MAST10006 · Calculus 2
Second-Order Differential Equations
A constant-coefficient second-order linear ODE a y′′ + b y′ + c y = g(x) is the workhorse of oscillation, circuits and the mass-spring system. The full solution always splits as y = yh + yp — the homogeneous part (the natural behaviour) plus a particular part (the forced response). The homogeneous part is found from the characteristic (auxiliary) equation aλ² + bλ + c = 0, whose discriminant decides one of three cases: real distinct roots, a repeated root (with its extra factor of x), or complex roots (written in the real eαx(A cos βx + B sin βx) form). The particular part is found by undetermined coefficients — guess a form matching g(x), with the resonance modification rule when the guess already solves the homogeneous equation. The chapter closes with the mass-spring-damper and its under-, critically- and over-damped regimes.
What this chapter covers
- 011 From the ODE to the auxiliary equation
- 022 The three cases of y_h (real distinct / repeated / complex)
- 033 The homogeneous procedure, one worked example per case
- 044 Particular solutions by undetermined coefficients
- 055 The resonance / modification rule
- 066-7 The mass-spring-damper and the damping regimes
Worked example: a homogeneous second-order ODE (complex roots)
- +1Auxiliary equation: λ² + 2λ + 5 = 0, so λ = (−2 ± √(4 − 20))/2 = −1 ± 2i (here α = −1, β = 2).
- +1Write yh in the complex-root form: y = e−x(A cos 2x + B sin 2x).
- +1Apply y(0) = 0: A = 0, so y = B e−x sin 2x and y′ = B e−x(2 cos 2x − sin 2x).
- +1Apply y′(0) = 4: 2B = 4, so B = 2. Hence y = 2 e−x sin 2x.
Key terms
- Characteristic (auxiliary) equation
- The quadratic aλ² + bλ + c = 0 obtained by trying y = eλx in the homogeneous equation a y′′ + b y′ + c y = 0. Its discriminant Δ = b² − 4ac decides which of the three homogeneous solution forms applies.
- Homogeneous solution
- The part yh solving a y′′ + b y′ + c y = 0: A eλ₁x + B eλ₂x for real distinct roots, (A + Bx)eλx for a repeated root, and eαx(A cos βx + B sin βx) for complex roots α ± iβ.
- Undetermined coefficients
- The method for the particular solution yp: guess a form whose shape matches the forcing g(x) (a general polynomial for a polynomial, C ekx for an exponential, C cos ωx + D sin ωx — both terms — for a sinusoid), then pin the coefficients by substituting back.
- Resonance / modification rule
- When the trial yp already appears in yh, substituting it gives zero, so multiply the trial by x (or x² if the matching root is repeated). In the undamped mass-spring this is mechanical resonance: the amplitude grows like t.
- Damping regimes
- For the free mass-spring m x′′ + c x′ + k x = 0, the sign of Δ = c² − 4mk gives over-damped (two real roots, decay, no oscillation), critically damped (repeated root, fastest decay), and under-damped (complex roots, decaying oscillation).
Second-Order Differential Equations FAQ
How do I choose the form of the homogeneous solution?
Compute the discriminant Δ = b² − 4ac of the auxiliary equation. Δ > 0 gives two real distinct roots and yh = A eλ₁x + B eλ₂x; Δ = 0 gives a repeated root and yh = (A + Bx)eλx; Δ < 0 gives complex roots α ± iβ and the real form eαx(A cos βx + B sin βx). The discriminant is the only decision; the procedure is otherwise identical.
Why does the repeated-root case need an extra x?
Because eλx alone supplies only one independent solution when the root is repeated. Writing (A + B)eλx collapses two constants into one and loses a solution; the correct second independent solution is x eλx, giving yh = (A + Bx)eλx.
When does the resonance modification kick in?
When your trial particular solution is already a homogeneous solution — for example forcing 4e2x on an equation whose roots include λ = 2. Substituting the unmodified trial gives 0, which cannot match the forcing, so multiply the trial by x (or x² if the matching root is repeated).
What is the 'both terms' trap in undetermined coefficients?
Even when the forcing is cos ωx alone, the trial must carry both cos ωx and sin ωx, because differentiating cosine produces sine, so a one-term guess cannot balance. The same applies to polynomial forcing: include every lower-degree term, not just the top power.
Exam move
Run the same four-step procedure every time, because the only real decision is which of the three boxes the discriminant lands you in. Read off a, b, c; write the auxiliary equation; compute Δ to identify the case at a glance; write yh in the matching form (don't forget the extra x for a repeated root, and convert complex roots to the real cos/sin form); then add yp from undetermined coefficients and apply BOTH initial conditions last. For the particular solution, match the trial to g(x), always include both sinusoid terms and every polynomial term, and apply the resonance modification (multiply by x, or x²) when the trial duplicates a homogeneous solution. For the mass-spring application, read the regime straight off the sign of Δ = c² − 4mk: over-, critically- or under-damped. Drill these until the recipe runs on autopilot — the signature long-answer is worth a lot and is pure procedure.
