What is inverse oflaplace 1/(s^2+4s+3) ?

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

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inverse oflaplace 1/(s^2+4s+3)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

(- 2, - 2), (0, 0)

x \to 4 lim (x 32 - x^{2})

\frac{\partial}{\partial x} (\frac{1}{x ^{2} y ^{2}})

\int t 2^{t^{2}} d t

\frac{d y}{d t} + y = e^{- t}

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