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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\frac{d}{d x} ((\frac{3 x - 1}{x ^{2} + 3})^{2})

x \to 0 lim (\frac{1 - cos ( x )}{5 x ^{2}})

y^{^{'}} - y = 10 t e^{2 t}

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

\frac{d}{d x} (e^{y} + \frac{y}{x})

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