# Find laplace transform of the following functions. f(t)=t \sin h(2t)

Find laplace transform of the following functions
$f\left(t\right)=t\mathrm{sin}h\left(2t\right)$
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Step 1
The given function is in the form of $f\left(t\right)={t}^{k}g\left(t\right)$. The Laplace transform of that is $L\left[{t}^{k}g\left(t\right)\right]={\left(-1\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}\left(L\left[g\left(t\right)\right]\right)$
Compare the function $f\left(t\right)=t\mathrm{sin}h\left(2t\right)$ with the $f\left(t\right)={t}^{k}g\left(t\right)$ we found,
k=1
$g\left(t\right)=\mathrm{sin}h\left(2t\right)$
Since, $L\left[{t}^{k}g\left(t\right)\right]={\left(-1\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}\left(L\left[g\left(t\right)\right]\right)$ therefore $L\left[t\mathrm{sin}h2t\right]={\left(-1\right)}^{1}\frac{d}{ds}\left(L\left[\mathrm{sin}h2t\right]\right)\stackrel{˙}{s}\left(1\right)$
Step 2
Since $L\left[\mathrm{sin}h\left(at\right)\right]=\frac{a}{{s}^{2}-{a}^{2}}$ , So use this find $L\left[\mathrm{sin}h\left(2t\right)\right]$
$L\left[\mathrm{sin}h\left(2t\right)\right]=\frac{2}{{s}^{2}-{2}^{2}}$
$=\frac{2}{{s}^{2}-4}\stackrel{˙}{s}\left(2\right)$
From (1) and (2) , $L\left[t\mathrm{sin}\left(2t\right)\right]=-\frac{d}{ds}\left[\frac{2}{{s}^{2}-4}\right]$.

Calculate $\frac{d}{ds}\left[\frac{2}{{s}^{2}-4}\right]$
$\frac{d}{ds}\left[\frac{2}{{s}^{2}-4}\right]=\frac{-2}{{\left({s}^{2}-4\right)}^{2}}\cdot 2s$
$=-\frac{4s}{{\left({s}^{2}-4\right)}^{2}}$
Therefore