Hello All,

I have been working at the following for a while, trying to find the inverse of it, my working to date is shown below, although I suspect I may be approaching it wrong.

$\displaystyle

F(s) = \frac{3e^{-3s}}{s(s+3)}

$

My train of thought so far is as follows;

By using the convolution theorem and treating

$\displaystyle

\mathcal{L}^{-1}(F(s)G(s)) = \int^{t}_{0} f(t)g(\tau-t)dt

$

$\displaystyle

G(s) = \frac{3}{s}

$

$\displaystyle

F(s) = \frac{e^{-3s}}{s+3}

$

from identity

$\displaystyle

\mathcal {L}[H(t-a)y(t-a)]=e^{-as}Y(s)

$

let $\displaystyle Y(s)=\frac{1}{s+3} $

therefore

$\displaystyle {L}[H(t-3)(e^{-3t}-3)]=\frac{e^{-3s}}{s+3}

$

$\displaystyle

{L}^{-1}\left\{\frac{3e^{-3s}}{s\left(s+3\right)}\right\}= 3\int_0^t H(t-3)(e^{-3t}-3)dt

$

Any solutions or help would be appreciated, or pointers as to any possible errors.

Thanks, Alex.