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

The answer to inverse oflaplace ((s+2))/(s(s+4)) is 1/2 \H(t)+1/2 e^{-4t}

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

inverse laplace

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

\frac{1}{2} H (t) + \frac{1}{2} e^{- 4 t}

Solution steps

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

Take the partial fraction of

\frac{s + 2}{s ( s + 4 )} : \frac{1}{2 s} + \frac{1}{2 ( s + 4 )}

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

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 + 4 )}}

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

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

= \frac{1}{2} H (t) + \frac{1}{2} e^{- 4 t}

\frac{d}{d x} (tan (5 x + 5))

\int_{0}^{3} \frac{x}{9 - x ^{2}} d x

\int 45 t - \frac{21}{t} d t

\frac{d}{d x} (\frac{2}{x ^{3} + 1})

\frac{d}{d x} (\frac{- 5}{x ^{5}})

What is inverse oflaplace ((s+2))/(s(s+4)) ?
The answer to inverse oflaplace ((s+2))/(s(s+4)) is 1/2 \H(t)+1/2 e^{-4t}