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

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

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

inverse laplace

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

δ (t) - 3 e^{- t} + e^{- t} \cdot 3 t

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

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

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

= δ (t) - 3 e^{- t} + e^{- t} \cdot 3 t

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