find the inverse Laplace transform of the given function

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

Question

find the inverse Laplace transform of the given function

(2s+1)/(s^2-2s+1)

kainui · Answer

Can you do partial fraction decomposition?

mimi_x3 · Answer

cant you just break it up

mimi_x3 · Answer

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

mimi_x3 · Answer

then use the table of laplaces

babalooo · Answer

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?

kainui · Answer

I think this is on a laplace transform table.

babalooo · Answer

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

anonymous · Answer

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

babalooo · Answer

how'd you get that?

babalooo · Answer

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

mimi_x3 · Answer

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