42 Second-order ODEs
Oscillation, damping, and resonance
In November 1940, four months after it opened, the Tacoma Narrows Bridge began twisting in a 65 km/h wind. The oscillations grew. Within hours the deck was flexing through metres of arc. Then it collapsed.
The engineers had missed a resonance condition. The wind’s forcing frequency matched the bridge’s natural frequency, and the amplitude grew without bound. A second-order ODE — mass, damping, stiffness, forcing — contained the answer. The question is whether you know how to read it.
This chapter develops every tool needed to answer that question. By the end you will be able to look at any spring, circuit, or structure, write down its governing equation, and classify its response before solving it.
42.1 What this chapter helps you do
Symbols to keep handy
These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.
r: r
_0: omega-zero
y_p: y-p
Q: Q
, : alpha, beta
y_c: y-c
Definitions to keep handy
These are the words we keep coming back to. If one feels slippery, come back here and steady it before you push on.
second-order ODE: A differential equation involving y and its second derivative y’’.
characteristic equation: The quadratic you get by trying y = e^{rt}; its roots tell you the solution shape.
natural frequency: The rate an undamped system wants to oscillate at.
damping: A mechanism that bleeds energy away so oscillations shrink over time.
resonance: When forcing hits the system’s natural rhythm and the response grows very large.
This chapter gives you a fast way to read the behaviour of oscillating systems from the structure of a second-order ODE. You will learn to:
- split solutions into complementary + particular parts (y=y_c+y_p)
- use the characteristic equation to classify decay, oscillation, and resonance
- connect the algebra of the roots to physical parameters (natural frequency, damping ratio, quality factor)
- recognise when forcing creates a bounded steady-state response and when it creates growth
Watch for this
- The behaviour comes from the roots of a quadratic, but the meaning comes from the model parameters. Keep translating back to the system.
- Resonance is not a math curiosity; it is a design constraint. The mathematics is telling you when energy builds instead of dissipating.
- When you write y=y_c+y_p, y_p is not unique. Any valid particular solution works; the initial conditions are absorbed by y_c.
42.2 Structure of the general equation
A second-order linear ODE with constant coefficients has the form
ay'' + by' + cy = f(x)
where a \neq 0. The coefficients a, b, c are constants; f(x) is the forcing function (or input, or source term).
How to read the core symbols
Symbol: y''
Reads as: “y double-prime”
Means: the second derivative (often acceleration when y is position)
Use when: a system’s dynamics depend on both position and velocity
Common misread: second order means you will need two initial conditions
Symbol: y=y_c+y_p
Reads as: “y equals y-sub-c plus y-sub-p”
Means: complementary (homogeneous) response + forced response
Use when: the equation is linear, so superposition applies
Common misread: y_p is not the whole solution; it is only the forced part
Why second order? Newton’s second law says F = ma, where a = \ddot{x} is the second derivative of position. Any system governed by Newton’s second law — masses on springs, charges in circuits, beams under load — produces a second-order ODE. The second derivative is not an accident of notation; it encodes the physics.
Linearity and superposition. The equation is linear because y, y', y'' each appear to the first power and are not multiplied together. Linearity buys you the superposition principle: if y_1 and y_2 are both solutions to the homogeneous equation (f = 0), then so is c_1 y_1 + c_2 y_2 for any constants c_1, c_2. This principle is the engine of everything that follows.
The decomposition: complementary + particular. The general solution is
y = y_c + y_p
where y_c (the complementary solution) solves the homogeneous equation ay'' + by' + cy = 0, and y_p (the particular solution) is any one solution to the full equation with f(x) present. The two free constants in y_c absorb the initial conditions.
Why two initial conditions? The ODE is second order, so its general solution contains two arbitrary constants. To pick a unique solution from the family, you need two pieces of information — typically y(x_0) and y'(x_0). Specify position and velocity at time zero, and the subsequent motion is determined.
