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

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

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

inverse laplace

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

- e^{- t} + 2 e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= - e^{- t} + 2 e^{- 2 t}

f (x) = \frac{4 x}{5 - cot ( x )}

\int x e^{\frac{1}{x}} d x

x \to - 2 lim (\frac{x ^{2} - 4}{x + 2})

e^{\frac{6}{x}}

x \to 3.01 lim (x^{2} + 3 x)

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