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

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

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

inverse laplace

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

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

Solution steps

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

\frac{d}{d x} (3 x - 2)

\frac{\partial}{\partial x} (2 y - 2)

\frac{\partial}{\partial x} (x^{6} + x y^{5} + 7)

\int (cos (x) + x cos (y) - y) d y

(6 + x) y^{^{'}} = 7 y

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