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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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}{2 ( s + 1 )}}

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

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

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

\frac{d ^{2}}{d x ^{2}} (x y e^{(- x + 3 y)})

\frac{\partial}{\partial z} (3 x y z^{2})

e^{2 x} ln (2 x)

x \to - 2 lim (\frac{8}{x})

\frac{\partial}{\partial x} (ln (4 x - 7 y))

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