Find the inverse Laplace transform of (4s-6)/(s^2-2s+2). (By looking at the table, I got 4e^t*cos(t)-6e^t*sin(t) but the answer in the textbook says 4e^t*cos(t)-2e^t*sin(t))

Question

loser66 · Answer

can you show your work?

anonymous · Answer

(4s-6)/((s-1)^2+1^2), then you look at the table.

loser66 · Answer

yes, what formula you pick?

anonymous · Answer

e^at*cos(bt)

loser66 · Answer

ok, for the numerator, you have 4s -6 = 4s-4-2 then break them into 2 as 4s -4 and -2  ,the first term combine with denominator, you have 4 e^t cos t, the second term is 2 e^t sin t that matches to your book, right?

loser66 · Answer

$$4L^{-1}\left(\frac{s-1}{(s-1)^2+1}\right)=4e^tcost$$

anonymous · Answer

I get it. Thanks.

loser66 · Answer

why don't you comment here to signal me you get it? just medal, I don't like medal at all