Topic: Second-order linear homogeneous ODE with constant coefficients , when the characteristic equation has a complex conjugate pair of roots.
We study:
a y ′ ′ + b y ′ + c y = 0 , a , b , c ∈ R , a ≠ 0 ay'' + by' + cy = 0, \qquad a,b,c \in \mathbb{R},\ a\neq 0 a y ′′ + b y ′ + cy = 0 , a , b , c ∈ R , a = 0
with characteristic equation a m 2 + b m + c = 0 am^2 + bm + c = 0 a m 2 + bm + c = 0
Definition Three cases of the characteristic equationFor a m 2 + b m + c = 0 am^2+bm+c=0 a m 2 + bm + c = 0 Δ = b 2 − 4 a c \Delta = b^2-4ac Δ = b 2 − 4 a c
Δ > 0 \Delta>0 Δ > 0 real distinct roots (Case 1).Δ = 0 \Delta=0 Δ = 0 repeated real root (Case 2).Δ < 0 \Delta<0 Δ < 0 complex conjugate roots ==m = α ± i β m = \alpha \pm i\beta m = α ± i β
Intuition WHY complex roots appear in pairsThe coefficients a , b , c a,b,c a , b , c real . A polynomial with real coefficients cannot have a lone complex root — if α + i β \alpha+i\beta α + i β α − i β \alpha-i\beta α − i β 0 0 0 + , × +,\times + , × conjugate pair .
From the quadratic formula:
m = − b ± b 2 − 4 a c 2 a = − b 2 a ⏟ α ± i 4 a c − b 2 2 a ⏟ β m = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \underbrace{\frac{-b}{2a}}_{\alpha} \pm i\,\underbrace{\frac{\sqrt{4ac-b^2}}{2a}}_{\beta} m = 2 a − b ± b 2 − 4 a c = α 2 a − b ± i β 2 a 4 a c − b 2
So α = − b 2 a \alpha = -\dfrac{b}{2a} α = − 2 a b β = 4 a c − b 2 2 a > 0 \beta = \dfrac{\sqrt{4ac-b^2}}{2a}>0 β = 2 a 4 a c − b 2 > 0 Δ < 0 \Delta<0 Δ < 0 4 a c − b 2 > 0 4ac-b^2>0 4 a c − b 2 > 0
Mechanically, the exponential method still gives two solutions:
y 1 = e ( α + i β ) x , y 2 = e ( α − i β ) x . y_1 = e^{(\alpha+i\beta)x}, \qquad y_2 = e^{(\alpha-i\beta)x}. y 1 = e ( α + i β ) x , y 2 = e ( α − i β ) x .
But these are complex-valued . For a real-world ODE we want real solutions. HOW do we extract them? This is where Euler's formula rescues us.
e i θ e^{i\theta} e i θ define e i θ e^{i\theta} e i θ e x e^x e x d d θ e i θ = i e i θ \frac{d}{d\theta}e^{i\theta}=i\,e^{i\theta} d θ d e i θ = i e i θ i i i i 0 = 1 , i 1 = i , i 2 = − 1 , i 3 = − i , i 4 = 1 , … i^0=1,\ i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1,\dots i 0 = 1 , i 1 = i , i 2 = − 1 , i 3 = − i , i 4 = 1 , …
e i θ = ∑ n = 0 ∞ ( i θ ) n n ! = ∑ i n θ n n ! e^{i\theta} = \sum_{n=0}^\infty \frac{(i\theta)^n}{n!} = \sum \frac{i^n\theta^n}{n!} e i θ = ∑ n = 0 ∞ n ! ( i θ ) n = ∑ n ! i n θ n
Split into even n = 2 k n=2k n = 2 k i 2 k = ( − 1 ) k i^{2k}=(-1)^k i 2 k = ( − 1 ) k odd n = 2 k + 1 n=2k+1 n = 2 k + 1 i 2 k + 1 = ( − 1 ) k i i^{2k+1}=(-1)^k i i 2 k + 1 = ( − 1 ) k i
e i θ = ∑ k = 0 ∞ ( − 1 ) k θ 2 k ( 2 k ) ! ⏟ = cos θ + i ∑ k = 0 ∞ ( − 1 ) k θ 2 k + 1 ( 2 k + 1 ) ! ⏟ = sin θ e^{i\theta} = \underbrace{\sum_{k=0}^\infty \frac{(-1)^k\theta^{2k}}{(2k)!}}_{=\cos\theta} + i\underbrace{\sum_{k=0}^\infty \frac{(-1)^k\theta^{2k+1}}{(2k+1)!}}_{=\sin\theta} 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
Write each complex solution using Euler. Since e ( α ± i β ) x = e α x e ± i β x e^{(\alpha\pm i\beta)x}=e^{\alpha x}e^{\pm i\beta x} 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 ) . y_1 = e^{\alpha x}\big(\cos\beta x + i\sin\beta x\big), \qquad y_2 = e^{\alpha x}\big(\cos\beta x - i\sin\beta x\big). 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 WHY we can recombine solutionsThe ODE is linear and homogeneous , so any linear combination of solutions is again a solution (superposition). We choose combinations that cancel the i i i
Build two real solutions (these are themselves solutions by superposition):
u = 1 2 ( y 1 + y 2 ) = e α x cos β x u = \tfrac{1}{2}(y_1+y_2) = e^{\alpha x}\cos\beta x u = 2 1 ( y 1 + y 2 ) = e α x cos β x v = 1 2 i ( y 1 − y 2 ) = e α x sin β x v = \tfrac{1}{2i}(y_1-y_2) = e^{\alpha x}\sin\beta x v = 2 i 1 ( y 1 − y 2 ) = e α x sin β x
These are real , non-proportional (linearly independent — check the Wronskian below), so they form a basis.
