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

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

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

inverse laplace

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

2 H (t) - e^{- t}

Solution steps

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

Take the partial fraction of

\frac{s + 2}{s ( s + 1 )} : \frac{2}{s} - \frac{1}{s + 1}

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

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

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

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

= 2 H (t) - e^{- t}

\int 2 cot^{3} (x) csc^{3} (x) d x

x \to \infty lim (- 2 x^{4})

\int \frac{1}{1 + r ^{2}} d r

\frac{\partial}{\partial x} (x e^{x^{3} y})

\frac{d}{d x} (π x^{2} h)

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