What is inverse oflaplace 8/(s^2+8s) ?

The answer to inverse oflaplace 8/(s^2+8s) is \H(t)-e^{-8t}

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inverse oflaplace 8/(s^2+8s)

inverse laplace

\frac{8}{s ^{2} + 8 s}

H (t) - e^{- 8 t}

Solution steps

L^{- 1} {\frac{8}{s ^{2} + 8 s}}

Take the partial fraction of

\frac{8}{s ^{2} + 8 s} : \frac{1}{s} - \frac{1}{s + 8}

= L^{- 1} {\frac{1}{s} - \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{1}{s}} - L^{- 1} {\frac{1}{s + 8}}

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

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

= H (t) - e^{- 8 t}

What is inverse oflaplace 8/(s^2+8s) ?
The answer to inverse oflaplace 8/(s^2+8s) is \H(t)-e^{-8t}