What is laplacetransform e^{-t+1} ?

The answer to laplacetransform e^{-t+1} is e/(s+1)

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laplacetransform e^{-t+1}

laplace transform

e^{- t + 1}

\frac{e}{s + 1}

Solution steps

L {e^{- t + 1}}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= e^{1} L {e^{- t}}

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

= e^{1} \frac{1}{s + 1}

Refine

e^{1} \frac{1}{s + 1} : \frac{e}{s + 1}

= \frac{e}{s + 1}

\frac{d}{d x} (\frac{1}{5 x ^{5}})

\frac{- 1}{x ^{2}}

\int \frac{4}{1 + 5 x} d x

\frac{d y}{d x} = - (1 + y)^{4}, y (1) = 2

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

What is laplacetransform e^{-t+1} ?
The answer to laplacetransform e^{-t+1} is e/(s+1)