# find the inverse Laplace transform of the given function. {F}{left({s}right)}=frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}

find the inverse Laplace transform of the given function.
$F\left(s\right)=\frac{2s-3}{{s}^{2}-4}$
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Step 1
Given function is:
$F\left(s\right)=\frac{2s-3}{{s}^{2}-4}$
Since the Laplace transform of:
$L\left(\mathrm{sinh}at\right)=\frac{a}{{s}^{2}-{a}^{2}}$
And
$L\left(\mathrm{cosh}at\right)=\frac{s}{{s}^{2}-{a}^{2}}$
Step 2
Now,
Using the above transform:
$F\left(s\right)=\frac{2s-3}{{s}^{2}-4}$
$=\frac{2s}{{s}^{2}-4}-\frac{3}{{s}^{2}-4}$
$=2\cdot \frac{s}{{s}^{2}-{2}^{2}}-\frac{3}{2}\cdot \frac{2}{{s}^{2}-{2}^{2}}$
$=2L\left(\mathrm{cosh}2t\right)-\frac{3}{2}L\left(\mathrm{sinh}2t\right)$
$=L\left(2\mathrm{cosh}2t-\frac{3}{2}\mathrm{sinh}2t\right)$
inverse Laplace transform of $L\left(2\mathrm{cosh}2t-\frac{3}{2}\mathrm{sinh}2t\right)$:
${L}^{-1}\left(L\left(2\mathrm{cosh}2t-\frac{3}{2}\mathrm{sinh}2t\right)\right)=2\mathrm{cosh}2t-\frac{3}{2}\mathrm{sinh}2t$
THEREFORE Inverse Laplace transform of given function is $2\mathrm{cosh}2t-\frac{3}{2}\mathrm{sinh}2t$