# Find the inverse laplace trans. displaystyle{F}{left({s}right)}=frac{10}{{{s}{left({s}^{2}+{9}right)}}}

Find the inverse laplace trans.
$F\left(s\right)=\frac{10}{s\left({s}^{2}+9\right)}$
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Step 1
Given,
$F\left(s\right)=\frac{10}{s\left({s}^{2}+9\right)}$
Find the inverse Laplace transform of this function.
Step 2
$F\left(s\right)=\frac{10}{s\left({s}^{2}+9\right)}$
$=\frac{10}{9s}-\frac{10s}{9\left({s}^{2}+9\right)}$
Taking inverse Laplace transform of both sides,
${L}^{-1}\left[F\left(s\right)\right]={L}^{-1}\left[\frac{10}{9s}-\frac{10s}{9\left({s}^{2}+9\right)}\right]$
Then, $f\left(t\right)={L}^{-1}\left[\frac{10}{9s}\right]-{L}^{-1}\left[\frac{10s}{9\left({s}^{2}+9\right)}\right]$
$=\frac{10}{9}{L}^{-1}\left[\frac{1}{s}\right]-\frac{10}{9}{L}^{-1}\left[\frac{s}{{s}^{2}+9}\right]$
$=\frac{10}{9}{L}^{-1}\left[\frac{1}{s}\right]-\frac{10}{9}\cdot {L}^{-1}\left[\frac{s}{{s}^{2}+{3}^{2}}\right]$
Step 3
Use the formula such that
${L}^{-1}\left[\frac{1}{s}\right]=t$
${L}^{-1}\left[\frac{s}{{s}^{2}+{a}^{2}}\right]=\mathrm{cos}\left(at\right)$
$f\left(t\right)=\frac{10}{9}\cdot t-\frac{10}{9}\cdot \mathrm{cos}\left(3t\right)$
$=\frac{10}{9}\left[t-\mathrm{cos}\left(3t\right)\right]$
Step 4
Hence,
$f\left(t\right)=\frac{10}{9}\left[t-\mathrm{cos}\left(3t\right)\right]$