Given F(s) = 1/((s-1)^(2)(s+1))
After using partial fractions, I got (1/4)[(2/((s-1)^(2)))+(1/(s+1))-(1/(s-1))] before applying Laplace inverse.
Apparently, the answer is (1/4)[2te^(t) + e^(-t) - e^(t)], however I don't know where the t in 2te^(t) came from. Can someone please explain? Thanks in advance! :D
Recall that L{tf(t)) = -F'(s)
Since L^-1{1/(s-1)} = e^t
and d/ds 1/(s-1) = 1/(s-1)^2
L^-1{1/(s-1)^2} = te^t
To explain where the t in 2te^(t) came from, we need to understand the inverse Laplace Transform of the term (2/((s-1)^2)). When applying the inverse Laplace Transform to this term, we can observe that it corresponds to a derivative in the time domain.
In general, the inverse Laplace Transform of (n!/((s-a)^(n+1))) is given by ((t^n)/n!)*e^(at), where n is a non-negative integer and a is a constant.
For our specific case, (2/((s-1)^2)) can be written as (2*(1!)/((s-1)^(1+1))), which matches the above pattern with n = 1 and a = 1. Therefore, when we apply the inverse Laplace Transform to this term, it gives us (t^1/1!)*e^(1t), which simplifies to te^(t).
Hence, the term (2/((s-1)^2)) in the partial fraction expansion becomes 2te^(t) after applying the inverse Laplace Transform.