What is inverse oflaplace 2/(s^3+3s^2-4) ?

The answer to inverse oflaplace 2/(s^3+3s^2-4) is 2/9 e^t-2/9 e^{-2t}-(2e^{-2t}t)/3

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inverse oflaplace 2/(s^3+3s^2-4)

inverse laplace

\frac{2}{s ^{3} + 3 s ^{2} - 4}

\frac{2}{9} e^{t} - \frac{2}{9} e^{- 2 t} - \frac{2 e ^{- 2 t} t}{3}

Solution steps

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

Take the partial fraction of

\frac{2}{s ^{3} + 3 s ^{2} - 4} : \frac{2}{9 ( s - 1 )} - \frac{2}{9 ( s + 2 )} - \frac{2}{3 ( s + 2 ) ^{2}}

= L^{- 1} {\frac{2}{9 ( s - 1 )} - \frac{2}{9 ( s + 2 )} - \frac{2}{3 ( s + 2 ) ^{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{2}{9 ( s - 1 )}} - L^{- 1} {\frac{2}{9 ( s + 2 )}} - L^{- 1} {\frac{2}{3 ( s + 2 ) ^{2}}}

L^{- 1} {\frac{2}{9 ( s - 1 )}} : \frac{2}{9} e^{t}

L^{- 1} {\frac{2}{9 ( s + 2 )}} : \frac{2}{9} e^{- 2 t}

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

= \frac{2}{9} e^{t} - \frac{2}{9} e^{- 2 t} - \frac{2 e ^{- 2 t} t}{3}

\int_{0}^{2} 36 x + 73 d x

\frac{\partial}{\partial x} (I n (3 x^{2} y))

x \to 0 + lim (3 - x)

f (x) = \frac{r}{3}

\int 1 0^{u} d u

What is inverse oflaplace 2/(s^3+3s^2-4) ?
The answer to inverse oflaplace 2/(s^3+3s^2-4) is 2/9 e^t-2/9 e^{-2t}-(2e^{-2t}t)/3