Exercise · ZW-INVERSE-LAPLACE-TRANSFORM-010 · ★★★ Advanced

What is \(\mathcal L^{-1}\left\{\dfrac{1}{s^2(s+1)}\right\}\)?

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  • D. \(t-1+e^{-t}\)

Explanation

Partial fractions: \(\frac1{s^2(s+1)}=\frac1{s^2}-\frac1s+\frac1{s+1}\), whose inverse is \(t-1+e^{-t}\).

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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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