Intuition Physical reading (dual coding)e α x e^{\alpha x} e α x envelope ; cos β x , sin β x \cos\beta x,\sin\beta x cos β x , sin β x wiggle . If α < 0 \alpha<0 α < 0 damped oscillation (springs, RLC circuits). If α = 0 \alpha=0 α = 0 pure oscillation (simple harmonic). If α > 0 \alpha>0 α > 0 growing oscillation .
u , v u,v u , v u = e α x cos β x u=e^{\alpha x}\cos\beta x u = e α x cos β x v = e α x sin β x v=e^{\alpha x}\sin\beta x v = e α x sin β x W = ∣ u v u ′ v ′ ∣ W = \begin{vmatrix} u & v \\ u' & v'\end{vmatrix} W = u u ′ v v ′ W = β e 2 α x W = \beta\, e^{2\alpha x} W = β e 2 α x Why this step? Since β ≠ 0 \beta\neq0 β = 0 e 2 α x > 0 e^{2\alpha x}>0 e 2 α x > 0 W ≠ 0 W\neq0 W = 0
Worked example Example 1 — pure oscillation:y ′ ′ + 9 y = 0 y''+9y=0 y ′′ + 9 y = 0 Step 1. Characteristic: m 2 + 9 = 0 m^2+9=0 m 2 + 9 = 0 Why? Replace y ′ ′ → m 2 , y → 1 y''\to m^2,\ y\to1 y ′′ → m 2 , y → 1 Step 2. m 2 = − 9 ⇒ m = ± 3 i m^2=-9 \Rightarrow m=\pm 3i m 2 = − 9 ⇒ m = ± 3 i α = 0 , β = 3 \alpha=0,\ \beta=3 α = 0 , β = 3 Why? − 9 = 3 i \sqrt{-9}=3i − 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 y=e^{0\cdot x}(C_1\cos3x+C_2\sin3x)=C_1\cos3x+C_2\sin3x y = e 0 ⋅ x ( C 1 cos 3 x + C 2 sin 3 x ) = C 1 cos 3 x + C 2 sin 3 x Why? α = 0 \alpha=0 α = 0 3 3 3
Worked example Example 2 — damped oscillation:y ′ ′ + 2 y ′ + 5 y = 0 y''+2y'+5y=0 y ′′ + 2 y ′ + 5 y = 0 Step 1. m 2 + 2 m + 5 = 0 m^2+2m+5=0 m 2 + 2 m + 5 = 0 Step 2. m = − 2 ± 4 − 20 2 = − 2 ± − 16 2 = − 1 ± 2 i m=\dfrac{-2\pm\sqrt{4-20}}{2}=\dfrac{-2\pm\sqrt{-16}}{2}=-1\pm2i m = 2 − 2 ± 4 − 20 = 2 − 2 ± − 16 = − 1 ± 2 i Why? Δ = − 16 < 0 \Delta=-16<0 Δ = − 16 < 0 − 16 = 4 i \sqrt{-16}=4i − 16 = 4 i Step 3. α = − 1 , β = 2 \alpha=-1,\ \beta=2 α = − 1 , β = 2 y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) y=e^{-x}(C_1\cos2x+C_2\sin2x) y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) Why? α = − 1 < 0 \alpha=-1<0 α = − 1 < 0 e − x e^{-x} e − x 2 π / 2 = π 2\pi/2=\pi 2 π /2 = π
Worked example Example 3 — with initial conditions (IVP)Solve y ′ ′ + 4 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 6 y''+4y=0,\ y(0)=2,\ y'(0)=6 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 m=\pm2i \Rightarrow y=C_1\cos2x+C_2\sin2x m = ± 2 i ⇒ y = C 1 cos 2 x + C 2 sin 2 x Step 2. y ( 0 ) = C 1 = 2 y(0)=C_1=2 y ( 0 ) = C 1 = 2 Why? cos 0 = 1 , sin 0 = 0 \cos0=1,\sin0=0 cos 0 = 1 , sin 0 = 0 Step 3. y ′ = − 2 C 1 sin 2 x + 2 C 2 cos 2 x y'=-2C_1\sin2x+2C_2\cos2x y ′ = − 2 C 1 sin 2 x + 2 C 2 cos 2 x y ′ ( 0 ) = 2 C 2 = 6 ⇒ C 2 = 3 y'(0)=2C_2=6\Rightarrow C_2=3 y ′ ( 0 ) = 2 C 2 = 6 ⇒ C 2 = 3 Why? differentiate then plug x = 0 x=0 x = 0 Answer: y = 2 cos 2 x + 3 sin 2 x y=2\cos2x+3\sin2x y = 2 cos 2 x + 3 sin 2 x
