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

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

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

inverse laplace

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

H (t - 1) (t - 2)

Solution steps

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

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}} - \frac{1}{s})} (t - 1)

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

= H (t - 1) (t - 1 - H (t - 1))

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

= {0 t - 2 t < 1 t \geq 1

= H (t - 1) (t - 2)

\frac{d y}{d x}, 5 e^{x y} - 4 x = y + 30.9

\frac{d}{d x} ((8 x)^{\frac{1}{2}})

\frac{\partial}{\partial x} (3 \cdot (y + 1) \cdot (x - 3)^{3})

\int (x^{2} - 4)^{3} x d x

n = 1 \sum \infty \frac{- 4 ^{n}}{4 ^{n}}

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