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

Question

asked Jan 16, 2017 119k views

1 Answer

Boris Kotov · Answer 1 · 2017-01-20T03:56:30+0000

(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

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

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

Please log in or register to add a comment.

Related questions

Categories

Other Questions