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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

x \to 0 lim (2 x + c)

\int cos^{3} (π x - 1) d x

f (x) = \frac{1}{2 - 3 x}

\frac{d}{d x} ((\frac{x ^{2} - 9}{4 x + 5})^{4})

\frac{1}{x} \frac{d y}{d x} - \frac{2}{x ^{2}} y = x^{2} e^{- x}

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