What is inverse oflaplace (-s-4)/(s^2+4s+3) ?

The answer to inverse oflaplace (-s-4)/(s^2+4s+3) is -3/2 e^{-t}+1/2 e^{-3t}

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inverse oflaplace (-s-4)/(s^2+4s+3)

inverse laplace

\frac{- s - 4}{s ^{2} + 4 s + 3}

- \frac{3}{2} e^{- t} + \frac{1}{2} e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{- s - 4}{s ^{2} + 4 s + 3} : - \frac{3}{2 ( s + 1 )} + \frac{1}{2 ( s + 3 )}

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

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

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

= - \frac{3}{2} e^{- t} + \frac{1}{2} e^{- 3 t}

\int 1 + y^{3} d y

\frac{\partial}{\partial x} (4 x^{3} y - 6 x^{2} y^{2})

\int 4 sin (2 x + 3) d x

y = e^{2 x} arcsin (x)

16 cos (2 x)

What is inverse oflaplace (-s-4)/(s^2+4s+3) ?
The answer to inverse oflaplace (-s-4)/(s^2+4s+3) is -3/2 e^{-t}+1/2 e^{-3t}