4.6.13
Topic: Second-order linear homogeneous ODE with constant coefficients , jab characteristic equation mein complex conjugate pair of roots ho.
Hum study karte hain:
a y ′′ + b y ′ + cy = 0 , a , b , c ∈ R , a = 0
jiska characteristic equation hai a m 2 + bm + c = 0
Definition Characteristic equation ke teen casesa m 2 + bm + c = 0 Δ = b 2 − 4 a c
Δ > 0 real distinct roots (Case 1).Δ = 0 repeated real root (Case 2).Δ < 0 complex conjugate roots ==m = α ± i β
Intuition Complex roots pairs mein kyun aate hainCoefficients a , b , c real hain. Real coefficients wale polynomial mein akela complex root nahi ho sakta — agar α + i β α − i β 0 + , × conjugate pair ke roop mein saath aate hain.
Quadratic formula se:
m = 2 a − b ± b 2 − 4 a c = α 2 a − b ± i β 2 a 4 a c − b 2
Toh α = − 2 a b β = 2 a 4 a c − b 2 > 0 Δ < 0 4 a c − b 2 > 0
Mechanically, exponential method phir bhi do solutions deta hai:
y 1 = e ( α + i β ) x , y 2 = e ( α − i β ) x .
Lekin ye complex-valued hain. Real-world ODE ke liye hume real solutions chahiye. Inhe kaise nikale? Yahaan Euler's formula hamare kaam aata hai.
e i θ e i θ define karte hain (yahi e x d θ d e i θ = i e i θ i i 0 = 1 , i 1 = i , i 2 = − 1 , i 3 = − i , i 4 = 1 , …
e i θ = ∑ n = 0 ∞ n ! ( i θ ) n = ∑ n ! i n θ n
Even n = 2 k i 2 k = ( − 1 ) k odd n = 2 k + 1 i 2 k + 1 = ( − 1 ) k i
e i θ = = c o s θ k = 0 ∑ ∞ ( 2 k )! ( − 1 ) k θ 2 k + i = s i n θ k = 0 ∑ ∞ ( 2 k + 1 )! ( − 1 ) k θ 2 k + 1
Euler use karke har complex solution likho. Kyunki e ( α ± i β ) x = e α x e ± i β x
y 1 = e α x ( cos β x + i sin β x ) , y 2 = e α x ( cos β x − i sin β x ) .
Intuition Solutions ko recombine kyun kar sakte hainODE linear aur homogeneous hai, isliye solutions ka koi bhi linear combination phir se ek solution hoga (superposition). Hum aisi combinations choose karte hain jo i
Do real solutions banao (ye khud superposition se solutions hain):
u = 2 1 ( y 1 + y 2 ) = e α x cos β x v = 2 i 1 ( y 1 − y 2 ) = e α x sin β x
Ye real hain, non-proportional hain (linearly independent — neeche Wronskian check karo), isliye ye ek basis banate hain.
Intuition Physical reading (dual coding)e α x envelope hai; cos β x , sin β x wiggle hain. Agar α < 0 damped oscillation (springs, RLC circuits). Agar α = 0 pure oscillation (simple harmonic). Agar α > 0 growing oscillation .
