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

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

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

inverse laplace

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

2 e^{- t} - 2 e^{- 2 t} - e^{- 2 t} \cdot 2 t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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