Answer:Given $\dfrac{d^2y}{dt^2}+4 \dfrac{dy}{dt}+8y=1$...............................................................................(1)

Taking Laplace transform on both sides of (1), we get

$L[y''+4y'+8y]=L[1] \Rightarrow L[y'']+4L[y']+8L[y]=\dfrac{1}{s}$

or,

$[s^2L\{y\}-sy(0)-y'(0)]+4[sL\{y\}-y(0)]+8L[y]=\dfrac{1}{s}$............................(2)

Substituting the given conditions in (2), we get                   $[ y(0)=0, y'(0)=1]$

$[s^2L\{y\}-s.0-1]+4[sL\{y\}-0]+8L[y]=\dfrac{1}{s}$

$\Rightarrow L\{y\}(s^2+4s+8)-1=\dfrac{1}{s} \Rightarrow L\{y\}=\dfrac{s+1}{s(s^2+4s+8)}$

Separating it into partial fraction, we get

$L\{y\}=\dfrac{1}{8s}-\dfrac{1}{8}\dfrac{(s-4)}{s^2+4s+8}$

         $=\dfrac{1}{8s}-\dfrac{1}{8}\Bigg[\dfrac{s+2}{(s+2)^2+2^2}-3.\dfrac{2}{(s+2)^2+2^2}\Bigg].$

Taking inverse Laplace transform, we have

$y(t)=\dfrac{1}{8}-\dfrac{1}{8}e^{-2t}cos2t+\dfrac{3}{8}e^{-2t}sin2t $

       $=\dfrac{1}{8}[1-e^{-2t}(cos2t-3sin2t)]$                   Answer