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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{2}{s + 3}} : 2 e^{- 3 t}

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

\int_{0}^{2 π} x d x

t + t

\frac{d}{d x} (cos (x^{4}) \cdot 4 x^{3})

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

y^{^{'}} = \frac{3 x}{y + x ^{2} y}, y (0) = - 3

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