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

The answer to inverse oflaplace 1/(s^2-2s-3) is -1/4 e^{-t}+1/4 e^{3t}

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

inverse laplace

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

- \frac{1}{4} e^{- t} + \frac{1}{4} e^{3 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{2} - 2 s - 3} : - \frac{1}{4 ( s + 1 )} + \frac{1}{4 ( s - 3 )}

= L^{- 1} {- \frac{1}{4 ( s + 1 )} + \frac{1}{4 ( 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}{4 ( s + 1 )}} + L^{- 1} {\frac{1}{4 ( s - 3 )}}

L^{- 1} {\frac{1}{4 ( s + 1 )}} : \frac{1}{4} e^{- t}

L^{- 1} {\frac{1}{4 ( s - 3 )}} : \frac{1}{4} e^{3 t}

= - \frac{1}{4} e^{- t} + \frac{1}{4} e^{3 t}

x \to 0 + lim (\frac{x - [ x ]}{[ x ] - 2 x})

\frac{d}{d x} (x^{2} lo g_{2} (7 - 6 x))

\int 6 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

\int_{0}^{4} 4 x - 2 x d x

\frac{d y}{d x} = e^{3 x} + 9 y

What is inverse oflaplace 1/(s^2-2s-3) ?
The answer to inverse oflaplace 1/(s^2-2s-3) is -1/4 e^{-t}+1/4 e^{3t}