# 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.) Lleft{t^2 cdot te^tright}

Efan Halliday 2020-12-05 Answered
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.)
$L\left\{{t}^{2}\cdot t{e}^{t}\right\}$
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Mayme
Step 1
Consider the provided question,
We have to find $L\left\{{t}^{2}\cdot t{e}^{t}\right\}$
We have to use the convolution theorem:
$L\left\{f\left(t\right)\cdot g\left(t\right)\right\}=L\left\{f\left(t\right)\right\}\cdot L\left\{g\left(t\right)\right\}$
Step 2
Now, the given Laplace transform is find as,
$L\left\{{t}^{2}\cdot t{e}^{t}\right\}=L\left\{{t}^{2}\right\}\cdot L\left\{t{e}^{t}\right\}$
$=\frac{2}{{s}^{3}}\cdot \frac{1}{\left(s-1{\right)}^{2}}$
$\left(L\left\{{t}^{n}\right\}=\frac{n!}{\left({s}^{n+1}\right)},\left(L\left\{{t}^{k}f\left(t\right)=\left(-1{\right)}^{k}\frac{{d}^{k}}{d{s}^{k}}\left(L\left(f\left(t\right)\right)\right)\right\}\right)$
$=\frac{2}{{s}^{3}\left(s-1{\right)}^{2}}$