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

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

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

inverse laplace

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

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

Solution steps

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

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

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

H (t - 1) = {01 t < 1 t \geq 1

= {0 - \frac{1}{4} + \frac{t - 1}{2} + \frac{1}{4} e^{- 2 (t - 1)} t < 1 t \geq 1

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

\int 1 + e^{2 x} d x

\frac{d}{d x} (25 x^{2} e^{2 x} - 16)

0, x^{2}, x + 2

\frac{- 8}{s ( s + 2 )}

n = 1 \sum \infty \frac{1}{3 ( n ^{3} + 4 )}

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