Step-by-step explanation:
Given:
[tex]x = e^{-t}\sin t[/tex]
Taking the 1st and 2nd derivatives of the above expression,
[tex]\dfrac{dx}{dt} = -e^{-t}\sin t + e^{-t}\cos t[/tex]
[tex]\dfrac{d^2x}{dt^2} = e^{-t}\sin t - e^{-t}\cos t -e^{-t}\cos t[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- e^{-t}\sin t[/tex]
[tex]\:\:\:\:\:\:\:\:\:= -2e^{-t}\cos t[/tex]
Therefore,
[tex]\dfrac{d^2x}{dt^2} + 2\dfrac{dx}{dt} + 2x[/tex]
[tex]= -2e^{-t}\cos t + 2(-e^{-t}\sin t + e^{-t}\cos t)[/tex]
[tex]\:\:\:\:+ 2e^{-t}\sin t[/tex]
[tex]= -2e^{-t}\cos t - 2e^{-t}\sin t + 2e^{-t}\cos t + 2e^{-t}\sin t[/tex]
[tex]= 0[/tex]
This shows that [tex]x = e^{-t}\sin t[/tex] is the solution to the differential equation
[tex]\dfrac{d^2x}{dt^2} + 2\dfrac{dx}{dt} + 2x = 0[/tex]
