What is inverse oflaplace (s+3)/((s+2)(s+1)) ?

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

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inverse oflaplace (s+3)/((s+2)(s+1))

inverse laplace

\frac{s + 3}{( s + 2 ) ( s + 1 )}

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

Solution steps

L^{- 1} {\frac{s + 3}{( s + 2 ) ( s + 1 )}}

Take the partial fraction of

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

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

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

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

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

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

\frac{\partial}{\partial x} (cos (x - y) - x e^{y})

f (x) = 3 x - x^{2}

x \to 1 lim (\frac{1}{x - 2})

f (x) = 9 (5^{x})

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