What is laplacetransform e^{1-t} ?

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

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laplacetransform e^{1-t}

laplace transform

e^{1 - t}

\frac{e}{s + 1}

Solution steps

L {e^{1 - t}}

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}

x y^{^{'}} sin^{2} (\frac{y}{x}) = x + y sin^{2} (\frac{y}{x})

\frac{\partial}{\partial x} (8 y^{7} cos (8 x))

\frac{1}{b ^{\frac{2}{7}}}

x \to 2 lim (\frac{3}{x ^{2} - 4})

\frac{d x}{x} - y^{2} d y = 0

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