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

The answer to inverse oflaplace (-8)/(s(s+2)) is -4\H(t)+4e^{-2t}

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

inverse laplace

\frac{- 8}{s ( s + 2 )}

- 4 H (t) + 4 e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{- 8}{s ( s + 2 )} : - \frac{4}{s} + \frac{4}{s + 2}

= L^{- 1} {- \frac{4}{s} + \frac{4}{s + 2}}

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{4}{s + 2}}

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

L^{- 1} {\frac{4}{s + 2}} : 4 e^{- 2 t}

= - 4 H (t) + 4 e^{- 2 t}

n = 1 \sum \infty \frac{1}{3 ( n ^{3} + 4 )}

n = 1 \sum \infty n^{2} (\frac{1}{3})^{n}

\int 1 - \frac{x}{y - x ^{2}}

\frac{d y}{d x} = \frac{1}{10} \cdot y \cdot (500 - y)

\frac{d}{d x} (sin^{2} (x + 5))

What is inverse oflaplace (-8)/(s(s+2)) ?
The answer to inverse oflaplace (-8)/(s(s+2)) is -4\H(t)+4e^{-2t}