Linear equations, oscillation and forcing

An oscillator needs position, velocity and a force law

A small-displacement mass–spring model about an equilibrium position is m x″+c x′+k x=F(t), with m>0, k>0 and viscous damping coefficient c≥0. The restoring force is −kx and the drag force is −cx′; F is an external force measured with the same sign convention as x. Initial displacement x(t₀)=x₀ and velocity x′(t₀)=v₀ specify a state. Position alone does not determine the motion.

In SI units, m is in kilograms, c in kilograms per second or newton-seconds per metre, and k in newtons per metre. Each term has units of force. The displacement is measured from the chosen static equilibrium, so a constant gravitational force can already be balanced in that reference position. Large displacements, nonlinear springs, dry friction or memory effects require a model beyond the linear constant-coefficient assumptions used here.

The mathematical framework is broader: y″+p(t)y′+q(t)y=g(t), with p,q,g continuous on an interval containing the initial time, has a unique solution there for specified y and y′. If a leading coefficient a(t) is divided out, choose an interval where it is nonzero. Singularity of that coefficient cannot be dismissed simply because a characteristic polynomial is convenient elsewhere.

Lessons 137–138 supplied initial-value meaning, local uniqueness and linear first-order methods. Here the two-dimensional initial state connects those ideas to oscillation. Lesson 140 will make the state-system geometry explicit. Throughout this article an exact solution, a long-time approximation and a finite plot remain different claims.

Linearity fixes how homogeneous and forced parts combine

Define L[y]=y″+p(t)y′+q(t)y. The product and sum rules give L[A y₁+B y₂]=A L[y₁]+B L[y₂] for constants A,B. Therefore linear combinations of homogeneous solutions, with L[y]=0, are again homogeneous. If y_p solves L[y_p]=g, adding any homogeneous solution gives another solution of that same forced equation.

Conversely, subtracting one particular solution from any forced solution gives a homogeneous solution. Thus the whole forced solution set is one particular solution plus the homogeneous space. Adding two solutions with the same nonzero forcing generally gives forcing 2g rather than g. It is the differential operator that is linear, with its right side transformed consistently.

For continuous p and q, two homogeneous solutions whose initial state vectors (y₁(t₀),y₁′(t₀)) and (y₂(t₀),y₂′(t₀)) are independent span every homogeneous solution. To prove this, choose their two coefficients to match arbitrary initial position and velocity. The difference from the desired solution has zero initial state, and the uniqueness theorem makes that difference zero throughout the interval.

This argument explains why two independent modes are enough for a regular second-order scalar problem. It also explains why repeating the same exponential twice does not create another degree of freedom. Dependence is a relation between whole functions on an interval with constant coefficients, not whether their graphs happen to have different values at one sampled time.

The Wronskian measures independence of the initial states

For two differentiable functions, define W=y₁y₂′−y₁′y₂. If W(t₀)≠0, their initial state vectors form an invertible 2×2 matrix. The two coefficients needed to match position and velocity are then uniquely determined. For solutions of the same continuous homogeneous second-order equation, this yields a fundamental pair across the interval.

Differentiate W for two C² homogeneous solutions: W′=y₁y₂″−y₁″y₂. Substitute y_i″=−p y_i′−q y_i. The q terms cancel and W′=−pW. The first-order integrating-factor argument gives Abel’s identity W(t)=W(t₀)exp(−∫_(t₀)^t p(s) ds). A nonzero Wronskian at one point therefore stays nonzero everywhere in the continuous-coefficient interval.

For such solutions, a zero Wronskian at one point instead gives dependent initial states and hence dependent solutions by uniqueness. This conclusion uses their common differential equation. For arbitrary functions without that common solution structure, blanket converses based on a vanishing Wronskian require more care; they are not being used here.

