What is inverse oflaplace 4/(s(s+2)) ?

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

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inverse oflaplace 4/(s(s+2))

inverse laplace

\frac{4}{s ( s + 2 )}

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

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{2}{s}} : 2 H (t)

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

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

\int 9 cot^{4} (x) d x

\frac{1}{s ^{2} ( s + 1 )}

\int \frac{t ^{3}}{t ^{2} + 9} d t

y^{^{'''}} - y^{^{''}} + 4 y^{^{'}} - 4 y = 0

\frac{d}{d x} (\frac{x - 2}{x + 2})

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