Answer
See below
Work Step by Step
Given: $y''+y=e^{-t}$
Since $y(0)=0\\
y'(0)=1$
obtain: $L[y]=L[y''+y]+L[e^{-t}]\\
=(s^2+1)L[y]-1+\frac{1}{s+1}\\
=\frac{1}{(s+1)(s^2+1)}+\frac{1}{s^2+1}$
Using the Convolution Theorem
$y(t)=L^{-1}[\frac{1}{s+1}.\frac{1}{s^2+1}]+L^{-1}[\frac{1}{s+1}]\\
=\int^t_0 e^{-\tau} \sin (t-\tau) d \tau +\sin t$