If a known nonzero homogeneous solution y₁ is available, reduction of order can construct a second locally. Put y₂=y₁v; substitution gives v″+(2y₁′/y₁+p)v′=0. Integrating yields y₂=y₁∫exp(−∫p)/y₁² dt. This formula is used only on an interval where y₁≠0, and the integral’s constants can add a multiple of y₁. A zero of y₁ may make this particular representation unsuitable even when the original equation is regular.

Distinct characteristic roots give independent exponential modes

For constant coefficients a y″+b y′+d y=0 with a≠0, try y=e^(rt). Substitution gives (a r²+b r+d)e^(rt)=0, so roots of the characteristic polynomial supply solutions. The exponential is nonzero, making this division legitimate. This technique relies on constant coefficients; substituting one fixed exponential does not generally solve a variable-coefficient equation.

If the roots r₁ and r₂ are distinct real numbers, y₁=e^(r₁t), y₂=e^(r₂t). Their Wronskian is (r₂−r₁)e^((r₁+r₂)t), which never vanishes. The basis argument therefore proves that every homogeneous solution is C₁e^(r₁t)+C₂e^(r₂t), rather than merely observing that two guessed functions happen to work.

A negative root gives a decaying mode in forward time, a positive root a growing mode, and a zero root a constant mode. The full motion depends on which coefficients the initial data excite. Two negative roots do not require displacement to decrease monotonically; opposing modal coefficients can produce a turning point or a sign change.

For y″+3y′+2y=0, the roots are −1 and −2. Initial values y(0)=1, y′(0)=0 give y=2e^(−t)−e^(−2t). Direct differentiation verifies the equation and both data. The roots describe the available modes, while the data determine their particular combination.

A repeated root needs a genuinely new solution

If the characteristic polynomial has one repeated root r, the homogeneous equation can be written y″−2r y′+r²y=0 after dividing by its nonzero leading coefficient. The function e^(rt) is one solution, but writing C₁e^(rt)+C₂e^(rt) merely combines two constants into one. A second independent solution must be found.

Set y=e^(rt)u(t). Differentiating twice and substituting makes the u and u′ terms cancel, leaving e^(rt)u″=0. Therefore u=C₁+C₂t, and y=(C₁+C₂t)e^(rt). The Wronskian of e^(rt) and t e^(rt) is e^(2rt), never zero. This proves both the new solution and completeness of the pair.

For y″+4y′+4y=0, the repeated root is −2. With y(0)=1 and y′(0)=0, the coefficients are C₁=1 and C₂=2, giving y=(1+2t)e^(−2t). Although t grows, the exponential decay dominates it as t→∞. A polynomial factor does not automatically mean unbounded future motion.

The repeated-root case is also the limit of nearby distinct-root bases after a suitable rescaling: [e^((r+ε)t)−e^(rt)]/ε tends to t e^(rt). Simply setting the two roots equal in an unrescaled basis loses that independent direction. Near coincidence, this observation also helps explain why a poorly chosen numerical basis can suffer cancellation.

Complex roots give real oscillations with an envelope

For real constant coefficients, nonreal characteristic roots come in a conjugate pair α±iβ, with β>0. Euler’s formula writes e^((α+iβ)t)=e^(αt)(cos βt+i sin βt). Because the operator has real coefficients, its real and imaginary parts are real solutions. They form the pair e^(αt)cos βt and e^(αt)sin βt.

Their Wronskian is βe^(2αt), which never vanishes. Hence every real homogeneous solution is y=e^(αt)(A cos βt+B sin βt). The real part α controls the exponential envelope, while β is the angular oscillation frequency. A negative α gives a decaying envelope, zero α gives persistent oscillation, and positive α gives a growing envelope.

For nonzero (A,B), write A=R cos φ, B=R sin φ with R=√(A²+B²)>0 and φ=atan2(B,A). Then A cos βt+B sin βt=R cos(βt−φ). The phase is defined modulo 2π, and the zero solution has no uniquely defined phase. The period of the sinusoidal factor is 2π/β. The full damped function is not periodic when α≠0.

