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

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

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

inverse laplace

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

\frac{1}{3} e^{- t} - \frac{1}{3} e^{- 4 t}

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{1}{3 ( s + 4 )}} : \frac{1}{3} e^{- 4 t}

= \frac{1}{3} e^{- t} - \frac{1}{3} e^{- 4 t}

n = 0 \sum \infty \frac{2 n}{n + 1}

t e^{-, at}

\int_{0}^{2} x 2 x - x^{2} d x

x \to 0 + lim (ln^{2} (x) - ln (x))

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

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