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

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

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

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

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