What is inverse oflaplace F(s)=(s^2-2s-6)/(s-2) ?

The answer to inverse oflaplace F(s)=(s^2-2s-6)/(s-2) is \delta'(t)-6e^{2t}

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inverse oflaplace F(s)=(s^2-2s-6)/(s-2)

inverse laplace

F (s) = \frac{s ^{2} - 2 s - 6}{s - 2}

δ' (t) - 6 e^{2 t}

Solution steps

L^{- 1} {\frac{s ^{2} - 2 s - 6}{s - 2}}

Take the partial fraction of

\frac{s ^{2} - 2 s - 6}{s - 2} : s - \frac{6}{s - 2}

= L^{- 1} {s - \frac{6}{s - 2}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {s} - L^{- 1} {\frac{6}{s - 2}}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {\frac{6}{s - 2}} : 6 e^{2 t}

= δ^{^{'}} (t) - 6 e^{2 t}

x \to 3 + lim (\frac{x + 2}{x - 2})

\frac{d}{d x} (x^{l n (x)})

\frac{d y}{d x} = \frac{y}{2}

\int \frac{4}{x ^{4}} d x

\frac{d}{d x} (16 + x^{2})

What is inverse oflaplace F(s)=(s^2-2s-6)/(s-2) ?
The answer to inverse oflaplace F(s)=(s^2-2s-6)/(s-2) is \delta'(t)-6e^{2t}