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

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

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

laplace transform

e^{- 1 (t + 1)}

\frac{1}{e ( s + 1 )}

Solution steps

L {e^{- 1 \cdot (t + 1)}}

Expand

e^{- 1 (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{1}{e ( s + 1 )}

= \frac{1}{e ( s + 1 )}

\int e^{0.5 x} d x

\int e^{- 2 x + 3} d x

x^{2} e^{2 x}

\frac{7}{4} x^{2} - 3 x + 12

4 + 7 x

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