Finding inverse of a Laplace transform by convolution

[Pouyan]

Homework Statement

find the inverse Laplace transform of the given function by
using the convolution theorem

Homework Equations

F(s) = s/((s+1)(s²)+4)

The theorem : Lap{(f*g)(t)} = F(s)*G(s)

The Attempt at a Solution

I know how to find it the answer is :
we have 1/(s+1) * s/(s+4) and the inverse of each of these functions are : e^-t * cos(2t)
further the answer is : ∫(e^(-(t-τ))*cos(τ)dτ)
But if I try to solve this problem without convolution theorem; and with partial fraction I get :

s/((s+1)(s²+4)) = (1/5) ( (1/(s+1) + s/(s²+4) + 4/(s²+4) )

and the inverse of this function is :

(1/5) (cos(2t) - e^-t +2sin(2t))

MY QUESTION IS :

∫(e^(-(t-τ))*cos(τ)dτ) = (1/5) (cos(2t) - e^-t +2sin(2t)) is this right ?

 

[Ray Vickson]

Homework Statement

find the inverse Laplace transform of the given function by
using the convolution theorem

Homework Equations

F(s) = s/((s+1)(s²)+4)

The theorem : Lap{(f*g)(t)} = F(s)*G(s)

The Attempt at a Solution

I know how to find it the answer is :
we have 1/(s+1) * s/(s+4) and the inverse of each of these functions are : e^-t * cos(2t)
further the answer is : ∫(e^(-(t-τ))*cos(τ)dτ)
But if I try to solve this problem without convolution theorem; and with partial fraction I get :

s/((s+1)(s²+4)) = (1/5) ( (1/(s+1) + s/(s²+4) + 4/(s²+4) )

and the inverse of this function is :

(1/5) (cos(2t) - e^-t +2sin(2t))

MY QUESTION IS :

∫(e^(-(t-τ))*cos(τ)dτ) = (1/5) (cos(2t) - e^-t +2sin(2t)) is this right ?

What you wrote is not the convolution; the convolution is a definite integral, and you wrote an indefinite integral. In this method especially, limits are crucial.

Anyway, in your integral you should have $\cos(2 \tau) \, d \tau$, not $\cos(\tau) \, d \tau$.

After fixing things up, you will be able to answer your own question, by either (i) doing the integral; or (ii) differentiating both sides to see if the derivatives match.