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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

y = 4 + 9 x

y^{^{'}} = y - y^{3}

sin (3 t) cos (5 t)

(2 x - 1) d x + (5 y + 8) d y = 0

\int_{0}^{\frac{1}{3}} \frac{3}{1 + 9 x ^{2}} d x

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