Find the Laplace transform of \(H(t) = e^{-t} + 2te^{-t}\).
DSSSB AE - 2023,
Answer: C
Use the standard transforms \(\mathcal{L}\{e^{-t}\} = \dfrac{1}{s + 1}\) and \(\mathcal{L}\{te^{-t}\} = \dfrac{1}{(s + 1)^2}\).
\[ \begin{aligned} H(s) = \frac{1}{s + 1} + \frac{2}{(s + 1)^2} \\ &= \frac{(s + 1) + 2}{(s + 1)^2} \\ &= \frac{s + 3}{s^2 + 2s + 1} \end{aligned} \]
That is the third option. The denominator \(1 + 2s + s^2\) is \((s + 1)^2\), so the options with \(2 + s + s^2\) are wrong, and the last option, with numerator 3, lacks the \(s\) from the first term. The answer is option C.
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