For y″+2y′+5y=0, the roots are −1±2i. The solution with y(0)=1 and y′(0)=0 is e^(−t)(cos 2t+(1/2)sin 2t). The extra sine coefficient is required by the derivative of the envelope; setting it to zero from the initial displacement alone would give the wrong initial velocity.

Damping regimes compare a decay scale with a restoring frequency

For m x″+c x′+k x=0 with m,k>0 and c≥0, define ω₀=√(k/m), δ=c/(2m) and ζ=δ/ω₀=c/(2√(mk)). Here ω₀ and δ have units inverse time, while ζ is dimensionless. The characteristic roots are −δ±√(δ²−ω₀²). These definitions make the discriminant’s physical scales explicit.

If 0≤δ<ω₀, the roots are −δ±iω_d with ω_d=√(ω₀²−δ²). The motion is e^(−δt)(A cos ω_dt+B sin ω_dt), after a convenient choice of initial time origin. For c=0 this is undamped periodic motion. For 0<c<2√(mk), the oscillation has a decaying envelope and ω_d<ω₀.

If δ=ω₀, the repeated negative root gives critical damping, x=(A+Bt)e^(−ω₀t). If δ>ω₀, the roots are distinct and negative, giving overdamping and two real decaying exponential modes. Nontrivial free responses in these latter cases do not have a sinusoidal oscillation factor. Initial conditions can still produce a turning point or an equilibrium crossing; the root classification alone does not prescribe a monotone displacement history.

Critical damping is a boundary between these root types. Claims about a fastest settling time need a specified measure, initial state and comparison family. A late-time modal decay rate, first crossing time and time to stay within a tolerance are different quantities. The classification gives precise formulas from which such a chosen comparison can be calculated.

An energy balance checks signs and separates loss from motion

For the mass–spring model, define E=(1/2)m(x′)²+(1/2)kx². These are kinetic and elastic potential energy, with units joules. Differentiation and the equation give E′=x′(mx″+kx)=F(t)x′−c(x′)². External force contributes instantaneous power F x′, and viscous damping removes power c(x′)² when c≥0.

For free undamped motion, F=0 and c=0, so E is constant. For free damped motion with c>0, E is nonincreasing. It can have zero derivative at isolated turning points where x′=0 even while the object remains displaced. Nonincreasing energy does not mean displacement is monotone or velocity never changes sign.

With m,k>0, positive energy controls both |x| and |x′|: |x|≤√(2E/k) and |x′|≤√(2E/m). For the constant-coefficient free damped model with c>0, the root formulas show both state components approach zero in future time. The energy identity provides a separate check on the force and damping signs, while the explicit modes establish that long-time conclusion here.

Under forcing, even a damped oscillator can gain energy during part of a cycle if F x′ exceeds the damping loss. A sustained periodic response balances input and loss on average over a period; energy can still vary within the cycle. A drawing with decreasing amplitude is not a substitute for checking this balance, particularly if a numerical method introduces artificial energy growth.

A trial particular solution must account for an overlapping mode

For a constant-coefficient forced equation, undetermined coefficients uses a finite family of functions closed under differentiation, such as polynomials times an exponential, or sine and cosine factors. Substitute a trial with unknown coefficients, match the forcing and verify the result. This is a useful method for those structures, not an algorithm for every possible forcing or every variable-coefficient operator.

If a proposed trial already belongs to the homogeneous space, applying the operator gives zero and cannot produce a nonzero forcing. The trial must be modified. For y″+3y′+2y=e^(−t), the trial A e^(−t) overlaps a root. Taking y_p=A t e^(−t) instead gives L[y_p]=A e^(−t), so A=1 works.

The root multiplicity determines the needed polynomial factor for exponential forcing. To see why, write P(D)[e^(λt)u]=e^(λt)P(D+λ)u, with D denoting differentiation. This shift identity follows by differentiating the product. If λ is a root of multiplicity s, P(D+λ) has a factor D^s, explaining why the exponential trial must acquire a suitable t^s factor.

