What is inverse oflaplace (e^{-s})/(((s-3)(s+1))) ?

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

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inverse oflaplace (e^{-s})/(((s-3)(s+1)))

inverse laplace

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

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

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

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

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

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

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

\frac{d}{d x} (4 x^{\frac{- 1}{8}})

\frac{\partial}{\partial x} (5 x^{0.75} y^{0.25})

\int \frac{( ln ( x ) ) ^{24}}{x} d x

f (x) = ln (x^{l n (x)})

\int_{- \infty}^{- 1} e^{- 20 t} d t

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