Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y"+5y=2t^2-9 , y(0)=0 , y'(0)=-3

$y"+5y=2t^2-9$
$y(0)=0,y^{\prime }(0)=-3$
$L\left\{y"\right\}+sL\left\{y^{\prime }\right\}=L\{2t^2-9\}$
$s^{2}Y(s)-sy(0)-y^{\prime }(0)+5Y(s)=L\left\{2t^2\right\}-L\left\{9\right\}$
${\displaystyle =\frac{4}{s^3}-\frac{9}{s}}$
${\displaystyle \Rightarrow s^{2}Y\left(s\right)-sy\left(0\right)+3+5Y\left(s\right)=\frac{4}{s^3}-\frac{9}{s}}$
${\displaystyle \Rightarrow Y\left(s\right)[s^2+5]=\frac{4}{s^3}-\frac{9}{s}-3}$
${\displaystyle \Rightarrow Y\left(s\right)=\frac{\frac{4}{s^3}-\frac{9}{s}-3}{s^2+5}=\frac{4-9s^2-3s^3}{s^3(s^2+5)}}$
$\Rightarrow y(t)=L^{-1}\left\{\frac{4-9s^2-3s^3}{s^3(s^2+5)}\right\}$
${\displaystyle =\frac{2t^2}{5}-\frac{3\sqrt{5}\sin \left(\sqrt{5}t\right)}{5}+\frac{49\cos \left(\sqrt{5}t\right)}{25}-\frac{49}{25}}$
$[\because L^{-1}\left\{\frac{4-9s^2-3s^3}{s^3(s^2+5)}\right\}=\frac{4}{5s^3}-\frac{15}{s(s^2+25)}+\frac{49s}{25(25+5^2)}-\frac{49}{25s^2}]$
${\displaystyle \therefore \frac{2t^2}{5}-\frac{49}{25}-\frac{3\sqrt{5}}{25}\sin \left(\sqrt{5}t\right)+\frac{49}{25}\cos \left(\sqrt{5}t\right)}$