Exercise · ZW-INVERSE-LAPLACE-TRANSFORM-002 · ★☆☆ Basic
What is \(\mathcal L^{-1}\left\{\dfrac{5}{s^2+25}\right\}\)?
Choose one option.
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- B. \(\sin 5t\)
Explanation
Since \(\mathcal L\{\sin bt\}=\frac{b}{s^2+b^2}\) and \(b=5\), the numerator is already right: \(\sin 5t\).
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-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-007 What is mathcal L^-1\(1)/(s^2+2s+5)\?
- 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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