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

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

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inverse oflaplace 1/(s(s+4))

inverse laplace

\frac{1}{s ( s + 4 )}

\frac{1}{4} H (t) - \frac{1}{4} e^{- 4 t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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

= \frac{1}{4} H (t) - \frac{1}{4} e^{- 4 t}

(6 e^{x} cos (x)) (0.6)

sin (\frac{1}{2} x)

\frac{\partial}{\partial x} (\frac{12 x}{6 x ^{2} + y ^{2} + 4})

\frac{d y}{d x}, x^{3} + y^{3} = 4 x y + 1

\frac{\partial}{\partial x} (e^{3 x^{2}})

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