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Differential Equations
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Determine the Laplace transform.
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\[\frac{ e ^{-2s}-3e ^{-4s} }{ s+2 }\]
I meant inverse Laplace
\[\mathcal L^{-1}\left\{\frac{ e ^{-2s}-3e ^{-4s} }{ s+2 }\right\}\]
\[=\mathcal L^{-1}\left\{\frac{ e ^{-2s} }{ s+2 }\right\}-\mathcal L^{-1}\left\{\frac{3e ^{-4s} }{ s+2 }\right\}\]
\[=\mathcal L^{-1}\left\{\frac{ e ^{-2s} }{ s+2 }\right\}-3\mathcal L^{-1}\left\{\frac{e ^{-4s} }{ s+2 }\right\}\]
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now use \[\mathcal L\big\{f(t-a)h(t-a)\big\}=e^{as}F(s)\]
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