 # Find the Laplace transforms of the following L {e^((-6t) sinh(2t))/(e^(-6t) cosh(2t))} Tahmid Knox 2020-10-27 Answered

Find the Laplace transforms of the following
$L\left\{\frac{{e}^{-6t}\mathrm{sinh}\left(2t\right)}{{e}^{-6t}\mathrm{cosh}\left(2t\right)}\right\}$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it grbavit
Step 1
We have to evaluate the Laplace transformation $L\left\{\frac{{e}^{-6t}\mathrm{sinh}\left(2t\right)}{{e}^{-6t}\mathrm{cosh}\left(2t\right)}\right\}$
Use the property of Laplace transformation
$L\left\{\frac{{e}^{-6t}\mathrm{sinh}\left(2t\right)}{{e}^{-6t}\mathrm{cosh}\left(2t\right)}\right\}=\frac{L\left({e}^{-6t}\mathrm{sinh}\left(2t\right)\right)}{L\left({e}^{-6t}\mathrm{cosh}\left(2t\right)\right)}$
Use the property

Step 2
Hence,
$L\left({e}^{-6t}\mathrm{sinh}\left(2t\right)\right)=\frac{2}{{\left(s-\left(-6\right)\right)}^{2}-{2}^{2}}$
$=\frac{2}{{\left(s+6\right)}^{2}-4}$
Hence, $L\left({e}^{-6t}\mathrm{sinh}\left(2t\right)\right)=\frac{2}{{\left(s+6\right)}^{2}-4}$
And,
$L\left({e}^{-6t}\mathrm{cosh}\left(2t\right)\right)=\frac{s-\left(-6\right)}{{\left(s-\left(-6\right)\right)}^{2}-{2}^{2}}$
$=\frac{s+6}{{\left(s+6\right)}^{2}-4}$
Hence, $L\left(e6\left(-6t\right)\mathrm{cosh}\left(2t\right)\right)=\frac{s+6}{{\left(s+6\right)}^{2}-4}$
Step 3
Hence,$L\left\{\frac{{e}^{-6t}\mathrm{sinh}\left(2t\right)}{{e}^{-6t}\mathrm{cosh}\left(2t\right)}\right\}=\frac{\left(\frac{2}{{\left(s+6\right)}^{2}}-4\right)}{\frac{s+6}{{\left(s+6\right)}^{2}}-4}$
$=\frac{2}{s+6}$
Hence, $L\left\{\frac{{e}^{-6t}\mathrm{sinh}\left(2t\right)}{{e}^{-6t}\mathrm{cosh}\left(2t\right)}\right\}=\frac{2}{s+6}$