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

The answer to inverse oflaplace (4s)/((s+2)^2) is 4e^{-2t}-e^{-2t}*8t

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

inverse laplace

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

4 e^{- 2 t} - e^{- 2 t} \cdot 8 t

Solution steps

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

Take the partial fraction of

\frac{4 s}{( s + 2 ) ^{2}} : \frac{4}{s + 2} - \frac{8}{( s + 2 ) ^{2}}

= L^{- 1} {\frac{4}{s + 2} - \frac{8}{( s + 2 ) ^{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} {\frac{4}{s + 2}} - L^{- 1} {\frac{8}{( s + 2 ) ^{2}}}

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

L^{- 1} {\frac{8}{( s + 2 ) ^{2}}} : e^{- 2 t} \cdot 8 t

= 4 e^{- 2 t} - e^{- 2 t} \cdot 8 t

\frac{d}{d y} (\frac{x}{y})

\int_{0}^{\frac{π}{12}} tan (6 x) d x

\frac{x + 6}{x ^{3} + x - 7}

\int x (x + 4)^{6} d x

\int_{0}^{1} cos^{2} (x) d x

What is inverse oflaplace (4s)/((s+2)^2) ?
The answer to inverse oflaplace (4s)/((s+2)^2) is 4e^{-2t}-e^{-2t}*8t