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

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

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inverse oflaplace (e^{-2s})/(s(s-2))

inverse laplace

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

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

Solution steps

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

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

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

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

H (t - 2) = {01 t < 2 t \geq 2

= {0 - \frac{1}{2} + \frac{1}{2} e^{2 (t - 2)} t < 2 t \geq 2

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

\frac{d}{d x} (- 2 e^{5 y} x - 2 sin (2 x))

\int_{- 2}^{2} (4 - x^{2}) d x

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

6 x^{\frac{1}{2}}

\int_{0}^{6} (1497.6) (x) d x

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