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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

f (x) = \frac{4}{x ^{3}}

\frac{x}{x ^{2} - x + 1}

\int \frac{e ^{2 x} + 1}{e ^{2 x} - 3} d x

\frac{1}{x - 1}, 25

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