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QUESTION

Find the Laplace transform of the function $te^{-2t}$.

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ANSWER

$\ds{\cal{L}}\{t\}=\frac{1}{s^2},$ therefore
$\ds{\cal{L}}\{te^{-2t}\}=\left.\frac{1}{s^2}\right|_{s\to
s-(-2)}=\frac{1}{(s+2)^2}$




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