The structure in one line
\underbrace{ay'' + by' + cy = 0}_{\text{homogeneous: find } y_c} \quad + \quad \underbrace{ay'' + by' + cy = f(x)}_{\text{find any } y_p} \quad \Longrightarrow \quad y = y_c + y_p
42.3 Homogeneous equations: the characteristic equation
How to read the characteristic-equation symbols
Symbol: r
Reads as: “r”
Means: the growth/decay (and possibly oscillation) behaviour in the trial solution e^{rx}
Use when: you turn ay''+by'+cy=0 into an algebraic quadratic
Common misread: r is not “time” or a measured parameter; it is a root that encodes behaviour
Symbol: \Delta=b^2-4ac
Reads as: “Delta” (the discriminant)
Means: whether the roots are real, repeated, or complex
Use when: you want to classify the system without solving in detail
Common misread: the sign of \Delta matters more than its magnitude
Symbol: \alpha,\beta (when roots are \alpha\pm i\beta)
Reads as: “alpha, beta”
Means: \alpha controls decay/growth; \beta is the oscillation frequency
Use when: you rewrite complex exponentials into sines and cosines
Common misread: \beta is not the same as the natural frequency \omega_0 unless the model is undamped
Take the homogeneous equation
ay'' + by' + cy = 0
and try a solution of the form y = e^{rx}. Substituting:
y = e^{rx}, \quad y' = re^{rx}, \quad y'' = r^2 e^{rx}
a r^2 e^{rx} + b r e^{rx} + c e^{rx} = 0
Factor out e^{rx}, which is never zero:
e^{rx}(ar^2 + br + c) = 0 \implies ar^2 + br + c = 0
This is the characteristic equation. An ODE has become an algebraic problem: find the roots of a quadratic. The nature of those roots determines the form of y_c.
The quadratic formula gives
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
The discriminant \Delta = b^2 - 4ac controls everything.
42.3.1 Case 1: Two distinct real roots (\Delta > 0)
Two distinct roots r_1 \neq r_2, both real. The general solution is
y_c = C_1 e^{r_1 x} + C_2 e^{r_2 x}
These are two linearly independent solutions; their linear combination spans the solution space.
Physical behaviour: If both roots are negative, both exponentials decay. The system returns to equilibrium without oscillating. In a spring-mass context this is the overdamped regime — the mass creeps back slowly, never crossing the equilibrium.
Example. Solve y'' - 5y' + 6y = 0.
Characteristic equation: r^2 - 5r + 6 = 0, which factors as (r-2)(r-3) = 0.
Roots: r_1 = 2, r_2 = 3.
General solution: y = C_1 e^{2x} + C_2 e^{3x}.
42.3.2 Case 2: Repeated real root (\Delta = 0)
One root r = -b/(2a) repeated. The naive approach gives only e^{rx}, which is one solution. A second independent solution is xe^{rx} (derivable by reduction of order — see §6 below, or verify by substitution).
General solution:
y_c = (C_1 + C_2 x)e^{rx}
Physical behaviour: The system returns to equilibrium as fast as possible without oscillating. This is critical damping — the boundary between decaying and oscillating behaviour. Car suspension systems are often tuned close to critical damping.
Example. Solve y'' + 4y' + 4y = 0.
Characteristic equation: r^2 + 4r + 4 = (r+2)^2 = 0.
Root: r = -2 (repeated).
General solution: y = (C_1 + C_2 x)e^{-2x}.
42.3.3 Case 3: Complex conjugate roots (\Delta < 0)
The roots are r = \alpha \pm \beta i where \alpha = -b/(2a) and \beta = \sqrt{4ac - b^2}/(2a) > 0.
Using Euler’s formula e^{i\theta} = \cos\theta + i\sin\theta, the complex exponentials can be rewritten as real-valued functions:
y_c = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))
Physical behaviour: The e^{\alpha x} factor is the envelope. If \alpha < 0, the oscillation decays — this is underdamped. If \alpha = 0, oscillation continues forever at constant amplitude — undamped simple harmonic motion. If \alpha > 0, the oscillation grows (unstable system).
