What is laplacetransform 5e^{-t} ?

The answer to laplacetransform 5e^{-t} is 5/(s+1)

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laplacetransform 5e^{-t}

laplace transform

5 e^{- t}

\frac{5}{s + 1}

Solution steps

L {5 e^{- 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)}

= 5 L {e^{- t}}

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

= 5 \cdot \frac{1}{s + 1}

Refine

5 \frac{1}{s + 1} : \frac{5}{s + 1}

= \frac{5}{s + 1}

f (x) = 6 x - x^{2}, at x = 2

f (x) = - csc (x)

\int (\frac{x ^{2}}{4} + y^{2} + z^{2})^{- \frac{3}{2}} d y

\int \frac{1}{x ^{2} + 4} d x

\frac{d}{d x} (sin (8 x) cos (9 y))

What is laplacetransform 5e^{-t} ?
The answer to laplacetransform 5e^{-t} is 5/(s+1)