For y″+2y′+y=e^(−t), λ=−1 is a double root. Substituting y=e^(−t)u gives u″=1, so a particular solution is (t²/2)e^(−t). Add the complete homogeneous solution before fitting both initial data. The particular solution alone need not meet those data, and any homogeneous terms inadvertently included in it can be absorbed into the two homogeneous coefficients.

Variation of parameters works from any known fundamental pair

For the normalized equation y″+p(t)y′+q(t)y=g(t), suppose a known homogeneous fundamental pair y₁,y₂ has W=y₁y₂′−y₁′y₂≠0 on the continuous-coefficient interval. Seek y_p=u₁y₁+u₂y₂, allowing the coefficients to vary. Impose u₁′y₁+u₂′y₂=0 so the first derivative simplifies. Substitution leaves u₁′y₁′+u₂′y₂′=g.

Solve this 2×2 system: u₁′=−y₂g/W and u₂′=y₁g/W. Definite integrals from t₀ give y_p(t)=−y₁(t)∫_(t₀)^t y₂(s)g(s)/W(s) ds+y₂(t)∫_(t₀)^t y₁(s)g(s)/W(s) ds. The resulting particular solution has y_p(t₀)=y_p′(t₀)=0. Nonzero initial data can therefore be supplied entirely by the homogeneous part.

Equivalently, the zero-state response is ∫_(t₀)^t K(t,s)g(s) ds with K(t,s)=[y₂(t)y₁(s)−y₁(t)y₂(s)]/W(s). The signs and normalization follow from the solved linear system; they should be checked rather than memorised without the Wronskian convention. If the original equation has leading coefficient a(t), its normalized forcing is g=F/a, which must be used here.

The method allows variable coefficients once a fundamental pair is known. Its integrals need not have elementary antiderivatives. For a constant undamped oscillator y″+ω₀²y=g, the kernel reduces to sin(ω₀(t−s))/ω₀ when ω₀>0. This integral response is another way to derive forced oscillation, and it retains the initial time explicitly.

A damped harmonic response has an amplitude and a phase lag

For m x″+c x′+k x=F₀ cos(ωt), take m,k,c,F₀,ω>0. Seek a periodic particular solution x_p=A cos(ωt)+B sin(ωt). Put K_ω=k−mω² and C_ω=cω. Matching the cosine and sine coefficients gives K_ω A+C_ω B=F₀ and −C_ω A+K_ω B=0.

The determinant D=K_ω²+C_ω² is positive under these assumptions. Thus A=F₀K_ω/D and B=F₀C_ω/D. Write x_p=X cos(ωt−φ), with X=F₀/√D and φ=atan2(C_ω,K_ω). The two-argument angle function preserves the correct quadrant when K_ω is negative; an unqualified one-argument arctangent can lose that phase information.

The full solution is x=x_h+x_p, with x_h fixed by the initial displacement and velocity after the particular values are subtracted. Since c>0 with m,k>0, x_h decays and x_p is the long-time periodic response. The full finite-time motion is not generally identical to x_p. As c becomes small the transient may last a long time, so a short displayed window can be misleading.

X has units of length, ω has units inverse time, and φ is a dimensionless phase angle. At ω=ω₀=√(k/m), the phase lag is π/2 and X=F₀/(cω₀), finite when c>0. This forcing frequency is not generally the frequency of largest displacement amplitude; that separate question depends on the whole response curve.

Ideal resonance differs from two nearby finite oscillations

For an undamped oscillator m x″+k x=F₀ cos(ωt), assume m,k,F₀,ω>0 and zero initial position and velocity at 0. When ω≠ω₀=√(k/m), a solution is x(t)=F₀[cos(ωt)−cos(ω₀t)]/[m(ω₀²−ω²)]. Substitution and the initial data verify it. The denominator cannot be used at equal frequencies.

