What is inverse oflaplace 6/(s^2+5s+4) ?

The answer to inverse oflaplace 6/(s^2+5s+4) is 2e^{-t}-2e^{-4t}

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inverse oflaplace 6/(s^2+5s+4)

inverse laplace

\frac{6}{s ^{2} + 5 s + 4}

2 e^{- t} - 2 e^{- 4 t}

Solution steps

L^{- 1} {\frac{6}{s ^{2} + 5 s + 4}}

Take the partial fraction of

\frac{6}{s ^{2} + 5 s + 4} : \frac{2}{s + 1} - \frac{2}{s + 4}

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

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

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

= 2 e^{- t} - 2 e^{- 4 t}

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What is inverse oflaplace 6/(s^2+5s+4) ?
The answer to inverse oflaplace 6/(s^2+5s+4) is 2e^{-t}-2e^{-4t}