Exercise · ZW-INVERSE-LAPLACE-TRANSFORM-005 · ★★☆ Intermediate
What is \(\mathcal L^{-1}\left\{\dfrac{1}{s(s+1)}\right\}\)?
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- C. \(1-e^{-t}\)
Explanation
By partial fractions, \(\frac1{s(s+1)}=\frac1s-\frac1{s+1}\), whose inverse is \(1-e^{-t}\).
Review the concept: Inverse Laplace transform
The inverse transform \(\mathcal L^{-1}\) of a function \(F(s)\) is the function \(f(t)\), defined for \(t\ge0\), whose Laplace transform is \(F(s)\).
\[f(t)=\mathcal L^{-1}\{F(s)\}\iff \mathcal L\{f\}=F\]
More practice problems: inverse laplace transform
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- ZW-INVERSE-LAPLACE-TRANSFORM-008 What is mathcal L^-1\(s+2)/((s+1)(s+3))\?
- ZW-INVERSE-LAPLACE-TRANSFORM-009 The solution of y''-3y'+2y=0, y(0)=1, y'(0)=0 has Y(s)=(s-3)/((s-1)(s-2)). What is y(t)?
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