You will learn how to solve second-order linear homogeneous equations when the characteristic equation has complex roots, and interpret the solutions as oscillatory motion with or without damping.
When the characteristic equation a r 2 + b r + c = 0 ar^2 + br + c = 0 a r 2 + b r + c = 0 has complex roots α ± β i \alpha \pm \beta i α ± β i , the general solution is:
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 )
The term e α x e^{\alpha x} e α x controls growth or decay (damping).
The term C 1 cos β x + C 2 sin β x C_1 \cos \beta x + C_2 \sin \beta x C 1 cos β x + C 2 sin β x produces oscillation with frequency β \beta β .
Special cases:
α = 0 \alpha = 0 α = 0 : undamped simple harmonic motion (pure oscillation forever)
α < 0 \alpha \lt 0 α < 0 : damped oscillation (most real systems)
α > 0 \alpha \gt 0 α > 0 : growing oscillation (unstable)
t y Undamped (α = 0) Light damping (α = −0.3) Heavy damping (α = −1.2)
The visual shows three cases: undamped oscillation (constant amplitude), lightly damped (decaying envelope), and heavily damped (rapid decay with minimal oscillation).
Example 1: Undamped Oscillation
Solve y ′ ′ + 4 y = 0 y'' + 4y = 0 y ′′ + 4 y = 0 .
Solution :
Characteristic equation: r 2 + 4 = 0 ⟹ r = ± 2 i r^2 + 4 = 0 \implies r = \pm 2i r 2 + 4 = 0 ⟹ r = ± 2 i
General solution: y = C 1 cos 2 x + C 2 sin 2 x y = C_1 \cos 2x + C_2 \sin 2x y = C 1 cos 2 x + C 2 sin 2 x
Example 2: Damped Oscillation
Solve y ′ ′ + 6 y ′ + 13 y = 0 y'' + 6y' + 13y = 0 y ′′ + 6 y ′ + 13 y = 0 .
Solution :
Characteristic equation: r 2 + 6 r + 13 = 0 ⟹ r = − 3 ± 2 i r^2 + 6r + 13 = 0 \implies r = -3 \pm 2i r 2 + 6 r + 13 = 0 ⟹ r = − 3 ± 2 i
General solution: y = e − 3 x ( C 1 cos 2 x + C 2 sin 2 x ) y = e^{-3x} (C_1 \cos 2x + C_2 \sin 2x) y = e − 3 x ( C 1 cos 2 x + C 2 sin 2 x )
Example 3: Initial Value Problem
Solve y ′ ′ + 2 y ′ + 2 y = 0 y'' + 2y' + 2y = 0 y ′′ + 2 y ′ + 2 y = 0 with y ( 0 ) = 1 y(0) = 1 y ( 0 ) = 1 , y ′ ( 0 ) = 0 y'(0) = 0 y ′ ( 0 ) = 0 .
Solution :
Characteristic roots: − 1 ± i -1 \pm i − 1 ± i
General solution: y = e − x ( C 1 cos x + C 2 sin x ) y = e^{-x} (C_1 \cos x + C_2 \sin x) y = e − x ( C 1 cos x + C 2 sin x )
Applying initial conditions: C 1 = 1 C_1 = 1 C 1 = 1 , C 2 = 1 C_2 = 1 C 2 = 1
Final solution: y = e − x ( cos x + sin x ) y = e^{-x} (\cos x + \sin x) y = e − x ( cos x + sin x )
This mathematics describes real oscillating systems everywhere: car suspensions absorbing bumps, electrical circuits in radios and phones, bridges swaying in the wind, and even the motion of tall buildings during earthquakes. Engineers use these equations to design systems that either resonate at specific frequencies (like musical instruments) or dampen vibrations quickly (like shock absorbers).
You’ve Got This
Complex roots bring oscillation into the picture, and suddenly differential equations start describing things you can actually see and feel in the real world. You’re now equipped to model springs, circuits, and vibrations. This is where the subject gets really fun.
Complex roots in the characteristic equation produce solutions involving: A. Pure exponentials B. Linear polynomials C. Sine and cosine multiplied by an exponential D. Logarithmic terms
When $\alpha = 0$ in complex roots $\alpha \pm \beta i$, the motion is: A. Overdamped B. Critically damped C. Growing exponentially D. Undamped simple harmonic
The value of $\beta$ in the complex roots determines: A. The damping rate B. The initial amplitude C. The frequency of oscillation D. The direction of motion
For the equation $y'' + 4y' + 5y = 0$, the solution will: A. Grow without bound B. Oscillate with increasing amplitude C. Approach a constant D. Decay while oscillating
Complex roots with negative real part model: A. Explosive growth B. Pure exponential decay without oscillation C. Chaotic behavior D. Stable damped oscillations like car suspensions
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