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

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

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

inverse laplace

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

- 2 e^{- 2 t} + 2 e^{- t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= - 2 e^{- 2 t} + 2 e^{- t}

\int e^{8 x + e^{8 x}} d x

\frac{d}{d x} (x^{x} (ln (x) + 1))

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

\frac{d}{d x} (x^{3} (x - 8)^{2})

\frac{d}{d x} (\frac{1}{x - y})

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