Example. Solve y'' + 2y' + 5y = 0.
Characteristic equation: r^2 + 2r + 5 = 0.
Roots: r = (-2 \pm \sqrt{4 - 20})/2 = (-2 \pm \sqrt{-16})/2 = -1 \pm 2i.
So \alpha = -1, \beta = 2.
General solution: y = e^{-x}(C_1 \cos 2x + C_2 \sin 2x).
Summary: three cases
| Discriminant | Roots | Solution | Behaviour |
|---|---|---|---|
| \Delta > 0 | Two distinct real | C_1 e^{r_1 x} + C_2 e^{r_2 x} | Overdamped / exponential |
| \Delta = 0 | Repeated real | (C_1 + C_2 x)e^{rx} | Critically damped |
| \Delta < 0 | Complex conjugate | e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x) | Underdamped / oscillatory |
42.4 The spring-mass-damper system
How to read the oscillation parameters
Symbol: \omega_0
Reads as: “omega-zero” (natural frequency)
Means: the system’s preferred oscillation rate in the undamped model
Use when: you compare forcing frequency to what the system naturally wants to do
Common misread: units matter; \omega_0 is radians per second, not cycles per second
Symbol: \zeta
Reads as: “zeta” (damping ratio)
Means: how strong damping is relative to critical damping
Use when: you classify overdamped/critical/underdamped behaviour
Common misread: small \zeta does not mean “no effect”; it can still produce very large resonance peaks
Symbol: Q=\dfrac{1}{2\zeta}
Reads as: “Q” (quality factor)
Means: selectivity/sharpness of resonance (large Q means a sharp, tall peak)
Use when: you talk about filters and frequency response
Common misread: Q is not “quality” in the everyday sense; it is a ratio that measures amplification near resonance
The physical archetype for every second-order linear ODE is a mass m on a spring with stiffness k, connected to a dashpot (viscous damper) with damping coefficient c:
m\ddot{x} + c\dot{x} + kx = 0
Here x(t) is displacement from equilibrium, dots denote time derivatives. This maps directly onto the standard form ay'' + by' + cy = 0 with a = m, b = c, c_{\text{coeff}} = k.
The characteristic equation is mr^2 + cr + k = 0, with roots
r = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m}
Two natural parameters emerge from dividing the equation by m:
\ddot{x} + \frac{c}{m}\dot{x} + \frac{k}{m}x = 0
Natural frequency (undamped):
\omega_0 = \sqrt{\frac{k}{m}}
Damping ratio:
\zeta = \frac{c}{2\sqrt{km}} = \frac{c}{2m\omega_0}
The equation becomes \ddot{x} + 2\zeta\omega_0 \dot{x} + \omega_0^2 x = 0, and the discriminant condition translates cleanly:
| \zeta | Regime | Behaviour |
|---|---|---|
| \zeta > 1 | Overdamped | Two negative real roots; slow return to equilibrium |
| \zeta = 1 | Critically damped | Repeated negative real root; fastest return without oscillation |
| \zeta < 1 | Underdamped | Complex roots; decaying oscillation at frequency \omega_d = \omega_0\sqrt{1-\zeta^2} |
| \zeta = 0 | Undamped | Pure imaginary roots; oscillation at \omega_0 forever |
42.4.1 Worked numerical example
A spring-mass-damper system has m = 1 kg, c = 5 N·s/m, k = 6 N/m. Find x(t) given x(0) = 1, \dot{x}(0) = 0.
Step 1: characteristic equation.
r^2 + 5r + 6 = 0 \implies (r+2)(r+3) = 0 \implies r_1 = -2,\ r_2 = -3
Step 2: classify. Both roots real and negative. Discriminant = 25 - 24 = 1 > 0.
Check via \zeta: \omega_0 = \sqrt{6} \approx 2.449, \zeta = 5/(2\sqrt{6}) \approx 1.02 > 1. Overdamped.
