$(D^2+2D+5)y=e^{-t} \sin t$, when  $y(0)=0, y'(0)=1 $

Taking Laplace on both sides.

$L [(D^2+2D+5)y]=L[e^{-t} \sin t]$

$L[y'']+2L[y']+5L[y]= \dfrac {1} {(S+1)^2+1}$

$[S^2\overline y-Sy(0)-y'(0)]+2[S\overline y-y(0)]+5\overline y= \dfrac {1} {(S+1)^2+1}$

$S^2\overline y-0-1+2S\overline y-0+5\overline y= \dfrac {1} {(S+1)^2+1}$

$S^2\overline y+2S\overline y+5\overline y= \dfrac {1} {(S^2+2S+1+1)}+1$

$\overline y(S^2+2S+5)= \dfrac {1+S^2+2S+2} {(S^2+2S+2)}$

$\overline y= \dfrac {S^2+2S+3} {(S^2+2S+2)(S^2+2S+5)}$

Taking Laplace Inverse on both sides.

$L^{-1}[\overline y]=L^{-1} \Bigg[\dfrac {S^2+2S+3} {(S^2+2S+2)(S^2+2S+5)}\Bigg]$

$y=L^{-1} \Bigg[\dfrac {(S+1)^2+2} {[(S+1)^2+1][(S+1)^2+4]}\Bigg]$

$y=e^{-t}L^{-1} \Bigg[\dfrac {S^2+2} {[S^2+1][S^2+4]}\Bigg] \cdots(Equation \ 1)$

Now using Partial Fraction Method

Let $S^2 =t$

$\therefore \dfrac{(t+2)} {(t+1) (t+4)}=\dfrac{A} {t+1}+\dfrac {B} {t+4} \cdots (Equation \ 2)$

put $t=-1$

$\therefore -1+2=A(-1+4)+B(-1+1)$

$\therefore 1=A(3)+B(0)$

$\therefore A=\dfrac {1} {3}$

$t=-4$

$\therefore -4+2=A(-4+4)+B(-4+1)$

$\therefore -2=A(0)+B(-3)$

$\therefore B=\dfrac {-2} {-3}$

$\therefore B=\dfrac {2} {3}$

put values of t,A&B in equation (2)

$\therefore \dfrac{(S^2+2)} {(S^2+1) (S^2+4)}=\dfrac{\dfrac {1} {3}} {(S^2+1)}+\dfrac {\dfrac {2}{3}} {(S^2+4)}$

Applying Laplace Inverse to both the sides

$\therefore L^{-1} \Bigg[ \dfrac{(S^2+2)} {(S^2+1) (S^2+4)} \Bigg]= L^{-1} \Bigg[\dfrac{\dfrac {1} {3}} {(S^2+1)}\Bigg]+L^{-1}\Bigg[\dfrac {\dfrac {2}{3}} {(S^2+4)} \Bigg]$

$\therefore L^{-1} \Bigg[ \dfrac{(S^2+2)} {(S^2+1) (S^2+4)} \Bigg]= \dfrac {1} {3}L^{-1} \Bigg[\dfrac{1} {(S^2+1)}\Bigg]+\dfrac {2}{3}L^{-1}\Bigg[\dfrac {1} {(S^2+4)} \Bigg]$

$\therefore L^{-1} \Bigg[ \dfrac{(S^2+2)} {(S^2+1) (S^2+4)} \Bigg]= \dfrac {1} {3}\sin t+\dfrac {2}{3}\times\dfrac {1}{2} \sin 2t$

$\therefore L^{-1} \Bigg[ \dfrac{(S^2+2)} {(S^2+1) (S^2+4)} \Bigg]= \dfrac {1} {3}\sin t+\dfrac {1}{3} \sin 2t$

put this value in equation (1)

$y=e^{-t}\Bigg( \dfrac {1} {3}\sin t+\dfrac {1}{3} \sin 2t\Bigg)$

$y=\dfrac {e^{-t}}{3}\Bigg( \sin t+ \sin 2t\Bigg)$