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

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

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inverse oflaplace 6/(s(s+2))

inverse laplace

\frac{6}{s ( s + 2 )}

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

3 x^{2} - 8 x - x

f (x) = ln (3 x^{3} + 2)

\int \frac{x}{( x ^{2} + z ^{2} ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial y} (x e^{x y} + y^{2} sin (x y))

\frac{\partial}{\partial x} (ln (x^{7} + y^{4}))

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