What is inverse oflaplace (e^{-2s})/(s(3s+2)) ?

The answer to inverse oflaplace (e^{-2s})/(s(3s+2)) is \H(t-2)(1/2-1/2 e^{-(2(t-2))/3})

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inverse oflaplace (e^{-2s})/(s(3s+2))

inverse laplace

\frac{e ^{- 2 s}}{s ( 3 s + 2 )}

H (t - 2) (\frac{1}{2} - \frac{1}{2} e^{- \frac{2 ( t - 2 )}{3}})

Solution steps

L^{- 1} {\frac{e ^{- 2 s}}{s ( 3 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 ( 3 s + 2 )}} (t - 2)

L^{- 1} {\frac{1}{s ( 3 s + 2 )}} : \frac{1}{2} H (t) - \frac{1}{2} e^{- \frac{2 t}{3}}

= H (t - 2) (\frac{1}{2} H (t - 2) - \frac{1}{2} e^{- \frac{2 ( t - 2 )}{3}})

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

= {0 \frac{1}{2} - \frac{1}{2} e^{- \frac{2 ( t - 2 )}{3}} t < 2 t \geq 2

= H (t - 2) (\frac{1}{2} - \frac{1}{2} e^{- \frac{2 ( t - 2 )}{3}})

\frac{\partial}{\partial y} (y - x)

\frac{x ^{2}}{9}

\int_{0}^{1} \frac{x}{x ^{2} + 1} d x

\frac{d}{d x} (1 + 2 sin (\frac{x ^{2}}{2}))

\int \frac{4}{x ^{2} + 16} d x

What is inverse oflaplace (e^{-2s})/(s(3s+2)) ?
The answer to inverse oflaplace (e^{-2s})/(s(3s+2)) is \H(t-2)(1/2-1/2 e^{-(2(t-2))/3})