What is inverse oflaplace 5/s-5/s e^{-5s} ?

The answer to inverse oflaplace 5/s-5/s e^{-5s} is 5\H(t)-5\H(t-5)

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inverse oflaplace 5/s-5/s e^{-5s}

inverse laplace

\frac{5}{s} - \frac{5}{s} e^{- 5 s}

5 H (t) - 5 H (t - 5)

Solution steps

L^{- 1} {\frac{5}{s} - e^{- 5 s} \frac{5}{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{5}{s}} - L^{- 1} {\frac{5}{s} e^{- 5 s}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

L^{- 1} {\frac{5}{s} e^{- 5 s}} : 5 H (t - 5)

= 5 H (t) - 5 H (t - 5)

n = 0 \sum \infty - (- 1)^{n + 1}

\frac{d}{d x} (\frac{3}{4 x ^{2}})

f (x) = x^{4} - 13 x^{2} + 36

\int_{0}^{1} \frac{45}{x ^{5}} d x

(1 + \frac{4}{x})^{x}

What is inverse oflaplace 5/s-5/s e^{-5s} ?
The answer to inverse oflaplace 5/s-5/s e^{-5s} is 5\H(t)-5\H(t-5)