What is inverse oflaplace {4/s+6/(s^5)-1/(s+8)} ?

The answer to inverse oflaplace {4/s+6/(s^5)-1/(s+8)} is 4\H(t)+(t^4)/4-e^{-8t}

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inverse oflaplace {4/s+6/(s^5)-1/(s+8)}

inverse laplace

{\frac{4}{s} + \frac{6}{s ^{5}} - \frac{1}{s + 8}}

4 H (t) + \frac{t ^{4}}{4} - e^{- 8 t}

Solution steps

L^{- 1} {\frac{4}{s} + \frac{6}{s ^{5}} - \frac{1}{s + 8}}

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{4}{s}} + L^{- 1} {\frac{6}{s ^{5}}} - L^{- 1} {\frac{1}{s + 8}}

L^{- 1} {\frac{4}{s}} : 4 H (t)

L^{- 1} {\frac{6}{s ^{5}}} : \frac{t ^{4}}{4}

L^{- 1} {\frac{1}{s + 8}} : e^{- 8 t}

= 4 H (t) + \frac{t ^{4}}{4} - e^{- 8 t}

\int e^{2 z} d z

\frac{d}{d x} (arctan (x t))

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

\int \frac{11}{1 - x ^{2}} d x

\frac{d}{d x} (8 x^{2} sin (x) tan (x))

What is inverse oflaplace {4/s+6/(s^5)-1/(s+8)} ?
The answer to inverse oflaplace {4/s+6/(s^5)-1/(s+8)} is 4\H(t)+(t^4)/4-e^{-8t}