solve the initial value problem with using laplace Y'+3Y=6e^{-3t}cos 6t+24 Y(0)=0

Step 1
The given IVP is as follows.
$Y^{\prime }+3Y=6e^{-3t}\cos 6t+24$
$Y(0)=0$
Apply Laplace transform on both sides of the given differential equation as follows.
$L{Y}^{\prime }+3Y=L6e^{-3t}\cos 6t+24$
$L{Y}^{\prime }+3Y=L6e^{-3t}\cos 6t+L24$
$sLY-Y(0)+3LY=6L{e}^{-3t}\cos 6t+L24$
$(s+3)LY=6\frac{s+3}{(s+3{)}^2+6^2}+\frac{24}{s}$
$LY=\frac{6}{(s+3{)}^2+6^2}+\frac{24}{s(s+3)}$
$LY=\frac{6}{(s+3{)}^2+6^2}+\frac{8}{s}-\frac{8}{(s+3)}$
Step 2
Apply inverse Laplace transform on both sides as follows.
$L^{-1}\left\{L\left\{Y\right\}\right\}=L^{-1}\{\frac{6}{(s+3{)}^2+6^2}+\frac{8}{s}-\frac{8}{(s+3)}\}$
$Y(t)=L^{-1}\left\{\frac{6}{(s+3{)}^2+6^2}\right\}+8L^{-1}\left\{\frac{1}{s}\right\}-8L^{-1}\left\{\frac{1}{(s+3)}\right\}$
$Y(t)=e^{-3t}\sin 6t+8-8L^{-1}\left\{\frac{1}{(s+3)}\right\}$