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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\frac{e ^{- 0.1 s}}{s + 3}

\int \frac{sin ( x )}{1 + sin ( x )} d x

y^{^{'}} + (x + 2) y^{2} = 0

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

\frac{\partial}{\partial x} (e^{a x} cos (b x + y))

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