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

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

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

inverse laplace

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

e^{- t} \cdot 3 t - \frac{e ^{- t} t ^{2}}{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

= e^{- t} \cdot 3 t - \frac{e ^{- t} t ^{2}}{2}

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