Question

(2s+5)/(s^2+6s+34) inverse Laplace transform

Answer 1

I have the answer. thank you very much! :)

Answer 2

.

Answer 3

are you transferring it back into function F(t)

Answer 4

someone help me lol

Answer 5

(2s+5)/(s^2+6s+9+25) = (2s+6-1)/((s+3)^2+25) = (2(s+3)-1)/((s+3)^2+5^2) = 2[(s+3)/((s+3)^2+5^2) - 1/5 (5 / ((s+3)^2+5^2))

Hence inverse Laplace transform is 2e^(-3t)cos 5t - (1/5) e^(-3t)sin 5t

Answer 6

Does the answer match?

Answer 7

\[e^{-3t}(2\cos(5t)-\frac{\sin5t}{5})\]

Answer 8

cool it matches :)

Answer 9

are making me doubt my answer a moment and rectify