What is inverse oflaplace ((s+5))/((s+2)(s+3)) ?

The answer to inverse oflaplace ((s+5))/((s+2)(s+3)) is 3e^{-2t}-2e^{-3t}

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inverse oflaplace ((s+5))/((s+2)(s+3))

inverse laplace

\frac{( s + 5 )}{( s + 2 ) ( s + 3 )}

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

Solution steps

L^{- 1} {\frac{( s + 5 )}{( s + 2 ) ( s + 3 )}}

Take the partial fraction of

\frac{s + 5}{( s + 2 ) ( s + 3 )} : \frac{3}{s + 2} - \frac{2}{s + 3}

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

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

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

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

\int_{0}^{\frac{1}{3}} 2 x - 6 x^{2} d x

\int y^{7} 1 - y^{4} d y

\frac{d}{d x} (- x e^{2 x})

x \cdot y^{2} \cdot \frac{d y}{d x} = y^{3} - 4 \cdot x^{3}, y (1) = 2

x^{2} - 8, 7 x

What is inverse oflaplace ((s+5))/((s+2)(s+3)) ?
The answer to inverse oflaplace ((s+5))/((s+2)(s+3)) is 3e^{-2t}-2e^{-3t}