What is inverse oflaplace 1/(s*(s^2+5s+4)) ?

The answer to inverse oflaplace 1/(s*(s^2+5s+4)) is 1/4 \H(t)-1/3 e^{-t}+1/12 e^{-4t}

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inverse oflaplace 1/(s*(s^2+5s+4))

inverse laplace

\frac{1}{s \cdot ( s ^{2} + 5 s + 4 )}

\frac{1}{4} H (t) - \frac{1}{3} e^{- t} + \frac{1}{12} e^{- 4 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s ^{2} + 5 s + 4 )} : \frac{1}{4 s} - \frac{1}{3 ( s + 1 )} + \frac{1}{12 ( s + 4 )}

= L^{- 1} {\frac{1}{4 s} - \frac{1}{3 ( s + 1 )} + \frac{1}{12 ( 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}{4 s}} - L^{- 1} {\frac{1}{3 ( s + 1 )}} + L^{- 1} {\frac{1}{12 ( s + 4 )}}

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

L^{- 1} {\frac{1}{3 ( s + 1 )}} : \frac{1}{3} e^{- t}

L^{- 1} {\frac{1}{12 ( s + 4 )}} : \frac{1}{12} e^{- 4 t}

= \frac{1}{4} H (t) - \frac{1}{3} e^{- t} + \frac{1}{12} e^{- 4 t}

\frac{d}{d x} (- 6 cos (x))

\int \frac{x ^{2}}{( 4 - x ^{2} ) ^{\frac{1}{2}}} d x

\frac{d}{d x} (x^{2} + x^{4})

f (x) = \frac{1}{1 - x ^{2}}

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

What is inverse oflaplace 1/(s*(s^2+5s+4)) ?
The answer to inverse oflaplace 1/(s*(s^2+5s+4)) is 1/4 \H(t)-1/3 e^{-t}+1/12 e^{-4t}