Find the laplace transform of the following f(t)=tu_2(t) Ans. F(s)=left(frac{1}{s^2}+frac{2}{s}right)e^{-2s}

Question

Find the laplace transform of the following

f (t) = t u_{2} (t)

Ans.

F (s) = (\frac{1}{s^{2}} + \frac{2}{s}) e^{- 2 s}

Answer 1

Step 1
first calculate Laplace transform of $u_{2} (t)$
$F_{u} (s) = \int_{-} \infty^{\infty} u_{2} (t) e^{- s t} d t$
$= \int_{2}^{\infty} e^{- s t} d t$
$= \frac{e^{- 2 s}}{s}$
Step 2
therefore, Laplace transform of $t u_{2} (t)$ is :
$- \frac{d}{d s} F_{u} (s) = - \frac{d}{d s} \cdot \frac{e^{- 2 s}}{s}$
$= - \frac{- 2 s e^{- 2 s} - e^{- 2 s}}{s^{2}}$
$= (\frac{1}{s^{2}} + \frac{2}{s}) e^{- 2 s}$
Therefore, Laplace transform of $t u_{2} (t)$ is
$= (\frac{1}{s^{2}} + \frac{2}{s}) e^{- 2 s)}$

Find the laplace transform of the following f(t)=tu_2(t) Ans. F(s)=left(frac{1}{s^2}+frac{2}{s}right)e^{-2s}

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