Step 3: general solution.
x(t) = C_1 e^{-2t} + C_2 e^{-3t}
Step 4: apply initial conditions.
x(0) = 1: C_1 + C_2 = 1
\dot{x}(0) = 0: -2C_1 - 3C_2 = 0 \implies C_1 = -3C_2/2
Substituting: -3C_2/2 + C_2 = 1 \implies -C_2/2 = 1 \implies C_2 = -2, C_1 = 3.
Solution:
x(t) = 3e^{-2t} - 2e^{-3t}
The mass starts at x = 1, initially moves slightly before the two exponentials pull it back to zero without oscillating. Physical interpretation: heavy damping prevents overshoot.
Now change the damping: m = 1, c = 2, k = 1. Characteristic equation: r^2 + 2r + 1 = (r+1)^2 = 0. Root r = -1 repeated. Critically damped. With x(0) = 1, \dot{x}(0) = 0:
x = (C_1 + C_2 t)e^{-t}; x(0) = C_1 = 1; \dot{x}(0) = C_2 - C_1 = 0 \implies C_2 = 1.
x(t) = (1 + t)e^{-t}
Now change to m = 1, c = 1, k = 4. Characteristic equation: r^2 + r + 4 = 0.
Roots: r = (-1 \pm \sqrt{1-16})/2 = -1/2 \pm i\sqrt{15}/2. Underdamped. General solution: x(t) = e^{-t/2}(C_1\cos\frac{\sqrt{15}}{2}t + C_2\sin\frac{\sqrt{15}}{2}t).
With x(0) = 1, \dot{x}(0) = 0:
x(0) = C_1 = 1.
Differentiate using the product rule: \dot{x}(t) = -\tfrac{1}{2}e^{-t/2}(C_1\cos\frac{\sqrt{15}}{2}t + C_2\sin\frac{\sqrt{15}}{2}t) + e^{-t/2}(-C_1\tfrac{\sqrt{15}}{2}\sin\frac{\sqrt{15}}{2}t + C_2\tfrac{\sqrt{15}}{2}\cos\frac{\sqrt{15}}{2}t).
At t = 0: \dot{x}(0) = -\tfrac{1}{2}C_1 + C_2\tfrac{\sqrt{15}}{2} = 0, so C_2 = \tfrac{C_1}{\sqrt{15}} = \tfrac{1}{\sqrt{15}}.
x(t) = e^{-t/2}\!\left(\cos\frac{\sqrt{15}}{2}t + \frac{1}{\sqrt{15}}\sin\frac{\sqrt{15}}{2}t\right)
The oscillation is visible; the exponential envelope e^{-t/2} decays it.
Engineering rule of thumb
\zeta > 1: overdamped — sluggish, no oscillation. \zeta = 1: critically damped — fastest non-oscillatory return. \zeta < 1: underdamped — fast but overshoots; for \zeta \ll 1 the oscillation persists for many cycles.
Automotive suspension targets \zeta \approx 0.3–0.4: some overshoot is acceptable; too much damping makes the ride stiff and slow to recover.
42.5 Method of undetermined coefficients
When f(x) \neq 0, we need a particular solution y_p. The method of undetermined coefficients works when f(x) belongs to a standard family: polynomials, exponentials, sines and cosines, and products of these.
The strategy: guess y_p has the same form as f(x), with unknown coefficients. Substitute into the ODE; match coefficients on both sides.
42.5.1 Standard guess table
| f(x) | Guess for y_p |
|---|---|
| P_n(x) (polynomial, degree n) | A_n x^n + \cdots + A_1 x + A_0 |
| e^{\alpha x} | Ae^{\alpha x} |
| \cos(\beta x) or \sin(\beta x) | A\cos(\beta x) + B\sin(\beta x) |
| e^{\alpha x}\cos(\beta x) or similar | e^{\alpha x}(A\cos(\beta x) + B\sin(\beta x)) |
| P_n(x)e^{\alpha x} | e^{\alpha x}(A_n x^n + \cdots + A_0) |
The modification rule. If the guess for y_p duplicates a term already present in y_c, multiply the guess by x (or x^2 if y_c contains both e^{rx} and xe^{rx}, as in the repeated-root case). Failure to apply this rule produces a contradiction when you substitute.
