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

2020-11-01
Step 1
first calculate Laplace transform of $$u_2(t)$$
$$F_u(s) =\int_-\infty^\infty u_2(t)e^{-st}dt$$
$$=\int_2^\infty e^{-st}dt$$
$$=\frac{e^{-2s}}{s}$$
Step 2
therefore, Laplace transform of $$tu_2(t)$$ is :
-\frac{d}{ds}F_u(s)=-\frac{d}{ds}\cdot \frac{e^{-2s}}{s}\)
$$=\frac-{-2se^{-2s}-e^{-2s}}/{s^2}$$
$$=(\frac{1}{s^2}+\frac{2}{s})e^{-2s}$$
Therefore, Laplace transform of $$tu_2(t)$$ is
$$=(\frac{1}{s^2}+\frac{2}{s})e^{-2s)}$$

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