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)}}

Expand

e^{- (t - 1)} : e^{- t + 1}

= 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} (x - ln (sec (x) + tan (x)))

(- 2 x^{3} + x^{2} - 2 x - 1) x^{3}

\frac{d}{d x} (2 e^{5 x})

\int 2 x y + 2 x d x

t \to 0 lim (\frac{1 - e ^{- 2 t}}{t})

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