What is inverse oflaplace 1/(2s^2+10s+12) ?

The answer to inverse oflaplace 1/(2s^2+10s+12) is 1/2 e^{-2t}-1/2 e^{-3t}

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inverse oflaplace 1/(2s^2+10s+12)

inverse laplace

\frac{1}{2 s ^{2} + 10 s + 12}

\frac{1}{2} e^{- 2 t} - \frac{1}{2} e^{- 3 t}

Solution steps

L^{- 1} {\frac{1}{2 s ^{2} + 10 s + 12}}

Take the partial fraction of

\frac{1}{2 s ^{2} + 10 s + 12} : \frac{1}{2 ( s + 2 )} - \frac{1}{2 ( s + 3 )}

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

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

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

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

= \frac{1}{2} e^{- 2 t} - \frac{1}{2} e^{- 3 t}

\int x^{\frac{1}{3}} d x

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

\int \frac{5 x}{1 + x ^{4}} d x

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 12 x + 5)

\frac{d}{d x} (\frac{x ^{2} + 4 x}{x + 2})

What is inverse oflaplace 1/(2s^2+10s+12) ?
The answer to inverse oflaplace 1/(2s^2+10s+12) is 1/2 e^{-2t}-1/2 e^{-3t}