inverse laplace transform of

2/(s+2)^2 - s/(s^2-2s+2)

Question

inverse laplace transform of

2/(s+2)^2 - s/(s^2-2s+2)

experimentx · Answer

make use of latex ... it's confusing.

hartnn · Answer

$\large \dfrac{2}{(s+2)^2}-\dfrac{s}{s^2-2s+2}$
right ?

anonymous · Answer

yes

experimentx · Answer

i think you can do the second .. the first one, it is shifted by +2

anonymous · Answer

well in the second one, how do i get rid of the s on top?

experimentx · Answer

when you have cosine  ,, you get s at the top.

anonymous · Answer

ohhh yeah, i was thinking sin and cos both had beta

anonymous · Answer

your right, so the second one will be e^x cos x + e^x sin x

experimentx · Answer

http://www.wolframalpha.com/input/?i=laplace+transform++e%5Ex+cos+x

experimentx · Answer

yeah .. you will get that one. what about the first ...

anonymous · Answer

i need to take the 2 out so its would be 1/(s+2)^2, so then it would be 2xe^-2x ??

experimentx · Answer

looking at the table ... that seems okay too .

anonymous · Answer

well back to the second on, because i might of confused myself....

s/s^2-2s+2 is 

s/(s-1)^2 +1 then is becomes

s-1+1/(s-1)^2+1

and finally s-1/(s-1)^2+1 + 1/(s-1)^2+1

okay nevermind, it made sense when i typed it in... thank you again.

experimentx · Answer

yw