What is inverse oflaplace ((3s-17))/(s^2-2s-3) ?

The answer to inverse oflaplace ((3s-17))/(s^2-2s-3) is 5e^{-t}-2e^{3t}

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inverse oflaplace ((3s-17))/(s^2-2s-3)

inverse laplace

\frac{( 3 s - 17 )}{s ^{2} - 2 s - 3}

5 e^{- t} - 2 e^{3 t}

Solution steps

L^{- 1} {\frac{( 3 s - 17 )}{s ^{2} - 2 s - 3}}

Take the partial fraction of

\frac{3 s - 17}{s ^{2} - 2 s - 3} : \frac{5}{s + 1} - \frac{2}{s - 3}

= L^{- 1} {\frac{5}{s + 1} - \frac{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{5}{s + 1}} - L^{- 1} {\frac{2}{s - 3}}

L^{- 1} {\frac{5}{s + 1}} : 5 e^{- t}

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

= 5 e^{- t} - 2 e^{3 t}

n = 0 \sum \infty (\frac{2}{e})^{n}

(x^{2} + 4)^{5}

\int (e^{2 x}) d x

x^{x}

\frac{d}{d x} (\frac{1}{10 x + 11})

What is inverse oflaplace ((3s-17))/(s^2-2s-3) ?
The answer to inverse oflaplace ((3s-17))/(s^2-2s-3) is 5e^{-t}-2e^{3t}