42.5.2 Worked example 1: exponential forcing
Find a particular solution of y'' + 3y' + 2y = e^{-3x}.
Homogeneous equation: r^2 + 3r + 2 = (r+1)(r+2) = 0, roots r = -1, -2.
y_c = C_1 e^{-x} + C_2 e^{-2x}.
f(x) = e^{-3x}: the exponent -3 does not match any root, so no modification is needed. Guess y_p = Ae^{-3x}.
Substitute:
y_p'' + 3y_p' + 2y_p = 9Ae^{-3x} - 9Ae^{-3x} + 2Ae^{-3x} = 2Ae^{-3x}
Set equal to e^{-3x}: 2A = 1 \implies A = 1/2.
y_p = \tfrac{1}{2}e^{-3x}
General solution: y = C_1 e^{-x} + C_2 e^{-2x} + \frac{1}{2}e^{-3x}.
42.5.3 Worked example 2: sinusoidal forcing
Find a particular solution of y'' + 4y' + 5y = \sin(2x).
Homogeneous: r^2 + 4r + 5 = 0 \implies r = -2 \pm i. So y_c = e^{-2x}(C_1\cos x + C_2\sin x).
f(x) = \sin(2x): guess y_p = A\cos(2x) + B\sin(2x).
Compute derivatives:
y_p' = -2A\sin(2x) + 2B\cos(2x) y_p'' = -4A\cos(2x) - 4B\sin(2x)
Substitute:
(-4A\cos 2x - 4B\sin 2x) + 4(-2A\sin 2x + 2B\cos 2x) + 5(A\cos 2x + B\sin 2x) = \sin 2x
Collect \cos(2x): -4A + 8B + 5A = (A + 8B)\cos 2x
Collect \sin(2x): -4B - 8A + 5B = (B - 8A)\sin 2x
Match to 0\cdot\cos 2x + 1\cdot\sin 2x:
A + 8B = 0 \implies A = -8B B - 8A = 1 \implies B + 64B = 1 \implies B = \tfrac{1}{65},\quad A = -\tfrac{8}{65}
y_p = -\frac{8}{65}\cos(2x) + \frac{1}{65}\sin(2x)
42.5.4 Modification rule in action
Find a particular solution of y'' + 2y' + y = e^{-x}.
Characteristic equation: (r+1)^2 = 0, root r = -1 repeated.
y_c = (C_1 + C_2 x)e^{-x}.
Naive guess Ae^{-x} is in y_c; modified guess Axe^{-x} is also in y_c (the C_2 x e^{-x} term). So multiply again: guess y_p = Ax^2 e^{-x}.
y_p' = (2Ax - Ax^2)e^{-x}, \quad y_p'' = (2A - 4Ax + Ax^2)e^{-x}
Substitute (the x^2 e^{-x} and xe^{-x} terms cancel):
y_p'' + 2y_p' + y_p = 2Ae^{-x}
Match: 2A = 1 \implies A = 1/2.
y_p = \tfrac{1}{2}x^2 e^{-x}
42.6 Forced oscillation and resonance
Now force the undamped spring-mass system:
m\ddot{x} + kx = F_0 \cos(\omega t), \quad x(0) = 0,\ \dot{x}(0) = 0
The homogeneous solution is x_c = C_1\cos(\omega_0 t) + C_2\sin(\omega_0 t) where \omega_0 = \sqrt{k/m}.
Case 1: \omega \neq \omega_0 (off resonance). Guess x_p = A\cos(\omega t).