If ω is near but unequal to ω₀, the identity cos a−cos b=−2sin((a+b)/2)sin((a−b)/2) shows a fast oscillation multiplied by a slowly varying envelope. This is beating. For each fixed unequal frequency pair, the response remains bounded in the ideal model, although its bound can be large. A finite observation window with a growing beat envelope is not by itself unbounded resonance.

At exact equality ω=ω₀, use a modified trial or take the fixed-time limit of the correctly combined zero-state formula. The result is x(t)=F₀t sin(ω₀t)/(2mω₀). Differentiating gives x″+ω₀²x=(F₀/m)cos(ω₀t), and both initial data are zero. Its oscillation envelope grows linearly without bound as time increases in this ideal undamped model.

The limiting statement keeps t fixed while ω approaches ω₀. Long-time behaviour at a fixed small detuning is a different limit, so the two need not describe the same envelope over arbitrarily long times. Real systems can include damping, nonlinear response and changing parameters that invalidate the ideal large-amplitude extrapolation. The derived resonance result belongs to the stated linear undamped assumptions.

A displacement peak and a velocity peak occur at different frequencies

For the damped harmonic response, maximizing displacement amplitude X(ω)=F₀/√[(k−mω²)²+c²ω²] means minimizing its squared denominator D. Differentiating gives D′=2ω[2m²ω²+c²−2mk]. For positive forcing frequencies, an interior displacement maximum exists only if c²<2mk, equivalently ζ<1/√2.

In that case the peak frequency is ω_r=√(k/m−c²/(2m²))=√(ω₀²−2δ²), below the undamped natural frequency. If c²≥2mk, D is increasing for every ω>0 and displacement amplitude decreases from its zero-frequency limiting value F₀/k. There is then no positive-frequency displacement peak, even though some of these damping values still belong to the underdamped free-motion regime.

The velocity amplitude is ωX(ω), a different function. For c>0, its maximum occurs at ω=ω₀ and equals F₀/c. One way to check this is to write its squared reciprocal, apart from F₀², as D/ω²=k²/ω²−2km+m²ω²+c². Its derivative vanishes at ω²=k/m and changes from negative to positive there.

A measured resonance claim must identify the response being measured: displacement, velocity, power or another observable. The damped free oscillation frequency ω_d, undamped natural frequency ω₀ and displacement peak frequency ω_r are also different quantities. Their explicit formulas and assumptions prevent one symbol called frequency from hiding three different questions.

Extend the method only where its assumptions extend

For a variable-coefficient equation, the characteristic polynomial with fixed roots usually does not apply. Variation of parameters and the Wronskian framework still work on a continuous normalized interval once a fundamental pair is found. Some special variable forms admit their own substitution. For example, on t>0, an Euler equation can be tested with y=t^r rather than e^(rt).

For t²y″+2ty′−2y=0 on t>0, substitution gives r(r−1)+2r−2=(r−1)(r+2), so t and t^(−2) are solutions. Their Wronskian is −3t^(−2), never zero on that interval, proving completeness there. The singular point t=0 needs original-equation and branch analysis; the basis on t>0 is not a theorem through that point.

For a finite jump in applied force F with constant positive m and no impulse, the natural matched state keeps x and x′ continuous. Integrating the force balance over a shrinking interval around the switch gives no finite velocity jump when the forces and state remain bounded. Acceleration can have unequal one-sided limits, so the matched trajectory is C¹ and piecewise C² rather than globally C² at that switch.

An ideal impulse is a different input model and can produce a velocity jump; it requires a precisely defined impulse formulation. A real friction law may also cease to be viscous or smooth near zero velocity. The linear formulas remain exact for their mathematical assumptions, while applying them to a changed law requires a newly stated problem.

