Question

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))

Answer 1

can you show your work?

Answer 2

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

Answer 3

yes, what formula you pick?

Answer 4

e^at*cos(bt)

Answer 5

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?

Answer 6

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

Answer 7

I get it. Thanks.

Answer 8

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