What is inverse oflaplace 1/(s*(s+2)) ?

The answer to inverse oflaplace 1/(s*(s+2)) is 1/2 \H(t)-1/2 e^{-2t}

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inverse oflaplace 1/(s*(s+2))

inverse laplace

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

\frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s + 2 )} : \frac{1}{2 s} - \frac{1}{2 ( s + 2 )}

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

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

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

= \frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

x \to 0 lim (\frac{x ^{3}}{x ^{2}})

\int \frac{cos ( x ) - 1}{x ^{2}} d x

\int \frac{cos ( x )}{sin ^{2} ( x ) - 2} d x

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

x \to 0 + lim (x^{- \frac{17}{l n ( x )}})

What is inverse oflaplace 1/(s*(s+2)) ?
The answer to inverse oflaplace 1/(s*(s+2)) is 1/2 \H(t)-1/2 e^{-2t}