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

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

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

inverse laplace

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

e^{- 2 t} - e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{1}{s + 3}} : e^{- 3 t}

= e^{- 2 t} - e^{- 3 t}

\frac{d}{d x} (e^{4 x + 1})

\int_{- 1}^{0} 1.5 x^{2} d x

y - x y^{^{'}} = 1 x^{2} y^{^{'}}

\frac{\partial}{\partial y} (2 (y + 4)^{3} (x - 5))

x \to - 1 lim (7 x - 2 x^{2})

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