-m\omega^2 A\cos(\omega t) + kA\cos(\omega t) = F_0\cos(\omega t)
A(k - m\omega^2) = F_0 \implies A = \frac{F_0}{k - m\omega^2} = \frac{F_0/m}{\omega_0^2 - \omega^2}
Applying initial conditions x(0) = 0, \dot{x}(0) = 0: the general solution is x = C_1\cos\omega_0 t + C_2\sin\omega_0 t + \frac{F_0/m}{\omega_0^2-\omega^2}\cos\omega t. From x(0) = 0: C_1 + \frac{F_0/m}{\omega_0^2-\omega^2} = 0, so C_1 = -\frac{F_0/m}{\omega_0^2-\omega^2}. From \dot{x}(0) = 0: C_2\omega_0 - \frac{F_0/m}{\omega_0^2-\omega^2}\omega\sin 0 = 0, giving C_2 = 0. Substituting:
x(t) = \frac{F_0/m}{\omega_0^2 - \omega^2}\bigl(\cos(\omega t) - \cos(\omega_0 t)\bigr)
Beating. When \omega is close to but not equal to \omega_0, write \omega_0 = \omega + \epsilon for small \epsilon. The difference \cos(\omega t) - \cos(\omega_0 t) can be rewritten using a product-to-sum identity (from \cos A - \cos B = 2\sin\frac{A+B}{2}\sin\frac{B-A}{2}):
\cos(\omega t) - \cos(\omega_0 t) = 2\sin\!\left(\frac{\omega_0 + \omega}{2}t\right)\sin\!\left(\frac{\omega_0 - \omega}{2}t\right)
This is a rapid oscillation at frequency (\omega_0 + \omega)/2 \approx \omega_0 modulated by a slow envelope at frequency (\omega_0 - \omega)/2 \approx \epsilon/2. The amplitude swells and collapses periodically — beating. It is audible between two tuning forks of slightly different pitch.
Case 2: \omega = \omega_0 (resonance). The guess A\cos(\omega_0 t) duplicates a homogeneous solution; apply the modification rule and guess x_p = t(A\cos(\omega_0 t) + B\sin(\omega_0 t)).
Substituting and simplifying (the cosine terms cancel; only the sine survives in the second derivative’s cross-terms):
x_p = \frac{F_0}{2m\omega_0} t\sin(\omega_0 t)
The particular solution at resonance is
\boxed{x_p = \frac{F_0}{2m\omega_0}\, t\sin(\omega_0 t)}
The factor of t in front of the sine is critical. The amplitude \frac{F_0}{2m\omega_0} t grows linearly without bound as t \to \infty.
Resonance is an unbounded response
At \omega = \omega_0, the forcing and the natural oscillation are perfectly in phase. Each cycle, the force adds energy to the system. With no damping to dissipate it, amplitude grows without limit. In a real system, something fails first — yielding, fracture, or collapse.
The Tacoma Narrows Bridge, Broughton Suspension Bridge (1831), and numerous aircraft structures have failed under resonance conditions. In engineering design, knowing \omega_0 and ensuring forcing frequencies avoid it — or adding sufficient damping — is not optional.
With damping present. Add a damping term c\dot{x} to the forced equation. The steady-state amplitude response at driving frequency \omega becomes
X(\omega) = \frac{F_0/k}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}}, \quad r = \frac{\omega}{\omega_0}
(This formula is derivable by applying undetermined coefficients to m\ddot{x}+c\dot{x}+kx=F_0\cos\omega t — the calculation is straightforward but lengthy, and we state the result here.)
As \zeta \to 0, the peak amplitude at r = 1 becomes unbounded. For \zeta > 0, the peak is finite but can be very large for small \zeta. The peak occurs near r = 1 and grows sharper and taller as damping decreases — the amplification factor Q = 1/(2\zeta) is called the quality factor in electrical engineering.
42.7 Reduction of order
Suppose you know one solution y_1 to ay'' + by' + cy = 0 — perhaps from inspection or from the characteristic equation in a non-constant coefficient problem — but cannot find a second independent one directly. The method of reduction of order produces y_2.
