What is inverse oflaplace 8e^{-3s} ?

The answer to inverse oflaplace 8e^{-3s} is 8\delta(t-3)

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inverse oflaplace 8e^{-3s}

inverse laplace

8 e^{- 3 s}

8 δ (t - 3)

Solution steps

L^{- 1} {8 e^{- 3 s}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= 8 L^{- 1} {e^{- 3 s}}

L^{- 1} {e^{- 3 s}} : δ (t - 3)

= 8 δ (t - 3)

\int_{1}^{2} \frac{- 7 ln ( x )}{x ^{2}} d x

\int \frac{- 7}{cos ( x ) - 1} d x

\int tan^{3} (2 x) sec^{4} (2 x) d x

\int_{x}^{x + 2} (4 t + 1) d t

\frac{\partial}{\partial x} (2 cos (x y))

What is inverse oflaplace 8e^{-3s} ?
The answer to inverse oflaplace 8e^{-3s} is 8\delta(t-3)