What is inverse oflaplace ((1-e^{-3s}))/s ?

The answer to inverse oflaplace ((1-e^{-3s}))/s is \H(t)-\H(t-3)

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inverse oflaplace ((1-e^{-3s}))/s

inverse laplace

\frac{( 1 - e ^{- 3 s} )}{s}

H (t) - H (t - 3)

Solution steps

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

Expand

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

= H (t) - H (t - 3)

\frac{d}{d x} (csc^{4} (2 x))

\int \frac{y ^{n - 1}}{( a y ^{n} + b ) ^{m}} d y

f (x) = 10 x

(9, - 2), (- 3, 2)

\int (sin (x) + cos (x)) d x

What is inverse oflaplace ((1-e^{-3s}))/s ?
The answer to inverse oflaplace ((1-e^{-3s}))/s is \H(t)-\H(t-3)