u , v u = e α x cos β x v = e α x sin β x W = u u ′ v v ′ W = β e 2 α x Ye step kyun? Kyunki β = 0 e 2 α x > 0 W = 0
Worked example Example 1 — pure oscillation:y ′′ + 9 y = 0 Step 1. Characteristic: m 2 + 9 = 0 Kyun? y ′′ → m 2 , y → 1 Step 2. m 2 = − 9 ⇒ m = ± 3 i α = 0 , β = 3 Kyun? − 9 = 3 i Step 3. y = e 0 ⋅ x ( C 1 cos 3 x + C 2 sin 3 x ) = C 1 cos 3 x + C 2 sin 3 x Kyun? α = 0 3
Worked example Example 2 — damped oscillation:y ′′ + 2 y ′ + 5 y = 0 Step 1. m 2 + 2 m + 5 = 0 Step 2. m = 2 − 2 ± 4 − 20 = 2 − 2 ± − 16 = − 1 ± 2 i Kyun? Δ = − 16 < 0 − 16 = 4 i Step 3. α = − 1 , β = 2 y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) Kyun? α = − 1 < 0 e − x 2 π /2 = π
Worked example Example 3 — initial conditions ke saath (IVP)Solve karo y ′′ + 4 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 6 Step 1. m = ± 2 i ⇒ y = C 1 cos 2 x + C 2 sin 2 x Step 2. y ( 0 ) = C 1 = 2 Kyun? cos 0 = 1 , sin 0 = 0 Step 3. y ′ = − 2 C 1 sin 2 x + 2 C 2 cos 2 x y ′ ( 0 ) = 2 C 2 = 6 ⇒ C 2 = 3 Kyun? differentiate karo phir x = 0 Answer: y = 2 cos 2 x + 3 sin 2 x
Common mistake "Roots complex hain, toh mainy = C 1 e ( α + i β ) x + C 2 e ( α − i β ) x Kyun sahi lagta hai: ye exponential method ka literal output hai, aur ye ek valid complex general solution hai . Fix: real ODE ke liye hume real solutions chahiye; Euler apply karo aur combine karo taaki e α x ( C 1 cos β x + C 2 sin β x ) C 1 , C ˉ 1
Common mistake Imaginary part ko exponential ke growth rate mein daalna.Kyun sahi lagta hai: "m = α + i β e m x = e α x e i β x β Fix: ∣ e i β x ∣ = 1 real part α amplitude change karta hai; β
α = 0 e α x y = C 1 cos β x + C 2 sin β x α = 0 Fix: ye sirf tab valid hai jab α = 0 e α x
β Fix: cos sin β C 2 → − C 2 β = 2 a 4 a c − b 2 > 0
Δ = b 2 − 4 a c Δ < 0
Real-coefficient quadratic mein complex roots conjugate pairs mein kyun aane chahiye? Real coefficients conjugation ke saath commute karte hain, isliye agar α + i β α − i β
Euler's formula batao. e i θ = cos θ + i sin θ
Roots m = α ± i β y = e α x ( C 1 cos β x + C 2 sin β x )
y = e α x ( C 1 cos β x + C 2 sin β x ) α Growth/decay envelope e α x
β Oscillation frequency; period = 2 π / β
e ( α ± i β ) x u = 2 1 ( y 1 + y 2 ) = e α x cos β x v = 2 i 1 ( y 1 − y 2 ) = e α x sin β x
y ′′ + 9 y = 0 m = ± 3 i y = C 1 cos 3 x + C 2 sin 3 x
y ′′ + 2 y ′ + 5 y = 0 m = − 1 ± 2 i y = e − x ( C 1 cos 2 x + C 2 sin 2 x )
∣ e i β x ∣ = 1 cos 2 + sin 2 = 1
e α x cos β x e α x sin β x W = β e 2 α x = 0
α = 0 Undamped pure oscillation (simple harmonic motion).
Recall Feynman: 12-saal ke bachche ko explain karoEk jhule ki imagine karo. Agar ek baar dhakka do aur chod do, toh wo aage-peeche jhulta hai — yahi cos sin e α x α i i e i θ
"Alpha Amplitude, Beta Beat."
α A mplitude envelope e α x β B eat (oscillation frequency cos / sin Euler ne i e i θ = cos + i sin
Characteristic am^2+bm+c=0
Complex roots alpha ± i beta
Complex solutions e^ alpha±ibeta x
Maclaurin series of e^i theta
Euler e^i theta = cos + i sin
Linear recombination cancels i
y = e^ alpha x C1 cos beta x + C2 sin beta x