Common mistake "Roots are complex, so I'll writey = C 1 e ( α + i β ) x + C 2 e ( α − i β ) x y=C_1e^{(\alpha+i\beta)x}+C_2e^{(\alpha-i\beta)x} y = C 1 e ( α + i β ) x + C 2 e ( α − i β ) x Why it feels right: it's the literal output of the exponential method, and it is a valid complex general solution. Fix: for a real ODE we want real solutions; apply Euler and combine to get e α x ( C 1 cos β x + C 2 sin β x ) e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x) e α x ( C 1 cos β x + C 2 sin β x ) C 1 , C ˉ 1 C_1,\bar C_1 C 1 , C ˉ 1
Common mistake Putting the imaginary part into the exponential's growth rate.Why it feels right: "m = α + i β m=\alpha+i\beta m = α + i β e m x = e α x e i β x e^{mx}=e^{\alpha x}e^{i\beta x} e m x = e α x e i β x β \beta β Fix: ∣ e i β x ∣ = 1 |e^{i\beta x}|=1 ∣ e i β x ∣ = 1 real part α \alpha α changes amplitude; β \beta β
Common mistake Forgetting thee α x e^{\alpha x} e α x α ≠ 0 \alpha\neq0 α = 0 y = C 1 cos β x + C 2 sin β x y=C_1\cos\beta x+C_2\sin\beta x y = C 1 cos β x + C 2 sin β x α ≠ 0 \alpha\neq0 α = 0 Fix: that's only valid when α = 0 \alpha=0 α = 0 e α x e^{\alpha x} e α x
β \beta β Fix: cos \cos cos sin \sin sin β \beta β C 2 → − C 2 C_2\to-C_2 C 2 → − C 2 β = 4 a c − b 2 2 a > 0 \beta=\frac{\sqrt{4ac-b^2}}{2a}>0 β = 2 a 4 a c − b 2 > 0
What condition on Δ = b 2 − 4 a c \Delta=b^2-4ac Δ = b 2 − 4 a c Δ < 0 \Delta<0 Δ < 0 (negative discriminant) → complex conjugate roots.
Why must complex roots of a real-coefficient quadratic come in conjugate pairs? Real coefficients commute with conjugation, so if
α + i β \alpha+i\beta α + i β is a root, conjugating the equation shows
α − i β \alpha-i\beta α − i β is too.
State Euler's formula. e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ .
For roots m = α ± i β m=\alpha\pm i\beta m = α ± i β y = e α x ( C 1 cos β x + C 2 sin β x ) y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x) y = e α x ( C 1 cos β x + C 2 sin β x ) .
In y = e α x ( C 1 cos β x + C 2 sin β x ) y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x) y = e α x ( C 1 cos β x + C 2 sin β x ) α \alpha α The growth/decay envelope
e α x e^{\alpha x} e α x (amplitude over time).
What does β \beta β The oscillation frequency; period
= 2 π / β =2\pi/\beta = 2 π / β .
