# The Laplace transform of the function {left({2}{t}-{3}right)}{e}^{{frac{{{t}+{2}}}{{3}}}} is equal to: a) {e}^{{frac{2}{{3}}}}{left(frac{2}{{left({s}-

The Laplace transform of the function $\left(2t-3\right){e}^{\frac{t+2}{3}}$ is equal to:
a) ${e}^{\frac{2}{3}}\left(\frac{2}{{\left(s-\frac{1}{3}\right)}^{2}}-\frac{3}{s-\frac{1}{3}}\right)$
b) ${e}^{\frac{2}{3}}\left(\frac{2}{{\left(s-\frac{1}{3}\right)}^{2}}\cdot \frac{3}{s-\frac{1}{3}}\right)$
c) ${e}^{\frac{2}{3}}\left(\frac{6}{{\left(s-\frac{1}{3}\right)}^{2}}\right)$
d) $\left(\frac{2}{{\left(s-\frac{1}{3}\right)}^{2}+\frac{2}{3}}\right)$
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

hosentak
Step 1
Given that
$\left(2t-3\right){e}^{\frac{t+2}{3}}$
To find the Laplace transformation of given function
Step 2
Given
$\left(2t-3\right){e}^{\frac{t+2}{3}}$
$=2t{e}^{\frac{t+2}{3}}-3{e}^{\frac{t+2}{3}}$
Apply Laplace transform on both side
$L\left\{\left(2t-3\right){e}^{\frac{t+2}{3}}\right\}=L\left\{2t{e}^{\frac{t+2}{3}}-3{e}^{\frac{t+2}{3}}\right\}$
$=L\left\{2t{e}^{\frac{t+2}{3}}\right\}-L\left\{3{e}^{\frac{t+2}{3}}\right\}$
$=2L\left\{t{e}^{\frac{t}{3}}\cdot {e}^{\frac{2}{3}}\right\}-L\left\{3{e}^{\frac{t}{3}}\cdot {e}^{\frac{2}{3}}\right\}$
$=2{e}^{\frac{2}{3}}L\left\{t{e}^{\frac{t}{3}}\right\}-3{e}^{\frac{2}{3}}L\left\{{e}^{\frac{t}{3}}\right\}$
$=2{e}^{\frac{2}{3}}\frac{1}{{\left(s-\frac{1}{3}\right)}^{2}}-3{e}^{\frac{2}{3}}\frac{1}{s-\frac{1}{3}}$

$={e}^{\frac{2}{3}}\left[\frac{2}{{\left(s-\frac{1}{3}\right)}^{2}}-\frac{3}{s-\frac{1}{3}}\right]$
$L\left\{\left(2t-3\right){e}^{\frac{t+2}{3}}\right\}={e}^{\frac{2}{3}}\left[\frac{2}{{\left(s-\frac{1}{3}\right)}^{2}}-\frac{3}{s-\frac{1}{3}}\right]$