# Find the inverse laplace transform

How to find the inverse laplace transform of $$\frac{s(c-F(s))}{s-a}$$ where $a$ , $c$ are constants and $L^{-1}\{F(s)\}=f(t).$ I have found different answers by different approaches, where is the mistake I don't know.

-
I have found different answers by different approaches, where is the mistake I don't know... And how could we unless you show the details of those different approaches? –  Did Mar 11 '13 at 7:53
$$A(s)=\frac{s(c-F(s))}{s-a} = \frac{sc}{s-a}-\frac{sF(s)}{s-a}=c+\frac{ca}{s-a}-\frac{sF(s)}{s-a}$$ $$\mathcal{L}\left[ g(t) \right] = G(s)= \frac{sF(s)}{s-a}$$ We know: $$\mathcal{L}[e^{-at}g(t)]=G(s+a)$$ $$\mathcal{L}\left[\int_0^tg(t) dt \right]=\frac{G(s)}{s}$$ therefor : $$\mathcal{L} \left[ e^{-at} g(t) \right] =\frac{(s+a)F(s+a)}{s}=F(s+a)+\frac{a}{s}F(s+a)$$ $$e^{-at} g(t)= e^{-at} f(t)+a \int_0^t e^{-at} f(t) dt$$ $$g(t)= f(t)+a e^{at} \int_0^t e^{-at} f(t) dt$$ Also We know : $$\mathcal{L}[\delta(t)]=1$$ $$\mathcal{L}[e^{at}]=\frac{1}{s-a}$$ therefor : $$\mathcal{L^{-1}} \left[ \frac{s(c-F(s))}{s-a} \right]=c \delta(t) + ca e^{at} -f(t)-a e^{at} \int_0^t e^{-at} f(t) dt$$