What is inverse oflaplace 6/(s(s+4)) ?

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

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inverse oflaplace 6/(s(s+4))

inverse laplace

\frac{6}{s ( s + 4 )}

\frac{3}{2} H (t) - \frac{3}{2} e^{- 4 t}

Solution steps

L^{- 1} {\frac{6}{s ( s + 4 )}}

Take the partial fraction of

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

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

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

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

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

= \frac{3}{2} H (t) - \frac{3}{2} e^{- 4 t}

\frac{d}{d x} (a x^{2} + b x + c)

(x^{2} + 1) \frac{d y}{d x} + 2 x y = 4 x^{2}

\int 10 sin^{2} (x co) s^{2} x d x

\frac{d}{d x} (k x^{4} (x_{0} - x))

\frac{d}{d x} ((sin (5 x^{2}))^{2})

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