# Find the inverse Laplace transform of (any two) i) frac{(s^2+3)}{s(s^2+9)} ii) logleft(frac{(s+1)}{(s-1)}right)

Joni Kenny 2021-02-20 Answered
Find the inverse Laplace transform of (any two)
i) $\frac{\left({s}^{2}+3\right)}{s\left({s}^{2}+9\right)}$
ii) $\mathrm{log}\left(\frac{\left(s+1\right)}{\left(s-1\right)}\right)$
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Clelioo
(i) The given Laplace transformation is,$\frac{\left({s}^{2}+3\right)}{s\left({s}^{2}+9\right)}$
We can write it,
$\frac{{s}^{2}+3}{s\left({s}^{2}+9\right)}=\frac{{s}^{2}}{s\left({s}^{2}+9\right)}+\frac{3}{s\left({s}^{2}+9\right)}$
$=\frac{s}{\left({s}^{2}+9\right)+\frac{{s}^{2}-\left({s}^{2}+9\right)}{s\left({s}^{2}+9\right)\right)}\left(\frac{3}{-9}\right)}$
$=\frac{s}{\left({s}^{2}+9\right)+\frac{1}{3}\cdot \frac{s}{{s}^{2}+9}+\frac{1}{3}\cdot \frac{1}{s}}$
$=\frac{2}{3}\cdot \frac{s}{{s}^{2}+9}+\frac{1}{3}\cdot \frac{1}{s}$
$=\frac{2}{3}\cdot \frac{s}{{s}^{2}+{3}^{2}}+\frac{1}{3}\cdot \frac{1}{s}$
Now taking inverse Laplace of both sides we have,
${L}^{-1}\left[\frac{{s}^{2}+3}{s\left({s}^{2}+9\right)}\right]=\frac{2}{3}{L}^{-1}\left[\frac{s}{\left({s}^{2}+{3}^{2}\right)}\right]+\frac{1}{3}{L}^{-1}\left[\frac{1}{s}\right]$
$=\frac{2}{3}\mathrm{cos}3t+\frac{1}{3}$
ANSWER:${L}^{-1}\left[\frac{{s}^{2}+3}{s\left({s}^{2}+9\right)}\right]=\frac{2}{3}\mathrm{cos}3t+\frac{1}{3}$
Step 3
(ii)
The given Laplace transformation is,
$\mathrm{log}\left(\frac{s+1}{s-1}\right)$
We can write it,
$\mathrm{log}\left(\frac{s+1}{s-1}\right)=\mathrm{log}\left(s+1\right)-\mathrm{log}\left(s-1\right)$
Now,the property of Laplace transformation is,
$tf\left(t\right)↔-\frac{d}{ds}\left[F\left(s\right)\right]$

$tf\left(t\right)↔-\frac{d}{ds}\left[\mathrm{log}\left(s+1\right)-\mathrm{log}\left(s-1\right)\right]$
$⇒tf\left(t\right)↔-\frac{1}{s+1}+\frac{1}{s-1}$
$⇒tf\left(t\right)↔-{e}^{-t}+{e}^{t}$
$⇒f\left(t\right)↔\frac{{e}^{t}-{e}^{-t}}{t}=2\left(\frac{{e}^{t}-{e}^{-t}}{2t}\right)=2\left(\frac{\mathrm{sin}ht}{t}\right)$
Therefore,
${L}^{-1}\left[\mathrm{log}\frac{s+1}{s-1}\right]=2\left(\frac{\mathrm{sin}ht}{t}\right)$
Answer: ${L}^{-1}\left[\mathrm{log}\frac{s+1}{s-1}\right]=2\left(\frac{\mathrm{sin}ht}{t}\right)$