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}

\frac{d}{d x} (- \frac{2}{5} x^{\frac{5}{2}} + 32 x^{\frac{1}{2}})

\frac{d}{d x} (x x + 8)

\int x^{3} - 8 x^{2} + 16 x + 1 d x

3 x^{\frac{1}{3}} - 5 x^{\frac{4}{3}}

\int e^{7 x + 1} d 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}