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

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

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

inverse laplace

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

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}}}

Take the partial fraction of

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

\frac{d}{d x} (ln (4 x - 3))

\int 7 e^{3} d x

\frac{d}{d x} (3 x^{2} + 6 x)

\int_{0}^{\frac{π}{2}} sin^{10} (x) d x

\frac{\partial}{\partial y} (- 4 x y + x^{4} + y^{4})

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