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

The answer to inverse oflaplace 2/(s(s^2+4s)) is -1/8 \H(t)+t/2+1/8 e^{-4t}

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

inverse laplace

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

- \frac{1}{8} H (t) + \frac{t}{2} + \frac{1}{8} e^{- 4 t}

Solution steps

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

Take the partial fraction of

\frac{2}{s ( s ^{2} + 4 s )} : - \frac{1}{8 s} + \frac{1}{2 s ^{2}} + \frac{1}{8 ( s + 4 )}

= L^{- 1} {- \frac{1}{8 s} + \frac{1}{2 s ^{2}} + \frac{1}{8 ( 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}{8 s}} + L^{- 1} {\frac{1}{2 s ^{2}}} + L^{- 1} {\frac{1}{8 ( s + 4 )}}

L^{- 1} {\frac{1}{8 s}} : \frac{1}{8} H (t)

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

L^{- 1} {\frac{1}{8 ( s + 4 )}} : \frac{1}{8} e^{- 4 t}

= - \frac{1}{8} H (t) + \frac{t}{2} + \frac{1}{8} e^{- 4 t}

\int_{6}^{8} \frac{84}{( x - 6 ) ^{3}} d x

\frac{d y}{d x} = 2 y sin (x)

2 x y \frac{d y}{d x} + x^{2} - y^{2} = 0

f (x) = x, (9, 3)

y = \frac{3}{x}, y = 12 x, y = \frac{1}{12} x

What is inverse oflaplace 2/(s(s^2+4s)) ?
The answer to inverse oflaplace 2/(s(s^2+4s)) is -1/8 \H(t)+t/2+1/8 e^{-4t}