Exercise · ZW-INVERSE-LAPLACE-TRANSFORM-007 · ★★☆ Intermediate
What is \(\mathcal L^{-1}\left\{\dfrac{1}{s^2+2s+5}\right\}\)?
Choose one option.
Show answer
- B. \(\tfrac12e^{-t}\sin 2t\)
Explanation
Complete the square: \(s^2+2s+5=(s+1)^2+4\). Then \(\frac1{(s+1)^2+4}=\frac12\cdot\frac2{(s+1)^2+2^2}\), with inverse \(\frac12e^{-t}\sin2t\).
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-003 What is mathcal L^-1\(s)/(s^2+9)\?
- ZW-INVERSE-LAPLACE-TRANSFORM-004 What is mathcal L^-1\(6)/(s^4)\?
- ZW-INVERSE-LAPLACE-TRANSFORM-005 What is mathcal L^-1\(1)/(s(s+1))\?
- ZW-INVERSE-LAPLACE-TRANSFORM-006 What is mathcal L^-1\(2s+3)/(s^2+4)\?
- 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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