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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

\frac{d}{d x} (4 sin (\frac{x}{2}))

\int_{- \infty}^{0} \frac{1}{7 - 9 x} d x

\frac{d}{d x} (5 e^{- 2 x} + 4)

\frac{d}{d x} (8^{4 - x^{2}})

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