Set y = v(x)\,y_1(x) where v(x) is an unknown function. Compute y' and y'', substitute into the ODE. Terms involving v (without derivatives) cancel — this is guaranteed because y_1 itself satisfies the equation. The result is a first-order ODE in w = v', which can be solved by the integrating factor method from Chapter 1.
For the constant-coefficient case, this is how the repeated-root solution xe^{rx} is derived. Set y_1 = e^{rx}, y = ve^{rx}:
y' = (v' + rv)e^{rx}, \quad y'' = (v'' + 2rv' + r^2v)e^{rx}
Substituting into ay'' + by' + cy = 0 and using ar^2 + br + c = 0:
a(v'' + 2rv') + bv' = 0 \implies av'' + (2ar + b)v' = 0
At the repeated root r = -b/(2a), we have 2ar + b = 0, so av'' = 0, giving v'' = 0 and thus v = C_1 + C_2 x. Therefore y = (C_1 + C_2 x)e^{rx} — the result we stated earlier, now derived.
Reduction of order becomes more important when dealing with variable-coefficient equations (Euler–Cauchy type) or when one solution is known from a series method. It will reappear in context.
42.8 Where this leads
You now have the complete toolkit for second-order linear ODEs with constant coefficients: classify by roots of the characteristic equation; add a particular solution via undetermined coefficients; understand why resonance is structurally different from off-resonance forcing.
Three threads open from here:
Laplace transforms (Chapter 4) handle initial-value problems algorithmically, absorb discontinuous forcing naturally, and are the standard industrial tool for control systems analysis. The Laplace transform of y'' introduces the transfer function, which encodes the characteristic equation in a different algebraic form.
Systems and phase plane (Chapter 3) extend the ideas to pairs of first-order equations — which is what a second-order ODE becomes when you introduce v = y' as a second unknown. Phase-plane portraits make the overdamped/underdamped/critically damped classification visual.
Fourier series (Volume 7, Fourier–PDEs section) answer the question: what if f(x) is not a simple sinusoid but a periodic function of arbitrary shape? Decompose it into harmonics, solve for each, and superpose. The resonance analysis extends directly.
Where this shows up
- A structural engineer computing the natural frequencies of a building must keep them away from the frequencies of earthquakes and wind — both sources of periodic forcing.
- An electrical engineer designing an RLC filter tunes \omega_0 and \zeta to pass or reject particular frequency bands. The quality factor Q = 1/(2\zeta) measures selectivity.
- A control systems engineer designing a PID controller must ensure the closed-loop characteristic equation has roots with negative real parts — the same stability condition as \alpha < 0 for the underdamped spring.
- A physicist modelling a damped quantum harmonic oscillator writes down the same equation. The mathematics is unchanged; only the interpretation of x differs.
42.9 What you can do now
You can now classify a second-order linear system by its characteristic roots and predict whether it will decay smoothly, decay while oscillating, or respond dangerously to forcing. You can construct y=y_c+y_p, apply the standard forced-solution methods, and recognise why resonance is structurally different from the off-resonance case.
42.10 Exercises
1. Find the general solution of y'' - 5y' + 6y = 0.
2. Solve y'' + 4y' + 4y = 0, y(0) = 1, y'(0) = 0.
3. Solve y'' + 9y = 0, y(0) = 0, y'(0) = 3.
4. Find a particular solution of y'' + 3y' + 2y = e^{-3x}.
5. Solve y'' + 4y = \cos(2x).
Note: the driving frequency \omega = 2 equals \omega_0 = \sqrt{4} = 2 — this is resonance. The standard guess fails; the modification rule applies.
6. An RLC circuit has L = 1 H, R = 2 Ω, C = \tfrac{1}{3} F with no forcing voltage. The governing equation for charge q(t) is L\ddot{q} + R\dot{q} + q/C = 0. Find q(t) given q(0) = 1 C, \dot{q}(0) = 0. Classify the response.