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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\frac{d}{d x} (4 sec (5 x))

f (t) = \frac{t ^{2}}{t ^{3} + 1}

\frac{d}{d x} (2 arctanh (x))

\frac{y ^{\frac{3}{2}}}{3} - y^{\frac{1}{2}}

f (x) = ln (sinh (x^{2}))

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