Question

Inverse laplace of 1/{[(s-1)^2] + 1}?

Answer 1

Partial fractions:
1/[s(s+1)^2] = A/s + B/(s+1) + C/(s+1)^2
1 = A(s+1)^2 + Bs(s+1) + Cs
1 = As^2 + 2As + A + Bs^2 + Bs + Cs
Equate like powers of s:
1 = A
0 = 2A + B + C
0 = A + B
Together these imply that B = -1 and C = -1.

1/[s(s+1)^2] = 1/s - 1/(s+1) - 1/(s+1)^2
Now go to a Laplace transform table, and you get
f(t) = 1 - e^(-t) - te^(-t)

Answer 2

do this help

Answer 3

do this help

Answer 4

yeah it helps but the equation i wrote is totally different..
I solved it
inverse laplace of 1/[(s-1)^2 +1]=e^tcost
any way thanks alot ! :)

Answer 5

=( e^t ) cos (t) **

Answer 6

now click best response