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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

n = 5 \sum \infty \frac{n ^{2}}{n !}

\int_{- 1}^{1} 1 + x^{- \frac{2}{3}} d x

t \to 1 lim (\frac{1}{t ^{2} - 1})

(f (t) + g (t))^{'}

\int x^{2} + 6 d x

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