What is inverse oflaplace (-2e^s)/((s^2-2s-8)) ?

The answer to inverse oflaplace (-2e^s)/((s^2-2s-8)) is \H(t+1)(1/3 e^{-2(t+1)}-1/3 e^{4(t+1)})

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inverse oflaplace (-2e^s)/((s^2-2s-8))

inverse laplace

\frac{- 2 e ^{s}}{( s ^{2} - 2 s - 8 )}

H (t + 1) (\frac{1}{3} e^{- 2 (t + 1)} - \frac{1}{3} e^{4 (t + 1)})

Solution steps

L^{- 1} {\frac{- 2 e ^{s}}{( s ^{2} - 2 s - 8 )}}

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{- 2}{s ^{2} - 2 s - 8}} (t + 1)

L^{- 1} {\frac{- 2}{s ^{2} - 2 s - 8}} : \frac{1}{3} e^{- 2 t} - \frac{1}{3} e^{4 t}

= H (t + 1) (\frac{1}{3} e^{- 2 (t + 1)} - \frac{1}{3} e^{4 (t + 1)})

x \to \infty lim ((x)^{x})

\int \frac{u}{1 - u} d u

\int x^{- 3} ln (5 x) d x

n = 6 \sum \infty e^{5 - 2 n}

\int 2 + \frac{1}{1 - x} d x

What is inverse oflaplace (-2e^s)/((s^2-2s-8)) ?
The answer to inverse oflaplace (-2e^s)/((s^2-2s-8)) is \H(t+1)(1/3 e^{-2(t+1)}-1/3 e^{4(t+1)})