38 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.
38.1 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).
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
\[\underbrace{ay'' + by' + cy = 0}_{\text{homogeneous: find } y_c} \quad + \quad \underbrace{ay'' + by' + cy = f(x)}_{\text{find any } y_p} \quad \Longrightarrow \quad y = y_c + y_p\]
38.2 Homogeneous equations: the characteristic equation
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.
38.2.1 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: \(y = C_1 e^{2x} + C_2 e^{3x}\).
38.2.2 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: \(y = (C_1 + C_2 x)e^{-2x}\).
38.2.3 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 |
38.3 The spring-mass-damper system
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 |
38.3.1 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.
38.4 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.
38.4.1 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.
38.4.2 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: \(y = C_1 e^{-x} + C_2 e^{-2x} + \frac{1}{2}e^{-3x}\).
38.4.3 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 \(0\cdot\cos 2x + 1\cdot\sin 2x\):
\[A + 8B = 0 \implies A = -8B\] \[B - 8A = 1 \implies B + 64B = 1 \implies B = \tfrac{1}{65},\quad A = -\tfrac{8}{65}\]
\[y_p = -\frac{8}{65}\cos(2x) + \frac{1}{65}\sin(2x)\]
38.4.4 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: \(2A = 1 \implies A = 1/2\).
\[y_p = \tfrac{1}{2}x^2 e^{-x}\]
38.5 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.
38.6 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.
38.7 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.
38.8 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.