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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.

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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
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$$ A(s)=\frac{s(c-F(s))}{s-a} = \frac{sc}{s-a}-\frac{sF(s)}{s-a}=c+\frac{ca}{s-a}-\frac{sF(s)}{s-a}$$ $$\mathcal{L}\left[ g(t) \right] = G(s)= \frac{sF(s)}{s-a} $$ 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 : $$ \mathcal{L^{-1}} \left[ \frac{s(c-F(s))}{s-a} \right]=c \delta(t) + ca e^{at} -f(t)-a e^{at} \int_0^t e^{-at} f(t) dt $$

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