# Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.) {L}{leftlbrace{e}^{{-{t}}}cdot{e}^{t} cos{{left({t}right)}}rightrbrace}

Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
$L\left\{{e}^{-t}\cdot {e}^{t}\mathrm{cos}\left(t\right)\right\}$
You can still ask an expert for help

## Want to know more about Laplace transform?

• 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

Step 1
Consider the provided question,
We have to find $L\left\{{e}^{-t}\cdot {e}^{t}\mathrm{cos}\left(t\right)\right\}$ Step 2
Now, the given Laplace transform is find as,
$L\left\{{e}^{-t}\cdot {e}^{t}\mathrm{cos}\left(t\right)\right\}=L\left\{{e}^{-t}\right\}+L\left\{{e}^{t}\mathrm{cos}\left(t\right)\right\}$
$=\left(\frac{1}{s+1}\right)\cdot \frac{s-1}{{\left(s-1\right)}^{2}+{1}^{2}}\left[Use,L\left({e}^{at}\right)=\frac{1}{s-a}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}L\left({e}^{at}\mathrm{cos}bt\right)=\frac{s-a}{{\left(s-a\right)}^{2}+{b}^{2}}\right]$
$=\frac{s-1}{\left(s+1\right)\left[{\left(s-1\right)}^{2}+1\right]}$
Thus , $L\left\{{e}^{-t}\cdot {e}^{t}\mathrm{cos}\left(t\right)\right\}=\frac{s-1}{\left(s+1\right)\left[{\left(s-1\right)}^{2}+1\right]}$