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Second-Order Differential Equations

A practical hub for second-order ODEs: constant-coefficient methods, forcing, damping, and how these equations model oscillations in physics.

  • University Physics Year 1
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Second-order ordinary differential equations (ODEs) involve a second derivative like y'' or d²y/dt². They appear whenever a system’s dynamics depends on acceleration: oscillations, circuits, waves, and many stability problems.

Why physics needs this (examples you can click)
What you’re usually solving
  • Solve the homogeneous equation (the “natural” behavior)
  • If there’s a forcing term, find a particular solution
  • Apply two conditions (e.g. y(0) and y'(0)) to fix constants
  • Check limiting cases and whether the result matches the physics

Start here (quick classification)

When you see a second-order ODE, ask:

  1. Is it linear?

    Linear looks like:

    a(x) y'' + b(x) y' + c(x) y = f(x).

  2. Are coefficients constant?

    If a,b,c are constants, the characteristic-equation method is usually fastest.

  3. Homogeneous or forced?

    • Homogeneous: f(x) = 0
    • Forced (nonhomogeneous): f(x) ≠ 0

  4. What does the forcing look like?

    Polynomials / exponentials / sines-cosines often suggest undetermined coefficients.

Core method (with worked examples)

  • Basics of second-order differential equations

    Constant-coefficient linear ODEs, forcing, and the standard solution patterns.

    Open lesson

Method templates (what to do in practice)

Template A: constant-coefficient homogeneous ODE

For

ay'' + by' + cy = 0,

try y = e^rt. You get the characteristic equation

ar² + br + c = 0.
  • Two distinct real roots: y = C₁e^r₁t + C₂e^r₂t
  • Repeated real root: y = (C₁ + C₂t)e^rt
  • Complex roots r = α± iβ: y = e^(α t)(C₁ cos β t + C₂ sin β t)

Template B: apply initial conditions

Second-order means you need two conditions (typically y(0) and y'(0)). Differentiate your general solution, plug in t = 0, and solve for C₁,C₂.

Template C: forcing (undetermined coefficients)

If

ay'' + by' + cy = f(t)

and f(t) is built from polynomials/exponentials/sines/cosines, guess a particular solution of the same “shape”, substitute, then solve for the constants.

If your guess duplicates a term in the homogeneous solution, multiply your guess by t (or t², etc.) until it becomes independent.

Pitfalls that show up in physics
  • Two initial conditions: forgetting y'(0) (or using only one condition) leaves constants undetermined.
  • Damping frequency: in underdamped motion the oscillation is ω_d = square root of (ω₀²-β²), not ω₀.
  • Resonance in forcing: if the forcing frequency matches the natural frequency, your usual sinusoid guess must be multiplied by t.

Worked example: SHM with initial conditions

Solve:

y'' + ω²y = 0,

with y(0) = y₀ and y'(0) = v₀.

Characteristic equation:

r² + ω² = 0 ⇒ r = ± iω.

So the general solution is:

y(t) = A cos(ω t) + B sin(ω t).

Apply y(0) = y₀:

y(0) = A = y₀.

Differentiate:

y'(t) = -Aω sin(ω t) + Bω cos(ω t),

so y'(0) = Bω = v₀ ⇒ B = v₀/ω.

Final answer:

y(t) = y₀ cos(ω t) + v₀/ω sin(ω t).
Physics meaning

This is the same math behind a mass–spring oscillator and many small-angle oscillations.
The constants y₀ and v₀ are fixed by the initial displacement and initial velocity.

Worked example: underdamped oscillator form

The standard damped-oscillator equation is:

y'' + 2β y' + ω₀² y = 0.

If β < ω₀ (underdamped), define

ω_d = square root of (ω₀²-β²) .

Then the general solution is:

y(t) = e^(-β t)(A cos(ω_d t) + B sin(ω_d t)).

If y(0) = y₀ and y'(0) = v₀, then:

A = y₀, B = (v₀ + β y₀)/ω_d.

So:

y(t) = e^(-β t)(y₀ cos(ω_d t) + (v₀ + β y₀)/ω_d sin(ω_d t)).

This is exactly the same mathematics used in a mass-spring-damper model and in the free transient of a series RLC circuit.

Practice (with hints + answers)

1) Solve: y'' - 3y' + 2y = 0

Hint: Use y = e^rt so r²-3r + 2 = 0.

Answer: (r-1)(r-2) = 0 so r = 1,2:

y = C₁e^t + C₂e^2t.
2) Classify the damping: y'' + 6y' + 25y = 0

Hint: Compare with y'' + 2β y' + ω₀²y = 0.

Answer: 2β = 6 ⇒ β = 3, and ω₀² = 25 ⇒ ω₀ = 5. Since β < ω₀, the motion is underdamped.

3) How many conditions do you need to fully determine a second-order solution?

Hint: Count the constants in the general solution.

Answer: Two independent conditions (e.g. y(0) and y'(0)).

4) When does your forced-sinusoid guess need an extra factor of t?

Hint: Think “overlap with the homogeneous solution”.

Answer: When the forcing has the same functional form as a term in the homogeneous solution (resonance/duplication). Multiply the guess by t (or higher powers) until it becomes independent.

Quick reference: forms you should recognize

Simple harmonic motion (SHM)

y'' + ω² y = 0 ⇒ y = A cos(ω x) + B sin(ω x).

Damped oscillator (standard form)

y'' + 2β y' + ω₀² y = 0.

Depending on β vs ω₀, you get underdamped, critically damped, or overdamped motion.

Forced oscillator (one common template)

y'' + 2β y' + ω₀² y = f(x).

Physics connections

Back to Mathematics for Undergraduate Physics