1 Answer

Adrian Krebs · Answer 1 · 2022-02-05T18:42:53+0000

Explanation:

Given:

$x = e^(-t)\sin t$

Taking the 1st and 2nd derivatives of the above expression,

$(dx)/(dt) = -e^(-t)\sin t + e^(-t)\cos t$

$(d^2x)/(dt^2) = e^(-t)\sin t - e^(-t)\cos t -e^(-t)\cos t$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- e^(-t)\sin t$

$\:\:\:\:\:\:\:\:\:= -2e^(-t)\cos t$

Therefore,

(d^2x)/(dt^2) + 2(dx)/(dt) + 2x

$= -2e^(-t)\cos t + 2(-e^(-t)\sin t + e^(-t)\cos t)$

$\:\:\:\:+ 2e^(-t)\sin t$

$= -2e^(-t)\cos t - 2e^(-t)\sin t + 2e^(-t)\cos t + 2e^(-t)\sin t$

= 0

This shows that
$x = e^(-t)\sin t$ is the solution to the differential equation

(d^2x)/(dt^2) + 2(dx)/(dt) + 2x = 0

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