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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

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