Deragz

2021-12-10

I have a Laplace Transform problem which Im

MoxboasteBots5h

The definition of the Laplace transform is $L\left(f\right)\left(s\right)\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dtask\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{sinh}thefollow\in gholds:PSK\mathrm{sin}hx=\frac{1}{2}\left({e}^{x}-{e}^{-x}\right)$. Now you put these two things together and compute
$\frac{1}{2}{\int }_{0}^{\mathrm{\infty }}x{e}^{4x}{e}^{-xs}dx-\frac{1}{2}{\int }_{0}^{\mathrm{\infty }}x{e}^{-4x}{e}^{-xs}dx$
Hope this helps

levurdondishav4

You got $t\mathrm{sin}h4t=t\frac{{e}^{4t}-{e}^{-4t}}{2}$ ,so:
$L\left\{t\mathrm{sin}h4t\right\}=\frac{1}{2}\left(L\left\{t{e}^{4t}\right\}+L\left\{t{e}^{-4t}\right\}\right)$
We know that $L\left\{{e}^{at}f\left(t\right)\right\}=F\left(s-a\right)$. Furthermore $L\left\{t\right\}=\frac{1}{{s}^{2}}$.So in our case $L\left\{t{e}^{4t}\right\}=F\left(s-4\right)=\frac{1}{{\left(s-4\right)}^{2}}$. Finally:
$L\left\{t\mathrm{sin}h4t\right\}=\frac{0.5}{{\left(s-4\right)}^{2}}+\frac{0.5}{{\left(s+4\right)}^{2}}$

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