What is laplacetransform 4e^{-t} ?

The answer to laplacetransform 4e^{-t} is 4/(s+1)

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laplacetransform 4e^{-t}

laplace transform

4 e^{- t}

\frac{4}{s + 1}

Solution steps

L {4 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)}

= 4 L {e^{- t}}

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

= 4 \cdot \frac{1}{s + 1}

Refine

4 \frac{1}{s + 1} : \frac{4}{s + 1}

= \frac{4}{s + 1}

\int_{- 2}^{5} (x^{2} - 4) d x

\frac{d}{d x} ((5 x^{4} - 6 x^{3} + x)^{5})

\int_{0}^{2} \frac{4}{x ^{2}} d x

\frac{d}{d s} (\frac{- s ^{2} + 1}{( s ^{2} + 1 ) ^{2}})

\frac{\partial}{\partial x} (x^{2 + 2 y^{2} + 3 z^{2}})

What is laplacetransform 4e^{-t} ?
The answer to laplacetransform 4e^{-t} is 4/(s+1)