Question

find the inverse Laplace transform of the given function

(2s+1)/(s^2-2s+1)

Answer 1

Can you do partial fraction decomposition?

Answer 2

cant you just break it up

Answer 3

s^2 - 2s + 1 = (s-1)(s-1)

Answer 4

then use the table of laplaces

Answer 5

yeah, I got to the (s-1) part and was planning to do the partial derivatives.. but wasn't too sure how to approach that since it's same root (s-1)^2... would I just do A/(s-1) + B/(s-1)=2s+1?

Answer 6

I think this is on a laplace transform table.

Answer 7

but don't you have to find the coefficients first?

Answer 8

e^t (3 t+2) would be the closest u can get

Answer 9

how'd you get that?

Answer 10

I'm confused.. how'd you get that...

Answer 11

\[\frac{2s+1}{s^2-2s+1} = \frac{2}{s-1} + \frac{3}{(s-1)^2} = 2\frac{1}{s-1} + 3\frac{1}{(s-1)^2}\]
\[=>2e^{t} + 3e^{t}t\]