Check the full state and the claimed response

For a candidate solution, calculate y′ and y″ and substitute into the original equation, then check both initial data. Confirm the complete fundamental pair, coefficient domain and connected interval. A correct characteristic root or one matching initial displacement is only part of that verification.

For a forcing calculation, distinguish a particular periodic solution from the transient needed to meet the initial state. At resonance, replace an invalid zero denominator with an admissible derivation. For near resonance, label beat time scales and detuning. For damped response, specify whether the plotted curve is displacement amplitude or a different observable before identifying its peak.

Use the energy equation as an independent physical check where the mass–spring assumptions apply. Free undamped motion should conserve the stated energy exactly in the mathematical solution; free positive damping should not create it. Numerical sampling or finite time stepping can violate these properties, which is an implementation or approximation issue rather than a new physical prediction.

The explorer should retain initial position and velocity, parameter units, chosen time direction and exact-versus-approximate labels. Diagrams can illustrate envelopes, state trajectories and forcing phase, but a finite drawing does not establish a long-time stability or resonance theorem. The analytic assumptions and derivations explain those conclusions; the interactive cases test their displayed implementation.

Worked examples

Try it yourself

1. Why does initial displacement alone not normally select a unique solution of a regular second-order linear oscillator?

   Show the answer and reasoning

   Initial velocity is another independent state component.

   A fundamental homogeneous pair has two constants. Position and velocity at the same time fix them under the continuous-coefficient uniqueness assumptions.

2. If L[y_p]=g and L[y_h]=0, what is L[y_p+y_h]? What if two solutions with forcing g are added?

   Show the answer and reasoning

   The first gives g; adding two forced solutions gives 2g.

   Linearity applies to the operator and right sides together. A homogeneous correction preserves the given forcing.

3. Compute the Wronskian of e^(−t) and e^(−3t).

   Show the answer and reasoning

   W=−2e^(−4t), nonzero for all real t.

   Use y₁y₂′−y₁′y₂. Its nonzero value shows independent initial states; both functions can form a fundamental pair for the corresponding constant-coefficient equation.

4. For y″+p(t)y′+q(t)y=0, what first-order equation does the Wronskian of two solutions satisfy?

   Show the answer and reasoning

   W′=−pW.

   Differentiate W and substitute the equation for each second derivative. The q terms cancel, giving the integrating-factor identity.

5. Give a homogeneous basis for y″+5y′+6y=0.

   Show the answer and reasoning

   e^(−2t) and e^(−3t).

   The polynomial (r+2)(r+3) has two distinct real roots. Their Wronskian is nonzero, so the two modes span the solution space.

6. Why are e^(−2t) and another copy of e^(−2t) insufficient for a double root −2? Give the missing basis member.

   Show the answer and reasoning

   They are dependent; the second independent member is t e^(−2t).

   Two copies combine into one coefficient. Substitution y=e^(−2t)u reduces the double-root equation to u″=0.

7. For roots −2±3i, state the real homogeneous form and the period of its sinusoidal factor.

   Show the answer and reasoning

   y=e^(−2t)(A cos(3t)+B sin(3t)); sinusoidal period 2π/3.

   The real part controls the envelope and the imaginary part the angular frequency. The full decaying function is not periodic.

8. For m=1,k=9, classify c=2,c=6,c=8.

   Show the answer and reasoning

   Underdamped, critically damped and overdamped, respectively.

   The critical value is 2√(mk)=6. Compare c with that value, keeping m,k positive.

9. What is E′ for a free viscously damped mass–spring oscillator with c>0?

   Show the answer and reasoning

   E′=−c(x′)²≤0.

   There is no forcing power. Energy can have zero derivative at a turning point without implying zero displacement or zero energy.

10. What quantity gives instantaneous input power from a force F(t) acting along displacement x?

    Show the answer and reasoning

    F(t)x′(t).

    Force times velocity has units of energy per time. The sign determines whether the force supplies or extracts mechanical energy at that instant.

