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Mathematics · University

Second-order equations and resonance

For constant coefficients, an exponential trial function reduces the homogeneous ODE to a polynomial.

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Prerequisites

Complex roots and elementary differential equations.

Study notes

Characteristic equation

For constant coefficients, an exponential trial function reduces the homogeneous ODE to a polynomial.

Initial data

A second-order equation usually requires two initial conditions for a unique local solution under standard hypotheses.

Repeated roots

A repeated root r needs both e^(rt) and te^(rt).

Resonance

A forcing term that duplicates a homogeneous solution requires an extra power of t in a particular-solution trial.

Worked examples

Solve y″+4y=0 with y(0)=3,y′(0)=0.

The characteristic roots are ±2i, and the initial conditions remove the sine term.

y(t)=3cos(2t).

For y″−3y′+2y=0, give the larger characteristic root.

r²−3r+2=(r−1)(r−2).

2

Check your understanding

  1. What pair spans the solutions of y″−2y′+y=0?

    1. sin t and cos t
    2. e^t and te^t
    3. 1 and t²
    4. e^t and 2e^t

  2. Give a valid form for a particular solution of y″+y=cos t.

  3. For y(t)=2cos(3t), calculate y″(0).

  4. Which initial data are sufficient for a standard second-order IVP?

    1. Three arbitrary values of y at the same point
    2. y(t₀) and y′(t₀)
    3. Only y(t₀) for every equation
    4. Only an equation title

Answers and marking guidance

Answers and marking notes
  1. e^t and te^t

    The repeated root r=1 needs the generalized solution te^t.

  2. A suitable trial is A t sin t; substitution gives A=1/2, so y_p=(t/2)sin t.

    Credit the resonance adjustment and substitution, not an ordinary A cos t trial alone.

  3. -18

    Differentiating twice gives −18cos(3t).

  4. y(t₀) and y′(t₀)

    Two independent initial conditions determine the two constants under the usual existence hypotheses.

Further reading and review references

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