How do you get real solutions from e ( α ± i β ) x e^{(\alpha\pm i\beta)x} e ( α ± i β ) x Take
u = 1 2 ( y 1 + y 2 ) = e α x cos β x u=\frac12(y_1+y_2)=e^{\alpha x}\cos\beta x u = 2 1 ( y 1 + y 2 ) = e α x cos β x and
v = 1 2 i ( y 1 − y 2 ) = e α x sin β x v=\frac{1}{2i}(y_1-y_2)=e^{\alpha x}\sin\beta x v = 2 i 1 ( y 1 − y 2 ) = e α x sin β x by superposition.
Solve y ′ ′ + 9 y = 0 y''+9y=0 y ′′ + 9 y = 0 m = ± 3 i m=\pm3i m = ± 3 i ,
y = C 1 cos 3 x + C 2 sin 3 x y=C_1\cos3x+C_2\sin3x y = C 1 cos 3 x + C 2 sin 3 x .
Roots of y ′ ′ + 2 y ′ + 5 y = 0 y''+2y'+5y=0 y ′′ + 2 y ′ + 5 y = 0 m = − 1 ± 2 i m=-1\pm2i m = − 1 ± 2 i ;
y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) y=e^{-x}(C_1\cos2x+C_2\sin2x) y = e − x ( C 1 cos 2 x + C 2 sin 2 x ) .
Why is ∣ e i β x ∣ = 1 |e^{i\beta x}|=1 ∣ e i β x ∣ = 1 cos 2 + sin 2 = 1 \cos^2+\sin^2=1 cos 2 + sin 2 = 1 ; it's a rotation on the unit circle, not a scaling.
What is the Wronskian of e α x cos β x e^{\alpha x}\cos\beta x e α x cos β x e α x sin β x e^{\alpha x}\sin\beta x e α x sin β x W = β e 2 α x ≠ 0 W=\beta e^{2\alpha x}\neq0 W = β e 2 α x = 0 , confirming independence.
α = 0 \alpha=0 α = 0 Undamped pure oscillation (simple harmonic motion).
Recall Feynman: explain to a 12-year-oldImagine a swing. If you push it once and let go, it swings back and forth — that's the cos \cos cos sin \sin sin e α x e^{\alpha x} e α x α \alpha α i i i i i i e i θ e^{i\theta} e i θ
"Alpha Amplitude, Beta Beat."
α \alpha α A mplitude envelope e α x e^{\alpha x} e α x β \beta β B eat (oscillation frequency cos / sin \cos/\sin cos / sin Euler ate the i i i e i θ = cos + i sin e^{i\theta}=\cos+i\sin 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
Intuition Hinglish mein samjhoDekho, jab humara second-order ODE a y ′ ′ + b y ′ + c y = 0 ay''+by'+cy=0 a y ′′ + b y ′ + cy = 0 a m 2 + b m + c = 0 am^2+bm+c=0 a m 2 + bm + c = 0 b 2 − 4 a c b^2-4ac b 2 − 4 a c negative aa jaaye, tab roots real nahi milte — wo aate hain complex conjugate pair mein: m = α ± i β m=\alpha\pm i\beta m = α ± i β e ( α + i β ) x e^{(\alpha+i\beta)x} e ( α + i β ) x Euler's formula ka jaadu kaam aata hai: e i θ = cos θ + i sin θ e^{i\theta}=\cos\theta+i\sin\theta e i θ = cos θ + i sin θ
Euler use karke dono complex solutions ko mix (superposition) karke hum do real solutions nikaalte hain: e α x cos β x e^{\alpha x}\cos\beta x e α x cos β x e α x sin β x e^{\alpha x}\sin\beta x e α x sin β x y = e α x ( C 1 cos β x + C 2 sin β x ) y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x) y = e α x ( C 1 cos β x + C 2 sin β x ) Alpha = Amplitude, Beta = Beat . Matlab α \alpha α e α x e^{\alpha x} e α x β \beta β
Physical example samjho: spring ya RLC circuit. Agar α < 0 \alpha<0 α < 0 damped oscillation . Agar α = 0 \alpha=0 α = 0 α > 0 \alpha>0 α > 0 ∣ e i β x ∣ = 1 |e^{i\beta x}|=1 ∣ e i β x ∣ = 1 β \beta β
Exam tip: pehle Δ < 0 \Delta<0 Δ < 0 α = − b / 2 a \alpha=-b/2a α = − b /2 a β = 4 a c − b 2 / ( 2 a ) \beta=\sqrt{4ac-b^2}/(2a) β = 4 a c − b 2 / ( 2 a ) e α x e^{\alpha x} e α x α ≠ 0 \alpha\neq0 α = 0