11. For y″+3y′+2y=e^(−t), why does a trial A e^(−t) fail, and what trial works?

    Show the answer and reasoning

    It is homogeneous and gives zero forcing; use A t e^(−t), with A=1.

    The forcing exponent equals a simple characteristic root, so one factor t removes the overlap.

12. For y″+2y′+y=e^(−t), what simple particular solution handles the double overlap?

    Show the answer and reasoning

    y_p=(t²/2)e^(−t).

    Substitution y=e^(−t)u leaves u″=1. A single t factor would still be homogeneous.

13. In a(t)y″+b(t)y′+d(t)y=F(t), what forcing enters the normalized variation-of-parameters formulas where a≠0?

    Show the answer and reasoning

    g=F/a.

    The formulas are for y″+py′+qy=g. Omitting the leading-coefficient division gives the wrong particular response.

14. For positive damping and harmonic forcing, what is the displacement phase lag at ω=√(k/m)?

    Show the answer and reasoning

    π/2.

    The in-phase stiffness-minus-inertia term is zero, while cω is positive. atan2(cω,0) is π/2, and the amplitude remains finite.

15. State the zero-state solution at exact resonance for m x″+k x=F₀ cos(ω₀t), with ω₀=√(k/m)>0.

    Show the answer and reasoning

    x=F₀t sin(ω₀t)/(2mω₀).

    The usual nonresonant denominator vanishes. A resonant trial or a fixed-time limit gives the t-weighted sine response.

16. Does every underdamped free oscillator have a positive-frequency displacement-amplitude maximum under harmonic forcing?

    Show the answer and reasoning

    No; the displacement peak requires ζ<1/√2, stronger than ζ<1.

    The peak condition is c²<2mk. Otherwise the displacement amplitude decreases from its zero-frequency limit, even for some free underdamped parameters.

17. For c>0,m>0,k>0, where does the harmonic velocity amplitude peak?

    Show the answer and reasoning

    At ω=√(k/m), with amplitude F₀/c for F₀>0.

    Minimize D/ω²=k²/ω²−2km+m²ω²+c². Its minimum occurs at ω²=k/m, which need not equal the displacement peak frequency.

18. With a finite force jump and no impulse, which oscillator state quantities are matched continuously? Can acceleration jump?

    Show the answer and reasoning

    Position and velocity are continuous; acceleration can jump.

    A bounded force acting across a shrinking time interval gives no finite velocity impulse. The matched trajectory is C¹ and piecewise C², with the equation checked on the smooth pieces and in its integrated balance.

What to carry forward

• A regular second-order initial-value problem needs both position-like and velocity-like data.
• A fundamental pair is justified by independent initial states and uniqueness, not only by guessing two formulas.
• The Wronskian obeys W′=−pW for solutions of the same continuous normalized homogeneous equation.
• Distinct, repeated and complex characteristic roots require different independent solution bases.
• A decaying exponential envelope differs from a periodic function, and modal coefficients determine the actual initial motion.
• The damping ratio separates underdamped, critical and overdamped root regimes under positive mass and stiffness.
• The energy identity E′=F x′−c(x′)² independently checks force and damping signs.
• Decreasing energy does not imply monotone displacement.
• A particular solution supplies forcing; the homogeneous correction supplies the remaining initial state.
• Overlapping forcing modes require the appropriate polynomial factor, with multiplicity retained.
• Variation of parameters uses a known fundamental pair and normalized forcing on a valid coefficient interval.
• A damped harmonic particular solution has amplitude and quadrant-sensitive phase, with a separate transient.
• Beats at fixed detuning are bounded in the ideal undamped model; exact resonance has a linearly growing envelope.
• Natural frequency, damped free frequency and displacement peak frequency are distinct quantities.
• Velocity and displacement peaks answer different response questions.
• Variable coefficients, singular points and forcing switches need their own interval and solution-regularity checks.