What is inverse oflaplace ((s/2-1))/(s^2+3s+2) ?

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

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inverse oflaplace ((s/2-1))/(s^2+3s+2)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{\frac{s}{2} - 1}{s ^{2} + 3 s + 2} : - \frac{3}{2 ( s + 1 )} + \frac{2}{s + 2}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

= - \frac{3}{2} e^{- t} + 2 e^{- 2 t}

\frac{\partial}{\partial y} (\frac{x}{3 + y})

\frac{\partial}{\partial t} (k e^{- \frac{t}{R ( t ) C}})

\frac{d}{d t} (8 t^{3})

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

\int \frac{1}{1 - 2 x ^{2}} d x

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