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UnkleRhaukus
Group Title
\[
\mathcal L ^{1}\left\{\frac{e^{\pi p}}{p^2+1}\right\}\]
 2 years ago
 2 years ago
UnkleRhaukus Group Title
\[ \mathcal L ^{1}\left\{\frac{e^{\pi p}}{p^2+1}\right\}\]
 2 years ago
 2 years ago

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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[ \begin{align*} \mathcal L ^{1}\left\{\frac{e^{\pi p}}{p^2+1}\right\} &=h(t\pi)\times\mathcal L^{1} \left\{\frac1{p^2+1}\right\}_{t\rightarrow t\pi}\\ &=h(t\pi)\times\sin(t)\Big_{t\rightarrow t\pi}\\ &=h(t\pi)\times\sin(t\pi)\\ &=h(t\pi)\sin(t)\\ \end{align*}\]
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
@AccessDenied
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
have i got this one right/?
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
This appears correct to me. :)
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
thanks!, would you more likely leave the answer in the final form of \[=h(t−π)\sin(t−π)\]or \[=−h(t−π)\sin(t)\]
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.1
I would probably use h(tpi) sin(t) just because I prefer to see the simplest forms of functions..
 2 years ago
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