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

Find the inverse laplace transform of
$\frac{6s+9}{{s}^{2}+17}s>0$
$y\left(t\right)=\dots$
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Step 1
Given the Laplace transform of a function y(t) such that
$L\left\{y\left(t\right)\right\}=\frac{6s+9}{{s}^{2}+17}$
Find f(t) using inverse Laplace transform.
Step 2
First, separate the numerator.
$\frac{6s+9}{\left({s}^{2}+17\right)}=6\frac{s}{\left({s}^{2}+17\right)}+9\frac{1}{\left({s}^{2}+17\right)}$
Since ${L}^{-1}\left\{\frac{a}{\left({s}^{2}+{a}^{2}\right)}\right\}=\mathrm{sin}\left(at\right)$ , multiply the numerator and denominator on the second sum by $\sqrt{17}$
$\frac{6s+9}{\left({s}^{2}+17\right)}=6\frac{s}{\left({s}^{2}+\left(\sqrt{17}{\right)}^{2}}+\frac{9}{\left(\sqrt{17}\right)}\cdot \frac{\sqrt{17}}{{s}^{2}+\left(\sqrt{17}{\right)}^{2}}$
Step 3
Now, take inverse Laplace transform on both sides.
${L}^{-}1\left\{\frac{6s+9}{\left({s}^{2}+17\right)}\right\}=6{L}^{-1}\left\{\frac{s}{{s}^{2}+\left(\sqrt{17}{\right)}^{2}}\right\}+\frac{9}{\sqrt{17}}{L}^{-}1\left\{\frac{\sqrt{17}}{{s}^{2}+\left(\sqrt{17}{\right)}^{2}}\right\}$
Since
${L}^{-}1\left\{\frac{6s+9}{\left({s}^{2}+17\right)}\right\}=6\mathrm{cos}\left(\sqrt{17t}\right)+9/\left(\sqrt{17}\right)sin\left(\sqrt{17t}\right)$
Therefore, $y\left(t\right)=6\mathrm{cos}\left(\sqrt{17t}\right)+\frac{9}{\left(\sqrt{17}\right)}\mathrm{sin}\left(\sqrt{17t}\right)$