What is inverse oflaplace (5(1-s^2))/((s+2)^3) ?

The answer to inverse oflaplace (5(1-s^2))/((s+2)^3) is -5e^{-2t}+e^{-2t}*20t-(15e^{-2t}t^2)/2

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inverse oflaplace (5(1-s^2))/((s+2)^3)

inverse laplace

\frac{5 ( 1 - s ^{2} )}{( s + 2 ) ^{3}}

- 5 e^{- 2 t} + e^{- 2 t} \cdot 20 t - \frac{15 e ^{- 2 t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{5 ( - s ^{2} + 1 )}{( s + 2 ) ^{3}}}

Take the partial fraction of

\frac{5 ( 1 - s ^{2} )}{( s + 2 ) ^{3}} : - \frac{5}{s + 2} + \frac{20}{( s + 2 ) ^{2}} - \frac{15}{( s + 2 ) ^{3}}

= L^{- 1} {- \frac{5}{s + 2} + \frac{20}{( s + 2 ) ^{2}} - \frac{15}{( 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{5}{s + 2}} + L^{- 1} {\frac{20}{( s + 2 ) ^{2}}} - L^{- 1} {\frac{15}{( s + 2 ) ^{3}}}

L^{- 1} {\frac{5}{s + 2}} : 5 e^{- 2 t}

L^{- 1} {\frac{20}{( s + 2 ) ^{2}}} : e^{- 2 t} \cdot 20 t

L^{- 1} {\frac{15}{( s + 2 ) ^{3}}} : \frac{15 e ^{- 2 t} t ^{2}}{2}

= - 5 e^{- 2 t} + e^{- 2 t} \cdot 20 t - \frac{15 e ^{- 2 t} t ^{2}}{2}

\int \frac{( 2 x + 3 )}{( x + 1 ) ^{2}} d x

(e^{y} + 1)^{2} \cdot e^{- y} d x + (e^{x} + 1)^{3} \cdot e^{- x} d y = 0

\int_{- 1}^{1} π (6 - 6 x^{2})^{2} d x

\int_{0}^{2} (4 - x^{2}) d x

What is inverse oflaplace (5(1-s^2))/((s+2)^3) ?
The answer to inverse oflaplace (5(1-s^2))/((s+2)^3) is -5e^{-2t}+e^{-2t}*20t-(15e^{-2t}t^2)/2