

α > 0

: envelope

\to \infty

(blow-up oscillation).

IVP

y^{''} + 4 y = 0

y (0) = 0

y^{'} (0) = 0

ke liye unique solution kya hai?

Trivial solution

y \equiv 0

y (0) = C_{1} = 0

aur

y^{'} (0) = 2 C_{2} = 0

dono constants ko zero force karte hain, toh kuch bhi oscillate nahi karta.

Kya

y = e^{α x} cos β x

akela ek general solution hai?

Nahi — yeh bina free constant ke ek single solution hai jo do initial conditions match nahi kar sakta; general solution ko dono

cos

aur

sin

terms chahiye independent constants

C_{1}, C_{2}

ke saath.

C_{1} cos β x + C_{2} sin β x

ko single-amplitude form

R cos (β x - ϕ)

mein swap karne se solution set change hota hai?

Nahi — yeh wahi family hai

R = C_{1}^{2} + C_{2}^{2}

ke saath rewrite ki gayi; lekin

ϕ

ko sahi quadrant mein choose karna chahiye, sirf

tan ϕ = C_{2} / C_{1}

se nahi (agle item dekhein).

Phase angle

ϕ

ko sahi tarah se (sirf

tan ϕ = C_{2} / C_{1}

se nahi) kaise pin down karte hain?

Dono signs use karo:

cos ϕ = C_{1} / R

aur

sin ϕ = C_{2} / R

ke saath

R = C_{1}^{2} + C_{2}^{2}

. Kyunki

tan

har

π

pe repeat karta hai, akela

tan ϕ = C_{2} / C_{1}

18 0^{\circ}

ki ambiguity chhodta hai;

C_{1}, C_{2}

ke signs (equivalently, atan2(C_2,C_1) use karna) sahi quadrant select karte hain — jaise

C_{1} < 0, C_{2} > 0

ϕ

ko quadrant II mein rakhta hai.

Recall Ek-line summary

Complex roots $\Rightarrow$ $α = - \frac{b}{2 a}$ envelope hai (real part, growth/decay decide karta hai), $β = \frac{4 a c - b ^{2}}{2 a}$ wiggle hai (imaginary part, frequency decide karta hai), aur Euler woh bridge hai jo scary $i$ ko ordinary sines aur cosines mein badal deta hai.