What is inverse oflaplace 2/(((s-4)s(s+1))) ?

The answer to inverse oflaplace 2/(((s-4)s(s+1))) is 1/10 e^{4t}-1/2 \H(t)+2/5 e^{-t}

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inverse oflaplace 2/(((s-4)s(s+1)))

inverse laplace

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

\frac{1}{10} e^{4 t} - \frac{1}{2} H (t) + \frac{2}{5} e^{- t}

Solution steps

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

Take the partial fraction of

\frac{2}{( s - 4 ) s ( s + 1 )} : \frac{1}{10 ( s - 4 )} - \frac{1}{2 s} + \frac{2}{5 ( s + 1 )}

= L^{- 1} {\frac{1}{10 ( s - 4 )} - \frac{1}{2 s} + \frac{2}{5 ( 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}{10 ( s - 4 )}} - L^{- 1} {\frac{1}{2 s}} + L^{- 1} {\frac{2}{5 ( s + 1 )}}

L^{- 1} {\frac{1}{10 ( s - 4 )}} : \frac{1}{10} e^{4 t}

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

L^{- 1} {\frac{2}{5 ( s + 1 )}} : \frac{2}{5} e^{- t}

= \frac{1}{10} e^{4 t} - \frac{1}{2} H (t) + \frac{2}{5} e^{- t}

\int x^{2} + 4 x - 8 d x

\int_{- 1}^{- 2} (\frac{16 - x ^{3}}{x ^{3}}) d x

f (x) = x - 7

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

x 18 - x^{2}

What is inverse oflaplace 2/(((s-4)s(s+1))) ?
The answer to inverse oflaplace 2/(((s-4)s(s+1))) is 1/10 e^{4t}-1/2 \H(t)+2/5 e^{-t}