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

The answer to inverse oflaplace 1/(s(s^2+4s+3)) is 1/3 \H(t)-1/2 e^{-t}+1/6 e^{-3t}

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

inverse laplace

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

\frac{1}{3} H (t) - \frac{1}{2} e^{- t} + \frac{1}{6} e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s ^{2} + 4 s + 3 )} : \frac{1}{3 s} - \frac{1}{2 ( s + 1 )} + \frac{1}{6 ( s + 3 )}

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

L^{- 1} {\frac{1}{3 s}} : \frac{1}{3} H (t)

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

L^{- 1} {\frac{1}{6 ( s + 3 )}} : \frac{1}{6} e^{- 3 t}

= \frac{1}{3} H (t) - \frac{1}{2} e^{- t} + \frac{1}{6} e^{- 3 t}

\int 12 x^{\frac{1}{2}} d x

\frac{d}{d x} (lo g_{e} (\frac{1}{x}))

d er i v a t i v e f (x) = x^{2} + x - 3

(17 + x^{2}) y^{^{'}} - 2 x y = 0

\frac{d}{d x} (\frac{cos ( x )}{1 + x ^{2}})

What is inverse oflaplace 1/(s(s^2+4s+3)) ?
The answer to inverse oflaplace 1/(s(s^2+4s+3)) is 1/3 \H(t)-1/2 e^{-t}+1/6 e^{-3t}