What is inverse oflaplace 1/(s(s+1)(s+5)) ?

The answer to inverse oflaplace 1/(s(s+1)(s+5)) is 1/5 \H(t)-1/4 e^{-t}+1/20 e^{-5t}

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inverse oflaplace 1/(s(s+1)(s+5))

inverse laplace

\frac{1}{s ( s + 1 ) ( s + 5 )}

\frac{1}{5} H (t) - \frac{1}{4} e^{- t} + \frac{1}{20} e^{- 5 t}

Solution steps

L^{- 1} {\frac{1}{s ( s + 1 ) ( s + 5 )}}

Take the partial fraction of

\frac{1}{s ( s + 1 ) ( s + 5 )} : \frac{1}{5 s} - \frac{1}{4 ( s + 1 )} + \frac{1}{20 ( s + 5 )}

= L^{- 1} {\frac{1}{5 s} - \frac{1}{4 ( s + 1 )} + \frac{1}{20 ( s + 5 )}}

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}{5 s}} - L^{- 1} {\frac{1}{4 ( s + 1 )}} + L^{- 1} {\frac{1}{20 ( s + 5 )}}

L^{- 1} {\frac{1}{5 s}} : \frac{1}{5} H (t)

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

L^{- 1} {\frac{1}{20 ( s + 5 )}} : \frac{1}{20} e^{- 5 t}

= \frac{1}{5} H (t) - \frac{1}{4} e^{- t} + \frac{1}{20} e^{- 5 t}

- 3 (t - 1)

y^{^{'}} = - 4 y + 57

e^{x} d x - y d y = 0, y (0) = 1

\int_{1}^{3} \frac{4}{x ^{2} + 1} d x

\int_{- 2}^{2} 4 (4 - x^{2}) d x

What is inverse oflaplace 1/(s(s+1)(s+5)) ?
The answer to inverse oflaplace 1/(s(s+1)(s+5)) is 1/5 \H(t)-1/4 e^{-t}+1/20 e^{-5t}