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Forcing the oscillator \( y'' + y \) with \( \sin 2x \) produces the particular integral

  • A.\( -\tfrac{1}{5}\sin 2x \)
  • B.\( -\tfrac{1}{3}x\sin 2x \)
  • C.\( -\tfrac{1}{3}\sin 2x \)
  • D.\( \tfrac{1}{3}\sin 2x \)

Answer: C

For a sine of angular frequency two, the operator \( d^{2}/dx^{2} \) replaces \( \sin 2x \) by \( -4\sin 2x \):

\[ \left(-4 + 1\right)A\sin 2x \;=\; \sin 2x \quad \Longrightarrow \quad A \;=\; -\tfrac{1}{3}. \]

The negative sign is essential and follows from \( -4 + 1 = -3 \). Resonance, which would bring in a factor of \( x \), happens only if the forcing frequency equals one

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