alexmahone
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Find the inverse Laplace transform of $\displaystyle \frac{2s+7-e^{-2s}}{(s+1)^2}$.
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The inverse Laplace transform of the expression $\displaystyle \frac{2s+7-e^{-2s}}{(s+1)^2}$ can be computed by separating the terms into $\displaystyle \frac{2s+7}{(s+1)^2}$ and $\displaystyle -\frac{e^{-2s}}{(s+1)^2}$. The first term can be solved using standard inverse Laplace techniques, while the second term utilizes the shifting theorem for Laplace transforms. The relevant formula for the second piece is $\displaystyle \mathfrak{L}^{-1}\left[\frac{e^{-\tau s}}{(s+\alpha)^{n+1}}\right]$, which applies to the exponential decay component.
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Alexmahone said:Find the inverse Laplace transform of $\displaystyle\frac{2s+5-e^{-2s}}{s^2+s+1}$.
dwsmith said:$s^2 + s + 1/4 + 1 - 1/4 = (s + 1/2)^2 + 3/4$
Then break up the numerator.
Alexmahone said:I changed the question. (Sorry about that.)