What is inverse oflaplace (e^{-2s})/(s^2-3s-4) ?

The answer to inverse oflaplace (e^{-2s})/(s^2-3s-4) is \H(t-2)(-1/5 e^{-t+2}+1/5 e^{4(t-2)})

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inverse oflaplace (e^{-2s})/(s^2-3s-4)

inverse laplace

\frac{e ^{- 2 s}}{s ^{2} - 3 s - 4}

H (t - 2) (- \frac{1}{5} e^{- t + 2} + \frac{1}{5} e^{4 (t - 2)})

Solution steps

L^{- 1} {\frac{e ^{- 2 s}}{s ^{2} - 3 s - 4}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 2) L^{- 1} {\frac{1}{s ^{2} - 3 s - 4}} (t - 2)

L^{- 1} {\frac{1}{s ^{2} - 3 s - 4}} : - \frac{1}{5} e^{- t} + \frac{1}{5} e^{4 t}

= H (t - 2) (- \frac{1}{5} e^{- (t - 2)} + \frac{1}{5} e^{4 (t - 2)})

= H (t - 2) (- \frac{1}{5} e^{- t + 2} + \frac{1}{5} e^{4 (t - 2)})

What is inverse oflaplace (e^{-2s})/(s^2-3s-4) ?
The answer to inverse oflaplace (e^{-2s})/(s^2-3s-4) is \H(t-2)(-1/5 e^{-t+2}+1/5 e^{4(t-2)})