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Use partial fractions to find the inverse Laplace transform of $F(s) =\large \frac{s+1}{s^3+5s^2+6s}$. Then use $u = s+1$ to show that $F(s) = G(u) = \large\frac{u}{(u-1)(u+1)(u+2)}$.

Use partial fractions to find the inverse Laplace transform of $G(s)$, and then use the translation theorem to show that the inverse transforms of $F(s)$ and $G(s+1)$ are the same.

I found the inverse Laplace transforms of $F(s)$ and $G(s)$ using partial fractions as:

$$\mathcal{L}^{-1}\{F(s)\} = f(t) = \frac{1}{6}-\frac{2}{3}e^{-3t}+\frac{1}{2}e^{-2t}$$

$$\mathcal{L}^{-1}\{G(s)\} = g(t) = \frac{1}{6}e^{t}+\frac{1}{2}e^{-t}-\frac{2}{3}e^{-2t}$$

Now I'm having trouble using the translation theorem to show that $F(s)$ and $G(s+1)$ are the same. Can some explain exactly what the translation theorem allows us to conclude and how it can possibly relate to this problem?

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You are missing a cube on the deffinition of $F(s)$ I think. –  Pragabhava Dec 5 '12 at 23:52
    
Yes! Thanks for the catch. –  user1038665 Dec 5 '12 at 23:55

1 Answer 1

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What I assume is the so called Translation Theorem is simply a property of the Laplace transfom, i.e $$ \mathcal{L}\{e^{-at} f(t)\} = \int_0^\infty e^{-s t} e^{-a t}f(t) dt = \int_0^\infty e^{-(s+a)t}f(t) dt $$ and denoting $$ F(s) = \int_0^\infty e^{-s t} f(t) dt $$ then $$ F(s+a) = \mathcal{L} \{e^{-a t} f(t)\}. $$

In your case, you have that $F(s) = G(s+1)$, and then $$ f(t) = \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\{G(s+1)\} = e^{-t} g(t) $$

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