What is inverse oflaplace (e^{-2s})/(s^2(s-1)) ?

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

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inverse oflaplace (e^{-2s})/(s^2(s-1))

inverse laplace

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

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

Solution steps

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

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

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

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

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

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

\frac{\partial}{\partial x} (e^{- 2 x} \cdot cos (2 π y))

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

\frac{d y}{d x} + y = 9

\frac{\partial}{\partial z} (x^{4} y + x z - y z^{2} - 7)

\int x^{y} d x

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