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The unilateral Laplace transform of $f(t)$ is $\frac{1}{s^2+s+1}$. The unilateral Laplace transform of $t\,f(t)$ is

1. $-\frac{s}{(s^2+s+1)^2}$
2. $-\frac{2s+1}{(s^2+s+1)^2}$
3. $\frac{s}{(s^2+s+1)^2}$
4. $\frac{2s+1}{(s^2+s+1)^2}$

Correct answer: $-\frac{2s+1}{(s^2+s+1)^2}$

Solution

For unilateral Laplace transforms, $\mathcal{L}\{t f(t)\}=-\frac{d}{ds}F(s)$. Here $F(s)=\frac{1}{s^2+s+1}$, so differentiating gives $-\frac{d}{ds}F(s)=\frac{2s+1}{(s^2+s+1)^2}$ with a negative sign from the property, resulting in $-\frac{2s+1}{(s^2+s+1)^2}$.

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