Find the inverse laplace transform of frac{6s+9}{s^2+17} s>0 y(t)=dots

Step 1
Given the Laplace transform of a function y(t) such that
$L\{y(t)\}=\frac{6s+9}{s^2+17}$
Find f(t) using inverse Laplace transform.
Step 2
First, separate the numerator.
$\frac{6s+9}{(s^2+17)}=6\frac{s}{(s^2+17)}+9\frac{1}{(s^2+17)}$
Since $L^{-1}\left\{\frac{a}{(s^2+a^2)}\right\}=\sin (at)$ , multiply the numerator and denominator on the second sum by $\sqrt{17}$
$\frac{6s+9}{(s^2+17)}=6\frac{s}{(s^2+(\sqrt{17}{)}^2}+\frac{9}{(\sqrt{17})}\cdot \frac{\sqrt{17}}{s^2+(\sqrt{17}{)}^2}$
Step 3
Now, take inverse Laplace transform on both sides.
$L^{-}1\left\{\frac{6s+9}{(s^2+17)}\right\}=6L^{-1}\left\{\frac{s}{s^2+(\sqrt{17}{)}^2}\right\}+\frac{9}{\sqrt{17}}{L}^{-}1\left\{\frac{\sqrt{17}}{s^2+(\sqrt{17}{)}^2}\right\}$
Since $L^{-}1\left\{\frac{a}{(s^2+a^2)}\right\}=\sin (at)\text{ and }{L}^{-}1\{s/(s^2+a^2)\}=\cos (at)$
$L^{-}1\left\{\frac{6s+9}{(s^2+17)}\right\}=6\cos (\sqrt{17t})+9/(\sqrt{17})sin(\sqrt{17t})$
Therefore, $y(t)=6\cos (\sqrt{17t})+\frac{9}{(\sqrt{17})}\sin (\sqrt{17t})$