What is inverse oflaplace e^{-2s}(1/(s+1)) ?

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

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inverse oflaplace e^{-2s}(1/(s+1))

inverse laplace

e^{- 2 s} (\frac{1}{s + 1})

H (t - 2) e^{- t + 2}

Solution steps

L^{- 1} {e^{- 2 s} (\frac{1}{s + 1})}

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 + 1}} (t - 2)

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

= H (t - 2) e^{- (t - 2)}

= H (t - 2) e^{- t + 2}

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