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The differential equation obtained by eliminating the arbitrary constants A and \(\phi\) from :

\[ y = Ae^{-\alpha t} \sin(\omega t + \phi) \]

is :

  • A.\(\dfrac{d^{2}y}{dt^{2}} + 2\alpha\dfrac{dy}{dt} + (\alpha^{2} + \omega^{2})y = 0\)
  • B.\(\dfrac{d^{2}y}{dt^{2}} + 2t\dfrac{dy}{dt} + (\alpha^{2} - \omega^{2})y = 0\)
  • C.\(\dfrac{d^{2}y}{dt^{2}} + 2\alpha t\dfrac{dy}{dt}\) \(+ (\alpha^{2} + \omega^{2})y = 0\)
  • D.\(\dfrac{d^{2}y}{dt^{2}} + 2\alpha t\dfrac{dy}{dt}\) \(+ (\alpha^{2} - \omega^{2})y = 0\)

Answer: A

The equation is \(\dfrac{d^{2}y}{dt^{2}} + 2\alpha\dfrac{dy}{dt} + (\alpha^{2} + \omega^{2})y = 0\).

Write \(y = \text{Im}(Ce^{(-\alpha + i\omega)t})\) with \(C = Ae^{i\phi}\). So y is a solution of the constant-coefficient equation whose characteristic roots are \(-\alpha \pm i\omega\). Those roots satisfy

\[ \begin{aligned} (m + \alpha)^{2} + \omega^{2} &= 0 \\ m^{2} + 2\alpha m + \alpha^{2} + \omega^{2} &= 0 \end{aligned} \]

which corresponds to \(y'' + 2\alpha y' + (\alpha^{2} + \omega^{2})y = 0\). The two arbitrary constants A and \(\phi\) are exactly the two constants of the general solution of this second-order equation, so eliminating them gives it.

The options with a factor t in the middle term have variable coefficients and do not have damped sinusoids as solutions. \(\alpha^{2} - \omega^{2}\) would correspond to real roots \(-\alpha \pm \omega\), giving exponentials rather than oscillations.

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