What is inverse oflaplace 1+4/(s(s+2)) ?

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

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inverse oflaplace 1+4/(s(s+2))

inverse laplace

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

δ (t) + 2 H (t) - 2 e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

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

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

= δ (t) + 2 H (t) - 2 e^{- 2 t}

\int \frac{x + 3}{x ^{2} + 8 x - 9} d x

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

\int x 3 x^{2} + 4 d x

\int \frac{5 x}{x ^{2} + 1} d x

n = 1 \sum \infty \frac{1}{n 3 ^{n}}

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