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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

\int \frac{e ^{- x}}{( 9 e ^{- 2 x} + 1 ) ^{\frac{3}{2}}} d x

1 - x^{2}

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

\int sin (13 x) d x

\int_{0}^{1} ln (4 x) d x

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