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

The answer to inverse oflaplace (s-2)/(s^2+5s+6) is -4e^{-2t}+5e^{-3t}

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{s - 2}{s ^{2} + 5 s + 6} : - \frac{4}{s + 2} + \frac{5}{s + 3}

= L^{- 1} {- \frac{4}{s + 2} + \frac{5}{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{4}{s + 2}} + L^{- 1} {\frac{5}{s + 3}}

L^{- 1} {\frac{4}{s + 2}} : 4 e^{- 2 t}

L^{- 1} {\frac{5}{s + 3}} : 5 e^{- 3 t}

= - 4 e^{- 2 t} + 5 e^{- 3 t}

\frac{x ^{3}}{4} + \frac{1}{3 x}

\frac{\partial}{\partial t} (3 t)

f (x) = sin (θ) cos (θ)

\frac{d v}{d t} = - 2 v - 32

y^{^{''}} = - 16 y

What is inverse oflaplace (s-2)/(s^2+5s+6) ?
The answer to inverse oflaplace (s-2)/(s^2+5s+6) is -4e^{-2t}+5e^{-3t}