Question

Help with num 12 please. thanks​

Answer 1

Step-by-step explanation:

Given:

$x = e^{-t}\sin t$

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

$\dfrac{dx}{dt} = -e^{-t}\sin t + e^{-t}\cos t$

$\dfrac{d^2x}{dt^2} = e^{-t}\sin t - e^{-t}\cos t -e^{-t}\cos t$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- e^{-t}\sin t$

$\:\:\:\:\:\:\:\:\:= -2e^{-t}\cos t$

Therefore,

$\dfrac{d^2x}{dt^2} + 2\dfrac{dx}{dt} + 2x$

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

$\:\:\:\:+ 2e^{-t}\sin t$

$= -2e^{-t}\cos t - 2e^{-t}\sin t + 2e^{-t}\cos t + 2e^{-t}\sin t$

$= 0$

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

$\dfrac{d^2x}{dt^2} + 2\dfrac{dx}{dt} + 2x = 0$