What is inverse oflaplace F(s)=(s+9)/(s^2-2s-3) ?

The answer to inverse oflaplace F(s)=(s+9)/(s^2-2s-3) is -2e^{-t}+3e^{3t}

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inverse oflaplace F(s)=(s+9)/(s^2-2s-3)

inverse laplace

F (s) = \frac{s + 9}{s ^{2} - 2 s - 3}

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

Solution steps

L^{- 1} {\frac{s + 9}{s ^{2} - 2 s - 3}}

Take the partial fraction of

\frac{s + 9}{s ^{2} - 2 s - 3} : - \frac{2}{s + 1} + \frac{3}{s - 3}

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

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

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

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

\int (sin (θ)) d θ

3 x (18 x^{4} + \frac{13}{x + 1})

\int 1 + cos (2 x - 2) d x

- 8 sin (t)

(12 y^{2} - x y) d x + x^{2} d y = 0

What is inverse oflaplace F(s)=(s+9)/(s^2-2s-3) ?
The answer to inverse oflaplace F(s)=(s+9)/(s^2-2s-3) is -2e^{-t}+3e^{3t}