Answer
Given below.
Work Step by Step
Take the derivatives of the function.
$y(x)=e^{-x}sin2x$
$y'(x)=-e^{-x}sin2x+2e^{-x}cos2x$
$y''(x)=e^{-x}sin2x-2e^{-x}cos2x-2e^{-x}cos2x-4e^{-x}sin2x=-3e^{-x}sin2x-4e^{-x}cos2x$
Substituting these functions into the differential equation yields $\implies y''(x)+2y'(x)+5y=0$
$\implies -3e^{-x}sin2x-4e^{-x}cos2x+2(-e^{-x}sin2x+2e^{-x}cos2x)+5(e^{-x}sin2x)=0$
$\implies 0=0$
This means that the solution is valid for all real values. The interval linked to this solution is $(−∞,∞)$.
