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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

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

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

x \cdot \frac{d y}{d x} = - y + 2 x^{2} y

\frac{\partial}{\partial y} (x^{2} + 2 y^{2} - 1)

\int \frac{9 x ^{8} + 8}{x ^{9} + 8 x} d x

\frac{\partial}{\partial x} (2 x e^{x^{2} + y^{2}})

8 y^{^{''}} + 32 y^{^{'}} = 0, y (2) = 1, y^{^{'}} (- 1) = 1

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