What is inverse oflaplace {1/(S^2+3S)} ?

The answer to inverse oflaplace {1/(S^2+3S)} is 1/3 \H(t)-1/3 e^{-3t}

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inverse oflaplace {1/(S^2+3S)}

inverse laplace

{\frac{1}{S ^{2} + 3 S}}

\frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

Solution steps

L^{- 1} {\frac{1}{S ^{2} + 3 S}}

Take the partial fraction of

\frac{1}{S ^{2} + 3 S} : \frac{1}{3 S} - \frac{1}{3 ( S + 3 )}

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

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}{3 S}} - L^{- 1} {\frac{1}{3 ( S + 3 )}}

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

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

= \frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

\int (3 x^{2} + 2 x + 1) d x

\frac{\partial}{\partial z} (2 x z)

\int \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

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

y = x^{2} - 3 x, y = 4 x

What is inverse oflaplace {1/(S^2+3S)} ?
The answer to inverse oflaplace {1/(S^2+3S)} is 1/3 \H(t)-1/3 e^{-3t}