# Find the inverse Laplace transform of the given function by using the convolution theorem. {F}{left({s}right)}=frac{s}{{{left({s}+{1}right)}{left({s}^{2}+{4}right)}}}

Find the inverse Laplace transform of the given function by using the convolution theorem. $F\left(s\right)=\frac{s}{\left(s+1\right)\left({s}^{2}+4\right)}$
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Step 1
the given function is $F\left(s\right)=\frac{s}{\left(s+1\right)\left({s}^{2}+4\right)}$
Let
Since
step 2
That implies,
${L}^{-1}\left\{G\left(s\right)\right\}={L}^{-1}\left\{\frac{1}{s+1}\right\}={e}^{-t}$ and
$L\left\{H\left(s\right)\right\}={L}^{-1}\left\{\frac{1}{{s}^{2}+4}\right\}=\mathrm{cos}2t$

Now, $F\left(s\right)=\frac{s}{\left(s+1\right)\left({s}^{2}+4\right)}$
Step 3
By using the convolution theorem,
If  both exists for $s>a\ge 0$ ,then
$H\left(s\right)=F\left(s\right)G\left(s\right)$
$=L\left\{h\left(t\right)\right\},s>a$
Where $h\left(t\right)={\int }_{0}^{t}f\left(t-\tau \right)g\left(\tau \right)d\tau ={\int }_{0}^{t}f\left(\tau \right)g\left(t-\tau \right)d\tau$
Step 4
That implies,
$F\left(s\right)=G\left(s\right)H\left(s\right)$
$=\frac{s}{\left(s+1\right)\left({s}^{2}+4\right)}$
$=L\left\{f\left(t\right)\right\}$
$=L\left\{{\int }_{0}^{t}g\left(t-\tau \right)h\left(\tau \right)d\tau \right\}$
Thus , $F\left(s\right)=L\left\{{\int }_{0}^{t}g\left(t-\tau \right)h\left(\tau \right)d\tau \right\}$
Step 5
Taking the inverse Laplace transform,
${L}^{-1}\left\{F\left(s\right)\right\}={\int }_{0}^{t}g\left(t-\tau \right)h\left(\tau \right)d\tau$
$={\int }_{0}^{t}{e}^{-\left(t-\tau \right)}\mathrm{cos}2\tau d\tau$
Therefore ,