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}

\int \frac{e ^{- s t}}{s} d t

\int ln (6 x + 1) d x

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

4 x^{3} - 4 x^{2} + 3

n = 1 \sum \infty \frac{n ^{3}}{e ^{n}}

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}