What is inverse oflaplace 1/(s(s+3)) ?

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

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inverse oflaplace 1/(s(s+3))

inverse laplace

\frac{1}{s ( s + 3 )}

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

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s + 3 )} : \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}

x^{^{''}} - k x = 0

\frac{\partial}{\partial t} (e^{- 2 t} cos (π x))

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

x \to - 2 lim (3 x^{3} + 6 x^{2} + 4)

5 x \cdot \frac{d y}{d x} = 2 x e^{x} - 5 y + 6 x^{2}

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