What is inverse oflaplace 10(e^{-3s})/s ?

The answer to inverse oflaplace 10(e^{-3s})/s is 10\H(t-3)

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inverse oflaplace 10(e^{-3s})/s

inverse laplace

10 \frac{e ^{- 3 s}}{s}

10 H (t - 3)

Solution steps

L^{- 1} {10 \cdot \frac{e ^{- 3 s}}{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)}

= 10 L^{- 1} {\frac{e ^{- 3 s}}{s}}

L^{- 1} {\frac{e ^{- 3 s}}{s}} : H (t - 3)

= 10 H (t - 3)

x \to 4 - lim (x - 4)

\int_{0}^{π} (x + sin (x)) d x

a^{5} + cos^{5} (x)

n = 1 \sum \infty (- 3)^{- 3 n - 1}

f (x) = e^{3 x}, (0, 1)

What is inverse oflaplace 10(e^{-3s})/s ?
The answer to inverse oflaplace 10(e^{-3s})/s is 10\H(t-3)