Solve the following initial value problem by using Laplace transform. y"+4y=4t+12 , y(0)=4 , y'(0)=0

$y^{″}+4y=4t+12,y(0)=4,y^{\prime }(0)=0$
Taking the Laplace transform on both side
$L\left\{y"\right\}+4L\left\{y\right\}=4L\left\{t\right\}+12L\left\{1\right\}$
$s^{2}L\left\{y\right\}-sy(0)-y^{\prime }(0)+4L\left\{y\right\}=4\frac{1}{s^2}+12\frac{1}{s}$
$(s^2+4)L\left\{y\right\}-4s-0=4\left(\frac{3s+1}{s}\right)$
$(s^2+4)L\left\{y\right\}=4(\frac{3s+1}{s}+s)$
$(s^2+4)L\left\{y\right\}=4\left(\frac{3s+1+s^2}{s}\right)$
$L\left\{y\right\}=4\left(\frac{s^2+3s+1}{s(s^2+4)}\right)$
${\displaystyle =4\left(\frac{s^2+4+3s-3}{s(s^2+4)}\right)}$
${\displaystyle =4[\frac{1}{s}+\frac{3s-3}{s(s^2+4)}]}$
${\displaystyle =4[\frac{1}{s}+\frac{1}{4}\left(\frac{12s-12}{s(s^2+4)}\right)]}$
${\displaystyle =4[\frac{1}{s}+\frac{1}{4}\left(\frac{3s^2-3s^2+12s-12}{s(s^2+4)}\right)]}$
${\displaystyle =4\frac{1}{s}+\frac{s(3s+12)-3(s^2+4)}{s(s^2+4)}}$
$L\left\{y\right\}=\frac{4}{s}+\frac{3s+12}{s^2+4}-\frac{3}{s}$
$L\left\{y\right\}=\frac{1}{s}+3\frac{s}{s^2+4}+\frac{12}{s^2+4}$
$y(t)=L^{-1}\left\{\frac{1}{s}\right\}+3L^{-1}\left\{\frac{s}{s^2+4}\right\}+6L^{-1}\left\{\frac{2}{s^2+4}\right\}$
${\displaystyle y\left(t\right)=1+3\cos \left(2t\right)+6\sin \left(2t\right)}$