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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\frac{2}{s ^{2} + 6 s + 13}

3 sin (2 t + 5)

\frac{d}{d θ} (sec (θ))

\frac{y ^{^{'}} - ( x y )}{( 1 + x )} = x

f (x) = \frac{9 + 4 x - 8 x}{x}

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