Laplace transform

[limber zephyr]

hekp

[limber zephyr]

im just trying to simply the function, (second pic is the answer) how do they get that?

[zenith current]

Can you describe your question further?

[limber zephyr]

the indicies

"-st" "t" --> -(s-1)t

[jaunty moss]

well $e^ae^b = e^{a+b}$ right

[tacit igloo]

esca | এস্কা

[limber zephyr]

fuck sakei thought it was *

ok im going insane

[jaunty moss]

do you see it now

[limber zephyr]

kinda

[half monolith]

exponent rules e^(-st)*e^t = e^((-st)+t) = e^(t-st) = e^(t(1-s)) = e^(-(s-1)t)

[limber zephyr]

yeah this can be closed now

!close

!help

[acoustic kernel]

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[limber zephyr]

uhh

[half monolith]

.close