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

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

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inverse oflaplace 6/(s(s+3))

inverse laplace

\frac{6}{s ( s + 3 )}

2 H (t) - 2 e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= 2 H (t) - 2 e^{- 3 t}

\frac{d}{d x} (\frac{1}{1 + 5 x})

\frac{d}{d x} (a sin^{3} (x))

x \to 0 - lim (\frac{14 x ^{3} + 15 x}{x})

ln (\frac{1}{3} (2 + e^{- 3 x}))

y^{^{''}} - 4 y + 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

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