yes, sorry. is:
$$
\mathcal L^{-1}\frac 1 {(s^2+1)^2}$$
so
$$
\mathcal L^{-1}\frac 1 {(s^2+1)} * \frac 1 {(s^2+1) }$$
then
f(τ) = sen τ
g(t-τ) = sen (t-τ)
∫ sen τ * sen (t-τ) dτ
Integration by parts
u= sen (t-τ)
du = -cos(t-τ)
dv= sen τ dτ
v= -cos τ
∫ sen τ *...
Homework Statement
Hi.
I need help to resolve the inverse laplace transform of {1/((x^2)+1)^2}2. The attempt at a solution
I have tried to do:
{(1/((x^2)+1) * (1/((x^2)+1)}
then, convolution, sen x
But, isn't working
Thanks for your help :)