What is inverse oflaplace 4/((s+1)(s+2)^3) ?

The answer to inverse oflaplace 4/((s+1)(s+2)^3) is 4e^{-t}-4e^{-2t}-e^{-2t}*4t-e^{-2t}*2t^2

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inverse oflaplace 4/((s+1)(s+2)^3)

inverse laplace

\frac{4}{( s + 1 ) ( s + 2 ) ^{3}}

4 e^{- t} - 4 e^{- 2 t} - e^{- 2 t} \cdot 4 t - e^{- 2 t} \cdot 2 t^{2}

Solution steps

L^{- 1} {\frac{4}{( s + 1 ) ( s + 2 ) ^{3}}}

Take the partial fraction of

\frac{4}{( s + 1 ) ( s + 2 ) ^{3}} : \frac{4}{s + 1} - \frac{4}{s + 2} - \frac{4}{( s + 2 ) ^{2}} - \frac{4}{( s + 2 ) ^{3}}

= L^{- 1} {\frac{4}{s + 1} - \frac{4}{s + 2} - \frac{4}{( s + 2 ) ^{2}} - \frac{4}{( s + 2 ) ^{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{4}{s + 1}} - L^{- 1} {\frac{4}{s + 2}} - L^{- 1} {\frac{4}{( s + 2 ) ^{2}}} - L^{- 1} {\frac{4}{( s + 2 ) ^{3}}}

L^{- 1} {\frac{4}{s + 1}} : 4 e^{- t}

L^{- 1} {\frac{4}{s + 2}} : 4 e^{- 2 t}

L^{- 1} {\frac{4}{( s + 2 ) ^{2}}} : e^{- 2 t} \cdot 4 t

L^{- 1} {\frac{4}{( s + 2 ) ^{3}}} : e^{- 2 t} \cdot 2 t^{2}

= 4 e^{- t} - 4 e^{- 2 t} - e^{- 2 t} \cdot 4 t - e^{- 2 t} \cdot 2 t^{2}

What is inverse oflaplace 4/((s+1)(s+2)^3) ?
The answer to inverse oflaplace 4/((s+1)(s+2)^3) is 4e^{-t}-4e^{-2t}-e^{-2t}*4t-e^{-2t}*2t^2