What is inverse oflaplace (3s)/(s^2+5s+6) ?

The answer to inverse oflaplace (3s)/(s^2+5s+6) is -6e^{-2t}+9e^{-3t}

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inverse oflaplace (3s)/(s^2+5s+6)

inverse laplace

\frac{3 s}{s ^{2} + 5 s + 6}

- 6 e^{- 2 t} + 9 e^{- 3 t}

Solution steps

L^{- 1} {\frac{3 s}{s ^{2} + 5 s + 6}}

Take the partial fraction of

\frac{3 s}{s ^{2} + 5 s + 6} : - \frac{6}{s + 2} + \frac{9}{s + 3}

= L^{- 1} {- \frac{6}{s + 2} + \frac{9}{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{6}{s + 2}} + L^{- 1} {\frac{9}{s + 3}}

L^{- 1} {\frac{6}{s + 2}} : 6 e^{- 2 t}

L^{- 1} {\frac{9}{s + 3}} : 9 e^{- 3 t}

= - 6 e^{- 2 t} + 9 e^{- 3 t}

8 cos (x)

\int \frac{y}{y ^{2} + 1} d y

(x + y) d x + (x + y - 4) d y = 0

\int x^{- 1} e^{x} d x

\frac{d}{d x} (\frac{8 x}{x + 5})

What is inverse oflaplace (3s)/(s^2+5s+6) ?
The answer to inverse oflaplace (3s)/(s^2+5s+6) is -6e^{-2t}+9e^{-3t}